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ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 1 19 August 2010
JC 2 PRELIMINARY EXAMINATION
Time allowed: 3 hours
Additional Materials: List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in the question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages.
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Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 2 of 6
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010
MATHEMATICS 9740 Higher 2 Paper 1
Calculator model: _____________________
Arrange your answers in the same numerical order.
Place this cover sheet on top of them and tie them together with the string provided.
Question no. Marks 1
2
3
4
5
6
7
8
9
10
11
12
13
14
/ 100 Index No: Form Class: ___________ Name: _________________________
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 3 of 6
1 The nth term of a sequence is given by ( ) 1 21 n
nu n+= − , for 1n ≥ . The sum of the first n terms is denoted by Sn. Use the method of mathematical induction to show that
( ) ( )1 11
2n
nn n
S + += − for all positive integers n. [4]
2 The 3 flavours of puddings produced by a dessert shop are mango, durian and strawberry. A
mango pudding requires 5g of sugar and 36ml of water. A durian pudding requires 6g of sugar and 38ml of water. A strawberry pudding requires 4g of sugar and 40ml of water. The puddings are sold in pairs of the same type at $1.60, $2.20 and $1.80 for mango, durian and strawberry respectively. On a particular day, 754g of sugar and 5972ml of water were used to make the puddings and all the puddings made were sold except for a pair of strawberry puddings. The collection from the sale of puddings was $142.40. Formulate the equations required to determine the number of each type of pudding made on that day. [4]
3 The diagram below shows the graph of f( )y x= . The curve passes through the origin and
has a maximum point at ( )4 , 4A and asymptotes 2x = − and 2y = . Sketch on separate diagrams, the graphs of
(i) 1f( )y x= [3]
(ii) f '( )y x= [3] showing clearly asymptotes, intercepts and coordinates of turning points where possible.
4 The complex number w has modulus 3 and argument 23π . Find the modulus and argument of
*i
w− , where w* is the complex conjugate of w. Hence express
*i
w− in the form a ib+ , where
a and b are real, giving the exact values of a and b in non-trigonometrical form. [4]
Find the possible values of n such that *
niw−
is purely imaginary. [2]
[Turn Over
y
x
f( )y x=
2
2−
( )4 , 4 A
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 4 of 6
5 Solve the equation 4 0z i− = , giving the roots in the form ire α , where 0r > and π α π− < ≤ . [3]
The roots represented by 1z and 2z are such that ( ) ( )1 2arg arg 0z z> > . Show 1z , 2z and
1 2z z+ on an Argand diagram. Deduce the exact value of ( )1 2arg z z+ . [3]
6 An economist is studying how the annual economic growth of 2 countries varies with time. The annual economic growth of a country is measured in percentage and is denoted by G and the time in years after 1980 is denoted by t. Both G and t are taken to be continuous variables.
(i) Country A is a developing country and the economist found that G and t are can be modeled
by the differential equation 12
dG Gdt
+= . Given that, when 0t = , 0G = , find G in terms of t.
[4] (ii) Comment on the suitability of the above differential equation model to forecast the future
economic growth of Country A. [1] (iii) Country B is a developed country and the economist found that G and t can be modeled by
the differential equation 12
dG Gdt
+ = −
.
Given that Country B has been experiencing decreasing economic growth during the period of study, sketch a member of the family of solution curves of the differential equation model for Country B. Hence, comment on the economic growth of Country B in the long term. [2]
7 (i) Given that ( ) 1cosf xx e
−
= , where 1 1x− ≤ ≤ , find ( )f 0 , ( )f 0′ and ( )f 0′′ . Hence write down
the first three non-zero terms in the Maclaurin series for ( )f x . Give the coefficients in terms
of ke π , where k∈ . [4] (ii) Given that ( )g tan secx x x= + , where x is sufficiently small for 3x and higher powers of x
to be neglected. Deduce the first three non-zero terms in the series expansion of ( )g x .
Hence, show that ( ) ( ) 22 2 2f g 2x e x e x eπ π π
+ ≈ + . [3]
(iii) Explain clearly why it is inappropriate to state that ( ) ( ) 22 2 2f g 2a a
a ax e x dx e x e dx
π π π
− −+ ≈ +∫ ∫ ,
where a∈ . [1]
8 (i) Show that ( ) ( ) ( )21 2 1
! 1 ! 2 ! 2 !Ar Br C
r r r r+ +− + =
+ + +, where A, B and C are constants to be
found. [2]
(ii) Hence find ( )2
1
3 3 32 !
n
r
r rr=
+ −+∑ . [3]
(iii) Give a reason why the series ( )2
0
3 3 32 !r
r rr
∞
=
+ −+∑ converges, and write down its value. [2]
y
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 5 of 6
9 Relative to the origin O, two points A and B have position vectors given by 3 3a i j k= + + and 5 4 3b i j k= − + respectively.
(i) Find the length of the projection of OA
on OB
. [2] (ii) Hence, or otherwise, find the position vector of the point C on OB such that AC is
perpendicular to OB. [2] (iii) Find a vector equation of the reflection of the line AB in the line AC. [3] 10(a) The first 2 terms of a geometric progression are a and b ( b < a ). If the sum of the first n
terms is equal to twice the sum to infinity of the remaining terms, prove that 3n na b= . [3]
(b) The terms 1 2 3, , ,...u u u form an arithmetic sequence with first term a and having non-zero common difference d. (i) Given that the sum of the first 10 terms of the sequence is 105 more than 510u , find
the common difference. [3] (ii) If 26u is the first term in the sequence which is greater than 542, find the range of
values of a. [3] 11 The region R in the first quadrant is bounded by the curve ( )22 2y a a x= − , where 0a > ,
and the line joining ( )2 ,0a and ( )0,a . The region S, lying in the first quadrant, is bounded
by the curve ( )22 2y a a x= − and the lines 2x a= and y a= . (i) Draw a sketch showing the regions R and S. [1] (ii) Find, in terms of a, the volume of the solid formed when S is rotated completely about the x-axis. [4]
(iii) By using a suitable translation, find, in terms of a, the volume of the solid formed when R is rotated completely about the line 2x a= . [4]
12 The curve C has the equation 23 2x axy x a+ += + where a is a constant.
(i) Find dydx and the set of values of a if the curve has 2 stationary points. [4]
(ii) Sketch the curve C for a = 1, stating clearly the exact coordinates of any points of intersection with the axes and the equations of any asymptotes. [3]
Hence, find the range of values of k such that the equation 23 2 ( 1)x x k x+ + = + has exactly 2 real roots. [2]
13 The curve has the parametric equations
25
1x
t=
+ , 1tany t−=
(i) Sketch the curve for 2 2t− ≤ ≤ . [1] (ii) Find the cartesian equations of the tangent and the normal to the curve at the point
where 1t = . [5] (iii) Find the area enclosed by the x-axis, the tangent and the normal at the point where t = 1. [3]
[Turn Over
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 6 of 6
14 The functions f and g are defined as follows:
f : sinx x , ,2 2
x π π ∈ − ,
( )( )g : 1 38π
+ −x x x , x∈ .
(i) Sketch the graph of the function g, labeling clearly the exact values of the coordinates of turning point(s) and intersections with the axes, if any. [1] State the range of the function g in exact values. [1]
(ii) Given that gf exists as a function. By considering the graphs of f and g, explain why
gf ( ) gf ( )α β≠ if 2 2π πα β− ≤ < ≤ . [2]
Hence what can be said about the function gf ? [1] Without sketching the graph of gf , find the range of gf in the form [ ],a b , giving the exact values of a and b. [1]
(iii) (a) Give a reason why fg does not exist as a function. [1]
(b) Find the greatest exact value of k for which fg is a function if the domain of g is restricted to the interval [ ]1,k . [2]
- End of Paper -
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 2 23 August 2010
JC 2 PRELIMINARY EXAMINATION
Time allowed: 3 hours
Additional Materials: List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in the question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 5 printed pages.
[Turn Over
9740 / 02
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 2 of 5
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010
MATHEMATICS 9740 Higher 2 Paper 2
Calculator model: _____________________
Arrange your answers in the same numerical order.
Place this cover sheet on top of them and tie them together with the string provided.
Question no. Marks
1
2
3
4
5
6
7
8
9
10
11
12
13
14
/ 100 Index No: Form Class: ___________ Name: _________________________
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 3 of 5
Section A: Pure Mathematics [40 marks]
1 Find the exact value of ( )
12
2 11
1 .xxe dx
e −−
−∫ [4]
2 The variable complex numbers z and w are such that 2 3z i− − = and ( )arg 5 3w i π− + = .
(i) Illustrate both of these relations on a single Argand diagram. [2] (ii) State the least value of z w− . [1]
(iii) Find the greatest and least possible values of ( )arg 3z + , giving your answers in radians correct to 3 decimal places. [4]
3 Find in terms of a, the range of values of x that satisfy the inequality ln 2 0axx
− ≥
,
where 1a> . [4] 4 (a) State the derivative of 3cos x . Hence, find 5 3sin .x x dx∫ [4]
(b) Find the exact value of ( )2 5 3
2 2
10
5x dx−
−∫ , in the form 2 3a b+ , using the
substitution 5 secx θ= . [6]
5 The plane p1 has equation 2 3x y z+ − = . The line l1 has equation 1 12 4
x zy+ −= = .
(i) Show that the line 1l is parallel to, but not contained in the plane p1. [2] (ii) Find the cartesian equation of the plane p2 which contains the line l1 and is perpendicular to the plane p1. [3]
(iii) Find, in scalar product form, the vector equation of the plane p3 which contains the point ( )4,1, 1− and is perpendicular to both p1 and p2. [2]
Another line l2 which is parallel to the vector 203
−
intersects the line l1 at the
point ( )1,0,1−A . (iv) Given that the line l2 meets the plane p1 at the point B, find the coordinates of B. [4] (v) Find the sine of the acute angle between the line l2 and the plane p1, and hence, find the
length of the projection of the line segment AB on the plane p1, giving your answer in surd form. [4]
Section B: Statistics [60 marks]
6 Mr Raju, who owns a supermarket wishes to find out what customers think about the goods
that he sells. He has been advised that he should take a random sample of his customers for this purpose. State, with reasons, which of the following sampling procedures is preferable. A. Select every 10th customer on each day in a typical week. B. Select the first 20 customers on each day in a typical week [2]
[Turn Over
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 4 of 5
7 One day, Pinocchio went shopping and bought a pair of size 30 Wood Shoes. The right
shoes have lengths which are normally distributed with mean 20 cm and standard deviation 0.14 cm. The left shoes have lengths which are normally distributed with mean 20.1 cm and standard deviation 0.11 cm. The length of the right shoe is independent of the length of the left shoe. When wearing the pair of shoes, Pinocchio takes six steps, heel to toe as shown in the diagram. Calculate the probability that the distance AB, from the back of the first step to the front of the sixth step, exceeds 120 cm. [3]
8 On Ulu Island the weights of adult men and women may both be taken to be independent
normal random variables with means 75kg and 65 kg and standard deviations 4 kg and 3 kg respectively. Find the probability that the weight of a randomly chosen man and the weight of a randomly chosen woman differ by more than 1 kg. [3] Explain if this is equal to the probability that the difference in weight between a randomly chosen married woman and her husband is more than 1kg. [1]
9 Research has shown that before using an Internet service, the mean monthly family
telephone costs is $72. A random sample of families which had started to use an Internet service was taken and their monthly telephone costs were :
$70, $84, $89, $96, $74 Stating a necessary assumption about the population, carry out a test at the 5% significance level, whether there is an increase in the mean monthly telephone costs. [5] If the assumption stated above still holds, and if the standard deviation of the monthly telephone costs is $9.89, find the range of values of the mean monthly family telephone costs 0µ that would lead to a reverse in the decision to the above test. [3]
10 Six overweight men registered at a slimming centre for a slimming programme. The
following table records x, the height (to the nearest cm) and y, the weight (to the nearest 0.1 kg) of these six men.
(i) Given that the least square regression line of x on y line is 103.6 0.726x y= + , show that the value of k to the nearest 0.1 kg is 80.3. Hence or otherwise, find the least square regression line of y on x in the form y ax b= + , giving the values of a and b to the nearest 3 decimal places. [5] (ii) Based on the data given, use an appropriate regression line to predict the weight of an overweight man who is 165 cm tall. [2] (iii) Find the value of the product moment correlation coefficient between x and y and sketch the scatter diagram of y against x. A particular man among the 6 men who registered for the slimming programme is unusually overweight. Indicate who this man is. [3]
Man A B C D E F x (height in cm) 150 157 160 162 167 170 y (weight in kg) 65.1 73.2 85 k 80.9 89.9
A B
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 5 of 5
11 In a hotel, large number of cups and saucers are washed each day. The number of cups that
are broken each day while washing averages 2.1. State in context, a condition under which a Poisson distribution would be a suitable probability model. [1] Assume that the number of broken cups and saucers follow a Poisson distribution.
(i) Show that on any randomly chosen day, the probability that at least 3 cups are broken is 0.350 correct to 3 significant figures. [1] The probability that there will be at least two days in n days with at least 3 broken cups is more than 0.999. Find the least value of n. [3]
(ii) The number of saucers broken each day averages 1.6, independently of the number of cups broken. The total number of cups broken and saucers broken during a week of 7 days is denoted by T. State a possible model for the distribution of T. [2]
A random sample of 100 weeks is chosen. Using a suitable approximation, find the probability that the average weekly total number of broken cups and saucers does not exceed 26. [3]
12 Fish are bred in large batches and allowed to grow until they are caught at random for sale.
When caught, only 20% of the fish measure less than 8 cm long. (i) What is the probability that the 10th fish caught is the sixth fish that is less than 8 cm long? [2] (ii) A large number, n, of fish are caught and the probability of there being 10 or fewer fish in
the catch which measures less than 8 cm long is at most 0.0227 . Using a suitable approximation, derive the approximate inequality 10.5 0.2 0.8n n− ≤ − . [4] Hence find the least possible number of fish to be caught. [2]
13 An automated blood pressure machine is being tested. Members of the public, %p of whom
have high blood pressure (hypertension), try it out and are then seen by a doctor. She finds that 80% of those with hypertension and 10% of those with normal blood pressure have been diagnosed as hypertensive by the machine. The probability that a randomly chosen patient who was diagnosed as hypertensive by the machine actually has hypertension is
32 .
(i) Find the value of p [3] (ii) Hence, find the probability that a randomly chosen patient does not have hypertension,
given that the machine diagnosed him as having normal blood pressure. [2] Comment on the usefulness of the machine. [1]
14 Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labeled ‘1’ and ‘2’, the three blue balls are labeled ‘1’, ‘2’ and ‘3’, and the five green balls are labeled ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’.
(i) Find the number of ways of choosing 2 balls of identical colour. [2] (ii) Find the number of ways of choosing 6 balls if it includes at least one ball of each colour. [4] (iii) A person arranged 3 balls in a row with the numbered sides facing him forming a 3-digit
number. Among these 3 balls, none of them are green. Find the number of possible 3-digit numbers facing that person.
[The number formed is independent of the colours of the balls used. i.e. the number 112 is counted as one number whether the colour of the ball labeled ‘2’ is red or blue.] [3]
- End of Paper -
1
Anglo-Chinese Junior College H2 Mathematics 9740
2010 JC 2 PRELIM Marking Scheme
Paper 1:
1 Let Pn denote the statement ( ) ( )1 1
12
n
n
n nS
+ += −
For n = 1, LHS = ( )2 21 1 1 1 1S u= = − = RHS = ( ) ( )2 1 2
1 12
− =
LHS = RHS
∴ P1 is true.
Assume Pk true for some k +∈ℤ , i.e. ( ) ( )1 11
2
k
k
k kS
+ += −
Prove that Pk+1 is true, i.e. ( ) ( )( )2
1
1 21
2
k
k
k kS
++
+ += −
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) [ ]
( ) ( )( )
1 1
1 2 2
1
1
2
LHS
11 1 1
2
11 2 1
2
11 2
2
1 21 RHS
2
k k k
k k
k
k
k
S S u
k kk
kk k
kk
k k
+ +
+ +
+
+
+
= = +
+= − + − +
+= − − +
+= − − −
+ += − =
Since P1 is true, and Pk is true ⇒ Pk+1 is true,
by the principle of mathematical induction, Pn is true n +∀ ∈ℤ .
2
Let x be the no of mango puddings produce.
Let y be the no of durian puddings produce.
Let z be the no of strawberry puddings produce.
5 6 4 754x y z+ + = ……………………………(1)
36 38 40 5972x y z+ + = ………………………(2)
0.8 1.1 0.9( 2) 142.4x y z+ + − = ……………….(3)
Solving (1), (2) & (3) using GC,
46, 42 and 68.x y z= = =
3
(i)
1f( )
yx
=
2−
y
x
A’ (4, 0.25)
0.5y =
0x =
2
3
(ii)
f '( )y x=
4 1
* 3
i
w
−=
2arg arg( ) arg( *)
* 2 3 6
ii w
w
π π π− = − − = − − − =
1 1 3 1 3 1cos sin
* 3 6 6 3 2 2 6 6
ii i i
w
π π − = + = + = +
1cos sin
* 3 6 6
n ni n n
iw
π π− = +
*
ni
w
−
is purely imaginary, cos6
nπ=0
( )2 1 , 6 2
n k kπ π
= + ∈ℤ ,
( )3 2 1 ,n k k∴ = + ∈ℤ .
5 4 4 4 20 1
i
z i z i z e
π
− = ⇒ = ⇒ =
2
4 2 , 2, 1,0,1i k
z e k
ππ +
= = − −
12
4 2
7 3 5
8 8 8 8
, 2, 1,0,1
, , ,
i k
i i i i
z e k
z e e e e
ππ
π π π π
+
− −
= = − −
=
z1
z2
z2 + z2
x
y
8
π
5
8
π
O
y
x 2− A’
(4,0)
3
5
81
82
i
i
z e
z e
π
π
=
=
( )1 2
1 4 3arg
8 2 8 8z z
π π π + = + =
6 1
2
dG G
dt
+=
1 1
1 2
ln 1 0.5
dG dtG
G t C
=++ = +
∫ ∫
0.5
0.5
1
1 , where
t C
t C
G e
G Ae A e
++ = ±
= − + = ±
When 0t = , 0G = ,
1A = 0.51 tG e∴ = − +
Examples of possible comments:
The model is not suitable because …
� The economist is assuming that that there are no fluctuations in the economic
growth in the future.
� The economist is assuming that the country will enjoy perpetual economic
growth in the long term.
� The economist is assuming Country A is always experiencing positive and
increasing economic growth in the future.
� Factors affecting economic growth remains unchanged.
In the long term, Country B is expected to be still in recession with an economic
growth decreasing towards -1%.
7(i)
( )1cosf xx e
−
= ( ) 2f 0 eπ
⇒ =
( )1
cos
2
1f
1
xx ex
−−′ =−
( ) 2f 0 eπ
′⇒ = −
t
G
1−
0.51 , 1tG Be B−= − + =
0
4
(ii)
(iii)
( ) ( ) ( ) ( )
( )( ) ( )
1 1
1 1
3cos cos 2 2
2 2
3cos cos 2 22
2
1 1 1f 1 2
21 1
11 f 0
1
x x
x x
x e e x xx x
e xe x ex
π
− −
− −
−
−
− − − ′′ = + − − − − −
′′= − − ⇒ =−
( )2
22 2f ...2
ex e e x x
ππ π
∴ = − + +
( )
( )1
2 2
2
g tan sec
1tan 1 1 1 ...
cos 2 2
12
−
= +
= + ≈ + − = + + − − +
≈ + +
x x x
x xx x x
x
xx
( ) ( )22
22 2 2 2
22 2
f g 12 2
2 (Shown)
e xx e x e e x x e x
e e x
ππ π π π
π π
+ ≈ − + + + +
= +
The statement ( ) ( ) 22 2 2f g 2a a
a ax e x dx e x e dx
π π π
− −+ ≈ +∫ ∫ is inappropriate as
( ) ( ) 22 2 2f g 2x e x e x eπ π π
+ ≈ + only when a is sufficiently small.
8
(i)
(ii)
( ) ( )( )( ) ( )
( )
( )2
1 2 1! 1 ! 2 !
1 2 2 2 1
2 !
12 !
r r r
r r r
r
r rr
− ++ +
+ + − + +=
+
+ −=+
1, 1, 1.A B C= = = −
( )
( )
( ) ( )
2
1
2
1
1
3 3 32 !
132 !
1 2 13! 1 ! 2 !
n
r
n
r
n
r
r rr
r rr
r r r
=
=
=
+ −+
+ −=+
= − + + +
∑
∑
∑
5
(iii)
( ) ( )
( ) ( )
( ) ( )
1 2 1
1! 2! 3!
1 2 1
2! 3! 4!
1 2 1
3! 4! 5!
1 2 1
4! 5! 6!3
1 2 1
2 ! 1 ! !
1 2 1
1 ! ! 1 !
1 2 1
! 1 ! 2 !
n n n
n n n
n n n
− +
+ − +
+ − +
+ − + =
+ − + − − + − + − + + − + + +
⋮
( ) ( )1 1 132 1 ! 2 !n n
= − + + +
( )
( ) ( )
2
0
2
1
3 3 32 !
3 3 3 32! 2 !
n
r
n
r
r rr
r rr
=
=
+ −+
+ −= − ++
∑
∑
( ) ( ) ( )3 1 1 132 2 1 ! 2 !n n
= − + − + + +
As ( ) ( )1 1, 0 and 0.1 ! n+2 !
nn
→ ∞ → →+
the series converges to 0.∴
9(i)
(ii)
Length of the projection of OA����
on OB����
=
3 51 20
1 4 2 250 50
3 3
⋅ − = =
or 4
2
Method 1:
From (i), 2 2OC =����
5 51 2
2 2 4 4550
3 3
OC
= − = −
����
Method 2:
Line OB:
0 5
0 4
0 3
λ = + −
r
5 3 5 3
4 1 4 1
3 3 3 3
AC OC OA
λλ λ
λ
− = − = − − = − − −
���� ���� ����
Since AC OB⊥���� ����
,
A
C B O
6
(iii)
5 3 5
4 1 . 4 0
3 3 3
2
5
λλλ
λ
− − − − = − −
=
52
45
3
OC
= −
����
Since 2
5OC OB=���� ����
, : 2 : 3OC CB =
51 1
' 45 5
3
OB OB
= − = − −
����� ����
Or use midpoint theorem,
( )1'
2OC OB OB= +���� ����� ����
5 5 52 1
' 2 2 4 4 45 5
3 3 3
OB OC OB
= − = − − − = − −
����� ���� ����
5 3 201 1
' ' 4 1 15 5
3 3 18
AB OB OA
= − = − − − = −
����� ����� ����
Vector equation of line 'AB is
3 20
1 1 ,
3 18
r α α = + ∈
ℝɶ
10
(a)
(b) (i)
GP : a = a, r = ba
The sum to infinity of the remaining terms, 1
narSr∞ =
−
( )
( )
2
1 21 1
1 2
3 1
13
13
3
n
nn
n n
n
n
n
n n
S S
a r arr r
r r
r
r
ba
b a
∞=
−=
− −
− =
=
=
=
=
7
(ii)
[ ] [ ]10 5105 10
10 2 9 105 10 42
10 45 105 10 40
21
S u
a d a d
a d a d
d
= +
+ = + +
+ = + +
=
25 542
24(21) 542
38
u
a
a
≤
+ ≤
≤
and
26 542
25(21) 542
17
u
a
a
>
+ >
>
17 38a∴ < ≤
11
(i)
(ii)
(iii)
When S is rotated completely about the x-axis,
( ) ( )
( )
( )
22
0
22
3
0
3 2
3
Required volume 2 22
22
2 2
2 22
cu. units
a
a
aa a a x dx
a xaa
aa a
a
π π
ππ
ππ
π
= − −
−= −
−
= −
=
∫
After a translation of 2a units in the negative x-direction,
New equation is ( )2
2 22 2 ( 2 )
yy a a x a x
a= − + ⇒ = −
When R is rotated completely about the line 2x a= ,
( ) ( )2
22
0
53
2
0
3 3 3
1 2Required volume 2
3
4 4
3 5
4 4 8 cu. units
3 5 15
a
a
ya a dy
a
ya
a
a a a
π π
π π
π π π
= − −
= −
= − =
∫
12
x
y
a
2a O
R S ( )22 2y a a x= −
8
(i)
(ii)
( )( ) ( )( )
( )( )
2
2
2
2 2
2
3 2
6 3 2
3 6 2
x axyx a
x a x a x axdydx x a
x ax a
x a
+ +=+
+ + − + +=
+
+ + −=
+
For stationary points,
( )2 2
0
3 6 2 0
dydx
x ax a
=
+ + − =
For 2 stationary points,
( ) ( )( )2 2
2
2
6 4 3 2 0
24 24
1
a a
a
a
− − >
> −
> −
2 is always positive.
a .
a
∴ ∈ ℝ
23 2 43 2
1 1x xy x
x x+ += = − ++ +
Points of intersection with the axes : (0,2)
Asymptotes : 3 2 and 1y x x= − = − .
The range of values of k such that the equation 23 2 ( 1)x x k x+ + = + has exactly 2
real roots :
1.93 or 11.9k k> < −
13
(i)
9
(ii)
(iii)
( )
( )
( )
( )
2 22
22
2
2
-5 2t1 and = 1 1
.
1+t1 .
101
1=
10
dy dxdt dtt t
dy dy dtdx dt dx
tt
t
t
=+ +
=
=−+
+−
When 1,
15
t
dydx
=
= −
Equation of tangent :
( )
145 52
1 15 2 4
πy
x
πy x
−= −
−
= − + +
Gradient of normal is 5
Equation of normal:
( )
4 552
2552 4
πy
x
πy x
−=
−
= − −
( ) ( )( )2
1 1 1 25Area = 52 2 4 5 2 4 4
13 1.60 (3s.f )80
π π π
π
× + − − ×
= =
14
Functions
Given : sin ,f x x֏ ,2 2
xπ π ∈ −
5
10
(i)
(ii)
and : ( 1)(3 ),8
g x x xπ
+ −֏ x∈ℝ .
( , ]2
gRπ
= −∞
From the graph of ( )y f x= ,
( ) ( )f fα β≠ if 2 2
π πα β− ≤ < ≤ .
In fact, 1 ( ) ( ) 1f fα β− ≤ < ≤ .
From the graph of ( )y g x= ,
( ) ( )gf gfα β<
or ( ) ( )gf gfα β≠
0 x
y
-1 1 3
(1, )2
π
2π
38
π
( )y g x=
0 x
y
2
π
2
π−
-1
1
α β
( )f α
( )f β
( )y f x=
0 x
y
1 -1
2π
3 8π
( )y g x=
11
(iii)
(a)
(iii)
(b)
if 1 ( ) ( ) 1f fα β− ≤ < ≤ .
gf is one-one (or increasing).
[0, ]2
gfRπ
= .
( , ] [ . ]2 2 2
g fR Dπ π π
= −∞ ⊄ − = .
Alternative:
5(4) [ (4)] [ ]
8fg f g f
π= = − is undefined because
5[ , ]
8 2 2fD
π π π− ∉ − = .
Hence, fg does not exist as a function.
( )2
g xπ
=
( )( )1 38 2
x xπ π
+ − = −
( )( )1 3 4x x+ − = −
2 2 3 4x x− + + = − 2 2 7 0x x− − =
2 32
2x
±=
1 8x = ±
Since 1,x ≥ 1 8x = + .
1 1 8k≤ ≤ +
Greatest value of k is 1 8+ .
0 x
y
-1 1 3
(1, )2
π
( )y g x=
2y
π= −
12
1
Anglo-Chinese Junior College H2 Mathematics 9740
2010 JC 2 PRELIM Marking Scheme
Paper 2:
1
( )
( ) ( )
( ) ( )
( )
1
2
2 1
1
112
2 2
2 1 2 111
2
1 12
2 1 2 12 2
112
4 2 2
1
1 1
1 1 1 1
2 2 2 2
14 1
2
x
x
x x
x x
x xx x
e dxe
e dx e dxe e
e e e e
e e e e
−−
− −−
− − − −
−
−
−
= − − + −
=− + + +
= + − + +
∫
∫ ∫
2 (i)
2 (ii)
2 (iii)
Least value of z w− = 1
Greatest ( )arg 3z + = 1 11 3tan sin 0.826 (3 dp)
5 26
− − + =
Least ( )arg 3z + = 1 11 3tan sin 0.432 (3 dp)
5 26
− − − = −
3
ln 2 0a
xx
− ≥
where 1a> .
2
2 1
20
1 1 8 ) 1 1 8 )2
4 40
1 1 8 1 1 80
4 4
axx
x x a
x
a ax x
x
a ax or x
− ≥
− −≥
+ + − +− −
≥
− + + +≤ < ≥
4 (a) ( ) ( )3 2 3cos 3 sind
x x xdx
= −
x
y
i (2,1)
O −3
(5,−3)
o
2
3
26
3
3
z
w
2
5 3
3 2 3
3 3 3 2
33 2 3
33 3
sin
( sin )
1 1cos cos 3
3 3
cos cos3
1cos sin
3 3
x x dx
x x x dx
x x x x dx
xx x x dx
xx x c
=
= − − −
=− +
=− + +
∫∫
∫
∫
4 (b) 5 sec 10,
4
5 sec tan 2 5,3
x when x
dxwhen x
d
πθ θ
πθ θ θ
θ
= = =
= = =
( )
( )
[ ]
2 5 32 2
10
332 2
4
3
2
4
3 3
2
4 4
3
3
4
4
5
5sec 5 5 sec tan
1 sec
5 tan
1 cos 1cot cos
5 sin 5
1 1 1cos
5 sin 5
1 2 1 22 3 . .
5 15 5 15
x dx
d
d
d or ec d
ec
i e a b
π
π
π
π
π π
π π
ππ
ππ
θ θ θ θ
θθ
θ
θθ θ θ θ
θ
θθ
−
−
−
= −
=
=
= − = −
= − = = −
∫
∫
∫
∫ ∫
5
(i)
(ii)
1
1
: 2 3
1
p r
⋅ = − ɶ
1
1 2
: 0 1 ,
1 4
l r λ λ− = + ∈
ℝɶ
2 1
1 . 2 2 2 4 0
4 1
= + − = −
1 1
0 . 2 1 0 1 3
1 1
− = − + − ≠ −
⇒ l1 is parallel to p1. ⇒ l1 is not contained in p1.
Alternative method:
1 2 1
2 2 3
1 4 1
λλλ
− + = − ≠ + −
i
Since no solution for λ , ∴ l1 is parallel and not contained to p1
1 2 3
2 1 3 2
1 4 1
− × = − −
3
(iii)
(iv)
(v)
2
3 1 3
: 2 0 2 3 0 1 4
1 1 1
p r
− − − ⋅ = = + + =
iɶ
2 : 3 2 4p x y z− + + =
3
2 4 2
: 1 1 1 8 1 4 5
4 1 4
p r
⋅ = ⋅ = + − = − ɶ
3
2
: 1 5
4
p r
⋅ = ɶ
2
1 2
: 0 0 ,
1 3
l r µ µ− = + ∈ −
ℝɶ
1
1
: 2 3
1
p r
⋅ = − ɶ
( )1 2 1
0 . 2 3 2 5 3 1 point is 1,0, 2
1 3 1
µµ µ
µ
− + = ⇒ − + = ⇒ = ∴ − − −
B
2 11 1 5
sin 0 213 6 78
3 1
θ = ⋅ = − −
Length of the projection of AB on p1 = cosAB ���
=
225 53 53 1
0 1 13 31878 78 6 6
3
− = = = −
6 Procedure A is preferable as it is unbiased.
Early customers may not be typical customers in general.
7 Let R be the r.v for the length of a right shoe and L for the left shoe
R N(20, 0.142) and L N(20.1 , 0.11
2)
Method 1
X = R + L N(40.1, 0.0317)
3X N(40.1 x 3, 0.0317 x 32)
P(X > 120) = 0.713
Method 2
R + L N(40.1, 0.0317)
P(R + L >3
120 )
= 0.713
8 Let M and W be the rv for the weight of an adult man and woman respectively.
M N(75,42) and W N(65,3
2)
W M ~ N(-10, 52)
P( 1>−MW ) = P(W – M > 1) + P(W – M< -1) = 0.978
Or P( 1>−MW ) = 1 - P( 1<−MW ) = 1 – P(-1<W – M<1) = 0.978
4
No , (i) The weight of a husband and wife may not be independent
Or (ii) Randomness is not there ( a randomly chosen women but spouse is not
randomly chosen)
Or (iii) Distribution of weight of married woman is different from distribution of
adult woman.
Etc
9 Telephone costs are assumed to be normally distributed.
To test H0 : µ = 72
against H1: µ > 72 at 5% level of significance
Under H0 , T= ns
x 0
_
µ−t(5-1)
Test statistics : T= ns
x 0
_
µ−=
82.6 72
10.6677 / 5
−= 2.2219
p value = P(T > 2.2219) = 0.0452 < 0.05
Reject H0 at the 5% level of significance. We conclude that there is sufficient
evidence at the 5% level of significance that there is evidence of an increase in
mean monthly costs.
To test H0 : µ = 0µ
against H1: µ > 0µ at 5% level of significance
Under H0 , Z = n
x
89.9
0
_
µ− N(0,1)
Test statistics : Z = n
x
σµ0
_
−= 082.6
9.89 / 5
µ−= (82.6 - 0µ )
89.9
5
Do not reject H0 if P(Z > (82.6 - 0µ )89.9
5) > 0.05
(82.6 - 0µ )89.9
5< 1.64485...........(1)
µ0 > 75.3
10
(i)
161x = (from calculator or computation)
5
(ii)
(iii)
when 161x = , 103.6 0.726x y= +
(161 103.6) / 0.726y = −
79.06336088=
using y y n=∑
179.06336088 (65.1 73.2 85 80.9 89.9)
6k= + + + + +
80.3k =
Use G.C. to find regression line of y on x:
97.593 1.097y x= − +
Use y on x line to predict weight.
When 165x = , 97.593 1.097(165)y = − +
83.4y = (1 d.p.) – using 3 d.p. of a and b to compute.
or
83.5y = - using full accuracy of a and b to compute.
Using G.C., 0.893r =
C is unusually overweight.
11
(i)
Breakages occur randomly or
Breakages occur independently or
Mean number of breakages is a constant
Let A be the r.v for the number of broken cups per day
A Po(2.1)
P(A≥3) = 1 – P(A≤2) = 0.350369 = 0.350 (3 sig figs)
Let X be the r.v for the number of days with a least 3 broken cups out of n cups.
X
y
65.1
73.2
80.3 80.9
85.0
89.9
150 157 160 162 167 170
x
6
(ii)
P(X≥2) > 0.999
1 – P(X≤1) > 0.999............(1)
P(X≤1) < 0.001
Using GC,
n P(X≤1)
21 0.00145
22 0.000984
least n is 22
T Po(2.1x7 + 1.6 x 7)
T Po(25.9)
Method 1:
Since n is large, _
T (25.9, 100
9.25) approx by central limit theorem
P(_
T ≤ 26) = 0.578 (to 3 sig fig)
Method 2:
100 weeks, Y Po(2590)
λ = 2590 > 10 . Normal approx to Poisson
Y N(2590,2590) approx
P(Y≤2600) = P( Y < 2600.5) (With cc)
= 0.578 (to 3 sig fig)
12
(i)
(ii)
Let X be the r.v for the number of fish which measures less than 8 cm long.
9 5 4
5
1 4 1( ) ( ) ( )5 5 5
C = 0.00330 ( to 3 sf)
X B(n,0.2)
Since n large and p= 0.2 , X N(0.2n,(0.2)(0.8)n) approx
P(X≤10) ≤0.0227
P(X < 10.5) ≤0.0227
P(Z < n
n
4.0
2.05.10 −) ≤0.0227
n
n
4.0
2.05.10 −≤ -2.000929..........(1)
10.5 – 0.2n ≤ -0.800372 n
Hence 10.5 – 0.2n ≤ -0.8 n approx Method 1
Using GC Y1 = 10.5 -0.2x + 0.8 x →Table Ans : 91
Method 2 : Use GC and graph
7
Method 3:
Hence (10.5 – 0.2n)2 ≥ (-0.8 n )
2
Hence 4n2 - 484n + 11025 ≥0 approx……(2)
From GC : n≥90.6 or n≤ 30.4
n≥91 or n≤30 (NA because does not satisfy (1) )
Least n = 91
13
(i)
(ii)
Let H be the event that the member of public has hypertension
Let D be the event that the machine diagnosed hypertension
(i)
P(H/D) = 3
2=
)(
)(
DP
DHP ∩
3
2=
)1.0)(1(8.0
8.0
kk
k
−+ (1)
k = 0.2
p% = 20 %
p = 20
P(H’/D’) = )(
)(
DP
DHP
′
′∩′=
)9.0)(1(2.0
)9.0)(1(
kk
k
−+−
= 0.947
Examples of possible comments:
(1) If it does not find you hypertensive then you can be reasonably confident
that your blood pressure is normal.
(2) If it diagnoses hypertension, then you should consult your doctor for further
tests.
(3) Any other logical comments with reference to the context of the question
14
(i)
(ii)
Case 1: 2 red balls - 2
12
=
Case 2: 2 blue balls - 3
32
=
Case 3: 2 green balls - 5
102
=
No. of ways = 1 3 10 14+ + =
Case 1: No red ball. Choose 6 balls from a total of 8 (blue and green)
balls: 8
6
.
Case 2: No blue ball.
Choose 6 balls rom a total of 7 (red and green)
H
H’
D
D
D
’
D’
0.8
0.2
0.1
0.9
k
1-k
8
(iii)
balls: 7
6
[Note: We can’t exclude green balls because total
number of red and blue balls is only 5.]
No. of ways = 10 8 7
210 (28 7) 1756 6 6
− + = − + =
Alternative Method:
Case Green Blue Red No. of ways
1 4 1 1 5 3 2
304 1 1
× × =
2 3 2 1 5 3 2
603 2 1
× × =
3 3 1 2 5 3 2
303 1 2
× × =
4 2 3 1 5 3 2
603 2 1
× × =
5 2 2 2 5 3 2
202 2 2
× × =
6 1 3 2 5 3 2
51 3 2
× × =
Total 175
Excluding the green balls, we only have
1, 1, 2, 2, 3. Since we are ignoring the colours of
the balls, we are forming 3-digit numbers from the
5 digits 1, 1, 2, 2, 3.
Case 1: All 3 digits are distinct. The 3 digits are 1, 2, 3 and the number of ways of
arranging them are 3!
Case 2: 2 digits are identical.
Step 1: Choose 2 digits that are identical
(1, 1 or 2, 2): 2
Step 2: Choose a digit from the remaining digits
(1, 3 or 2, 3): 2
Step 3: Arrange the 3 chosen digits in a row:
3!
32!=
No. of ways = ( )( )3!3! 2 2 6 12 18
2!
+ = + =
Page 1 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
Anderson Junior College Preliminary Examination 2010
H2 Mathematics Paper 1 (9740/01)
1 Without using a graphic calculator, find the exact solution of the inequality
2
12 29 45x
x+
≤−
, 5x ≠ ± .
Hence solve the inequality 2
(12 29 ) 45 1
x xx+
≤−
. [7]
2 It is given that f(x) = tan for 0 ,
2 22 for .
2
x x
x x
π
π ππ
< ≤ < ≤
(i) Sketch the graph of y = f(x) for 0 x π< ≤ . [2]
(ii) Find the exact volume of revolution when the region bounded by the curve in (i), the line 2y = and the y - axis is rotated completely about the x-axis. [5]
3 Given that xy 2sinln 1−= , show that yxdxdy 241 2 =− . [1]
By repeated differentiation of this result,
(i) Find the series expansion of y in ascending powers of x, up to and including the term in 3x . [4]
(ii) Hence deduce the approximate value of 3eπ
, giving your answer in the form of
( )1 3 , ,8
a b a b+ ∈Ζ where a and b are to be determined. [2]
4 The curve C is given by the equation y = 1xx where x > 0. The variable point P
moves along the x−axis from x = 10 to x = 1 at a constant rate of 1 unit per second.
Another variable point Q moves along the curve C so that the x−coordinates of both
P and Q are the same at any point in time.
(i) Find dydx
in terms of x. [3]
(ii) The area of triangle OPQ is represented by A. Show that the rate of change of
A is 1 3ln 22 2
− unit2/s when t = 6. [4]
Page 2 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
5 Show that d 1tan .d 2 1 cos
xx x = +
[2]
Hence, or otherwise, show that sin d f ( )1 cosx x x x x C
x+
= ++∫ , where f (x) is a single
trigonometric function to be determined and C is an arbitrary constant. [5]
6 The function f is defined by f : 4 sec ,2
x x xπ π→ + < ≤ .
(i) Show that the inverse function f -1 exists and find f
-1 in similar form. [3] (ii) On a single, clearly labelled diagram, sketch the graphs of f and 1f − . [2]
Another function g is defined by 1g : ln , 2
2x x
x → > −
.
Find the maximal domain of the function f for the composite function gf to be
defined. Find the corresponding range of gf . [3]
7 A sequence of real numbers 1u , 2u , 3u , ….. satisfies the recurrence relation
14 2n nu au+ = − , a∈ for all positive integers n and 1 1u = . Express nu in terms of a and n . [4]
Find the range of values of a for which the sequence converges. [1]
If the sequence converges, find the limit in terms of a . [1]
8 The curve G has equation 1
ca y x bx
= + +−
where x ≠ 1 and a, b and c are
constants with a > 0.
Given that the curve G has two stationary points P and Q,
(i) Find the range of all possible values of c. [2]
(ii) Find the coordinates of P and Q in terms of a, b and c. [2]
The point (0,1) lies on one of the asymptotes of G and the lines 1y = − and 5y = are tangents to G. Find the values of a, b and c. [4]
9 The non-zero complex numbers 1z and 2z satisfy the equation 2 2
2 1 2 1 0z z z z− + =
Find the complex number 2
1
zz
, given that its imaginary part is positive. [3]
(i) The points P and Q represent 1z and 2z respectively in an Argand diagram. Describe, with reasons, the geometrical properties of triangle OPQ where O is the origin. [3]
(ii) Find the values of n for which 1 2 0n nz z+ = . [2]
Page 3 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
10 (a) The diagram below shows the graph of y = f(x). The curve cuts the x-axis at
0x = , 3x = and 5x = . It has asymptotes 4x = and 1y = − . There is a
minimum at the point A(-2, -3) and a maximum at the point 5 ,12
C
.
Sketch the graph of 2 f ( )y x= − , indicating the asymptotes and corresponding points for A, B and C clearly. [4]
(b) Describe a series of linear transformations to show how the graph of
2
1ln6 9
yx x
= − + , 0 < x < 3
can be obtained from the graph of lny x= , x > 0. [4] 11 Part of an isosceles triangle, formed by numbers following a particular pattern is
shown below: 1 1 3 5 5 7 9 11 13 21 23 25 27 29 31 33 73 75 77 79 81 83 85 87 89
(i) State the middle term in the nth row. [1] (ii) Find the first and last term of the nth row. [2]
(iii) Show that the sum of the terms in the nth row is 1(2 1)3nn −− . [2]
(iv) Find the minimum number of rows in the triangle for the number of terms in the triangle to exceed 500. [3]
……
.
B (0, 0)
C ( )52 ,1
A (-2, -3)
y = f(x)
4 3 5
-1
x
y
×
×
Page 4 of 4 AJC / 2010 Preliminary Examination / 9740 / P1
12(a) By using the substitution u = 1t
, or otherwise, show that
5 21
3 3 3 31 0(1 ) (1 )t udt dut u
∞=
+ +∫ ∫ .
Hence evaluate the exact value of this integral. [4]
(b) The diagram below shows the part of the curve C with parametric equations
2
3 3, 1 1
t tx yt t
= =+ +
, for t 1≥ .
(i) C approaches to a point when t → ∞. Find the coordinates of this point. [1]
(ii) The region bounded by the curve, the x-axis and the line x = 12
is denoted by R.
Using the result found in (a), find the exact area of R. [4] 13 The equations of a plane 1π and a line l are shown below:
1
2: 1 6
2
π ⋅ = −
r
1 4: 53 2− +
= = −x yl z
The point A has position vector 3 4− +i j k .
(i) Find the distance between point A and the plane 1π . [2]
(ii) B is another point such that 5 2→
= − −AB j k . Find the length of projection of →
AB onto the plane 1π . [2]
(iii) Using your answers in (i) and (ii), find the area of triangle ABC, where C is the reflection of A in the plane 1π . [2]
(iv) Find the equation of the plane 2π which contains the line l and the origin. Hence, find the line of intersection between the planes, 1π and 2π . [4]
END OF PAPER
12
x
y
R
0
Page 1 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
Anderson Junior College Preliminary Examination 2010
H2 Mathematics Paper 2 (9740/02)
Section A: Pure Mathematics (40 marks)
1 Find the values of a and b if the expansion of 2
91
axbx+
+ in ascending powers of x up
to and including the term in 2x is 23536
x x+ + . With these values of a and b, state
the range of values of x for which expansion is valid. [6]
2 Express 3 2
2
3 2 1( )3 2
x x xf xx x+ + +
=+ +
in partial fractions. [2]
Hence, or otherwise, show that
( )2
1
1 1f ( ) 12 2
N
xx N N
N=
= + + −+∑ . [3]
Using the sketch of f ( ) where 0y x x= > shown below, or otherwise, show that
1
11f ( ) f ( )
N N
xx x dx
+
=
<∑ ∫ . [3]
3 A complex number z satisfies ,z a a a +− = ∈ .
(i) The point P represents the complex number w, where 1wz
= , in an Argand
diagram. Show that the locus of P is a straight line. [2]
(ii) Sketch both loci on the same diagram and show that the two loci do not
intersect if 102
a< < . [4]
(iii) For 12
a = , find the range of values of 1arg za
−
, give your answer correct
to 0.1°. [3]
State the limit of 1arg za
−
when a approaches zero. [1]
0 x
y y= f(x)
Page 2 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
4 Relative to the origin O , the points A and B have position vectors acossin
1
tt
= −
and
bcos 2sin 21 2
tt
= −
respectively, where t is a real parameter such that 0 t π≤ < .
(i) Show that ⋅a b = cos( )m nt+ where m and n are constants to be determined. [2]
(ii) Hence, find the exact value of t for which AOB∠ is a maximum. [3]
(iii) Another point C has position vector c0
rs
=
where r and s are real constants.
Given that A, B and C are collinear when 2
t π= , find the values of r and s. [3]
5 An underground storm canal has a fixed capacity of 36000 m and is able to
discharge rainwater at a rate proportional to V , the volume of rainwater in the storm canal.
On a particular stormy day, rainwater is flowing into the canal at a constant rate of
3300 m per minute. The storm canal is initially empty. Let t be the time in minutes for which the rainwater had been flowing into the storm canal,
(i) show that 300(1 )kteVk
−−= , where k is a positive constant. [4]
A first alarm will be sounded at the control room when the volume of rainwater in the storm canal reaches 34500 m and a second alarm will be sounded when the storm canal is completely filled. Given that the first alarm was sounded 20 minutes after the rainwater started flowing into the storm canal.
(ii) Find the time interval between the first and second alarm. (Assuming the
weather condition remains unchanged). [3]
(iii) Briefly discuss the validity of the model for large values of t . [1]
Page 3 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
Section B: Statistics (60 marks) 6 Mary Lim has 7 cousins. In how many ways can she invite some or all of them to
her birthday party? [2] At her birthday party, Mary sets up a round table of 8 seats with a different
welcome gift at each seat. If all her cousins turn up for the party and given that 4 of them are from the Lee family, 2 are from the Yeo family and 1 is from the Tan family, find the number of ways they can be seated with Mary for a meal at the table if family members of the same surname are seated together but members of the Lee and Yeo families are not adjacent to each other. [3]
After the meal, Mary and her cousins start to play a game. The game requires a
formation of 2 facilitators and 2 teams of 3 members each. Find the number of possible formations if not all the members in each team have the same surname. [3]
7 Chemical X will react with chemical Y to form compound Z. A scientist at a
chemical plant wants to study the relation between chemicals X and Y by varying the amount of chemical Y (in milligrams) placed in a reaction flask containing 50mg of chemical X. When chemical Y is completely used up, the amount of chemical X (in milligram) left is recorded as shown in the table below.
Chemical Y used (y mg) 20 40 60 80 100 Chemical X left (x mg) 29.1 16.2 8.9 5.1 3.8
(a) (i) Find the equations of the least square regression lines of y on x and x on y. [2] (ii) Using the appropriate regression line found in (i), estimate the least amount of
chemical Y used given that chemical X has completely used up. Comment on the validity of your estimation. [2]
(b) (i) A group of scientists propose alternative models of the form byaw += ,
where w is a function of x, to describe the relation between chemicals X and Y.
Model A: 2xw =
Model B: x
w 1=
Model C: xw ln=
State, with a reason, which model is the most appropriate and find the corresponding least square regression line by performing a suitable transformation. [2]
(ii) Hence, estimate the change in w when y increases by 5. [1]
(iii) In the same experiment conducted earlier, the amount of compound Z formed, z (in miligrams), is measured. From the data collected, the linear product moment correlation coefficient between z and x is found to be 0.9− and the least square regression line of z on x is given by 125.1 +−= xz . Find the least square regression line of x on z. [4]
Page 4 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
8 In year 2009, the average length of time for cars parked at the Integrated Resort
Haven (IRH) was 12 hours. The facilities are upgraded in 2010 and the management of IRH wants to find out if there is difference in the mean length of time for cars parked at IRH.
Assuming that the resort and the car park are open at all times. The length of time,
x hours, for cars parked at IRH were recorded for 200 randomly selected cars on a particular day and the following results were obtained:
2( 12) 80 , ( 12) 1425x x− = − =∑ ∑ .
(i) Denoting the population mean and variance of the parking times by µ and 2σ respectively, find unbiased estimates of µ and 2σ . [2]
(ii) Given that an appropriate hypothesis test carried out could not provide sufficient evidence to indicate a difference in the mean length of parking time, find the range of values of the significance level of this test. [3]
(iii) State, with a reason, the range of the significance level of another test (without carrying out the test) such that the sample could not provide sufficient evidence that there is an increase in the mean length of parking time. [1]
(iv) The sample of 200 cars could also be obtained using systematic sampling. Describe how this can be done. [2]
9 A multiple-choice question (MCQ) consists of 5 suggested answers, only one of
which is correct. For each of the questions set for a particular topic, there is a probability of p that a student, Alice, knows the correct answer, and whenever she knows the correct answer she selects it. If she does not know the correct answer, she randomly selects one of the 5 suggested answers. The events K and C are defined as follows:
K: Alice knows the correct answer.
C : Alice selects the correct answer.
(i) Find the probability, in terms of p, that Alice selects the correct answer. [2] (ii) Describe what the event K’∩ C represents in the context of this question. [1]
(iii) Given that P(K’|C) = 116
, find the value of p. [2]
Taking p to be 0.3 and Alice answers 1 MCQ daily from Monday to Friday. (iv) Given that she answers 3 MCQ correctly, find the probability that this
happens in 3 consecutive days. [2] (v) Alice scores 3 marks for each correct answer, but loses 1 mark for each
incorrect answer. Find the probability that Alice obtains a negative score for the 5 MCQ she attempted. [2]
Page 5 of 5 AJC / 2010 Preliminary Examination / 9740 / P2
10 At a newly opened shop, the number of orders for herbal chicken soup received in a
randomly chosen 30-minute interval follows a Poisson distribution with mean 2.3. The shop is opened for 8 hours daily, from 11 am to 7 pm. (Assume that the orders received are independent.)
(i) Find the probability that there are at least 6, but less than 10 orders received in a randomly chosen one-hour period. [2]
(ii) Find the probability that in 100 randomly chosen one-hour periods, the shop receives an average of more than 5 orders in a one-hour period. [2]
(iii) The shop owner incurs a fixed operating cost of $250 per day. The cost price and selling price of a bowl of herbal chicken soup are $8 and $20 respectively. By using a suitable approximation, find the probability that he makes a profit of at least 40% of his total cost incurred per day. [4]
(iv) After operating the shop for half a year, the shop owner wishes to assess if he should continue with the business. He decides to observe the lunch time crowd from 12 pm to 2 pm for 25 days selected at random. If there are less than 14 days with more than 10 orders during the lunch period, he will close down the shop. Comment on whether he should close down the shop. [3]
Explain whether the Poisson distribution is a good model for the number of orders for herbal chicken soup in a day. [1]
11 Peter bought an ice-cream machine. The amount of time taken by the machine to
produce a large tub of ice-cream follows a normal distribution with mean µ minutes and standard deviation σ minutes. It is found that there is a 88% chance that the machine will take less than 60 minutes and a 70% chance that it will be more than 50 minutes. The amount of time taken by the machine to produce a small tub of ice-cream also follows a normal distribution with mean 20 minutes and standard deviation 2 minutes. The amount of time taken by the machine to produce a large tub of ice-cream and a small tub of ice-cream are independent of each other.
(i) Find µ and σ . [3]
(ii) Find the probability that the difference between the amount of time taken by the machine to produce 5 large tubs of ice-cream and thrice the amount of time taken to produce 3 small tubs of ice cream is more than 1 hour. [4]
(iii) After using the machine for a year, Peter decides to test the functionality of the machine by using it to produce n large tubs of ice-cream (where n is large) and observing the time taken for each production. He will consider buying a new machine if there are more than 20 times that it takes at least 60 minutes to produce a large tub of ice-cream. Using a suitable approximation, find the greatest value of n such that the probability that he needs to consider buying a new machine is less than 0.2. [5]
END OF PAPER
Page 1 of 6
AJC H2 Maths _Prelim 2010_P1 (Solutions)
1
2
12 294
5
x
x
+≤
− ⇒
2
2
(12 29) 4(5 )0
5
x x
x
+ − −≤
−
⇒ 2( 5 )( 5 )(2 3) 0x x x− + + ≤
⇒ x= -3/2, 5x ≤ − or 5x ≥ x= -3/2, 5x < − or 5x > ( 5x ≠ ± )
2
(12 29 )4
5 1
x x
x
+≤
−
Replace x by 1/x, ⇒ x= -2/3, 1
5x
< − or 1
5x
>
⇒ x= -2/3, 1
05
x− < < or 1
05
x< <
⇒ x= -2/3, 1 1
5 5x− < < as x = 0 is a solution to the inequality
2 (i)
(ii) Volume required = 2 22
02
2( 2) ( ) tan
2
xdx x dx
ππ
ππ π π ππ
− −∫ ∫
= 2
2 2
20
2 2 tan2
xx x
ππππ π − − −
= 272
4π π −
units
3
3 xy 2sinln 1−=
241
21
xdx
dy
y −=
⇒ (shown) 241 2 yx
dx
dy=− ------- (1)
dx
dy
x
x
dx
dyx
dx
yd 2
41
441
2
2
2
2
=
−
−+− ⇒ ( ) y
dx
dyxx
dx
yd4441 2
2
2
=
−− ---- (2)
( ) ( )dx
dy
dx
ydx
dx
dyx
dx
ydx
dx
yd444841
2
2
2
22
3
3
=
−
−−+−
( ) ( ) 0812412
22
3
3
=
−−+−dx
dyx
dx
ydx
dx
yd ---------------- (3)
1,0When == yx , 16 ,4 ,23
3
2
2
===dx
yd
dx
yd
dx
dy
0
2
π
x
2
π
Page 2 of 6
...!3
16
!2
421
32
++++=∴xx
xy 3
8221
32 xxxy +++≈∴
xee 2sin31
Let −
=π
x2sin3
1−=π
⇒ 4
3=x ( )3511
8
1
4
3
3
8
4
32
4
321
32
3 +=
+
+
+≈
π
e
5 ,11 ==∴ ba
4 (i) y = 1/ xx ⇒
1ln lny x
x=
⇒ 3/ 2 3/ 2
1 1 1ln
2
dyx
y dx x x= − +
⇒ 3/ 2
ln1
2
dy y x
dx x
= −
= 1/
3/ 2
ln1
2
xx x
x
−
(ii) A = 1
2xy ⇒
1 1
2 2
dA dyy x
dx dx= +
When t = 6, x = 4 so y = 2 and ( )11 ln 2
4
dy
dx= −
so ( ) ( ) ( )1 1 12 4 1 ln 2
2 2 4
dA
dx
= + − =
3 1ln 2
2 2−
Given that 1dx
dt= − ,
we have dA dA dx
dt dx dt= × = ( )3 1
ln 2 12 2
− −
= 1 3
ln 22 2
− units/sec
5 21
tan sec 2 2 2
d x x
dx
= 2
1 1.
1 cos2cos
2
x x= =
+
Using integration by parts,
sin ( sin ) tan - tan (1 cos )
1 cos 2 2
x x x xdx x x x dx
x
+ = + + + ∫ ∫
= 2( sin ) tan - tan (2cos ) 2 2 2
x x xx x dx
+ ∫
tan sin tan sin 2 2
x xx x x dx
= + − ∫
= tan 2sin cos tan cos2 2 2 2
x x x xx x A
= + + +
2 2tan 2sin 1 2sin2 2 2
x x xx A
= + + − +
tan , where = 1 + 2
xx C C A
= +
Page 3 of 6
6
Function f is a one-one function from
the sketch of ( )y f x= , hence f-1
exist.
Let 4 secy x= +
1
sec 4cos
x yx
⇒ = = −
Hence -1 1 1
f : cos ( ) , 34
x xx
−→ ≤−
.
1g : ln ln( 2) , 2
2x x x
x→ = − − >
−
For Rf to be a subset of Dg , range of f need to be
restricted to (2,3] , therefore the maximal domain of f is 2
( , ]3
ππ .
Corresponding range of gf is [0, )∞
7
14 2n nu au+ = − 1
1
4 2n n
au u+⇒ = −
1 1
1 1
4 4 2 2n n
a au u+ −
⇒ = − −
2
1 1
11
4 2 4n n
a au u+ −
⇒ = − +
2
1 2
1 11
4 4 2 2 4n n
a a au u+ −
⇒ = − − +
3 2
1 2
11
4 2 4 4n n
a a au u+ −
⇒ = − + +
( 1) 1 2 1
1 ( 1)
11 ...
4 2 4 4 4
n n
n n n
a a a au u
− + −
+ − −
⇒ = − + + + +
1 1
1(11 4
4 21
4
n
n
n
a
au u
a+
− ⇒ = − −
1
6 2
4 4 4
n
n
a au
a a+
− ⇒ = − − −
16 2
4 4 4
n
n
a au
a a
−− ∴ = − − −
If the sequence converges then 14
a< 4a⇒ < . The sequence converges to
2
4 a−
−.
8 (i) ay = x + b +
1
c
x −
⇒ ( )2
11
1
dy c
dx a x
= −
− = 0 when (x – 1)
2 = c.
This equation has two real and distinct solutions only when c > 0.
0 |
3
|
π
2
π
( )y f x=
1( )y f x−=
y x=
Page 4 of 6
(ii) From (i), the stationary points of C are at x = 1 c± .
When x = 1 c− , ay = 1 c− + b − c ⇒ 1 2b c
ya
+ −=
When x = 1 c+ , ay = 1 c+ + b + c ⇒ 1 2b c
ya
+ +=
(iii) The equations of the asymptotes of C are x = 1 and y = x b
a
+.
Given (0, 1) is a point on y = x b
a
+, we get a – b = 0 - - - (1)
Given that the line y = k does not pass through any point on C only for values of k in the
interval (0, 2), the min turning pt corresponds to y = 0 and the max turning pt corresponds to
y = 2:
1 2b cy
a
+ −= = −1 ⇒ a + b – 2 c = –1
1 2b cy
a
+ += = 5 ⇒ 5a – b – 2 c = 1
Hence a = 1, b = 1 and c = 3
2, i.e. c =
9
4.
9 2 2
2 1 2 1 0z z z z− + =
2
2 2
1 1
1 0z z
z z
⇒ − + =
2
1
1 1 4 1 3
2 2
z i
z
± − ±⇒ = =
Since imaginary part is positive, 2
1
1 3
2
z i
z
+⇒ =
(i) 2 2 12 2
1 1
(1/ 2) ( 3 / 2) 1 arg tan 3 / 3z z
andz z
π− = + = = =
|z2| = | z1| and arg (z2) = arg(z1) + π/3
Triangle OPQ form an equilateral triangle as OP = OQ and ∠POQ = 60o
(ii) 21 2
1
0 1
n
n n zz z
z
+ = ⇒ = −
( 2 )3 (1 2 ) 3(1 2 )3
ni
i k ne e k n k
ππ π π
π+ = ⇒ = + ⇒ = +
for any integer k
10
a
10
b ( ) 2
2
1ln ln 3 2ln(3 )
6 9y x x
x x
− = = − = − − − +
Page 5 of 6
The graph can be obtained from lny x= by
1. translate of 3 unit in the negative x direction ln( 3)y x= +
2. reflect in the y-axis ln( 3)y x= − +
3. scale by factor 2, parallel to the y-axis
4. reflect in the x-axis
11 ) i) For the nth
row, middle term = 13n−
ii) For the nth
row, first term = 13 2( 1)n n− − − , last term = 13 2( 1)n n− + −
iii) Sum of terms in nth
row
( ) ( )1 1
1 1
2 13 2( 1) 3 2( 1)
2
2 1(2)(3 ) (2 1)3
2
n n
n n
nn n
nn
− −
− −
− = − − + + −
−= = −
iv) No. of terms in each nth
row row = (2n-1) Therefore, total no. of terms in n rows 1
(2 1)n
r
r=
= −∑
1
(2 1) 500n
r
r=
− >∑
1
2
2 500
( 1)2 500 500 0
2
n
r
r n
n nn n
=
− >
+ − > ⇒ − >
∑
Min no. of rows = 23
12 (i)
5
3 31 (1 )
tdt
t
∞=
+∫ 50
21 3
3
1
1
1(1 )
u duu
u
− +
∫=
21
3 30 ( 1)
udu
u +∫ (shown)
= 1
3
1 2 3 3
03 (1 )u u du−+∫ =
1
3
13 2
0
(1 )
2
u − +
− =
1 1 1
24 6 8− + =
(ii) As t → ∞ , x, y → 0. The point is (0, 0)
(iii) Area required = 1 2 3
12
3 3 20
(1 2 )
1 (1 )
t ty dx dt
t t∞
−= •
+ +∫ ∫ = 5 2
3 31
2
(1 )
t tdt
t
∞ −
+∫
= 5 2
3 3 3 31 1
2
(1 ) (1 )
t tdt dt
t t
∞ ∞−
+ +∫ ∫
= 2(1
8) − 2 3 3
1
13 (1 )
3t t dt
∞ −+∫
= 3 2
1
1 1 (1 )
4 3 2
t∞− +
− −
= 1 1 1 50
4 6 4 24
+ − =
units2
13 i) distance between point A and the plane 1π
Page 6 of 6
2 2 2 2 2 2
3 2
1 1
4 2 6 6 1 82 3
32 1 2 2 1 2
− ⋅ − − − = − = − =
+ + + + unit
ii)
0
5
2
= − −
����
AB , 2 2 2
2 21 1
1 132 1 ( 2) 2 6
n
= =
+ + − − −
⌢
ɶ
length of projection
2 2 2
0 21
ˆ 5 13
2 2
12 61 2 2 2
4 2 6 2 5 65 units3 3 3 3
10 5
= × = − × − −
= − = − = + + =
����
ɶAB n
iii) 1 2
Area of triangle = ( )( 65)2 3
ABC AC
21 2(3 2)( 65) 2 65 units
2 3= × =
iv)
1 3
: 4 2 ,
5 1
= − + λ λ ∈ℜ
l r , Two vectors // to 2π are
1
4
5
−
and
3
2
1
,
normal to 2π is // to
1 3 14 1
4 2 14 14 1
5 1 14 1
− − − × = =
Equation of 2π :
1
1 0
1
− ⋅ =
r , ie. 0− + + =x y z
To find line of intersection:
0
2 2 6
− + + =
+ − =
x y z
x y z
The augmented matrix, M is 1 1 1 0
2 1 2 6
− −
, RREF (M) = 1 0 1 2
0 1 0 2
−
2
2
− =
=
x z
y
2 2 1
2 2 0
0 1
+ = = +
x z
y z
z z
ie. Equation of line:
2 1
2 0 ,
0 1
= + µ µ∈ℜ
r
[Alternatively, Cartesian equation of line: 2 , 2− = =x z y ]
Alternatively,
use ˆ����
iɶ
AB n ,
followed by
Pythagoras thm
Page 1 of 5
AJC H2 Maths _Prelim 2010_P2 (Solutions)
1
( ) ( )1 1
22
2
99 1
1
axax bx
bx
−+= + +
+ ( )
2
2
1 1
2 23 1 ... 1 ...
18 2! 9
a ax x bx
− = + + + − +
( )2
2 23 1 ... 1 ...
18 648
a ax x bx
= + − + − +
22
3 36 216
a ax b x
≈ + + − −
By comparing coefficients, a = 6 and b = -2
The valid range for expansion of 2
9
1
ax
bx
++
is 2
1<x .
2 3 2
2
3 2 1 1 1( )
3 2 1 2
x x xf x x
x x x x
+ + += = + −
+ + + +
3 2
21 1 1
3 2 1 1 1( )
3 2 1 2
N N N
x x x
x x xf x x
x x x x= = =
+ + + = = + − + + + + ∑ ∑ ∑
1
1 1
1 1
1 2
1 1
1 2
N
x
N N
x x
xx x
xx x
=
= =
= + − + +
= + − + +
∑
∑ ∑
1 1( 1)
2 2 2
NN
N= + + −
+
1
( )N
x
f x=∑ = sum of N rectangles under the curve.
1
1
( )
N
f x dx
+
∫ = sum of the area under the curve
As shown in the diagram, as the curve is concave
upwards,
Area of N rectangles < area bounded by the curve 1
1 1
( ) ( )
NN
x
f x f x dx
+
=
∴ <∑ ∫
3 Let w =
1
z,
|z – a | = a ⇒ 1 1 aw
a a aw w
−− = ⇒ =
1w w
a⇒ − =
The locus of w is a straight line (the perpendicular bisector of 1
1w
a= & w2=0
If 1
02
a< < ⇒ 0 2 1a< < and 1
12a
>
Hence 1
22
aa
<
Therefore the 2 loci do not intersect.
1
2a =
0 x
y y= f(x)
1 2 3 N+1 N
a 2a
1
a 1
2a
Page 2 of 5
0.5 1sin
1 2 0.5 3
19.5 o
AP a
ABa
a
θ
θ
= = = =−−
∴ =
Hence 1
160.5 arg 180o oza
≤ − ≤
or 1
180 arg 160.5o oza
− < − ≤ −
As a approaches 0, 1
arg 180o
za
− →
4i
4ii
4iii
a b⋅ =ɶ ɶ
cos cos 2
sin sin 2
1 1 2
⋅ − −
t t
t t1
cos cos 2 sin sin 22
= − −t t t t1 1
cos( 2 ) cos32 2
= + − = −t t t
( )22 2 2 2
1cos3
2 12cos cos35 21cos sin 1 cos 2 sin 2
2
−⋅ ∠ = = = − + + + +
ta b
AOB ta b
t t t t
For maximum ∠AOB , since 0 ≤ ∠ ≤ πAOB and cosθ is a decreasing function over [ ]0,π , we aim
to minimize cos∠AOB , ie. 2 1
cos35 2
−
t .
Thus, cos3 1= −t , ie. 3
π=t (since 0 t π≤ < ).
When 2
tπ
= ,
0 1
1 , 0
1 1 2
− = = −
a b
1 0 1
0 1 1
1 2 1 3/ 2
AB
− − = − = − −
����
and
0
1 1
0 1 1
r r
AC s s
= − = − −
����
Since A, B and C are collinear, AB k AC=���� ����
⇒
1
1 1
3/ 2 1
r
k s
− − = −
Looking at the z component, k= 3/2, so r = -2/3 and s = 1/3
5i
300dV
kVdt
= −1
1300
dV dtkV
⇒ =−∫ ∫
1ln 300 kV t C
k⇒ − − = +
( )300 k t CkV e− +⇒ − =
300 ktkV Ae−⇒ − =
When 0t = , 0V = 0300 0 Ae⇒ − = 300A⇒ = 300(1 )kte
Vk
−−⇒ = (Shown).
When 20t = , 4500V = , 20300(1 )
4500ke
k
−−⇒ = 2015 (1 )kk e−⇒ = −
From the GC, 0.030293k =
2nd alarm : when 6000V =
0.030293300(1 )6000
0.030293
te−−∴ = 30.7t⇒ =
The residents will have 10.7 minutes between the 1st and 2
nd alarm.
33009903
0.030293t V m→ ∞ ⇒ → = which is impossible as the canal has only a fixed volume of
36000m . The model is not valid for large values of t.
1/2 2 θ
P
BA
Page 3 of 5
6 No of ways to invite her guests = (2)7 – 1 = 127
No of ways to arrange a round table = 2! X 4! X 2! X 8 = 768
No of ways to select 2 facilitators and 2 teams (Without any restrictions)
= 8C2 x (6C3 x 3C3) ÷ 2! = 280 No of ways to select the 2 facilitators and 2 teams with Lee family forming 1 team
= 4C3 x 5C2 x 3C3 = 40
No. of possible formations = 280 – 40 = 240
7 (a)(i) For x on y, s.f.) (3 1.31309.013.313085.0 +−=⇒+−= yxyx
For y on x, (3s.f.) 0.9685.2999.958526.2 +−=⇒++−= xyxy
(ii) Since chemical Y is the controlled variable, use regression line of x on y.
91.10013.313085.00 =⇒+−= yy
The estimation is not valid as this is an extrapolation, linear relation may not hold outside the
range of data.
1(b)(i) By comparing the linear product moment correlation for the 3 models, Model C is the
most appropriate with the highest value of 993.0=r as it best describes the data given.
Using linear transformation xw ln= ,
Regression line of w on y is 0.026136 3.8294 0.0261 3.83 (3 s.f.)w y w y= − + ⇒ = − +
(ii) Change in w = 0.026136(5) 0.13068 0.131 (3 s.f.)− = − ≈ −
w decreases by 0.131.
(b)(iii) ( ) 5.1b where9.022 −==−= bdr
( )54.0
5.1
9.0 Hence
2
−=−−
=d
Since ( ) 125.1 line reg on the lies , +−= xzzx ,
125.1 +−= xz
29.1 16.2 8.9 5.1 3.83.3220655
5x
+ + + += =
1.5(3.3220655) 12 7.0169z = − + =
Hence, reg line of x on z is (3.3220655) 0.54( 7.0169)x z− = − −
0.54 7.11 (3.s.f)x z∴ = − +
8 (i) Unbiased estimate of µ is
8012 12.4
200x = + =
Unbiased estimate of 2σ is s2 2200 1425 80
( ) 7199 200 200
= − =
(ii) H0 : µ = 12
H1 : µ 12≠ Test Statistic : 12
/ 200
XZ
s
−= ~ N(0,1) under H0 by CLT
Using GC, p -value = 0.0325 Given that H0 is not rejected, 3.25α <
(iii) Since we will be using a one-tailed test in stead of a two-tailed test,
1(3.250944) 1.625472 1.63
2α α< = ⇒ <
Page 4 of 5
(iv) The management must have a sampling frame (the list of all cars parked). If there are N cars,
choose a random number, k, from 1 to N/200 (take the nearest integer value), then select
every (N/200)th car until a sample of 200 is obtained.
9 (i) P(C) = ( ) 1
1 15
p p× + − ×
(ii) K’∩C represents the event where Alice does not know the correct answer but she
answers correctly.
(iii) P(K’|C) = ( ' ) 1
( ) 16
P K C
P C
∩=
( )
( )
11
151 16
1 15
p
p p
− ×=
× + − × solving, get p =0.75.
When p = 0.3, P(C) = 0.3+0.7×0.2=0.44
P(3 consecutive correct | 3 correct answers) = 3 2
3 2 5
3
(0.44) (0.56) 3 3
(0.44) (0.56) 10C
×=
×
P(negative score) = P(0 correct or 1 correct) = 5 5 4
1(0.56) (0.56) (0.44)C+ = 0.271
10 Let X be the r.v. denoting the number of orders for herbal chicken soup in 30 min.
~ (2.3)X Poi
Let Y be the r.v. denoting the number of orders for herbal chicken soup in 1 hour.
~ (2.3 2)Y Poi × i.e. ~ (4.6)Y Poi
(i) ( ) ( ) ( )6 10 9 5P Y P Y P Y≤ < = ≤ − ≤ 0.29471 0.295(3. . )s f= ≈
(ii) ~ (4.6)Y Poi , Mean = variance = 4.6
Since n = 100 is large, by CLT, 4.6
(4.6, )100
Y N∼ approximately
( 5) 0.031090 0.0311(3. . )P Y s f> = ≈
(iii) Let T be the r.v. denoting the number of orders for herbal chicken soup in a day (8 hours).
~ (2.3 16)T P × i.e. ~ (36.8)T P
Since 10λ > , ~ (36.8,36.8)T N approximately.
( )( )20 1.4 8 250P T T≥ +
( ) 3508.8 350 0 ( 39.773)
8.8P T P T P T
= − ≥ = ≥ = ≥
= P(T≥40) = P(T>39.5) cc = 0.328
(iv) Let A be the r.v. denoting the number of orders for herbal chicken soup in 2 hours
(lunch time)
~ (2.3 4)A P × i.e. ~ (9.2)A P
Let H be the r.v. denoting the number of days with more than 10 orders during lunch period
from 12 to 2pm.
( ) ( )10 1 10 1 0.68202 0.31798P A P A> = − ≤ = − =
( )( )~ 25, 10H B P A > i.e. ( )~ 25,0.31798H B
( 14) ( 13) 0.98947 0.989 (3.s.f)P H P H< = ≤ = =
Since the probability of having less than 14 days with more than 10 orders during the lunch
period is quite high, he should close down his business.
Page 5 of 5
(iv) The use of the Poisson model is not suitable in this context as the number of orders during
lunch and dinner period will likely be higher than the rest of the hours in a day, thus it is
unlikely that the mean number of orders is the same for each 30 minute period.
11 (i) Let A be the r.v. denoting the amount of time taken by the machine to produce a large tub of
ice- cream. 2( , )A N µ σ∼
( 60) 0.88
600.88
601.17499
60 1.17499 (1)
P A
P Zµ
σµ
σµ σ
< =
− < = −
=
= − − − − −
Equating (1) and (2), 5.8845 5.88 (3.s.f) and 53.08583 53.1 (3.s.f)σ µ∴ = = = =
(ii) Let B be the r.v. denoting the amount of time taken by the machine to produce a small tub of
ice- cream. 2(20, 2 )B N∼
Required prob = ( )( )1 2 3 4 5 1 2 33 60P A A A A A B B B+ + + + − + + >
Let ( )1 2 3 4 5 1 2 33T A A A A A B B B= + + + + − + + .
( )85.42915,281.1367T N∼
( )( )1 2 3 4 5 1 2 33 60P A A A A A B B B+ + + + − + + > ( )60P T= >
( )1 60 60P T= − − < < 0.935317 0.935 (3.s.f)= =
(iv) Let C be the number of times the machine takes at least 60 minutes to produce a large
tub of ice-cream.
( )( ), 60C Bin n P A ≥∼ i.e. ( ),1 0.88C Bin n −∼
Since n is large, np > 5 and n(1-p) > 5, ( )0.12 ,0.1056C N n n∼
( )20 0.2P C > <
( )20 0.5 0.2 (cc)P C > + <
20.5 0.12
0.20.1056
nP Z
n
− > <
Using GC, ( )0.84162 0.2P Z > =
Thus, 20.5 0.12
0.841620.1056
n
n
−> 0.12 0.2735 20.5 0n n⇒ + − <
Solving -14.26 < n < 11.98 greatest n = 143.
( 50) 0.70
500.70
500.52440
50 0.52440 (2)
P A
P Zµ
σµ
σµ σ
> =
− > = −
= −
= + − − −−
1
CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2
MATHEMATICS 9740/01 Paper 1 31 August 2010 3 hours Additional Materials: Answer Paper
Graph Paper List of Formulae (MF 15)
READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this cover sheet in front and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
Name : __________________________________________ Class : ___________
Question No. Marks Question No. Marks
1 / 4 7 / 8
2 / 7 8 / 10
3 / 9 9 / 10
4 / 8 10 / 15
5 / 8 11 / 12
6 / 9 TOTAL / 100
------------------------------------------------------------------------------------------------------------------------- This document consists of 5 printed pages. [Turn Over]
2
1. Three sisters Audrey, Catherine and Gladys went on a shopping spree while on holiday. They bought items for their sister Eliza, who could not go due to her ‘A’ level examinations. After returning home, they could not recall the prices of the items they had bought, but they had their credit card statements with the total amounts they each had spent. The list of items that each sister had bought and the amount on their statements are summarised in the table below:
Type of Item (Price) Catherine Audrey Gladys High heels ($x per pair) 5 2 3 Facial Mask ($y per box) 10 7 15 Handbags ($z per piece) 3 8 5 Credit Card Statement ($) 1298.20 1158.30 1837.70
(i) Find the values of x, y and z. (ii) Hence evaluate the total cost of their gift to Eliza, which consists of 1 pair of high
heels, 5 boxes of facial mask and 2 handbags.
[3] [1]
2. A sequence is defined by 11 =u and nn un
nu
+=+ 21
1 for 1≥n .
(i) Write down the values of 432 ,, uuu and 5u . (ii) Hence make a conjecture for nu in terms of n. Prove your conjecture using
mathematical induction.
[2] [5]
3. In an East Asia ancient culture, there is a war monument which is formed by placing
concentric circular slabs of granite on top of each other. The slabs of granite are of decreasing measurements. (i) The first slab of the monument has diameter 200 cm and the diameter of each
subsequent slab is three-quarters the diameter of the previous slab. Show that the total sum of the cross-sectional circular area of the slabs used will be less than 72 000 cm2, no matter how many slabs there are.
(ii) The first slab of the monument has thickness 50 cm and each subsequent slab has
thickness d cm less than of that for the previous slab. Given that the maximum possible number of slabs is 14, find the largest integer value of d.
(iii) Given that d = 3, write down an expression in terms of n for the volume of the nth
slab and hence evaluate the total volume of the monument which consist of 14 slabs.
[4] [2] [3]
[Turn Over]
3
4. The diagram below shows the graph of xxy −+= 2ln2 . The two roots of the equation 2ln2 =− xx are denoted by α and β , where βα < .
α βx
y
α βx
y
(i) Find the values of α and β , correct to 3 decimal places. A sequence of real numbers x1, x2, x3, … satisfies the recurrence relation 2)ln( 2
1 +=+ nn xx for 1≥n . (ii) Prove that if the sequence converges, it must converge to either α or β . (iii) By considering nn xx −+1 and the graph above, prove that
βα <<>+ nnn xxx if1 , βα ><<+ nnnn xxxx or if1 .
(iv) Hence deduce the value that xn converges to for 21 =x , giving your answer correct to 3 decimal places.
[2]
[2] [2] [2]
5. The curve C has equation 1
2
−+
=x
xaxy where 1≠x and a is a constant.
(i) Show that C does not have any stationary points when 01 <<− a . (ii) Sketch C when 01 <<− a . Show on your diagram, the equations of the asymptotes
and the coordinates of any points of intersection with the axes.
[3] [5]
[Turn Over]
4
6.
(a)
The diagram above shows the graph of f ( )y x= . On separate diagrams, sketch the graphs of (i) y2 = f (x) (ii) y = f ′(x) showing clearly in each case the axial intercepts, the asymptotes and the coordinates of the turning points, where possible.
(b) A graph with equation y = g(x) undergoes in succession, the following
transformations: A: A reflection in the y-axis B: A translation of 4 units in the direction of the positive x-axis C: Scaling parallel to the y-axis by a factor of 2
The equation of the resulting curve is 32 −= xy . Determine the equation y = g(x).
[3] [3] [3]
7. Relative to the origin O, the points A, B and C have position vectors
−
321
and312
,121
respectively.
The point P lies on the line AB produced such that AP:PB = 3: 1. (i) Find the vector OP .
(ii) The line l has equation r =
−+
11
2
210
λ , where ℜ∈λ .
Determine whether l and the line through A and B are intersecting. (iii) Find the shortest distance from C to AB. Hence or otherwise, find the area of triangle
ABC.
[2] [3] [3]
[Turn Over]
y
x 0 −2
2
(−1, −3)
1
5
~End of Paper~
8. The equation of a closed curve is (x + 2y)2 + 3(x − y)2 = 27. (i) Show, by differentiation, that the gradient of the curve at the point (x, y) may be
expressed in the form dydx =
y − 4x7y − x
(ii) The normal to the curve at the point (−2, 1) cuts the x-axis at P and the y-axis at Q. Denoting the origin as O, find the area of the triangle OPQ.
(iii) Find the equations of the tangents to the curve that are parallel to the y-axis.
[3]
[4] [3]
9. Let y = ln (1 − sin x).
(i) Show that ey dydx = − cos x. Hence, show
d2ydx2 +
dy
dx2
= e−y − 1
(ii) Find the Maclaurin’s series for y, up to and including the term in x3.
(iii) Using the result in part (ii), find the Maclaurin’s expansion for cos x
sin x − 1 up to and
including the term in x2.
(iv) Given that x is small, use binomial expansion to expand cos x
sin x − 1 up to and
including the term in x2. State the range of x for the expansion to be valid.
[3]
[3] [1] [3]
10. (a) Find xxx dsine2∫ .
(b) A curve C is defined by the parametric equations ( )ttx sin2 −= , ( )ty cos12 −= for ππ ≤≤− t .
(i) Sketch the curve C. (ii) Find the exact area of the region bounded by C and the line y = 4.
(c) The region R in the second quadrant is bounded by the y-axis, the line y = x + 5, and the curve
2
e xy = . Find the volume of the solid formed when R is rotated through 4 right angles about the y-axis.
[4] [2]
[6]
[3] 11. (a) (i) Solve the equation iz )318(186 +−= , expressing the solutions in the form
θire , where 0>r and πθπ ≤<− . (ii) Show all the roots on an Argand diagram and write down an equation of the
form kyx =+ 22 which contains all the roots. (b) The complex number z satisfies the relations
i32i23 +−≤+−z and ( )4
3i56arg0 π≤+−< z .
(i) Illustrate both of these relations on a single Argand diagram. (ii) Hence write down the exact area represented by the given relations.
[4] [3]
[4] [1]
Page 1
CATHOLIC JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2
MATHEMATICS 9740/02 Paper 2 15 September 2010 3 hours Additional Materials: Answer Paper Graph Paper List of Formulae (MF 15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this cover sheet in front and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Name : ____________________________________________ Class : _______________
Question No. Marks Question No. Marks
1 / 5 7 / 12
2 / 10 8 / 10
3 / 11 9 / 6
4 / 14 10 / 8
5 / 7 11 / 8
6 / 9 TOTAL / 100
------------------------------------------------------------------------------------------------------------------------------------ This document consists of 6 printed pages. [Turn Over
Page 2
Section A: Pure Mathematics [40 marks]
1 Solve algebraically the inequality
017747042
2
2
≥−−+−+
xxxx .
Hence, solve the inequality
017747042
2
2
≥−−+−+
xx
xx
eeee .
[3] [2]
2 The equations of the line l1 and the plane p1 are respectively,
−+
=
111
001
µr and
−+
=
11
2
111
tsr , where µ, s, t ∈ ℜ
(i) Find the acute angle between l1 and p1.
(ii) A second plane p2 has equation 11 =
⋅β
αr . Given that the two planes p1 and p2
intersect at the line l2: 25;
2154
==−− zyx , find the values of α and β.
(iii) The plane p3 with equation 12 =++ zbyx is parallel to l2. Find the value of b.
Hence find the distance between l2 and p3.
[3] [3] [4]
3 The functions f and g are defined as follow:
xxx 2:f + , }0{\ℜ∈x
)3ln(:g −xx , 3>x
(i) Show that fg does not exist. Let the function h be defined as follows:
)3ln(:h −xx , ax >for (ii) Find the least value of a such that fh exists. Define fh in similar form. (iii) Sketch the graph of f, indicating the turning points, asymptotes and axial intercepts, if
any. State the range of f and determine if f -1 exists. The function f has an inverse if its domain is restricted to kx > . Find the least value of k.
[2] [3] [4] [2]
[Turn Over
Page 3
4 (a) Verify that y = x is a particular solution of the differential equation
0,,2d
d 22
≠+
= yxxy
yxxy
(b) Show that the substitution y = ux reduces the differential equation
xyyx
xy
2dd 22 +
=
to the differential equation
uu
xux
21
dd 2−
= .
Hence find the general solution of the differential equation
xyyx
xy
2dd 22 +
= .
(c) Due to a rapid disease outbreak, the population of fish in a river, x (in thousands), is
believed to obey the differential equation tae
tx 22
2
4dd −=
where t is the time in days, and a > 0 is a constant. Given that the entire population of fish is wiped out by the disease eventually, show that the general solution of the differential equation is taex 2−= . Explain the meaning of a, in the context of the question. Sketch the family of solution curves of the differential equation for a = 1 and 2.
[2] [3] [4] [3] [2]
Section B: Statistics [60 marks]
5 (a) Tourists visiting an amusement park in Singapore can purchase a day pass. Each day pass allows one tourist to take any of the 6 designated rides only once.
(i) How many different ways can the order of the rides be chosen if a tourist goes on
all 6 rides?
(ii) The attendants in the park keep track of the rides each tourist has taken by marking on the ticket as shown in the diagram on the right. Filling in a square indicates that the tourist has taken that particular ride. If a tourist can choose to go on a ride or not, how many different ways can the ticket be marked?
(iii) If a tourist goes on at least one ride, find how many different selections he can
make? (b) From 10 students, including Vera and Daen, a group of 5 students from the
Mathematics Society will be selected to attend an Enrichment Camp. Daen will not join the group without Vera, but Vera will join the group without Daen. In how many ways can the group be formed?
[1] [1] [1] [4]
[Turn Over
Amusement Park Crazy Monkey Swing Merry Go Bush A Ride to Heaven Nightmare to Remember Round about the Park Foodie Food
Page 4
6 The fish-balls used by a popular noodle stall are supplied by two companies, supplier A
and supplier B. On average, 1 out of 50 fish-balls from supplier A is deformed and 3 out of 100 fish-balls from supplier B are deformed. It is known that the noodle stall gets a proportion, denoted by p of their fish-balls from supplier A. (i) Represent the situation using a probability tree.
(ii) Given that31
=p , show that the probability that a randomly chosen fish-ball from the
noodle stall is deformed will be 752 .
(iii) For a general value of p, the probability that a fish-ball is supplied by B given that it is
deformed is denoted by f( p). Show thatppp
−−
=3
)1(3)f( .
Prove by differentiation that f is a decreasing function for 10 ≤≤ p , and explain what this statement means in the context of the question.
[2] [1] [3] [3]
7 The occurrences of floods per year at a particular residential area in Singapore follow a Poisson distribution with mean 4. (i) Find the probability that in 4 months, this particular residential area is flooded at least
twice. (ii) A random sample of 12 periods of 4 months is taken. Find the probability that in at
most 5 of these 12 periods, this particular residential area is flooded at least twice. (iii) Using a suitable approximation, find the probability that in 5 years, this particular
residential area is flooded at least 11 times. (iv) In a long term study of the flooding problem in this particular residential area, it is
proposed that the residential area be observed for 40 years. The probability that in at least n years out of these 40 years, this residential area is flooded at most thrice each year is found to be less than 0.8. Using a suitable approximation, find the least value of n.
[2] [2] [4] [4]
[Turn Over
Page 5
8 The masses of snapper fish and pomfret fish sold by a fishmonger are normally distributed and independent of each other. The mean mass, standard deviation and selling price of snapper fish and pomfret fish are given in the following table:
Snapper fish Pomfret fish Mean mass in kg 1 0.6
Standard deviation in kg 0.1 0.05 Selling price per kg in $ 12 7
Find the probability that the (i) total mass of 3 snapper fish and 2 pomfret fish is more than 4.5 kg (ii) mass of 3 snapper fish exceeds twice the mass of a pomfret fish by more than 1.85 kg. (iii) total selling price of a snapper fish and 2 pomfret fish is more than $21. A customer buys 15 fish, out of which n are snapper fish and the rest are pomfret fish. The probability that the customer pays more than $150 is less than 0.7. Find the largest value of n.
[2] [2] [2] [4]
9 CJC has 800 JC2 students in 2010 as follows:
Arts Science Total Boys 75 320 395 Girls 145 260 405 Total 220 580 800
The Student Council (SC) wishes to conduct a survey on a sample of JC2 students to find out their preferences for Graduation Night. (a) Explain how stratified sampling can be used to select a sample of 30 students.
Suggest another set of strata that may be suitable for this survey. (b) The amount that a sample of 50 students are willing to pay for Graduation Night can
be summed up as follows: 62.4825)(,4537 2 =−= ∑∑ xxx
Assume that the mean and variance for this sample are good estimators of the population mean and variance. Another sample of 60 students is surveyed. Find the probability that the mean of this second sample lies between 90 and 100.
[3] [3]
[Turn Over
Page 6
10 A manufacturer claims that the average length of its metal rods is 14 cm. (i) A customer complains that the manufacturer understated the average length of its
metal rods. A random sample of 8 metal rods is taken and the length, x cm, of each metal rod is summarised as follows:
72.1607,40.113 2 == ∑∑ xx Test, at the 4% significance level, whether the customer’s complaint is valid. State any assumptions made in carrying out the test. Explain, in the context of the question, the meaning of ‘at the 4% significance level’.
(ii) Another random sample of 9 metal rods with standard deviation 0.200 cm is taken.
What range of values should the mean length of the sample be, in order for the customer’s complaint to be not valid, at the 4% significance level? Give your answer correct to 2 decimal places.
[4] [1] [3]
11 The table below gives the proportion of people (in %) in an occupation earning more than $5000 and the proportion of university graduates (in %) in that occupation.
Occupation Teacher Chemist Accountant Lecturer Engineer Electrician Police Plumber Taxi
Driver % of graduates, x 97 87 75 84 52 36 22 10 8
% earning more than $5000, y 66 65 62 45 53 43 33 18 10
A (i) Calculate the product moment correlation coefficient between x and y.
(ii) Draw a scatter diagram for the data. One of the values of y appears to be incorrect. (iii) Indicate the corresponding point on your diagram by labelling it P and explain why
the scatter diagram for the remaining points may be consistent with a model of the form xbay ln+= .
(iv) Omitting P, calculate the least squares estimates of a and b for the model
xbay ln+= . (v) Assume that the value of x at P is correct. Estimate the value of y for this value of x. (vi) Comment on the use of the model in (iv) in predicting the value of y when x =100.
~End of Paper~
[1] [1] [2] [2] [1] [1]
This question paper consists of 6 printed pages (including this cover page).
Name Reg. No.
Year 6C( )
MATHEMATICS 9740/01 Paper 1 21 September 2010 3 hours Additional Materials: Answer Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Name, Class and Register Number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, attach the question paper to the front of your answer script. The total number of marks for this paper is 100.
For teachers’ use: Qn Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Total
Score
Max Score 4 6 6 8 8 9 9 10 12 14 14 100
DUNMAN HIGH SCHOOL, SENIOR HIGH PRELIMINARY EXAMINTION 2010 Higher 2
2
DHS 2010 Year 6 H2 Math Preliminary Examination
1 Expand 12−+
xx
in ascending powers of x up to and including the term in 2x and state
the range of values of x for which the expansion is valid. [4]
2 Prove by induction that1
1 (2 1)(2 3) 3(2 3)
n
r
nr r n=
=+ + +∑ . [5]
Hence state the value of 1
1 (2 1)(2 3)r r r
∞
= + +∑ . [1]
3 (i) Show that 2 24 5 ( )n n n a b− + = − + , where a and b are constants to be
determined. [1]
(ii) Show that ( )2 2 2 2
31 4 5 1 ( 1) 1 5 2.
N
nn n n N N
=
+ − − + = + + − + − −∑ [3]
(iii) Without the use of a graphic calculator, deduce that the sum in (ii) is strictly less
than 2 1.N + [2]
4
(i) The diagram above shows the curve with equation 3 1.y x= + Given that the two shaded areas R and S have the same value, find the value of b. [4]
(ii) Find the volume of the solid generated when S is rotated completely about the
x-axis. [4]
[Turn over
3 1y x= +
O
2
x
y
S
R
b
3
DHS 2010 Year 6 H2 Math Preliminary Examination
5 With respect to the origin O, the position vectors of three points A, B and C are given
by2 3
, 7 and .3 52 1
bOA OB OC
a
→ → → = = =
(i) Find the values of a and b if A, B and C are collinear. [2]
(ii) If given instead that OA→
is perpendicular to ,OB→
find a relationship between a
and b. Furthermore, if angle AOC is ο60 , find possible values for a and b, giving your answers correct to the nearest integer. A student claimed that since angle AOC is ο60 , angle BOC must be ο30 . Without performing any calculation, state, with a reason, whether his claim is necessarily true. [6]
6 (a) Find 2e d .
(1 3e )
t
t t+∫ [2]
(b) Write down ( )2d tan( ) .d
xx
Hence find ( )3 2 2sec d .x x x∫ [4]
(c) Evaluate 2
4
0 3 dx x x−∫ without the use of the graphic calculator. [3]
7 (i) Explain whether the following statement is always true: “If i, , ,z x y x y= + ∈ is a root of the equation
11 1 0P( ) 0, ,n n
n nz a z a z a z a n− +−= + + ⋅⋅⋅ + + = ∈ then iz x y= − is also a root.”
[1]
(ii) Solve the equation 4 3 i 0,z + + = expressing the roots in the form ie ,r α where
0r > and .π α π− < ≤ [5]
(iii) Show the roots on an Argand diagram. [1]
(iv) The points representing the roots in (iii) form a quadrilateral. Find the area of
this quadrilateral. [2]
[Turn over
4
DHS 2010 Year 6 H2 Math Preliminary Examination
8 The diagram shows a hexagon PQRSTU inscribed in a circle with radius 6 cm. The sides QR and UT are parallel, and 2 cmQR UT x= = .
(i) Show that A, the area of the hexagon PQRSTU, is ( ) 2 22 6 36 cm .x x+ − [3]
(ii) Using differentiation, find the value of x when A is maximum. (You need not verify that it gives a maximum value.) [4]
Initially 6 cm,x = and the lengths of the parallel sides QR and UT are each decreasing
at a constant rate of 11 cm s .10
− Find the rate of change of A at the instant when
2 cm.x = [3]
[ Area of a trapezium = 1 sum of the lengths of the parallel sides height2× × ]
9 (a) It is given that ln(1 e )= + xy .
(i) Show that 2
2
d d d1d d d
= −
y y yx x x
. [3]
(ii) Find the Maclaurin’s series for y up to and including the term in 2.x [2] (iii) Verify that the same result is obtained if the standard series expansions for
ex and ln(1 )x+ are used. [4]
(b) Given that x is sufficiently small for x3 and higher powers of x to be neglected, and that 10 tan 3 cos 2− =x x , form a quadratic equation in x and hence find the value of x, leaving your answer in exact form. [3]
[Turn over
O 6 6
2x 2x U T
P S
Q R
•
5
DHS 2010 Year 6 H2 Math Preliminary Examination
10 (a) The curve C has parametric equations
e cos , sin cos , 0 .4
x yθ πθ θ θ θ= = + ≤ ≤
(i) Show that the equation of the tangent at 42 e , 22
π
is given by
4 2e .2
y xπ
−= + [4]
(ii) Show that the area bounded by the curve C and the x-axis can be
expressed as 40
e cos 2 d .π
θ θ θ∫ Hence, evaluate the area, leaving your
answer correct to 2 decimal places. [4]
(b) The diagram shows the graph of ( )f ,y x= which has a turning point at
( )2,2 .A −
Sketch, on separate diagrams, the graphs of
(i) f '( ) ,y x= [3]
(ii) 1 ,
f( )y
x= [3]
showing clearly all relevant asymptotes, intercepts and turning point(s), where possible.
[Turn over
y
O x
( )2,2A −
x = 2
6
DHS 2010 Year 6 H2 Math Preliminary Examination
11 (a) A geometric series has first term a. Find the range of values of a if the sum to
infinity of the series is 1 .2
[3]
(b) The rth term of a series is given by 22
1 ln 3 .3
rr rT = + Find the sum of the first N
terms. [4]
(c) Jon wants a gigantic cake prepared for his mother’s birthday. The cake is to
consist of 5 layers, where each layer has a square base with length that is 910
of
the layer beneath it and a constant height of h units. The cost of the cake is proportional to its volume and the largest layer costs $200.
(i) Given that the largest layer has length a units, find the cost of the whole
cake, rounded to the nearest dollar. [4]
(ii) Jon also wants candles to be placed at each layer such that the total number of candles used is 75. If the largest layer has d2 candles and each subsequent layer has d candles less than the one directly below, find the number of candles placed at the top layer. [3]
END OF PAPER
This question paper consists of 7 printed pages (including this cover page).
Name Reg. No.
Year 6C( )
MATHEMATICS 9740/02 Paper 2 23 September 2010 3 hours Additional Materials: Answer Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Name, Class and Register Number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, attach the question paper to the front of your answer script. The total number of marks for this paper is 100.
For teachers’ use: Qn Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Total
Score
Max Score 8 10 11 11 4 7 8 9 9 11 12 100
DUNMAN HIGH SCHOOL, SENIOR HIGH PRELIMINARY EXAMINTION 2010 Higher 2
DHS 2010 Year 6 H2 Math Preliminary Examination
2
Section A: Pure Mathematics [40 marks] 1 Show that the equation *( 1 i)( 1 i) 2z z− + − − = represents a circle with centre C(1,–1)
and radius 2 and sketch the locus on an Argand diagram. On the same diagram, sketch the locus represented by * 3.z z+ = [3]
The points of intersection of these two loci are represented by A and B. Find, in any order, (i) the length AB, [1]
(ii) the complex numbers represented by A and B. [2]
State the cartesian equation of the perpendicular bisector of the line segment joining A and B. Explain also why the perpendicular bisector of the line segment joining any two distinct points on the circumference of the circle must pass through the centre of the circle, C. [2]
2 For 0a > , the functions f and g are defined as follows:
22f : , ,axx x a
x a≠ −
+
2 2g : , .x x a x− ∈
(i) Show that the inverse of f exists. [2]
(ii) Define 1f − in a similar form. [3] (iii) Show that the composite function fg does not exist. [2]
(iv) Solve the equation 1f( ) = f ( ) ,x x− expressing your answer(s) in terms of .a [3]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
3
3 In an experiment, Andy and Bob attempt to devise a formula to describe how the volume of water, V m3, in a tank, changes with time at t hours.
(i) Andy gives his formula as 2
d 1 8 .d 60V Vt V
= −
Given that 1V = when 0,t =
show that 3
208 7e .t
V = − Sketch this solution curve. [4]
(ii) However, Bob believes that it is more likely to be 2
22
d 12 2.d
V tt
= − Given that
d 0dVt= when 0t = , show that the general solution for V can be expressed as
2 21 1( ) ,2 4
V t C= − + − where C is a constant.
Hence, or otherwise, sketch on a single diagram, two distinct members of the family of solution curves. [5]
(iii) It is also given that 1V = when 0t =
in Bob’s model. Suppose that the water
in the tank does not overflow, explain, using your diagrams in parts (i) and (ii), why Andy’s model is more appropriate compared to Bob’s model. [2]
4 The planes 1Π
and 2Π are defined by
1Π2
: 4 10,1
=
r
2Π
1: 3 8.
1
=
r
(i) Find the acute angle between the two planes. [3]
(ii) Obtain a vector equation of l1, the line of intersection of the two planes. [4]
The cartesian equation of another line, l2, is given by 2 7 , ,6 3
x z y m− −= =
where m is a real constant. (iii) If the plane 1Π and line l2 intersect at the point (6, m, 5), find the value of m. [2] (iv) Show that the lines l1 and l2 are perpendicular for all values of m. [2]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
4
Section B: Statistics [60 marks] 5 A group of students plan to collect data from students’ parents for a project.
(i) Describe how a quota sample of size 80 might be obtained based on the parents’ educational qualifications. [2]
Subsequently, they decide to invite 80 parents to respond to an online survey. The table below shows the profile of educational qualifications among the students’ parents:
“O” Level “A” Level University Total
Father 180 360 660 1200 Mother 300 420 480 1200 Total 480 780 1140 2400
(ii) Explain why stratified sampling is preferred over the method in (i). [1]
(iii) How many mothers with an “A” Level qualification should be included in the
stratified sample? [1]
6 To assess the level of work satisfaction in relation to the time spent at work, the
human resource department of an organisation polls nine of its officers. The number of work hours per week (t) of each officer and the level of job satisfaction (x) are recorded, where a higher value of x indicates a higher level of satisfaction. The data are shown in the table below:
Work hours (per week), t 20.1 22.0 24.4 25.3 28.8 36.5 40.6 46.0 55.1 Satisfaction level, x 24.5 16.3 18.6 12.5 5.2 4.7 1.4 1.8 0.8
(i) Draw a scatter diagram for the data and find the product moment correlation
coefficient for the sample. [3]
(ii) A suggested model is of the form .bx at
= + Find a and b. Justify why this
model provides a better fit than a linear model between x and t. [2] (iii) Use the model in (ii) to predict the satisfaction level when an officer works
5.0 hours per week. Comment on the reliability of your prediction. [2]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
5
7 For every Monday of the week, the probability that Mylo wears a tie is 0.4. The probability that he wears a jacket is 0.2. If he wears a jacket, the probability that he wears a tie is 0.6. Find the probability that, on a randomly selected Monday, (i) Mylo wears a tie and a jacket, [2]
(ii) Mylo wears neither a tie nor a jacket. [2]
For Tuesday and Wednesday, the probability that Mylo wears a jacket is
• twice the probability he wears a jacket on the previous day if he wears a jacket on the previous day,
• the same as the probability he wears a jacket on the previous day if he does not wear a jacket on the previous day.
(iii) By constructing a tree diagram, find the probability that Mylo wears a jacket
on the third day given that he wears a jacket on exactly two of the three days. [4]
8 A test consists of five Pure Mathematics questions, A, B, C, D and E, and six
Statistics questions, F, G, H, I, J and K. (i) The examiner plans to arrange all eleven questions in a random order,
regardless of topic. Find the number of ways to arrange all eleven questions such that
(a) the last question is a Pure Mathematics question, [2]
(b) a Pure Mathematics question must be separated from another with exactly one Statistics question. [2]
(ii) Later, the examiner decides that the questions should be arranged in two
sections, Pure Mathematics followed by Statistics. Find the number of ways to arrange all eleven questions such that (a) question A is followed by question F, [2] (b) questions B and K are separated by more than seven questions. [3]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
6
9 A health food company claims that a breakthrough product that it launches, Zenobrain, has the benefit of maintaining good mental health and will help its consumers acquire a mean IQ score of not less than 115.
To verify the claim, a random sample of 12 consumers is taken and their IQ scores, x, are recorded and summarised as follows:
( )100 50x − =∑ and ( )2100 4008x − =∑ . (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) State the null and alternative hypotheses and carry out an appropriate test at
5% level of significance. [4] (iii) State, with a reason, whether
(a) any assumption is needed for the test in (ii) to be valid, [1] (b) the conclusion would be the same if a two-tailed test is used with the
same level of significance. [1]
10 In a randomly chosen week, the numbers of unsolicited text messages and phone calls
received by a mobile line subscriber follow independent Poisson distributions with means 5 and 3 respectively.
(i) Find the probability that the subscriber receives exactly 2 unsolicited text
messages in a day. [2]
(ii) Show the probability that the subscriber receives at most 10 unsolicited text messages or phone calls in a week is 0.816. [2]
(iii) The subscriber decides to terminate his mobile line subscription if, in the next
10 weeks, there are more than 3 weeks where he receives more than 10 unsolicited text messages or phone calls in a week. Find the probability that he terminates his mobile line subscription. [3]
(iv) Another subscriber intends to terminate her subscription if she receives a total
of 20 or more unsolicited text messages or phone calls in the next 2 weeks. By using a suitable approximation, calculate the probability that she terminates her subscription. [4]
[Turn over
DHS 2010 Year 6 H2 Math Preliminary Examination
7
11 (a) The distance, X km, covered by a school athlete during a regular training session has mean 4 km. During competition season, training increases in intensity and the distance, Y km, covered during the training session increases to a mean of 6 km. Given that X and Y are independent normal distributions with same variance 2,σ and P( 3) 0.4,Y X− > = find the probability that the athlete covers a total distance between 8 km and 12 km in two randomly chosen regular training sessions. [5]
(b) The amount of time that a student spends online each day has mean 120
minutes and standard deviation 45 minutes. A random sample of 60 students is taken and they are surveyed on the amount of time that they spend online each day. Find the probability that
(i) the total time spent online each day by the 60 students is at least 7000
minutes, [3] (ii) the sample mean time spent online each day by the 60 students is
within 5 minutes of the population mean time of 120 minutes. [3] Explain whether you need to assume that the amount of time spent online by a student each day follows a normal distribution in your calculations above. [1]
END OF PAPER
1
Dunman High School 2010 Year 6 H2 Mathematics (9740) Preliminary Examination Paper 1
Suggested Solutions
Qn Suggested Solution 1 1
2
2 2
2
1 1(1 ) 12 2 2
1 1 ...2 2 2 2
1 1 1 1 1 ...2 2 4 4 81 3 3 ...2 4 8
−− ⎛ ⎞= − × +⎜ ⎟+ ⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= − − + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
= − − + + +
= − + +
x xxx
x x x
x x x x
x x
Valid values of x: 1 2 22< ⇒ − < <
x x
2 Let Pn be the proposition
1
1(2 1)(2 3) 3(2 3)
=
=+ + +∑
n
r
nr r n
for n Z +∈ .
When n = 1:
LHS = 1
1
1 1 1(2 1)(2 3) (2(1) 1)(2(1) 3) 15
=
= =+ + + +∑
r r rRHS = 1 1
3(2(1) 3) 15=
+
1Since LHS RHS, P is true.= ∴ Assume Pk is true for some ,k Z +∈
i.e. 1
1 ,(2 1)(2 3) 3(2 3)
k
r
kr r k
=
=+ + +∑
to prove 1P +k is true,
i.e. 1
1
1 1 .(2 1)(2 3) 3(2 5)
k
r
kr r k
+
=
+=
+ + +∑
1
1
1
1LHS(2 1)(2 3)
1 1(2 1)(2 3) (2 3)(2 5)
13(2 3) (2 3)(2 5)
k
rk
r
r r
r r k k
kk k k
+
=
=
=+ +
= ++ + + +
= ++ + +
∑
∑
2
2
(2 5) 33(2 3)(2 5)
2 5 33(2 3)(2 5)
(2 3)( 1)3(2 3)(2 5)
1 RHS (shown)3(2 5)
k kk kk kk kk kk k
kk
+ +=
+ +
+ +=
+ ++ +
=+ ++
= =+
1P is true P is true+∴ ⇒k k
Since 1P is true, 1P is true P is true+⇒k k , by mathematical induction, P is truen for +∈n .
133(2 3) 3 2
1 1as ,3 63 2
=+ ⎛ ⎞+⎜ ⎟
⎝ ⎠
∴ →∞ →⎛ ⎞+⎜ ⎟⎝ ⎠
nn
n
n
n
3(i) 2 2
24 5 ( 2) 4 5
( 2) 1 =
n n n
n
− + = − − +
− +
3(ii) 2 2
3
2 2
32 2
2 2
2 2
1 4 5
1 ( 2) 1
3 1 1 1
4 1 2 1
5 1 3 1
N
nN
n
n n n
n n
=
=
⎛ ⎞+ − − +⎜ ⎟⎝ ⎠
⎛ ⎞= + − − +⎜ ⎟⎝ ⎠
= + − +
+ + − +
+ + − +
∑
∑
M M
M
2 2
2 2
2 2
2 2
( 2) 1 ( 4) 1
( 1) 1 ( 3) 1
1 ( 2) 1
1 ( 1) 1 5 2
N N
N N
N N
N N
+ − + − − +
+ − + − − +
+ + − − +
= + + − + − −
M
3
4(i) 0 0 3
1 1
3Area of d 1 d4
R y x x x− −
⎡ ⎤= = + =⎣ ⎦∫ ∫ 1
3 31 ( 1)y x x y= + ⇒ = −
( )
( )
1 43 3
2 22
43
3Area of d ( -1) d 14
3 1 14
bb b
S x y y y y
b
⎡ ⎤= = = −⎢ ⎥⎣ ⎦
⎡ ⎤= − −⎢ ⎥⎣ ⎦
∫ ∫
Equating and solve for b:
( )
43
34
3 31 14 4
1 2 2. 68 (3 s.f.)
b
b
⎡ ⎤− − =⎢ ⎥⎣ ⎦
⇒ = + =
4(ii) ( )13For = , 1 1. 1892 (say)y b x b k= − = =
Volume required
= 2 2 3 2
12 (1) ( 1) d
kb k x xπ ⎡ ⎤− − +⎢ ⎥⎣ ⎦∫
= 3. 53π (or 11.1) (unit cube)
5(i) If A, B and C are collinear, then
2 3
7 3 5 72 1 2
i.e. 2, 0, 4
AB BCb b
aa b
λ
λ
λ
⎯⎯→ ⎯⎯→=
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
= − = =
5(ii) If OA
⎯⎯→ is perpendicular to ,OB
⎯⎯→ then
023 7 0
2i.e. 2 21 2 0
OA OBb
ab a
⎯⎯→ ⎯⎯→=
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
+ + =
4
ο
2
2
2
2 33 5
1cos60
13 35
2(6 15 ) 13 3531 168 1309 0
10 (nearest int.) or 4 (nearest int.)20 (nearest int.) 6 (nearest int.)
a
a
a aa a
a ab b
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ =+
+ + = +
− − == = −= − = −
His claim is not necessarily true since points O, A, B and C may not be coplanar.
6(a) 2
2
e 1 d 3e (1 3e ) d(1 3e ) 3
tt t
t t t−= ++∫ ∫
1(1 3e ) 1
3 3(1 3e )
t
tc c−+
= + = − +− +
6(b)
( )( ) ( )2 2 2d tan 2 secd
x x xx
=
( ) ( )
( ) ( )
( ) ( )
3 2 2 2 2 2
2 2 2
2 2 2
1sec d 2 sec d2
1 tan 2 tan d21 tan ln sec2
x x x x x x x
x x x x x
x x x c
⎡ ⎤= ⎣ ⎦
⎡ ⎤= −⎢ ⎥⎣ ⎦
⎡ ⎤= − +⎣ ⎦
∫ ∫∫
6(c) 42
0 3 dx x x−∫
3 42 2
0 3 ( 3) d ( 3) dx x x x x x= − − + −∫ ∫
3 44 4
3 3
0 34 4x xx x
⎡ ⎤ ⎡ ⎤= − − + −⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ 27 or 13.52
=
7(i) No. The statement is not always true. It applies only for (polynomial) equation in z with real coefficients.
7(ii) 4 3+ i 0z + = ⇒ 4 3 iz = − −
⇒
5i4 62ez
π⎛ ⎞− ⎜ ⎟⎝ ⎠=
1 14 4
1 5 (12 5)i ( 2 ) i4 6 242 e 2 e
kkz
π ππ −− += = , 0,1,2,3k =
1 1 1 14 4 4 4
5 7 19 17-i i i -i24 24 24 242 e or 2 e or 2 e or 2 ezπ π π π
∴ =
5
7(iii)
7(iv) The quadrilateral is a square. Let the length of each side be L
Pythagoras Theorem: 1
2 2 242| | =2(2 ) 2 2L z= =
8(i) 236ON x= −
( )
( )
2
2
12 12 2 362
2 6 36
A x x
x x
⎡ ⎤= × × + −⎢ ⎥⎣ ⎦
= + −
8(ii) ( )2
2
d 1 22 36 2 6d 2 36A xx xx x
⎛ ⎞⎛ ⎞= − + + −⎜ ⎟⎜ ⎟⎝ ⎠ −⎝ ⎠
( )
( )( )
2 2
2
2
2
2
72 2 12 236
4 18 3
364 6 3
36
x x xx
x x
xx x
x
− − −=
−
− −=
−+ −
=−
dFor maximum , 0 : 0 3 cmd
= > ⇒ =AA x xx
( ) ( ) ( )d d d 1 d 12 2 d d d 10 d 20
xQR x xt t t t
= = = − ⇒ = −
( )( )4 8 1dWhen 2, 32 4 2d 32Axx
= = = =
•P S
Q R
U T
2x 2x
6 6 O
N
6 6
Im
L
ReL
2Z 3Z
4Z
O1Z
6
2 1
d d dd d d
1 4 220
2 cm s5
A A xt x t
−
= ×
⎛ ⎞= × −⎜ ⎟⎝ ⎠
= −
A is decreasing at the rate of 12 cm s .5
−
9(a) (i)
ln(1 e )= + xy e 1 e⇒ = +y x
d d: e ed d
d ed
−
=
⇒ =
y x
x y
yx x
yx
2
2
2
2
d d d: e 1d d d
d d d 1 (shown)d d d
− ⎛ ⎞= −⎜ ⎟⎝ ⎠
⎛ ⎞⇒ = −⎜ ⎟⎝ ⎠
x yy yx x x
y y yx x x
9(a) (ii)
2
2
d 1 d 1When 0, ln 2, , d 2 d 4y yx yx x
= = = =
2
2
11 4ln 2 ...2 21 1ln 2 ...2 8
= + + +
= + + +
xy x
x x
9(a) (iii)
2
2
2
22
2
2 2
2
ln(1 e ) ln 1 1 ...2
ln 2 ...2
ln 2 ln 1 ...2 4
2 4ln 2 ...
2 4 2
1ln 2 ...2 4 2 41 1ln 2 ... (verified)2 8
⎛ ⎞⎛ ⎞+ = + + + +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎛ ⎞
= + + +⎜ ⎟⎝ ⎠
⎛ ⎞= + + + +⎜ ⎟
⎝ ⎠
⎛ ⎞+⎜ ⎟⎛ ⎞ ⎝ ⎠= + + − +⎜ ⎟
⎝ ⎠⎛ ⎞⎛ ⎞= + + − +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= + − +
x xx
xx
x x
x xx x
x x x
x x
7
9(b) 10 tan 3 cos 2− =x x ( )2
2
2
210 3 1
25 2 0
5 5 4( 2) 5 33 (rej -ve as is small)2 2
xx
x x
x x
− = −
⇒ + − =
− ± − − − +∴ = =
10(a) (i) e cos , sin cos , 0
4x yθ πθ θ θ θ= = + ≤ ≤
d de (cos sin ), cos sin ,d d
x yθ θ θ θ θθ θ= − = −
-d d d/ ed d dy y xx
θ
θ θ= =
At (e cos ,sin cos ),θ θ θ θ+ the equation of the tangent
is -( sin cos ) e ( e cos ),y xθ θθ θ θ− − = −
Set ,6πθ =
at (63e 3 1,
2 2
π
+ ) , the equation of the tangent is6-
63 1 3e( ) e ( ),2 2 2
ππ
− − = −y x
6 1e2
y xπ
−= +
10(a) (ii)
Area under the curve C is
40
2 240
40
(sin cos )e (cos sin )d
e (cos sin )d
e cos 2 d ( )
0.68 (2d.p.)
A
shown
πθ
πθ
πθ
θ θ θ θ θ
θ θ θ
θ θ
= + −
= −
=
=
∫
∫
∫
10(b) (i)
y
f '( )y x=
Ox
x = 2
-2
8
10(b) (ii)
11(a) 11 22 1
1 2 1 2 1 1 1 2 1
10 1, (since 0)2
ar
a rr a
a a
a a r
=−= −
= −
⇒ − < ⇒ − < − <
< < ≠ ≠
11(b)
( )
( )
21 1 1
1 2 ln 33
111 9 1 ln 319 1
91 1 1 1 ln 38 9
rN N N
rr r r
N
N
T r
N N
N N
= = =
⎛ ⎞= +∑ ∑ ∑⎜ ⎟⎝ ⎠
⎛ ⎞− ⎜ ⎟⎝ ⎠= × + +−
⎛ ⎞⎛ ⎞= − + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
11(c) (i)
Volume of whole cake 2 2 2 2 3 2 4 2
2 52 4 6 8 2 2 2
2
(0.9 ) (0.9 ) (0.9 ) (0.9 )[1 (0.9 ) ](1 0.9 0.9 0.9 0.9 ) 3.4280
1 0.9
a h a h a h a h a h
a h a h a h
= + + + +
−= + + + + = =
−
Cost of whole cake $3.4280 200 $686 (nearest dollar)= × =
11(c) (ii)
2
2
5 2( ) (5 1)( ) 752
2 15( 5)( 3) 0
5 or 3 (rej. since >0)
d d
d dd d
d d d
⎡ ⎤+ − − =⎣ ⎦
− =− + == = −
No. of candles at top layer 25 (5 1)( 5) 5= + − − =
O x2
y
A’(-2, 0.5)
1f( )
yx
=
1
Dunman High School 2010 Year 6 H2 Mathematics (9740)
Preliminary Examination Paper 2 Suggested Solutions
SECTION A
1 *
2( 1 i)( 1 i) 2
1 i 2
1 i 2
z z
z
z
− + − − =
− + =
− + =
i.e. A circle with centre (1,–1) and radius 2. (shown)
(i)
( )22
2
12 22
7
AB AD=
⎛ ⎞= −⎜ ⎟⎝ ⎠
=
(ii) Complex numbers represented are
3 7 1 i2 2
⎛ ⎞+ −⎜ ⎟⎜ ⎟⎝ ⎠
and 3 7 1 i.2 2
⎛ ⎞− +⎜ ⎟⎜ ⎟⎝ ⎠
Cartesian equation of the perpendicular bisector of the line segment joining A and B is 1.y = − Since any two distinct points on the circumference of the circle are equidistance from the centre C, hence perpendicular bisector of the line segment joining these points must pass through C.
2 Re O
C(1,-1)
A
x=1.5
Im
B
D
2
O x
2(i)
Recommended (1)From the graph above, any horizontal line ,y b b= ∈ cuts the graph of f at most once, therefore f is one-one and the inverse of f exists. OR (2)From the graph above, any horizontal line ,y b b a= ≠ , cuts the graph of f exactly once, therefore f is one-one and the inverse of f exists.
(ii)
22
2
2
2 31 1 2
, ,
, ,
f : , , or f : , ,
axy x ax a
xy a y axa yx y a
a ya x ax x a x a x a
a x a x− −
= ≠ −+
+ =
= ≠−
≠ − ≠− −
a a
(iii) 2 2g f
g f
[ , ), \{ },
,
R a D a
R D
= − ∞ = −
∴ ⊄ thus fg does not exist
2x a= −
y a=
y
3
(iv)
Method 1(Recommended) 1
2
2 2
2
2
f( ) = f ( )f( ) =
0
( ) 0
0 or .
x xx xax x
x ax a x ax
x x a a
x x a a
−
⇒
⇒ =+
⇒ + − =
⇒ + − =
⇒ = = − Method 2 (not preferred)
1
2
2
2 2
3
3
3 22
f( ) = f ( )
( ) ( )
[( 1) ( )] 0
[( 1) ( )] 0
( 1)0 or ( 1) .1 1
x x
ax a xa xx a
ax a x a x x a
ax a x a a
ax a x a a
a a a ax x a a a aa a
−
⇒ =−+
⇒ − = +
⇒ + + − =
⇒ + + − =
− − −⇒ = = = = − − = −
+ +
3(i)
3
2 2
2
3
3
''3 20
3 ''20
320
d 1 8 1 8( ) ( )d 60 60
d 18 d 60
1 1ln | 8 | '3 60
| 8 | e , '' 3 '
8 e , e
When 0, 1,7,
= 8 7e
t C
tC
t
V VVt V VV V
V t
V t C
V C C
V A A
t VA
V
+
−= − =
⇒ =−
⇒ − = +
⇒ − = =
⇒ − = = ±
= =⇒ = −
⇒ −
1 V
t 820ln( )or 2.677
4
4(i)
2 14 . 31 1. 15cos
| || | 21 11 21 11
9.3 .
θ
θ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = =
∴ =
1 1
1 2
n nn no
(ii)
22
2
31
1
3
4 2
d 12 2dd 4 2d
dWhen 0, 0, 0d
d 4 2d
, is a constant.
V ttV t t Ct
Vt Ct
V t tt
V t t C C
= −
⇒ = − +
= = ∴ =
⇒ = −
⇒ = − +
4 2
2 21 1( ) ( )2 4
V t t C
t C
= − +
= − + −
(iii)
When 0, 1,t V= = then 211 .4
C = > Therefore given the above initial condition, Bob’s model corresponds to solution curve type (I) in part (ii). Therefore in Bob’s model, the volume of water approaches infinity in the long run (not realistic) whereas in Andy’s model, the volume of water reasonably diminishes to zero in the long run/after some time. Thus, Andy’s model is more appropriate than Bob’s model.
C
t
C
V 1(I)4
C >
1(II)4
C ≤
5
(ii)
11
2
⎛ ⎞⎜ ⎟= × = −⎜ ⎟⎜ ⎟⎝ ⎠
1 2d n n
Set z=0,
2 4 103 8
1, 3
x yx y
x y
+ =+ =
⇒ = − =
1
1 13 1 , .0 2
l α α α−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∴ = + − ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1: r a + d =
Alternative
1
2 4 103 8
Let ,2 4 10
3 8
1 , 3 ,2 2
1 1: r 3 1 ,
20 2
x y zx y z
z tx y t
x y tt tx y
tl α α
+ + =+ + =
= ∈⇒ + = −
+ = −
⇒ = − + = −
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∴ = + − = ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(iii)
Since the point with co-ordinates (6,m.5) lies on the first plane, 1 1
6 24 10
5 112 4 5 10
D
m
m
=
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⇒ + + =
a d
7 .4
m⇒ = −
(iv)
2
2 20 , .
7 1l mβ β β
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
2 2: r a + d =
1 2
1 0 2 2 0 ( )2 1
independent of thevalueof m⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
1 2d d
Therefore lines l1 and l2 are perpendicular for all real values of m.
6
SECTION B
5(i)
To obtain a quota sample of size 80: Identify and categorise the parents into mutually exclusive sub-groups according to education levels. Set a quota, i.e. a target number of respondents for each group where the total adds up to 80. Poll respondents on a first-come-first-serve basis, say, when the parents arrive at school in the morning with their children, until the number for each category is filled.
(ii)
Stratified sampling is more representative in terms of the proportion of parents’ educational qualifications in each category.
(iii)
420 80 142400
× =
6(i)
From GC, 0.860r = −
(ii)
260.56From GC, regression line 37.612xt
= − +
ie, ( )37.6 , 261 3 sig figa b= − =
Suggested model between x and 1t
is a better fit with |r |= 0.930 > |r|= 0.860 for the linear
model between x and t.
(iii) 260.5637.612 78.95.0
x = − + =
5t = lies outside the data range of t , thus model may not be valid and estimate not likely to be
reliable.
x
t
7
7(i) Let J be the event where Mylo wears a jacket and T be the event where Mylo wears a tie.
( ) 0.6( ) 0.6
( )( ) 0.6
0.2( ) 0.6 0.2 0.12
P T JP T J
P JP T J
P T J
=
∩=
∩=
∩ = × = (ii)
[ ]( )
( ) '1 ( )1 ( ) ( ) ( )
1 0.4 0.2 0.121 0.480.52
P T JP T JP T P J P T J
∪= − ∪
= − + − ∩
= − + −
= −=
(iii)
Let nJ be the nth day where Mylo wears a jacket.
3 1 3 2
3 1 3 2 1 2
Required Probability( ) ( )
( ) ( ) ( )(0.2)(0.6)(0.4) (0.8)(0.2)(0.4)
(0.2)(0.6)(0.4) (0.8)(0.2)(0.4) (0.2)(0.4)(0.2)0.048 0.064
0.048 0.064 0.0160.1120.1280.875
P J J P J JP J J P J J P J J
∩ + ∩=
∩ + ∩ + ∩
+=
+ ++
=+ +
=
=
'J
J
J
J
J 'J
0.2
0.8
0.4
0.6
0.2
0.8
0.2
J
J
J
'J
'J
'J
'J
0.8
'J
0.4
0.6
0.6
0.4
0.2
0.8
Mon Tue Wed
8
8(i) (a)
Number of ways 10!(5) 18144000= =
(b) S S S S S S
( )( )( )
Number of ways= 6! 5! 3 259200=
(ii) (a) ( )( )
Number of ways= 4! 5! 2880=
(ii) (b)
Case One: 8 Questions (4M and 4S) between B and K B K
Case Two: 8 Questions (3M and 5S) between B and K B K
Case Three: 9 Questions between B and K B K
Pure Mathematics Questions Statistics Questions
( )( ) ( )( ) ( )( )
Number of ways4 5
= 3! 5! 4! 4! 5! 4!3 4
2880 2880 28808640
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠= + +=
4! ( )5
4!4⎛ ⎞⎜ ⎟⎝ ⎠
( )4
3!3⎛ ⎞⎜ ⎟⎝ ⎠
5!
4! 5!
or=(5!)6 5 4 3 (3!) 259200⋅ ⋅ ⋅ ⋅ =
9
9(i)
2
Let 10050 , 4008
u xu u= −
∴ = =∑ ∑
Unbiased estimate of population mean:
100 4.16667 100 104.17 104 (3 s.f .)x u= + = + = ≈ Unbiased estimate of population variance:
22 1 504008
11 12s
⎛ ⎞= −⎜ ⎟
⎝ ⎠ = 345.42 345 (3 s.f .)≈
(ii)
To test H0: 115μ = against H1: 115μ <
One-tail test at 5% level (α = 0.05)
Use t-test since σ2 is unknown and sample size of 12 is small
under H0, T = 115345.42/12X − ~ t (11).
From GC, p-value = 0.0342 Since p-value =0.0342< 0.05, there is sufficient evidence to reject H0 at the 5% level of significance and conclude that the mean IQ score is less than 115, hence the manufacturer’s claim is disputable.
(iii) (a) The IQ score of customers is normally distributed. (b) For 2-tailed test, p-value =2(0.0324) =0.0684 > 0.05. H0 will not be rejected. The
conclusion would be different. 10 (i)
Let X be the number of unsolicited text messages received in a day.
( )5~ Po 7P( 2) 0.125 (3 s.f.)
X
X = =
(ii) Let Y be the number of unsolicited text messages or phone calls received in a week. ( )~ Po 8
P( 10) 0.816 (3 s.f.) (shown)Y
Y ≤ =
(iii) Let W be the number of weeks where receives more than 10 unsolicited text messages or phone calls in a week out of 10 weeks.
( )~ B 10,0.184P( 3) 1 P( 3) 0.0944 (3 s.f.)
WW W> = − ≤
=
(iv) Let T be the total number of unsolicited text messages or phone calls received in the next 2 weeks.
10
( )~ Po 16Since =16>10, ~ N(16,16) approximately.
P( 20) P( 19.5) (apply c.c.) 0.191 (3 s.f.)
TT
T T
λ ∴
≥ = >=
11 (a)
2~ N(2,2 )Y X σ− P( 3) 0.4
3 2P( ) 0.42
From GC,1 = 0.253352
= 2.7910
Y X
Zσ
σσ
− > =−
> =
2 2
1 22
1 2
1 2
Var( ) 2 2.7910 3.9471
~ (8,3.9471 )P(8 12) 0.345 (3 s.f.)
X X
X X NX X
+ = × =
+< + < =
b(i) Let X min be the amount of time spent by a student online each day.
( )1 2 60
221 2 60
E( ) 60 120 7200
Var( ) 60 45 90 15
X X X
X X X
+ + = × =
+ + = × =
L
L
Since n=60 is large, by Central Limit Theorem,
( )21 2 60 ~ 7200, 90 15X X X N ⎛ ⎞+ + ⎜ ⎟⎝ ⎠
L approximately.
1 2 60P( 7000)
0.717 (3 s.f.)X X X+ + ≥
=
L
(ii) Since n=60 is large, by Central Limit Theorem, 245~ 120,
60X N
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
approximately.
P( 120 5)
P( 5 120 5)
P(115 125)0.611 (3 s.f.)
X
X
X
− <
= − < − <
= < <=
We do not need to assume that the amount of time spent online follows a normal distribution since by the Central Limit Theorem, the sample mean follows a normal distribution approximately when n is large.
[Turn Over 2
1 Solve the inequality
2
42 ( 2)
xx x
≤− −
,
giving your answer in an exact form. [3]
Hence solve 2
e 4e 2 (e 2 )
x
x x≤+ +
. [2]
2 The sequence of numbers nu , where n = 0, 1, 2, 3, …, is such that 0u = 2− and
1
1
( 2)2 1
nn
n
n uuu n
−
−
+=
+ +.
Proof by induction that, for 0,n ≥
22 1nnun+
=−
. [5]
3 The functions f and g are defined as follows: ( ) ,212:f 2 −−xx 1−<x , ( )g : ln ,x x a+ 1−>x .
(a) Define 1f − in a similar form. [3] (b) State the value of a such that the range of g is (0, )∞ . [1]
(c) Show that the composite function gf exists, and find the range of gf, giving
your answer in terms of a. [2] 4 A curve is defined by the parametric equations
21txt
=+
, 21tyt
=−
, where 1, 1.t ≠ −
(i) Show that the tangent to the curve at any point with parameter t has equation
( ) ( )3 32 2 31 1 4 .t y t x t− = + − [3]
(ii) Find the gradient of the tangent to the curve at 12
t = . Hence determine the
acute angle between this tangent and the line 3y x= + . [3]
[Turn Over 3
5 Robert took a study loan of $100 000 from a bank on 1st January 2010. The bank charges an annual interest rate of 10% on the outstanding loan at the end of each year. After his graduation, Robert pays the bank $x at the beginning of each month. The first payment is made on 1st January 2014. Let nu denote the amount owed by Robert at the end of nth year after 2013, where 0
+∈n .
(i) Find 0u . [1]
(ii) Show that nu = 01.1 (1.1 1)n nu kx− − , where k is a constant to be determined. [4]
(iii) Given that Robert owes the bank less than $1000 at the end of 2020, find the minimum value of x, giving your answer to the nearest dollar. [3]
6 (a) Find 2
1 d3 4
tt−∫ . [3]
(b) Use the substitution 5xu = to find ( )25 cos 5 dx x x∫ . [5]
7 It is given that the function ( )fy x= has the Maclaurin’s series 21 4 ...x ax+ + + and
satisfies ( ) ( )2 2d1 1dyx b yx
+ = + , where a and b are real constants.
(i) Show that b = 2 and find the value of a. [4]
(ii) Find the series expansion of f( )4
xx+
in ascending powers of x, up to and
including the term in 2x . [3]
(iii) State the equation of the normal to the curve f( )4
xyx
=+
at x = 0. [1]
[Turn Over 4
8 (i) Express 4( 1) ( 2)
rr r r
−− +
in the form 1 2
A B Cr r r
+ +− +
. [2]
(ii) Hence find 2
4( 1) ( 2)
n
r
rr r r=
−− +∑ . [3]
Give a reason why the series is convergent, and state its limit. [2]
(iii) Use your answer to part (ii) to find 2
3( 1)( 3)
n
r
rr r r=
−+ +∑ . [2]
9 On a single Argand diagram, sketch the loci given by
(i) 21 i 2z − − ≥ ,
(ii) 1arg123 i
z π+ ≥ + ,
(iii) 1z z> − . [7] Hence, or otherwise, find the range of values of iz − and arg ( i).z − [3]
10 A file is downloaded at r kilobytes per second from the internet via a broadband connection. The rate of change of r is proportional to the difference between r and a constant. The initial value of r is 348. If r is 43, it remains at this constant value.
(i) Show that d ( 43)dr k rt= − . [2]
(ii) Hence obtain an expression for r in terms of k and t. [4]
The total amount of data downloaded, I kilobytes, in time t seconds, is given by
dd
I rt= .
(iii) Given that there is no data downloaded initially, find I in terms of k and t. [2]
(iv) It is given that a file with a size of 5700 kilobytes takes 90 seconds to
download. Find the value of k . [2] (v) Explain what happens to the value of r in the long run. [1]
[Turn Over 5
11
The diagram above shows part of the structure of a modern art museum designed by Marcus, with a horizontal base OAB and vertical wall OADC. Perpendicular unit vectors i, j, k are such that i and k are parallel to OA and OC respectively. The walls of the museum BCD and ABD can be described respectively by the equations
15 36
6
− ⋅ − =
r and 14 5 10 4 00 0 4
λ µ−
= + +
r , where ,λ µ∈ .
(i) Write down the distance of A from O. [1]
(ii) Find the vector equation of the intersection line of the two walls BCD and
ABD. [3] (iii) Marcus wishes to repaint the inner wall ABD. Find the area of this wall. [3] Suppose Marcus wishes to divide the structure into two by adding a partition such that it intersects with the walls BCD and ABD at a line. This partition can be described by the equation 2 7x y zα β− + = , where ,α β ∈ .
(iv) Find the values of and α β . [2] (v) Another designer, Jenny, wishes to construct another partition which is
described by the equation 2 7x y zα γ− + = , where γ β≠ . State the relationship between Jenny’s and Marcus’ partitions. [1]
Deduce the number of intersection point(s) between the walls BCD, ABD, and Jenny’s partition. [1]
B
k O
C
D
i j
A
[Turn Over 6
12 The curves 1C and 2C have equations 2 2 2( 2) (1 )x a y− = − and 2 4
1xyx−
=+
, where
1 2,< <a respectively. Describe the geometrical shape of C1. [1]
(a) State a sequence of transformations which transforms the graph of 2 2 1x y+ = to the graph of 1C . [3]
(b) (i) Sketch 1C and 2C on the same diagram, stating the coordinates of any
points of intersection with the axes and the equations of any asymptotes. [6]
(ii) Show algebraically that the x-coordinates of the points of intersection
of 1C and 2C satisfy the equation
( ) ( )2 2 2 2 2 2 21 2 ( 1) ( 4)x x a x a x+ − = + − − . [2] (iii) Deduce the number of real roots of the equation in part (ii). [1]
[Turn Over
2
Section A: Pure Mathematics [40 marks] 1 A tin has a fitting cylindrical lid which overlaps its cylindrical body by 5 cm. When
completely closed, it has base radius x cm and height y cm, as shown in the diagram. The body and the lid are made from thin metal sheet such that the difference in their radii is negligible. The total area of metal sheet used to make the tin and its lid is 400π cm2.
Show that the volume V cm3 of the tin is given by
( )2200 5V x x xπ= − − .
If x varies, find the values of x and y for which V has its maximum value. [6]
2 (i) Solve the equation
5 32 0z − = ,
expressing the answers in the form ier θ , where 0r > and .π θ π− < ≤ [2]
(ii) Explain why the equation 52 1 32 0w
w+ − =
has four roots. [1]
The roots of the equation are denoted by w1, w2, w3 and w4. By finding 1w
,
show that 4
1
1i iw=∑ is a real number. [4]
[Turn Over
3
3 The nth term and the sum of the first n terms of a sequence are denoted by nu and Sn
respectively. Given that Sn is a quadratic polynomial in n and 1 100u = , 2 90u = and 10
3360r
ru
=
=∑ , find Sn in terms of n. [4]
Hence show that this sequence is an arithmetic progression. [3] 4 The points A, B, C have position vectors i, 2j − tk, tk relative to an origin O
respectively, where t is a fixed constant. The points X and Y lie on AB and BC respectively such that
1AX BYXB YC
µµ
= =−
,
where µ is a parameter such that 0 1.µ< <
(i) Find the vector XY
in terms of t and µ . [2]
(ii) Prove that O, X, Y are non-collinear. [2]
(iii) Determine if XOY can be 90ο , justify your answer. [3]
(iv) Find the projection vector of XY
onto 4i + j. [2]
5 A point P(x, y) moves in the x-y plane such that the distance from the line 3x = − is
always equal to the distance from the point ( )6, 2 .−
(i) Prove that the locus of P can be represented by a curve C with equation ( ) ( )22 9 2 3 .y x+ = − [2]
(ii) Sketch the curve C, making clear the main relevant features of C. [2]
The region R is bounded by the curve C, the lines 7y = and 2x = . Find
(iii) the exact area of R , showing your working clearly, [4]
(iv) the volume of revolution formed when R is rotated through 2π radians about the x-axis. [3]
[Turn Over
4
Section B: Statistics [60 marks] 6 An airline wishes to assess its in-flight service for a specific flight and employs a
marketing research company to administer a survey. The seats on this flight are divided into classes as follows:
First Class Business Class Economy Class Total
No. of Seats 20 80 300 400 An employee of the company proposes using stratified sampling method to select 80 seats and ask the passengers occupying these seats to complete a questionnaire. Describe how this sample can be obtained. Suggest a practical difficulty which may be encountered in carrying out this proposal. [3]
Another employee suggests using simple random sampling method to choose a sample of 80 passengers from the list of passengers who have checked in. Explain why it may not be appropriate to use this sampling method. [1]
7 A boy intends to arrange a set of coloured square tiles flat on the floor as shown
below with each row being labelled.
There are four identical blue tiles, three identical yellow tiles, two identical green tiles and one red tile. Find the number of ways to arrange the tiles if (i) there are no restrictions, [1] (ii) exactly one yellow tile and exactly two blue tiles are in the same row, [3] (iii) there are less than 3 yellow tiles in the fourth row . [2]
All the ten tiles are now placed in a row. Find the number of ways he can arrange the tiles such that all the blue tiles are separated. [2]
First row Second row Third row Fourth row
[Turn Over
5
8 The life span, x (in hours), of a certain electronic component is known to follow a normal distribution with mean 9000 hours and standard deviation σ hours. Following a change in the manufacturing procedure, a batch of components is produced and a random sample of 10 components is taken from this batch. (a) The life spans of the 10 components are summarized by
Σ(x – 9000) = 2010, Σ(x – 9000)2 = 911 157.
Test at the 5% level of significance whether the mean life span of the components has increased after the change in the manufacturing procedure. [3]
(b) Let x denote the mean life spans (in hours) of the 10 components in the sample.
If σ = 25, find the set of values of x so that we can conclude at the 1% level of significance that the population mean life span of the components has increased. State an assumption for the above test to be valid. [5]
9 An orchard owned by Mark produces oranges which have masses (in grams) that follow a normal distribution N(190, 576). Visitors to this orchard can buy the oranges at $0.10 per 100 g. Find the probability that the payment made by a visitor buying 20 oranges will differ from 2 times the payment made by another visitor buying 10 oranges by at most $0.15. [4]
The orchard produces apples which are graded according to their mass. Apples with a mass exceeding 150 g are graded as 'large' while apples with a mass less than 70 g are graded as 'small'. Mark finds that the proportion of apples graded as 'large' is the same as that of the apples graded as 'small'. It is given that the mass of a randomly chosen apple from Mark's orchard follows a normal distribution with mean µ g and standard deviation 30 g. (i) Find the value of µ . [1] (ii) What is the probability that Mark will get at least 5 apples graded as 'large'
when he randomly selects 65 apples from his orchard? [3]
[Turn Over
6
10 A public opinion poll surveyed a sample of 1000 voters. The table below shows the number of males and females supporting Party A, Party B and Party C.
Party A Party B Party C
Male 200 130 70 Female 250 300 50
(a) One of the voters is chosen at random. Events A, C and M are defined as
follows: A : The voter chosen supports Party A. C : The voter chosen supports Party C. M : The voter chosen is a male.
Find (i) P(A M ), (ii) P(M ' ∩ C '). Determine whether A and M are independent. [4]
(b) It is given that in the sample, 20% of Party A supporters, 30% of Party B
supporters and 5% of Party C supporters are immigrants.
(i) One of the voters selected from the sample at random is an immigrant. What is the probability that this voter supports Party A? [2]
(ii) Three voters are chosen from the sample at random.
Find the probability that there is exactly one immigrant voter who supports Party C or exactly one female who supports Party A (or both). [4]
11 At the counselling centre CareforSociety, the average number of call-ins received in a
month regarding alcohol abuse problem is denoted by λ . The probability of receiving at most 9 such calls in a week is 0.701. Assuming that there are 4 weeks in a month, (i) show that the value of λ is 32.5, correct to 3 significant figures. State a
condition under which the distribution used is valid. [3] (ii) by using a suitable approximation, find the probability that in a month, the
number of call-ins received is more than 25 but not more than 40. [4]
The centre also has 70 support groups, each consisting of n people, which help one another to deal with alcohol abuse problems. It is known that, on average, 3 in 20 people in such support groups will be successful in correcting their alcohol abuse problem. Given that in the 70 support groups, the probability of having an average of at least 4 people per group successfully correcting their alcohol abuse problem is more than 0.7, determine the minimum value of n. [4]
[Turn Over
7
12 With the implementation of the new bus fare system, Jasmine wanted to know how the new system would affect her. She identified 12 common locations and used a map to measure the straight line distance, x km, of each location from her home. She also measured the road distance, y km, of each location from her home and the corresponding bus fare, s cents. The data are shown below.
Location A B C D E F G H I J K L Straight line distance, x 7.7 3.0 24.1 13.2 9.3 9.0 10.4 3.5 17.6 4.5 2.0 2.5
Road distance, y 8.8 3.3 28.0 16.1 9.4 8.9 12.5 15.8 22.5 5.0 2.2 2.8
Bus fare, s 121 81 181 149 125 121 137 149 173 91 71 71
(i) By considering the values of x and y, explain why Location F should be omitted from any further analysis. State, with a reason, another location that should be omitted. [2]
Omit the data for the two locations in (i). (ii) Use a suitable regression line to give an estimate of the straight line distance
when the road distance is 20.0 km. [2]
(iii) Draw a scatter diagram of s against y. State, with a reason, which of the following models is more appropriate to describe the relationship between y and s:
Model I: 2s a by= + , Model II: lns a b y= + , where a and b are positive constants. [3]
(iv) Using the more appropriate model found in (iii), calculate the equation of the corresponding regression line. [2]
(v) Comment on the use of the regression line found in (iv) to estimate the road
distance travelled if the bus fare is 170 cents. [2]
[Turn Over 2
2010 HCI Prelim Paper 1 Solutions
Qtn Solutions 1.
2
4 02 ( 2)
xx x
− ≤− −
2
2
2 4 0( 2)
x xx− −
⇒ ≤−
( )2
2
1 50
( 2)x
x− −
⇒ ≤−
2( 1) 5 0,x⇒ − − ≤ , 2( 2)x − is always positive for all real values of x. 1 5 1 5 , 2x x⇒ − ≤ ≤ + ≠
For 2
e 4e 2 (e 2)
x
x x≤+ +
Replace x by ex− ,
1 5 e 1 5x⇒ − ≤ − ≤ + ( )ln 5 1x⇒ ≤ −
2. Let P(n) be the proposition 22 1nnun+
=−
.
When n = 0, LHS of P(0) = 0 2u = − (given)
RHS of P(0) = 2 21= −
−
∴ P(0) is true. Assume P(k) is true for some { }0k +∈ ∪
i.e. 22 1kkuk+
=−
.
Show that P(k+1) is true
i.e. 13
2 1kkuk++
=+
.
When n = k + 1,
LHS of P(k+1) = 1( 3)
2 2k
kk
k uuu k+
+=
+ +
=
( )( )3 22 122 2
2 1
k kk
k kk
+ +−
+⎛ ⎞ + +⎜ ⎟−⎝ ⎠
= ( 3)( 2)
(2 1)( 2)k kk k
⎡ ⎤+ +⎢ ⎥+ +⎣ ⎦
[Turn Over 3
= 3
2 1kk++
= RHS of P(k+1)
Since P(0) is true & P(k) is true ⇒ P(k + 1) is also true, hence by mathematical induction P(n) is true for all { }0 .n +∈ ∪
3 (a) (b) (c)
x
( )
7,221
21:f
1221
21
122
212
1
2
>+−∴
−<+−=
−=+±
−−=
− xxx
xyx
xy
xy
a
Q
a = 2
( )∞= ,7fR , ( )∞−= ,1gD . Since gf DR ⊆ , gf exists. ( ) ( ) ( )( )f g, 1 7, ln 7 ,a−∞ − ⎯⎯→ ∞ ⎯⎯→ + ∞ Df Rf Rgf
4(i)
( )2
22
d 1d 1
x tt t
−=
+
( )2
22
d 1d 1
y tt t
+=
−
( )fy x=
x a= −
1 a−
( )gy x=
7
-1
[Turn Over 4
(ii)
32
2
d 1d 1y tx t
⎛ ⎞+= ⎜ ⎟−⎝ ⎠
Equation of tangent:
( ) ( ) ( ) ( )( )
32
2 2 2
3 3 2 22 2 2 2
32 3
11 1 1
1 1 1 1
1 4
t t ty xt t t
t y t x t t t t
t x t
⎛ ⎞+ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟− − +⎝ ⎠⎝ ⎠
− = + − + + −
= + −
When 12
t = , d 27dyx=
Let α be the acute angle between the two lines.
( )
1
1
tan 27 87.879tan 1 45
42.9
AB
A Bα
−
−
= = °
= = °
= − = °
Alternative Solution
( )( )1
27 1 26tan1 27 1 28
26tan 42.928
α
α −
−= =
+
= = °
5(i) 4
0 $1.1 (100 000) $146 410= =u (ii) 1 01.1( 12 )= −u u x
[ ]2 0
2 20
1.1 1.1( 12 ) 12
1.1 1.1 (12 ) 1.1(12 )
= − −
= − −
u u x x
u x x
α
A B
Note: α = A - B tan A = 27 , tan B = 1
[Turn Over 5
2 23 0
3 3 20
1.1 1.1 1.1 (12 ) 1.1(12 ) 12
1.1 1.1 (12 ) 1.1 (12 ) 1.1(12 )
⎡ ⎤= − − −⎣ ⎦= − − −
u u x x x
u x x x
: :
( )( )
( )
10
10
0
0
1.1 1.1 (12 ) 1.1 (12 ) ... 1.1(12 )
1.1 12 1.1 1.1 ... 1.1
1.1 1.1 1 1.1 12
0.1
1.1 132 1.1 1
−
−
= − − − −
= − + + +
⎛ ⎞−⎜ ⎟= −⎜ ⎟⎝ ⎠
= − −
n n nn
n n n
nn
n n
u u x x x
u x
u x
u x
(iii) n = 7 at end of 2020 7 7
01.1 132 (1.1 1) 1000u x− − < $2270.30x >
Least x to the nearest dollar = $2271
6(a)
2
2
2
2
1 d3 41 1 d34
41 1 d4 3
2
31 1 2ln4 3 3
2
3 2 3ln12 2 3
tt
tt
t
t
tC
t
t Ct
−
= −−
= −⎛ ⎞
− ⎜ ⎟⎝ ⎠
⎡ ⎤−⎢ ⎥
⎢ ⎥= − +⎢ ⎥+⎢ ⎥⎣ ⎦
−= − +
+
∫
∫
∫
(b)
d 5 ln 5d
d 1 1d 5 ln 5 ln 5
5x
x
x
ux
xu u
u
=
∴ = =
=
[Turn Over 6
2
2
2
5 cos (5 ) d
1cos dln 5
1 cos dln 5
1 (1 cos 2 ) d2ln 5
1 sin 22ln 5 2
1 sin 2(5 )52ln 5 2
x x
xx
x
u u uu
u u
u u
uu C
C
= ⋅
=
= +
⎡ ⎤= + +⎢ ⎥⎣ ⎦⎡ ⎤
= + +⎢ ⎥⎣ ⎦
∫
∫
∫
∫
7i
( ) ( )2 2
dWhen 0, 1, 4.d
d1 1d
d (2) 4d
2 (Shown)
yx yx
yx b yx
y bx
b
= = =
+ = +
⇒ = =
⇒ =
( )2
22
2
2
d d d1 2 2 2d d d
d 16d
16 82!
y y yx x yx x x
yx
a
⎛ ⎞+ + = ⎜ ⎟⎝ ⎠
=
= =
ii
12
12 2
2 2
2 2
2
f( )(4 )
1 (1 4 8 ...)(1 )2 4
1 3( )( )1 1 2 2(1 4 8 ...)(1 ( )( ) ( ) ...)2 2 4 2! 41 3(1 4 8 ...)(1 ...)2 8 1281 31 963(1 ...)2 8 128
−
−
+
= + + + +
− −= + + + + − + +
= + + + − + +
= + + +
x x
xx x
x xx x
xx x x
x x
iii Gradient of normal = 16
31−
Equation of normal:
[Turn Over 7
1 162 31
y x= −
8(i) 4
( 1) ( 2) 1 2r A B C
r r r r r r−
= + +− + − +
4 ( 2) ( 1)( 2) ( 1)r Ar r B r r C r r− = + + − + + − 1, 2, 1A B C= = − =
(ii)
2
2
4( 1) ( 2)
1 2 11 211 14
1 2 12 3 51 2 13 4 61 2 14 5 7
1 2 14 3 1
1 2 13 2
1 2 12 1 1
1 2 11 2
1 2 1 1 1 2 12 3 3 1 21 1 1 16 1 2
n
r
n
r
rr r r
r r r
n n n
n n n
n n n
n n n
n n n n
n n n
=
=
−− +
= − +− +
= − +
+ − +
+ − +
+ − +
+
+ − +− − −
+ − +− −
+ − +− − +
+ − +− +
= − + + + − ++ +
= − + ++ +
∑
∑
M
(iii) 1 1 1lim 01 2n n n n→∞
⎛ ⎞− + + =⎜ ⎟+ +⎝ ⎠, hence the series in (ii) converges.
2
4 1( 1) ( 2) 6r
rr r r
∞
=
−=
− +∑
[Turn Over 8
(iv)
2
3( 1)( 3)
n
r
rr r r=
−+ +∑
1 3 ( 1) 3...(2)(3)(5) ( 1)( )( 2) ( 1)( 3)
n nn n n n n n
− − −= + + +
− + + +
1
2
4 2( 1) ( 2) (1)(2)(4)
n
r
rr r r
+
=
−= −
− +∑
1 1 1 1 16 1 2 3 12n n n
= − + + −+ + +
1 1 1 112 1 2 3n n n
= − − + ++ + +
9.
21 i 2z − − ≥
(1 i) 2z⇒ − + ≥
1arg12 3 i
zπ π+⎛ ⎞≤ ≤⎜ ⎟+⎝ ⎠
arg( 1)12 6 6
7arg( 1)4 6
z
z
π π ππ
π π
⇒ + ≤ + ≤ +
⇒ ≤ + ≤
Method 1: 1 724 2
QR = − = ; 7 1 24 4
PQ = + =
Method 2: QR is the perpendicular bisector, so PQ = 2 (radius) i 2z⇒ − >
( )arg i4 2
zπ π≤ − <
- 1 x
O
(1,1) P
1
1 R
Q
4π
6π
y
[Turn Over 9
10(i)
Let the constant be a. d ( ) , where is a constant.dr k r a kt
∴ = −
Given 43r = when d 0drt= ,
0 (43 )Since 0, then 43
k ak a
∴ = −≠ =
d ( 43) dr k rt
∴ = − (shown)
(ii)
1
1 d d43
ln 43
r k trr kt C
=−− = +
∫ ∫
1
1
43 e
43 e where e
kt C
Ckt
r
r A A
+− =
= + =
When 0t = , 348r = . 305A∴ = . 43 305ektr∴ = +
(iii)
2
d
(43 305e ) d
30543 e
kt
kt
I r t
t
t Ck
=
= +
= + +
∫∫
When 0t = , 0I = .
2305Ck
∴ = −
30543 (e 1)ktI tk
∴ = + −
(iv) Given 5700I = and 90t = , 90
90
90
3055700 43(90) (e 1)
3051830 (e 1)
6 e 1
k
k
k
k
kk
∴ = + −
= −
= −
Solving using GC, 0.167k = − or 0k = (NA)
(v) 1643 305e
tr
−= +
If t becomes larger, 16305e 0
t−→ , 43r →
Hence r would be reduced to a steady 43 kilobytes per second in the long run.
[Turn Over 10
11i OA = 14 ii Plane ABD
5 1 164 0 200 4 4
4 14 4. 5 0 . 5 56
1 0 1r
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟× = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
%
4 5 565 6 36
x y zx y z− + =⎧
⇒ ⎨− − + =⎩
Using GC to solve:4 18 1 ,
0 1r γ γ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∴ = − + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
%
OR 14 5 1
4 . 5 364 6
14 5 20 24 3625 25 50
2
λ μλμλ μ λ μ
μ λμ λ
+ − −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− − + − + =
= += +
14 5 10 4 ( 2) 00 0 4
12 40 48 4
12 10 1 ,8 1
r λ λ
λ
γ γ
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
iii 4 128 8 0
8(Reason: j is zero.)
48
0
12 14 20 0 08 0 8
12 4 80 8 88 0 8
+⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − + ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuuv uuuv
%
uuuv
uuuv
uuuv
OD OD
OB
AD
BD
γγ γ
γ
[Turn Over 11
1Area =2
8 2 1 1 41 8 0 8 1 0 8 52
8 8 1 4 1
8 42 51.8 (3 s.f.)
×
− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= × = × = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= =
uuuv uuuvABD BD AD
iv 2(4) 7( 8) (0)2(12) 7(0) 8
64, 5
α βα β
β α
− − + =− + =
⇒ = =
OR The 3 planes intersect at the line
4 18 1 ,
0 1r γ γ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
%
2 17 . 1 0
12 7 0
5
αα
α
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− + ==
4 28 . 7 8 56 64
0 564β
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− − = + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=
v Their partitions are parallel to each other.
There is no intersection point.
12 Ellipse
(a) 2 2 2( 2) (1 )x a y− = − 2
22
( 2) 1x ya−
⇒ + =
Method 1: Sequence of transformations:
1) Scale // to x-axis by factor a. 2) Translate in the positive x-direction by 2 units.
[Turn Over 12
Method 2: Sequence of transformations:
1) Translate in the positive x-direction by 2a
units.
2) Scale // to x-axis by factor a.
(bi) 2
2
11 4
( )
4 ( 1)
3
−+ −
− +
− −− − −
−
xx x
x x
xx
x
y
(bii) Sub
2 41−
=+
xyx
into 2 2 2( 2) (1 )x a y− = − :
222 2 4( 2) 1
1
⎛ ⎞⎛ ⎞−⎜ ⎟− = − ⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠
xx ax
( ) ( ) ( )22 22 2 2 21 ( 2) 1 4⇒ + − = + − −x x a x a x --- (*) (shown) Hence the x-coordinate of the points of intersection of
1C and 2C satisfy equation (*). (b) (iii)
From (ii), number of intersection points between 1C and 2C gives the number of real roots of the equation (*). From the graphs, there are 2 points of intersection between 1C and 2C . Hence 2 real roots.
1= −x
1= −y x
2
2 41−
=+
xyx
4−2
22
( 2) 1x ya−
+ =
2− a 2 a+ 2−
HCI Prelim H2 Mathematics P2 Solutions
1
2010 HCI H2 Mathematics Preliminary Examination Paper 2 Solution Qn Solutions 1
Surface area of the tin and lid = 22 2 10x xy xπ π π+ + = 400π
2200 5x xyx
− −=
Volume of the container
( )
22
3 2
200 5
200 5
x xxx
x x x
π
π
⎛ ⎞− −= ⎜ ⎟
⎝ ⎠
= − −
( )2d 200 3 10dV x xx
π= − −
d 200d 3V xx
= ⇒ = or 10x = − (rejected)
( )2
2
d 6 10 0d
V xx
π= − − < when 203
x =
V is maximum when 203
x = .
When 203
x = , 553
y =
(or x = 6.67, y = 18.3).
2(i) 5 32 0z − = 5 32z⇒ = ei0 = 32ei2k π 2 i
52ek
zπ
⇒ = where 0, 1, 2.k = ± ± (ii)
The highest power in the equation 52 1 32w
w+⎛ ⎞ =⎜ ⎟
⎝ ⎠ is four since the terms with w5 are
canceled out. Hence the equation has only four roots.
512 32w
⎛ ⎞+ =⎜ ⎟⎝ ⎠
2 i512 2
kz e
wπ
⇒ + = =
( )2 i 2 i5 51 2 2 2 1
k ke e
wπ π
⇒ = − = −
( ) ( ) ( ) ( )1 2 3 4
2 i 2 i 4 i 4 i5 5 5 5
1 1 1 1
2 1 1 1 1
w w w w
e e e eπ π π π− −
⇒ + + +
⎡ ⎤= − + − + − + −⎢ ⎥⎣ ⎦
HCI Prelim H2 Mathematics P2 Solutions
2
= 2 42 2cos 2cos 4
5 5π π⎡ ⎤+ −⎢ ⎥⎣ ⎦
= 2 44 cos cos 2 .5 5π π⎡ ⎤+ − ∈⎢ ⎥⎣ ⎦
Or use GC, 1 2 3 4
1 1 1 1w w w w
+ + + = −10.
3 2nS an bn c= + +
1 1 100U S a b c= = + + =
2 4 2 190S a b c= + + =
10 100 10 360 100 90 550S a b c= + + = + + = Using GC,
5a = − , 105b = , 0c = Thus 25 105nS n n= − +
1n n nU S S −= −
( ) ( )( )225 105 5 1 105 1n n n n= − + − − − + −
110 10n= −
( )1
110 10 110 10 10n nU U
n n−−
= − − − +
10 (a constant)= − Hence sequence is an AP.
4i 0 1 12 (1 ) 0 2
0
0 0 00 (1 ) 2 2 2
2
0 1 12 2 2 2 4
2 3
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − +⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + − − +⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuuv
uuuv
uuuv
OXt t
OYt t t t
XYt t t t t
μμ μ μ
μ
μ μ μμ
μ μμ μ μμ μ μ
OR
12ABt
−⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠
uuur
02
2BC
t
⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠
uuur
HCI Prelim H2 Mathematics P2 Solutions
3
( )1 1
1 2 2 2XBt t t
μμ μ
μ
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
uuur
0 02 2
2 2BY
t tμ μ
μ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
uuur
1 0 12 2 2 2 4
2 3XY XB BY
t t t t t
μ μμ μ μ
μ μ μ
− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + = − + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur uuur uuur
ii Suppose O, X, Y are collinear. Then
1 02 2 2
21 0 1 (Out of range)
=
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − +⎝ ⎠ ⎝ ⎠− = ⇒ =
uuuv uuuvOX kOY
kt t t
μμ μμ μ
μ μ
Thus O, X, Y are not collinear.
iii 1 02 2 2
2OX OY
t t t
μμ μμ μ
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − +⎝ ⎠ ⎝ ⎠
uuuv uuuv
= μ(4 – 4μ + t2 – 2μt2) = 0
2
2 2
4 1 10 (reject) or = + 4 2 2 2
tt t
μ μ +⇒ = =
+ +
For all t ∈ℝ\{0}, 0 < μ < 1. Thus ∠XOY can be 90° when 0≠t .
HCI Prelim H2 Mathematics P2 Solutions
4
iv 12 4
3projection vector
1 4 42 4 . 1 . 1
3 0 017
44 4 2 4 1
170
42 1
170
XYt t
t t
μμμ
μμμ
μ μ
−⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟− +⎝ ⎠
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠ ⎝ ⎠=
⎛ ⎞− + − ⎜ ⎟= ⎜ ⎟
⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟= − ⎜ ⎟⎜ ⎟⎝ ⎠
uuuv
5(i) ( ) ( )2 26 2 3x y x− + + = + ( ) ( ) ( )
( )
2 2 22 3 6
9 2 3
y x x
x
+ = + − −
= −
(ii)
(iii)
For the equation ( ) ( )22 9 2 3y x+ = − , When 2x = , 1y = .
HCI Prelim H2 Mathematics P2 Solutions
5
When 7y = , 6x = . Method 1: Using dx y∫
( ) ( )( ) ( )
2
2 2
2 9 2 3
2 23 1 32 18 2 9
y x
y yx
+ = −
⎛ ⎞+ += + = +⎜ ⎟
⎜ ⎟⎝ ⎠
( )
( ) ( )
27
1
73
1
2
21 3 d 2(6)2 9
1 ( 2)3 122 27
1 21 27 3 1 122
10 units
yR y
yy
⎛ ⎞+⎜ ⎟= + −⎜ ⎟⎝ ⎠
⎧ ⎫⎡ ⎤+⎪ ⎪= + −⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭
⎧ ⎫⎡ ⎤= + − + −⎨ ⎬⎣ ⎦⎩ ⎭=
∫
Method 2: Using dy x∫
( ) ( )22 9 2 3
2 3 2 3 [ 2 3 2 3 N.A.]
y x
y x y x
+ = −
= − + − = − − −
( )6
2
632
2
4(7) 2 3 2 3 d
28 2 (2 3)
R x x
x x
= − − + −
⎧ ⎫⎡ ⎤⎪ ⎪= − − + −⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭
∫
= { 28 – [(–12 + 27) – (–4 + 1)]} = 10 units2
(iv) Volume required
= vol. of cylinder – (vol. generated by curve from y = –2 to y = 1)
( )6 2
2
2(7) (4) 2 3 2 3 dx xπ π⎡ ⎤= − − + −⎢ ⎥⎣ ⎦∫
= 196π – 92π = 327 unit3 (3 s.f.)
6 Use random sampling method to select a sample from each class. The number of seats from each class would be proportional to the size of each stratum. Any 1 of the answers below: Some passengers have booked a flight ticket but did not turn up or changed flight so some of the seats in the sample may not have a passenger. OR
First Class Business Class Economy Class 4 16 60
HCI Prelim H2 Mathematics P2 Solutions
6
The flight is not fully booked so the chosen seat could be empty. OR The passenger may ignore the questionnaire. It is not appropriate to use simple random sampling as passengers from different classes may have different opinions on the service. The number of passengers in the first class is very small, so the passengers from the first class may not be chosen at all using the simple random sampling method.
7(i) No. of ways = 10! 12600
4!3!2!=
7(ii) Case 1: The 2 blue tiles and 1 yellow tile are in the 4th row with the 4th tile being red or
green. No. of ways = no. of ways with B, B, Y, G in 4th row + no. of ways with B, B, Y, R in 4th row
= 4! 6! 4! 6! 32402! 2!2! 2! 2!2!2!
× + × =
Case 2: The 2 blue tiles and 1 yellow tile are in the third row.
No. of ways = 3! 7! 18902 2!2!2!
× =
Total no. of ways = 3240 + 1890 – 3!2!
3!2!
4!2! –
3!2!
4!2! 3!
= 5130 – 108 – 216 = 4806
7(iii) No. of ways such that less than 3 yellow tiles are in the fourth row
= 12600 − 43C 7!
4!2!=12600 − 420 =12180
7 last part
No. of ways
= 6!3!2!
× 7C4 = 2100
8(i) 2010 10(9000) 9201,10
x += =
( )( )( )2
22 90001 50714790009 10 9
xs x
⎡ ⎤−⎢ ⎥= − − =⎢ ⎥⎣ ⎦
∑∑
H0 : 9000μ = H1 : 9000μ >
Test Stat:9000 ~ (9)
5071479 10
x t−=
×
p–value = 0.01265 < 0.05 Since the p –value = 0.01265 < 0.05, we reject 0H and conclude that there is sufficient evidence, at 5% level of significance, that the mean life span of the electronic component has increased.
HCI Prelim H2 Mathematics P2 Solutions
7
8(ii) H0: 9000μ = vs H1: 9000μ >
Under H0, X ~ N(9000, 252
10 ) = N(9000, 62.5).
Test Statistic = X – 9000
62.5 ~ N(0, 1).
Level of significance = 1% P(Z > 2.326347877) = 0.01 At the 1% significance level, reject 0H if z ≥ 2.326347877.
z = x – 9000
62.5 ≥ 2.326347877
x ≥ 9018.391395 = 9020. Assumptions: The standard deviation of the life span remains unchanged after the change in process.
9 First part
~ N(190, 576)X
1 20 21 300.001( ... ) 0.001(2)( ... ) ~ N(0, 0.03456)T X X X X= + + − + + P( | | 0.15) P( 0.15 0.15)T T≤ = − ≤ ≤ = 0.580 OR A = X1 +... + X20 – 2(X21 +... + X30) ~ N(0, 34560)
0.15P( | | ) P( 150 150)0.001
A A≤ = − ≤ ≤
= 0.580 9(i) Let Y be the r.v. denoting the mass of a randomly chosen apple from Mark's orchard.
2~ N( , 30 )Y μ Since the shaded area is the same, using the symmetric property of the normal curve,
110μ =
9(ii) Probability that Mark will get an apple graded as 'large' chosen at random = ( 150)P Y > =
HCI Prelim H2 Mathematics P2 Solutions
8
0.09121128 Let A be the r.v. denoting the number of apples graded as large out of 65 randomly chosen apples.
~ (65, 0.09121128)A B
( 5) 1 ( 4)0.718
P A P A≥ = − ≤=
10(a) (i) P(A ⎜M ) = 200 1400 2
=
(ii) P(M ' ∩ C ') 250 300 111000 20
+= =
( ) ( ) ( )9 1,20 2
P A P A M P A= = ≠
A and M are not independent. 10(b) (i) No. of immigrants in the sample
= ( ) ( ) ( )0.2 200 250 0.3 130 300 0.05 120 225+ + + + =
P(voter supports Party A given voter is an immigrant) 0.2 450 0.4225×
= =
(ii) Number of immigrants supporting Party C = ( )0.05 120 =6 P(exactly one immigrant voter supporting Party C or exactly one female voter supporting Party A (or both))
( )( ) ( )
exactly 1 immigrant voted for C
exactly 1 female voted for A both
P
P P
=
+ −
6 994 250 750 250 6 7441 2 1 2 1 1 1
10003
0.434C C C C C C CC
+ −= =
Alternative method:
Required Probability = 6 994 993 250 750 749 6 250 7443 3 3!1000 999 998 1000 999 998 1000 999 998
× + × − ×
= 0.434
HCI Prelim H2 Mathematics P2 Solutions
9
11(i)
Let X be the r.v. denoting the number of call–ins in a week. Hence ~ Po4
X λ⎛ ⎞⎜ ⎟⎝ ⎠
.
We are looking for the λ such that P(X ≤ 9) = 0.701 From graph, the value of 32.5λ = (to 3 sig.fig). The condition is that the rate of call–ins received by the centre is constant throughout a month / the call–in occurs randomly / The call–ins occur in a month are independent of one another
11(ii) Let Y be the r.v. denoting the number of call–ins in a week. ~ Po(32.5)Y
Since the mean is bigger than 10, hence ~ (32.5, 32.5)Y N approximately.
c.c(25 40) (25.5 40.5)P Y P Y< ≤ ⎯⎯→ < < = 0.810
11(iii)
Let S be the r.v. denoting the number of successful cases out of the n people in a support group.
3~ ( , )20
S B n
Since the number of groups concerned, which is 70, is large, therefore by applying CLT,
3 313 20 20~ ( , )20 70
nS N n
⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ approximately.
EITHER n ( 4)P S ≥ 27 0.589 28 0.812
Hence minimum value of n is 28. OR
( 4) 0.7P S ≥ >
λ
( 9)P X ≤
HCI Prelim H2 Mathematics P2 Solutions
10
( 4) 0.3P S < < 4 0.15( ) 0.30.1275
704 0.15 0.52440.1275
700.12754 0.15 ( 0.5244 )
700.12750.15 (0.5244 ) 4 0
705.23912 or 5.0899(reject)
27.45
nP Zn
nn
n n
n n
n nn
−< <
−< −
− < −
− − >
> < −>
Least n = 28.
12(i) Location F should be omitted as the road distance cannot be smaller than the straight line distance, indicating that it is an incorrect data entry.
From the scatter diagram, another location that should be omitted is location H, as it is an outlier based on the scatter diagram.
12(ii)
The suitable regression line is the regression x on y:
0.3936554 0.81702935x y= + When y = 20.0,
16.7x = km
12(iii)
y
s
302
70
180
HCI Prelim H2 Mathematics P2 Solutions
11
Since the graph of lns a b y= + is concave downwards whereas the graph of 2s a by= + is concave upwards, the graph of lns a b y= + will be more suitable to describe the scatter diagram of s and y. Hence model II is more suitable.
12(iv)
The appropriate regression line of s on ln y is 25.9499647 (45.24427905) lns y= + , i.e. 25.9 (45.2) lns y= + (to 3 s.f.)
12(v)
Since r for s and ln y is 0.992 close to 1, the linear correlation is strong between s and ln y. Furthermore, 170 cents is within the data range of the sample. Therefore the estimation using the line in (iv) is reliable. Since y is the independent variable, the line found in (iv) is also suitable for the estimation.
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 1 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE NAME
CLASS INDEX NUMBER
MATHEMATICS
Paper 1
Additional Materials: Answer Paper Graph Paper List of Formulae (MF15)
9740/01
15 September 2010
3 hours
READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
You are expected to use a graphic calculator.
Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
Innova Junior College [Turn over
IJC/2010/JC2 9740/01/S/10
2
1 For any given mass of gas, the volume V cm3 and pressure p (in suitable units) satisfy
the relationship
1 nV pk
,
where k and n are constants.
For a particular type of gas, 2.3n . At an instant when volume is 32 cm3, the pressure is 105 units and the pressure is increasing at a rate of 0.2 units s1.
Calculate the rate of decrease of volume at this instant. [4]
2 Given that the coefficient of 2x in the series expansion of
2
1 3n
x is 108, find the
value of n, where n is a positive integer. [4]
3 The sequence of numbers , where nu 1, 2, 3,...,n is such that
19
8u and 18 7n nu u n 8 .
Use the method of mathematical induction to show that
32 nnu n for . [
Deter
1n 5]
(i) mine if the sequence converges. [1]
(ii) Find in terms of n. [2]
1
rr
u
n
3
4 C
O
A
M B The diagram shows a quadrilateral OABC with OA AB and . OC BC
Points A and B have position vectors
1
2
and
1
0
1
respectively.
(a) Find cosine of angle OAB in terms of . [2]
(b) M is the midpoint of OB and 4AM MC
. By considering the area of the quadrilateral OABC, show that
5
2OA OC OA OB
. [4]
5 Illustrate, on a single Argand diagram, the locus of the point representing the complex
number z that satisfies both the inequalities
3arg 3 3i
4 2z
and 3 3i 2z . [4]
(i) Find the greatest and least values of 3iz . [2]
(ii) Find the least possible value of zarg , giving your answer in radians. [3]
IJC/2010/JC2 9740/01/S/10 [Turn over
4
6 Show, by means of the substitution , that the differential equation 2w x y
d
d2 3
y
xx y xy 0
can be reduced to the form
d
d3
w
xw . [2]
Hence find y in terms of x, given that 1
2y when 2x . [4]
7 A curve is defined parametrically by
2cosx t , 2 1y t ,
where 0 t .
(i) Find the equation of the normal to the curve at the point P where 3
t
. [5]
(ii) The normal at P meets the y-axis at N and the x-axis at M. Given that the curve meets the y-axis at Q, find the area of triangle MNQ, correct to 1 decimal place. [5]
8 A curve has equation
2 2( 4)1
4 9
x y .
(i) Sketch the curve, stating the equations of the asymptotes and the coordinates of the vertices. [3]
(ii) The region enclosed by the curve 2 2( 4)
14 9
x y , the x-axis and the line
is rotated through 4 right angles about the y-axis to form a solid of
revolution of volume V. Find the exact value of V, giving your answer in terms
of
2x
. [4]
IJC/2010/JC2 9740/01/S/10
5
9 In a medical research centre, a particular species of insect is grown for treatment of open wounds. The insects are grown in a dry and cool container, and they are left to multiply. The increase in the number of insects at the end of each week is at a constant rate of 4% of the number at the beginning of that week. At the end of each week, 10 of the insects would die due to space constraint and are removed from the container.
A researcher puts y insects at the beginning of the first week and then a further y at the beginning of the second and each subsequent week. He also decides that he will not take any insect out of the container.
(i) How many insects will there be in the container at the end of the first week? Leave your answer in terms of y. [1]
(ii) Show that, at the end of n weeks, the total number of insects in the container is
26 250 1.04 1n
y . [4]
(iii) Find the minimum number of complete weeks for the population of the insects to exceed . [4] 12513 y
10 The functions f and g are defined as
f : 1x x for 1x
g : for 0x e 1xx
(i) Define in a similar form, including its domain. [3] 1f
(ii) State the relationship between f and 1f , and sketch the graphs of f and 1f on the same diagram. [3]
(iii) Find the exact solutions of the equation
1f ( ) f ( )x x . [2]
(iv) Show that the composite function fg exists. [2]
(v) Given that h ( ) fg( )x x for , show that h is an increasing function for . [2]
0x 0x
IJC/2010/JC2 9740/01/S/10 [Turn over
IJC/2010/JC2 9740/01/S/10
6
11 (a) Write 2cos3 cosx x in the form cos cospx qx , where p and q are
positive integers. [1]
Hence find
(i) cos3 cos dx x x , [2]
(ii) the exact value of 40
cos3 cos dx x x
x . [4]
(b) State a sequence of transformations which transform the graph of to
the graph of
siny x3
sin 22
y x
cos3 cos
. [2]
(c) Find the numerical value of the area of the region bounded by the curves
x x and 3
sin 22
y x
for 02
x
. [3] y
12 A plane has equation and a line has equation .
1 9 1
2
1
3
r.
3 4
0 1
0 2
r
(i) Find the coordinates of P, the point of intersection of and . [4] 1 1 Hence, or otherwise, find the shortest distance from point A (3, 0, 0) to 1 .
[2] The equations of planes and 2 3 are given as
2 : , and
1 2
1 0
1 3
s t
r
2
2
1
3 : x y z , where , .
(ii) Find the equation of plane 2 in the form d . Explain why the planes
1 an 2
r n =
d intersect. [4]
(iii) The line of intersection of planes 1 and 2 is . The line has equation
2 2
3 2
3 1
2 1
r =
.
Given that the three planes 1 , 2 and 3 do not have any points in common,
find the conditions satisfied by and . [3]
1
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level Higher 2
CANDIDATE NAME
Civics Group INDEX NUMBER Mathematics Paper 2 Additional materials: Answer Paper Graph paper List of Formulae (MF15)
9740/02
16 September 2010
3 hours
READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
You are expected to use a graphic calculator.
Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages.
Innova Junior College [Turn over
2
IJC/2010/JC2 9740/02/S/10
Section A: Pure Mathematics [40 marks]
1 Three space probes, Probe A, Probe B and Probe C with mass 600 kg, 630 kg and 900 kg respectively were sent into space to find the gravitational pull on the planets Venus, Mars and Saturn. The sum of the weights of Probe B on the three planets is 13860 N. The weight of Probe C on Saturn is 2880 N more than the weight of Probe A on Venus. The average weight of Probe A on Saturn, Probe B on Venus and Probe C on Mars is 4870 N.
Find the gravitational pull on each of the planets. [3]
Hence find the weight of Probe D, which has mass 500 kg, on planet Saturn. [1]
[Weight (in N) = Mass (in kg) gravitational pull 2ms ]
2 Express
2
2
6 1f ( )
3 1 3
x xx
x x
in the form
23 1 3
A Bx
x
C
x
,
where A, B and C are constants. [3]
Hence find the exact value of 1
3f dx x
. [4]
3 Given that 1 2cosy x , show that
dsin 2 0
d
yy x
x .
Show that 3
3
d0
d
y
x when . Hence write down the first two non-zero terms in the Maclaurin
series for
0x
1 2cos x . [8]
4 The curve C has equation 42 4
2
ay ax a
5
x
, where a is a constant.
(i) Given that curve C has turning points, show that 5
4a or . [3] 0a
(ii) Sketch curve C for the case when 1a , indicating clearly the coordinates of the turning
points and the equations of any asymptotes. [4]
(iii) Hence by sketching an appropriate graph on the same diagram, solve
9
62
x xx
. [3]
3
5 (a) The complex number z is given by
3
2
3 i
1 iz
p
, where . 0p
Given that 2z , find the value of p. [3]
Show that 5arg
6z
. [3]
(b) Find the exact roots of the equation 5 2 0z
in the form ier , where and 0r . [5]
Section B: Statistics [60 marks]
6 A company’s director wants to obtain his employees’ views on flexible working hours. The company has 600 employees. Describe clearly how you would choose a systematic random sample of 30 employees. Describe briefly one disadvantage of this sampling method in this context. [3]
7 Box A contains 6 balls numbered 1, 2, 2, 2, 5, 7.
Box B contains 4 balls numbered 1, 4, 4, 7.
Box C contains 3 balls numbered 3, 4, 6.
One ball is removed at random from each box.
(i) Show that the probability that each of the three numbers obtained is greater than 3 is 1
6.
[1]
(ii) Find the probability that each of the three numbers obtained is greater than 3 or the sum
of the three numbers obtained is 13 (or both). [4]
All the balls are now placed in a single container.
A game is played by a single player. The player takes balls, one by one and with replacement, from the container, continuing until either a number 1 results or a prime number results. The player wins if the number on the last ball chosen is 1 and loses otherwise.
Find the probability that a player wins a particular game. [3]
IJC/2010/JC2 9740/02/S/10 [Turn over
4
8 Magnolia cows are milked by hand and Daisy cows are milked by machine. The time taken to milk a randomly chosen Magnolia cow may be taken to have a normal distribution with mean 30 minutes and standard deviation 2 minutes. The time taken to milk a randomly chosen Daisy cow may be taken to have an independent normal distribution with mean 5.5 minutes and standard deviation 0.5 minutes.
(i) The probability that it will take less than a minutes to milk a randomly chosen Magnolia cow is 0.85. Find a. [1]
(ii) Using an appropriate approximation, find the probability that out of 50 randomly chosen Magnolia cows, there are more than 10 but at most 40 which take less than a minutes to milk. [4]
(iii) Find the probability that the total time taken to milk two randomly chosen Magnolia cows exceeds eleven times the time taken to milk a randomly chosen Daisy cow by at least 3 minutes. [3]
9 An electronic game called ‘Wishful Thinking’ is played with 5 boxes arranged in a row. When a button is pressed, each of the five boxes displays a picture of a fruit – either an apple, orange or pear. A possible result of the game is shown below.
Box 1 Box 2 Box 3 Box 4 Box 5
Events ‘Success’ and ‘Windfall’ are defined as follows.
‘Success’ : Exactly 3 boxes display the same fruit.
‘Windfall’ : All 5 boxes display the same fruit.
Find
(i) the total number of possible results, [1]
(ii) the number of ways of obtaining ‘Windfall’, [1]
(iii) the total number of ways of not obtaining ‘Success’. [3]
Ten electronic game machines with distinct serial numbers are sent to 3 different game centres. In how many ways can the game machines be distributed if each centre must have at least 3 machines? [3]
IJC/2010/JC2 9740/02/S/10
5
10 Observations of a cactus graft were made under controlled environmental conditions. The table gives the observed heights x cm of the graft at t weeks after grafting.
t 1 2 3 4 5 6 8 10 x 2.0 2.4 2.5 5.1 6.7 9.4 18.3 35.1
(i) Calculate the product moment correlation coefficient between t and x. [1]
(ii) Draw a scatter diagram for the data. [2]
(iii) Using your answer in part (i) and the scatter diagram in part (ii), explain why it is advisable to draw a scatter diagram first before interpreting the value of the product moment correlation coefficient. [1]
(iv) Explain why the scatter diagram may be consistent with a model of the form eat bx . [1]
(v) For the model , show that the relation between lneat bx x and t is linear. Hence calculate the equation of the appropriate regression line. [3]
(vi) Use the regression line in part (v) to predict the height of the cactus graft 20 weeks after grafting. Hence explain in the context of the question why it is unwise to extrapolate. [2]
11 A company claims that the guitar strings that the company manufactures have a tensile strength of 430 kpsi (kilo-pounds per square inch) on average. An engineer obtained a sample of 8 guitar strings and the tensile strength of each guitar string, x kpsi, is measured. The data is summarised by
2430 23, 430 211x x .
Test, at the 2% significance level, whether the company has overstated the average tensile strength of a guitar string. State any assumptions that you have made. [7]
The engineer will be able to conclude that the company has overstated the average tensile
strength of a guitar string if he conducts the same test at % significance level. State the smallest possible integer value of . [1]
The engineer takes a sample of 20 guitar strings manufactured by a rival company, whose guitar strings have tensile strength that is normally distributed with mean kpsi and standard
deviation 4.7 kpsi. The null hypothesis 430 is being tested against the alternative hypothesis
430 at 5% level of significance. Find the range of values of the sample mean for which the
null hypothesis is rejected, giving 2 decimal places in your answer. [3]
IJC/2010/JC2 9740/02/S/10 [Turn over
6
IJC/2010/JC2 9740/02/S/10
12 A roller-coaster ride has a safety system to detect faults on the track.
State a condition under which a Poisson distribution would be a suitable probability model for the number of faults detected on the track on a randomly chosen day. [1]
Faults on the track are detected at an average rate of 0.16 per day. Find the probability that on a randomly chosen day, the number of faults detected on the track is between 2 and 6 inclusive. [2]
Find the probability that in a randomly chosen period of 20 days, there are not more than 4 faults detected on the track. [2]
There is a probability of at least 0.15 that the mean number of faults detected on the track per day over a randomly chosen long period of n days is at least 0.2. Find the greatest value of n. [3]
There is also a separate safety system to detect faults on the roller-coaster train itself. Faults are detected by this system at an average rate of 0.05 per day, independently of the faults detected on the track.
Find the probability that in a randomly chosen period of 20 days, the number of faults detected on the track is at most 1 given that the total number of faults detected is 5. [4]
2010 IJC JC 2 PRELIMINARY EXAMINATION 2 Paper 1 (Solutions)
1 Given 1 nV pk
=
1d 1d
nV npp k
−=
1 npnk p
⎛ ⎞= ⎜ ⎟
⎝ ⎠
1 nn pp k
⎛ ⎞= ⎜ ⎟⎝ ⎠
nVp
=
Given 2.3n = − , when V = 32, p = 105, d 0.2dpt
=
d d dd d dV V pt p t
= ×
Since dV nVdp p
= , d dd dV nV pt p t
= ×
( )( )2.3 320.2
105−
= ×
= − 0.140 (to 3 s.f.)
Thus, the rate of decrease of volume at the instant is 3 10.140cm s− .
2 ( )
( )2 2 1 31 3
nn x
x−= +
+
Consider ( ) ( )( ) ( ) ( )211 3 1 3 3 ...
2!n n n
x n x x− − − −+ = + − + +
( ) ( )211 3 9 ...
2n n
nx x+
= − + +
( )( ) ( )22 1
2 1 3 9 ...21 3 n
n nnx x
x
+⎛ ⎞= − + +⎜ ⎟
+ ⎝ ⎠
( ) 21... 2 9 ...
2n n
x+⎛ ⎞
= + +⎜ ⎟⎝ ⎠
Given: coefficient of 2 108x = ⇒ ( )9 1 108n n + =
2 12 0n n+ − = ( )( )4 3 0n n+ − =
4n = − (rejected since n +∈ ) or 3n =
Thus, value of n = 3
3 Let nP denote the statement 32 nnu n −= + , for 1n ≥ .
When 1n = ,
LHS = 198
u = (given)
RHS = 3 1 91 2 18 8
−+ = + = = LHS
Thus 1P is true.
Assume that kP is true for some 1k ≥ , i.e.,
32 kku k −= +
Want to show that 1kP + is true, i.e.,
( )3 11 1 2 k
ku k − ++ = + +
LHS = 1ku +
= ( )1 7 88 ku k+ +
= ( )31 2 7 88
kk k−+ + +
= ( )31 8 8 28
kk −+ +
= ( ) ( )33
1 18 8 28 2
kk −+ +
= 3 31 2 kk − −+ + = ( )3 11 2 kk − ++ + = RHS Thus kP true ⇒ 1kP + is true
Since P(1) is true, and kP true ⇒ 1kP + is true,
by mathematical induction, 32 nnu n −= + , for 1n ≥ .
3i 32 nnu n −= +
The sequence does not converge because as n → ∞ , nu → ∞ . 3ii
1
n
rr
u=
∑ = ( )3
12
nr
rr −
=+∑
= 3
1 12
n nr
r rr −
= =+∑ ∑
= ( ) ( )3 6 31 2 2 ... 22
nn n − − −+ + + + +
= ( )( )3 3
3
2 1 21
2 1 2
n
n n
− −
−
⎛ ⎞−⎜ ⎟⎝ ⎠+ +
−
= ( )
1 118 81
12 18
nn n
⎛ ⎞−⎜ ⎟⎝ ⎠+ +
−
= ( ) 1 11 12 7 8nn n ⎛ ⎞+ + −⎜ ⎟
⎝ ⎠
4a 1
2OA α
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
uuur
1 1 20
2 1 1BA α α
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
ˆcosOA BA OA BA OAB• =uuur uuur uuur uuur
⇒ 2 21 2
ˆ1 4 4 1cos2 1
OABα α α α•
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ = + + + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⇒ 2
24ˆcos5
OAB αα
+=
+
4b Area of quadrilateral OABC = 2 × Area of ΔAOC
= 2 × 12
OA OC×uuur uuur
= OA OC×uuur uuur
Area of ΔAOB
= 12
OA OB×uuur uuur
Area of ΔBOC = 12
OB MCuuur uuuur
= ( )1 42
OB AMuuur uuuur
since 4AM MC=uuuur uuuur
= 142
OB AM⎛ ⎞⎜ ⎟⎝ ⎠
uuur uuuur
= ( )4 area of AOB
= 142
OA OB⎛ ⎞×⎜ ⎟⎝ ⎠
uuur uuur= 2 OA OB×
uuur uuur
Thus, area of quadrilateral OABC = Area of ΔAOB + Area of ΔBOC
B
O
A
M
C
B
O
A
1
4
= 12
OA OB×uuur uuur
+ 2 OA OB×uuur uuur
= 52
OA OB×uuur uuur
⇒ OA OC×uuur uuur
= 52
OA OB×uuur uuur
(Shown)
5
5i
The least value of 3iz + = 45 2− The greatest value of 3iz + = 45 2+
0
Im
Re 1 2 3
1
2
3 (3,3)
0
Im
Re 1 2 3
1
2
3 (3,3)
−3
2
2
5ii
1
2sin18
2sin 0.4908818
θ
θ −
=
= =
arg( ) 0.490884
z π= −
0.295 rad= (to 3 s.f.)
6 Given: 2w x y= Differentiate w.r.t x
2d d2d dw yxy xx x
= +
dd
2 3 0yx
x y+ + = ------- (1)
(1) x× : 2 2dd
2 3 0yx
x xy x y+ + =
d 3 0dw wx
+ =
d 3dw wx
= − (shown)
d 3dw wx
= −
1 1 d 1 d3
w xw
− =∫ ∫
1 ln3
w x c− = +
ln 3 3w x c= − −
3 3e x cw − −=
0
Im
Re 1 2 3
1
2
3 (3,3)
2 18
θ
3 3e ec xw − −= ± 3e xw A −= 2 3e xx y A −=
Given that 12
y = − when 2x = ,
2 612 e2
A −⎛ ⎞− =⎜ ⎟⎝ ⎠
⇒ 62eA = − Thus, 2 6 32e e xx y −= −
⇒ 6 32
2 e xyx
−= −
7i Given: 2cosx t= 2 1y t= −
d 2sindx tt
= − d 2dyt
=
d 2 1 d 2sin sinyx t t
∴ = = −−
Gradient of tangent at the point P is
1 1 23 3sin
3 2π
− = − = −
Gradient of normal at the point P is 1 32 23
− =⎛ ⎞−⎜ ⎟⎝ ⎠
When 3
t π= ,
⇒ 22 1 13 3
y π π⎛ ⎞= − = −⎜ ⎟⎝ ⎠
and 2cos 13
x π⎛ ⎞= =⎜ ⎟⎝ ⎠
Hence equation of normal at the point
( )2 31 13 2
y xπ⎛ ⎞− − = −⎜ ⎟⎝ ⎠
3 3 212 2 3
y x π= − − +
0.86603 0.22837y x= + 0.866 0.228y x= +
ii Equation of normal at P: 3 3 21
2 2 3y x π
= − − +
To find y-intercept:
When x = 0, y = 3 212 3
π− − +
or 0.22837y =
To find x-intercept:
When y = 0, x = 2 3 212 33
π⎛ ⎞+ −⎜ ⎟⎜ ⎟
⎝ ⎠
or x = −0.26370
To find y-intercept of the curve
When x = 0, 2
t π⇒ = , 1y π= −
Area of triangle MNQ 1 3 2 2 3 21 1 12 2 3 2 33
π ππ⎛ ⎞⎡ ⎤ ⎛ ⎞
= − − − − + + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟⎜ ⎟⎣ ⎦ ⎝ ⎠⎝ ⎠
0.252257=
0.3= (to 1 d.p.)
8
342
y x= −
vertex ( )2, 4−
x
y
vertex ( )2,4
centre ( )0, 4
342
y x= +
2− 2
0
x
y
4
2− 20
( )( )4 2 2
0d 2 4V x yπ π= −∫
( )4 2
0
44 4 d 169
V y yπ π⎛ ⎞= + − −⎜ ⎟⎝ ⎠∫
( )4
3
0
4 4 4 169
y yπ π⎡ ⎤= + − −⎢ ⎥⎣ ⎦
( )34 4 64 16 0 0 169 9 3
π π⎡ ⎤⎛ ⎞⎛ ⎞= + − + − −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦
688 256 1627 27
π π π⎡ ⎤= − =⎢ ⎥⎣ ⎦
9i Number of insects in the container at the end of 1st week =1.04y − 10
ii The total number of insects in the container are 3 2
1 2
1.04 ..... 1.04 1.04 1.04
10 1.04 1.04 ......... 1.04 1
n
n n
y y y y− −
= + + + +
⎡ ⎤− + + + +⎣ ⎦
( ) ( )1.04 1.04 1 1.04 110
1.04 1 1.04 1
n n
y− −
= −− −
( ) ( )26 1.04 1 250 1.04 1n ny= − − −
= ( )26 250 1.04 1ny ⎡ ⎤− −⎣ ⎦
iii ( )( )26 250 1.04 1 13 125
11.04 12
n
n
y y− − > −
− >
Wk Start End 1 2 3 n
y 1.04 10y y− +
21.04 1.04 1.04(10) 10y y y+ − − +
1.04y -10
( )1.04 1.04 10 10y y− + − = 21.04 1.04 1.04(10) 10y y+ − −
( )21.04 1.04 1.04 1.04(10) 10 10y y y+ − − + − 3 2
2
1.04 1.04 1.041.04 (10) 1.04(10) 10
y y y= + +
− − −
3 2
1 2
1.04 ..... 1.04 1.04 1.04
10 1.04 1.04 ......... 1.04 1
n
n n
y y y y− −
+ + + +
⎡ ⎤− + + + +⎣ ⎦
31.042
3ln1.04 ln2
10.3
n
n
n
>
>
>
Therefore the minimum number of complete weeks is 11.
10i Let f ( ) 1y x x= = − , 1x ≤ ⇒ 21 x y− = ⇒ 21x y= − ⇒ 1 2f ( ) 1x x− = −
1 ffD R− = = [ )0,∞
Thus, 1 2f : 1x x− → − , 0x ≥
ii The graph of 1f − is the reflection of the graph of f in the line y x= .
11a i
( ) ( )2cos3 cos cos 4 cos 2x x x x= +
1cos3 cos d 2cos3 cos d2
x x x x x x=∫ ∫
( )1 cos 4 cos 2 d2
x x x= +∫
1 1 1sin 4 sin 22 4 2
x x c⎛ ⎞= + +⎜ ⎟⎝ ⎠
1 1sin 4 sin 28 4
x x c= + +
ii 4
0cos3 cos dx x x x
π
∫
=4
40
0
1 1 1 1sin 4 sin 2 sin 4 sin 2 d8 4 8 4
x x x x x xπ
π⎡ ⎤ ⎛ ⎞+ − +⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠∫
1
1
x
y
f
1f −
4
0
1 1 1 1sin 4 sin 2 cos 4 cos 24 8 4 4 4 32 8
x xπ
π π π⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎡ ⎤= + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦
1 1 1 1 1 1sin sin cos 4 cos 24 8 4 2 32 4 8 4 32 8π π π ππ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦
1 1 116 32 32 8π ⎡ ⎤= + − − −⎢ ⎥⎣ ⎦
316
π −=
b Method 1: (Translation – Stretching)
1: Translation by 32
π in the direction of the x-axis.
2: Stretching parallel to the x-axis with scale factor of 12
.
Method 2: (Stretching − Translation)
1: Stretching parallel to the x-axis with scale factor of 12
.
2: Translation by 34
π in the direction of the x-axis.
c Area bounded by the curves
1.0471976
0
3sin 2 cos3 cos d2
x x x xπ⎡ ⎤⎛ ⎞= − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫
= 0.32476 = 0.325 units2 (to 3 sf)
12i
1π : 2
r 1 93
⎛ ⎞⎜ ⎟− =⎜ ⎟⎜ ⎟⎝ ⎠
.%
1l : 3 4
r 0 10 2
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + λ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
%
For point of intersection,
3 4 20 1 1 90 2 3
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ λ − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥−⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
.
⇒ 3 2 4 20 1 1 1 90 3 2 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + λ − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
. .
⇒ 9 6 38 1 6
−λ = =
− −
⇒ OPuuur
=3 4 150 3 1 30 2 6
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Thus, coordinates of the P are ( )15, 3, 6− .
15 3 123 0 36 0 6
AP⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
Shortest distance from A to 1π
=
21
3
21
3
AP⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠
.uuur
12 23 16 3
4 1 9
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠=
+ +
.
= 24 3 18 3 3 14
1414 14− −
= = =
Alternative solution for shortest distance: Line AN:
3 2r 0 1
0 3
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + λ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
%
N is the point of intersection of line AN and plane 1π .
3 2 20 1 1 90 3 3
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ λ − − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
.
⇒ 3 2 2 20 1 1 1 90 3 3 3
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + λ − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
. .
⇒ 6 (4 1 9) 9λ+ + + =
⇒ 314
λ =
Thus, 3 2 48
3 10 1 314 14
0 3 9ON
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
1π
A (3, 0, 0)
P (15, 3, −6)
21
3
⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠
1l
1π
A (3, 0, 0)
P
21
3
⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠
1l
N
48 3 21 33 0 1
14 149 0 3
AN⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
Thus, shortest distance AN
= 3 3 144 1 914 14
AN = + + =uuur
ii
1π : 2
r 1 93
⎛ ⎞⎜ ⎟− =⎜ ⎟⎜ ⎟⎝ ⎠
.%
2π : 1 2 2
r 1 0 21 3 1
s t⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
%
A normal to plane 2π = 2 2 6 30 2 (8) 2 43 1 4 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟× = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2π : 3 1 3
r 4 1 4 3 4 2 12 1 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− = − = − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
%. .
⇒ 3
r 4 12
⎛ ⎞⎜ ⎟− =⎜ ⎟⎜ ⎟⎝ ⎠
%.
Since 3 24 12 3
k⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟≠ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
for any k ∈ ,
the normal of planes 1π and 2π are not parallel to each other ⇒ The planes are not parallel to each other. ⇒ The planes will intersect in a line.
iii
2l : 3 23 12 1
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ μ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
r =
3π : x y zα + − = β
⇒ 11
α⎛ ⎞⎜ ⎟ = β⎜ ⎟⎜ ⎟−⎝ ⎠
r.
1π , 2π and 3π do not have any points in common ⇒ 2l is parallel to 3π and does not lie on 3π
3π
11
α⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
211
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
(3, 3, 2)
2l
⇒ 2l is perpendicular to 11
α⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
and 33 12 1
α⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ≠ β⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
.
Thus, 21 1 01 1
α⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
.
⇒ 2 1 1 0α + + = ⇒ 1α = −
and 33 12 1
α⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ≠ β⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
. ⇒ 3( 1) 3 2− + − ≠ β
⇒ 2β ≠ −
2010 IJC JC 2 PRELIMINARY EXAMINATION 2 Paper 2 (Solutions)
1 Let V 2ms− , M 2ms− , S 2ms− be the gravitational pull on each planet Venus, Mars and Saturn respectively.
630S + 630V + 630M = 13860 ---------- (1)
900S − 600V = 2880 ---------- (2)
600S + 630V + 900M = 3(4870) = 14610 --------- (3)
630 630 630 13860900 600 0 2880600 630 900 14610
SVM
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟− =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠
From GC,
M = 3.8, V = 9, S = 9.2
Hence the weight of Probe D on Saturn is ( )500 9.2 N× = 4600 N
2
( )( )2
2
6 1f ( )3 1 3
x xxx x
− +=
+ +
23 1 3A Bx C
x x+
= ++ +
( ) ( )( )
( )( )2
2
3 3 1
3 1 3
A x Bx C x
x x
+ + + +=
+ +
2 2 26 1 3 3 3x x Ax A Bx Cx Bx C− + = + + + + + ( ) ( )23 3 3A B x C B x A C= + + + + +
Comparing coefficients, A + 3B = 1 B + 3C = −6 3A + C = 1
From GC, A = 1, B = 0, C = −2
( )( ) ( ) ( )
2
2 2
6 1 1 2f ( )3 13 1 3 3
x xxxx x x
− +∴ = = −
++ + +
( )1
3f dx x
−
−∫ =1
23
1 2 d3 1 3
xx x
−
−
⎡ ⎤−⎢ ⎥+ +⎣ ⎦∫
1
1
3
1 2ln 3 1 tan3 3 3
xx−
−
−
⎡ ⎤= + −⎢ ⎥⎣ ⎦
( ) ( )1 11 2 1 1 2 3ln 3 1 1 tan ln 3 3 1 tan3 33 3 3 3
1 2 1 2ln 2 ln83 6 3 33 3
π π
− −⎡ ⎤⎛ ⎞ ⎛ ⎞= − + − − − − + − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞ ⎛ ⎞= − − − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
1 1 2 2ln3 4 6 33 3
π π⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1 1ln3 4 3 3
π= −
3 ( )1 2cosy x−= 2cos xy =
( ) dsin 2dyy xx
− =
( ) dsin 2 0dyy xx+ = (shown)
( )22
2d dsin cos 2
ddy yy y
xx⎛ ⎞+ = −⎜ ⎟⎝ ⎠
( ) ( )
( )
3 2
3 2
22
2
d d dsin cosdd d
d d d dcos 2 sin 0d d dd
y y yy yxx x
y y y yy yx x xx
⎛ ⎞+ ⎜ ⎟⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
( ) ( )33 2
3 2d d d dsin 3 cos sin 0
d dd dy y y yy y y
x xx x⎛ ⎞ ⎛ ⎞+ − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
When 0=x ,
2
0cos 1 π== −y
d 0dyx=
2
2
d 2d
yx
= −
3
3
d 0d
yx
=
4 ( ) 4 52 42ay ax a
x+
= − − + +−
⇒ ( )2
d (4 5)( 1)d 2y aax x
+ −= − −
−
For turning points, d 0dyx=
⇒ ( )24 5 02
aax+
− + =−
⇒ ( )22 4 5a x a− = +
⇒ ( )24 4 4 5 0a x x a− + − − =
⇒ 2 4 5 0ax ax− − = Since there are turning points, there are two distinct real roots. Thus, 0D > ⇒ ( )24 4( )( 5) 0a a− − − >
⇒ 24 5 0a a+ > ⇒ ( )4 5 0a a + >
Thus, 54
a < − or 0a > . (Shown)
Alternatively
For turning points, d 0dyx=
⇒ ( )24 5 02
aax
+− + =
−
⇒ ( )2 4 52 axa+
− =
Since there are turning points, there are two distinct real roots.
Thus, 4 5 0aa+
>
Thus, 54
a < − or 0a > . (Shown)
4(ii) When 1a = , 962
y xx
= − − +−
54
− 0
54
− 0
++ −
−6
−1
−2
−6
−14
5 x
y
6y x= − −
2x =
962
y xx
= − − +−
4(iii) 962
x xx
+ + >−
⇒ 962
xx
x > − − +−
From GC, the curves intersect at x = 0.823 (to 3 s.f) From graph, solution is 2x > or 0.823x < .
5(a) ( )( )
3 3
2 2
3 i 3 i
1 i 1 iz
p p
− −= =
+ +
( )( )
3
22
3 1
1 p
+=
+
2
81 p
=+
2z =
⇒ 2
8 21 p
=+
24 1
3 or 3 (rejected)
p
p
= +
= −
−6
−1
−2
−6
−14
5 x
y
6y x= − −
2x =
962
y xx
= − − +−
y x=
( )( )
3
2
3 iarg arg
1 iz
p
⎛ ⎞−⎜ ⎟= ⎜ ⎟+⎜ ⎟⎝ ⎠
( ) ( )3 2arg 3 i arg 1 3i= − − +
( ) ( )3arg 3 i 2arg 1 3i= − − +
73 26 3 6π π π⎛ ⎞ ⎛ ⎞= − − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
For arg( )zπ π− < ≤ , 7 5arg( ) 26 6
z π ππ= − = (Shown)
5(b) 5 2z = −
( )
i
2 i
2e
2e , 0, 1, 2k k
π
π π+
=
= = ± ±
By De Moivre’s Theorem
21510
i2 e , 0, 1, 2
k
z kπ π+⎛ ⎞⎜ ⎟⎝ ⎠= = ± ±
( )3 31 1 1 1 1
5 5 5 510 10 10 10 10i i i ii2 e , 2 e ,2 e , 2 e , 2 ez
π π π ππ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠= 1 1 3 1 1 1 3
10 5 10 5 10 10 5 10 5i i i i
2 e , 2 e , 2 , 2 e , 2 ezπ π π π
− −= −
6 - Use a list with all the employees’ names arranged in alphabetical order and number them from 1 to 600
- Determine the sampling interval : 600 2030
k = =
- Randomly select the first person from the first 20 people on the list, then select every 20th person subsequently until a sample of 30 employees is obtained. The sample obtained might be over-represented by a particular department (or gender or ethnic group) of the company which has the greatest proportion of employees, hence the systematic random sample obtained is not a good representative of the population.
7(i) P(all are greater than 3)
= 2 3 2. .6 4 3
= 16
7(ii) Let X be the event : “each of the 3 numbers is greater than 3” Y be the event : “sum of the 3 numbers is equal to 13” P(Y) = P((2,7,4), (5,4,4))
56π
76π
−
= 3 1 1 1 2 1. . . .6 4 3 6 4 3
+
= 572
Required probability = P( )X Y∪ = P( ) P( ) P( )X Y X Y+ − ∩
= 1 5 1 2 1. .6 72 6 4 3+ −
= 524
P(a player wins a particular game)
= 22 4 2 4 2. . ......
13 13 13 13 13⎛ ⎞+ + +⎜ ⎟⎝ ⎠
= 22 4 41 ......
13 13 13
⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= 4
13
2 1.13 1
⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠
= 29
8 Let M minutes be the time taken to milk a Magnolia cow, and D minutes be the time taken to milk a Daisy cow.
( ) ( )2 2~ N 30,2 , ~ N 5.5,0.5M D
8(i) ( )P 0.85M a< = From GC, a = 32.073 = 32.1 (3 s.f.)
8(ii) Let X be the number of cows which take less than a minutes to milk, out of 50 Magnolia cows. ( )~ B 50,0.85X Since n = 50 is large, 42.5 5, 7.5 5np nq= > = > , ∴ ( )~ N 42.5,6.375X approx. Reqd prob = ( )P 10 40X< ≤
= ( )P 10.5 40.5X< ≤ (using c.c.) = 0.21415 = 0.214 (3 s.f.)
8(iii) ( )1 2E 11 2 30 11 5.5 0.5M M D+ − = × − × = −
( ) 2 2 21 2Var 11 2 2 11 0.5 38.25M M D+ − = × + × =
( )1 2 11 ~ N 0.5, 38.25M M D+ − −
Reqd prob = ( )1 2P 11 3M M D+ − ≥ = 0.28573 = 0.286 (3 s.f.)
9(i) Total number of possible results = 53 243=
9(ii) Number of ways of obtaining ‘Windfall’ = 3
9(iii) No. of ways to obtain a success
= 5! 5!2 3 1203!2 ! 3!
⎛ ⎞× + × =⎜ ⎟
⎝ ⎠
Total number of ways 243= −120 123=
9 Reqd no. of ways
= 10 7 43 3 4
3! 126002!
C C C× × × =
10(i) From GC, r = 0.92378 = 0.924 (3 s.f.)
10(ii)
10(iii) From (i) and (ii), we see that though r is close to +1, the scatter diagram indicates a curvilinear relationship between t and x, instead of a positive linear relationship. Thus it is advisable to draw a scatter diagram first before interpreting the value of the product moment correlation coefficient.
10(iv) As t increases, x increases but by increasing amounts. Thus the scatter diagram may be consistent with a model of the form eat bx += .
10(v) eat bx += ln x at b= + , where a and b are constants Thus the relation between ln x and t is linear. Find regression line of ln x on t.
t 1 2 3 4 5 6 8 10
ln x 0.69315 0.87547 0.91629 1.6292 1.9021 2.2407 2.9069 3.5582 From GC, eqn of regression line of ln x on t is: ln 0.20723 0.33498x t= + ln 0.207 0.335x t= + (3 s.f.)
10(vi) When t = 20, ( )( )ln 0.20723 0.33498 20 6.90683x = + = x = 999 Height of the cactus graft 20 weeks after grafting is 999 cm. It is impossible that the cactus can grow to that height after 20 weeks. Thus it is unwise to extrapolate.
x
t1 10
2.0
35.1
11 The t-test will be used. Assume a normal distribution for tensile strength of guitar strings.
( )2
2231 211
7 8
20.696
s⎡ ⎤−
= −⎢ ⎥⎢ ⎥⎣ ⎦
=
23 4308
427.125
x −= +
=
0
1
: 430: 430
HH
μμ=
<
Level of significance = 2%
Under 0H : 7430 ~XT t
S n−
=
From GC, 1.79t = − p-value 0.0585 0.02= > Do not reject 0H and conclude that there is insignificant evidence at the 2% significance level that the manufacturer has overstated its claim.
11 6α =
11 4.7σ =
0
1
: 430: 430
HH
μμ=
≠
Level of significance = 5%
Under 0H : 430 ~ (0,1)4.7 20XZ N−
=
To reject 0H ,
430 1.964.7 20x −
< − or 430 1.964.7 20x −
>
427.94x < or 432.06x >
12 Suitable condition
• The mean number of faults detected is a constant from day to day. • The faults are detected independently of one another.
Let X be the number of faults detected on the track in a day. ( )~ Po 0.16X
( )P 2 6X≤ ≤ = ( ) ( )P 6 P 1X X≤ − ≤ = 0.011513 = 0.0115 (3 s.f.)
Let Y be the number of faults detected on the track in 20 days. ( )~ Po 3.2Y
( )P 4Y ≤ = 0.78061 = 0.781 (3 s.f.)
1 2 nX X XXn
+ +=
L
Since n is large, by Central Limit Theorem, 0.16~ N 0.16,Xn
⎛ ⎞⎜ ⎟⎝ ⎠
approx.
( )P 0.2 0.15X ≥ ≥ ⇒ 0.2 0.16P 0.150.16
Z
n
⎛ ⎞⎜ ⎟−⎜ ⎟≥ ≥⎜ ⎟⎜ ⎟⎝ ⎠
⇒ ( )P 0.1 0.85Z n≤ ≤
From GC, 0.1 1.0364n ≤ ⇒ n ≤ 107.41
Greatest value of n is 107. 12 Let W be the number of faults detected on the train in 20 days.
( )~ Po 1W
( )~ Po 4.2Y W+
( ) ( ) ( ) ( ) ( )( )
P 0 P 5 P 1 P 4P 1| 5
P 5Y W Y W
Y Y WY W
= = + = =≤ + = =
+ =
= 0.002124370.163316
= 0.0130 (3 s.f.)
JURONG JUNIOR COLLEGE J2 Preliminary Examination
MATHEMATICS 9740/01 Higher 2 20 August 2010
Paper 1 3 hours
Additional materials: Answer Paper Graph paper List of Formulae (MF15) Cover Page
READ THESE INSTRUCTIONS FIRST
Write your name and civics class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.
The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together, with the cover page in front.
This document consists of 6 printed pages.
[Turn over
2
1 (i) Express 3
11 x−
in the form 21 1A Bx C
x x x+
+− + +
, where A, B, and C are constants to be
found. [2] (ii) Hence find the coefficient of rx in the expansion of 2
2 ,1
xx x+
+ + in ascending powers
of .x [4]
2 (i) Show that 1
,2 3 3k k k+
= −
where ,k +∈ 3k ≥ . [2]
(ii) Hence find
3 2
n
r
r
=
∑ . [4]
3 A curve is defined by the parametric equations
3(sin 2 cos 2 ),x t t= − 3(2 cos 2 ).y t t= −
Find the equations of the tangent and the normal to the curve at the point P, where4
t π= .
The tangent and normal to the curve at P meet the y-axis at R and S respectively. Find the area of the triangle PRS. [6]
4 A sequence 1u , 2u , 3 , u is such that 1 0u = and
( )1 3 4 1 ,n nu u n n+ = + − for all .n +∈
(i) Prove by induction that 23 2 1nnu n= − − for all n +∈ . [4]
(ii) Given 0 0u = , find 2
0
N
nn
u=∑ in terms of .N
( )( )2
1
1Given 1 2 16
n
rr n n n
=
= + +
∑
[4]
3
5 Given that 1 ln(1 )y x= + + , show that
(i)
22
2 2
d d 1 0.d d 2(1 )
y yyx x x
+ + = +
[2]
(ii) the Maclaurin’s series for y in ascending powers of x, up to and including the term in 3,x is
2 31 3 171 .2 8 48
x x x+ − +
[3]
(iii) Expand
22 31 3 171
2 8 48x x x + − +
in powers of x up to and including 3,x simplifying
your answer. Explain briefly how the result can be used as a check on the correctness
of the first four terms in the series for y. [3]
6 (a) A geometric progression has first term 1 and common ratio r. The sum of the first four
terms is less than the sum to infinity of the remaining terms. Without the use of a
graphic calculator, find the range of values of r. [4]
(b) Adam decided to save some money each day to buy his favorite toys. On the first day,
he saved one twenty-cent coin; on the second day, he saved two twenty-cent coins;
and so on. Find the total amount of money saved at the end of one year.
(Assume 365 days in a year). [3]
After spending all his savings in the first year to buy his favorite toys, Adam started
saving in the same manner again in the following year. As an encouragement, his
mother contributed the same amount that he saved every Saturday and Sunday.
Assuming the second year started on a Monday, find the total amount of money saved
at the end of the second year. [3]
7 The line l passes through the points A and B with coordinates ( )1, 2, 3− and
( )5, 14, 11 respectively. The plane p has equation 2 3 6 7.x y z+ − = −
(i) Show that the line l is parallel but not contained on the plane p.
[3]
(ii) Find the distance of the line l from the plane p.
[3]
(iii) Find a cartesian equation of the plane which contains l and is perpendicular to p. [2]
[Turn over
4
8 (a) The complex numbers p and q are such that
2 ip a= + , iq b= − ,
where a and b are real numbers.
Given that 13 13ipq = + , find the possible values of a and b .
[4]
(b) Sketch the locus of z which satisfies
4 3i 2 and Re(z) 4z − − = ≥ .
(i) Find the least and greatest value of z .
(ii) Show that the greatest value of arg( i)z − is 4π .
[2] [2] [1]
9 The function f is defined by 2
f: xxxλ −
for , 0,x xλ∈ − < < where λ is a
positive constant.
(i) Show by differentiation that ( )f x increases as x increases. [2]
(ii) Find ( )1f ,x− stating the domain of 1f .− [4]
(iii) Find the value ofλ such that 1 1 1 1f f .2 2
− − − = −
[3]
10
(i) Find the integral 2
1 d .1 4
x xx
−+∫
[3]
(ii) By sketching the graphs of 3exy = and 3,y x= + or otherwise, solve the inequality
3e 3.x x> + [3]
Hence find 22 3e 3 d ,x x x− − −∫
giving your answer in an exact form.
[4]
5
11 (a)
The function f is defined by f(x) = (ln x)2 for x > 0. The diagram shows a sketch of the graph of
y = f(x). The minimum point of the graph is A.
(i) State the x-coordinate of A. [1]
(ii) Use the substitution x = eu to show that the area of the region bounded by the
x-axis, the line x = e and the curve is given by 1 20
d .uu e u∫
Hence, find the exact value of this area. [5]
(b)
Find the volume of the solid formed when the region R, bounded by the lines 3 22
y x= − + ,
1y = and the curve cos ,y x= is rotated 2π radians about the y-axis, giving your answer correct to 3 decimal places. [4]
x
y
O A
( )2lny x=
[Turn over
x
y
O
1y =
3 22
y x= − +
cosy x=
R
6
12 The curve C has equation
2
2 ,ax kyx kx k
−=
− +
where a, k are constants and 0a > . (i) Find the values of k such that C has two asymptotes.
[2]
(ii) The diagram shows the graph of 2
2
5x kyx kx k
−=
− + for some 0.k >
On separate diagrams, draw sketches of the graphs of
(a) 2
2
25 ,25 5
x kyx kx k
−=
− +
(b) 2
22
5 ,k xyx kx k
−=
− +
(c) 2
2 ,5
x kx kyx k− +
=−
including the coordinates of the points where the graphs cross the axes and the equations of any asymptotes. [8]
y
x 2
5k
k−
O
JURONG JUNIOR COLLEGE J2 Preliminary Examination
MATHEMATICS 9740/02 Higher 2 26 August 2010
Paper 2 3 hours
Additional materials: Answer Paper Graph paper List of Formulae (MF15) Cover Page
READ THESE INSTRUCTIONS FIRST
Write your name and civics class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.
The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together, with the cover page in front.
This document consists of 6 printed pages.
[Turn over
2
Section A: Pure Mathematics [40 marks]
1 (i)
The diagram shows a rectangle inscribed in a semicircle of centre, O, and fixed radius a. The length OP is denoted by .x Show that, as x varies, the perimeter of the rectangle is a maximum when its sides are in the ratio 4 : 1. [6]
(ii) Variables x and y are related by the equation 2 2 2 3y xy x
x+ = − + , where 0y > .
Given that x is increasing at the rate of 15
units 1s− , find the rate of increase of y when x is 1.
[4]
2 A particular solution of a differential equation is given by 2 32( ) 2 .3
x y xy y+ = − Show that
2( ) dyy y xdx
+ = − [2]
A second, related, family of curves is given by the differential equation
2ddyx y yx
= +
By means of the substitution y ux= , show that the general solution for y, in terms of x, is
xyx c−
=+
,
where c is an arbitrary constant. [3] Sketch, on a single diagram, three distinct members of the second family of solution curves, stating clearly the coordinates of the points where the curves cross the axes and the equations of any asymptotes. [5]
O x P
3 3 Relative to the origin O, the position vectors of A, B and C are
α i + 3j + 4k, 2i – 2k and 4i + β j – k respectively.
(i) Point M lies on the line segment AB such that : 1: 2AM MB = . Given that the position
vector of M is 2j + 2k, findα . [2]
(ii) Given that the length of projection of BC
onto the line OM is 32
units, find β , where β is
a positive constant. [3]
(iii) The line l has vector equation 20 02 1
aµ
= + −
r , ,aµ ∈ ∈ . Using the values of α and β
found above, determine the value of a if l makes an angle of 6π radians with the plane ABC.
[4]
4 (i) Given that 2 2z = and arg( i )4
z π− = , find w in the form ia b+ , where ,a b ∈ , if
2 2wz = and 2 5arg
6zw
π
= −
. [4]
(ii) Solve the equation
4 1 3 i 0z + − = ,
giving the roots in the form re iθ where r > 0 and − π < θ ≤ π. [3] Show the roots on an Argand diagram. [2] Describe the geometrical shape formed by the points representing the roots and justify your
answer. [2]
[Turn over
4
Section B: Statistics [60 marks] 5 From the letters of the word DISTRIBUTION, find
(i) the number of 4-letter code-words that can be formed if the code-word contains exactly three ‘I’s. [2]
(ii) the number of code-words that can be formed using all the letters such that all the three ‘I’s
are separated. [2]
6 In a badminton team of 8 players, 5 are boys and 3 are girls. Boy A and Girl B are the only 2
left-handed players in the team. In a particular practice, 4 players are chosen to play doubles.
Find the probability that
(i) exactly 1 left-handed player is chosen, [2]
(ii) 2 girls are chosen given that exactly 1 left-handed player is chosen, [3]
(iii) either Boy A or Girl B is chosen (or both). [2]
7 The number of guitars sold by a music shop per day follows a Poisson distribution with mean λ .
It is known that on 2 in 7 days, there are no guitars sold. Show that 1.253λ = , correct to 3
decimal places. [2]
(i) Calculate the probability that less than 4 guitars are sold in a day. [2]
(ii) Using a suitable approximation, find the probability that, in a random sample of 100 days,
there will be more than 95 days in which less than 4 guitars are sold per day. [4]
(iii) Calculate the probability that in a period of 90 days, the mean number of guitars sold per
day is more than 1.5. [3]
5
8 The random variable X has a normal distribution with mean 15 and variance 5. The random
variable T is the sum of 2 independent observations of X.
(i) Find ( 2 3 )P T X> + . [3]
(ii) Three independent observations of X are obtained. Find the probability that exactly two of
the observations have value less than 20. [3]
The random variable Y has a normal distribution with mean µ and variance 2σ .
(iii) If 22.5σ = , find the greatest probability of (15.1 29.9)P Y< < , stating the value of µ . [2]
(iv) If 10µ = and ( 27) 0.25P X Y+ > = , calculate the value of σ and state an assumption
needed to carry out the calculation. [4]
9 A sample of 60 customers is to be chosen to take part in a survey conducted by a restaurant
owner.
(i) Explain briefly how the restaurant owner could use quota sampling. [1]
(ii) The purpose of the survey is to investigate customers’ opinions about the different lunch
and dinner menus.
Give a reason why a stratified sample might be preferable in this context. [1]
Explain clearly how the restaurant owner could use stratified sampling from his list of
regular customers if the ratio of regular customers for lunch and dinner is 2 : 3. [2]
[Turn over
6
10 Water in a reservoir undergoes a purification process before it can be consumed. The effectiveness, y %, of the process for various flow rates, x m3 s-1, is shown below.
x 1 2 4 6 8 10 20 30 40 y 80 60 45 40 30 25 18 15 10
The variables x and y are thought to be related by the equation ey bax= , where a and b are
constants.
(i) Give a sketch of the scatter diagram of y against ln x . Comment on whether a linear model would be appropriate referring to the scatter diagram. [2]
(ii) Find the value of the product moment correlation coefficient between y and ln x and explain
whether it supports your comment in part (i). [2] (iii) Find the least squares regression line of y on ln x and estimate the values of a and b . [3] (iv) Predict the effectiveness of the process when water flows at 0.5 m3s-1. Comment on the
reliability of your prediction. [2]
(v) Explain why in this context, the above model would not be appropriate for large values of x . [1]
11 The past records of a supermarket show that the mean amount spent per customer was $59 with standard deviation $8. The supermarket’s management suspects that the mean amount spent per customer has decreased. A random sample of 8 customers was taken and the amount spent per customer , $x, was recorded. The following result was obtained.
432x =∑ Stating a necessary assumption about the population, test the supermarket’s management’s
suspicion at the 5% significance level. [6] To encourage customers to spend more at the supermarket, the management initiated a
promotional campaign whereby each customer will receive a voucher which can be used to redeem products at the supermarket. A week after the start of the campaign, the manager of the supermarket took a sample of 9 customers and the amount spent per customer, $y, is summarised by
2( 70) 72, ( 70) 1234y y− = − − =∑ ∑ . The actual mean amount spent per customer is $ µ . In a test at the 5% level of significance, the
hypotheses are: Null hypothesis : 0µ µ=
Alternative hypothesis : 0µ µ≠ .
Given that the null hypothesis is rejected in favour of the alternative hypothesis, find the set of
possible values of 0µ . [6]
1
2010 JJC H2 Mathematics Prelim Exam P1 Solutions 1(i)
Let 3 2
11 1 1
A Bx Cx x x x
+≡ +
− − + +
Hence, ( ) ( )( )21 1 1A x x Bx C x≡ + + + + −
Using cover-up rule, 13
A =
Let 0x = , 1 A C= + ,
23
C =
Comparing coefficients of x , 0 A B C= + − , 13
B =
( ) ( )3 2
1 1 21 3 1 3 1
xx x x x
+∴ ≡ +
− − + +.
1(ii) 2
21
xx x+
+ + 3
3 11 1x x
= −− −
( ) ( )1 133 1 1x x− −= − − −
( ) ( )3 6 23 1 1x x x x= + + + − + + +K K
2 3 4 52 2x x x x x= − − + − − +K
Hence, coefficients of 2 if is a multiple of 31 otherwise
r rx ⎧= ⎨−⎩
2.
3 2
n
r
r
=
⎛ ⎞⎜ ⎟⎝ ⎠
∑1 2 4 3
2 2 2 2 2n n n− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
K
13 3
13 3
1 23 3
5 43 3
4 33 3
n n
n n
n n
+⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− −⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞
+ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
M
1
13
n +⎛ ⎞= −⎜ ⎟⎝ ⎠
or ( 1)! 13!( 2)!
nn+
−−
2
3. 3(sin 2 cos 2 )x t t= − , 3(2 cos 2 )y t t= −
3(2cos 2 2sin 2 )dx t tdt
= +
3(2 2sin 2 )dy tdt
= +
3(2 2sin 2 ) 1 sin 23(2cos 2 2sin 2 ) cos 2 sin 2
dy t tdx t t t t
+ += =
+ +
When 4
t π= , at point P , 3x = , 3
2y π= and 2dy
dx=
Equation of tangent at P is 3 2( 3)2
y xπ− = −
When x = 0, 3 62
y π= − . R is (0 , 3 6
2π− )
Equation of normal at P is 3 1( 3)2 2
y xπ −− = −
When x = 0, 3 32 2
y π= + . S is (0 , 3 3
2 2π+ )
As PR ⊥ PS,
Area of the triangle PRS = 1 .(3)2
RS = 1 3 1( 6)(3) 11 2 2 4
+ =
4. (i) Let nP be the statement 23 2 1n
nu n= − − for all 1n ≥ . When n = 1, LHS = 1 0u = (By definition)
RHS = ( )223 2 2 1 0− − = 1P∴ is true. Assume that kP is true for some k +∈ , 1.k ≥ That is, 23 2 1k
ku k= − − -----------------------(1)
We want to prove 1+kP , ie ( )211 3 2 1 1.k
ku k++ = − + −
LHS ( )( )3 4 1ku k k= + −
( ) ( )23 3 2 1 4 1k k k k= − − + −
1 2 23 6 3 4 4k k k k+= − − + − ( )1 23 2 2 1 1k k k+= − + + −
( )213 2 1 1k k+= − + − kP∴ is true 1kP +⇒ is true.
Since 1P is true, and kP is true 1kP +⇒ is true. By Mathematical Induction, 23 2 1n
nu n= − − is true for all 1n ≥ .
3
4. (ii) ( )
2 22
0 03 2 1
N Nn
nn n
u n= =
= − −∑ ∑
( )( ) ( ) ( )
( )
( ) ( )
2 1
2 1
2 12
1 3 1 22 2 1 2 2 1 2 12 6
3 1 2 1 2 4 1 32 3
2 13 1 8 2 32 3
N
N
N
N N N N
N N N
NN N
+
+
+
−= − + + − +⎡ ⎤⎣ ⎦
− += − + +⎡ ⎤⎣ ⎦
+−= − + +
5(i) Method 1
2 1 ln(1 )12
11
2 (1 )
y xdyydx xdydx y x
= + +
=+
=+
d 1d 2(1 )yyx x=
+
Method 2
12d 1 1 (1 ln(1 ))
d 2 11
2 (1 )d 1 d 2(1 )
y xx x
y xyyx x
− ⎛ ⎞= + + ⎜ ⎟+⎝ ⎠
=+
=+
222
2 2
22
2 2
d d 1 1 ( 1)(1 )d d 2 2(1 )
d d 1 0d d 2(1 )
y yy xx x x
y yyx x x
− −⎛ ⎞+ = − + =⎜ ⎟ +⎝ ⎠
⎛ ⎞+ + =⎜ ⎟ +⎝ ⎠
5(ii)
3 2 2
3 2 2 3
2 3
2 3
2 3
2 3
d d d d d 1. 2 0d d d d d (1 )
when 0,d 1 d 3 d 17 1, , ,d 2 d 4 d 8
1 3 1712 4 2! 8 3!1 3 17 12 8 48
y y y y yyx x x x x x
xy y yyx x x
x xy x
x x x
+ + − =+
=
= = = − =
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞∴ ≈ + + − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
= + − +
4
5(iii)
22 31 3 171
2 8 48x x x⎛ ⎞+ − +⎜ ⎟
⎝ ⎠
2 3 2 3
2 3
1 3 17 1 3 171 12 8 48 2 8 48
1 11 ...2 3
x x x x x x
x x x
⎛ ⎞⎛ ⎞= + − + + − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
= + − + +
Given 1 ln(1 )y x= + + 2 1 ln(1 )y x⇒ = + +
2 2 3
2 3
1 1LHS 1 ...2 3
RHS 1 ln(1 )1 11 ...2 3
y x x x
x
x x x
= = + − + +
= + +
⎛ ⎞= + − + +⎜ ⎟⎝ ⎠
Since LHS = RHS, the first four terms in the series for y is correct. 6(a) 4 4S S S∞< −
42S S∞<
( )42 1 11 1
rr r
−<
− −
( )42 1 1r− < (Since 1 0r− > )
4 12
r >
2 2
2 2
1 1 02 2
1 1 (no solution) or 2 2
r r
r r
⎛ ⎞⎛ ⎞− + >⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
< − >
1 14 41 1 0
2 2r r⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟− + >⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
1411
2r ⎛ ⎞− < < −⎜ ⎟
⎝ ⎠ or
141 1
2r⎛ ⎞ < <⎜ ⎟
⎝ ⎠.
6(b) Total sum ( ) ( ) ( )20 2 20 3 20 365 20= + + +K
( ) ( )( )365 1 36520
2+⎛ ⎞
= ⎜ ⎟⎝ ⎠
= 1335900 cents
Total sum ( ) ( ) ( ) ( ) ( ) ( )1335900 6 20 7 20 13 20 14 20 ... 363 20 364 20= + + + + + + +
( ) ( )( ) ( )( )( )1335900 20 13 13 14 13 2 14 13 51 14= + + + + + + +K
( )521335900 20 13 7272
⎛ ⎞= + +⎜ ⎟⎝ ⎠
1720700= cents
5
7(i)
5 114 211 3
6 312 2 68 4
AB−⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
uuur
3 26 3 6 18 244 6
0
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟• = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
=
⇒ l is parallel to p.
1 22 3 2 6 183 6
14 7
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟• = − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
= − ≠ −
∴ l is parallel but not contained on the plane p. 7(ii) Method 1
Let C(c,0,0) be a point on π .
20 3 70 6
c⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟• = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
72
c = −
23
7 / 2 16
0 2490 3
5 / 2 21 2 3 17
3 6
d
⎛ ⎞⎜ ⎟⎜ ⎟⎛ ⎞− −⎛ ⎞ ⎛ ⎞ ⎜ ⎟−⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠= − •⎜ ⎟⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − • =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
l
π
A
d
C
6
Method 2
Let m be the line perpendicular to p and passing through A.
Vector equation of line 1 2
: 2 3 ,3 6
m μ μ−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
r
Let F be the foot of perpendicular of A to p.
( )
1 2 22 3 3 73 6 6
2 1 2 3(2 3 ) 6(3 6 ) 717
μμμ
μ μ μ
μ
− +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ • = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
− + + + − − = −
=
2 2 2
1 212 37
3 6
1 2 112 3 27
3 6 3
21 37
61 2 3 ( 6)71
OF
AF
AF
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠
= + + −
=
uuur
uuur
uuur
∴Distance of the line l from the plane p=1 unit
7(ii) Let the plane required be 1p .
1
3 2 48Normal of = 6 3 26
4 6 3
1 482 26 48 52 9 913 3
p−⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟× =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟• = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
1
48Equation of : 26 91 48 26 3 91
3p r x y z
−⎛ ⎞⎜ ⎟∴ • = ⇒ − + − =⎜ ⎟⎜ ⎟−⎝ ⎠
7
8(a) 13 13pq i= +
( )( )2 13 13
2 2 13 13( 2 ) ( 2) 13 13
ia b i ib i abi a i
a b i ab i
+ − = +
− + + = ++ + − = +
Comparing real and imaginary parts,
2 13 - (1)2 13 - (2)
a bab+ =− =
(2): 15ba
=
Subst. 15ba
= into (1):
( )( )
2
152 13
13 30 03 10 0
3 or 10 35 or 2
aa
a aa a
a
b
⎛ ⎞+ =⎜ ⎟⎝ ⎠
− + =
− − =
=
=
8b(i)
2 2
Least
4 1
17
z OP=
= +
=
2 2
Greatest
4 3 27
z OQ=
= + +=
8b(ii)
1
Max arg( )4tan4
4
z i PAR
π
−
− = ∠
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
(4, 3)
x
y
O 4
3
Locus of z
P (4, 1)
Q
A(0, 1)
R (4,5)
8
9(i) ( )2
f xxxλ
=−
( )( ) ( )
( )
12 2 2
'3 22 2
1 22f 0
x x x xx
x x
λ λλ
λ λ
−⎛ ⎞− − − −⎜ ⎟⎝ ⎠= = >
− −
f is strictly increasing, thus f(x) increases as x increases.
9(ii) domain of 1f − = range of f = ( ),0−∞ .
( )( )
2
2 2 2
2 2 2
2
2
1
1
xyx
y x x
x y y
yxy
λλ
λ
λ
=−
− =
+ =
= ±+
Since 0,xλ− < < that is, 0,y <21
yxy
λ=
+.
That is, 1
2f ( )
1xxx
λ− =+
, for 0.x <
9(iii)
( )1 1
2f f ( )
1 1xx
xλλ
− − =+ +
( )
( )( )
( )( )
2
2
0.50.5
1 1 0.5
4 5 04 5 1 0
λ
λ
λ λλ λ
−= −
+ + −
− − =− + =
Since 0,λ >54
λ = .
Alternatively: ( )f 0.5 0.5− = −
( )( )2
0.50.5
0.5
1 0.25
λ
λ
−=
− −
= −
54
λ =
9
10(i)
( )
2
2 2
2 1
1 1 4
1 8 1 8 1 4 1 41 1ln 1 4 tan (2 )8 2
x dxxx dx dxx x
x x c−
−+
= −+ +
= + − +
∫
∫ ∫
(ii) From the graphs, 2.82x < − or 0x > .
For 2.82x < − or 0x > , 3 3xe x> + i.e.3 3 0xe x− − > if 2.82x < − or 0x > For 2.82 0x− < < , 3 3xe x< + i.e.3 3 0xe x− − < if 2.82 0x− < <
22 3 3 xe x dx− − −∫
= 0 22 0 3 3 3 3 x xe x dx e x dx− − − + − −∫ ∫
= 2 2
0 22 0[3 3 ] [3 3 ]
2 2x xx xe x e x−− − + − −
= 2 21 3 3 11e e−− − + − (need to check positive or negative within each modulus to
remove the modulus sign) = 2 21 3 3 11e e−+ + − = 2 23 3 10e e− + −
x
y
O
3y x= +
3 xy e=
-2.82
3
10
11a(i) x-coordinate of A = 1
(ii) Area of region bounded = ∫e
dxx1
2)(ln
Let x = eu ⇒ ln x = u, uedudx
=
When x = 1, u = 0 & When x = e, u = 1
∴ Area of region bounded = ∫1
02 dueu u
Area of region bounded = ∫1
02 dueu u
= 1 1200 2[ ] uu ue duu e − ∫
= 1 1
00[ 2 2 ]u ue ue e du⎡ ⎤− −⎣ ⎦ ∫
= e – 2e +1
02 ue⎡ ⎤⎣ ⎦
= 222 −=−+− eee 11(b)
Let P be the point of intersection of 3 22
y x= − + and cosy x= .
Using GC, the coordinates of point of intersection = (0.94031, 0.58954) Volume formed about the y –axis
= ∫ −−⎥⎦⎤
⎢⎣⎡ −
1
58954.0
212
)(cos)2(32 dyyyπ
= π (0.0907) = 0.285 (3 d.p)
12(i) Horizontal asymptote is 0y = . Discriminant of ( )2 20 4 0 4 0x kx k k k k k− + = ⇒ − = ⇒ − = . Therefore when 0k = or 4, C has two asymptotes.
(ii)(a)
y
x 2 25k
k−
O
11
(ii)(b)
(c)
y
O 2 5k
k−
k
2 5k
215 25 5
k ky x= + −
1 k− O
x
1
2010 JJC Prelim P2 Solutions 1(i)
Let the length and the width of the rectangle be 2x and y The perimeter S = 2y + 4x ……(1) 2 2 2y a x= − …………………(2) Subst (2) into (1) 2 22 4S a x x= − +
2 2
2 4 0dS xdx a x
= − + =−
2 2
2 4 0xa x
− + =−
2 2
2xa x
=−
2 245
x a=
2 21 y5
a⇒ =
⇒ 4 1 and 5 5
x a y a= =
32
2 2
2 2 2
2 0( )
d s adx a x
= − <−
Length : width = 2x : y = 4:1
a
x
y
2
1(ii) 2 2 2 3y xy x
x+ = − + ……(1)
Differentiate w.r.t x
2
22 2dy dyy x y xdx dx x
+ + = +
2
2(2 ) 2dy y x x ydx x
+ = + − …….(2)
Subst. x = 1 into (1): 2 2 y y+ = ⇒ y= 1 and y ≠ −2 (rejected)
(2): 1dydx
=
1
.
1(1)( )5
1 units 5
dy dy dxdt dx dt
s−
=
=
=
2
2 3
2 3
2 2 3
2
2
2( ) 23
2( ) 23
23
2 2 2
( ) (shown)
x y xy y
x y xy y
y x y
dy dyy x ydx dx
dyy y xdx
+ = −
+ − = −
+ = −
+ = −
+ = −
Alternative: 2
2
2( ) 1 2 2 2
( ) (shown)
dy dy dyx y x y ydx dx dx
dyy y xdx
⎛ ⎞+ + = + −⎜ ⎟⎝ ⎠
+ = −
3
2
( )
2
2
2
2
2
d d
use d dd d
dd
dd1 d 1
d1 d 1 d
1 , where c is an arbitrary constant
1
yx y yxy ux
y u x ux x
ux x u ux uxx
u x u u x ux
uu x
u xu
x cux x cy
x cyx c x c
= +
=
= +
⎛ ⎞∴ + = +⎜ ⎟⎝ ⎠
+ = +
=
=
− = +
− = +
= − = − ++ +
∫ ∫
2 Family of solution curves:
11, 11
0, 111, 1
1
c yx
c y
c yx
= − = − −−
= = −
= = − ++
y
1y = −
x
0c =
1x = − 1x =
1c =
1c = −
4
3(i) Using ratio theorem, 2
33
20 2
1 3 2 02
2 2
1341
OA OBOM
OM OBOA
α
+=
−=
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟= −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥−⎝ ⎠ ⎝ ⎠⎣ ⎦
−⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
= −
uuur uuuruuuur
uuuur uuuruuur
3(ii) 4 20
1 2
2
1
BC β
β
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
uuur
Length of projection of BCuuur
onto the line OM = 32
2 2
2
32
2 01
1 1 321 1
1 3
2 8 02 or 4 (rejected)
2
BC OM
OM
β
β
β ββ
β
•=
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟•⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ =
++ =
+ − == −∴ =
uuur uuuur
uuuur
Alternatively, 1 3β + = 1 3 or 1 3 β β+ = + = − 2 or 4 (rejected)
2β
β= −∴ =
1 2 A M C
5
3(iii) 2 1 1
0 3 3 12 4 2
4 1 52 3 11 4 5
1 5Normal of plane 1 1
2 5
3 5
4
AB
AC
ABC
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − × −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎛⎜= −⎜⎜⎝
uuur
uuur
( ) ( )
2
2 2
2
3 30 5 0 5 sin
61 4 1 4
1 3 4 1 502
6 8 50 1
7 48 7 01 7 or - (rejected since angle is acute, 3 +4>0)7
7
a a
a a
a a
a a
a a
a
π
⎞⎟⎟⎟⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟• − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞+ = + ⎜ ⎟⎝ ⎠
+ = +
− − =
=
∴ =
6
4(i) 2 2 2z z= ⇒ =
( ) ( )arg arg arg( ) 4 4
arg( ) 4 234
iz i z
z
π π
π π
π
− = ⇒ − + =
⎛ ⎞= − −⎜ ⎟⎝ ⎠
=
2 2
2 2
2
wz
w z
w
=
=
∴ =
2 5arg
6
52arg( ) arg( )63 5arg( ) 24 6
73
(principle value)3
2 cos sin3 3
1 3
zw
z w
w
w i
i
π
π
π π
π
π
π π
⎛ ⎞= −⎜ ⎟⎜ ⎟
⎝ ⎠
− = −
⎛ ⎞= +⎜ ⎟⎝ ⎠
=
=
⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
= +
7
4(ii)
( )
4
4
2 23
16 24
1 1 364
2 51 1 1 16 3 3 64 4 4 4
1 3 0
1 3
2
2
2 , 0, 1, 2
2 ,2 ,2 ,2
k i
k i
k i
i i i i
z i
z i
e
z e
e k
e e e e
π π
π π
π
π π π π
⎛ ⎞+⎜ ⎟⎝ ⎠
⎛ ⎞+⎜ ⎟⎝ ⎠
+
− −
+ − =
= − +
=
=
= = ± −
=
The points form a square since the diagonals are perpendicular and of equal length.
x
y
z2
z1
z3
3π
− 6π
z4
8
5(i) The different letters left are D S T R B U O N
No of different code words 81
4! 323!
C= =
5(ii) To permute the remaining 9 letters 1st, 9!
2 ways.
To slot in the 3Is, 103C ways.
Thus number of code words that can be formed is
= 9!2
103C
= 21772800 6(i)
P(1 left-handed player) = 2 6
1 38
4
47
C CC
=
6(ii) R L R R R R L RP(G G B B )+P(G G B B )P(2 G given exactly 1 left-handed)=P(exactly 1 left-handed)
2 4 2 41 2 2 1
84
C C + C CC= 47
8
235= =4 57
6(iii) P(Boy A or Girl B is chosen) = 1 – P (Both Boy A and Girl B are not chosen)
= 6
48
4
11114
CC
− =
Alternative: P(Boy A or Girl B is chosen) = P(Boy A is chosen) + P(Girl B is chosen) −P(Boy A and Girl B are chosen)
= 7 7 6
3 3 28 8 8
4 4 4
C C CC C C
+ −
= 1 1 3 112 2 14 14+ − =
9
7 Let X denote the number of guitars sold by a music shop in a day.
( )0X P λ
Given that 2( 0)7
P X = =
27
e λ−⇒ =
2ln 1.2537
λ⇒ = − =
7(i) ( 4) ( 3)P X P X< = ≤ 0.96145 0.961= ≈
7(ii) Let Y be the number of days out of 100 in which at least 4 guitars were sold per day.(100,1 0.96145) (100,0.03855)Y B B− =
Since n is large and 5np < , we use a Poisson approximation. 0 (3.855) approxY P P(more than 95 days in which less than 4 guitars were sold per day) = P( at most 4 days in which at least 4 guitars were sold per day ) = ( 4)P Y ≤ = 0.657 (to 3 sf)
7(iii) ( )0 1.253X P For a large sample of size 90, by Central Limit theorem,
1.2531.253, approx90
X N ⎛ ⎞⎜ ⎟⎝ ⎠
.
( )1.5
0.0182
P X >
=.
Alternative : ( ) ( )1 90 0 01.253 90 112.77X X P P+ + × =K
( ) ( )1 901.5 135P X P X X> = + + >K
( )1 901 135P X X= − + + ≤K 0.0183=
8(i) (15,5)X N
Let T = 1 2X X+
1 22
1 2
( 3 ) ( ) 3 ( ) 15 15 3(15) 15
( 3 ) ( ) 9 ( ) 5 5 3 (5) 55
E T X E X X E X
Var T X Var X X Var X
− = + − = + − = −
− = + + = + + =
3 ( 15,55)T X N− − ( 2 3 ) ( 3 2)P T X P T X> + = − > = 0.0109
10
8(ii) ( 20) 0.9873 0.987P X < = ≈
Probability = [ ]23 ( 20) ( 20)P X P X< >
23(0.9873) (1 0.9873)= − = 0.0371 Alternatively, use binomial distribution. Let W be r.v. “no. of observations with value less than 20 out of 3”
(3,0.987)W B ( 2) 0.0371P W = =
8(iii) 2( , 22.5 )Y N μ
For greatest probability, 15.1 29.9 45 22.52 2
μ += = =
Greatest (15.1 29.9) 0.258P Y< < =
8(iv) (iv) (15,5)X N 2(10, )Y N σ 2 2(15 10, 5 ) (25, 5 )X Y N Nσ σ+ + + = + ( 27) 0.25P X Y+ > =
2
27 25 0.255
P Zσ
⎛ ⎞−> =⎜ ⎟
+⎝ ⎠
2
2
2 0.255
2 0.755
P Z
P Z
σ
σ
⎛ ⎞> =⎜ ⎟
+⎝ ⎠⎛ ⎞
< =⎜ ⎟+⎝ ⎠
2
2 0.67455 σ
=+
1.95σ = Assumption : The random variables X and Y are independent of each other.
11
9(i) To obtain a quota sampling of 60, divide the diners into two subgroups : male and
female. Select the first 30 males and 30 females who leaves the restaurant. Or any other relevant answers.
9(ii) By drawing random samples according to the proportion in each stratum, lunch and dinner, the sample will be a better representation of the population. Draw random samples from each stratum with sample size proportional to the size of the strata as follows :
Lunch Dinner Number of customers to be sampled
2 60 245× = 3 60 36
5× =
10(i) ey bax= ln lny a b x= + The scatter diagram is plotted with y against ln x . y
From the scatter diagram, the points lie close to a straight line, so the linear model is appropriate.
10(ii) From GC, since 0.982r ≈ − which is very close to -1, it supports the claim in part (i).
10(iii) 73.3 18.4 lny x= − 316.89 10 (3sf )a = × 18.4b = − (3sf)
10(iv) 0.5, 73.3 18.4 ln(0.5) 86.0x y= = − = 86.0% (3sf) Since 0.5x = is out of the given data range of 1 40x≤ ≤ , the prediction is unreliable.
10(v) For large values of x , the model gives 0y < . So the model is not valid for large values of x .
ln x
12
11 Assume that the amount spent per customer follows a normal distribution. Let X be the actual amount spent per customer per visit. and μ be the actual mean amount spent per customer per visit. 0H : 59μ = 1H : 59μ <
Under 0H , 28~ N(59, )
8X and test statistic XZ
n
μσ−
= ~ N(0, 1)
where 432 54, 59, 8, 8.8
x nμ σ= = = = =
At 5% level of significance, we use a left-tailed z- test and reject H0 if p-value < 0.05. From the GC, p value 0.0385− = . Since p value 0.0385 0.05− = < , we reject 0H and conclude that at 5% level of significance there is sufficient evidence to suggest that the mean amount spent per customer per visit has decreased in recent months.
11 0 0H :μ μ=
1 0H : μ μ≠ ( 70) 70 62
9yy −∑= + =
( )22 721 1234 82.25
9 1 9s
⎡ ⎤−= − =⎢ ⎥
− ⎢ ⎥⎣ ⎦
Under 0H , test statistic 0YT sn
μ−= ~ t(8)
where 062, , 82.25, 9y s nμ μ= = = = At 5% level of significance, we use a 2-tailed t- test and reject H0 if 2.306T ≥ Since 0H is rejected at 5% level of significance,
0 2.306ysn
μ−< − or 0 2.306y
sn
μ−>
082.2562 2.306
9μ > + or 0
82.2562 2.3069
μ < −
{ }0 0 0: 55.0 or >69.0 μ μ μ∈ <
Class Adm No Candidate Name:
This question paper consists of 6 printed pages.
[Turn over
2010 Preliminary Examination II
Pre-university 3
MATHEMATICS 9740
PAPER 1 9740/1
Thursday 16 September 3 hours
Additional materials:
Cover page Answer papers List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST
Write your name, admission no. and class in the spaces at the top of this page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
2
1 (i) Express ( ) 2
12
f rr r
=−
in partial fractions. [2]
(ii) Hence find 23
22
n
r r r= −∑ in terms of n. [3]
(There is no need to express your answer as a single algebraic fraction.)
(iii) Deduce the value of 23
22r r r
∞
= −∑ . [2]
2 Functions f and g are defined by
2
: 10, for ,: 3 , for .
xf x e xg x x x x
→ + ∈ℜ
→ − ∈ℜ
(i) Sketch the graphs of f and g on the same diagram, indicating clearly the equations of any asymptotes and the coordinates of any turning points. Hence, or otherwise, solve ( ) ( )g x f x= . [4]
(ii) Show that the composite function gf exists and find an expression for ( )gf x . [2]
(iii) If the domain of g is restricted to the set { }:x x a∈ℜ ≥ , find the least value of a for
which 1g − exists. Hence, find 1g − and state its domain. [4]
3 The fourth, ninth and nineteenth term of an arithmetic progression are consecutive terms of a geometric progression.
(i) Show that the common ratio of the geometric progression is 2. [3]
(ii) The twentieth term of the arithmetic progression is 63. Find its first term and common difference. [3]
(iii) The sum of the first n terms of the arithmetic progression is denoted by nS . Using
the results in (ii), find the least value of n for which nS exceeds 200. [3]
3
[Turn over
4 The curve C has equation
( )22 2 21 1a x b y+ − =
where a and b are positive constants.
Given that the curve passes through the point 3 ,02
−
and the equations of its
asymptotes are 2 2y x= + and 2 2y x= − − , show that 2a = and 1b = . [4]
Hence sketch C, stating the equations of any asymptotes and the coordinates of any points of intersection with the axes. [3]
5 The equation of a curve C is 3 32 3x xy y p− + = , where p is a constant. Find dydx
. [2]
It is given that C has a tangent which is parallel to the y-axis. Show that the y-coordinate of the point of contact of the tangent with C must satisfy
6 32 2 0y y p− − = .
Hence show that 12
p ≥ − . [3]
Find the values of p in the case where the line 4x = is a tangent to C. [3]
It is given instead that C has a tangent which is parallel to the x-axis. Show that 12
p ≥ −
in this case also. [2]
6 A disease is spreading through a population of N individuals. It is given that the rate of increase of the number of infected individuals at any time is proportional to the product of the number of infected individuals and the number of uninfected individuals at that time. At any time t, x is the number of infected individuals.
Given that initially only one person is infected, show that 1
N k t
N k tN ex
N e=
− +, where k is a
positive constant. [7]
4
7 (i) Given that ( ) ( )ln 2 1f x x= + , find ( ) ( ) ( ) ( )0 , 0 , 0 and 0f f f f′ ′′ ′′′ . Hence obtain the
first three non-zero terms in the Maclaurin’s series for ( )f x . [5]
(ii) Hence, or otherwise, show that the first three non-zero terms in the expansion of
( )2ln 2 3 1x x+ + are 2 3ax bx ax+ + , where a and b are constants to be found. [3]
8 The diagram below shows the graph of ( )y f x= with a vertical asymptote 12
x = − . The
points ( )2,0A − and ( )0,2B are the point of inflexion and the minimum point
respectively.
Sketch, on separate diagrams, the graphs of
(i) 12xy f = +
, [3]
(ii) ( )1y
f x= . [3]
O
y = f (x)
12
x = −
y
A B
x
5
[Turn over
9 Given that the plane π : d⋅ =r n and the line l : λr = a + b intersect at a point,
show that dλ − ⋅=
⋅a n
b n. [2]
Find
(i) the value of λ when l lies in π, [2]
(ii) the position vector of the point of intersection of l and π. [2]
10 (a) Given that the equation 4 3 29 29 60 0z z z z− − + − = has a root of the form 1 iz k= + , where k is a non-zero real number, find the possible values of k.
Hence solve the equation 4 3 29 29 60 0z z z z− − + − = . [5]
(b) In an Argand diagram, the point P represents the complex number z such that
2 2 2z i− − ≤ and ( ) 3arg 2 24 4
z iπ π≤ − − ≤ .
(i) Sketch the locus of P. [3]
(ii) Hence, or otherwise, show that ( ) 1 2 2arg tan4 2 2
zπ − +≤ ≤ −
. [3]
11 (a) Use the substitution cosx θ= to find the exact value of
2 /2
sin 1 cos
dπ
π
θ θθ+∫ . [5]
(b) (i) Find sin 2 xe x dx∫ . [4]
(ii) Hence find the exact value of /2
05 sin 2 xe x dx
π∫ . [2]
6
12
The diagram shows the curve C with parametric equations
1 12 , 2 , 0x t y t tt t
= + = − > .
(i) Find gradient of the curve at the point where t = 1. [3]
(ii) Show that the cartesian equation of C is 2 2 8x y− = . [2]
Two points, P and Q, lie on the curve C with coordinates ( )3,1 and ( )8,0
respectively. Point R ( )4,0 lie on the x-axis. The region S is bounded by the lines QR
and PR and the arc PQ of the curve C.
(iii) Find the exact value of the volume of revolution when S is rotated completely about the x–axis. [3]
End of Paper
y
O Q ( 8 , 0) R (4, 0)
P (3, 1) C
x
S
1
Class Adm No Candidate Name:
2010 Preliminary Examination II
Pre-University 3
MATHEMATICS Higher 2 9740/02 Friday 17 September 2010 3 hours Additional materials: Cover Page, Writing paper List of Formulae (MF 15) READ THESE INSTRUCTIONS FIRST Write your name, admission number and class in the spaces at the top of this page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This question paper consists of 6 printed pages.
[Turn over]
2
Section A: Pure Mathematics [40 marks]
1 The rth term of a sequence is given by ur = r(3r + 1), r = 1, 2, 3, ….
(i) Write down the values of ∑=
n
rru
1
for n = 1, 2, 3, and 4. [2]
(ii) Make a conjecture for a formula for∑=
n
rru
1
, giving your answer in the form
nf(n), where f(n) is a function of n. [1]
(iii) Use the method of mathematical induction to prove your conjecture for∑=
n
rru
1
.
[4] 2 (i) Solve the equation 5 32z = , expressing the solutions in the form θire , where 0>r and πθπ ≤<− . [3]
(ii) Show all the solutions on an Argand diagram. [2]
(iii) Show that the roots of the equation 5
5 32 1 02ww − − =
are
1 cot5
ki π− where k = 0, 1, 2± ± . [4]
3 (a) The points M and N have position vectors 963
and12pq
respectively, where p
and q are constants. (i) The straight line which passes through M and N has equation
9 16 3 where 3 0
Rλ λ = + ∈
r . Find the values of p and q. [3]
(ii) The line MN is extended to point S such that N divides MS in the ratio 3:2.
Find the position vector of S. [2]
[Turn over]
3
(b) The planes 1p and 2p have equations 1
3 7a
− =
r and 2
1 41
− =
r respectively,
where a is a constant. Both planes pass through the point (1, ,3)b and meet in the line l where b is a constant.
(i) Find the values of a and b. [3] (ii) Hence find the vector equation of line, l. [3]
4 (a) A conical water tank with its axis vertical and downwards has a radius of
r m at the top and is 52
r m high (see diagram). If water flows into the tank
at a rate of 25 3 1m s− , find the rate at which the depth of water is increasing
when the water is 15 m high. [3]
(b) A chocolate maker is interested in using containers in the shape shown below to package her chocolates. The container is made up of an open cylinder of height h cm and radius r cm, with a hollow hemispherical cover of radius r cm. In order to minimise production cost in this economic recession, the chocolate maker wants to use containers with the least surface area while maintaining the volume of each container at 500 cm3. The material used to construct the container costs $0.02 per cm2
.
(i) Show that the surface area, S is given by 21000 53
S rr
π= + . [4]
(ii) Find the radius that gives the minimum surface area. [4] (iii) Find the minimum cost of one container, leaving your answer to 2
decimal places. [2]
[Volume of sphere, 3
34 rV π= ; Surface area of sphere, 24 rS π= ]
r
h
52
r
r
4
Section B: Statistics [60 marks]
5 The Millennia Institute Library is planning to organize a campaign to promote reading among students. The librarian conducted a survey on 50 randomly chosen students to collect ideas for the campaign. However, she realized that most of these students happen to be from a certain level.
(i) Advise the librarian on a more appropriate sampling method. [3]
(ii) Give an advantage and disadvantage of the sampling method selected in part (i).
[2]
6 Two players M and N regularly play each other in chess. When M has the first move
in a game, the probability of M winning the game is 0.4 and the probability of N winning the game is 0.2. When N has the first move in a game, the probability of N winning the game is 0.3 and the probability of M winning the game is 0.2. Any game that is not won by either player ends in a draw.
(a) To decide who has the first move in a game, a fair coin is tossed. The player
who gets heads will make the first move. Find the probability that the game ends in a draw. [2]
(b) Find the probability that N had made the first move given that the game ends in a draw. [3]
(c) To make the game more interesting, M and N change the procedure for deciding who has the first move in the game. As a result of the new procedure, the probability of M making the first move in a game is p. Find the value of p such that M and N have an equal chance of winning the game. [2]
7 Two families are invited to a party. The first family consists of a man and both his
parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if (i) there is no restriction, [1] (ii) the men and women are seated alternately, [2] (iii) members of the same family are seated together and the two other women must
be seated separately, [3] (iv) members of the same family are seated together and the seats are numbered.
[2]
5
8 In a particular study, a medical student recorded the age and blood pressure of 8 men. The regression line of y on x is determined to be 1.144 79.914y x= + .
Unfortunately, the student then lost the record of one of the men. He urgently needs the missing data, b in his report.
(i) Show that b is 130. [3] (ii) Draw a scatter diagram for the data in the table. [2]
(iii) Find the linear product moment correlation coefficient between y and x. [1]
(iv) Explain why a linear product moment correlation coefficient that is close to 1
alone is not sufficient to conclude a strong linear relationship between two variables. [1]
(v) Use the given regression line to estimate the age of a man when his blood
pressure is 150, giving your answer to the nearest integer. [1]
(vi) Explain whether this choice of the regression line is appropriate [1]
9 Durians and mangoes are sold by mass. The masses, in kg, of durians and mangoes are modelled as having independent normal distributions with means and standard deviations as shown in the table.
Mean Standard Deviation Durians 1.6 0.2 Mangoes 0.3 0.05
Durians are sold at $8 per kg and mangoes at $3 per kg.
(i) Find the mass, m that will be exceeded by 80% of the durians. [1]
(ii) Find the probability that the total mass of 3 randomly chosen durians and 4 randomly chosen mangoes exceeds 6.5 kg. [3]
(iii) The mean mass of n randomly chosen durians is D kg. Given that ( 1.45) 0.0122P D < = , find the value of n. [3]
(iv) Find the probability that the total selling price of 3 randomly chosen durians and 4 randomly chosen mangoes is less than $45. [3]
Age (x) 49 58 44 75 37 57 63 66 Blood Pressure (y) 133 148 b 166 123 147 154 152
6
10 There is an outbreak of an infection caused by a new strain of a virus in City X. The probability, p, of a randomly chosen person being infected is 0.2. Find the probability that, in a random sample of 12 people chosen from City X, at least 4 are infected. [3]
After a certain time, the virus mutates and the value of p increases to 0.6. By using a suitable approximation, find the probability that, in a random sample of 100 people, there are at least 60 but not more than 80 who are infected. [4]
The virus then mutates to a much more deadly form and the value of p is now 0.97.
By using a suitable approximation, find the probability that, in a random sample of
100 people, at most 5 are not infected. [3]
11 A beverage producer claims that each packet of soya bean milk he produces contains 250 ml of the drink. A consumer group took a sample of 50 packets and recorded the volume, x in ml of each packet. The results are summarized by
12349x =∑ , 2 3054283x =∑ .
(i) Find the unbiased estimates of the population mean and variance. [2] (ii) Test, at 5% significance level, whether the producer is overstating the mean
volume. [4] (iii) In conducting the test in part (ii), explain if there is a need to assume that the
volume of a packet of soya bean milk follows a normal distribution. [1] (iv) In conducting the test in part (ii), explain the meaning of ‘a significance level of
5%’ in the context of the question. [1] (v) Given that the population variance of the volumes of packets of soya bean milk is
85, find the range of values of the sample size for which the producer’s claim will not be rejected at a significance level of 5%. [3]
End of Paper
1
2010 MI PU3 Prelim Exam II 9740/02 H2 Mathematics Paper 1 Suggested Solutions
1 (i) ( ) ( )2
1 12 2 2
A Bf rr r r r r r
= = = +− − −
By cover-up rule, 1 1, 2 2
A B= − =
( ) ( )1 12 2 2
f rr r
∴ = − +−
(ii)
( )23 3 3
2 1 1 12 22 2 2
1 13
1 12 41 13 51 14 6...
1 14 2
1 13 1
1 12
3 1 12 1
n n n
r r rf r
r r r r
n n
n n
n n
n n
= = =
⎡ ⎤= × = × −⎢ ⎥− −⎣ ⎦
= −
+ −
+ −
+ −
+
+ −− −
+ −− −
+ −−
= − −−
∑ ∑ ∑
(iii) 2
3
2 3 1 1lim2 2 1
32
nr r r n n
∞
→∞=
⎡ ⎤= − −⎢ ⎥− −⎣ ⎦
=
∑
2
(i)
For ( ) ( )g x f x= , by GC, 2.02x ≈ −
(ii) Rf = (10, ∞) ⊆ Dg = ℜ gf∴ exists. (shown)
( ) ( ) ( )210 3 10x xgf x e e= + − +
(3/2, -9/4)
3 0
11 A
( ) 10xf x e= +
y = 10
x
y 2( ) 3g x x x= −
2
(iii) From the graph of g above, 1g − exits if 3 2
x ≥ .
Hence, the least value of 32
a = .
Let 23 9( )
2 4y g x x⎛ ⎞= = − −⎜ ⎟
⎝ ⎠
3 92 4
x y∴ = + + , since 32
x ≥
1 3 9( )2 4
g x x−∴ = + + , 1
9 ,4g
D −
⎡ ⎞= − ∞⎟⎢⎣ ⎠
3 (i) Let the AP with T1 = a, common difference = d.
T4, T9, T19 are consecutive terms of a GP :
19 9
9 4
18 88 3
T T a d a dr rT T a d a d
+ +⇒ = = ⇒ = =
+ +
( )( ) ( )2 218 3 8 10 5a d a d a d d ad⇒ + + = + ⇒ =
Since 5 10, ---------(1)10 2
ad d a≠ = =
188 2 2 (shown)
13 32
a aa dra d a a
⎛ ⎞+ ⎜ ⎟+ ⎝ ⎠∴ = = =+ ⎛ ⎞+ ⎜ ⎟
⎝ ⎠
(ii) Given 20 19 63 ---------(2)T a d= + =
Substitute (1) into (2), 119 632
a a⎛ ⎞+ =⎜ ⎟⎝ ⎠
6 & 3 a d∴ = =
(iii) [ ]2(6) ( 1)3 200
2nnS n= + − >
23 9 400 0n n+ − >
13.14n < − (NA) or 10.14n >
Hence, least 11n =
3
4 ( ) ( )2 2
22 2 22 2
11 1 1
1 1x ya x b y
a b
++ − = ⇒ − =
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Asymptotes : ( )1ay xb
= ± +
Comparing with 2 2y x= + and 2 2y x= − − ,
2 2a a bb= ⇒ = ------(1)
And the curve passes through the point 3 , 02
⎛ ⎞−⎜ ⎟⎝ ⎠
,
2
2 3 1 0 12
a ⎛ ⎞− + − =⎜ ⎟⎝ ⎠
2a⇒ = (shown), since a is positive.
Substitute into (1), 1b = (shown)
5 3 3
2 2
2 3 -------(1)
6 3 3 3 0
x xy y pdy dyx x y ydx dx
− + =
⇒ − − + =
2
2
2dy y xdx y x
−⇒ =
−
If tangent is parallel to the y-axis, 2y x= .
Substitute into (1), ( ) ( )32 2 32 3 0y y y y p− + − =
6 32 2 0y y p⇒ − − = (shown) For the point of contact of the tangent with C,
y = 2x + 2
y = − 2x −2
(−1.5, 0) (−0.5, 0)
(0, −√3)
(0, √3)
y
x
4
( ) ( )( )22 4 2 4 2 0b ac p− = − − − ≥
12
p∴ ≥ − (shown)
When 4x = is a tangent to C, 2 4 2y y= ⇒ = ± When y = 2, 112p = .
When y = −2, 144p =
If the tangent is parallel to the x-axis, 22y x= . Substitute into (1),
( ) ( )33 2 2 6 32 3 2 2 0 8 4 0x x x x p x x p− + − = ⇒ − − =
For the point of contact of the tangent with C,
( ) ( )( )22 4 4 4 8 0b ac p− = − − − ≥
12
p∴ ≥ − (shown)
6 ( )dx kx N x
dt= − , k is a positive constant.
1( )
dx kdtx N x
=−∫ ∫
1 1 1 dx kt CN x N x
⎡ ⎤+ = +⎢ ⎥−⎣ ⎦∫
ln ln( )x N x Nkt NC− − = +
ln x Nkt NCN x
⎛ ⎞ = +⎜ ⎟−⎝ ⎠
, Nkt NCx Ae A eN x
= =−
When t = 0, x = 1, 1
1A
N=
−
11
Nktx eN x N
=− −
( )1 Nkt Nktx N e Ne⇒ − + =
∴1
Nkt
Nkt
NexN e
=− + (shown)
5
7 (i) ( ) ( )ln 2 1f x x= +
( )
( ) ( )( ) ( )
1
2 2
3 3
2( ) 2 2 12 1
( ) 2 2 1 (2) 4 2 1
( ) 4( 2) 2 1 (2) 16 2 1
f x xx
f x x x
f x x x
−
− −
− −
⎫′ = = + ⎪+ ⎪⎪′′ = − + = − + ⎬⎪
′′′ = − − + = + ⎪⎪⎭
(0) 0(0) 2(0) 4(0) 16
ffff
= ⎫⎪′ = ⎪⎬′′ = − ⎪⎪′′′ = ⎭
2 3
2 3
2 3
1 1( ) (0) (0) (0) (0) ...2 3!
1 10 2 ( 4) (16) ...2 6
8( ) 2 2 ...3
f x f xf x f x f
x x x
f x x x x
′ ′′ ′′′= + + + +
= + + − + +
∴ = − + +
(ii)
( )2
2 3 2 3
2 3
ln 2 3 1 ln(2 1)( 1)
ln(2 1) ln( 1)8 1 12 2 ...3 2 3
53 3 ...2
x x x x
x x
x x x x x x
x x x
+ + = + +
= + + +
= − + + − + +
= − + +
53, 2
a b⇒ = = −
8
(i)
( ) ( ) ( )( ) ( ) ( )
2,0 4,0 4,1
0, 2 0,2 0,31Asymptote : 12
A A
B B
x x
⎫′ ⎪− → − → −
⎪′→ → ⎬
⎪⎪= − → = −⎭
6
(ii)
( )
( )
2,0 asymptote at 21 1Asymptote at intercept at 2 2
10, 2 , min. pt. 0, , max. pt.2
"New" asymptote : y = 0
A x
x x x
B
⎫− → = −⎪⎪= − → − = − ⎪⎬
⎛ ⎞ ⎪→ ⎜ ⎟ ⎪⎝ ⎠⎪⎭
12xy f ⎛ ⎞= +⎜ ⎟
⎝ ⎠
1x = −
y
B′ A′
x
x = - 2
0 (- ½, 0)
(0, ½)
y
x
7
9 d⋅ =λ
r n ------ (1)r = a + b ------(2)
Substitute (2) into (1),
( ) d
dλ
λ
⋅
⇒ ⋅ ⋅
a + b n = a n + b n =
dλ − ⋅
∴ =⋅a n
b n (Shown)
(i) If l is in π , a is on π , d⋅ =a n . 0b . n =
Hence, λ has infinitely many solutions.
(ii) When l and π intersect at one point, a n
b ndλ − ⋅
=⋅
, from above.
Substitute this into the equation of the line l, the position vector
required is a nr a b
b nd − ⋅⎛ ⎞= + ⎜ ⎟⋅⎝ ⎠
10 (a) Given that 1 iz k= + is a root, so substitute into the given equation ( ) ( ) ( ) ( )4 3 21 i 1 i 9 1 i 29 1 i 60 0k k k k+ − + − + + + − =
( ) ( )2 3 4 2 3 21 4 i 6 4 i 1 3 i 3 i 9 18 i 9 29 29 i 60 0k k k k k k k k k k+ − − + − + − − − + − + + − =
Comparing the real or imaginary parts on both sides,
3 3 34 4 3 18 29 3 12 0 2 or 0 (N.A.)
k k k k k k k kk k
− − + − + = − + =⇒ = ± =
OR, 2 4 2 2 4 26 3 9 9 31 6 40 0
By GC, 2k k k k k k
k− + + − + − = + − =
⇒ = ±
Hence, ( )( ) ( )( ) 21 2i 1 2i 2 5z z z z− − − + = − + is a factor of the given equation
( )( )4 3 2 2 29 29 60 2 5 12 0z z z z z z z z− − + − = − + + − =
( ) ( )2 22 5 0 or 12 0
1 2i , 4 or 3
z z z z
z z z
− + = + − =
∴ = ± = − =
8
(b) (i)
(ii) min. (arg z) = argument of any complex numbers along CD
1 2tan2
4π
−=
=
At A, 2 2 and 2 2x y= − = + . -----(1)
max. (arg z) = argument of a, where A ≡ a
1
1
tan -----(2)
2 2tan2 2
ABOB
−
−
=
⎛ ⎞+= ⎜ ⎟⎜ ⎟−⎝ ⎠
Hence, ( ) 1 2 2arg tan4 2 2
zπ − ⎛ ⎞+≤ ≤ ⎜ ⎟⎜ ⎟−⎝ ⎠
(Shown)
11 (a) Let c o s s i nd xx
dθ θ
θ= ⇒ = −
When , 0 .2
xπθ = = When , 1 . xθ π= = −
1
2 2 /2 0
sin 1 1 cos 1
d dxx
π
π
θ θθ
− −=
+ +∫ ∫
= 11
0t a n x
−−⎡ ⎤− ⎣ ⎦
= 4π
( ) 3arg 2 24
z i π− − =
( )arg 2 24
z i π− − =
2
20 B
D A
x
y
C αα Note : α = π/4
9
(b)(i)
[ ]
sin 2 sin 2 .2cos 2
sin 2 2 cos 2 .2sin 2
5 sin 2 sin 2 2 cos 2
1sin 2 sin 2 2cos 25
x x x
x x x
x x x
x x
e x dx e x e x dx
e x e x e x dx
e x dx e x e x C
e x dx e x x C
= −
⎡ ⎤= − +⎣ ⎦
= − +
∴ = − +
∫ ∫∫
∫
∫
(b)(ii)
( ) ( )
( )
/2 /2
00
/2
15 sin 2 5 sin 2 2cos 25
2 1
x xe x dx e x x
e
π π
π
⎡ ⎤= × −⎣ ⎦
= +
∫
12 (i) 2 21 12 , 2dy dx
dt dtt t= + = −
2
2
12
12
dy dy dx tdx dt dt
t
+= ÷ =
−
When t = 1, 2 1 32 1
d yd x
+= =
−
(ii) ( ) ( )1 12 ------ 1 ; 2 ------ 2x t y tt t
= + = −
(1) + (2), ( )4 ------ 3x y t+ =
(1) – (2), ( )2 ------ 4x yt
− =
(3) × (4), ( )( ) 8x y x y+ − = ⇒ 2 2 8x y− = (Shown)
(iii) Volume of the solid =
32
8y dxπ ∫ + volume of cone
= ( ) ( )( )3
2 2
8
18 1 13
x dxπ − + π∫
= ( )4 4 8 1 13
− π or ( )4 8 2 1 13
π− (exact value)
1
2010 MI PU3 Prelim Exam II 9740/02 H2 Mathematics Paper 2 Suggested Solutions
1
(i) 1 2 3 4
2 2 2 2
1 1 1 11 2 4, 2 3 18, 3 4 48, 4 5 100r r r r
r r r ru u u u
= = = =
= × = = × = = × = = × =∑ ∑ ∑ ∑
(ii)
2
1( 1)
n
rr
u n n=
= +∑
(iii)
Let Pn be the statement that 2
1( 1)
n
rr
u n n=
= +∑ for n Z +∈
1
2
1
Prove is true:(1)(3 1 1) 4(1)(1 1) 4
is true
PLHSRHSP
= × + =
= + =
2
1
Assume is true for some i.e.
( 1)
k
k
rr
P k Z
u k k
+
=
∈
= +∑
( )( )( )( )( )( )
12
11
12
2
2
2
1 1
Prove is true i.e. ( 1)( 2)
( 1)(3 3 1)
( 1) ( 1)(3 4)
1 3 4
1 4 4
1 2
Since is true and is true is true,
k
k rr
k
rr
k k
P u k k
LHS
u k k
k k k k
k k k k
k k k
k kRHS
P P P
+
+=
=
+
= + +
= + + + +
= + + + +
= + + + +
= + + +
= + +
=⇒
∑
∑
by the Principle of Mathematical Induction, nP is true for n Z +∈
2
2
(i) 5 (2 )
25
32 32
2 where 2, 1,0
i k
ki
z e
z e k
π
π⎛ ⎞⎜ ⎟⎝ ⎠
= =
= = ± ±
(ii) (iii)
55 32 1 0
2ww ⎛ ⎞− − =⎜ ⎟
⎝ ⎠
5
321
2
ww
⎛ ⎞⎜ ⎟
=⎜ ⎟⎜ ⎟−⎝ ⎠
25
2 25 5
25
25
25
5 5 5
21
2
2
2
1
2
ki
k ki i
ki
ki
ki
k k ki i i
w ew
w we e
ewe
ew
e e e
π
π π
π
π
π
π π π
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=−
= −
=
−
=⎛ ⎞
−⎜ ⎟⎜ ⎟⎝ ⎠
5
5
2
2 Im
ki
ki
ew
e
π
π
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
=⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
5Z
Re (z)
Im (z)
25π
25π
−
25π
25π
25π
2Z
3Z
4Z
1Z
3
cos sin5 5
sin5
cot51
1 cot (shown)5
k kiw
ki
k
wi
kw i
π π
π
π
π
⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠= +
⎛ ⎞= − ⎜ ⎟⎝ ⎠
3 (a)(i)
12 9 16 33 0
3 16 33 0
pq
pq
λ
λ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
336 (3)(3) 15
qp p
λ ==− = =
(a)(ii)
2 35
2 53 3
9 12 142 56 15 213 3
3 3 3
OM OPON
OP OM ON
OP
+=
= − +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuuur uuuruuur
uuur uuuur uuur
uuur
(b)(i)
Point (1, ,3) is on both planes:1 2
1 43 1
2 3 41
b
b
bb
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− + ==
1 11 3 73
1 3 3 73
aa
a
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− + ==
(b)(ii)
4
Line is perpendicular to the normals to both planes:
1 2 03 1 5
3 1 5
1 0 1 0: 1 1 where or 1 5 where
3 1 3 5
l
l r rα α α α
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− × − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + ∈ = + ∈⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
4 (a) 2
2 3
2
2
-1
1 1 2 43 3 5 75
425
At 15 :425= (15)25
25 0.221 ms36
V r h h h h
dV hdh
dV dV dhdt dh dt
h mdhdt
dhdt
π π π
π
π
π
⎛ ⎞= = =⎜ ⎟⎝ ⎠
=
=
=
= ≈
(b)(i)
2 3
3
2
2
2 5003
1500 23
500 23
V r h r
rhr
h rr
π π
ππ
π
= + =
−=
= −
2 2
22
2 2
2
2 2500 22 3
31000 4 3
31000 5
3
S rh r r
r r rr
r rr
rr
π π π
π ππ
π π
π
= + +
⎛ ⎞= − +⎜ ⎟⎝ ⎠
= − +
= +
5
(b)(ii)
2
2
2
1000 53
1000 103
1000 10For minimum surface area: 03
S rr
dS rdr r
dS rdr r
π
π
π
= +
= − +
= − + =
2
3
1000 103
300
4.5707814974.57 cm
rr
r
rr
π
π
=
=
=≈
2
2 3
1000 10 0 (verified 4.57 gives a minimum surface area)3
d S rdr r
π= + > =
(b)(iii)
2min
minimum
1000 5 (4.570781497) 328.17145184.570781497 3
(328.1714518)(0.02) $6.56
S
Cost
π= + =
= ≈
5
(i) Use stratified sampling. Divide the student population into different strata for example PU1, PU2 and PU3 (or any other logical strata) and select students randomly from each strata according to the strata’s proportion to the population until 50 students have been selected.
(ii) Advantage: Good representation of student population.
Disadvantage: Time-consuming
6 (a) (game ends in draw) (0.5)(0.4) (0.5)(0.5) 0.45P = + =
Win
Draw
Lose
Win
Draw
Lose
M starts first
M starts second
0.5
0.5
0.4 0.4
0.2
0.2 0.5
0.3
6
(b) ( made the 1st move game ends in draw )
( made the 1st move game ends in draw)(game ends in draw)
0.5 0.50.45
0.556
P NP N
P∩
=
×=
=
(c)
( wins game) ( loses game)0.4 0.2(1 ) 0.2 0.3(1 )
1 0.3333
P M P Mp p p p
p
=+ − = + −
= ≈
7 (i) Number of arrangements = (10 1)! 362880− =
(ii) Number of arrangements = (5 1)! 5! 2880− × = (iii) Number of arrangements = 4
2(4 1)! 3! 3! 2592P− × × × =
(iv) Number of arrangements = (6 1)! 3! 3! 10 43200− × × × =
8 (i)
449 56.1258
1.144(56.125) 79.914 144.1211023 144.1218130
xx
nyb
b
= = =
= + =+
=
≈
∑
(ii)
7
(iii) Using GC, 0.9888 0.989r = ≈ (iv) A r value that is close to 1 alone does not lead to the conclusion
that there is a strong linear relationship between 2 variables because a r value close to 1 could be a result of outliers.
(v) 150 1.144 79.914
61The age of the man is 61.
xx
= +≈ .
(vi) Age is the independent variable (can be precisely controlled) and the
estimate from the regression line of y on x is appropriate. (Also accept r is close to 1, using any of the 2 regression lines will result
in a close estimate. Thus, the estimate is appropriate).
9 2
2
Let D be the mass of a durian~ (1.6, 0.2 )
Let M be the mass of a durian~ (0.3, 0.05 )
D N
M N
(i) Let ( ) 0.8P D m> = Using GC, 1.43167 1.43m = ≈ kg. (ii)
2 2(3) (4) ~ (3 1.6 4 0.3, 3 0.2 4 0.05 ) (6, 0.13)D M N N+ × + × × + × =
(3) (4)( 6.5) 0.08275892 0.0828P D M+ > = ≈
(iii) 20.2~ 1.6,D N
n⎛ ⎞⎜ ⎟⎝ ⎠
( 1.45) 0.0122
1.45 1.6 0.01220.2
1.45 1.6 2.25077170.2
9
P D
P Z
n
nn
< =
⎛ ⎞⎜ ⎟−
< =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠−
= −
≈
8
(iii)
( )
2 2 2 2(3) (4)
(3) (4)
8 3 ~ (8 4.8 3 1.2, 8 3 0.2 3 4 0.05 )
(42, 7.77)
8 3 45 0.859
D M N
N
P D M
+ × + × × × + × ×
=
+ < ≈
10
Let be the number of people infected with the disease(12,0.2)
( 4)1 ( 3)1 0.7945680.205
XX BP X
P X≥
= − ≤= −≈
Let be the number of people infected with the disease
(100,0.6)60 5; (1 ) 40 5
(60,24) approximately
(60 80)
(59.5 80.5)0.541
cc
XX Bnp n p
X N
P X
P X
= > − = >
≤ ≤
= ≤ ≤=
Let be the number of people NOT infected with the disease
(100,0.03)3 5; 50, 0.1,
(3) approximately( 5)0.916
YY Bnp n p
Y PoP Y
= < > <
≤=
11 (i)
Unbiased estimates of the population mean , 12349 246.98 24750
XX
n= = = ≈∑
Unbiased estimates of the population variance,
9
( ) ( ) 222 1 88.30571429 88.3
1x
s x nn n
⎡ ⎤⎛ ⎞⎢ ⎥= − = ≈⎜ ⎟⎜ ⎟− ⎢ ⎥⎝ ⎠⎣ ⎦
∑∑
(ii) 0 : 250H μ = 1 : 250H μ < Use z-test, p-value = 0.01153 (z = -2.272) p-value = 0.01153 < 0.05, reject 0H At 5% significance level, the producer is overstating the mean volume. (iii) n is large (n = 50). By Central Limit Theorem, x-bar will
approximately follow a normal distribution. The z-test can still be conducted. There is no need to assume that the volume of a packet of soya bean follow a normal distribution.
(iv) The significance level of 5% means that 0.05 is the probability that
we conclude the producer has overstated the mean volume when the mean volume is actually 250 ml.
(v) 2 85σ = (given) Level of significance = 0.05. Use z-test.
Do not reject 0H if 246.98 250 1.6448536
85 /z
n−
= > −
25.21525
nn<≤
MJC/2010 JC2 Preliminary Examination/9740/01
2
1 Given that
( ) ( )1 2f ( )
! 1 ! 2 !r rrr r r−
= − +− −
,
use the method of differences to prove that
2
1f ( ) 2!
N
r
NrN=
+= −∑ . [3]
Hence, give a reason why the series is convergent and state the sum to infinity. [2]
2 (i) Find ( )2 1d ed
x
x+ . [1]
(ii) Hence, find 23 1e dxx x+∫ . [3]
(iii) Find the exact value of ( )21 3 1 2
0e e dxx x+ +∫ . [2]
3 A sequence of positive real numbers 1 2 3, , ,...x x x satisfies the recurrence relation
132
nn
n
xxx+ =
+
for 1.n ≥
(i) Given that as , ,nn x→∞ →α find the exact value ofα . [2]
(ii) Show that this sequence is
(a) strictly increasing for 0 ,nx< < α [2]
(b) strictly decreasing for nx > α . [2]
[Turn over
MJC/2010 JC2 Preliminary Examination/9740/01
3
4 (i) Express ( ) ( )( )4f
1 3 2xx
x x−
=+ +
in partial fractions.
Hence, expand ( )f x in ascending powers of x, up to and including the
term in 3x . [4]
(ii) State the range of values of x for which this expansion is valid. [1]
(iii) Find the coefficient of nx in this expansion. [2]
5 On a single Argand diagram, sketch the following loci.
(i) 5 3 7iz − = + , [1]
(ii) 6 3i 4 3iz z− − = − + . [1]
Two complex numbers that satisfy the above equations are represented by p and q,
where Re( ) Re( )p q< . By using the cartesian equations of the loci, find p and q.
Hence, determine the value of arg( )p q− . [5]
6 Newton’s Law of Cooling states that the rate at which the temperature of a body
falls is proportional to the amount by which its temperature exceeds that of its
surroundings. At time t minutes after cooling commences, the temperature of the
body is o Cθ . Assuming that the room temperature remains constant at o30 C
and the body has an initial temperature of o90 C , show that 30 60e ktθ −= + ,
where k is an arbitrary constant. [5]
Given that it takes 8 minutes for the temperature of the body to drop from o90 C
to o55 C , determine how much more time is needed for the body to cool to o35 C ,
leaving your answer to one decimal place. [3]
MJC/2010 JC2 Preliminary Examination/9740/01
4
7 (a) A geometric progression has first term a and common ratio 12
− . The first
two terms of the geometric progression are the first and fourth terms
respectively of an arithmetic progression. Find the sum of the first n even-
numbered terms of the arithmetic progression in terms of a and n. [4]
(b) A customer borrows $50000 from a bank at the beginning of a month. In
the middle of the month, the customer pays $x to the bank. On the last day
of every month, the bank adds interest at a rate of 3.5% of the remaining
amount owed after payment has been made. This repayment process is
repeated every month until the loan is repaid in full.
(i) Find, in terms of x, the amount owed at the beginning of the third
month. [1]
(ii) Show that, if the repayment of the loan is completed upon the nth
payment, then11750(1.035 )
1.035 1
n
nx−
≥−
. [4]
8 The diagram shows the graph of ( )fy x= , where ( )f x is a cubic polynomial and
C is a maximum point.
x
y
( )2, 75A −
( )1,12C
( )0, 3B
0
[Turn over
MJC/2010 JC2 Preliminary Examination/9740/01
5
It is also given that when 3y = , 0, or x α β= , where 0α β< < . On separate
diagrams, sketch the graphs of
(i) ( )f 1 2y x= − , [3]
(ii) ( )2 f 3y x= − . [3]
Determine the equation of the curve. [4]
9 A line l passes through the points A and B with coordinates ( )0, 1,2− and ( )1,0,1
respectively.
(i) Find the angle between OA and the line l. [2]
(ii) Hence, find the shortest distance from the origin to the line l. [1]
A plane 1π has equation ( ) 2 3 4.+ + =r i j k
(iii) Show that the line l lies on the plane 1.π [2]
A second plane 2π contains the line l and is perpendicular to the plane 1.π
(iv) Find a vector equation of 2.π [2]
A third plane 3π is perpendicular to both the planes 1π and 2 ,π and is at a
perpendicular distance of 3 units from the origin.
(v) Find possible vector equations of 3π . [3]
MJC/2010 JC2 Preliminary Examination/9740/01
6
10 The functions f, g and h are defined by
f : cosx x
for ,x∈
1g : ln1
xx
+
for 1 1,x− < ≤
h : x
1ln for 1 1,1
1 for 1.
xx
x
− < < + − =
(i) Show that the composite function fg exists. [2]
(ii) Find the series expansion for fg( )x up to and including the term in 3x . [4]
(iii) Write down the value of 1h ( 1)− − . Hence, find 1h ( )x− , expressing your
answer in the form
1h : x−
p( ) for ,for 1,
x x ac x
>= −
where a and c are constants, and p is a function of x. [4]
11 The complex numbers z and w are given by 3 iz = + and 1 i 3w = − + .
(a) Find the set of values of the positive integer n for which ( )* 0nnz z− = . [4]
(b) Sketch an Argand diagram, with origin O, showing the points Z, W and P
representing the complex numbers z, w and z w+ respectively. Show that
OWPZ is a square. [3]
By considering the argument of z w+ , deduce that 5tan 2 312π= + . [3]
[Turn over
MJC/2010 JC2 Preliminary Examination/9740/01
7
12 The curve C has equation 2
2
2 3 14
x xyx− +
=−
.
(i) Prove, using an algebraic method, that C cannot lie between
7 3 5 7 3 5 and 8 8
− + . [3]
(ii) Sketch the curve C, showing clearly all asymptotes, axial intercepts and
turning points. [4]
(iii) R is the point on C where 5x = − . The tangent and normal to the curve at
R cut the y-axis at P and Q respectively. Show, to 3 significant figures,
that the coordinates of P and Q are ( )0,5.15 and ( )0, 9.31− respectively.
Hence, determine the area of triangle PQR. [5]
Section A: Pure Mathematics [40 marks]
1 Without using a calculator, solve the inequality 2
7 1.4 3
xx x+
≤+ −
[4]
2 Three non-zero and non-parallel vectors p, q and r are such that 3× = ×p q p r .
Show that 3 λ− =q r p , where λ is a scalar. [2]
It is also given that p is a unit vector, 5q = , 2r = and the angle between q and r is 1 5cos6
− .
By considering ( ) ( )3 3− ⋅ −q r q r , find the exact values of λ . [4]
3 Prove by induction ( ) ( ) ( )1
1
33 2 3 2 1 62
nr n
rr n n+
=
− = − + −∑ for n +∈ . [5]
Hence, find the least value of n for ( )53 2 5800
nr
rr
=
− >∑ . [3]
4 (a) The variables x and y are related by
( )3lny x xy= +
Find the value of ddyx
when both x and y are equal to 1. [4]
(b)
Water is poured at a constant rate of 3 -11 cm s into a cone of semi-vertical angle 45
with its axis vertical and vertex downwards (see diagram). At the beginning, the cone is
partially filled with 330 cm of water. Find the rate at which the depth of water is
increasing after 2 minutes. [6]
45
[Turn Over
3
MJC/2010 JC2 Preliminary Examination/9740/02
5 (i) The region R is bounded by the ellipse 2 2
2 2 1x ya b
+ = and the line 1y xb a− = as shown in
the diagram below.
Show that the area of R is 0 2 21 d
2 a
bab a x xa−
− + −∫ . Hence, by substituting sinx a θ= ,
find the exact area of R in terms of a and b. [8]
(ii) Find the volume of revolution formed when R is rotated completely about the y-axis.
Give your answer in the form 2ka b , where k is a constant to be determined. [4]
Section B: Statistics [60 marks]
6 There are 800 students in PPQZ Secondary School where 65% of the students are boys. A
random sample of 200 students is selected to find out their preferred canteen stalls.
(a) Student A suggests selecting the sample by taking the first 120 boys and the first 80 girls who visit the canteen during recess time. Identify the sampling method used. State, in the context of the question, one disadvantage of this method used. [2]
(b) Student B suggests using systematic sampling using the following steps:
(1) Label all the students from 1 to 800. (2) Using a sampling interval of 5, he randomly selects an integer from 1 to 5 to
determine the first student to be chosen and selects every 5th student thereafter until 200 students are chosen.
Identify the mistake that student B made and write down the correct step. [2]
2 2
2 2 1x ya b
+ =
1y xb a− =
y
x − a a
b
−b
R
O
7 In a computer game played by a single player, the player has to find, within a fixed time, the path through a maze shown on the computer screen. On the first occasion that a particular player plays the game, the computer shows a simple maze, and the probability that the player
succeeds in finding the path in the time allowed is 57
. On subsequent occasions, the maze
shown depends on the result of the previous game. If the player succeeded on the previous occasion, the next maze is harder, and the probability that the player succeeds is one third of the probability of success on the previous occasion. If the player failed on the previous occasion, another simple maze is shown and the probability of the player succeeding is
again 57
.
The player plays three games. Find the probability that
(i) the player succeeds in exactly one of the games, [2]
(ii) the player succeeds in at least one of the games, [2]
(iii) the player has exactly two failures given that the player has at least one success. [2]
8 (a) Find the number of teams of 8 players that can be selected from a group of 13 players if
at least two of the three tallest players and at most one of the two shortest players are to be included. [4]
(b) Secret codes can be sent using 6 available letters X, Y, Y, Z, Z, Z. (i) How many different 5-letter secret codes can be sent? [3] (ii) How many different 5-letter secret codes will begin and end with Y? [1]
[Turn Over
5
MJC/2010 JC2 Preliminary Examination/9740/02
9 The time t minutes spent queuing at an ice-cream bar by each customer in a random sample of 50 customers was measured and it was found that 213.5t =∑ and ( )2 44.105t t− =∑ where
t is the mean time spent at the ice-cream bar by each customer in the sample.
Find unbiased estimates of the population mean and variance. [2] Test, at 1% level of significance, whether the mean time spent is less than 4.5 minutes. [4] Determine the smallest level of significance of the test where the null hypothesis is rejected. [1] State, giving a reason, whether any assumptions about the population are needed in order for the test to be valid. [1]
10 On average, a hospital and a police station receive 36 and 15 calls respectively in a three-week period. Calls are received at random times. The number of calls received by the hospital may be assumed to be independent of the number of calls received by the police station.
(i) Find the probability that the hospital and the police station receive a total of more than 11 calls in a randomly chosen week. [3]
(ii) Using a suitable approximation, find the probability that out of 50 randomly chosen weeks, the number of weeks in which the hospital and police station receive a total of more than 11 calls is greater than 45. [4]
(iii) A nurse records the number of calls received by the hospital per week for 100 randomly chosen weeks. She then calculates the average number of calls received per week based on her data. Another nurse independently goes through the same procedure. Find the probability that the sum of the two averages obtained by the nurses is at most 23. [3]
11 (a) King Crabs and Snow Crabs are sold by weight. The masses, in kg, of King Crabs and Snow Crabs are modelled as having normal distributions with means and variances as shown in the table.
Mean Mass Variance King Crabs 1.65 0.71 Snow Crabs 1.10 0.34
King Crabs and Snow Crabs are sold at $40 and $45 per kg respectively.
(i) Find the probability that the average mass of 3 King Crabs and 2 Snow Crabs is
less than 1.5 kg. [3]
(ii) Find the probability that the total selling price of a King Crab and a Snow Crab exceeds $140. [3]
(iii) State an assumption for your working in (i) and (ii) to be valid. [1]
(b) The masses of bars of chocolate are normally distributed with mean µ kg and standard deviation σ kg. It is known that 20% of the bars of chocolate have masses which differ from µ kg by at most m kg. Find the probability that a randomly chosen bar of chocolate has a mass which exceeds µ kg by at least 3m kg. [5]
[Turn Over
7
MJC/2010 JC2 Preliminary Examination/9740/02
12 The table below shows the number of monthly new car licences, x, issued by the government and the selling price of a car, $y in year 2009. A student wants to investigate how the selling price of a car depends on the number of monthly new car licences issued by the government.
(i) Plot a scatter diagram for the data and explain, in the context of the question, if a linear model is appropriate in the long run. [3]
(ii) State, with a reason, which of the following model is most appropriate to fit the data points.
(A) exy a b= +
(B)
2y ax b= +
(C) by ax
= + [2]
(iii) For the model chosen in (ii), calculate the product moment correlation and comment on its
value. [2]
(iv) Use an appropriate regression line to estimate the selling price of the car when the monthly number of new car licences issued is (a) 1300 and (b) 2000. Comment on the reliability of your answers. [4]
(v) The student concluded that the decrease in the number of new car licences issued causes the selling price of the car to rise. State, with a reason, whether you agree with this conclusion. [1]
Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec x 1432 1339 1664 1774 2055 2275 2221 2173 1749 1854 1360 2012
y 42000 48000 37000 36000 32850 31800 31800 32050 36000 34500 46000 33000
H2 MATHS (9740) JC2 PRELIMINARY EXAM 2010 PAPER 1 SUGGESTED SOLUTIONS
Qn Solution 1 Method of Differences
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )( ) ( )
2 2
2
1 2f ( )! 1 ! 2 !
1 2 22! 1! 0!2 3 23! 2! 1!3 4 24! 3! 2!4 5 25! 4! 3!
3 2 22 ! 3 ! 4 !2 1 21 ! 2 ! 3 !
1 2! 1 ! 2 !
2 12 2 21 ! 1 ! !
2 12
N N
r r
r rrr r r
N NN N NN NN N N
N NN N N
N N NN N N
N N N NN
= =
⎛ ⎞−= − +⎜ ⎟⎜ ⎟− −⎝ ⎠⎡= − +⎢⎣
+ − +
+ − +
+ − +
+− −
+ − +− − −
− −+ − +
− − −
⎤−+ − + ⎥− − ⎦
− −= − + + + − +
− −
− − + −= +
∑ ∑
M
!12
!NN+
= −
As 1, 0!
NNN+
→∞ → , hence the series is convergent.
2
f ( ) 2r
r∞
=
=∑
Qn Solution 2 Integration (by part)
(i)
(ii)
( )2 21 1d e 2 ed
x xxx
+ +=
23 1e dxx x+∫
= ( )( )2 1 21 2 e d2
xx x x+∫
(iii
Qn3 (i)
(ii)
i)
= ( 212
x
= (12
x
(1 3
0ex∫
= 12
x⎡ ⎡⎢ ⎣⎣
= (1 e2⎡⎢⎣
= 2e 0+
n SolutiRecurr
)
)
As n →
⇒α =
((
2
2
⇒α
⇒αThusα Metho
1nx + −
Sketch
From tFor 0 <Theref For nxTheref
22 1e 2 ex x+ − ∫2 22 1 1e ex xx + +−
)2 1 2e e dx x+ +
2 22 1e ex xx + +−
)2 2 2e e e ⎤− + ⎥⎦0.5e
on rence Relatio
, nx→∞ →α
32α
= ⇒+α
( )( )2 9
7 0
+α − α
α − = ⇒7α = sinceα
od 1 [Graph32
nn
n
xxx
=+
h 1n ny x x+= −
the graph: nx< < α : nx
fore the sequ
n > α : 1nx x+ −fore the sequ
α
)2 1e dx x+
) (2 1e2
x
c+
+ =
1
1 2
0
e x⎤⎤ + ⎥⎦ ⎦
( )1 e2
⎤ ⎡ ⎤− −⎢ ⎥⎦ ⎣ ⎦
on 1 & nx +α →α
22 9
2α
α =+α
2 0
0 or
=
⇒ α = α0.>
hical]
nn
x−
32
nn
n
xx
= −+
1 0n nx+ − > ⇒uence is stric
0n nx x +< ⇒uence is stric
α
( )2 1x c− +
α
7.α =
nx−
1n nx x+⇒ > ctly increasin
1 nx+ < ctly decreasi
ng.
ng. [Shownn]
Qn4
(i)
nx
1nx +
1nx +
nx
Metho
Sketch
From tFor 0 <Theref For nxTheref Metho
1nx + −
For 0 <
2 x+
Thus: Theref For nx
2 x+
Thus: Theref
n SolutioBinomi
Let (f x
4x⇒ − Using c
For x =
For x =
α
od 2 [Graph
h & y x y= =
the graph: nx< < α : nx
fore the sequ
n > α : 1nx + <fore the sequ
od 3 [Algebr32
nn
n
xxx
=+
nx< < α :
0, 0n nx x> >
1 0n nx x+ − >fore the sequ
n > α
0, 0n nx x> >
1 0n nx x+ − <fore the sequ
on ial Expansio
) ( )(1xx
x−
=+
(4 3 2A x= +
cover up rule23
= − , 23
− −
1= − , 1 4− −
hical] 32
xx
=+
1n nx+ > uence is stric
nx uence is stric
raic]
3 nn
n
xx
−− =
(3
2nx −
=
0 & 3 2− +
10 n nx x+⇒ >uence is stric
0 & 3 2− +
10 n nx x+⇒ <uence is stric
on
( )4
3 2A
x x−
=+ +
) ( )2 1B x+ +
e, 143
B ⎛ ⎞= ⎜ ⎟⎝ ⎠
⇒
( )1A= − ⇒
ctly increasin
ctly decreasi [Shown]
22
n n
n
x xx
− +
+
)2
2n
n
x
x
− +
+
0 sincenx >
ctly increasin
0 sincenx <
ctly decreasi [Show
1 3 2A B
x+
+ +
14B⇒ = −
5A =
ng.
ng.
e 2 nx+ <
ng.
e 2 nx+ >
ng. wn]
9 3=
9 3=
(ii)
(iii)
( ) 5 14f1 3 2
xx x
∴ = −+ +
( ) 5 14f1 3 2
xx x
= −+ +
( ) ( )1 15 1 14 3 2x x− −= + − +
( )1
1 35 1 14 2 12xx
−− ⎡ ⎤⎛ ⎞= + − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
( )1
1 14 35 1 12 2
xx−
− ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )2 35 1 x x x= − + − +K
( ) ( )( ) ( )( )( )2 31 2 1 2 33 3 37 1 12 2! 2 3! 2x x x⎛ ⎞− − − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
K
( )2 35 1 x x x= − + − +K2 33 9 277 1
2 4 8x x x⎛ ⎞
− − + − +⎜ ⎟⎝ ⎠
K
2 311 43 14922 4 8
x x x= − + − + +K
Expansion of ( ) 11 x −+ is valid for 1 1x− < <
Expansion of 131
2x −
⎛ ⎞+⎜ ⎟⎝ ⎠
is valid for 2 23 3
x− < <
Therefore, the range of values of x for the expansion of f(x) to be valid is
2 23 3
x− < < .
Coefficient of ( ) ( ) ( )1 3 31 5 1 7 1 5 72 2
n nn n nnx + ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + − = − −⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
Qn Solution 5 Complex Numbers (Loci)
5
Im
Re
4 3i−
6 3i+p
q
5 3 7iz− = +
6 3i 4 3iz z− − = − +
Cartesian equation of (i):
2 2 2( 5) 4x y− + = ---- (1) Gradient of line segment joining 6 3i+ and 4 3i−
= ( )3 33
6 4
− −=
−
Cartesian equation of (ii):
( )1 53
y x= − − ---- (2)
Using GC to solve (1) & (2), we get
1.5359 2ip = + 8.4641 2iq = −
1 4 5arg( ) sin 2.62
8 6p q ππ − ⎛ ⎞∴ − = − = =⎜ ⎟
⎝ ⎠(3 s.f.)
Alternative: 6.9282 4ip q− = − +
( ) 1 4arg tan 2.62 (3s.f.)
6.9282p q π −∴ − = − =
Or using GC,
( )arg 2.62 (3s.f.)p q∴ − =
Qn Solution 6 Differential Equations
( )d 30 , 0d
1 d d30
ln 30
30 e kt C
k kt
k t
kt C
θ θ
θθθ
θ − +
= − − >
= −−− = − +
− =
∫ ∫
30 e e e (where e )30 e
kt C kt C
kt
A AA
θ
θ
− −
−
− = ± = = ±
= +
when 0, 9090 30
6030 60e (shown)kt
tA
A
θ
θ −
= == +=
∴ = +
8
8
when 8, 5555 30 60e
5e121 5ln8 12
k
k
t
k
θ−
−
= =
= +
=
= −
1 5ln8 12
1 5ln8 12
when 35,
35 30 60e
1e12
22.707additional time needed 22.707 8 14.7 min (1d.p.)
t
t
t
θ⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
=
= +
=
=∴ = − =
Qn Solution 7 AP and GP
(a) 132
ar a d d a= + ⇒ = −
( )
[ ]
( )
Sum of first even-numbered terms
12 ( 1)2 2
22
22
n
n a n a
n a an
an n
⎡ ⎤⎛ ⎞= + − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
= −
−=
(b)(i)
( )2 2
Amount owed at the beginning of the third month
50000 (1.035) (1.035)
50000(1.035 ) (1.035 1.035 )
x x
x
= − −⎡ ⎤⎣ ⎦= − +
(b)(ii) 1 2 1
Amount owed at the beginning of the th month50000(1.035 ) (1.035 1.035 1.035 )n n
nx− −= − + + +L
For the repayment to be completed during the nth payment,
1 2 150000(1.035 ) (1.035 1.035 1.035 )n n x x− −− + + + ≤L 1 2 150000(1.035 ) (1 1.035 1.035 1.035 )
1.035 11.035 11.035 1
0.035
n n
n
n
x
x
x
− −≤ + + + +
−≤
−−
≤
L
Thus 11750(1.035 )
1.035 1
n
nx−
≥− .
Qn Solution 8 Transformation of Graphs + System of Linear Equations (i)
(ii)
Let the curve be 3 2y ax bx cx d= + + + . Since the points ( )2,75− , ( )0,3 and ( )1,12 lie on the curve.
Using ( )0,3 , 3d =
Using ( )2,75− , ( ) ( ) ( )3 22 2 2 75a b c d− + − + − + = 8 4 2 72 (1)a b c− + − = LLL
Using ( )1,12 , 12a b c d+ + + = 9 (2)a b c+ + = LLL
Since ( )1,12 is a maximum point, d 0d
yx= .
23 2 0ax bx c+ + = 3 2 0 (3)a b c+ + = LLL Using GC to solve (1),(2) and (3),
8, 7, 10a b c= − = = Thus the equation of the curve is 3 28 7 10 3y x x x= − + + + .
x
y
( )" 1, 3C
( )" 0, 0B
( )" 2, 72A −
( )"' 1, 3C −
α β
( )"' 2, 72A − −
0
x
y
( )f 1 2y x= −
1' , 32
B ⎛ ⎞⎜ ⎟⎝ ⎠
3' , 752
A ⎛ ⎞⎜ ⎟⎝ ⎠
( )' 0,12C
0
Qn Solution 9 Vectors (line and planes)
(i)
(ii)
Vector equation of the line l 1 1
= 0 1 ,1 1
Rλ λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
r
Angle between OA and the line l
( )o o
0 11 . 1
2 1 3 3cos 5 3 15 15
39.232 39.2 1 d.p.
θ
θ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠= = =
= ≈
Let the shortest distance from the origin to the line l be x.
( )
sin5
1.41 3 s.f.
x
x
θ =
= (iii)
Since
1
1
1 10 . 2 4 A point on lies on 1 3
1 1and 1 . 2 0 is parallel to
1 3
l
l
π
π
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ = ⇒⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ = ⇒⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
The line l lies on the plane (shown) Alternative Solution (1): Since
1
1
1
0 1-1 . 2 4 Point on lies on 2 3
1 10 . 2 4 Point on lies on 1 3line lies on
A l
B l
l
π
π
π
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ = ⇒⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ = ⇒⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∴
Alternative Solution (2): Since
1
1 1 1 1 1 0 1 . 2 . 2 1 2 3 3 4
1 1 3 1 3
line lies on l
λλ λ λ λ λ
λ
π
⎡ ⎤ +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ = = + + + − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
∴
(iv)
2
1 1 5normal of 1 2 4
1 3 1π
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= × = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2
2
5 1 5: . 4 0 . 4
1 1 1
5: . 4 6
1
π
π
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞⎜ ⎟− =⎜ ⎟⎜ ⎟⎝ ⎠
r
r
(v)
3
1 5 14 1normal of 2 4 14 14 1 OR
3 1 14 1π
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= × − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
3
1normal of director vector of 1
1lπ
⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟−⎝ ⎠
3
1: . 1
1dπ
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟−⎝ ⎠
r
3perpendicular distance from origin to 1 311
d dπ = =⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
3 33
d d= ± ∴ = ±
3
1: . 1 3 or
1π
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟−⎝ ⎠
r
3
1: . 1 3
1π
⎛ ⎞⎜ ⎟ = −⎜ ⎟⎜ ⎟−⎝ ⎠
r
Qn Solution 10 Functions + Maclaurin’s Series (i) gR [ ln 2, )= − ∞ and fD =
Since g fR D⊆ , the composite function fg exists. (ii)
( )( )2 3
2 42 3 2 3
1fg( ) cos ln1
cos ln 1
cos2 3
2 3 2 31
2! 4!
xx
x
x xx
x x x xx x
⎛ ⎞= ⎜ ⎟+⎝ ⎠= − +
⎛ ⎞≈ − + −⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞− + − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠≈ − + −L
( )
2 22 2 3 3
2 3
2 3
11 22 2 2 3 3
1121 112 2
x x x xx x
x x
x x
⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − − + − − + + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦
= − − +
= − + +
L
L
L
(iii) h(1) 1= − so 1h ( 1) 1− − = .
Let ln(1 )y x= − + . Then e 1yx −= − . For 1 1x− < < , h( ) ln(1 )x x= − + so h( ) ln 2x > − . Thus 1h ( ) e 1xx− −= − for ln 2x > − . Thus
1h : x− a
e 1 for ln 2,1 for 1.
x xx
− − > −= −
Qn Solution 11 Complex Numbers (a)
3 i 2 cos isin6 6
z π π⎛ ⎞= + = +⎜ ⎟⎝ ⎠
2 cos i sin6 6
n n n nz π π⎛ ⎞= +⎜ ⎟⎝ ⎠
( )* 2i Im( ) 2i 2 sin 06
nn n n nz z z π⎛ ⎞− = = =⎜ ⎟⎝ ⎠
sin 06
nπ=
,6
n k k Zπ π += ∈
The set of values of n is { }: 6 ,n n k k Z += ∈ (b)
( ) ( )3 1 i 1 3z w+ = − + +
Note that OWPZ is a parallelogram,
1 3 2arg( ) tan1 3
w ππ −= − =
Im ( )i 1 3+ P
W i 3 i Z Re 1− O 3 1− 3
23 6 2
WOZ π π π∠ = − =
OZ= 2z =
OW= 1 i 3 2w = − + =
Since2
WOZ π∠ = and OZ OW= , OWPZ is a square. (shown)
5arg( ) arg( )
6 4 12z w z POZ π π π+ = +∠ = + =
Also, 1 3 1arg( ) tan3 1
z w − ++ =
−
15 3 1tan
12 3 1π − +
∴ =−
( )23 15 3 1 4 2 3tan 2 3
12 3 1 23 1π ++ +
∴ = = = = +−−
(deduced)
Qn Solution 12 Graphing + tangent/normal (i)
Let 2
2
2 3 14
x xyx− +
=−
( ) ( )
2 2
2
4 2 3 12 3 4 1 0
yx y x xy x x y− = − +
− − + + =
For real roots of x, Discriminant 0≥ ( ) ( )( )
( )
2
2
2
2
3 4 2 4 1 0
9 4 8 2 4 0
16 28 1 0For 16 28 1 0
y y
y y y
y yy y
− − − + ≥
− + − − ≥
− + ≥
− + =
( ) ( )( )
228 28 4 162 16
28 72032
7 3 58
y± − −
=
±=
±=
216 28 1 0
7 3 5 7 3 5 or 8 8
y y
y y
∴ − + ≥
− +≤ ≥
Hence, C cannot lie between 7 3 5 7 3 5 and 8 8
− +
(ii)
(iii) At 225,7
x y= − =
Using GC, 5
d 0.40136d x
yx =−
=
Equation of tangent at 5x = − is
( )22 0.40136 570.40136 5.149657
y x
y x
− = +
= +
When 0, 5.149657 5.15x y= = = (to 3 s.f) The coordinates of P are ( )0,5.15 (shown) Equation of normal at 5x = − is
( )22 1 57 0.401362.4915 9.3148
y x
y x
−− = +
= − −
When 0, 9.3148 9.31x y= = − = − (to 3 s.f) The coordinates of Q are ( )0, 9.31− (shown)
Area of PQR = ( )( )1 5.15 9.31 52
+
= 36.2 units2
x
y
O
2x =
2y =
2x = −
1 121
4−
2
2
2 3 14
x xyx− +
=−
( )5.23,1.71( )0.764,0.0365
H2 MATHS (9740) JC2 PRELIMINARY EXAM 2010 PAPER 2 SUGGESTED SOLUTIONS
Qn Solution 1 Equations and Inequalities
2
7 14 3
xx x+
≤+ −
2
7 1 04 3
xx x+
− ≤+ −
( )2
2
7 4 30
4 3x x x
x x+ − + −
≤+ −
( )2
2
7 4 3 03 4
x x xx x
+ − − +≤
− − −
2
2
2 3 03 4
x xx x− +
≥− −
Since 2 22 3 ( 1) 2 0 for ,x x x x− + = − + > ∈ 2 3 4 0x x− − >
( )( )4 1 0x x− + > 1 or 4x x< − >
Qn Solution 2 Vectors
3× = ×p q p r ( )× − =p q 3r 0
( )// −p q 3r λ− =q 3r p , where λ is a scalar.
( ) ( )3 . 3 .λ λ− − =q r q r p p
2 2 226 . 9 λ− + =q q r r p
( ) ( )2 22 255 6 5 2 9 2 16
λ⎛ ⎞− × × + =⎜ ⎟⎝ ⎠
2 11
11
λ
λ
=
= ±
Qn Solution 3 Mathematical Induction (i) Let nP be the statement:
( ) ( ) ( )1
1
33 2 3 2 1 6,2
nr n
rr n n n+ +
=
− = − + − ∀ ∈∑ .
Check 1P ,
LHS of 1P : ( ) ( )1
1
13 2 3 2 1 3r
rr
=
− = − =∑
1− 4
− ++
RHS of 1P : ( ) ( ) ( )1 1 33 2 1 1 1 6 12 3 6 32
+ − + − = − − = = LHS
1P∴ is true. Assume kP true for some k +∈ .
( ) ( ) ( )1
1
33 2 3 2 1 62
kr k
rr k k+
=
− = − + −∑ ----(*)
Prove 1kP + true, i.e. ( ) ( ) ( )( )1
2
1
33 2 3 2 1 2 62
kr k
rr k k
++
=
− = − + + −∑
LHS of 1kP + :
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( )( )
11
1 1
1 1
1 1
1 1
1
2
3 2 3 2 3 2 1
33 2 1 6 3 2 1233 2 1 6 3 2 3 12
33 2 3 2 1 3 12
12 3 2 3 1 12
33 2 1 2 6 RHS2
k kr r k
r r
k k
k k
k k
k
k
r r k
k k k
k k k
k k k
k k
k k
++
= =
+ +
+ +
+ +
+
+
⎡ ⎤ ⎡ ⎤− = − + − +⎢ ⎥ ⎣ ⎦⎣ ⎦
⎡ ⎤= − + − + − +⎣ ⎦
= − + − + − +
= + − + − +
⎛ ⎞⎡ ⎤= − + +⎜ ⎟⎣ ⎦ ⎝ ⎠
= − + + − =
∑ ∑
1true truek kP P +∴ ⇒ Since 1P is true and 1true truek kP P +⇒ , therefore, by Mathematical Induction, is truenP n +∀ ∈ .
(iii) ( )
( ) ( )5
4
1 1
3 2 5800
3 2 3 2 5800
nr
rn
r r
r r
r
r r
=
= =
− >
− − − >
∑
∑ ∑
( ) ( )( ) ( ) ( )( )1 4 13 33 2 1 6 3 2 4 5 6 58002 2
n n n+ +⎡ ⎤ ⎡ ⎤− + − − − − >⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
( ) ( )( )
( ) ( )( )
( ) ( )( )
1
1
1
33 2 1 6 60 58002
33 2 1 66 5800233 2 1 58662
n
n
n
n n
n n
n n
+
+
+
⎡ ⎤− + − − >⎢ ⎥⎣ ⎦
− + − >
− + >
Using GC, using table method
When n = 9, ( ) ( )( )1 33 2 12
n n n+⎡ ⎤− +⎢ ⎥⎣ ⎦= 2737<5866
When n = 10, ( ) ( )( )1 33 2 12
n n n+⎡ ⎤− +⎢ ⎥⎣ ⎦= 5979>5866
∴Least n = 10
Qn Solution 4 Application of Differentiation (Rate of change)
(a)
( )( ) ( )
3ln
3ln 3ln
y x xy
x x y
= +
= + +
d 1 3 3 dd d2y yx x y xx= + +
when 1 and 1,d 7d 4
x yyx
= =
= −
(b) 2
3
131 (from diagram )3
V r h
h r h
π
π
=
= =
2ddV hh
π=
( )2
d d d.d d d
1 1
h h Vt V t
hπ
=
=
3
3
After 2 min, amt of water poured into cone 2 1 60120 cm
Amt of water in cone after 2 min 120 30150 cm
= × ×
== +
=
2
3
3
1 1503
1 1503
450
r h
h
h
π
π
π
∴ =
=
=
2
3
d 1d 450
0.0116
ht
ππ
=⎛ ⎞⎜ ⎟⎝ ⎠
=
Qn Solution 5 Integration (area and volume)
(i)
y
x- a a
b
-b
R
2 20 22
0 2 2
Area of Area of quadrant-Area of triangle
1d2
1 d (shown)2
a
a
R
b xb x ababab a x xa
−
−
=
= − −
= − + −
∫
∫
sin
d cosd
0 sin 0
sin2
x ax a
a
a a
θ
θθ
θ θπθ θ
=
=
= ⇒ =⎛ ⎞⎜ ⎟⎜ ⎟− = ⇒ = −⎝ ⎠
( ) ( )
( )
( )
0 2 2
0 22
22 0 2
2
0 2
2
0
20
2
1d2
1sin cos d2
11 sin cos d2
1cos d2
1 cos 2 1d2 2
1 sin 2 12 4 2
sin1 10 02 2 4 2
4
a
b a x x aba
b a a a abaa b aba
ab ab
ab ab
ab ab
ab ab
ab
π
π
π
π
π
θ θ θ
θ θ θ
θ θ
θ θ
θθ
ππ
π
−
−
−
−
−
−
− −
= − −
= − −
= −
+= −
⎡ ⎤= + −⎢ ⎥⎣ ⎦
⎡ ⎤−⎛ ⎞⎛ ⎞= + − − + −⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥⎝ ⎠⎣ ⎦
⎛ ⎞= ⎜⎝ ⎠
∫
∫
∫
∫
∫
( ) 2
12
2 units4
ab
ab π
−⎟
= −
(ii)
2 2
0
2 22 2
20
2 32 2
20
2 32 2
2
22 2
2 2
2
Volume generated1dy3
1dy3
13 3
13 3
13 3
2 13 3
13
b
b
b
x a b
a ya a bb
a ya y a bb
a ba b a bb
a ba b a b
a b a b
a b
π π
π π
π π
π π
π π
π π
π
= −
= − −
⎡ ⎤= − −⎢ ⎥
⎣ ⎦
⎡ ⎤= − −⎢ ⎥
⎣ ⎦⎡ ⎤
= − −⎢ ⎥⎣ ⎦⎛ ⎞= −⎜ ⎟⎝ ⎠
=
∫
∫
Qn Solution 6 Sampling Methods
(a)
(b)
Quota sampling. It will not be representative of the cohort as it misses out those who do not visit the canteen. OR It is not random. Step (2) is incorrect. The interval size should be 4 students. He should pick the random start student from the 1st-4th student and select every 4th student thereafter.
Qn Solution 7 Probability (i)
(ii)
(iii)
5.
5 16 2 2 5 16 2 2 5Required Prob7 21 7 7 7 21 7 7 7
380 1029
⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= + +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
=
( )Required Prob 1 P Player fails in all three games
2 2 2 335 17 7 7 343
= −
⎛ ⎞⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
Let Event A = Player has exactly two failures Let Event B = Player has at least one success
( ) ( )( )
PRequired Prob P |
P
380(i) 1029
335(ii)343
76 201
A BA B
B∩
= =
⎛ ⎞⎜ ⎟⎝ ⎠= =⎛ ⎞⎜ ⎟⎝ ⎠
=
Qn Solution 8 Permutations and Combinations
(a)
No of teams 3 tallest, no shortest 3 8
563 5⎛ ⎞⎛ ⎞
=⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
3 tallest, 1 shortest 3 2 8140
3 1 4⎛ ⎞⎛ ⎞⎛ ⎞
=⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
S F
S F S F
57
27
521
1621
57
27
563
5863
57
27
521
1621
57
27
S F S S SF F F
Set: S = Success; F = Failure
2 tallest, no shortest 3 884
2 6⎛ ⎞⎛ ⎞
=⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
2 tallest, 1 shortest 3 2 8336
2 1 5⎛ ⎞⎛ ⎞⎛ ⎞
=⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
Total no of teams= 56 140 84 336 616+ + + =
b(i)
(ii)
No of different 5-letter secret codes = No of codes using {X,Y,Z,Z,Z}+ No of codes using {X,Y,Y,Z,Z}+No of codes
using {Y,Y,Z,Z,Z} 5! 5! 5!3! 2!2! 2!3!
= + +
20 30 10 60= + + = Y {Z, Z, Z} Y or Y {X, Z, Z} Y No of 5-letter secret codes that will begin and end with Y
1 3 4= + =
Qn Solution 9 Hypothesis Testing
Unbiased estimate of μ is 213.5 4.2750
tt
n= = =∑
Unbiased estimate of 2σ is ( )22 11
s t tn
⎡ ⎤= −⎢ ⎥⎣ ⎦− ∑
[ ]1 44.105490.90010 0.900 (3 s.f)
=
= ≈
Since 50n = is large, by Central Limit Theorem, 2
N ,50
X σμ⎛ ⎞⎜ ⎟⎝ ⎠
approximately. H0: 4.5μ = H1: 4.5μ <
Test Statistic: XZ Sn
μ−=
Level of significance: 1% Critical Region: Reject H0 if -value 0.01p < Assuming H0 is true, from the GC, -value 0.0432439p = . Since -value 0.0432 0.01p = > , we do not reject H0 and conclude that there is no significant evidence, at 1% level, that the population mean time spent is less than 4.5 minutes. In order for H0 to be rejected in favour of H1, we require 0.0432439 0.01α< ⇒ 4.3244α > 4.33% is the smallest level of significance resulting in the rejection of H0.
It is not necessary to assume that the population follows a normal distribution for the test to be valid because since 50n = is large, by Central Limit Theorem,
2
N ,50
X σμ⎛ ⎞⎜ ⎟⎝ ⎠
approximately.
Qn Solution 10 Binomial + Poisson (incl. CLT) (i) Let H be the number of calls received by the hospital in a week.
Let S be the number of calls received by the police station in a week. Then ( ) ( )~ Po 12 , ~ Po 5H S . Also, ( )~ Po 17H S+ .
( )P 11 0.915330.915
H S+ > =
≈
(ii) Let W be the number of weeks, out of 50, in which the hospital and police
station receive a total of at most 11 calls. Then ( )~ B 50,1 0.91533W − . i.e. ( )~ B 50,0.08467W . Since 50n = is large, 0.08467p = is small, and 4.2335 5np = < ,
( )~ Po 4.2335W approximately. From GC, ( )P 4 0.58332
0.583W ≤ =
≈
(iii) Since 100n = is large, by the Central Limit Theorem,
12N 12, approximately100
H ⎛ ⎞⎜ ⎟⎝ ⎠
( )1 212 12N 12 12, approximately100 100
H H ⎛ ⎞+ + +⎜ ⎟⎝ ⎠
( )1 2P 23 0.020613
0.0206
H H+ ≤ =
≈
Qn Solution 11 Normal Distribution
(a)(i) Let the weight of 1 King Crab and 1 Snow Crab be K and S respectively.
( )( )
~ 1.65,0.71~ 1.10,0.34
K NS N
Let 1 2 3 1 2
5K K K S ST + + + +
=
( ) ( ) ( )
( ) ( )
1E 3E 2E51 3 1.65 2 1.1051.43
T K S= +⎡ ⎤⎣ ⎦
= +⎡ ⎤⎣ ⎦=
( ) ( ) ( )
( ) ( )2
1Var 3Var 2Var51 3 0.71 2 0.34250.1124
T K S= +⎡ ⎤⎣ ⎦
= +⎡ ⎤⎣ ⎦=
( )~ 1.43,0.1124T N∴
( )P T 1.5 0.583< =
(ii) Let the selling price of 1 King Crab and 1 Snow Crab be X and Y respectively.
( ) ( )( ) ( )
2
2
40 ~ 40 1.65, 40 0.71 ~ 66,113645 ~ 45 1.10, 45 0.34 ~ 49.5,688.5
X K N X NY S N Y N
= × × ⇒
= × × ⇒
Let the total selling price of an King Crab and a Snow Crab be C X Y= + ( ) ( )~ 66 49.5,1136 688.5 ~ 115.5,1824.5C N C N+ + ⇒
( )P C 140 0.283> =
(iii) The weight of all crabs are independent of one another. NOTE: Weight of snow crab and weight of king crab is independent of each other is insufficient, students need to bring out that the weight of all crabs are independent of one another.
(b) Let X be the mass of a bar of chocolate, 2~ ( , )X N μ σ . P( ) 0.2P( ) 0.2
P( ) 0.2P( ) 0.4
0.25335
m
m m
m
m
X mZ
ZZ
σ
σ σ
σ
σ
μ− ≤ =≤ =
− ≤ ≤ =
≤ − =
=
3( 3 )
( )( 0.76004)
0.224
mP X m
P ZP Z
σ
μ− ≥= ≥= ≥=
Qn Solution 12 Correlation & Regression (i)
The linear model is not valid in the long run, as it is impossible to have the selling price of the car to be $0 or negative when the number of new car licenses increases to a certain number.
48000
31800
1339 2275
y
x
(ii)
(iii)
(iv)
(v)
Model C is the most appropriate, as when x increases, y decreases at a decreasing rate.
0.97817 0.978r = = (3 s.f.) It indicates a strong positive linear correlation between y and1/ x . Regression line:
500120008452.3yx
= + (5.s.f)
(a) When x =1300,
500120008452.3 46923.07 $46900 (3s.f.)1300
y = + = =
Since x =1300 lies outside the data range, the linear relation may no longer hold, hence, the estimate is unreliable. (b) When x =2000,
500120008452.3 33458.30 $33500 (3s.f.)2000
y = + = =
Since x =2000, lies within the data range, and r = 0.978 is close to 1, the estimate is reliable. No, I do not agree, as there is no causal effect between the two variables. The rise in the selling price of the car could be due to other factors like the production cost of the car.
NATIONAL JUNIOR COLLEGE
PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS 9740/01 Paper 1 13 September 2010
3 hours
Additional Materials: Answer Paper List of Formulae (MF15) Cover Sheet 0815 – 1115 hours
READ THESE INSTRUCTIONS FIRST
Write your name, registration number, subject tutorial group, on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for diagrams or graphs.
Do not use paper clips, highlighters, glue or correction fluid.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of
angles in degrees, unless a different level of accuracy is specified in the question.
You are expected to use a graphic calculator.
Unsupported answers from a graphic calculator are allowed unless a question specifically states
otherwise. Where unsupported answers from a graphic calculator are not allowed in a question,
you are required to present the mathematical steps using mathematical notations and not calculator
commands.
You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.
The number of marks is given in the brackets [ ] at the end of each question or part question.
This document consists of 9 printed pages.
National Junior College
[Turn Over
2
1 The sum of the digits in a three-digit-number is 15. Reversing the digits in that number
decreases its value by 594. Also, the sum of the tenth digit and four times the unit digit is
five more than the hundredth digit. Find the number. [4]
2 Find the value of p such that
0
2 3 d
ln 3
p x x . [2]
The graph of 3xy for 0 1x , is shown in the diagram below. Rectangles, each of width
1
n, where n is an integer, are drawn under the curve.
3
n
2
n 11n
n
2n
n
3n
n
x
y
1
n0
3xy
…
(i) Show that the total area of all n rectangles, A, is given by
1
1
2 3
3 1
n
nn . [2]
(ii) State the limit of A as . [1] n
NJC 2010 9740/01/2010 [Turn Over
3
3 The points A and B have position vectors a and b respectively, relative to the origin O such
that =a b . The point P with position vector p lies on AB such that b • p = a • p .
(i) Show that AB is perpendicular to OP. [2]
(ii) Determine the position vector of the point D in terms of a and b, where D is the
reflection of O about the line AB. [2]
(iii) Give the geometrical meaning of a b . [1]
4 The diagram shows the graph of siny x for π
02
x . P is a fixed point on the curve
with coordinates 11,sinx x
and Q is a point on the curve with coordinates
1 1,sinx x , where is measured in radians.
y
siny x
Q
P
x O
(i) Prove that the gradient of PQ is 1sin (cos 1) cos sinx x1
. [2]
(ii) Given that is sufficiently small for 3 and higher powers of to be neglected, express the gradient of PQ as a linear expression in terms of . [2]
Verify that the gradient of PQ approximates to the gradient of the tangent at P when tends to zero. [2]
NJC 2010 9740/01/2010 [Turn Over
4
5 Given that 3 4i 5z , illustrate the locus of the point P representing the complex
number z in an Argand diagram. [1]
Hence, find the least exact value of 1z . [2]
The locus of point Q, representing the complex number w, satisfies the relation
2i iw w a , where a is a real number, 2a . Find the range of values of a such that
the loci of P and Q meet more than once. [3]
6 By using the substitution , find the general solution of the differential equation y vx
23d
d yx
x
yx . [4]
(i) State the equation of the locus where the stationary points of the solution curves lie.
[1]
(ii) Sketch, on a single diagram, the graph of the locus found in part (i) and two
members of the family of solution curves where the arbitrary constant in the solution
is non-zero. [3]
7(a) By considering 2 (2 2 )x A x B where are real constants, or otherwise, find and A B
2
2d .
2 8
xx
x x
[5]
(b) Show that 1
e cos 2 d e (cos 2 2 sin 2 )5
x xx x x x c , where c is an arbitrary constant.
Hence, find 2e cos d .x x x [5]
NJC 2010 9740/01/2010 [Turn Over
5
8 (a) The sum, , of the first n terms of an arithmetic progression is given by .
Write down the expression for S
nS 2 2nS n n
1n nS . Hence, find the value of the common difference.
[3]
(b) A metal screw of length L (measured in millimetre) is driven into a concrete wall by an
electrical screwdriver, such that its distance driven into the wall is proportional to the angle
turned by the screwdriver.
WallL
Due to some reasons, every subsequent turn by this electrical screwdriver can only achieve
an 80% of the angle turned previously.
(i) Given that the initial angle turned by the screwdriver is radians, write down the
expressions for the first 3 distances driven into the concrete wall, leaving your
answers in terms of and k, where k is the constant of proportionality. [2]
(ii) Find the total distance driven into the concrete walls after n turns, leaving your
answer in the form , where a and b are constants to be determined. [3] 1 nak b
(iii) Given that and assuming that the total distance driven could never exceed the
length of the screw, find the minimum length of the metal screw required, giving
your answer in terms of
2k
. [2]
NJC 2010 9740/01/2010 [Turn Over
6
9 Relative to the origin O, the point A has position vector given by and A lies on
the plane with equation defined by •
0
1
0
OA
1 r 3
1
3
2
. Another plane 2 has equation y x .
The planes and intersect at line l. 1 2
(i) Find the vector equation of the line l. [1]
(ii) Show that the cosine of the acute angle between the planes 1 and is 27
7. [2]
(iii) Find the position vector of the foot of perpendicular, OF
, from point A to the line l.
Hence, find the exact length of projection of AF
onto the plane . [5] 2
(iv) Another plane has equation 3 1px qy , where p and q are real constants. Find
the condition in which p and q must satisfy such that the planes , and 1 2 3
intersect at exactly one point. [2]
10(a) Given that is a solution to the equation 2 3i
2 *( i) 16 iz a z b 0 ,
where is the conjugate of the complex number z, find the values of a and b, where a and
b are real constants [3]
*z
(b) (i) Solve the equation , expressing the solutions in the form 5 1 0z ier , where 0r
and π π . Show the roots of the equation on an Argand diagram. [4]
(ii) For π π , show that ii 21 e 2cos e
2
. [2]
(iii) Deduce the roots of in the exponential form. [3] 51 1w 0
NJC 2010 9740/01/2010 [Turn Over
7
11 The curve C has equation
2 1f( ) ,
ax bxx
x c
where a, b and c are real constants.
Given that the line is an asymptote of C, find the value of a and show that
. [3]
12 xy
2b c 1
(i) For , using algebraic method, prove that the curve C cannot lie between 2 1c
values, which are to be determined. [3]
(ii) Sketch the graph of 22
f( ) ,1
x xx
x
1
showing clearly its asymptotes, the coordinates of
the axial intercepts, and turning point(s) (if any). [3]
Hence, state the range of x for which f( )x is concaving downwards. [1]
(iii) Given that the line 3y kx k ,where k is a real constant, passes through the
intersection of the asymptotes of C, deduce the range of where k
22 1 ( 3)(x x kx k x 1)
has 2 real solutions. [1]
NJC 2010 9740/01/2010 [Turn Over
8
12 A curve C has parametric equations
22x and 3 2y , where is a real parameter.
Sketch the curve C for 0 3 . [1]
The tangent to C at point 2 32 , 2P cuts the y-axis at point Q. Show that the equation
of the tangent at P may be written as
34 3 2 4y x . [2]
(a) (i) C cuts the y-axis at the point R. Find the area of triangle PQR, A, in terms of . [2]
(ii) If x increases at a rate of 4 units per second when 2 , find the rate of change of
A at that instant. [3]
(b) Calculate the exact area of the region bounded by the curve C, the tangent to C at
point P when 2 and the y-axis. [5]
End of Paper
NJC 2010 9740/01/2010 [Turn Over
9
NATIONAL JUNIOR COLLEGE
PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS 9740/01 Higher 2 Paper 1
13 September 2010, Monday 0815 – 1115 hours
Candidate Name: ______________________ Registration No.:_____________
Subject Class: 2ma______/ 2IPma2_______ Subject Tutor: _______________
For office use
NJC 2010 9740/01/2010 [Turn Over
Question No. Marks Obtained
TOTAL MARKS
1 4
2 5
3 5
4 6
5 6
6 8
7 10
8 10
9 10
10 12
11 11
12 13
Presentation – 1 / –2
TOTAL 100
GRADE
over Page
INSTRUCTIONS TO CANDIDATES
Write your name, registration number, subject tutorial group, subject tutor’s name and calculator model in the spaces provided on the cover sheet and attached it on top of your answer paper.
Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER.
Write your calculator’s model number(s) in the box below.
Scientific Calculator Model:
Graphic Calculator Model:
NATIONAL JUNIOR COLLEGE
PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS 9740/02 Paper 2 17 September 2010
3 hours
Additional Materials: Answer Paper List of Formulae (MF15) Cover Sheet 0815 – 1115 hours
READ THESE INSTRUCTIONS FIRST
Write your name, registration number, subject tutorial group, on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for diagrams or graphs.
Do not use paper clips, highlighters, glue or correction fluid.
Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of
angles in degrees, unless a different level of accuracy is specified in the question.
You are expected to use a graphic calculator.
Unsupported answers from a graphic calculator are allowed unless a question specifically states
otherwise. Where unsupported answers from a graphic calculator are not allowed in a question,
you are required to present the mathematical steps using mathematical notations and not calculator
commands.
You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.
The number of marks is given in the brackets [ ] at the end of each question or part question.
This document consists of 9 printed pages.
National Junior College
[Turn Over
2
Section A: Pure Mathematics [40 marks]
1 The functions g and h are defined as follows:
1
2
1g : 2 , 0,
2
h : 3 2, 0.
xx x
x x x x
(i) Justify why hg exists and find the range of hg. [3]
(ii) The function h has an inverse if its domain is restricted to x b . Find the value of b
for this domain to be maximal. Sketch the graph of h and its inverse on the same
diagram. [2]
(iii) Solve h exactly. [2] 1(x) x
2 (a) A sequence of positive real numbers 1x , 2x , satisfies the recurrence relation
for . Given that the sequence converges to L, find the value of L.
[3]
3 ,x
2
1 3 ln 1n nx x 1n
(b) (i) By expressing
2 7 11
4 !
r r
r
in the form 2 ! 4 !
A B
r r
where A and B are
real constants, show that
2
1
7 11 5 5
4 ! 4! 4 !
n
r
r r n
r n
. [3]
(ii) Use the method of mathematical induction to prove your result in (i). [5]
(iii) Hence, find
2
1
9 19
5 !r
r r
r
. [3]
NJC 2010 9740/02/2010 [Turn Over
3
3 Given that sin2 e ,xy show that d
2 cod
ysy x
x .
By further differentiation of this result, find the Maclaurin’s series for up to and
including the term in .
,y3x [5]
Deduce the Maclaurin’s series for e sin x up to and including the term in 2.x [3]
NJC 2010 9740/02/2010 [Turn Over
4 (i) Use the substitution 2 tanx to show
2
2 22
4d
44
x xx C
xx
, where C is an
arbitrary constant. [5]
x O – 2 2
15 3 2y x 31,
5
y (ii)
The diagram above shows the curve with equation
22
22
4
4
xy
x
with stationary
points at x . The line 0 15 3 2y x intersects the curve at and 2,0
31,
5
.
(a) The region bounded by the curve
2
22
4
4
xy and the line
x
15 3 2y x
is rotated through 4 right angles about the x-axis to form a
solid of revolution of volume V. Find the exact value of V, giving your
answer in the form bπ. [4]
(b) Sketch the gradient graph of
2
22
4.
4
xy
x
[2]
4
Section B: Statistics [60 marks]
5 The Head of Mathematics department of Holistics Junior College decides to take a survey
of opinions of 700 graduating students regarding the quality of teaching of the subject.
(i) What is the sampling frame in the context of the question? [1]
(ii) Describe clearly how a systematic sample of size 140 can be obtained. [3]
6 The data in the table below were obtained in an experiment to estimate the relation between
d, the duration of a television commercial (in seconds) and s, the average sales of a
particular brand of detergent (in thousands of bottles):
Duration, d 15 18 22 25 26 29 34 39
Average Sales, s 0.43 1.12 1.75 1.98 2.11 2.26 2.40 2.44
(i) Draw a scatter diagram to illustrate the data. [1]
(ii) Fit a model of the form s lnd to the data above and find the least squares
estimates of and . [1]
Hence state the product moment correlation coefficient between lnd and s, and explain
whether your answer suggests that a linear model is appropriate for the transformed
variables. [2]
(iii) Using the regression model in part (ii), predict the average sales if the duration of a
television commercial is 28 seconds. Comment on the reliability of your answer.
[3]
NJC 2010 9740/02/2010 [Turn Over
5
7 Four married couples attend a wedding dinner. One of the couples brought along two
children. Find the number of ways in which these ten people can be seated round a table if
(i) there are no restrictions, [1]
(ii) each couple must sit together. [3]
They are to take a photo with the bride and bridegroom. The twelve people are to arrange in
two rows of six and the bride and bridegroom must be together in the middle of the front
row. How many ways can the photographer arrange the twelve people such that the two
children must also be in the front row? [3]
In this question, give each of your answers as an exact fraction in its lowest term.
8 (a) In a certain sample space, it is known that events A and B are independent. Given that
3P
4A B and 2
P '15
A B , where 'A is the complement of event A, find
(i) P B , [3]
(ii) A B A B . [2] P
(b) A teacher is to form two groups of 4 students from a class of 3 boys and 5 girls for
Mathematics consultation session. Three of the girls, Ivy, Tamie and Cassy are good friends
from the class. Find the probability that
(i) the boys are together in the same group, [2]
(ii) either Ivy or Tamie is in the same group as Cassy. [3]
NJC 2010 9740/02/2010 [Turn Over
6
9 Suppose that the arrival and departure of aircrafts at a domestic airport follow two
independent Poisson distributions. In a one-hour period, it is expected that there are 4
arrivals and 3 departures.
(i) Show that the probability that there are at least 13 arrivals in a two-hour period is
0.0638, correct to 3 significant figures. [1]
(ii) Find the probability that, in a randomly selected one-hour period, there are less than
2 departures given that the airport handles a total of exactly 9 arrivals and departures.
[3]
A study of the domestic airport arrivals for 60 randomly selected two-hour periods is being
conducted to see if there are at least 13 arrivals for each two-hour period.
(iii) Giving two reasons, in this context, explain why the binomial distribution is a
suitable model for the study of the domestic airport arrivals. [2]
(iv) Using a suitable approximation, find the probability that there are at most 50 two-
hour periods with less than 13 arrivals each, explaining clearly why the
approximation is appropriate. [3]
NJC 2010 9740/02/2010 [Turn Over
7
10 The national average for monthly electricity usage measured in kilowatts hour (kWh), of
Housing Development Authority (HDA) units is 380 kWh. The monthly electricity usage of
3-room units follows a normal distribution with mean of 290 kWh and variance, 2 ,
whereas the monthly electricity usage of 5-room units follows an independent normal
distribution with mean of 450 kWh and variance 105 kWh2.
(i) Calculate the probability that the monthly electricity usage of two randomly chosen
3-room units exceeds 290 kWh each and a randomly chosen 5-room unit’s monthly
electricity usage is less than 450 kWh. [2]
(ii) Given that the probability of the total monthly electricity usage of four randomly
chosen 3-room units exceeds thrice the national average is 0.868, find 2 , correct to
the nearest integer, [3]
With effect from 1 July 2010, the monthly electricity bill is charged at 24 cents per
kilowatts hour.
(iii) Determine the value of a, correct to 2 decimal places, such that the probability of the
monthly electricity bill of a randomly chosen 5-room unit exceeding $a is 0.9. [2]
(iv) Eighty 5-room units are randomly selected. Using a suitable approximation, find the
probability that there are not less than seventy 5-room units with monthly electricity
bill exceeding $a. [3]
NJC 2010 9740/02/2010 [Turn Over
8
11 The drying time, X minutes, of Noppin brand paint under specified test conditions is known
to have mean value 75 minutes.
(a) On one occasion, in the manufacture of a large batch of Noppin paint, it was suspected that
an accidental chemical contamination had resulted in a change in the drying time of the
paint. Due to the high costs involved in discarding the entire batch of paint, the
manufacturer decided he will only do so if there was strong evidence from a test at 5% level
of significance to suggest that the drying time has changed. 50 random and independent
specimens of paint samples were taken from the batch and the drying time is summarised as
follows
3791x and 2
287959x .
(i) Determine whether the manufacturer will discard the entire affected batch of paint.
[6]
(ii) Assuming that the unbiased estimate of the population variance is the same as the
one found in the above sample, find the probability such that the total drying time of
another 60 randomly selected specimens of paint samples obtained from the same
batch is between 72 hours and 76 hours. [3]
(b) On another occasion, chemists proposed a new additive designed to decrease the drying
time. The mean and standard deviation from 20 random and independent specimens of paint
samples with the new additive are x minutes and 7.5 minutes respectively. A test is to be
carried out at the 5% level of significance to determine whether the new additive had been
effective.
(i) Determine the largest value of x for which the chemists can claim that their new
additive had been effective in decreasing the drying time of the paint. [3]
(ii) State a necessary assumption for validity of the test. [1]
End of Paper
NJC 2010 9740/02/2010
[Turn Over
9
NJC
NATIONAL JUNIOR COLLEGE
PRELIMINARY EXAMINATIONS
Higher 2
MATHEMATICS 9740/02 Higher 2 Paper 2
17 September 2010, Friday 0815 – 1115 hours
Name: _______________________________ Registration No. :_____________
Subject Class: 2ma______/ 2IPma2_______ Subject Tutor: _______________
For office use
ver Page o
2010 9740/02/2010 [Turn Over
Question No. Marks Obtained
TOTAL MARKS
1 7
2 14
3 8
4 11
5 4
6 7
7 7
8 10
9 9
10 10
11 13
Presentation – 1 / –2
TOTAL 100
GRADE
Write your calculator’s model number(s) in the box below.
Scientific Calculator Model:
Graphic Calculator Model:
Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER.
Write your name, registration number, subject tutorial group, subject tutor’s name and calculator model in the spaces provided on the cover sheet and attached it on top of your answer paper.
INSTRUCTIONS TO CANDIDATES
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
Page 1 of 16
1 Let the unit, tenth and hundredth digits be z, y, x respectively. 15 (1)x y z+ + = −
(100 10 ) (100 10 ) 59499 99 594 (2)x y z z y x
x z+ + − + + =
⇒ − = −
4 54 5 (3)
y z xx y z
+ = +⇒ − + + = −
Using GC to solve the equations simultaneously, 8, 5, 2x y z= = = .
Thus the number is 852.
2
( )
0
0
0
2 3 dln 3
3 2ln 3 ln 3
1 23 3ln 3 ln 3
3 1 23 3
1
p x
px
p
p
p
x
p
=
⎡ ⎤=⎢ ⎥
⎣ ⎦
− =
− =
=∴ =
∫
2(i) Total area of all the n rectangles, A
1 2 3
1 2 3
11
1
1
1
1 1 1 13 3 3 3
1 3 3 3 3
3 13
3 1
2 3
3 1
nn n n n
nn n n n
n
nn
n
n
n
n n n n
n
n
n
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞
= + + + +⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠= ⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠=
⎛ ⎞−⎜ ⎟
⎝ ⎠
L
L
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
2
2(ii) As n →∞ ,
limit of A = area under the curve 3xy = for 0 1x≤ ≤
1
03 d
2ln 3
x x=
=
∫
3(i) ( ) AB OP⋅ = −b auuuv uuuv
• p
= b • p – a • p = a • p – a • p (since b • p = a • p) = 0
Hence, AB is perpendicular to OP.
3(ii) Since =a b , then P must be the midpoint of AB.
Using ratio theorem, ( )12
OP = +a buuur
Thus, 2OD OP=uuuv uuuv
( )122
⎛ ⎞= +⎜ ⎟⎝ ⎠
a b
= +a b
3(iii) ×a b represents the
(1) area of rhombus OADB or OBDA. (or)
(2) magnitude of a vector which is perpendicular to a and b.
4(i) Gradient of PQ
1 1
1 1
1 1 1
1 1
sin( ) sin( )
sin cos cos sin sin
sin (cos 1) cos sin (shown)
x xx x
x x x
x x
θθθ θ
θθ θ
θ
+ −=
+ −+ −
=
− +=
4(ii) When θ is small,
Gradient of PQ
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
3
( )
1 1
21 1
1 1
sin (cos 1) cos sin
1sin 1 1 cos2
1cos sin2
x x
x x
x x
θ θθ
θ θ
θ
θ
− +=
⎡ ⎤⎛ ⎞− − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦≈
= −
4(iii) As θ tends to zero,
gradient of PQ = 1 10
1
1lim cos sin2
cos
x x
x
θθ
→
⎛ ⎞−⎜ ⎟⎝ ⎠
=
( )1
1
gradient of tangent at point d sind
cosx x
P
xx
x=
=
=
5
Least exact value of 1z −
2 2( 3 1) (4 0) 5
32 5 or 4 2 5
= − − + − −
= − −
Re (z)
Im (z)
× ( )3, 4−
O
9
–1
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
4
2i iw w a− = − represents the perpendicular bisector 22
ay +=
For the line to meet the circle more than once, 21 9
22 2 18
4 16 (ans)
a
aa
+− < <
− < + <∴− < <
6 d dvd dxyy vx v xx
= ⇒ = +
2
d 3 2 d
dv( ) 3 2dx
dv 3 2dx
yx x yx
x v x x vx
xx
= + −
⇒ + = + −
−⇒ =
2
3 2d d
23 ln | |
3 ln | | 2
v xx x
v x Cx
y x x Cx
⇒ = −
⇒ = + +
⇒ = + +
∫ ∫
Re (z)
Im (z)
× ( )3, 4−
O
9
–1
22
ay +=
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
5
6 (i) d 0 3 2 dy y xx= ⇒ = − +
6 (ii) (ii)
Note: (0,2) satisfies the differential equation. (0,2) is a singular solution.
7(a) 2 (2 2 )x A x B− = − +
By comparing coefficient of
1:1 22
x A A= − ⇒ = −
constant: 2 2 1A B B− = + ⇒ = −
2
2
2 2 2
2 1
2 d2 8
1 ( 2 2) 12 d
2 81 2 2 1 d d 2 2 8 3 ( 1)
1 1(2 2 8) sin 2 3
x xx x
xx
x xx x x
x x x
xx x C−
−
− + +
− − + −=
− + +− +
= − −− + + − −
−⎛ ⎞= − − + + − +⎜ ⎟⎝ ⎠
∫
∫
∫ ∫
2 1
12 8 sin 3
xx x C− −⎛ ⎞= − − + + − +⎜ ⎟⎝ ⎠
In absence of C, deduct from presentation marks.
2
23 +−= xy
0C < 0C >
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
6
7 (b) ( )e cos 2 d e cos 2 e 2 sin 2 d
e cos 2 2 e sin 2 2 e cos 2 d
e cos 2 2e sin 2 4 e cos 2 d
x x x
x x x
x x x
x x x x x
x x x x
x x x x
= − −
⎡ ⎤= + −⎣ ⎦
= + −
∫ ∫∫∫
5 e cos 2 d e cos 2 2e sin 2 x x xx x x x∴ = +∫
1e cos 2 d e (cos 2 2 sin 2 )5
x xx x x x c= + +∫ (shown)
2e cos d
1 e (1 cos 2 ) d21 1e d e cos 2 d 2 21 1e e (cos 2 2 sin 2 ) (ans) 2 10
x
x
x x
x x
x x
x x
x x x
x x c
= +
= +
= + + +
∫
∫
∫ ∫
8(a) Given 2S 2n n n= − .
1S Sn n−− ( ) ( )( )22 2 1 2 1n n n n= − − − − −
( )22 2 1 2 2n n n n= − − − + −
2 3n= −
Since the progression is AP,
common difference, d 1T Tn n−= −
( )( )2 3 2 1 3n n= − − − −
2= .
OR
( )2 1T T
1 12
d = −
= − −
=
8(b)
(i)
Let d be the distance driven for every turn.
1n = , d kθ=
2n = , 810
d kθ⎛ ⎞= ⎜ ⎟⎝ ⎠
or 0.8kθ
3n = , 28
10d kθ⎛ ⎞= ⎜ ⎟
⎝ ⎠ or 0.64kθ
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
7
(ii) Total distance driven into the wall after n turns 2 18 8 8...
10 10 10
n
k k k kθ θ θ θ−
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 18 8 81 ...10 10 10
n
kθ−⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
8110
8110
n
kθ
⎛ ⎞⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟=
⎜ ⎟⎛ ⎞− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
45 15
n
kθ⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
5a = , 45
b =
(iii) Distance driven in the long run
4lim 5 15
n
nkθ
→∞
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
5kθ=
Given 2k = , minimum length of the metal screw is 10θ unit.
9 (i)
r •13 32
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟⎝ ⎠
3 2 3x y z⇒ + + =
2Π : 0 0x y z− + + =
Using GC: l:
3 14 23 1 , 4 20 1
λ λ
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= + − ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
r R
OR 34 13 1 , 4
20
λ λ
⎛ ⎞⎜ ⎟
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= + ∈⎜ ⎟⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎝ ⎠
⎜ ⎟⎜ ⎟⎝ ⎠
r
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
8
9 (ii) For the plane 2Π : y x= 0 0x y z⇒− + + =
Let 2n be normal vector to plane 2Π , then 2
110
−⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
n
1 13 12 0 2 1 7cos (shown)
714 2 28 7θ
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∴ = = = =
9 (iii) Since point F lies on line l,
Let
3 34 413 314 4
20 2
λ
λ λ
λ
⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= + = +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎜ ⎟ ⎜ ⎟⎝ ⎠ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
uuurOF for some λ .
Then
3 34 403 114 4
02 2
λ λ
λ λ
λ λ
⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= + − = − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠− −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
uuurAF
Now, AF l AF⊥ ⇒uuur uuur
•11 02
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟−⎝ ⎠
341 42
λ
λ
λ
⎛ ⎞+⎜ ⎟⎜ ⎟⎜ ⎟− +⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎝ ⎠
• 11 02
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟−⎝ ⎠
112
λ = −
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
9
23 134 12
3 1 24 12 3
1 16 6
OF
⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= − = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
uuur
23 134 12
1 1 14 12 3
1 16 6
AF
⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= − − = −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
uuur
2 2 22 1 1 73 3 6 12
AF ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
Hence, exact length of projection from AFuuur
to the plane 2Π
cosAF θ=uuur
7 712 7
⎛ ⎞= ⎜ ⎟⎜ ⎟
⎝ ⎠
1 112 2 3
= = or 36
OR
exact length of projection from AFuuur
to the plane 2Π
2
2
AF= ×nn
uuur
2Π
1Π
A
l
F θ
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
10
9 (iv) 34 13 1 , for some 4
20
λ λ
⎛ ⎞⎜ ⎟
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟= + ∈⎜ ⎟⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎝ ⎠
⎜ ⎟⎜ ⎟⎝ ⎠
uuurOP
3Π has equation 1px qy+ = .
3 : ∏ r • 10
pq
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟⎝ ⎠
For the three planes to intersect exactly a point, l is not parallel to 3Π , then:
112
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
• 00
pq
⎛ ⎞⎜ ⎟ ≠⎜ ⎟⎜ ⎟⎝ ⎠
0⇒ + ≠p q
(ans)p q∴ ≠ −
10(a) 2 *( i) 16 i 0z a z b+ − + + =
( ) ( )
( )
22 3i ( i) 2 3i 16 i 05 12i 2 3 i 2i 3 16 i 0
8 2 10 3 i 0
a ba a b
a a b
+ + − − + + =
⇒ − + + − − − + + =
⇒ + + − + =
By comparing real and imaginary coefficient,
Real: 8 2 0 4 (ans)Im :10 3 0 22 (ans)
a aa b b
+ = ⇒ = −− + = ⇒ = −
10(b)
( )
( )
5
5
5 iπ
i 2 1 π5
i 2 1 π5
1 01
e
e
e , 0, 1, 2
k
k
zzz
z
z k
+
+
+ =
= −
=
=
= = ± ±
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
11
Re( )z
Im( )z
1
1
–1
–1 ×
×
×
×
×
O
iπ iπ i3π i3π5 5 5 51, e , e ,e ,ez
− −= −
(i) 2 2 2
2
2
i i ii
i
i
1 e e e e
e cos isin + cos isin2 2 2 2
2cos e (shown)2
θ θ θ
θ
θ
θ
θ θ θ θ
θ
−⎛ ⎞+ = +⎜ ⎟
⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
(ii) Replace complex number z with 1w− , ( )
( )
( )
i 2 1 π5
i 2 1 π5
i 2 1 π5
e
1 e
1 e
k
k
k
z
w
w
+
+
+
=
⇒ − =
∴ = +
From (i),
( ) ( ) ( )i 2 1 π5
2 1 πi 102 1 π
1 e 2cos e 10
k kkw
+ ++⎛ ⎞
∴ = + = ⎜ ⎟⎝ ⎠
for 0, 1, 2k = ± ± .
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
12
11 2 21 1f( ) ( )
2 1 2 (ans) and 1 2 1 (shown)
ax bx bc acx ax b acx c x c
ax b ac xa b ac b c
+ + − += = + − +
+ ++ − = −
∴ = − = − ⇒ = −
(i)
2
2Given 1, f( ) 2 11
2 Let 2 11
( 1) (2 1)( 1) 2 2 (1 ) (1 ) 0
c x xx
y xx
x y x xx y x y
= = − ++
= − ++
+ = − + +
+ − + − =
2
For all real values of , D 0
(1 ) 4(2)(1 ) 0( 7)( 1) 0
7 or 1 Hence, cannot lie between -7 and 1.
≥
⇒ − − − ≥⇒ + − ≥⇒∴ ≤ − ≥
x
y yy yy y
y
(ii)
f is concaving downwards for 1x < − .
2 1y x= −
1x = −
( )2, 7− −
( )0,1
x
y
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
13
(iii)
The line ( 3)kx k+ − passes through the point ( 1, 3)− − , which is the intersection of the
asymptotes. Since the oblique asymptote passes through the point ( 1, 3)− − and using the
graph in (ii), the gradient of the line 3kx k+ − has to be more than 2 for the above equation to
have 2 real solutions.
Hence, 2k > .
12
When λ = 0, ( ) ( )2 32 0 0 ; 0 2 2x y= = = + =
When λ = 3, ( ) ( )2 32 3 18 ; 3 2 29x y= = = + =
d 4d
x λλ= , 2d 3
dy λλ=
2d 3 3d 4 4yx
λ λλ
= =
Equation of tangent:
2
2
2 1 ( 3)( 1)2 1 3
1
x x kx k xx x kx k
x
+ + = + − +
+ +⇒ = + −
+
( )18,29
( )0 , 2
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
14
( ) ( )( )
( )
3 2
3 3
3 3
3
3
32 24
4 4 2 3 6
4 3 6 4 84 3 8 2
4 3 2 4 (shown)
y x
y x
y xy x
y x
λλ λ
λ λ λ
λ λ λ
λ λ
λ λ
− + = −
− + = −
= − + +
= + −
= + −
(a)
(i)
When x = 0, ( ) ( ) ( )3
3 3 44 3 0 2 4 2 42
y y λλ λ λ −= + − = − ⇒ =
Coordinates of Q is 340 ,
2λ⎛ ⎞−
⎜ ⎟⎝ ⎠
.
Area of triangle PQR,
A = ( )3 3 5
2 2 21 4 4 42 2 = units2 2 2 2
λ λ λλ λ⎛ ⎞ ⎛ ⎞− − +
− =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
(ii) Given: d 4dxt= when 2λ = .
5 4d 52 d 2
AA λ λλ
= ⇒ =
( )0,2R ( )2 32 , 2P λ λ +
340 ,2
Q λ⎛ ⎞−⎜ ⎟⎝ ⎠
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
15
By chain rule, d d dd d dA At t
λλ
= ⋅ .
(Find ddtλ ) For 2λ = , ( )d d d d4 4 2
d d d dx xt t t
λ λλ
= ⋅ ⇒ = ⋅
d 1d 2tλ
∴ =
( ) ( )4
25 2d 0.5 20 units /secd 2At= = .
(b)
For 2λ = ，equation of tangent at P is 2 3 4.y x= −
Coordinates of P is ( )8,10 .
Coordinates of Q is ( )0, 2− .
Method 1
Area
( )
( )( )
[ ]
8 8
0 0
822 3
00
2 4
0
1 d 3 4 d2
1 32 4 d 42 2
14 +8 d 96 32 02
y x x x
x xλ λ λ
λ λ λ
= − −
⎡ ⎤= + − −⎢ ⎥
⎣ ⎦
= − − −
∫ ∫
∫
∫
( )0,2R ( )8 , 10P
( )0, 2Q −
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 1
16
252
0
4 4 325
208 0 325
48 3 or 9 or 9.6 (ans)5 5
λ λ⎡ ⎤
= + −⎢ ⎥⎣ ⎦
⎡ ⎤= − −⎢ ⎥⎣ ⎦
=
OR Method 2
Area of region ( )10 10
2 2
1 2 4 d d3
y y x y−
= + −∫ ∫
Area bounded by the tangent at P and y-axis
( )
( )( )
10
2
102
2
1 2 4 d3
1 431 140 4348
y y
y y
−
−
= +
⎡ ⎤= +⎣ ⎦
= − −
=
∫
Area bounded by the curve and y-axis
( )
10
22 2 2
02 4
0
d
2 3 d
6 d
x y
λ λ λ
λ λ
=
=
=
∫∫∫
25
0
65λ⎡ ⎤= ⎢ ⎥⎣ ⎦
( )6 32 05192
5
= −
=
Area of region 192 48 348 or 9 or 9.65 5 5
= − = (ans)
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
1(i) g
1 5,2 2
R ⎛ ⎤= ⎜ ⎥⎝ ⎦ and [ )h 0,D = ∞ .
Since Rg ⊆ Dh , thus hg exists.
( ]g g hg1 5 1 3= ,0 , ,2 2 4 4hD D R⎛ ⎤ ⎡ ⎤= −∞ → → − =⎜ ⎥ ⎢ ⎥⎝ ⎦ ⎣ ⎦
OR any appropriate method such as graphical method
(ii) For inverse of function h to exist, 1.5b =
(ii)
(iii) To solve for exact value of 1h ( )x x− = is same as solving h( )x x= ,
2
2
3 24 2 0
x x xx x− + =
− + =
( ) ( )( )24 4 4 1 22
x± − −
∴ =
4 8
22 2 (rejected 1.5) or 2 2x
±=
= − ≥ +Q
From sketch above, for h−1(x) ≤ x , 2 2x ≥ + .
2(a) As n →∞ , nx L→ , 1nx L+ → .
( )( )21 3 ln 1n nx x+ = − +
0 x
y = h x( )
y x=y ( )1hy x−=
(1.5, –0.25)
( –0.25,1.5)
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 2 of 16
( )( )( )( )
2
2
3 ln 1
3 ln 1 0
L L
L L
⇒ = − +
⇒ − + − =
Using GC to solve ( )( )23 ln 1 0L L− + − = ,
0.860L = − (rejected as 0nx > )
or 1.8806225 1.88L = ≈ (3sf)
2 (b) ( ) ( ) ( )
( )( )
2
2
7 114 ! 2 ! 4 !
7 11 3 4
r r A Br r r
r r A r r B
+ += +
+ + +
+ + = + + +
When r = −3, B = −1
When r = 0, A = 1
Hence, ( ) ( ) ( )
2 7 11 1 14 ! 2 ! 4 !
r rr r r+ +
= −+ + +
.
( ) ( ) ( )2
1 1
7 11 1 14 ! 2 ! 4 !
1 13! 5!1 14! 6!1 15! 7!1 16! 8!
n n
r r
r rr r r= =
⎛ ⎞ ⎛ ⎞+ += −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠
= −
+ −
+ −
+ −
∑ ∑
( ) ( )
( )
( ) ( )
( ) ( )
1 1
1 ! 1 !1 1! 2 !
1 11 ! 3 !
1 12 ! 4 !
n n
n n
n n
n n
+
+ −− +
+ −+
+ −+ +
+ −+ +
M
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 3 of 16
( ) ( )
1 1 1 13! 4! 3 ! 4 !n n
= + − −+ +
( )
( )
4 1 4 14! 4 !
5 54! 4 !
nn
nn
+ + += −
+
+= −
+
(ii) Let Pn be the proposition that
( ) ( )2
1
7 11 5 54 ! 4! 4 !
n
r
r r nr n=
⎛ ⎞+ + += −⎜ ⎟⎜ ⎟+ +⎝ ⎠
∑ for all n +∈ .
Consider P1:
RHS of P1 = ( )
5 1 5 5 6 25 6 194! 1 4 ! 4! 5! 5! 5!
+ −− = − = =
+.
LHS of P1 = ( ) ( )
2 21
1
7 11 1 7 11 194 ! 1 4 ! 5!r
r rr=
⎛ ⎞+ + + += =⎜ ⎟⎜ ⎟+ +⎝ ⎠
∑ = RHS of P1
Hence, P1 is true.
Assume Pk is true for some k +∈ .
ie. ( ) ( )
2
1
7 11 5 54 ! 4! 4 !
k
r
r r kr k=
⎛ ⎞+ + += −⎜ ⎟⎜ ⎟+ +⎝ ⎠
∑
Consider Pk+1:
RHS of Pk+1 = ( ) ( )
5 1 5 5 64! 1 4 ! 4! 5 !
k kk k+ + +
− = −+ + +
.
LHS of Pk+1 = ( )
21
1
7 114 !
k
r
r rr
+
=
⎛ ⎞+ +⎜ ⎟⎜ ⎟+⎝ ⎠
∑
= ( )
( ) ( )( )
22
1
1 7 1 117 114 ! 1 4 !
k
r
k kr rr k=
⎛ ⎞ + + + ++ ++⎜ ⎟⎜ ⎟+ + +⎝ ⎠
∑
= ( ) ( )
25 5 2 1 7 7 114! 4 ! 5 !
k k k kk k+ + + + + +
− ++ +
= ( )
2 25 10 25 2 1 7 7 114! 5 !
k k k k kk
+ + − − − − − −−
+
= ( )
5 64! 5 !
kk+
−+
= RHS of Pk+1
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 4 of 16
Hence, Pk is true implies Pk+1 is also true.
Since P1 is true and Pk is true implies Pk+1 is also true, by Mathematical Induction, Pn is true for all n +∈ .
(iii) Let r = j – 1. Hence, we have
( )( ) ( )
( )
( )
( )
( )
22
1 1 1
2
2
2
2
2
2
1 9 1 199 195 ! 1 5 !
2 1 9 9 194 !
7 114 !
7 11 , since is a dummy variable4 !
r j
j
j
r
j jr rr j
j j jj
j jj
r r jr
∞ ∞
= − =
∞
=
∞
=
∞
=
⎛ ⎞⎛ ⎞ − + − ++ += ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟+ − +⎝ ⎠ ⎝ ⎠
⎛ ⎞− + + − += ⎜ ⎟⎜ ⎟+⎝ ⎠
⎛ ⎞+ += ⎜ ⎟⎜ ⎟+⎝ ⎠
⎛ ⎞+ += ⎜ ⎟⎜ ⎟+⎝ ⎠
∑ ∑
∑
∑
∑
( ) ( )
( )
( )
2 2
1
7 11 1 7 114 ! 1 4 !
5 5 194! 4 ! 5!
5 194! 5!5 194! 4! 56 1 or 5! 20
lim
r
n
r rr
nn
∞
=
→∞
⎛ ⎞+ + + += −⎜ ⎟⎜ ⎟+ +⎝ ⎠
⎛ ⎞+= − −⎜ ⎟⎜ ⎟+⎝ ⎠
= −
= −
=
∑
3 Let sin2 e xy =
2 sin
sin 2
4ed2 4e cos cos d
d2 cos (shown)d
x
x
yyy x y xx
y y xx
=
= =
=
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 5 of 16
2
2
3 2
3 2
3 2
3 2
d d2 cos sin d dd d d d2 cos sin sin cosd d d d
d d d2 cos 2 sin cos d d d
y y x y xx xy y y yx x x y x
x x x xy y yx x y x
x x x
⇒ = −
⇒ = − − −
∴ = − −
2
2
3
3
2 3
2 3
When 0, 2d 1dd 1 d 2d 3 d 4
1 32 4 22! 3!
2 (ans) 4 8
x yyxy
xy
x
x xy x
x xy x
= =
=
=
= −
⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∴ ≈ + + +
= + + −
Method 1
2 2 3sin 21e (2 )
4 4 4 8x y x xx= ≈ + + −
2 3sin sin( ) 2
2 32
22 2 3
2 2
1e e [2 ] (Replace by ) 4 4 8
1 = [2 ( )] 4 4 8
1 4 4 ...4 4 4 8
1 4 441 44
x x x xx x x
x xx
x x xx x
x x x
− −= ≈ − + + −
− − −
⎡ ⎤⎛ ⎞ ⎛ ⎞= − − + + − −⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤≈ − + +⎣ ⎦
= −( )2
2
4 2
1 (ans) 2
x x
xx
+
= − +
OR Method 2
2 2 3sin 21e (2 )
4 4 4 8x y x xx= ≈ + + −
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 6 of 16
( )
( ) ( )
1sin sin
12
2 32
2 32
22 3 2 3
2 2
2
e e
4
4(2 )4 8
(1 )2 8 16
2 31 2
2 8 16 2! 2 8 16
1 3 4 4
1 (ans) 2
x x
y
x xx
x x x
x x x x x x
x xx
xx
−−
−
−
−
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
= + + −
= + + −
− −⎛ ⎞ ⎛ ⎞= − + − + + − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎛ ⎞
= − − + +⎜ ⎟⎝ ⎠
= − +
L
L
4(i) 2d2 tan 2secd
xx θ θθ
= ⇒ =
( ) ( )( )
( )( )
2 22
2 22 2
22
22
2
2
2
2 2
2 2
4 4 4 tan d 2sec d4 4 4 tan
4 4 tan 2sec d4sec
4 4 tan d8sec
1 1 tan d2 sec sec1 cos sin d2
x xx
θ θ θθ
θ θ θθ
θ θθ
θ θθ θ
θ θ θ
− −=
+ +
−=
−=
= −
= −
∫ ∫
∫
∫
∫
∫
( )
2 2
2
1 cos 2 d21 1 sin 22 21 sin cos21 22 4 4
4
C
C
x Cx x
x Cx
θ θ
θ
θ θ
=
⎛ ⎞= +⎜ ⎟⎝ ⎠
= +
⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟
+ +⎝ ⎠⎝ ⎠
= ++
∫
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 7 of 16
(ii) (a)
( )( )
( )( )
( )
2221
22 2
221 222 2
13
2
2
Volume of region rotated about -axis,
4 3 2 d154
4 3π 2 d154
23π4 225 3
1 3 27 2π5 225 3 8
33 π (a100
x V
x x xx
x x xx
xxx
π−
−
−
⎛ ⎞ ⎡ ⎤−⎜ ⎟= − +⎢ ⎥⎜ ⎟+ ⎣ ⎦⎝ ⎠
⎛ ⎞−= − +⎜ ⎟⎜ ⎟+ ⎝ ⎠
⎡ ⎤+⎛ ⎞= −⎢ ⎥⎜ ⎟+ ⎝ ⎠⎢ ⎥⎣ ⎦
⎡ − ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞= − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦
=
∫
∫
ns)
33100
b∴ =
(ii) (b)
Section B: Statistics [60 marks]
5 (i) It is the list of all the 700 graduating students.
(ii) Arrange the list of all 700 students in some order (can be by surname, class, other reasonable category.)
O x
y
–2 2
ddyyx
=
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 8 of 16
Calculate the interval to take samples from to obtain the 140 samples. ie. 700 5140
= .
From the first group of 5 students, select the first student using random sampling. Then select every 5th student after that.
Note:
Students may say they use random sampling to pick anyone in the list as the first student, but they will then need to qualify that they will need to cycle back to the names at the front of the list if it reaches the end before getting all the 140 students.
6(i) s
(ii) Using GC,
4.975 4.98α = − ≈ − , 2.111 2.11β = ≈
0.951r = (to 3 sf)
Since the value of r is close to 1, it suggests an almost linear relationship. Hence a linear model is appropriate.
(iii) When d = 28 ,
( )4.97519 2.110569 ln 28s = − +
2.05766 2.06 or 2.058s = ≈
Thus average sales is 2060 or 2058 bottles.
The answer is reliable since based on the new model, the value of r is close to 1 and that suggests a linear relationship. Furthermore we are predicting s based on d which is within the range.
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 9 of 16
7(i) Number of ways = 9!= 362880 (ans)
(ii) Number of ways to arrange the couple among themselves = 2!
Number of ways to arrange 4 couples and 2 children = 6!6
Number of ways = 46!(2!) 19206
= (ans)
Number of ways to select 2 adults to in the front row =
82⎛ ⎞⎜ ⎟⎝ ⎠
Number of ways to arrange the bride and groom =2!
Number of ways to arrange the children and two adults in the front row =4!
Number of ways to arrange the 6 adults in the back row =6!
Number of arrangements = ( )( )( )8
4! 2! 6! 967680 (ans)2⎛ ⎞
=⎜ ⎟⎝ ⎠
OR
(iii)
Alternative Solution
Number of ways to arrange the children front row = ( )4
2!2⎛ ⎞⎜ ⎟⎝ ⎠
Number of ways to arrange the bride and groom =2!
Number of ways to arrange the 8 adults = 8!
Number of arrangements = ( )( )( )4
2! 2! 8! 967680 (ans)2⎛ ⎞
=⎜ ⎟⎝ ⎠
8(a)
(i)
A B
( )P A ( ) ( )P P ' A B A B= ∪ − ∩
3 24 15
= −
3760
=
215
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 10 of 16
Since events A and B are independent,
( ) ( ) ( ) 2P ' P ' P15
A B A B∩ = =
( )( ) ( ) 21 P P15
A B− =
( )37 21 P60 15
B⎛ ⎞− =⎜ ⎟⎝ ⎠
( ) 8P23
B =
OR Alternative solution
( ) ( ) ( ) ( )
( ) ( )
( )
( )
P P + P B P3 37 37+ P B P4 60 6023 2P60 15
8P (ans)23
A B A A B
B
B
B
∪ = − ∩
⇒ = −
⇒ =
∴ =
(ii) ( )P A B A B∩ ∪
( ) ( )( )( )
PP
A B A BA B
∩ ∩ ∪=
∪
( )( )
PP
A BA B∩
=∪
( ) ( )( )
P PP
A BA B×
=∪
(OR: ( ) ( )( )
P P ' P
B A BA B
− ∩=
∪)
37 860 23
34
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=
2961035
=
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 11 of 16
8(b)
(i)
P(all 3 boys in the same group)
3 5 43 1 4
2!8 44 4
2!
⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
=⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
5 1 (ans)70 14
= =
(ii) P(either Ivy or Tamie is in the same group as Cassy)
= P(Ivy together with Cassy but not Tamie)
+ P(Tammy together with Cassy but not Ivy)
+ P(Ivy and Tamie together with Cassy)
5 3 5 3 5 42 3 2 3 1 4
2! 2! 2!8 4 8 4 8 44 4 4 4 4 4
2! 2! 2!
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
= + +⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
25 5 (ans)70 14
= =
OR
(either Ivy or Tamie is in the same group as Cassy)
= P(Ivy together with Cassy)
+ P(Tammy together with Cassy)
– P(Ivy and Tamie together with Cassy)
6 4 6 4 5 42 4 2 4 1 4
25 52! 2! 2! (ans)8 4 8 4 8 4 70 144 4 4 4 4 4
2! 2! 2!
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
= + − = =⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 12 of 16
9(i) Let X be the number of arrivals at the airport in a two-hour period.
( )~ Po 8X
( ) ( )P 13 1 P 12 0.063797 0.0638X X≥ = − ≤ = ≈
(ii) Let W be the number of arrivals in a one-hour period.
Let Y be the number of departures in a one-hour period.
( )~ Po 4W ; ( )~ Po 3Y ; ( )~ Po 7W Y+
( )( )
( )( ) ( )
( )( ) ( ) ( ) ( )
( )
P 2 | 9
P 2 9
P 9
P 0, 9 P 1, 8
P 9
P 0 P 9 P 1 P 8
P 90.0051052524 0.1014046695
0.0503453384 0.0503 (ans)
Y W Y
Y W YW Y
Y W Y WW Y
Y W Y WW Y
< + =
< ∩ + ==
+ =
= = + = ==
+ =
= = + = ==
+ =
=
=≈
(iii) (1)There are two mutually exclusive outcomes – either there are at least 13 arrivals in each two-hour period or there isn’t.
(2)The probability of having at least 13 arrivals for each two hour period remains constant for each of the 60 two-hour periods.
(3)There is a fixed number of 60 two-hour periods independently selected under consideration.
(iv) Notice that we are unable to define the random variable as the number of two-hour periods, out of 60, with less than 13 arrivals each as it will not be possible to do any approximations.
np = 56.172 ( > 5) and nq = 3.828 ( < 5)
Let V be the number of two-hour periods, out of 60, with at least 13 arrivals each.
( )( )~ B 60, P 13V X ≥
Since n = 60 ( > 50, large) and ( )P 13 0.0638p X= ≥ = ( < 0.1, small), such that np = 3.828
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 13 of 16
( < 5), we have ( )~ Po 3.828V approximately.
At most 50 two-hour periods with less than 13 arrivals each means the same as at least 10 two-hour periods with at least 13 arrivals each.
( ) ( )P 10 1 P 9V V≥ = − ≤ = 0.0060899731 = 0.00609
Note:
Students may choose to use the more accurate value for
( )P 13 0.0637971966X ≥ = . If they do so, the following values will be obtained:
np = 3.827831796
( )~ Po 3.827831796V
( )P 10V ≥ = 0.0060881936 = 0.00609.
This will still be considered as correct and will be awarded the necessary marks.
10 Let X denote the monthly electricity usage of a 3-room unit.
( )2N 290,X σ
Let Y denote the monthly electricity usage of a 5-room unit.
( )N 450,105Y
(i) ( ) ( ) ( )P 290 P 290 P 450 0.5 0.5 0.50.125 (ans)
X X Y> × > × < = × ×
=
(ii) ( )21 2 3 4 N 1160, 4X X X X σ+ + +
( )( )1 2 3 4P 3 380 0.868 X X X X+ + + > =
( )1 2 3 4P 1140 0.132 X X X X⇒ + + + < =
2
1140 1160P 0.132 4
Zσ
⎛ ⎞−⇒ < =⎜ ⎟
⎝ ⎠
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 14 of 16
2
1140 1160 1.11698674σ−
⇒ = −
2
2021.1169867
8.952657980.1500827 80 (ans)
σ
σ
σ
⇒ =
⇒ =
∴ = ≈
(iii) Let C denote the monthly electricity charge of a 5-room unit.
( )20.24 N 0.24(450),0.24 (105)C Y=
( )N 108,6.048C∴
( )( )( )
P 0.9
1 P 0.9
P 0.1
C a
C a
C a
> =
⇒ − ≤ =
⇒ ≤ =
104.848321 104.85a∴ = ≈ (ans)
(iv) Let W denote the number of 5-room units with monthly electricity bill exceeding $a, out of eighty 5-room units.
( )B 80,0.9W
Since 80n = is large and 80(0.9) 72( 5), 80(0.1) 8( 5)np nq= = > = = > ,
( )N 72, 7.2 approxW∴
( ) ( )P 70 P 69.50.82425290.824 (ans)
W W≥ = >
=≈
11(a) 3791 75.8250
x = =
22 1 3791 525.38 26269287959 10.72204
49 50 49 2450s
⎡ ⎤= − = = =⎢ ⎥
⎣ ⎦
Let μ be the mean drying times of Noppin paint.
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 15 of 16
0.05
0 1.959961.95996−
1.77076
Test H0: 75μ = against H1: 75μ ≠
Level of Significance: 5% (2-tailed Z-test)
Test Statistic: ~ N ( 0 ,1)sn
XZ
μ−= approximately.
Method 1: Using critical region and observed test statistic, calculatedz
Critical region: 1.95996z >
calculated 10.7220450
75.82 75 1.77076z −= =
Since calculated 1.77076 1.95996z = < , we do not reject H0.
Method 2: Using p-value
p-value = 0.07660
Since p-value = 0.07660 > 0.05, we do not reject H0.
There is insufficient evidence at 5% level of significance to claim that the mean drying times of the Noppin paint has changed. Hence, the manufacturers will not discard the entire affected batch of paint.
11(a)
(ii)
Since X is unknown and n = 60 is large,
by Central Limit Theorem,
( )1 2 3 60 N 60 75, 60 10.72204082 approxX X X X+ + + + × ×L .
Suggested Solutions to 2010 NJC SH2 H2 Math Prelim Paper 2
Page 16 of 16
( ) ( )( )( )
1 2 3 60
1 2 3 60
P 60 72 60 76
P 4320 45600.9909990.991(ans)
X X X X
X X X X
< + + + + <
< + + + + <
=≈
L
L
(b)(i) Test H0: 75μ = against H1: 75μ <
Level of Significance: 5% (lower-tailed)
( )22 2 20 207.5 7.5 7.694841 19 19n
ns sn
σ= = ⇒ = =−
Test Statistic: ( )~ 19sn
XT t
μ−=
calculated 7.6948420
75xt −=
For the chemist to claim that the new additive had been effective, we need to reject H0 in favour of H1.
Hence, we need calculated 1.72913t < − .
7.6948420
75 1.72913
72.02482
x
x
−< −
<
Largest value of x = 72.0 (to 3 sig. fig.)
(ii) It is necessary to assume that the drying time of Noppin paint, X, follows a normal distribution.
0.05
0 1.72913−
NANYANG JUNIOR COLLEGE
This document consists of 6 printed pages.
NANYANG JUNIOR COLLEGE Internal Examinations
© NYJC 2010 [Turn Over
JC2 PRELIMINARY EXAMINATION Higher 2
MATHEMATICS 9740/01 Paper 1 16th September 2010 3 Hours Additional Materials: Answer Paper List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
2
NYJC 2010 JC2 Preliminary Examination 9740/01
1 Find the equation of the quadratic function which has a line of symmetry at x
1
3 and passes through
the points and . [4](1,9) (1,5)
2 The complex numbers z and w are 1 + ai and b – 2i respectively where a and b are real and a is
negative. Given that zw* = 8i, find the exact values of a and b. [3]
Find the smallest positive integer value of n such that wn is purely imaginary. [2]
3 Find the expansion of x4
1 in ascending powers of x, up to and including the term in 2x . Hence
find an approximate value of the integral 2
0
1d
4 2x
x , expressing your answer in the form
2a b , where a and b are constants to be determined. [6]
4 The differential equation 2
2
d9 3
d
xx t
t is such that x = 0,
d
d
x
t= 1 when t = 0. Find the Maclaurin’s
series for x up to and including the term. [3]3t
It is known that this differential equation has a general solution of the form x = Asin3t +Bt, where A
and B are constants. Assuming that t is sufficiently small for terms with powers of 4 and above to be
neglected, find the values of A and B. [4]
3
NYJC 2010 JC2 Preliminary Examination 9740/01 [Turn Over
5 With respect to the origin O, the points A, B and C have position vectors a, b and c respectively, and
are such that OACB is a parallelogram in an anti-clockwise sense.
(i) Express c in terms of a and b. [1]
(ii) Show that the area of parallelogram OACB is given by a b . [2]
(iii) Show that the maximum area of parallelogram OACB is a b . [2]
(iv) By considering ·a b where b2
, show that a
a1
a2
a3
and b
b1
b3
2 2 2 2 2 2 21 1 2 2 3 3 1 2 3 1 2 3)( b a b a b a a a b b ba . [3]
6 The diagrams below show the graphs of y = |f(x)| and y = f ( )x for x . The point A, B and A’ has
the coordinates of , and (1, 2) (1,1.5) ( 1 respectively. , 2)
On separate diagrams, sketch the graphs of,
(i) , [2] f ( )y x
(ii) 1
f ( )y
x , [3]
(iii) , [3] f '( )y x
showing clearly any asymptotes and the coordinates of any stationary point(s).
f ( )y x
A ( 1 , 2)
x
y
B ( 1,1.5)
A' (1, 2) x
y f ( )y x
4
NYJC 2010 JC2 Preliminary Examination 9740/01
7 Functions f and g are respectively defined on the domain of real numbers by
2f : 2 2xx x , x > 1,
g : 3x x , x –3.
(i) By considering the derivative of f(x), prove that f is a one-one function. [2]
(ii) Solve the equation f(x) = x1f where x1f denotes the inverse function of f(x). [2]
(iii) Deduce the solution set to the inequality 1f fx x . [2]
(iv) Show that the composite function gf exists and define it in a similar form. State also its
range. [3]
8 A curve C is defined by the parametric equations , where . 2
2 , 3etx t y t
(i) Sketch the curve C. [1]
(ii) Given that the point P lies on the curve with coordinates (2, 3e), show that the equation of the
tangent to the curve at point P is 3e( 1)y x . [2]
(iii) The normal to the curve at point P cuts the x-axis at point Q. Given also that the tangent to the
curve at point P cuts the y-axis at point R, find the coordinates of Q and R.
Hence, show that the area of PQR is units23e(1 9e ) 2. [5]
(iv) State the range of values of m for which the line 3ey mx intersects the curve at 2 distinct
points. [1]
9 Given that f ( ) 1 , , 01
4 1 4
ax ax x a
x
.
(i) Find the equation of the asymptote(s) of f ( )y x . [2]
(ii) Find the coordinates of the turning point(s) for f ( )y x . [4]
(iii) (a) Sketch the graph of f ( )y x , for 0 < a < 1, indicating clearly the equations of the
asymptote(s), and the coordinates of any turning point(s) and axial intercept(s). [3]
(b) Determine the range of values of m such that there will be intersection between the
line 1
1 = 4 4
y a m x
1
and y = f(x). [1]
5
NYJC 2010 JC2 Preliminary Examination 9740/01 [Turn Over
10 An innovation is introduced into a community of 100 farmers at time 0t . Let x denote the number
of farmers who have adopted the innovation at time t. Assume that x is a continuous function of time.
The rate at which the number of farmers in that community who adopted the innovation at a
particular instant is proportional to the product of the number of farmers who have already adopted
and the number of farmers who have not adopted the innovation.
Initially, one farmer adopted the innovation and the rate at which the number of farmers who adopted
the innovation is one farmer per unit time.
(i) Show that
dx
dt k(100x x2 ) , where k is to be determined. [1]
(ii) Find the particular solution of x , in terms of t. [5]
(iii) Sketch the graph of x versus t, for . [2] 0t
(iv) Using the graph in (iii) or otherwise, find the time taken for 75% of the population of farmers
to adopt the innovation, leaving your answer to 2 decimal places. [1]
(v) Give a reason why the model may not be suitable. [1]
11 Solve the equation , leaving your answers in the form 5( 2) 32 0 ier , where and 0r
. [4]
(i) State the equation of the circle in the form | |z a b that passes through all the points
represented by the roots.
Sketch this circle, showing clearly the relationship between the roots and the locus clearly. [3]
(ii) The roots 1 and 2 are such that 1 2arg( ) 0 arg(2 2
2 )2 . On the same
Argand diagram, sketch the locus of the points representing z given that
1
3)
5arg(z
. [2]
Find the complex number represented by the point of intersection between the two loci. [1]
(iii) Find the least value of 2z if 1
3)
5arg(z
. [2]
6
NYJC 2010 JC2 Preliminary Examination 9740/01
12 (a) (i) Write down the derivative of (1 . [1] x2)n
(ii) Find 3 2(1 ) dnx x x where 1, 2n . [3]
(b) Region A is bounded by the curve sin cosy x x , the lines y 2 and x
34
.
(i) Find the exact area of region A. [4]
(ii) Deduce the exact area of the region in the first quadrant bounded by the curve
4sin cosy x x 4 and the two axes. [2]
(iii) Find the volume generated when region A is rotated through four right angles about the
x-axis. [2]
-----END OF PAPER-----
NANYANG JUNIOR COLLEGE
This document consists of 6 printed pages.
NANYANG JUNIOR COLLEGE Internal Examinations
© NYJC 2010 [Turn Over
JC2 PRELIMINARY EXAMINATION Higher 2
MATHEMATICS 9740/02 Paper 2 17th September 2010 3 Hours Additional Materials: Answer Paper List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
2
NYJC 2010 JC2 Preliminary Examination 9740/02
Section A: (40 Marks)
1 Two scientists are studying the growth of a certain species on an island.
(a) The first scientist proposes the following model:
“Let nx be the population of the species in the nth year and dn x
n x
n1, n 2 with
. It is believed that is a geometric sequence with common ratio .” 1 1x 100d nd re
(i) Find an expression for nx in terms of r and n. [2]
(ii) Obtain an inequality for r in order for the population to stabilise at K. Find the value
of K in terms of r. [2]
(b) The second scientist proposes the following model:
“Let ny be the population of the species (in thousands) in the nth year. It can be hypothesised
that and 1y 0.1 1 2 (1 )n ny ny y .”
(i) Find the limiting population of this model. [2]
(ii) Find the value of r in order for both models to have the same limiting population. [2]
2 A circular cylinder is inscribed in a right circular cone of base radius 2c and vertical height c. One of
the circular ends of the cylinder lies on the base of the cone and the other end is in contact with the
inner surface of the cone. If the cylinder has base radius x, find expressions for the volume V and the
total surface area S of the cylinder in terms of c and x. If c is fixed and x varies, find the maximum
value of V. [8]
Prove that S cannot exceed 8πc2. [2]
3 The sequence of numbers {ur} where r = 1, 2, 3, …, is such that it satisfies the recurrence relation
21
1r
r
ruu r
r
and = 1. 1u
(i) By dividing the above recurrence relation by r and using the method of difference, show that
2 22n
nu n n for n = 1, 2, 3, …. [5]
(ii) Prove the result in (i) using mathematical induction. [4]
(iii) Find the exact value of 3nu
n as . [1]n
3
4 (a) The line l passes through the point A with coordinates (1,-1,1) and is parallel to the vector
2i + j – 2k. The plane p has equation r (i + j + k) = 3. Find, in exact form,
(i) the position vector of B, the point of intersection between l and p, [3]
(ii) the sine of the acute angle between l and p, [2]
(iii) the shortest distance from A to p, [1]
(iv) the length of the projection of AB onto p. [2]
(b) Given that the system of linear equations
x + y + 3z = (, )
3x + y + 4z = 9
x + y = 3
has infinite solutions, obtain the numerical values of and . [4]
Section B: (60 Marks)
5 (a) Out of the 29 basketball teams, 14 teams have 12 players and 15 teams have 13 players. A
sample of 58 players is to be chosen as follows. Each team will be asked to place cards with
its players’ names (1 card for 1 name) into a hat and randomly draw out two names. The two
names from each team will be combined to make up the sample. Explain why this procedure
will not result in a simple random sample of the 363 basketball players. Describe a procedure
that will result in a simple random sample in this situation. [2]
(b) A cable company plans to survey potential customers in a small city currently served by
satellite dishes. It intends to select a sample of families from each of the five non-overlapping
neighbourhoods that make up the city. Suggest, with justification, a sampling technique that
could be used in this case. [2]
NYJC 2010 JC2 Preliminary Examination 9740/02 [Turn Over
4
6 A coffee production factory claims that the average amount of coffee in a packet is at least 10 grams.
A consumer suspects the factory has overestimated the mean. To check the claim, a random sample
of 8 packets is weighed and the values are given as follows.
9.5 10.1 9.3 10.4 9.6 9.1 9.9 9.0
Making necessary assumption(s), test the factory’s claim at 3% level of significance. [4]
A second sample of 64 packets is obtained and the data are summarised as follows:
(x 10) 25.6, (x 10)2 151.99
Test the factory’s claim at 5% level of significance. [2]
If an observation of mass 9.6 grams is added to the second sample, without conducting the test,
decide whether the conclusion of the second test will still remain the same? [1]
7 An experiment was conducted to investigate the relationship between the amount of unreacted
chemical, x, and the time that elapsed since the start of the experiment, t.
x 25.5 28 31 33.5 41 43.5 45.5 51 57.5 58 73
t 46 44 35.5 30 20 15.6 17 12.3 11 8.3 6
(i) Obtain the scatter diagram for x against t and comment on any relationship between x and t.
Calculate the linear product moment correlation coefficient, r, between x and t. [4]
(ii) State with a reason (without any calculations) which of the following models is more
appropriate for the data:
(a) x = atb, where a > 0 and b < 0
(b) x = a + bt2, where a > 0 and b < 0. [1]
(iii) If there was an error in recording the t values and all the t-values must be increased by 3, what
would be the effect on
(a) t , [1]
(b) standard deviation of t, [1]
(c) the correlation coefficient, r? [1]
NYJC 2010 JC2 Preliminary Examination 9740/02
5
8 Four soccer players, three tennis players, two badminton players and one swimmer sit at a round
table with 10 seats. Find the number of possible seating arrangements
(i) if the two badminton players sit together, [2]
(ii) if the two badminton players sit directly opposite each other, [2]
(iii) if the two badminton players sit together and none of the four soccer players sit next to either
of the two badminton players, [2]
(iv) if the two people sitting beside the swimmer are from different sports. [2]
9 A vehicle rental company has 7 cars and 4 vans available for rental per day. It is known that the
request for cars has a mean of 4 per day; and independently, the request for vans has a mean of 2 per
day.
(i) Find the probability that the number of request for a vehicle exceeds 11 on a particular day.[2]
(ii) Find the probability that some requests for a vehicle have to be refused on a particular day. [2]
(iii) Explain why the value found in (ii) is larger than the value found in (i). [1]
(iv) Using a suitable approximation, find the least number of days such that the average number
of requests for vehicles exceeding 7 is less than 0.001. [3]
10 Chris takes a bus to school every day in a 5-day week. The bus journey consists of two intermediate
stops. The probability of delay at stop A is 4
5 while the probability of delay at stop B is
2
5. The
delays are independent of each other. If the bus is delayed at either stop, Chris will be late for
school.
(i) Given that Chris was late on Monday, find the probability he was late exactly three times in a
week. [3]
(ii) Given that Chris was late one day, find the probability that he was delayed at stop A. [2]
(iii) Given that Chris was delayed at exactly one stop, find the probability that he was delayed at
stop B. [3]
NYJC 2010 JC2 Preliminary Examination 9740/02 [Turn Over
6
NYJC 2010 JC2 Preliminary Examination 9740/02
11 On the tropical island of Stabletree, records show that the number of floods occurring each month
may be modelled using a Poisson distribution with mean 2. A “bad” month is a month where there
are at least 4 floods occurring, and a “bad” year is one where there are more than 2 bad months in the
year. Regarding a month as a twelfth part of a year, and assuming independence of flood
occurrences,
(i) show that the probability that there are at most 2 bad months in a year is 0.760. [3]
(ii) use a suitable approximation to find the probability that out of fifty years, there are less than 5
bad years. [4]
(iii) Comment on whether it is suitable to use the model to estimate the probability that the year
2060 (which is 50 years from now) will be a bad year. [1]
12 The mass of a Munchi pear is normally distributed with mean mass 120g and standard deviation 10g. (i) If two Munchi pears are chosen at random, find the probability that one of the pears will have
a mass between 100g and 126g while the other will have a mass of less than 115g. [2]
(ii) If ten Munchi pears are chosen at random, find the probability that exactly three of the pears will have a mass more than 122g each. [2]
The mass of a Fuchi apple is normally distributed with mean mass 115g and standard deviation 8g. (iii) Find the probability that the total mass of three randomly chosen Fuchi apples will be more
than three times the mass of a randomly chosen Munchi pear. [2]
(iv) A random sample consisting of n Fuchi apples is chosen. Find the least value of n such that there is a probability of not more than 0.3 that the sample mean differs from its mean mass by more than 4g. [3]
-----END OF PAPER-----
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 1 of 9
Qn
1 Let the equation of parabola be y = ax2 + bx + c
At points (-1, 9) and (1, 5)
9 = a − b + c
5 = a + b+ c
dy
dx= 2ax + b
At line of symmetry, d 1
0, d 3= =
yx
x
2
03a b= +
using GC, 3, 2, 4a b c= = − =
Equation of parabola is 23 2 4y x x= − +
2 (1+ai)(b+2i) = 8i
=> (b-2a)+(ab+2)i = 8i
Comparing real/imaginary parts, b-2a = 0 and ab+2 = 8
i.e. b = 2a and a(2a) + 2 = 8
i.e. 2a = 3
i.e. a = - 3 (since a < 0) and b = - 2 3
w = - 2 3 - 2i => arg(w) = 5
6
π−
Hence, arg(wn) =
5
6
nπ−
∴Least n = 3
(then argument is 5
2
π−, so that w
n is of form – ki with k > 0).
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 2 of 9
Qn
3
( )( )
1
2
1
2
1
2
2
2
1(4 )
4
(4) 14
311 1 2 2
1 ...2 2 4 (1)(2) 4
1 3 1
2 8 128
xx
x
x x
xx
−
−
−
= −−
= −
−− = + − − + − +
≈ + +
( )
( ) ( )
2 2 2
0 0
22
0
22
3
0
2
3
1 1 2 3d 1 2 d
2 8 1284 2
1 3 1 d
2 4 32
1 1
2 8 32
21 1 17 2 1 2 2
2 8 32 32 8
xx x x
x
xx x
xx x
≈ + + −
= + +
= + +
= + + = +
∫ ∫
∫
17 1
,32 8
b a∴ = =
4 Since
2
2
d9 3
d
xx t
t+ = ,
d3x
dt3+ 9
dx
dt= 3 .
When t = 0, x=0, d
d
x
t= 1,
d2x
dt2= 0 ,
d3x
dt3= −6 . Thus the Maclaurin’s series is
3x t t= − +⋯
x = Asin3t + Bt
≈ A 3t −(3t)3
6
+ Bt
= (3A+ B)t −9A
2t3
Equating coefficients:
3A + B = 19A
2= 1
⇒ A =2
9, B =
1
3
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 3 of 9
Qn
5 (i) c = a + b
(ii) ˆ ˆsin sinOB AOB OA OB AOB OA OBOA= × × = = × = ×���� ���� ���� ����
Area a b
(iii) Since ˆsin AOB has maximum value of 1, thus maximum area of OACB is
a b .
(iv) Let
a =
a1
a2
a3
and b =
b1
b2
b3
. We have cosθ= ≤ia b a b a b .
Thus 2 2 2≤ia b a b . Since 1 1 2 2 3 3a b a b a b= + +ia b , we have
( ) ( )2 2 2 2 2 2 2
1 1 2 2 3 3 1 2 3 1 2 3)( b a b a b a a a b b ba + + ≤ + + + + .
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 4 of 9
6 i
ii
iii
7 7(i) ( ) ( ) 01222f >−=−=′ xxx since x > 1
So f is strictly increasing on its domain and therefore one-one.
(ii) f(x) = ( )x1f − ⇒ f(x) = x since the graphs of y = f(x), y = ( )x1f − and y
= x all meet at the same point(s).
Thus 222 +− xx = x
( )( ) 021 =−− xx
x = 2 or x = 1 (reject since x > 1)
Hence x = 2.
x
y
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 5 of 9
(iii) y
y = f-1(x)
y = f(x)
Soln set = (1,2]
O 1 2 x
(iv) Since Rf = (1,∞) ⊂ [–3, ∞) = Dg, gf exists.
gf(x) = g( 222 +− xx )
= ( )3222 ++− xx
= ( )522 +− xx (x > 1)
Therefore, ( )2 2 5gf : , 1x xx x− + >֏
Rgf = (2,∞).
8 (i)
(ii) 2
2
2
2 , 3e
d d d 6 e3 e
d d d 2
t
tt
x t y
y y t tt
x t x
= =
= × = =
At point P, by observation, 1t = .
Equation of tangent at point P:
( )1 1
3e 3e( 2)
3e( 1) (shown)
y y m x x
y x
y x
− = −
− = −
∴ = −
(iii) Coordinates of point R: (0, -3e)
Equation of normal at point P:
( )1 1
1
13e ( 2)
3e
1 23e
3e 3e
y y x xm
y x
y x
− = − −
− = − −
∴ = − + +
Let y = 0: 22 9ex = +
x O
3
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 6 of 9
Coordinates of point Q: ( 22 9e+ , 0)
( )2 2 2 22 (6e) 4 36e 2 1 9ePR = + = + = +
( )22 2
4 2
2
2 9e 2 (0 3e)
81e 9e
3e 1 9e
PQ = + − + −
= +
= +
Area of PQR∆
( )( )
( )( )( )
2 2
2
1
2
12 1 9e 3e 1 9e
2
3e 1 9e (shown)
PR PQ =
= + +
= +
(iv) For the line 3ey mx= − to intersect the curve at 2 distinct points, the
absolute value of m must be greater than the tangent to the curve at
point P :
3em∴ >
9 (i) Asymptotes needed: x =
1
4 and
y = ax – 1.
(ii)
( )2
1 , , 04 1
d 4 =
d
1
4
4 1
ay ax x a
x
y aa
x x
= − + ≠ >−
−−
( )
( )
2
2
d 4 = 0 when = 0
d 4 1
i.e. 4 1 = 4
3 1 = or
4 4
y aa
x x
x
x
−−
−
−
Coordinates of turning points:
3 5 1 3, 1 or , 1
4 4 4 4a a
− − − −
(iii) (a)
x O 3
y
P
R
Q
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 7 of 9
(b) The line 1 1
1 = 4 4
y a m x − − −
passes through the point of
intersection between the 2 asymptotes. For the 2 graphs to intersect, m >
a.
10 (i) ( )(100 )
dxk x x
dt= −
At 0t = , 1, 1dx
xdt
= =
11 (1)(99)
99k k= ⇒ =
therefore 21(100 )
99
dxx x
dt= −
(ii) 21(100 )
99
dxx x
dt= −
1 1d d
(100 ) 99x t
x x=
−∫ ∫
1 1 1 1d d
100 100 99x t
x x+ =
−∫ ∫
100ln ln(100 )
99x x t C− − = +
100ln
100 99
xt C
x
= + −
100 100
99 99
100
t C txe Ae
x
+= =
−
When 0t = , 1x = , 1
99A =
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 8 of 9
100 100
99 99
100 100
99 99
100 100
99 99
100
100 100
1 99
t t
t t
t t
x A e xAe
A e ex
Ae e
= −
= =
+ +
(iii)
(iv) Using GC, 5.64 years.
(v) The farmers may be influenced by adoption of innovation from other
sources, eg mass media, besides farmers.
Or any other reasonable answer
11 5 5 5 2( 2) 32 0 ( 2) 2 k ie πω ω− − = ⇒ − =
( )25 5 5 5
5
2 2
4cos 0, 1, 2
2
5,
k i k i k i k i
k i
e
ke
e e e
k
π π π π
π
ω
π
−= +
= =
+
± ±
=
(i) 2 2z − =
(ii) 352 2i
c eπ
= + .
(iii) By symmetry, 25
BAWπ
=∡ . Thus 2 in5
2sBWπ
=
.
100
0 t
x
O Re
Im
4 2
1W
4
2W
4
35π
4
A
B
25π
4
C
1
2010 NYJC JC2 Prelim 9740/1 Solutions
Page 9 of 9
12 (a)(i)
d
dx(1− x2 )n = −2xn(1− x2 )n−1
(ii)
( ) ( )3 2 2 21(1 ) d 2 1 d
2
− = − − − ∫ ∫
nnx x x x x x x
( )
( )
2 2
12
dLet = - , 2 1
d
1d 2 ,
d 1
+
= − −
−= − =
+
n
n
vu x x x
x
xux v
x n
( )2 1
2 1
23 2 2 1
22 1
22 2
2 1
(1 11 2 (1 ) d
2( 1) 2( 1)
(1 1(1 ) d
2( 1) 1
(1 1 1(1 )
2( 1) 2 ( )( )
)
)
1 2
)
+
+
+
+
+
+
−∴ − = − − − ⋅ −
+ +
−= − + ⋅ −
+ +
−= − − − +
+ + +
∫ ∫
∫
n
n
n
nn
n
n
x xx x dx x x x
n n
x xx x x
n n
x xx C
n n n
(b)(i) Area of A =
23π4−π4
− sin x + cos x dx
π4
3π4
∫
=π 2
2− sin x − cos x[ ]π
4
3π4
=π 2
2− 2
(ii) By translation, req’d area = π 2
2− Area of A = 2
(iii) Volume = π ( 2 )2π2− π (sin x + cos x)2 dx
π4
3π4
∫
4.93=
2010 NYJC JC2 Prelim 9740/2 Solutions
Page 1 of 9
Qn
1 (a)(i) Since 1
1
(1 ( ) ) 100(1 )
1 1
r n rnn
n k r rk
d e ex d
e e=
− −= = =
− −∑ .
(a)(ii) In order for population to stabilise at K, the sum of the geometric
sequence must converge. Thus 1re < . Hence 0r < .
100
lim1
n rn eK x
→∞= =
−.
(b)(i) Let lim nn
y y→∞
= . Taking limits
1lim lim 2 (1 )
2 (1 )
10,2
n n nn n
y y y
y y y
y
+→∞ →∞= −
⇒ = −
⇒ =
Using GC, ny is an increasing sequence, thus 1
2y = .
Thus, the limiting population is 500.
(b)(ii) Need K = 500. Thus
100 4
500 ln1 5r
re
= ⇒ = − .
2
Using similar triangle,
2
2
c c h
c x
xh c
−=
= −
( )
2
2
Volume, V2
22
xx c
xV c x
π
π
= −
= −
( )
2
2 2
Surface area, S 2 22
2 2
2
xx x c
x xc x
S x x c
π π
π π π
π
= + −
= + −
= +
x
2c
c h
2010 NYJC JC2 Prelim 9740/2 Solutions
Page 2 of 9
Qn
dAt max/min V, 0 :
d
V
x=
( ) ( )
( )
2d1 2
d 2
22
4 32
V xx c x
x
xx c x
xc x
ππ
π
π
= − + −
= − −
= −
( ) 4 3 02
xc x
π∴ − =
4 or 0 (rejected)
3
cx x= =
( ) ( )
( )
( )
2
2
d3 4 3
d 2 2
3 4 32
2 3
V xc x
x
x c x
c x
π π
π
π
= − + −
= − + −
= −
2
2
4 d 4when , 2 3 0
3 d 3
c V cx c
xπ = = − <
Therefore, volume V is maximum when 4
3
cx = .
2
3
4 1 4max volume
3 2 3
16max
27
c cc
cV
π
π
= −
=
( )Surface area, S 2x x cπ= +
( ) ( )( ) 2
2
S is a strictly increasing function for 0
S is maximum when is maximum:
Since 2
S 2 2 2 2 8
S 8
x
x
x c
x x c c c c c
c
π π π
π
>
∴
<
= + < + =
∴ <
2010 NYJC JC2 Prelim 9740/2 Solutions
Page 3 of 9
Qn
3 3(i) Dividing the reurrence relation by r gives
1
1
r ru ur
r r
+ − =+
.
Summing both sides from r = 1 to r = n – 1, we have
1 1
1
1 11
n n
r r
r r
u ur
r r
− −+
= =
− = + ∑ ∑
2 1
2 1
u u−
+ 3 2
3 2
u u− = ( )1
1 12
nn
−+ −
⋮
+ 1
1
n nu u
n n
−−−
[Note that LHS is a telescopic series and RHS is an arithmetic series]
Thus 1
1
nu u
n− =
( )12
n n − which gives
( ) ( )21
1 22 2
n
n n nu n n n
− = + = − +
as desired.
(ii) Let Pn denotes the proposition in (i)
For n = 1, LHS = 1 1u =
RHS = ( )211 1 2 1
2− + = = LHS.
So P1 is true.
Assume Pk is true for some k = 1, 2, 3, ….
That is, ( )2 22
k
ku k k= − + ------ (IH)
To prove ( ) ( )2
1
11 1 2
2k
ku k k+
+ = + − + + .
For n = k + 1,
LHS = 1ku + = ( )2 1k
kk u
k
++ by the recurrence rel
n
= ( )2 2 12
2
k kk k k
k
+ + − + by (IH)
= ( )212
2
kk k
++ +
= ( ) ( )211 1 2
2
kk k
+ + − + + = RHS.
So Pk+1 is true.
Since P1 is true and Pk true ⇒ Pk+1 is true, by the principle of MI, Pn is
true.
2010 NYJC JC2 Prelim 9740/2 Solutions
Page 4 of 9
Qn
(iii) 3
nu
n =
2
2
1 2
2
n n
n
− +
= 2
1 1 2 11
2 2n n
− + →
as n→∞ .
4 4(a)(i) Equation of l is r = i – j + k + λ(2i + j – 2k).
Substitute r =
λ−
λ+−
λ+
21
1
21
into equation of p gives
λ−
λ+−
λ+
21
1
21
•
1
1
1
= 3
⇒ 1 + 2λ – 1 + λ + 1 – 2λ = 3 ⇒ λ = 2.
Therefore
−
=
3
1
5
OB .
(ii) A
φ θ B p (side view)
Let φ be the angle between l and the normal vector of p. Taking scalar product of the direction vector of l and the normal vector of
p gives
− 2
1
2
•
1
1
1
= 33 cos φ ⇒ cos φ = 33
1.
[Note that φ is acute since cos φ > 0] Let θ be the acute angle between l and p.
sin θ = sin(90° – φ) = cos φ = 33
1.
(iii) shortest distance from A to p
= AB sin θ =
4
2
4
−
1
3 3
= 6 2
3 3 3= .
2010 NYJC JC2 Prelim 9740/2 Solutions
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Qn
(iv) AB = ( )22 2
4
2 4 2 4 6
4
= + + − = −
length of the projection of AB onto p
=
2
2 2 266 2
33
− =
using Pythagoras’ Theorem
(b) By GC, the equation of the line of intersection of the planes given by
the last 2 equations is
r =
3 2
0 2
0 1
− + λ
.
For the system to have infinite solutions, the above line must lie on the
plane given by the first equation. So the line is perpendicular to the
normal vector of the plane. That is,
1
3
α
•
2
2
1
−
= 0 ⇒ -2 + 2α + 3 = 0 ⇒ α = 1
2−
Furthermore, any point of the line is also a point of the plane. Take
(3,0,0) and substitute into the first equation gives 3 + α(0) + 3(0) = β ⇒ β = 3
5 (a) - The first method does not give a simple random sample as each player
does not have the same chance of being selected.
- A simple random sample can be obtained by numbering all the 363 from
001 to 363 and then picking three digits at a time from a random number
table, ignoring numbers over 363 and ignoring repeats, until a group of
58 numbers is obtained. The players corresponding to these 58 numbers
will be a simple random sample.
(b) Stratified sampling, where the population is divided into homogeneous
groups called strata and random individuals from each stratum are
chosen. Stratified samples can give useful information about each stratum
(in this case, about each of the five neighbourhoods) in addition to
information about the whole population (the city population)
Quota sampling as the sampling frame is not available. The population is
divided into non-overlapping groups and the sample size is determined
without any basis.
6 To test 0 1: 10 vs : 10H Hµ µ= < at 3% level of significance
Assume that the distribution of the amount of coffee in a packet is normally
distributed.
2010 NYJC JC2 Prelim 9740/2 Solutions
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Qn
Under H0, /
10~ (7)T
Xt
ns=
−
Reject H0 if p-value < 0.03
Since 9.6125x = , 0.49117s = , p-value = 0.0304
Since p-value > 0.03, there is insufficient evidence at 3% level of significance
level to reject H0, and we conclude that the factory’s claim could be correct.
To test 0 1: 10 vs : 10H Hµ µ= < at 5% level of significance
Under H0, using Central Limit Theorem, 10
~ (0,1)/
approxX
Nn
Zs
=−
Reject H0 if p-value < 0.05
Since x = 10 −
25.6
64= 9.6 ,
22 1 (614.4 10 64)
151.99 2.2563 64
s − ×
= − =
Thus, p-value = 0.0164
Since p-value < 0.05, there is sufficient evidence at 5% level of significance
level to reject H0, and we conclude that the factory’s claim may not be correct.
Since sample mean remains unchanged and s2 decreases, the p-value will
decrease. Thus we will still reject H0.
7 1(i)
0
10
20
30
40
50
60
70
80
0 10 20 30 40 50
x
There appears to be a curvilinear/non-linear relationship between x and t. r ≈ - 0.920.
(ii) Model (a) is more appropriate because the graph of x = atb, where a > 0
and b < 0 is concave upwards whereas the graph of x = a + bt2, where a > 0
and b < 0 is concave downwards. The points on the scatter diagram fall on a
graph that is concave upwards.
(iii)(a) t will increase by 3 (b)standard deviation of t remains the same (a) r remains unchanged
8 (i) 8! x 2! = 80640 (ii)
8C4 x 4! x 4! or 7! x 8 = 40320
(iii) 4C2 x 2! x 2! x 6! = 17280
(iv) [4C1.
3C1 +
3C1.
2C1 +
4C1 .
2C1] x 2! x 7! = 262080
2010 NYJC JC2 Prelim 9740/2 Solutions
Page 7 of 9
Qn
9 Let X be the number of requests for cars on a particular day. ~ Po(4)X
Let Y be the number of requests for vans on a particular day. ~ Po(2)Y
(i) Let T be the number of requests for vehicles on a particular day. Thus
~ Po(6)T
prob.req d ( 11)
1 ( 11)
0.0201
P T
P T
′ = >
= − ≤
=
(ii) Either demand for a car or a van is not met. Thus
prob.req d ( 7 or 4)
1 ( 7 and 4)
1 ( 7) ( 4)
0.101
P X Y
P X Y
P X P Y
′ = > >
= − ≤ ≤
= − ≤ ≤
=
(iii) The event in (i) is a subset of the event in (ii). Thus the value obtained
in (i) will be smaller.
(iv) Let n be the number of days needed. Assume that n is large. By Central
Limit Theorem, 6
~ N 6, approx.Tn
6
7) 0.001
7 60.001
0.9996
(
n
T
P Z
P
P
nZ
> <
−⇒ > <
⇒ ≤ >
Thus 3.09023 57.36
nn> ⇒ >
Thus least number of days required is 58.
10 (i) P(Chris is late in a day) =
4
5+2
5 − (
4
5)(2
5) =
22
25
Let X be the number of days Chris is late out of 4 days. X~B(4, 22
25).
P(Chris was late exactly thrice in a week | Chris was late on Mon)
= P(Chris was late twice in the remaining four days of the week)
= P(X = 2)
≈ 0.0669 (ii) P(Chris was delayed at A | Chris was late)
= P(Chris was delayed at A)/P(Chris was late)
4
10522 11
25
=
2010 NYJC JC2 Prelim 9740/2 Solutions
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Qn
(iii) P(Chris was delayed at B | he was delayed at exactly one stop)
= P(Chris was delayed at stop B only)/P(delayed at exactly one stop)
=
1 2
15 54 3 1 2 7
5 5 5 5
( )( )
( )( ) ( )( )
=+
11 i) Let X be the no. of floods in a month.
Then, X ~ Po(2), and P(bad month) = P(X ≥4) = 0.142877
Let Y be the no. of bad months in 12 months (note: 1 year = 12 months)
Then, Y ~ B(12, 0.142877)
Hence, P(Y≤2) = 0.76006
ii) Let W be the no. of bad years, out of fifty years.
Then, W ~ B(50, 1 – 0.76006) ~ B(50, 0.23994)
Since n = 50 is large, np = 11.997 > 5 and nq = 38.003 > 5,
W may be approximated using N(11.997, 9.11844).
P(W < 5) → (c.c.) P(W < 4.5) = 0.00652 or 0.00650
iii) It is not suitable since the weather conditions may have changed over
such a long time period and the model may no longer be applicable.
12 (i) Let M be the mass of a randomly chosen Munchi pear.
( )2N 120,10M ∼
Using GC,
( )P 100 126 0.7030M< < ≈
( )P 115 0.3085M < ≈
Required probability ( ) ( )0.7030 0.3085 2! 0.434 (3 s.f.)= × ≈
(ii) Using GC,
( )P 122 0.4207M > ≈
Let S be the number of pears of mass that is more than 126g.
( )B 10,0.4207S ∼
( )P 3 0.196 (3 s.f.)S = ≈
(iii) Let F be the mass of a randomly chosen Fuchi apple.
( )2N 115,8F ∼
2010 NYJC JC2 Prelim 9740/2 Solutions
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Qn
( ) ( )( )( )
2 2
1 2 3
1 2 3
3 N 3 115 3 120 , 3 8 9 10
3 N 15,1092
F F F M
F F F M
+ + − × − × × + ×
⇒ + + − −
∼
∼
Using GC,
Required probability
( )( )
1 2 3
1 2 3
P 3
P 3 0 0.325 (3 s.f.)
F F F M
F F F M
= + + >
= + + − > ≈
(iv) ( )2N 115,8F ∼
28N 115,F
n
⇒
∼
( )P 115 4 0.3F − > ≤
2 2
P( 115 4) P( 115 4) 0.3
4 4P P 0.3
8 8
P 0.72 2
P 0.852
invNorm(0.85) 1.036
1.036 4.2962
Hence, least 5.
F F
Z Z
n n
n nZ
nZ
nn
n
⇒ − < − + − > ≤
< − + > ≤
∴ − < < ≥
⇒ < ≥
=
≥ ⇒ ≥
=
3
1 It is known that the number of diagonals that can be drawn in a polygon of n sides can be expressed as a quadratic polynomial in n. By considering the number of diagonals in a triangle, a quadrilateral and a pentagon, find the number of diagonals that can be drawn in a polygon of 200 sides. [4]
2 Given that f ( ) ln(cos )x x , show that
2f ( ) sec'' x x .
By further differentiation, find the first two non-zero terms in the Maclaurin’s series of f ( )x . Hence obtain an approximate value for
0 4
0ln cos d
.x x . [6]
3 Solve the inequality
2ln( 1) 9x x . Hence, solve the inequality
2ln( 1) 9x x . [5]
4 An athlete at point C is running towards the finishing line AB, at a constant speed
of 10 in a direction perpendicular to AB, as shown in the diagram above. 1ms
A
B
C 10 ms 1
x m 2 m
3 m
(i) Given that ACB is and x is the distance of the athlete from AB, show
that
1 13 2tan tan
x x . [2]
(ii) Find the exact rate of change of when the athlete is 10 m from the finishing line. [3]
[Turn over
4
5 (i) Sketch the graph of 1
y xx
, stating the equations of any asymptotes and
the coordinates of any points of intersection with the axes. [2]
(ii) The curve C has equation
2 2 1p x pqx
ypx q
,
where p and q are positive constants.
State a sequence of transformations which transform the graph in (i)
to the graph of C. [3]
6 Find the first three terms in the expansion, in ascending powers of 1
x, of 3 x
expressing the coefficients in their simplest form. State the set of values of x for which the expansion is valid.
Hence, by substituting 25x , find an approximate value for 7 as a fraction. [6]
7 Find the exact value of
0
6 2 sin dxe x x
. [5]
5 8
The diagram shows the graph of f ( )y x . The curve crosses the x-axis at the
points A and B ( 1,0)5
,02
, and has a minimum point at C and a
maximum point at D (3 . The lines
(1,4)
, 2)1
2y , 0x and 2x are asymptotes to
the curve. Sketch, on separate diagrams, the graphs of
(i) f ( )y x , [3]
(ii) f ( )'y x , [3]
labeling each graph clearly and showing the asymptotes and coordinates of the
points corresponding to A, B, C and D.
[Turn over
y
x 0
2x
1
2y
(3,2)(1,4)
1 5
2
f ( )y x
C
D
A B
6
9 Find the roots of the quartic equation 4 2 i2 3 0z . Give your answers
exactly, in the form ire , where and 0r . [3] The polynomial P(z) has degree eight and real coefficients. All the roots of the
equation 4 2 i2 3 0z are also the roots of the equation P(z) = 0. By considering P(z) as a product of two quartic factors, find P(z), expressing all the coefficients in real and non-trigonometrical form. [3] 10 A curve is defined by the parametric equations
2,x e y tt t .
Find an expression for d
d
y
x in terms of t. [1]
(a) Given that the tangent to the curve at the point with parameter p passes through the origin, find the exact values of p. [3]
(b) Given that 2
2
d d dy dd d dd
y xt x tx
, deduce the exact range of t for which the
curve is concave upwards. [3] 11 A seed from a type of tree was planted and the change in the height of the tree is noted at the end of each year. It is observed that the change in height of the tree follows an arithmetic progression in the first 10 years of growth, and subsequently, follows a geometric progression from the end of the 10th year. Given that the tree’s change in height was 40 cm and 70 cm at the end of the 3rd year and 5th year of its growth respectively, find (i) the change in height of the tree at the end of the 4th year of its growth, [1] (ii) the height of the tree at the end of the 10th year of its growth. [3] Given that the tallest the tree can ever grow is 20 m, find the year at which the tree is at least twice as tall as at the end of 10th year of its growth. [6]
7
12 By using substitution 4 xu e or otherwise, show that
ln
0
1 1 5 d ln
4 44
p
xp
xpe
. [4]
Hence, (i) find the exact volume of the solid of revolution formed when the region
bounded the curve 2 1
4 xye
, the y-axis and the line is rotated
through 2 right angles about the x-axis, [3]
ln 6x
(ii) by considering the area under the curve 1
4 xye
, find the exact value of
1 5
1 8
1 4ln d
yy
y
. [3]
13
treadmill O
jogging direction
x
A woman starts to jog at a rate proportional to the distance x from O, along a
treadmill which is running at a constant rate of a 1ms , as shown in the diagram above. At time t seconds, the distance she has moved is x m. When her distance from O is 1 m, her position remains stationary. Show that
d1
d
xa x
t .
[3]
Given that at t = 2s, she is moving at 21
e ms 1 , find the speed of the treadmill.
[7] Hence find the exact distance she has further moved after another 2 seconds. [2]
[Turn over
8
14 The planes 1p and 2p have equations
2
1
1
4
r. and
1
2
r. respectively,
where and are negative constants. 1p and 2p are inclined at to each
other. The point A with position vector i − 2j, where
60
is a constant, is in both
1p and 2p .
(i) Find the position vector of A. [1] (ii) The point B has position vector 3i − 3j + 2k. Find the shortest distance
from B to 1p . [3]
(iii) Find the equation of 2p in the form dr.n . [4]
The plane 3p has equation x y z , where and are constants.
(iv) Find the values of and if 3p passes through the origin and 1p ,
2p and 3p have only one common point, which is also the foot of
perpendicular from A to 3p . [3]
(v) What can be said about the values of and if 1p , 2p and 3p have no
common point and all points on 3p are equidistant from 1p ? [2]
3
Section A: Pure Mathematics [40 marks]
C
l cm
B
D
x cm
10 cm
5 cm A
1 A student wants to draw a straight line of length l cm that begins at the side AB, ends at the side BC and passes through the point D, as shown in the diagram above. (AB and BC may be assumed to be infinitely long.) (i) By using similar triangles, show that
22 2
225
10
xl x
x
. [3]
(ii) Hence find the length of the shortest straight line he can possibly draw. [3] 2 The function f is defined by
21
f : ,2
bx a x
x ,
where a and b are positive constants.
(i) Sketch the graph of f and state range of f. [2]
(ii) Find . [3] 1f
(iii) Given that
2
21
gf : ,2
bx x
x ,
find g( )x in terms of a and b. [2]
[Turn over
4
3 The diagram above shows a partial design of the roof of the gallery of an amphitheatre. ABCD is an inclined rectangular roof, where AB = 10 m, and . EFGH is a rectangle on a horizontal ground, where EF = 10 m and
5 mBC 4 mEH . The points A and B are 3 m directly above E and F respectively. The
points C and D are 6 m directly above G and H respectively. Point O is the centre of EFGH and is taken as the origin. Perpendicular unit vectors i, j, k are in the direction of EF
, EH
and EA
respectively.
(i) Show that and find a vector equation of the line passing
through A and C. [3]
10
4
3
AC
(ii) A spot light is fixed at the top of a vertical pillar erected at point Q with coordinates 15,6,0 . Find the height of the pillar if the top of the
pillar is collinear with A and C. [3] (iii) The main electrical supply to the gallery runs along the diagonal AC. Find the position vector of point X on AC such that the length of the electric cable that is required to provide electricity from the main electrical supply to an amplifier mounted at D, will be the least. [3]
A
10 m
O i
jk
D C
5 m6 m
10 m6 m
BH G
3 m 4 m
E F
3 m
5
4 A sequence { }nx of negative numbers is defined by 0 2x and
1
11
2
n
n nx x
for n . 1
(i) Prove by mathematical induction that 1
4 11
3 2
n
nx
for . [4] 0n
(ii) The sequence { }nx converges to l as n tends to infinity. State the exact
value of l. [1]
(iii) Determine, with a reason, whether 1
N
nn
x converges. [1]
(iv) By considering 1n nx x , deduce that 1n nx x if n is odd and 1n nx x
if n is even. [2] 5 Sketch on an Argand diagram the set of points representing all complex numbers
z satisfying both the inequalities
i 2i 2z 2 and Re 1 3 iz . [4]
Find
(i) the range of , [2] arg 2 2iz (ii) the complex number z where arg 2 2iz is a maximum. [2]
The locus of the complex number w is defined by 5 2iw k , where k is a
real and positive constant. Find the range of values of k such that the loci of w and z will intersect. [2]
[Turn over
6
Section B: Statistics [60 marks] 6 (a) The Royal Club has membership of 1600 people, of which 1000 are males and 600 are females. The boss wants to find out what members think of the new set meals menu offered by the Oiishi restaurant of the Club. A sample of 80 members is to be chosen. State a suitable sampling method and describe in detail how this can be carried out. [3] (b) The amount of tips a member gives after a meal in Oiishi restaurant of Royal Club has mean $5 and standard deviation $1. Estimate the probability that the amount of tips collected from 80 randomly chosen customers is between $350 and $410. [2] 7 In an experiment, a new computer game and a new mathematics quiz are given to a group of teenagers. It may be assumed that the teenagers are playing the computer game and attempting the mathematics quiz for the first time. The computer game score, u, and the mathematics quiz score, v, of 8 teenagers are given in the table below. Teenager 1 2 3 4 5 6 7 8
Computer game score, u 52 90 64 74 80 82 76 74 Mathematics quiz score, v 60 90 68 74 74 90 82 70
(i) Find the linear product moment correlation between v and u and the equation of the regression line of v on u. [2] (ii) Comment on the suitability of using the regression line in (i) to predict u given v. Use the appropriate regression line to predict u given that v is 85, giving your answer to the nearest whole number. [3] The scores are actually recorded for 9 teenagers. However, the scores for the last teenager are lost.
Teenager 1 2 3 4 5 6 7 8 9 Computer game score, u 52 90 64 74 80 82 76 74 p Mathematics quiz score, v 60 90 68 74 74 90 82 70 q
It is known that the inclusion of the scores of the last teenager does not alter the mean computer game score and one of the regression lines is given by . 45.556 0.395u v (iii) Find p and q, giving your answers to the nearest whole number. [3]
7
8 A taxi company wants to track the number of morning, afternoon and night trips made in a day. Of the total trips in a day, p% are morning trips, q% are afternoon trips and the remaining are night trips. Past records indicated that the probability that a randomly chosen morning, afternoon and night trip is paid by NETS is 0.85, 0.65 and 0.25 respectively. (a) If and , find the probability that a randomly chosen trip is paid by NETS. [3]
30p 50q
(b) Given that the probability of a randomly chosen trip paid by NETS is taken in the morning is 0.5, express q in terms of p. [4] 9 Defective spots are found randomly on a roll of ribbon. On average, 5 defective spots are found per 10 m of ribbon. The roll of ribbon is cut into short ribbons, each of length 1 m. A short ribbon is not discarded if it contains at most 1 defective spot. (i) State a condition under which a Poisson distribution would be a suitable probability model and find the percentage of short ribbons that do not need to be discarded. [3] (ii) Given 52 randomly chosen short ribbons, find an approximate probability that there are more than forty-five short ribbons that do not need to be discarded. [4] (iii) Determine the maximum number of short ribbons required for which the probability of at least 3 short ribbons discarded does not exceed 0.1. [2] 10 A six-digit number is to be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9. For each of the following cases, find how many different ways the six-digit number can be formed. (i) The even and odd digits of the number must alternate and any digit may appear more than once. [2] (ii) The number must be odd and is less than 600 000 and no digit may appear more than once. [3] (iii) The number is formed from four different digits, eg. 621313, 255567. [4]
[Turn over
8
11 A star fruit plantation claims that it has developed a method of producing star fruits that are larger in size with a mean weight of 200 grams. A wholesaler suspects that the plantation is overstating the weight of the star fruits produced and decides to test the plantation’s claim. The wholesaler obtains a random sample of 10 star fruits. The weights (to the nearest gram) of the 10 star fruits are as follows:
189 192 200 205 206 198 188 190 200 192 Find an inequality satisfied by the significance level that will confirm the wholesaler’s suspicion. State an assumption made in conducting the test. [5] It is later known that the standard deviation of the weight of the star fruits produced by the plantation is 6 grams. Another random sample of 10 star fruits is obtained. A second test is conducted at 2% significance level and the wholesaler is unable to confirm his suspicion. Find, to the nearest gram, the range of values for the mean weight of these 10 star fruits in order to arrive at this conclusion. [4] Explain, in the context of the second test, the meaning of ‘at 2% significance level’? [1] 12 The weight, Y, of a Yummy cereal bar is normally distributed with mean (120 – k) g and standard deviation 10 g. The weight, F, of a Fullness cereal bar is normally distributed with mean 180 g and standard deviation 20 g. (i) Given that P(Y < 80) = P(Y > 150), show that the value of k is 5. [1] (ii) 5 Yummy cereal bars are randomly chosen. Find the probability that exactly one bar weighs lesser than the lower quartile weight and exactly one bar weighs more than the median weight. [2] (iii) Find the probability that the weight of 2 randomly chosen Yummy cereal bars differs from one and a half times the weight of a randomly chosen Fullness cereal bar by at most 5g. State the assumption needed for your working. [4] The Fullness cereal bars are sold at $2 per 100g. (iv) Find the probability that a randomly chosen Fullness cereal bar costs more than $3.50. [2] (v) A random sample of 100 Fullness cereal bars is to be taken. Using a suitable approximation, find the probability that there are less than 65 bars with each costing more than $3.50. [3]
1
RI 2010 9740/01/10 [Turn over
2
RI 2010 9740/01/10
1 By writing i , ,z x y x y , solve the simultaneous equations
2 2 0z zw and *1 i
wz
, where *z is the conjugate of z . [3]
2 Sketch the graph of exy x . By adding a suitable graph to the sketch, find the set of
values of x that satisfies 2
22 5 2exx
x . [4]
3 Four friends returned from a trip to Europe and converted their foreign currencies back to
Singapore Dollars. The amounts of foreign currencies converted and the total amounts
received in Singapore Dollars are shown in the following table.
Donald Leonard Michael Raphael
Sterling Pound 36 55 40 k
Euro Dollar 77 18 31 59
Swiss Franc 42 63 26 24
Total amount in
Singapore Dollars
269.90 233.45 175.50 313.00
Assuming that, for each foreign currency, the exchange rate quoted for each of the friends
is the same, calculate the value of k. [4]
4 Find (a) 2
1 d ,
1 2x
x x [2]
(b) 2
2
3 d .
( 2)(1 )
x xx
x x
[3]
3
RI 2010 9740/01/10 [Turn over
5 A sequence 0 1 2, , ,u u u is defined by
0 1u and 1n nu ru d for 0n
where r and d are non-zero constants.
(i) If r = d = 1, write nu in terms of n. [1]
If 1,r
prove by induction that 11 1
nn
d du r
r r
for all non-negative integral
values of n. [4]
(ii) If 1,r state the limit of the sequence as n . [1]
6 The diagram shows the graph of f ( )y x . The curve has a maximum point at (0,2) and it
cuts the x-axis at the points ( ,0)a and ( ,0) where is a positive constant.a a The lines
2x , 2x and 2y are asymptotes to the curve.
Sketch, on separate diagrams, the graphs of
(i) 2 f ( ),y x [3]
(ii) 1
.f (2 )
yx
[3]
Your sketches should show clearly the equations of asymptotes, stationary points and
intersection with the axes, where applicable.
x
y
2y
2
2x 2x
f( )y x
a aO
4
RI 2010 9740/01/10
7 The curve C has equation f ( ) 2 11
ay x x
bx
where a and b are positive constants.
(i) Show that C has exactly two stationary points. [3]
(ii) Given that C passes through the point (0,3) , find the value of a. [1]
Hence sketch the graph of f '( )y x , indicating clearly any intersection with the axes
and the equation(s) of asymptote(s) of the curve in terms of b. [3]
8 A cylindrical water tank has a horizontal base with a fixed area of 2 mA and is initially
empty. Water is poured into the tank at a constant rate of 3 15 m s , and leaks out through a
small hole in the base at a rate which is proportional to the depth of water in the tank. The
depth of water in the tank is x metres at time t seconds. Show that
d
5d
x
A kxt
, where k is a positive constant. [2]
Solve the differential equation, expressing x in terms of , and A k t . [5]
9 (a) Solve the equation 6 4 2(1 i)z , expressing the solutions in the form ier , where
0r and . [3]
(b) (i) The two complex numbers 2 3 + 2i and 2 3 + 2i satisfy the cubic
equation 3 2
3 2 1 0 0a z a z a z a . Explain clearly whether it is possible for all
the coefficients 3 2 1 0, , ,a a a a to be real numbers. [1]
(ii) In an Argand diagram, the point A represents the complex number 2 3 i .
If A, B, and C are the vertices of an equilateral triangle taken in clockwise order,
and these three points lie on a circle with centre at the origin, find the complex
number represented by B in the form ip q where ,p q , giving the exact
values of p and q. [3]
5
RI 2010 9740/01/10 [Turn over
10 The curve C has equation e xy x for 0x and ( , e )aP a a is a point on C .
(i) Sketch the curve C . [1]
(ii) Find, in terms of a , the equation of the tangent to the curve at P . [2]
This tangent cuts the y axis at the point (0, )Q h .
Using differentiation, find, as a varies, the exact maximum value of h . [6]
11 The equations of two planes p1 , p2 are
2x 5y + 3z = 3,
3x + y + 6z = ,
respectively, where and are constants.
(i) Given that the two planes intersect in a line l, with a vector equation given by
r =
4 2
2 1 , ,
5 3
s s
show that the value of is 12 and find the value of . [3]
A third plane p3 has equation given by
5x + 8y + tz = 12,
where t is a constant.
(ii) With the values of and found in (i), find the exact value of t if the three planes
have no point in common. [2]
(iii) The plane p4 contains the line l and the point (1, 1, 2). Find the cartesian equation of
p4 and the acute angle between p1 and p4. [5]
6
RI 2010 9740/01/10
12 Relative to the origin O, three points A, B and P have position vectors given by
a = 14i + 8j + 6k, b = 11i + tj + 12k and p = 12i 4j + 10k respectively, where t is a
constant.
(a) If P divides the line segment AB in the ratio : 2, find the values of and t.
[3]
(b) It is given that t = 2.
(i) Find the exact length of projection of PB onto AB .
Deduce the exact shortest distance of P from line AB. [4]
(ii) BAPQ forms a parallelogram. Find the position vector of the point Q and the
area of the parallelogram correct to 2 decimal places. [4]
13 Given that 2
4 1f ( ) , .
3(1 3 ) 4x x
x x
(i) Express f ( )x as a series expansion in ascending powers of x, up to and including
the term in 3x . [4]
(ii) Find the range of values of x for which the expansion in (i) is valid. [2]
(iii) A curve is given by the equation
2tan f ( ).y x
By differentiation, find the series expansion for y in ascending powers of x, up to an
including the term in 2x . [5]
7
RI 2010 9740/01/10 [Turn over
14 (a) The sum of the first n terms of a series, Sn , is given by 55 1
n
np , where p is a non-
zero constant and 5p .
Obtain an expression for Tn , the nth
term of the series and prove that this is a
geometric series.
Find the range of values of p for the sum to infinity to exist. [5]
(b) An arithmetic progression is grouped into sets of numbers as follows:
{2} , {6 , 10} , {14 , 18 , 22} , {26 , 30 , 34 , 38} , …
where the number of terms is 1 for the first set, 2 for the second set, 3 for the third set,
4 for the fourth set and so forth.
(i) Show that the first term in the nth set is 22 2 2n n .
(ii) Find, in terms of n, a simplified expression for the last term in the nth set.
(iii) Hence, find the sum of all the terms in the nth set. [5]
1
RI 2010 9740/02/10 [Turn over
2
RI 2010 9740/02/10
Section A : Pure Mathematics [40 marks]
1 (i) Given that 2 2 2 2
1 1
2( 1) 2 ( 1)
n a
n n n n
, show that
1
2a . [1]
(ii) Given that 2 2
2 1
2 ( 1)
N
N
n M
nS
n n
, state the smallest possible value of M, where M
and M N , such that NS can be defined. [1]
(iii) If M = 3, find NS in terms of N. [3]
(iv) Deduce that the sum to infinity of the series
2 2 2
1 1 1...
(2)(3) (3)(4) (4)(5)
is less than 1
8. [3]
2 The point A represents a fixed complex number a where arg( )4 2
a . The complex
numbers ia and a are represented by the points B and C respectively.
On a single clearly labelled Argand diagram, show the points A, B, C and the set of points
representing all complex numbers z satisfying both relations
iz a and arg( ) arg( ) .
4z a a
[4]
Find
(i) the minimum value of | |z in terms of | |a , [2]
(ii) the range of values of arg( )z in terms of arg( )a . [2]
3
RI 2010 9740/02/10 [Turn over
3 (a) Functions f and g are defined by
f : ( 2)( 4)x x x , for x , 4x ,
g :3
xx
x , for 3x \ .
Determine if f is one-one, justifying your answer. [1]
Only one of the composite functions fg and gf exists. Give a definition (including the
domain) of the composite that exists, and explain why the other composite does not
exist.
Find the range of the composite that exists. [6]
(b) The function h is defined by
h : (1 e )xx a , for 0x , where a is a positive constant.
Find h 1 (x) and state the domain of h 1 . [3]
Given that the graphs of 1h( ) and h ( )y x y x intersect at the point where x = b, and
that
0 h( ) d
b
x x I , find the area bounded by the curves1h( ), h ( )y x y x and the
axes, giving your answer in terms of I and b. [2]
4 The curve C has equation f ( )y x , where 2
4f ( ) ,
1
xx x
x
.
(i) Find the exact value of 1
1f (| |) dx x
. [2]
(ii) Using the substitution 23 e 3u x , find 2 ln(3 e 3) dx x .
Hence find the exact area of region R bounded by C, the curve 2ln(3 e 3)y x and
the line 0y . [7]
(iii) Find the volume of solid formed when R is rotated completely about the x-axis,
giving your answer correct to 2 decimal places. [3]
4
RI 2010 9740/02/10
Section B : Statistics [60 marks]
5 (a) Find the number of three-letter code-words that can be formed from the letters of the
word WHYOGEE. [3]
(b) A country is invited to send a delegation of six youths selected from six badminton
players, six tennis players and five football players to participate in the opening
ceremony of the Singapore 2010 Youth Olympic Games. No youth plays more than
one game. The delegation is to consist of at least one, and not more than three players
selected from each sport.
(i) Find the number of ways in which the delegation can be selected. [3]
During the ceremony, the youths from the delegation are to be seated in six out of ten
chairs which are arranged in a row.
(ii) Find the number of ways this can be done if no two empty chairs are adjacent.
[2]
6 A six-sided die is thrown. The probability of getting a ‘1’ is p , where 1
015
p , and the
probabilities of getting ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’ are consecutive terms of an arithmetic
progression with common difference p .
Show the probability of getting ‘6’ is 1 15p . [2]
The die is thrown twice and events A and B are defined as follows:
A : the sum of the scores of the two throws is at least 10;
B : the score of the first throw is 5.
Find
(i) P(A) , [3]
(ii) P(A|B). [2]
Hence, show that there is no value of p for which A and B are independent. [2]
5
RI 2010 9740/02/10 [Turn over
7 A researcher investigates the relationship between the stress level of a person and his job
performance. The table below shows the findings for 8 different workers for the same job.
Stress level, x 2 2.5 3 3.5 4 5 6 7
Performance, y 3.30 3.10 3.00 2.86 2.72 2.60 2.55 2.50
(i) Draw a scatter diagram for the data and calculate the product moment correlation
coefficient between x and y. [2]
(ii) Comment on whether a linear model is appropriate. [2]
(iii) The researcher proposes that a model of the form
where and are positive constantsb
y a a bx
is more appropriate.
Explain why he thinks this is so. [1]
(iv) Calculate the least square estimates of a and b for the model in (iii). [2]
(v) Estimate the performance when the stress level is 4.5. Comment on the reliability of
the estimate. [2]
8 The mass of a male student in Aishan Secondary School is denoted by X kilograms. The
masses of a random sample of 150 male students are summarized by
8400x and 2( 56) 5555.x
(i) Calculate unbiased estimates of the mean and variance of X. [2]
Aishan Secondary School claimed that the mean mass of a male student in the school is 55
kilograms.
(ii) Test, at the 3% significance level, whether the school is understating the mean mass
of a male student. Does the Central Limit Theorem apply in this context? [5]
(iii) State what you understand by the expression ‘at the 3% significance level’ in the
context of the question in (ii). [1]
In a separate study, the opinions of female students in Aishan Secondary School are to be
collected to determine if they are satisfied with their own weights. Describe how a quota
sample of size 100 might be obtained. [2]
6
RI 2010 9740/02/10
9 Every weekday, the last train from Bedok Station leaves the station at 11.29 pm. Its travel
time X (in minutes) from Bedok Station to Redbridge Station may be taken to follow a
normal distribution with mean 30 and variance 9.
(i) Find the travel time exceeded by 90% of the trips made by the last train. [2]
At 11.40 pm every weekday, Abel walks from his workplace to Redbridge Station to catch
the last train. The time Y (in minutes) he takes follows a normal distribution with mean 16
and variance 4.
(ii) Find the probability that the train arrives at Redbridge Station after 12 am and Abel
arrives before 12 am. [2]
(iii) Find the probability that Abel will not miss the train. [3]
(iv) Explain why the answer to part (iii) is greater than the answer to part (ii). [1]
The time W (in minutes) taken by Dina to walk from her workplace to Redbridge Station
has a mean of 4 and a variance of 6.
(v) Explain why W is unlikely to be normally distributed. [1]
(vi) Find the probability that the mean time taken by Dina to walk from her workplace to
Redbridge Station, in a random sample of 40 trips, is less than 3.5 minutes. [2]
10 Incoming telephone calls to the management office of a shopping mall are received
randomly and independently, at an average rate of 6.75 per hour. The mall (including its
management office) is open from 10 am to 10 pm.
(i) State, in the context of the question, a condition required for a Poisson distribution to
be a suitable model for the number of incoming calls from 10 am to 10 pm. [1]
Assume that the condition in (i) is satisfied.
(ii) Find the probability of receiving no fewer than 8 calls in a particular hour. [2]
(iii) In 4 non-overlapping one-hour periods, find the probability of receiving at most 6
calls in a one-hour period, exactly 7 calls in another one-hour period, and no fewer
than 8 calls in each of the 2 remaining one-hour periods. [3]
(iv) The number of incoming calls from 10 am to 10 pm, Y , has mean and standard
deviation . Use a suitable approximation to find P Y , giving your
answer correct to 4 decimal places. [4]
(v) A day is considered busy if there are more than 90 incoming calls received from
10 am to 10 pm. Find the probability that the 3rd
busy day in a month occurs on the
15th
day of the month. [3]
__
Q1 [3
Q2 [4
Q3 [4
_________
] Let z =
2
2
( i2
z zx yx
+
+
+On com
22x −1x = ±
When When
]
From tUsing Theref
] Let x, yrespecand 1 S36x +55x +40x +Using
59kx +
k∴ =
RAFFLES H2 Mathem2010 Year ________
20
= x + iy. Su
2 2
2
( * (1 i))) (2i (
z zy x y
xy x
+ −
+ +
+ + +mparing real2 0, 2xy= +1, 1y = m .
1 iz = − , w 1 iz = − + ,
2x +
2
25x
⇒ −
2
5 12 x
⇒ −
the diagram,G.C., x ≤
fore solutiony, z be the extively (i.e. 1Swiss Franc77 42y z+ =18 63y z+ =31 26y z+ =the GC, x = 9 24 3y z+ =313 59(1.7
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ubstitute the s
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2e 2x x≤ −
2 ex x≤ −
, or x α≤2.18 or ≤ −
n set is ( ,−∞xchange rate Sterling Po = z Singap269.9= 233.45 175.5 2.15, y = 1.7
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__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 2 of 8
Q4 [5] (a)[2]
2 2
1
1 1d d
1 2 2 ( 1)
1sin
2
x xx x x
xc−
=+ − − −
−⎛ ⎞= +⎜ ⎟
⎝ ⎠
∫ ∫
(b)[3] 2
2 2
1
3 1 1d d
1( 2)(1 ) 21
tan ln |1 |2 2
x xx x
xx x xx
x c−
− += +
−+ − +
⎛ ⎞= − − +⎜ ⎟
⎝ ⎠
∫ ∫
Q5 [6] (i) [1]
1nu n= +
(ii) [4] Let nP be the statement 01 for
1 1n
nd du r n
r r+⎛ ⎞= + − ∈⎜ ⎟− −⎝ ⎠
Z .
When n = 0, LHS: 0 1u =
(given)
RHS: 0 1 11 1
d drr r
⎛ ⎞+ − =⎜ ⎟− −⎝ ⎠
0P is true. Assume kP is true for some ,
i.e. 11 1
kk
d du rr r
⎛ ⎞= + −⎜ ⎟− −⎝ ⎠
Prove that 1+kP is true,
i.e. to prove 11 1
1 1k
kd du r
r r+
+⎛ ⎞= + −⎜ ⎟− −⎝ ⎠
1
1
1
1
= 11 1
= 11 1
= 11 1
= 1 1 1
k k
k
k
k
k
u ru d
d dr r dr rd rdr d
r rd rd rd dr
r rd dr
r r
+
+
+
+
= +
⎛ ⎞⎛ ⎞+ − +⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠⎛ ⎞+ − +⎜ ⎟− −⎝ ⎠
− + −⎛ ⎞+ +⎜ ⎟− −⎝ ⎠⎛ ⎞+ −⎜ ⎟− −⎝ ⎠
Since kP is true implies 1+kP is true, and 0P is true, by mathematical induction, nP is true for all
0n +∈Z .
(iii) [1] As ,
1ndn u
r→∞ →
−
∴
k +∈Z
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 3 of 8
Q6 [6] (i) [2]
(ii)[2]
Q7[7] (i)[3]
22 11
y xbx
= + ++
( )2
d 22d 1y bx bx= −
+
When d 0,dyx=
( )2
2 21
bbx
=+
( )
( )
211 1
bx b
x bb
+ =
= − ±
Since b is a positive constant, there are 2 distinct real solutions for x.
Hence, C has exactly two stationary points. (shown)
(II)[3]
x
y
2y =2
2x = − 2x =
2 f( )y x=
a− aO
2y = −2−
x
y
12y =
12
2ax −= 2
ax =
( )1
f 2y
x=
O 11−||
x
y
2y =2
1xb
= −
'f ( )y x=
( )11b
b−O ( )1
1bb
− +
||
__2 2b−
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 4 of 8
8 [7] [2] [5]
d d d d
= ⇒ =V xV Ax At t
d 5d
= −V kxt
, where k is a positive constant.
d 5d
⇒ = −xA kxt
(shown)
1 1
d 5d
1 d d5
ln 5 , where is an arbitrary constant
ln 5 , where is an arbitrary constant
5 e
5 e , where is an arbitrary constant
1 (5 e
k t cA
k tA
xA kxt
A x tkx
A kx t c ck
kkx t c cA
kx
kx B B
x Bk
− +
−
−
= −
⇒ =−
⇒ − − = +
⇒ − = − +
⇒ − =
⇒ − =
⇒ = −
∫ ∫
)k tA
When 0, 0 5= = ⇒ =t x B 5 (1 e )
−∴ = −
k tAx
k
Q9 [7] (a)[3]
i 2i6 44
i (8 1)24
4 2(1 i) 8e 8e
2e , 3, 2, 1,0,1, 2
k
k
z
z k
ππ π
π
⎛ ⎞+⎜ ⎟⎝ ⎠
+
= + = =
= = − − −
(b)(i)[1] Since the two complex roots are non-real, and not conjugate, for all the coefficients 3 2 1 0, , ,a a a a to be real, the degree of the polynomial has to be at least four. Since the degree is only three, it is not possible.
(ii)[3] Since the points A, B and C are on a circle centered at the origin, and they form an equilateral
triangle, angle AOB is 23π . OA rotated by 2
3π clockwise about O will give OB. Let b be the
complex number represented by B. 2 i3e (2 3 i)
1 3i (2 3 i) 2 2
3 1 i(1 3)
b
b
b
π−
= − +
⎛ ⎞= − − − +⎜ ⎟⎜ ⎟⎝ ⎠
= − + −
(i)[1]
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 5 of 8
(ii)[8] [2] [6]
e−= xy x d ( e ) e e (1 )d
− − −⇒ = − + = −x x xy x xx
At P , d, e , e (1 )d
a ayx a y a ax
− −= = = −
Equation of tangent to the curve at P is
e e (1 )( )
e (1 )( ) e
a a
a a
y a a x ay a x a a
− −
− −
− = − −
⇒ = − − +
At Q , 0, = =x y h e (1 )(0 ) ea ah a a a− −⇒ = − − + 2e ( 1)( ) e ea a aa a a a− − −= − + =
2d 2 e ( e ) e (2 )d
a a ah a a a aa
− − −⇒ = + − = −
For stationary values of h , d 0d
ha=
0a⇒ = (N.A. since 0a > ) or 2a = Maximum value of 24e−=h
2− 2 2+
dd
ha
0> 0 0<
Tangent
Q11[10] (i)[3]
Since the two planes intersect in a line l, with a vector equation given by
r = 4 22 1 , ,5 3
s s−⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟− + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
R
The line l lies in p2.
So 2 3
1 0 6 18 0 123 6
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⋅ = ⇒ − + + = ⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
λ λ λ
425
⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠
satisfies equation of p2.
So 4 32 12 2 30 12 24 30 65 6
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− ⋅ = ⇒ = − − = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
λ μ μ λ
(ii)[2] Since p1 and p2 intersect in line l, and the 3 planes have no common point of intersection, l does
not intersect p3. Hence l is parallel to p3, but l does not lie in p3.
2 5
21 8 0 10 8 3 03
3t t
t
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⋅ = ⇒ − + + = ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Also 4 5
22 . 8 123
5 23
⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟− = ≠⎜ ⎟⎜ ⎟
⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎜ ⎟⎝ ⎠
which means (4, 2, 5)− − is not in 3p and so l does not lie in p3.
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 6 of 8
(iii)[5] Since p4 contains l,
213
−⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
is parallel to p4.
Since p4 contains the points (1, −1, 2) and (4, −2, −5).
Hence 4 1 32 1 15 2 7
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
is parallel to p4.
A vector normal to p4 is 3 2 41 1 57 3 1
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟− × =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
.
Equation of p4 is given by : 4 1 45 1 5 1 4 5 11 2 1
x y z⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⋅ = − ⋅ = ⇒ + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
r
Let θ be the angle between p1 and p4.
cos θ = 2 2 2 2 2 2
2 45 5
3 1 1438 422 5 3 4 5 1
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠ =
+ + + + ⇒ θ = 110.514
Acute angle between p1 and p4 = 180° − θ = 69.5° (1 d.p.)
Q12[11] (a) [3]
By Ratio Theorem,
22
12 14 1124 8
2 210 6 12
28 11So 12 = 12 24 28 11 4+2 2
16 16 4Also, 4 = 4 10+2 2 6 6
OA OBOP
t
t t t
λλ
λλ λ
λ λ λ λλ λ
λλ λ
+=
+⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⇒ − = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
+ ⇒ + = + ⇒ =+
− + ⇒ − = + ⇒ = −+
uur uuuruuur
(b) [8] (i)[4]
11 12 12 4 6
12 10 2
11 14 32 8 6
12 6 6
Length of projection of onto 1 3
6 62 6
PB OB OP
AB OB OA
PB AB
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− −⎛ ⎞ ⎛⎜ ⎟ ⎜• −⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝=
uur uuur uuur
uuur uuur uur
uur uuur
2 2 2
3 36 12 21 79 3813 6 6
⎞⎟⎟⎟ − +⎠ = = =
+ +
Shortest distance of P from AB2
2 7 49 320 8 5413 9 3 3
BP ⎛ ⎞= − = − = =⎜ ⎟⎝ ⎠
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 7 of 8
(ii)[4] BAPQ forms a parallelogram.
Hence 3 12 96 4 10
6 10 16
AB PQ
OQ AB OP
=−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⇒ = + = − + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur uuur
uuur uuur uuur
Area of parallelogram BAPQ
2 2 2
3 26 12
6 4
48024
48 0 24
AB AP= ×
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − × −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠
= + +
uuur uuur
= 53.67 (to 2 dec places)
13[11] (i)[4]
2
12 2
1
12 2
1
22 3
2 3
2 3
4f ( )(1 3 ) 4
4(1 3 ) 4 14
4 (1 3 ) 144
12 1 3 (3 ) (3 ) ... 1 ...2 4
1 32 1 3 9 27 ...8 8
71 2132 6 ...4 4
xx x
xx
xx
xx x x
x x x
x x x
−
−
−
−
=− +
⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
⎛ ⎞= − +⎜ ⎟
⎝ ⎠⎡ ⎤⎛ ⎞
⎡ ⎤= + + + + − +⎢ ⎥⎜ ⎟⎣ ⎦⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞ ⎛ ⎞= + + − + + − + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
= + + + +
(ii)[2] For expansion in (i) to be valid,
2
2
3 1 and 14
1 and 431 and 23
xx
x x
x x
< <
⇒ < <
⇒ < <
1 1Range of values of ,3 3
x ⎛ ⎞= −⎜ ⎟⎝ ⎠
__________________________________ 2010 H2 Maths Preliminary Exam Paper 1
Page 8 of 8
(iii)[5]
2
2 22 2
2
2 tan f ( ) ----- (1)Differentiate (1) with respect to ,
d2sec f '( ) ----- (2)d
Differentiate (2) with respect to ,
d d2 2sec tan sec f "( ) ----- (3)d d
y xx
yy xx
x
y yy y y xx x
=
=
⎡ ⎤⎛ ⎞ + =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
2 2 2
2 2
2
When 0,71from (i), f (0) 2, f '(0) 6 and f ''(0)2
(1) gives 2 tan 2 4
d d 3(2) gives 2(2) 6 d d 2
3 d 71 d 35(3) gives 2 2(2)(1) 2 2 d 2 d 8
3 35Hence ...4 2 8 2!
x
πy y
y yx x
y yx x
π xy x
=
= = =
= ⇒ =
= ⇒ =
⎡ ⎤⎛ ⎞ + = ⇒ =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
⎛ ⎞= + + + =⎜ ⎟⎝ ⎠
23 35 ...4 2 16π x x+ + +
Q14 [5]
Given 1 55
n
n n
pS .−= −
⇒ 1
1 2 55
n
n n
pS .−
− −= −
∴ 1
1
1 25 55 5
n n n
n n
n n
T S S
p p−
−
− −
= −
⎛ ⎞= − − −⎜ ⎟
⎝ ⎠
=
1
1
1 5 5
np p−
−
⎡ ⎤⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎢ ⎥−⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
= ( )1
55
npp−⎛ ⎞⎟⎜− ⎟⎜ ⎟⎜⎝ ⎠
Now, ( )2
1 55
n
npT p
−
−
⎛ ⎞⎟⎜= − ⎟⎜ ⎟⎜⎝ ⎠
∴ 1
n
n
TT −
= ( )
( )
1
2
55
55
n
n
pp
pp
−
−
⎛ ⎞⎟⎜− ⎟⎜ ⎟⎜⎝ ⎠⎛ ⎞⎟⎜− ⎟⎜ ⎟⎜⎝ ⎠
=
5p (constant)
⇒ series is a geometric series. (shown)
For the sum to infinity to exist, 15p<
5 5 0p , p∴− < < ≠
RAFFLES INSTITUTION
H2 Mathematics 9740 2010 Year 6 ____________________________
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 1 of 8
2010 H2 Maths Preliminary Exam Paper 2 Solutions
Q1 [8] (i)[1] 2 2
1 12( 1) 2n n
−−
2 2
2 2
2 12 ( 1)
n n nn n− + −
=−
12
2 2
1 (shown)( 1) 2n a
n n−
= ∴ = −−
(ii) [1] Smallest value of M = 2 (iii)[3]
2 23
12
2 23
2 23
2 12 ( 1)
( 1)1 1( )
2( 1) 2
N
Nn
N
n
N
n
nSn n
nn n
n n
=
=
=
−=
−−
=−
= −−
∑
∑
∑
2 2
2 2
2 2
2 2
2
1 12(2) 2(3)
1 1 +2(3) 2(4)
1 1 +
2( 2) 2( 1)1 1 +
2( 1) 21 18 2
N N
N N
N
= −
−
−− −
−−
= −
M
2 2 2
1 1 1 ...(2)(3) (3)(4) (4)(5)
+ + +
23
2 23
12
2 23
1 ( 1)
1( 1)
( 1)
n
n
n
n nn
n nn
n n
∞
=
∞
=
∞
=
=−
−=
−−
<−
∑
∑
∑12
2 23
2
= lim ( 1)
1 1= lim8 2
N
N n
N
nn n
N
→∞=
→∞
−−
⎡ ⎤−⎢ ⎥⎣ ⎦
∑
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 2 of 8
1= (shown) 8
Q2 [8] [4]
(i)[2] Minimum value will be the perpendicular distance from origin to the line segment CB. Let the point of intersection of this perpendicular and CB be D. By observing the right angle triangle
formed by O, C, and D, OD = | a | sin 4π = | |
2a .
(ii)[2] It is clear that the locus of z satisfying both relations is the line segment CB, not including C. The
argument of the point represented by B is arg( )2
a π− , while the argument of the point represented
by C is arg( )a − π . Hence the range of values is
arg( ) arg( ) arg( )2
a z a π− π < ≤ − .
Q3 [12] (a)[7]
Since f (2.5) f (3.5) 0.75= = − , f is not one-one. OR
D ( ,4) R [ 1, )f f= −∞ = − ∞ Since the horizontal line 0.75y = − does not cut the graph of f at one and only one point, f is not one-one.
{ } { }D \ 3 R \ 1g g= − =R R fg does not exist because { }R \ 1g = R ( ,4) D f⊄ −∞ = . Since R [ 1, )f = − ∞ { }\ 3 Dg⊆ − =R , gf exists.
2
2 2
( 2)( 4) 6 8 3gf ( ) 1( 2)( 4) 3 6 11 ( 3) 2
x x x xxx x x x x
− − − += = = −
− − + − + − + D D ( ,4)gf f= = −∞ .
g( )y x=
1y =
3x = −O
0.75y = −°42O
f ( )y x=
(3, 1)−
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 3 of 8
12
R R whose domain is restricted to R
R whose domain is restricted to [ 1, )
[ ,1)
gf g f
g
=
= − ∞
= −
OR sketch y = gf(x)
12D ( ,4) R [ ,1)gf gf= −∞ = −
(b)[5] Let (1 )xy a e−= +
( )1
ln 1
x
ya
y ea
x
−− =
= − −
h 1− (x) = ( )ln 1xa− −
1D R ( ,2 ]hha a− = =
0D R ( ,2 ]h h a a+= = Since 1h ( )y x−= can be obtained by reflecting h( )y x= about the line y = x, therefore at the point of intersection of h( )y x= and 1h ( )y x−= , y = x = b.
0 0 (1 ) d (1 ) d
b bx ya e x a e y I− −+ = + =∫ ∫ .
∴ Area bounded by 1h( ), h ( )y x y x−= = and the axes
0 02
(1 ) d (1 ) d ( )
2
b bx ya e x a e y b b
I b
− −= + + + − ×
= −
∫ ∫
OR
Area OAB = 2 2
0 00
h( ) d d2 2
bb b x bx x x x I I
⎡ ⎤− = − = −⎢ ⎥
⎣ ⎦∫ ∫
Since h( )y x= and 1h ( )y x−= are symmetrical about y x= , Area OAB = Area OBC = 2
2bI − .
∴ Area bounded by 1h( ), h ( )y x y x−= = and the axes
2
22 22bI I b
⎛ ⎞= − = −⎜ ⎟
⎝ ⎠
4
1y =gf ( )y x=
12(3, )−
O 2 °
h( )y x=
1h ( )y x−=
( , )B b b
O
(0,2 )A a
(2 ,0)C a
y a=
y x= x a=
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 4 of 8
Q4 [12] (i)[2]
1 1
2 1 0
1
2 0
120
4f (| |)d 2 d
1
22( 2) d
1
4 ln( 1)
4 ln 2
xx x x
x
xx
x
x
−
−=
+
= −+
⎡ ⎤= − +⎣ ⎦= −
∫ ∫∫
(ii)[7]
[ ]
[ ]
2
2 2 2
ln(3 e 3)d
1ln( ) d
3
1ln( )d
3
1 1ln 1d
3 31
ln31
(3 e 3) ln(3 e 3) (3 e 3)3
x x
u u
u u
u u u
u u u c
x x x c
+ +
=
=
= −
= − +
⎡ ⎤= + + + + − + + +⎣ ⎦
∫∫∫
∫
–6 –4 –2 2 4 6
–4
–3
–2
–1
1
2
3
4
x
y
2
1 02
2( 2) 13
4ln(3 3)d d
1e
xx e x x
x
−
− +−
−+ + +
+∫ ∫2
12 2 2( 2)
3
2 2 2
2
1(3 3) ln(3 3) (3 3) 2 ln 2
3
1[ ln 1] 2 ln 2
31
( 1) 2 ln 23
ex e x e x e
e e e
e
−+
−⎡ ⎤= + + + + − + + +⎣ ⎦
= − + +
= + +
(iii)[3]
( )2
21 0222( 2) 1
3
4ln(3 3) d d
1
20.55 (2d.p.)
e
xx e x x
x
−
− +−
−⎛ ⎞π + + + π ⎜ ⎟+⎝ ⎠
=
∫ ∫
2( 2)3
e− +
( 1, 2)−
24
1x
yx−
=+
2ln(3 3)y x e= + +
R
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 5 of 8
Q5[8] (a)[3]
W,Y,O,G – 1 letter each E – 2 letters Case 1: All 3 letters are different, i.e. W, Y, O, G, H, E. Number of ways = 6
3 3! 120C × = . Case 2: 2 letters are identical, i.e. E, E and one of the remaining 5 letters, W, Y, O, G, H.
Number of ways = 51
3! 152!
C × = .
Total number of ways = 120 + 15 = 135 (b) [5] (i) [3]
Ways to select the players Badminton
(6) Tennis
(6) Football
(5) No. of ways
1 2 3 6 6 51 2 3C C C× × =900
1 3 2 6 6 51 3 2C C C× × =1200
2 1 3 6 6 52 1 3C C C× × =900
2 3 1 6 6 52 3 1C C C× × =1500
2 2 2 6 6 52 2 2C C C× × =2250
3 1 2 6 6 53 1 2C C C× × =1200
3 2 1 6 6 53 2 1C C C× × =1500
Total number of ways 9450.=
(iii) [2] Arrange the 6 youths in 6! ways and slot the 4 empty chairs into the row in 74C ways.
Total number of ways = 746! 25200C× = .
Q6 [7] [2]
Probability of getting ‘6’ 1 ( 2 3 4 5 )1 15
p p p p pp
= − + + + += −
(i)[3]
2
2
P( ) P(sum 10)+P(sum 11)+P(sum 12) 2(4p)(1 15 )+(5 )(5 )+2(5p)(1 15 )+(1 15 ) 1 12 20
Ap p p p p
p p
= = = =
= − − −
= − −
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Second die
Firs
t die
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 6 of 8
(ii)[2]
P( ) P(2nd die is 5 1st die is 5)+P(2nd die is 6 1st die is 5) 5 1 15 1 10
A Bp p
p
=
= + −= −
Alt :
P( )P( )
P( )P(1st die is 5, 2nd die is 5)+P(1st die is 5, 2nd die is 6)
P(1st die is 5)(5 )(5 )+(5 )(1 15 )
51 10
A BA B
B
p p p pp
p
∩=
=
−=
= −
For A and B to be independent, P(A) = P(A|B)
2
2
1 12 20 1 1020 2 0
10 or 10
p p pp p
p
− − = −
+ =
∴ = −
Since 0p ,> there are no values of such that A and B are independent.
Q7[9] (i)[2]
The product moment correlation coefficient between x and y, r is -0.943(3 s.f.).
(ii)[2] Since r is close to 1, it suggests a linear model is appropriate. However the scatter diagram shows the relationship is non-linear.
(iii)[1] The scatter diagram shows that as x increases, the rate of decrease in y becomes smaller and y appears to approach a value. For a linear model, the rate of decrease is constant. OR The product moment correlation coefficient between 1/x and y, 2r is 0.993 where 2r is almost 1 as
compared to r =0.943.
Therefore the model where 0 0by a a ,bx
= + > > is better.
(iv) [2] 2.16 (3s.f.), 2.33 (3s.f.)a b= =
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 7 of 8
(v) [2] When 4 5x . , y= = 2.68 (3 s.f..)
Since 4.5x = is within the data range of the x values and value of 2r is close to 1, the estimate is reliable.
8 [10] (i)[2]
Unbiased estimate of the mean = 8400 56150
=
Unbiased estimate of the variance = 5555149
.
(ii)[5] (i) 0H : 55μ = vs 1H : 55μ >
Perform a 1-tail test at 3% level of significance.
Under 0H , 2
0~ N , sXn
⎛ ⎞⎜ ⎟⎝ ⎠μ approximately, by Central Limit Theorem, where
0555555, , 56, 150149
s x nμ = = = =
Using a z-test, p-value = 0.0224 (3 s.f.)
Since the p -value = 0.0224 < 0.03, we reject 0H and conclude that there is sufficient evidence, at 3% significance level, that the school is understating the mean mass of male students in the school.
Central Limit Theorem has to be applied here as X is not stated to be normally distributed and the sample size of 150 is large.
(iii)[1] It means that the probability of concluding that the mean mass is more than 55 kg when it actually is 55 kg is 0.03.
[2] Suppose that there are 4 different levels of female students in the school. Select 25 from each level and carry out the data collection by interviewing the first 25 female students of each level who enter the school gate in the morning.
Q9 [11] (i)[2]
( )P 0.9X t> =
( )P 0.126.2 (in minutes)
X tt
< =
=
(ii)[2] ( ) ( ) ( )P 31 and 20 P 31 P 20X Y X Y> < = > × <
0.36944 0.977250.361 (3sf)
= ×=
(iii)[3] ( ) ( ) ( )2P 11 P 11 where ~ N 14, 13X Y X Y X Y> + = − > −
0.797 (3sf)=
(iv) Part (iii) includes the case in (ii), as well as other cases in which Abel reaches the station before the train, e.g. train arrives after 12.01am and Abel arrives before 12.01am.
(v) If ( )~ N 4, 6W , then 99.7% of the values would lie within 4 3 6± , which contains a significant range of negative values.
__________________________________ 2010 H2 Maths Preliminary Exam Paper 2
Page 8 of 8
(vi) Since sample size 40 is large, by Central Limit Theorem,
26~ N 4,40
W⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
approximately.
( )P 3.5 0.0984W < = (3sf)
Q10[13] (i)[1]
The average number of incoming calls received per hour is constant throughout the opening hours of the mall. OR The probability of 2 or more incoming calls received in a very short interval of time is negligible.
(ii)[2] Let X be the number of incoming calls received in an hour. ( )~ Po 6.75X .
( ) ( )P 8 1 P 70.364 (3sf)
X X≥ = − ≤
=
(iii)[3] Required probability is
( ) ( ) ( )
( )
2
2
4!P 6 P 7 P 82!
4!0.48759 0.14832 0.36409 0.115 (3sf)2!
X X X⎡ ⎤≤ × = × ≥ ×⎣ ⎦
≈ × × × =
(iv)[4] ( )~ Po 81Y . So 81μ = , 9σ = .
Since 81 10μ = > , ( )2~ N 81, 9Y approximately.
( )( )( )
P 81 9 81 9
P 72 90
P 72.5 89.5 by continuity correction0.6551 (4dp)
Y
Y
Y
− < < +
= < <
= < <
=
(v)[3] Let W be the number of busy days in 14 days. ( )( ) ( )~ B 14, P 90 ,that is, ~ B 14, 0.14593 .W Y W> Required probability is ( ) ( )P 2 P 90 0.29191 0.14593W Y= × > ≈ ×
0.0426= (3sf)
Name ( ) Class RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
Paper 1
Additional Materials: Answer Paper List of Formulae (MF15)
Cover Page
9740/01
14 September 2010
3 hours
READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, class and index number in the space at the top of this page. Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, place the cover page on top of your answer paper and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
RIVER VALLEY HIGH SCHOOL 9740/01/2010
2
1 A shampoo manufacturing company produces shampoo that comes in three different sizes
of bottles (small, medium and large). The amount of plastic required to manufacture the bottles, and the profit made from the sale of each bottle of shampoo are shown in the table below:
Small Medium Large
Volume of shampoo (cm3) 200 450 600 Amount of plastic (cm2) 150 335 475 Profit ($) per unit sold 2.50 3.80 4.20
On a particular round of production, the volume of shampoo and the amount of plastic used were 370700cm3 and 280400cm2 respectively. When all the bottles of shampoo from this round of production were sold, the profit made from the sale of the medium bottles was twice the total profit made from the sale of the small and large bottles. Determine the number of each size of shampoo produced and state an assumption made about the bottles manufactured. [6]
2 The complex number z satisfies the relations i 1 iz + ≤ and * 2iz z− ≤ . (i) Illustrate both of these relations on a single Argand diagram. [3]
(ii) Find the greatest and least possible values of arg ( 1 i)z + + , giving your answers in radians correct to 3 decimal places. [3]
3 Given that θ is sufficiently small for 3θ and higher powers of θ to be neglected, express
4 cos2 tan
θθ
−+
as a quadratic expression in θ . [4]
By letting π6
θ = , show that 29 11π 55π3
7 56 2016≈ + − . [2]
RIVER VALLEY HIGH SCHOOL 9740/01/2010
3
4 (i) Find ( )1 2d sind
xx
− . [1]
(ii) Evaluate the integral 5
1
1ln tan d
10x x−
∫ numerically. [1]
(iii) Given that 0 1k< ≤ and
52 1 2 1
10sin d ln tan d
10
k xx x x x− − = ∫∫ ,
show that 4
3 1 2
40d sin
1
k x x ak k bx
−= +−∫ , where a and b are constants to be
determined. [4]
5 The sequence of numbers u u u1 2 3, , , is given by 1 2u = and 12
nn
nunu+
+= for all positive
integers.
(i) By writing down the terms 2u and 3u , make a conjecture for nu in terms of n . [2]
(ii) Prove your conjecture by mathematical induction. [4]
(iii) Write down the limit of 1n nu u − as n tends to infinity. [1]
6 (i) Verify that 4 2 2 21 ( 1)( 1).r r r r r r+ + = + + − + [1]
(ii) By considering 2
11r r+ +
− 2
11r r− +
, show that
4 2 21
1 111 2 1
N
r
rr r N N
=
= − + + + + ∑ . [4]
(iii) Evaluate 4 22 1r
rr r
∞
= + +∑ . [2]
RIVER VALLEY HIGH SCHOOL 9740/01/2010
4
7 The diagram below shows a sketch of the graph of 2ax bx cyx d+ +
=+
where a , b , c and
d are constants. The asymptotes, also shown in the diagram, are 2x = − and 4y x= − .
(i) Write down the value of d and find the values of a and b . [4]
(ii) Given that the curve passes through the point ( )0 4 5, .− , find the value of c . [1] (iii) By sketching the graph of ( )2 2 24x y k− + = , determine the range of values of k
such that the equation ( )22
2 24 ax bx cx kx d
+ +− + = +
, where a , b , c and d are
the values found above, has at least one negative root. [3] 8 (i) Solve the equation 3( 2) 8z + = − , giving the roots in the form ie 2r θ − , where 0r > and π πθ− < ≤ . [4] (ii) Express 3( 2) 8z + + as a product of three linear factors. [2] (iii) Hence, find the values of the real numbers a and b in the equation 2 0z az b+ + =
where the roots are the complex numbers found in (i). [2] (iv) Explain why the roots in (i) lie on a circle centre (−2,0) and radius 2. [1]
y
x O 2−
4
4−
RIVER VALLEY HIGH SCHOOL 9740/01/2010
5
9 It is given that ( ) ( )f ln 1 sinx x= + , where 3π 3π8 8
x− ≤ ≤ .
(i) Show that ( )(1 sin )f cosx x x′+ = . [1] (ii) By further differentiation of this result, find the Maclaurin series for ( )f x , up to
and including the term in 3x . Hence, write down the equation of the tangent of ( )fy x= at 0x = . [5]
Denote the Maclaurin series of ( )f x in (ii) by ( )g x .
(iii) On the same diagram, sketch the graphs of ( )fy x= and ( )gy x= for
3π 3π8 8
x− ≤ ≤ . [2]
(iv) Find, for 3π 3π8 8
x− ≤ ≤ , the set of values of x for which the value of ( )g x is
within 0.05± of the value of ( )f x . [2] 10 (a) Given that the first, third and fourth terms of an arithmetic progression are three
consecutive terms of a geometric progression, and that the sum of the first twenty even-numbered terms of the arithmetic progression is 960, find the common difference of the arithmetic progression. [4]
(b) Annie puts $x on 1 January 2010 into a bank account which pays compound
interest at a rate of 3% per month on the last day of each month. On the first day of each subsequent month, she puts in an amount which is $x more than the amount she puts in the previous month. For example, she puts in $x on 1 January, $2x on 1 February and so on.
(i) Show that the amount of money in the bank account on the last day of
March is 3103$ 103(1.03) 1129
x − . [4]
(ii) Find the least integer value of x so that the amount of interest earned for
the first three months of the year 2010 exceeds $100. [2]
RIVER VALLEY HIGH SCHOOL 9740/01/2010
6
11 The lines 1l and 2l meet at P. The line 3l lies on the same plane containing 1l and 2l , and
is perpendicular to 1l . Given that 1l and 2l are parallel to the vectors a and b respectively,
show that 3l is parallel to the vector 2
⋅ −
a bb aa
. [3]
The lines 1l and 2l have equations ( )3 5 2 4 2t= − + + − +r i j k i j k and
( )3 5 2 8 4 3s= − + + − +r i j k i j k respectively. The line 3l lies on the plane containing 1l and 2l and is perpendicular to 1l . Given that the three lines intersect, find the equation of
3l . Find also the angle between 2l and 3l . [3]
The plane 1Π contains the lines 1l , 2l and 3l , and the plane 2Π has equation
2 4 9kx y z+ − = . Given that 1Π is parallel to 2Π , show that 52
k = . Hence, find the exact
distance between the two planes. [6]
12 (i) Sketch the graph of ( )
32 2
2
4y
x=
−. [2]
(ii) The region R is bounded by the curve, the axes and the line 3x = . Using the
substitution 2cosx θ= , find the exact area of R. [5]
(iii) Find the exact volume of revolution when R is rotated completely about the y-axis. [6]
− End of Paper −
1
RIVER VALLEY HIGH SCHOOL 9740/01/2010
Name ( ) Class
Tutor Calculator Model RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
Paper 1
Cover Page
9740/01
14 September 2010
INSTRUCTIONS TO CANDIDATES Attach this cover page on top of your answer paper. Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER.
Question Mark Max. Mark Question Mark Max. Mark
1 6 7 8
2 6 8 9
3 6 9 10
4 6 10 10
5 7 11 12
6 7 12 13
Total 100 Grade
Name ( ) Class RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
Paper 2
Additional Materials: Answer Paper List of Formulae (MF15)
Cover Page
9740/02
15 September 2010
3 hours
READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so. Write your name, class and index number in the space at the top of this page. Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, place the cover page on top of your answer paper and fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
RIVER VALLEY HIGH SCHOOL 9740/02/2010
2
Section A: Pure Mathematics [40 marks] 1 The diagram shows the graph of y = f(x). The curve passes through the origin O, the
points A(–3a, 2) and B(–a, –2), where 0 1a< < .
Sketch on separate clearly labelled diagrams, the graphs of
(a) 1f 12
y x = − +
; [4]
(b) f '( )y x= . [2]
2 Points O, A and B are such that 2 3OA = + −i j k
and 4OB = −i j
.
(i) Use vector product to find the exact area of the triangle OAB. [2]
(ii) A point P on line AB is such that AP : AB = 2 : 3. Given that point Q lies on OP
produced, show that (3 )OQ c= −i k
, for some constant c. [2] (iii) Given also that triangle AQB has a right angle at Q, find the value of c. [3]
( , 2)B a− −
( 3 , 2)A a−
0
x a=
y
x
RIVER VALLEY HIGH SCHOOL 9740/02/2010
3
3 The curve C has parametric equations
22x t t= + , 2(1 )y t= − ,
where t∈ .
(a) Find the range of values of t for which C is decreasing. [3]
(b) Find the equation of the normal to the curve at 2t = . Determine if the normal meets C again. [4]
4 A long cylindrical metal bar is submerged into iced water. A researcher claims that the
rate at which the length, l cm, of the bar is shrinking at any time t seconds is proportional to the volume of the bar at that instant, assuming that the cross-sectional area of the bar remains constant during the shrinking process.
Formulate and integrate a differential equation to show that ktl Ae= , where A and k are constants. State the range of values of A and of k. [5] Given that the initial length of the bar is L, sketch a graph to show the relation between l and t. Comment on the suitability of the claim by the researcher. [3] It is later found that l and t are related by the equation ktl B Le= + , where B is some constant. Given that l = 0.5L when t = T, show that the length of the bar when t = 3T, is
given by ( )3
2
0.5L BB
L−
+ . [2]
5 The functions f and g are defined by
3f : 2 1, 0x x x+ > , g : ln( ),x x a x a− > .
(i) Give a reason why 1-f exists. Hence find 1-f ( )x and state the domain of 1-f . [3] (ii) Only one of the composite functions fg and gf exists. State the greatest possible
value of a such that this composite function exists and using this value of a, give a definition of this composite function. Write down the range of the composite function. Explain why the other composite function does not exist. [5]
(iii) By sketching suitable graphs or otherwise, find the set of values of x such that 1 1- -f f ( ) ff ( )x x= . [2]
RIVER VALLEY HIGH SCHOOL 9740/02/2010
4
Section B: Statistics [60 marks]
6 Research has shown that 5 out of 8 students of a particular school make use of the
school’s e-learning portal. Three random samples of 40 students are chosen. Using a suitable approximation, find the probability that more than 80 students make use of the school’s e-learning portal. [3]
7 The sales of movie tickets sales are known to be high during the weekend, low during the
beginning of the week (i.e. Monday and Tuesday) and somewhat moderate at midweek. A sample of 50 is to be chosen from 350 days of a year of movie ticket sales to estimate the ticket sales per day. Describe how the sample could be chosen using systematic random sampling. [2]
Describe any disadvantage of using systematic sampling and explain briefly whether it
would be more appropriate to use stratified sampling instead. [2] 8 A group of 10 students consisting of 6 males and 4 females are seated at a round table
with chairs of different colours for lunch. Find the number of arrangements such that all the female students are seated together. [3]
After lunch, the same group of 10 students bought tickets to attend a concert. If the tickets
are for a particular row of 10 adjacent seats, find the number of possible seating arrangement when
(i) the first and last seats are occupied by the same gender, [3] (ii) four particular students did not turn up for the concert. [1]
9 The average daily sales at a small food store is estimated to be $452.80. The daily sales,
$x, over a period of 12 days is summarized by
( 450) 622.8x − =∑ and 2( 450) 78798x − =∑ . (i) Test, at 1% significance level, whether the food store has underestimated its
average daily sales. State an assumption you made about the data. [6]
(ii) Suppose in a test at 5% significance level, it is found that there is significant evidence of an increase in average daily sales. Using only this information, and giving a reason, discuss whether there is significant evidence at 5% significance level that the average daily sales has changed. [2]
RIVER VALLEY HIGH SCHOOL 9740/02/2010
5
10 (a) A student threw a fair die four times and recorded the number obtained for each
throw. Find the probability that
(i) at least two of the four numbers obtained are the same, [2]
(ii) exactly two of the four numbers obtained are the same given that the number 6 appeared at least once. [4]
(b) Events A and B are such that 1P( )2
B′ = , 7P( )20
A B′ ′∩ = and 4P( )7
B A = . Find
the probability P( )A . [3] 11 An experiment was conducted to verify how the radiation intensity x from a radioactive
source varies with time t. The following data were obtained from a particular source.
x 1 4 6 8 9 10 12 15 t 15.3 6.9 4.4 2.7 2.1 1.7 1.0 0.49
(i) Plot a scatter diagram for the above data. [1] (ii) Calculate the product moment correlation coefficient for the above data and
comment on it, in relation to your scatter diagram in (i). [2] (iii) Calculate the equation of the least square regression line of x on t. Give a practical
interpretation of the coefficient of t. [2] (iv) The variable y is defined by y = ln x. For the variables y and t, calculate the
product moment correlation coefficient and hence compare the suitability of this logarithmic model and the linear model in (iii). [2]
(v) Use an appropriate regression line to give the best estimate that you can of the
time when x = 10. [2]
RIVER VALLEY HIGH SCHOOL 9740/02/2010
6
12 The two most common types of disciplinary offences in a particular boy school is keeping
long hair and failure to wear the school badge. The mean number of disciplinary offences recorded per day involving long hair is 1.12. Assuming that each school week consists of five school days, the mean number of disciplinary offences recorded per school week involving failure to wear the school badge is 4.2. The number of cases for each disciplinary offence is assumed to have an independent Poisson distribution.
(i) Find the probability that at most 9 cases of disciplinary offence are recorded in a
given school week. [3] (ii) In a school week in which there are more than 7 cases of disciplinary offence
involving long hair, find the probability that at most 9 cases of disciplinary offence are recorded. [3]
(iii) Calculate the probability that on a Thursday in a particular school week, it is the
third day in the school week in which the discipline master caught at least 4 students having long hair in a day. (You may assume that Monday is the first day of school for a school week.) [3]
(iv) Explain why the Poisson distribution may not be a good model for the number of
disciplinary cases involving long hair, in a school year. [1] 13 A farm in Kranji rears chickens and ducks for sale to the local market. (a) The mass of a randomly chosen chicken has mean 2.5 kg and standard deviation
0.4 kg. If the probability that the mean mass of a large sample of n chickens is greater than 2.45 kg exceeds 0.95, find the least value of n. [3]
(b) The variable X denotes the mass, in kg, of a randomly chosen duck which is
normally distributed with mean 2.7 and standard deviation σ . The ducks are sold at $4 per kg.
(i) Given that 50P( 2.5) P( 2.9)107
X X< = < , show that the value of σ is
0.424, correct to 3 significant figures. [2] (ii) A housewife brings $50 to the farm. Calculate the probability that she has
less than $27.50 after buying 2 ducks. [3]
(iii) A sample of 4 ducks is randomly chosen. Calculate the probability that the most expensive duck costs less than $11. [2]
− End of Paper −
1
RIVER VALLEY HIGH SCHOOL 9740/02/2010
Name ( ) Class
Tutor Calculator Model RIVER VALLEY HIGH SCHOOL 2010 Year 6 Preliminary Examinations Higher 2
MATHEMATICS
Paper 2
Cover Page
9740/02
15 September 2010
INSTRUCTIONS TO CANDIDATES Attach this cover page on top of your answer paper. Circle the questions you have attempted and arrange your answers in NUMERICAL ORDER.
Question Mark Max. Mark Question Mark Max. Mark
1 6 6 3
2 7 7 4
3 7 8 7
4 10 9 8
5 10 10 9
11 9
12 10
13 10
Total 100 Grade
RVHS 2010 H2 Math Prelim P1 1
RVHS 2010 Preliminary Exam: 9740 H2 Mathematics Paper 1 Solutions Question 1 [6 marks] Let the no. of small, medium and large bottles manufactured be denoted by s, m and l
respectively. So, 150 335 475 280400s m l+ + = , 200 450 600 370700s m l+ + = and 3.8 2(2.5 4.2 )m s l= + Using GC, solving the augmented matrix:
150 335 475 280400200 450 600 3707005 3.8 8.4 0
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
Ans: s = 166, m = 550 and l = 150 Assumption: The plastic bottles are of negligible thickness. OR The plastic bottles are of the same thickness.
Question 2 [6 marks] (i) (ii)
1
1 1
1
iz i
zi
z i
+ ≤
+ ≤
− ≤
* 2
( ) 2
2 21 1
z z
x iy x iy
iyy
− ≤
+ − − ≤
≤
− ≤ ≤
1
0
1
Im(z)
Re(z)
RVHS 2010 H2 Math Prelim P1 2
arg( 1 i) arg( ( 1 i))z z+ + = − − −
Max. arg( 1 i)z + + =
2π .
Min. arg( 1 i)z + + =α.
1 1 1tan 2 sin 0.6445
α β θ − −= − = − = (3 dec. pl.)
Question 3 [6 marks]
4 cos2 tan
θθ
−+
=
2
4 12
2
θ
θ
⎛ ⎞− −⎜ ⎟⎝ ⎠+
=
2
32
2
θ
θ
+
+
= 12
3 2 12 2θ θ
−⎛ ⎞ ⎡ ⎤⎛ ⎞+ ⋅ +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎝ ⎠
= 2 213 1 ...
2 2 2 4θ θ θ⎛ ⎞ ⎛ ⎞
+ ⋅ − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
≈ 23 3 52 4 8
θ θ− +
Let π6
θ = : 2
34 3 π 5π21 2 8 28823
−≈ − +
+
2(8 3) 3 3 π 5π
2 8 2882(2 3 1)−
≈ − ++
1 1
Re(z) 0
1 i− −α
Im(z)
RVHS 2010 H2 Math Prelim P1 3
2
2
(8 3 3)(2 3 1) 3 π 5π2(4(3) 1 ) 2 8 288− −
≈ − +−
251 14 3 3 π 5π
22 2 8 288−
≈ − +
∴ 29 11π 55π3
7 56 2016≈ + − (shown)
Question 4 [6 marks] (i) ( )1 2
4
d 2sind 1
xxx x
− =−
(ii)
5 1
1ln tan d 5.292774105
10x x−⎛ ⎞⎛ ⎞ = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫ = −5.29 (3 s.f.)
(iii)
2 1 2
0sin d
kx x x−∫ (letting 1 2sinu x−= and 2dv x
dx= )
3 41 2
400
2sin d3 3 1
kkx xx x
x−⎡ ⎤
= − ⋅⎢ ⎥−⎣ ⎦
∫
3 41 2
40
2sin d3 3 1
kk xk xx
−= − ⋅−
∫
3 41 2
40
2sin d 5.2927741053 3 1
kk xk xx
−∴ − ⋅ = −−
∫
Hence 4 3
1 2
40
3d sin 5.2927741052 31
k x kx kx
−⎡ ⎤= +⎢ ⎥
− ⎣ ⎦∫
= 3 1 21 sin 7.942
k k− + ,
where a = 0.5 and b = 7.94
Question 5 [7 marks] (i)
232
u = , 343
u =
Conjecture: 1n
nun+
=
(ii) Let P(n) be the statement “ 1
nnu
n+
= ”, for all positive integers n.
When n = 1: LHS = 1u = 2 = RHS (given)
∴ P(1) is true.
RVHS 2010 H2 Math Prelim P1 4
Assume P(k) is true for some integers k +∈Z , i.e 1k
kuk+
=
To show: P(k+1) is true, i.e. 121k
kuk+
+=
+.
Show: LHS 1ku +=
2
2 .1
21
k
kku
k kk k
k RHSk
+=
+=
++
= =+
Since P(1) is true and P(k) is true ⇒ P(k+1) is true,
∴ By Mathematical Induction, P(n) is true for all positive integers n.
(iii)
1
1
1( 1)
1 21 1 as ( 1) 1
nn
n n
nun u
nu u nn n
−
−
+=
−+
⇒ = = + → →∞− −
Question 6 [7 marks] (i) 2 2
4 3 2 3 2 2
4 2
( 1)( 1)1
1 =
RHS r r r rr r r r r r r rr r LHS
= + + − +
= − + + − + + − +
= + +
(ii)
2 2
2 2
2 2
2 2
4 2
1 11 1
1 ( 1)( 1)( 1)
2( 1)( 1)
21
r r r rr r r rr r r r
rr r r r
rr r
−+ + − +− + − + +
=+ + − +
−=
+ + − +−
=+ +
RVHS 2010 H2 Math Prelim P1 5
4 2 4 21 1
2 21
2 2
2
2
1 21 2 1
1 1 12 1 1
1 1 12 3
1 1 7 31 1
13 7...
1 1 1 1
1 112 1
1 112 1
N N
r rN
r
r rr r r r
r r r r
N N N N
N N
N N
= =
=
= −+ + + +
⎛ ⎞= − −⎜ ⎟+ + − +⎝ ⎠
⎛= − −⎜⎝
+ −
+ −
+
⎞+ − ⎟+ + − + ⎠⎛ ⎞= − − +⎜ ⎟+ +⎝ ⎠
⎛ ⎞= −⎜ ⎟+ +⎝ ⎠
∑ ∑
∑
(iii)
4 22 1r
rr r
∞
=+ +∑ = 4 2 4 2
1
11 1 1 1r
rr r
∞
=
−+ + + +∑ 2
1 1 1lim 12 1 3N N N→∞
⎛ ⎞= − −⎜ ⎟+ +⎝ ⎠
Since 2
1lim 01N N N→∞→
+ +,
2
1 1 1 1 1 1lim 12 1 3 2 3 6N N N→∞
⎛ ⎞∴ − − = − =⎜ ⎟+ +⎝ ⎠.
Question 7 [8 marks] (i) 2d =
Method 1: Using long division:
( ) ( )2 2 22
2c b aax bx cy ax b a
x d x− −+ +
= = + − ++ +
1a∴ = , 2b = − Method 2:
( )( ) 24 2 2 842 2 2
x x hh x x hy xx x x
− + + − − += − + = =
+ + +
1a∴ = , 2b = −
(ii) 2 22
x x cyx− +
=+
At ( )0, 4.5− : 4.52c
− =
9c∴ = −
RVHS 2010 H2 Math Prelim P1 6
(iii) ( )2 2 24x y k− + = is a circle centre (4, 0), radius k
2 24 5 4
145 145 or 2 2
k .
k k
> +
⇒ < − >
Question 8 [9 marks] (i) 3
3 3 ( 2 )
2( )3 3
2( )3 3
( 2) 8( 2) 2 e
2 2e , 0, 1
2e 2, 0, 1
k i
k i
k i
zz
z k
z k
π π
π π
π π
+
+
+
+ = −
+ =
+ = = ±
= − = ±
(ii) When 1k = , i2e 2 2(cos sin ) 2 2 2 4z iπ π π= − = + − = − − = − i
30, 2e 2 2 cos sin 23 3
1 32 2 1 32 2
k z i
i i
π π π⎛ ⎞= = − = + −⎜ ⎟⎝ ⎠
⎛ ⎞= − − = − −⎜ ⎟⎜ ⎟
⎝ ⎠
i31, 2e 2 2 cos sin 2
3 3
1 32 2 1 32 2
k z i
i i
π π π− ⎛ ⎞⎛ ⎞ ⎛ ⎞= − = − = − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞
= + − = − +⎜ ⎟⎜ ⎟⎝ ⎠
So the 3 roots are 1 3z i= − ± and 4− .
3( 2) 8z + + = [ ( 1 3 )z i− − + ][ ( 1 3 )z i− − − ][ 4z + ] = [ 1 3 ]z i+ − [ 1 3 ]z i+ + [ 4z + ]
y
x O 2−4
4−
4+k
(4, k)
RVHS 2010 H2 Math Prelim P1 7
(iii)
( ) ( )2
2
[ 1 3 ][ 1 3 ]
1 3 1 3 4
2 4
z i z i
z i z i z
z z
+ − + +
= + − + + +
= + +
Hence 2a = and 4b = .
(iv)
3
3
( 2) 8
2 8
2 2
z
z
z
+ = −
+ =
+ =
Thus, the roots in (i) lie on a circle centre (-2,0) and radius 2. (or deduce answer from the form in answers of (i))
Question 9 [10 marks] (i) ( ) ( )f ln 1 sinx x= +
( ) cosf1 sin
xxx
′∴ =+
( ) ( )1 sin f cosx x x′⇒ + = (shown) (ii) Differentiate with respect to x:
( ) ( ) ( ) ( )1 sin f cos f sinx x x x x′′ ′+ + = − Differentiate with respect to x: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 sin f cos f sin f cos f cosx x x x x x x x x′′′ ′′ ′ ′′+ + − + = −
At 0x = : ( )f 0x = , ( )f 1x′ = , ( )f 1x′′ = − , ( )f 1x′′′ =
( )2 3
f2 6x xx x∴ ≈ − +
Equation of tangent of ( )fy x= at 0x = is y x= .
(iii)
y
3 , 2.148π⎛ ⎞− −⎜ ⎟
⎝ ⎠
3 , 2.588π⎛ ⎞− −⎜ ⎟
⎝ ⎠
3 , 0.7578π⎛ ⎞
⎜ ⎟⎝ ⎠
3 , 0.6548π⎛ ⎞
⎜ ⎟⎝ ⎠
x O
( ) 2 31 1g2 6
x x x x= − +
( ) ( )f ln 1 sinx x= +
RVHS 2010 H2 Math Prelim P1 8
(iv) To find x such that ( ) ( )f g 0.05x x− <
Using GC, draw the graph of ( ) ( )f gy x x= − to find the intersection with 0.05y = .
−0.7745598 < x < 0.96923979 Range is −0.775 < x < 0.969.
Question 10 [10 marks] (a) Let the first term and common difference of the AP be a and d respectively.
2 32
a d a da a d+ +
=+
( ) ( )22 3a d a a d+ = +
2 2 24 4 3a ad d a ad+ + = + 4a d= − ------ (1)
( ) ( )20 2 19 2 9602
a d d+ + =⎡ ⎤⎣ ⎦
2 40 96a d+ = ------ (2) Subs. (1) into (2):
8 40 96d d− + = 3d⇒ =
(b) (i)
Mth Amt at beginning of mth
Amt at end of mth
Jan x 1.03x Feb 1.03x + 2x 1.032x +2(1.03)x Mar 1.032x +2(1.03)x +
3x 1.033x + 2(1.03)2x + 3(1.03)x
∴ Amt on the last day of March = $[1.033x + 2(1.03)2x + 3(1.03)x] = $x[1.033 + 1.032 + 1.03 + 1.032 + 1.03 + 1.03]
=$x ( ) ( ) ( )3 2 11.03 1.03 1 1.03 1.03 1 1.03 1.03 11.03 1 1.03 1 1.03 1
⎡ ⎤− − −⎢ ⎥+ +
− − −⎢ ⎥⎣ ⎦
= $x( )3 21.03 1.03 1.03 1.03 3
0.03+ + −
RVHS 2010 H2 Math Prelim P1 9
= $( )31.03 1.03 1103 3
3 0.03x ⎛ ⎞−⎜ ⎟−⎜ ⎟⎝ ⎠
= $( )3103 1.03 1 9103
3 3x ⎛ ⎞− −⎜ ⎟⎜ ⎟⎝ ⎠
= $ ( )12103 103(1.03) 1129
x− (shown)
(b) (ii)
( ) ( )3103 103(1.03) 112 2 3 1009
x x x x− − + + > 328.37x⇒ >
∴ least value of x is 329 Question 11 [12 marks] Let PB
uuur be b and F be the foot of perpendicular from B to l1.
Then ( )2 PF⎛ ⎞⋅⎜ ⎟ = ⋅ =⎜ ⎟⎝ ⎠
a b a b a aa
uuur$ $
2 PB PF FB⎛ ⎞⋅⎜ ⎟∴ − = − =⎜ ⎟⎝ ⎠
a bb aa
uuur uuur uuur,
which is parallel to l3.
Alternative: Use cross product.
( )2 2 2 2
4 81 4
8 42 3
4 14 1 23 2
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− ⋅ −⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎛ ⎞⋅ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟− = − − −⎜ ⎟ ⎜ ⎟⎜ ⎟ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
a bb aa
= 8 44 2 1
3 2
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 021
⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠
Since P(3, –5, 2) is a common point,
∴ 3
3 0: 5 2
2 1l μ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
r , μ∈ .
P B b
a F l1
l2
l3
RVHS 2010 H2 Math Prelim P1 10
Angle between 2l and 3l = 1
8 04 2
3 1cos
64 16 9 0 4 1−
⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟− ⋅ −⎜ ⎟⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎜ ⎟+ + + +⎜ ⎟
⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
= 76.3o
1
4 0 51 2 4
2 1 8
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − × − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
n
So, 5
2 44 8
kc
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
1252
c
k
⎧ =⎪⎪⇒ ⎨⎪ =⎪⎩
2
5: 4 18
8
⎛ ⎞⎜ ⎟Π =⎜ ⎟⎜ ⎟−⎝ ⎠
r
Since 2 50 4 181 8
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
, a point on 2Π is Q ( )2,0, 1− .
2 3 10 5 51 2 3
PQ−⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
Required distance = 1PQ ⋅nuuur
= 1 5
15 425 16 643 8
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⋅⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
= 13 10535
Alternative Method 1 to calculate distance:
RVHS 2010 H2 Math Prelim P1 11
548 21
1 105 105
548 18
2 105 105
: .
: .
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠ −
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
∏ =
∏ =
r
r
Thus, O is between the planes. So, distance = 21
105 + 18
105 = 39
105
Alternative Method 2 to calculate distance: Equation of line through P and perpendicular to 1 2and :Π Π
3 55 4
2 8μ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
r
Let N be the foot of perpendicular from P to 2Π .
3 5 5135 4 4 1835
2 8 8μ μ
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + = ⇒ =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
3 5135 435
2 8ON
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∴ = − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
uuur
Required distance = 5
13 13 105435 35
8PN
⎛ ⎞⎜ ⎟= =⎜ ⎟⎜ ⎟−⎝ ⎠
uuur
Question 12 [13 marks] (i) (ii)
Required area = ( )
3
302 2
2 d4
xx−
∫
2 –2 x
y
0
( )3
2 2
2
4y
x=
−
0.25
2
3
RVHS 2010 H2 Math Prelim P1 12
= ( )
π6π 3
22 2
2 2sin d4 4cos
θ θθ
⋅−−
∫
= ( )
π2π 3
26 2
2 2sin d8 sin
θ θθ
⋅∫
= π
22π6
1 cos ec d2
θ θ∫
= [ ]π2π6
1 cot2
θ−
= 1 0 32⎡ ⎤+⎣ ⎦
= 32
units2
(iii)
( )3
2 2
2
4y
x=
− ⇒ ( )
32 2 24 x
y− = ⇒
23
2 24xy
⎛ ⎞= − ⎜ ⎟
⎝ ⎠
Required Vol = ( ) ( )2 22 23 3
14
3 2 4 2 dy yπ π−⎛ ⎞
− − ⋅⎜ ⎟⎝ ⎠
∫
(or 2 2
03 (2)dyπ ⋅∫ )
= 22 1
3 3
14
6 4 2 3y yπ π⎡ ⎤
− − ⋅⎢ ⎥⎣ ⎦
= [ ]6 (8 6) (1 3)π π− − − − = 2π units3
RVHS 2010 H2 Math Prelim P2 Solutions 1
RVHS 2010 Preliminary Exam: 9740 H2 Mathematics Paper 2 Solutions Question 1 [6 Marks] (a) (b)
Question 2 [7 Marks] (i)
Area of triangle OAB = 12
OA OB×uuur uuur
= 1 4
1 2 12
3 0
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟× −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
= 3
1 122
9
−⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠
= 3 262
units2
x
dydx
− a −3a a 0
'( 2 2, 2)B a− −
0
2( 1)x a= − y
x2−
'( 6 2, 2)A a− − −
RVHS 2010 H2 Math Prelim P2 Solutions 2
(ii) ( )1 23
OP OA OB= +uuur uuur uuur
by ratio thm
= 1 4
1 2 2 13
3 0
⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟+ −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥−⎝ ⎠ ⎝ ⎠⎣ ⎦
= 301
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
Since //OQ OPuuur uuur
, 301
OQ c⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠
uuur for some constant c.
(iii) 0AQ BQ =
uuur uuur
3 1 3 42 1 0
3
c c
c c
− −⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
2 29 12 3 4 2 3 0c c c c c− − + − − + = 25 9 1 0c c− + =
Using GC, c = 1.68 or c = 0.119 (rej. since OQ > OP) Question 3 [7 Marks] (a) 22x t t= + , 2(1 )y t= −
2d d2 2 , 2(1 ) ( 1) 2(1 )d dd 2(1 ) 2(1 ) ( 1)d 2 2 2(1 ) 1
x yt t tt ty t t tx t t t
= + = − − = − −
− − − − −= = =
+ + +
For decreasing C, d 0dyx< .
( 1) 01
tt−
<+
Method 1:
So 1 1t− < < . Method 2:
Plot the graph of ( 1)1
tyt−
=+
and find the range of values of t such that 0y < .
+ +
1− 1
−
RVHS 2010 H2 Math Prelim P2 Solutions 3
So 1 1t− < < .
(b) At 2t = , 2 2 22(2) 2 8, (1 2) 1x y= + = = − = .
At ( )8, 1 , the equation of normal is
( )2
1 3gradient 31 1
1 3 83 25
t
tt
y xy x
=
+= − = − = −
−
− = − −
= − +
Substitute the parametric equations into the equation of the normal. 2 2
2 2
2
2
(1 ) 3(2 ) 251 2 6 3 254 4 24 0
6 01 4(1)( 6) 25 0
t t tt t t t
t tt t
− = − + +
− + = − − +
+ − =
+ − =− − = >
Thus, the normal will meet the C again. [Alternative: Factorise and solve the equation.]
Question 4 [10 Marks] d
dl klt=
1d dl k tl
=∫ ∫
ln l kt C= + (note: l > 0) kt Cl e +=
ktl Ae= (shown) 0A > and 0k <
When t = 0, l = L: A = L
∴ ktl Le=
l
t
L
0
RVHS 2010 H2 Math Prelim P2 Solutions 4
The model suggested by the researcher is not suitable as l → 0 when t → ∞ (i.e. the bar vanished).
, 0.5 : 0.5 kTt T l L L B Le= = = +
0.5 kTL B eL−
⇒ =
When t = 3T, 3kTl B Le= + = ( )3
2
0.5L BB
L−
+
Question 5 [10 Marks] (i)
Any horizontal line cuts the graph of f at most 1 point. Therefore, f is a one to one function and -1f exists.
3f : 2 1, 0x x x+ >a . 3
3
3-1
Let 2 1
12
1f : , 12
y x
yx
xx x
= +
−=
−→ >
(ii) For gf to exist,
(1, ) ( , )f gR D a∞ = ⊆ = ∞ . Hence largest possible 1a = . For 1a = ,
3
3
3
( )(2 1)
ln(2 1 1)ln(2 ), 0
gf xg x
xx x
= +
= + −
= >
( , )gfR = −∞ ∞ .
Since (0, )g fR D= ⊄ = ∞ , so fg does not exist.
RVHS 2010 H2 Math Prelim P2 Solutions 5
(iii) 1-f f ( ) , (0, )fx x x D= ∈ = ∞ 1
1-ff ( ) , (1, )fx x x D −= ∈ = ∞
The set of values of x such that 1 1- -f f ( ) ff ( )x x= is (1, )∞ .
Question 6 [3 Marks] Let A be the r.v. denoting number of students out of 120 students who make use of the
school’s e-learning portal. 5
8~ B(120, )A Since 120 50n = > is large, 120(5 / 8) 75 5np = = > and 120(3 / 8) 45 5nq = = > .
2258~ (75, )A N approximately.
P( 80)P( 81)P( 80.5) (by cc)0.150 (3.s.f)
AAA
>= ≥= ≥=
Question 7 [4 Marks] The 350 days of ticket sales volume are listed in order (numbered from 1 to 350) with
an interval of selection, 35050 7= . So every 7th day is chosen. The first sample is selected
randomly from the first 7 days, say the 2nd night is selected. Then the day 9th, 16th, 23rd,… up to a total of 50 days are chosen. To choose every 7th day means that a particular day sales will be chosen. E.g. if Monday is chosen, then the next 7th day will also be a Monday. But since the movie ticket sales are periodic in nature, the sample will be biased and not be representative of the actual ticket sales. Stratified sampling seems more appropriate here as we should choose samples from Weekends, Beginning of the week and Midweek as our strata so as to get a more representative sample.
y
x 0
(1, 1)
y
x 0
1-f f ( )y x=
1-ff ( )y x=
RVHS 2010 H2 Math Prelim P2 Solutions 6
Question 8 [8 Marks] (7 1)! 4! 10 172800− × × = (i) Case 1: First and last seats occupied by females
4 3 8!× × ways
Case 2: First and last seats occupied by males
6 5 8!× × ways
Total number of ways 4 3 8! 6 5 8! 1693440= × × + × × =
(ii) 10! 1512004!
=
Question 9 [8 Marks] (i) Let X be the random variable for the sales per day.
=n 12 ( 450) 622.8450 450 501.9
12 12x
x−
= + = + =∑
( )( )( )
( )
222
2
4501 45012 1 12
622.81 78798 4224.97090911 12
64.99977622
xs x
s
⎡ ⎤−⎢ ⎥= Σ − −
− ⎢ ⎥⎣ ⎦
⎡ ⎤= − =⎢ ⎥
⎢ ⎥⎣ ⎦=
∑
Test H0 : 452.80μ = against H1 : 452.80μ > at 1% level of significance. Test statistic:
Under H0, 452.80 ~ (11)
12
XT ts−
= since n = 12 is small.
p-value = calculatedP( ) 0.011983849 0.01T t> = > We do not reject H0. We conclude that there is insufficient evidence at 1% level of significance that the food store has underestimated its average daily sales. Assume that the daily sales follow a normal distribution.
RVHS 2010 H2 Math Prelim P2 Solutions 7
Since H0 is rejected at 5% level of significance, p-value < 0.05 for the one-tailed test. For the two-tailed test to reject H0 at 5% level of significance, 2 x p-value* < 0.05, which may or may not be true. So there may or may not be significant evidence. * the original p-value for the one-tailed test
Question 10 [9 Marks] (a) (i)
P(at least 2 of the 4 numbers are the same) = 1− P(all the 4 numbers are different)
( )( )( )( )6 5 346 6 6 61= −
1318=
Alternatively, P(at least 2 of the 4 numbers are the same) =P(exactly 2 same, and another 2 same) + P(exactly 2 same and 2 different) + P(exactly 3 same) + P(all 4 same)
( ) ( ) ( ) ( )4 4 4 44! 4! 4!1 1 1 12!2! 6 2! 6 3! 6 6
1318
6 6 6 6 2 62 3 1 2 1 1⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
=
(a) (ii)
P(exactly 2 of the 4 numbers are same at least one 6) P(2 out of the 4 numbers are same at least 1 six)
P(at least one six)∩
=
P(six appear twice) P(six appear once)P(at least one six)
+=
P(6,6, , ) P( , ,6, )1 P(no six)B C A A B+
=−
( ) ( )
( )
4 44! 4!1 12! 6 2! 6
456
5 5 22 2 1
1
⎛ ⎞ ⎛ ⎞⎛ ⎞+⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠⎝ ⎠=−
360 0.537671
or=
(b) 4( )7
P B A =( ) 4
( ) 7P A B
P A⇒ =
I 4( ) ( )7
P A B P A⇒ =I
( ) ( ) ( ) ( )P A B P A P B P A B= + −U I
[ ] 41 ( ) ( ) 1 ( ) ( )7
P A B P A P B P A′ ′ ′− = + − −I
40.65 ( ) 0.5 ( )7
P A P A= + −
30.15 ( )7
P A=
( ) 0.35P A =
RVHS 2010 H2 Math Prelim P2 Solutions 8
Question 11 [9 Marks] (i)
(ii) r = −0.885 This illustrates a moderate linear correlation between the radiation intensity and time which agrees with the scatter diagram, where the points do not seems to lie close to a straight line.
(iii) Equation of the regression line of x on t : 11.610575 0.806146x t= − 11.6 0.806x t= − (correct to 3 s.f.)
There is a decrease of 0.806 unit in the radiation intensity with every unit increase in time.
(iv) 0.995r = − (3 s.f) Since the value of the product moment correlation coefficient for ln x and t is closer to −1, the logarithmic model is more suitable than the linear model.
(v) Equation of the regression line of y on t : 2.624253 0.174726y t= − When x = 10, ln(10) 2.624253 0.174726y t= = − , ⇒ 1.84098 1.84t = ≈ (correct to 3 s.f.)
Question 12 [10 Marks] (i) Let X and Y be the r.v. denoting the number of disciplinary cases involving long hair
and faiure to wear school badge respectively in a given week. ~ (1.12 5)X Po × and ~ (4.2)Y Po
~ (9.8)X Y Po+ P( 9) 0.483188233 0.483X Y+ ≤ = ≈ (3.s.f)
(ii) P( 9 8 )X Y X+ ≤ ≥ P( 9 8)
P( 8)X Y X
X+ ≤ ∩ ≥
=≥
P( 8)P( 1) P( 9)P( 0)P( 8)
X Y X YX
= ≤ + = ==
≥
0.038145 0.0381= ≈ (3.s.f)
RVHS 2010 H2 Math Prelim P2 Solutions 9
(iii) Let W be the r.v. denoting the number of disciplinary cases involving long hair recorded in a given day.
~ (1.12)W Po ( 4) 1 ( 3)0.0272442116 0.0272
P W P W≥ = − ≤= ≈
Require Prob
[ ][ ]( )23( 3) ( 4) ( 4)
2P W P W P W⎛ ⎞
= ≤ ≥ ≥⎜ ⎟⎝ ⎠
55.87 10−= × (3 s.f )
(iv) The mean number of disciplinary cases involving long hair fluctuates throughout the year, usually having more such cases after the school vacation.
Question 13 [10 Marks] (a) Let C denotes the mass of a randomly chosen chicken.
Since n is large, 20.4~ (2.5, )nC N approximately by Central limit Theorem.
P( 2.45) 0.95C > > Method 1 1 P( 2.45) 0.95C− ≤ > P( 2.45) 0.05C ≤ <
Let 20.4
2.5 ~ (0, 1)
n
CZ N−=
0.42.45 2.5P( ) 0.05
n
Z −≤ <
18 1.64485n− < −
13.15882901n > 173.15n >
The least value of n is 174. Method 2 The least value of n is 174.
RVHS 2010 H2 Math Prelim P2 Solutions 10
(b) (i)
2~ (2.7, )X N σ 50
107P( 2.5) P( 2.9)X X< = <
[ ]50107P( 2.5) 1 P( 2.9)X X< = − ≥
[ ]50107P( 2.5) 1 P( 2.5)X X< = − <
157 50107 107P( 2.5)X < = 50
157P( 2.5)X⇒ < = 2.5 2.7 50
157P( )Z σ−< =
5015 157P( )Z σ< − =
15 0.4719777448σ− = −
0.424σ = (3.s.f) (Shown) Alternative method: Plot using GC: Y1=normalcdf(-E99,2.5,2.7,X) Y2=(50/107)normalcdf(-E99,2.9,2.7,X) Sketch the graphs and find point of intersection.
(b) (ii)
21 2 ~ (2(2.7), 2(0.424) )X X N+
2 21 24( ) ~ (4 2(2.7), 4 2(0.424) )X X N+ × ×
1 2P(50 4( ) 27.5)X X− + <
1 2= P(4( ) 22.5)X X+ > 0.354=
(b) (iii)
P(most expensive duck cost less than $11) = P(all ducks cost less than $11 each)
[ ] 44 114= P(4 11) P( )X X⎡ ⎤< = <⎣ ⎦
= 0.0895
SAINT ANDREW’S JUNIOR COLLEGE Preliminary Examination MATHEMATICS Higher 2 9740/01
Paper 1
Wednesday 15 September 2010 3 hours Additional materials : Answer paper List of Formulae(MF15) Cover Sheet
READ THESE INSTRUCTIONS FIRST Write your name, civics group and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Answer all the questions. Total marks : 100 Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically state otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematic steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages including this page. [Turn over
2
1. The first 4 terms of a sequence are given by 1 63u , 2 116u , 3 171u , . Given
that is a cubic polynomial in , find in terms of n . 4 234u
nu n nu
Hence write down the value of . 50u
[3] [1]
2. Without using a graphic calculator, solve 11
2 3
xx
x
.
Hence, solve 2
22
11
2 3
xx
x
exactly. [5]
3. Iron Will, the magician, is constructing a prism with an equilateral triangle base for his latest escape act.
The edges of the prism are made from iron rods, the rectangular faces of the prism are made of glass panels, and the triangular faces of the prism are made of wooden boards.
The volume of the prism is fixed at exactly 2 3 cubic metres.
Show that2
8h
x , where x is the length of the sides of the
triangular base and h is the height of the prism. The cost of the iron rods is $1 per metre, the cost of the
wooden boards is $ 2 3 per sq. metre and the cost of the glass panels is $2 per sq. metre. Show that the expression of the total costC of constructing the prism in terms of x is as follows:
2 13 6 48 24C x x x x 2 Using an analytical method, find the minimum cost of constructing this prism.
[2] [2]
[4]
4. (a) Solve the equation , expressing your answer(s) in the form 2 3 4w i x iy . [3]
(b) Find the fourth roots of –16, expressing your answers in the form ier , where and 0r . On an Argand diagram, mark the points that represent the
roots clearly.
[5]
[Turn over
3
5. (a) Functions f and g are defined by
2f : 3x x for 0x
g : 2 4x x for all Rx (i) Find in a similar form. 1f
(ii) Sketch the graphs of f, 1f and on the same diagram. 1f f
(iii) Given that exists, find 1gf 1gf ( ) x .
[2]
[1]
[1]
(b) The function h is defined by h(x) = 212 x for 0 3 x for 3 63 6x x and h(x) = h( 6x ) for all Rx (i) Find h(16) + h(25). (ii) Sketch y = h(x) for . 6 1x 2
[2] [3]
6. A curve is defined parametrically by the equations C
1 cos x t , sin y t t for 2 2
t
(i) State the exact range of values that x and y can take. Hence sketch the curve C . Label your graph, indicating the axes intercept(s) clearly if any. [3]
(ii) The tangent to the curve, C at the point where P2
t
is denoted by . l
Find the Cartesian equation of l . [3]
(iii) Find the range of values of m such that the line y mx intersects C at two points. [2]
7. The equations of two planes , are given by 1 2
1
2
: 2 4 8
: 2 6
x y z
x z
(i) Find the vector equation of the line of intersection l between the planes and1 2 . [2]
(ii) Find the foot of perpendicular, from the point 1F 6,9, 2 to the plane . 1[3]
[Turn over
4
Another plane contains the points and and is parallel to l . 3 1F 2F
(iii) Given that 2
26
59
2
5
OF 3
1)
, show that the Cartesian equation of the plane is given
by 15 . 8 40 22x y z
[3]
(iv) By moving plane by m units in the direction of vector v, all three planes will
meet at . Find m and v. 3
l [3]
8. In a “man vs machine” contest, an athlete competes against a robot to see who can pull himself up a 16 m rope faster. The athlete climbs up 0.8 m in his first pull; the height attained by each subsequent pull decreases by a factor of 1/20 as he grows more and more tired. The robot is able to pull itself up by 0.4 m each time as it never grows tired.
(i) The athlete and robot both start climbing at the same time from the base of the rope. Assuming they pull at the same rate, after how many pulls will the robot overtake the athlete?
[4]
(ii) Will the athlete be able to reach the top of his rope? Justify your answer. [2]
(iii) The contestants again start climbing from the base but the athlete changes his strategy. He continues pulling at the same rate, but now he climbs up x m in each of his first two pulls; (x – 0.02) m in each of his next two pulls; and every subsequent pair of pulls shows a similar decrease of 2 cm from the previous pair. Show that the distance he travels after 2n pulls is 2 0.02 ( nx n n . Given that he won the contest, find, correct to 2 decimal places, the minimum value of x.
[4]
16m
[Turn over
[Turn over
5
9.
It is given that
1
2
tand e
d 1
xy
x x
, where 1tan x denotes the principal value.
(i) Find an expression for y in terms of x given that y = 1 when x = 0. [2]
(ii) Show that
22
2
d d1 (2 1)
dd
y yx x
xx0 . [2]
(iii) By further differentiation of the result in (ii), find the Maclaurin’s series for y up to and including the term in x3.
[4]
(iv) Hence, deduce the series expansion for
(a) 1
2
tane
(1 )
x
x
(b)
12 tane x up to and including the term in x2.
[4]
x
10. (a) Prove by induction for all n that
1
! 1n
r
r r n
! 1 [5]
(b)
The diagram shows part of the graph of 3 22 5y x x x 6 , and
. The 3 real roots of
the equation are denoted by 3 22 5 6x x x 0 where < .
0
(i) Find the values of , and .
A sequence of real numbers 1 2 3 4, , , ,...x x x x satisfies the recurrence relation
231 1 5 8 5n n n nx x x for .
[1]
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6
(ii) Prove algebraically that if the sequence converges to L, then L is equal to either , or .
(iii) Show that if 1n nx x , then orn nx x .
[3] [3]
11. (a) Given that A(sin + cos ) + B(cos – sin ) 4 sin , Find the values of the constants A and B.
Hence find the exact value of 1
4
0
4sind
sin cos
.
[2] [3]
(b)
(i) Express 2
2
( 1)
t
t in partial fractions.
[2]
(ii) Hence, find the exact value of
5
1
1d
2 1x
x x , using the substitution
t = 2 1x .
[6]
End of Paper
SAINT ANDREW’S JUNIOR COLLEGE Preliminary Examination MATHEMATICS Higher 2 9740/02
Paper 2
Monday 20 September 2010 3 hours Additional materials : Answer paper List of Formulae(MF15) Cover Sheet
READ THESE INSTRUCTIONS FIRST Write your name, civics group and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Answer all the questions. Total marks : 100 Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically state otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematic steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages including this page. [Turn over
2
Section A: Pure Mathematics [40 marks]
1 The diagram below shows a rectangle tank where and
1
2
4
AG
A
C
is parallel to the
vector .
2
1
2
(i) Find the length of projection of AG
onto the line passing through A and C. [2]
(ii) Given that GI , find : 3:IC 2 AI
. [2]
(iii) Find the acute angle between AG
and GC
. [2]
2 In a chemical reaction a compound X is formed from a compound Y. The mass in grams of X and Y present at time t seconds after the start of the reaction are x and y respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time t, the rate of formation of X is proportional to the mass of Y at that time.
When t = 0, x = 5 and d
1.9d
x
t .
(i) Show that x satisfies the differential equation d
0.02(100 )d
xx
t . [2]
(ii) Solve this differential equation, obtaining an expression for x in terms of t. [4]
(iii) Calculate the time taken for the mass of compound Y to decrease to half its initial value.
[2]
(iv) Sketch the solution curve obtained in part (ii) and state what happens to compounds X and Y as t becomes very large.
[3]
G
B
C
E
D
F
H
I
A
[Turn over
3 (a) Indicate clearly on a single Argand diagram, the locus of z that satisfy
0 arg( 1 i)2
z
and i 2 z .
Hence, find the complex number , in the form x + iy, for which z arg( 1)4
z
,
leaving your answer in exact form.
[6]
(b) Consider the complex number
33 32 cos i sin cos i sin
4 4 6 6
w
.
(i) Find the exact values of the modulus and argument of w. Hence, express in modulus-argument form.
w[3]
(ii) Find the possible value(s) of , when n is a multiple of 4. nw [2]
4. The curve has equation , where C fy x 2
8f
1
xx
x
.
(i) Using a non-calculator method, find, in simplest form, the exact coordinates of the turning points of . C
[3]
(ii) Sketch the curve , indicating clearly any axis intercept(s) and asymptote(s). C [2]
(iii) Given that , show that 0n 2
0f d 2 ln 1
nx x n . Hence deduce the exact
value of 2
2f dx x
. [3]
(iv) Sketch on a separate diagram, the graph of fg x x , labeling clearly any
stationary point(s) and asymptote(s). The region R is bounded by the curve y = g(x), the lines 0x , 1x and the x-axis. Find, in exact form, the volume generated when R is rotated completely about the x-axis. [4]
Section B: Statistics [60 marks]
5 ABC College has a student population of 1800, of whom 60% are female students and 40% are male students. The College intends to get a sample of 100 student volunteers to take part in a survey on the College’s National Education Programme.
(i) Comment whether such a sample consisting of all the volunteers is likely to give a true picture of the opinions of all the students about the College’s National Education Programme.
[1]
(ii) Suggest a method of obtaining a more representative sample and describe how it may be carried out.
[3]
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4
6 James has 9 marbles, 4 of them are red, 3 of them are blue, and 2 of them are yellow. He arranged these marbles on a straight line. All marbles are identical except for their colour.
Find the number of ways that the marbles can be arranged if
(a) there are no restrictions, [1]
(b) the arrangement is symmetrical about the centre marble, (e.g. R-R-Y-B-B-B-Y-R-R) [2]
(c) each blue marble is between two marbles of the same colour. (e.g. Y-B-Y or R-B-R) [4]
7 (a) A & B are two events with non-zero probability. Explain if each of the following statements is necessarily true, necessarily false, or neither necessarily true nor necessarily false.
(i) If A & B are mutually exclusive, then they are independent. [1]
(ii) If A & B are independent, then they are mutually exclusive. [1]
(b) A teacher brings 4 black, 3 blue, 2 red and 1 green markers to the classroom for each of his lessons. Unknown to him, the probabilities that a black, blue, red or
green marker is out of ink are , p1
2p ,
1
4p and
1
8p respectively, where
. 0 1 p
(i) Find, in terms of p, the probability that a randomly chosen marker from his set of ten markers is out of ink. [2]
(ii) In the classroom, the first marker he tries out is out of ink. Find, in terms of p, the probability that the next marker he tries out is a red marker that is also out of ink. [3]
(iii) After numerous lessons, the teacher realised that, in general, the first marker he tries out works at least 7 out of 10 times. Find the range of possible values of p. [2]
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5
8 (a) A secret source claims that 0.3% of the residents in the suburbs of Zozoland are spies from Buzzland. If the claim were true, what is the probability, using a suitable approximation, that there are at least five spies living in a Zozoland suburb of 1200 residents?
[3]
(b) Another independent source (assumed to be reliable) claims that there is an average of 1.3 Buzzland spies and 0.4 Dodoland spies in a typical Zozoland city.
(i) The investigation bureau will only do extensive combing of a city where there are more than five spies present. Given that there are 10,000 Zozoland cities, estimate the number of cities that will be under investigation. [3]
(ii) The bureau investigated a certain number of cities and caught 23 Buzzland spies and 11 Dodoland spies. Find the most likely number of cities that had been investigated, stating an assumption needed for your calculation in the context of the question. [4]
9 Marine biologists discovered that in a remote island off Philippines, a type of algae is growing at an exponential rate and is threatening the marine life in the region. The growth of the algae was collected over a period of 10 years and was recorded as follows (measured as cell density in millions). Year (x) 1 2 3 4 5 6 7 8 9 10
Cell density(y) 1.21 1.66 2.83 4.35 4.91 6.55 8.01 9.66 12.72 18.01
(i) The relationship between the year, x, and the cell density recorded as y, are related
by , where a and b are unknown constants. By plotting a scatter diagram,
comment on the relationship between x and ln y.
bxy ae[2]
(ii) Find the estimated regression line of ln y on x. Hence calculate estimates of a and b.
[3]
(iii) Estimate the cell density at Year 12. Comment on the reliability of your answer. [2]
(iv) Estimate the year at which the cell density is at 7 million. Comment on the choice of your regression line. [2]
10 (a) In a manufacturing company, the mean salary of its employees is S$30,000. The salary structure of the company is such that only 20% of the employees earn higher than the mean salary. Explain whether the use of Normal distribution to model the salary distribution of the employees in this company is appropriate.
[2]
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6
(b) Anne travels to work each day by bus. The total time, TA in minutes for Anne’s journey to her office, is a normal random variable with a mean of 55 and a variance of 25. Ben drives to work every day. The total time, TB in minutes, that Ben spends driving to work is a normal random variable with a mean of 53 and a variance of 16.
(i) Find the probability that the average of the individual times taken by Anne and Ben to travel to work in a day is between 50 to 60 minutes.
[2]
(ii) Find the greatest value of a, if the probability that Ben’s travel time differs from 53 minutes by at most a minutes is not more than 0.6.
[2]
(iii) Using a suitable approximation, calculate the probability, that for a randomly chosen period of 60 days, there are at least 43 days on which Anne will take longer than Ben to travel to work.
[4]
11 (a) A water treatment plant monitors their drinking water from their storage tanks on
an hourly basis. The water must maintain a pH level of 8.5 (they try to maintain an
alkaline level) for it to be ‘drinkable’. At 0800 hours, they recorded the following
pH level from 11 of their tanks:
Tank 1 2 3 4 5 6 7 8 9 10 11
pH level 8.31 8.41 8.51 8.46 8.52 8.48 8.33 8.1 8.39 8.42 8.52
(i) Using an appropriate test, determine at 1% level of significance, if this sample provides sufficient evidence that the mean pH level of the water differs from 8.5?
[3]
(ii) State an assumption necessary for the test in a(i) to be valid. [1]
(b) At 1500 hours, a random sample was recorded from 80 tanks. Denoting the pH
level readings by x, the results are summarized as follows:
, ( 8.5) 15.2x 2( 8.5) 232.2x
(i) Find the unbiased estimate of the population mean and variance. [2]
(ii) Another sample of n (assume n is large) readings was recorded. Using the unbiased estimate of the population mean and variance found in (b)(i), find the least value of n so that the probability that this sample mean has a pH reading of less than 8.2 is less than 0.3. [5]
End of Paper
1
[Turn Over
SERANGOON JUNIOR COLLEGE
2010 JC2 PRELIMINARY EXAMINATION
MATHEMATICS
Higher 2 9740/1
Wednesday 18 August 2010
Additional materials: Writing paper
List of Formulae (MF15)
TIME : 3 hours
READ THESE INSTRUCTIONS FIRST
Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
Total marks for this paper is 100 marks.
This question paper consists of 6 printed pages and no blank pages.
2
Answer all questions (100 marks).
1 Without using the calculator solve the inequality
22 3x x+ < . [3]
Hence solve the inequality
122 3
xxe e+ < . [2]
2 (i) Show that 1
2
d 1sin
d 1
xx
x
− =
−
for 1x < . [2]
(ii) Given that ( )1
2 4 2
0 2
1d 2cos d
1a
x x x
x
π=
−∫ ∫ , show that
sin 2 2 1 0a a+ − = . [3]
3 A sequence of numbers }{ nx is defined by the relation
( )119 2n
n nx x−−= − for 2,3,4,...n = and 1 7x = .
Write down the values of 2 3 4 5, , and x x x x in the form of ( ) ( )1nna b − − , where a and b are
real numbers. [2]
Hence make a conjecture for nx and prove the conjecture using Mathematical Induction.[4]
4 Given that 3 sinye e x x= + + . Show that
223 3
2
d d3 9 sin 0.
dd
y yy ye e x
xx
+ + =
[3]
Hence, find in terms of e, the Maclaurin’s series for y, up to and including the term in x2. [4]
3
[Turn Over
r
5 The function f is defined by f 2: 3 2x x x+ −� for x ∈� , x k≥ .
(i) Find the least value of k such that f has an inverse. [2]
(ii) Using the value of k in part (i), find f –1(x) and state the domain of f –1 . [4]
(iii) Hence, find the exact solution of the equation f (x) = f –1 (x). [2]
6 (a) The diagram, not to scale, shows the graph of y = f(x). The vertical asymptotes are
x = 0 and x = 2. Sketch the graph of 'f ( )y x= , showing all corresponding
coordinates and asymptotes where possible. [3]
(b) A cylindrical silo with a hemispherical rooftop is constructed to store
rubbish.
The cost of each unit area of the rooftop is fixed at $21 and the cost of
each unit area of the curved surface of the cylinder is fixed at $7. The
total cost of the rooftop and the curved surface is fixed at $2100.
Given that the radius of the hemisphere is r, show that the volume V of
the structure can be expressed as
37150
3V r r
π= − . [3]
Find the exact cost of the hemispherical top when V is at its maximum. [4]
[Curved surface area of sphere = 24 rπ ; volume of sphere = 34
3rπ ]
-1/2
B(1, 3)
2
y
x
y= f(x)
A(-1, 2)
4
7 (a) A complex number w is such that 64 3 i 16i 0 and Im( ) 5ww w w+ + = <∗ , where w∗
is the conjugate of w .
(i) Find w in the form i, where , .x y x y+ ∈� [3]
(ii) Find the integer values of n such that is realnw . [3]
(b) Find the Cartesian equation of the locus of the point P representing the complex
number z where 1 i 3 1 iz z− − = + . Hence sketch the locus of point P. [5]
8 Lovewell the Frog is stuck at the bottom of a well. It tries to make a series of vertical upward
jumps on the well wall to get out. Its first jump is 1 m, but because of fatigue from over-
exertion, each subsequent jump is 9
10 of the distance of the previous jump.
However, as the wall is wet and slippery, Lovewell slips down some distance after every jump is made. It slips down 0.3 m after its first jump, but because it learns to grip better, each subsequent slip is 0.02 m less than that of the previous slip.
For example, its first jump gets it 1m up the well, and then it slides down 0.3 m.
From its new position it makes a second jump up 0.9 m, and then it slides down 0.28 m. Its third jump sees it go up a further 0.81 m, and it slides down 0.26 m; so on and so forth. (i) How many jumps are needed for Lovewell to clear the 4m mark in the well? [5]
(ii) How many jumps are needed for Lovewell to clear the 6.5m mark in the well? [4]
(iii) Determine the highest mark in the well that Lovewell is able to reach? [2]
9 The position vectors of the points A, B, C and D are given as i + 3j , 2 j + 4k , i + j + k and
4i + 5k respectively.
(i) Find the vector equation of plane π in the form r ⋅ n = p , that contains the points A, B
and C. [3]
(ii) Find the foot of the perpendicular of the point D to the planeπ . Hence find the
shortest distance from the point D to the plane π . [4]
A line l parallel to the vector +j k passes through point D and it meets the plane π at the
point N.
(iii) Find the position vector of the point N and hence find the vector equation of the
reflection of line l about the plane π . [5]
5
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10 (a) The curve C has equation ( )2
f9
x ax by x
x
+ += =
+ where a, b are real constants. If the
two asymptotes of the curve intersect at the point ( )9, 12− − and the x-axis is a tangent
to the curve, find the values of a and b. [4]
Sketch the curve ( )fy x= , stating clearly the equations of the asymptotes and the
coordinates of any points of intersection with the axes. Hence deduce the values of p
and q such that the equation ( )f x p q kx+ − = has no real roots for 1k ≤ . [3]
(b) The diagram below shows the sketch of the graph ( )fy x= for 0a > .
Sketch on separate diagrams the graphs of
(i) ( )1
fy
x= , [3]
(ii) ( )2 fy x= , [2]
showing clearly the equations of any asymptote(s), the coordinates of any turning point(s) and axial intercept(s).
( )3 , 2a a−
( )5 ,3a a
a− 2a
4x a=0x =
0y =
y ax=
x
y
0
6
11 (i) Solve the equation 5 32z = − , giving your answers in the form eir
θ ,where 0r > and
π θ π− < ≤ . [4]
(ii) Hence, show the roots of the equation 5 32iw = in an Argand diagram.
The roots represented by W1 and W2 are such that 2
π− < arg (w1) < arg (w2) <
2
π.
Find the area of triangle OW1W2 where O is the origin. [5]
(iii) Express 5 32z + in the form
( )( )( )2 2cos 4 cos 4z p z z p z z kα β− + − + +
where , , andp kα β are constants to be found. [4]
End of Paper
1
[Turn Over
SERANGOON JUNIOR COLLEGE
2010 JC2 PRELIMINARY EXAMINATION
MATHEMATICS
Higher 2 9740/2
Wednesday 25 August 2010
Additional materials: Writing paper
List of Formulae (MF15)
TIME : 3 hours
READ THESE INSTRUCTIONS FIRST
Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
Total marks for this paper is 100 marks.
This question paper consists of 9 printed pages and 1 blank page.
2
[Turn Over
Section A: Pure Mathematics [40 marks]
1 The computer company call “Orange” manufactures the latest electronic gadget in town
called the iBoard with 3 different storage capacities namely 16GB, 32GB and 64GB. The
profit earned from each unit sold is as shown in the table below.
Storage Capacity 16GB 32GB 64GB
Profit $x $y $z
Within the first week after it was officially launched, the sales from 3 of its outlets for the three different storage capacities is as shown below.
The total profit collected from outlets B and C are $38 750 and $8750 respectively. If the total profit earned due to both the sales of 16GB and 32GB iBoard is equal to 12 times the total profit earned from the sales of the 64GB iBoard, find the value of x, y and z.
[4] Find the total profit collected from outlet A. [1]
2 A curve C has parametric equations
x = a sin2 t, y = a cos t, where a > 0.
(i) Sketch the curve. [2]
(ii) Find the equation of the normal at the point P where t = . [3]
(iii) Using a non-calculator method, determine whether the normal at P will meet C
again. [3]
02
tπ
≤ ≤
3
π
Storage Capacity 16GB 32GB 64GB
Outlet A 75 120 20
Outlet B 180 230 70
Outlet C 45 50 10
3
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3
The diagram above shows the graph of ( )fy x= . On separate diagrams, sketch the graphs
of
(i) ( )f 1 2y x= − , [3]
(ii) ( )fy x= , [2]
(iii) ( )2f 1y x= + . [3]
showing in each case, the coordinates of the points corresponding to A, B, C and the
equations of the asymptotes.
C(3,0)
0 2
1
x
• A(4, 3)
B(1, – 1)
y
4
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4 (a) (i) Show that( )2ln1
ln d ln .2
ttt t t t t C
t
+ = − + +
∫ [2]
(ii) A curve C is defined by the parametric equations
ln
ln , 0.
x t t
y t t t
= +
= − >
The region R, which is bounded by the curve C, the line 1x = ,
2 ln 2x = + and the x – axis is as shown below. Find the exact area of R. [4]
(b) The region S is enclosed by the curve D with equation ( )2
2 4y x+ = − and the
line y x= .
Find the volume generated when S is rotated through 2π radian about the x – axis, giving your answer correct to 2 decimal places. [3]
y
x 1 2 ln 2+
C
R
0
5
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5 A disease is found to be present in a protected reserve containing 35 orangutans. The rate at which the number of infected orangutans, x, is increasing at any time t is proportional to the product of the number of infected orangutans and the number that have yet to be infected at that instant. Initially there were 5 animals infected.
Form a differential equation that describes this model and show that
35
35
35,
1
kt
kt
Aex
Ae=
+
where >0 is to be found. A [6] Deduce the total number of infected orangutans after a long period of time and represent the solution to this model on an appropriate graph. [4]
6
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Section B: Statistics [60 marks]
6 A survey on dining experience was undertaken in a small town with 10 three-star
restaurants, 60 two-star restaurants and 30 one-star restaurants.
The 100 restaurants in the small town are numbered from 1 to 100. A sample of 10 restaurants is selected by randomly choosing 10 numbers that are assigned to the restaurants. (i) Suggest one disadvantage of this sampling method. [1]
(ii) Suggest a better method of sampling and explain briefly how this could be done.
[3]
7 In a certain country, the probability that it rains on a given Tuesday is 1
5. For each of the
next 2 days, Wednesday and Thursday, the conditional probability that it rains, given that
it rained the previous day is α and the conditional probability that it rains, given that it
did not rain the previous day is β . The situation is illustrated in the uncompleted tree
diagram below.
(i) Complete the tree diagram to represent all the possible outcomes up to Thursday.
[2]
For 1
3α = and
2
3β = , find
(ii) the probability that it rains on a Thursday, [2]
(iii) the probability that it rains on at least two days given that it rains on a Thursday.
[3]
R
R’
1
5
Tuesday
R
R’
R 1
5
R’
R
α
R’
R
β
R’
R
Wednesday
7
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8 Paul ordered 9 plates of sushi, namely 3 plates of unagi sushi, 2 plates of uni sushi and 4
plates of ebi sushi.
(i) Find the number of ways Paul can eat the 9 plates of sushi. [2]
(ii) Find the number of ways Paul can eat the 9 plates of sushi such that the first plate
and last plate are different types of sushi. [4]
(iii) Find the number of ways of arranging the nine plates of sushi on a round table
such that no two plates of ebi sushi are placed together. [2]
9 Miss Curious wants to determine if there is any correlation between the amount of
preparation and the results obtained in a recently concluded exam.
She asked her friends how much time they spent preparing for the exam (x), with their
exam scores (y), and recorded her findings in the table below.
(i) Give a sketch of the scatter diagram for the data and find the equation of the least squares regression line of y on x. [2]
(ii) State, with a reason, which of the following would be an appropriate model
to represent the above data (where a and b are constants and b > 0).
b
y ax
= +A : x
y a be−= +B : lny a b x= +C : [2]
(iii) For the appropriate model chosen, find the values of a and b. [1]
Explain how this model is a better one than the equation found in part (i). [1]
(iv) Obtain a good estimate of the score of a student who spent 8 hours studying for the exam and comment on the reliability of your answer. [2]
x (hour) 10 15 22 27 38 46 53 64
y (score) 11 40 51 56 61 62 64 66
8
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10 You are an Intelligence Quotient (IQ) expert. While reading the newspaper, you become
interested in a newspaper advertisement that reads as follows:
As a concerned IQ expert, you would like to investigate the validity of this
advertisement. You know that for the general population of children, the mean IQ is 100.
Through close contacts in the industry, you confirmed that the scientific study as stated in
the advertisement is valid.
(i) Test at 5% significance level, whether the mean IQ points of children who
participated in the DDD program has increased. State any assumptions that you
have to make in carrying out the test. [6]
(ii) What do you understand by 5% significance level in this context? [1]
Suppose that the population standard deviation σ is now known and a larger sample size
of 50 children is taken.
(iii) Find the range of values of the sample mean in terms of σ if the conclusion at
5% significance level is now different from that concluded in part (i). [4]
Increase the IQ of your children by 10 points in just 16 weeks! �
Subscribe now to Dr. Dune’s Drill (DDD) program
and astound your children’s friends, teachers and grandparents!
Assure a university education for your children
(and security for you in your old age). �
A scientific study of 15 children from all over Singapore
showed an average IQ score of 108*
after only six weeks of the fantastic DDD program .
*standard deviation is 15
9
[Turn Over
11 In a certain country, it is found that on average the number of pairs of twins born weekly
are 2.
(i) Find the most probable number(s) of pairs of twins that are to be born weekly. [1]
(ii) Find the probability of having at most 7 pairs of twins to be born in a two-week
period. [2]
(iii) Assuming there are 26 two-week periods in a year, estimate the probability that there are less than 73 two-week periods with at most 7 pairs of twins born in 3 years. [3]
(iv) Using a suitable approximation, find the least number of consecutive weeks such that the probability of having at most 20 pairs of twins born falls below half. You may assume that the number of weeks is more than 5. [3]
(v) Find the probability that the mean number of pairs of twins born in 50 weeks is less than 1.8. [2]
12 An ornithologist, who studies the behavior of birds, captures one male and one female
hornbill from a forest in Osaka, Japan. The masses of hornbills in that forest are assumed
to follow normal distributions with male hornbills having mean 3500g and standard
deviation 150g while female hornbills having mean 3000g and standard deviation σ .
(i) It is found from research that 5% of the female hornbills from the forest have
masses exceeding 3.2kg. Show that 122σ = . [2]
(ii) Find the probability that the difference in mass between two randomly chosen
male hornbills is at least 0.1kg. [3]
(iii) Find the probability that the mass of 5 randomly chosen female hornbills exceeds
twice the mass of 2 randomly chosen male hornbills. [3]
(iv) Five male hornbills are randomly chosen. Find the probability that the fifth male
hornbill is the third hornbill with mass exceeding 3.6kg. [3]
End of Paper
10
[Turn Over
BLANK PAGE
1
[Turn Over
SERANGOON JUNIOR COLLEGE
2010 JC2 PRELIMINARY EXAMINATION
MATHEMATICS
Higher 2 9740/1
Solutions
1 22 3x x+ <
23 2 3x x− < + < 23 2x x− < + and 22 3x x+ <
22 3 0x x+ + > and 22 3 0x x+ − < 21 23 0
4 16x⎛ ⎞+ + >⎜ ⎟
⎝ ⎠ and ( )( )2 3 1 0x x+ − <
x∈ and 3 12
x− < <
3 12
x∴− < <
21 1 1
2 2 22 3 2 3x x xxe e e e
⎛ ⎞+ < ⇒ + <⎜ ⎟⎜ ⎟
⎝ ⎠
∴replace x with 12
xe
1 12 23 1 0 1
2x x
e e− < < ⇒ < <
1 ln12
x <
0x∴ <
2
[Turn Over
2(i) Now, − π2 ≤ sin−1 x ≤ π2 where x ≤ 1
Let y = sin−1 x ⇒ sin y = x ⇒ cos ydydx
= 1
⇒dydx
=1
cos y
⇒dydx
=1
± 1− sin2 y
⇒dydx
=1
± 1− x2
⇒
dydx
=1
+ 1− x2
(since from the graphdydx
≥ 0 for all x ≤1)
(ii) sin−1 x⎡⎣
⎤⎦0
12 = cos2x +1( )dx
a
π4∫
π4− 0 =
12
sin2x + x⎡⎣⎢
⎤⎦⎥a
π4
π4=
12+π4
⎛⎝⎜
⎞⎠⎟−
12
sin2a + a⎛⎝⎜
⎞⎠⎟
sin 2 2 1 0a a⇒ + − =
3 ( ) ( ) ( )22 1 22 19 2 18 7 11 3 2 1x x−= − = − = = − −
( ) ( ) ( )33 1 33 29 2 36 11 25 3 2 1x x−= − = − = = − −
( ) ( ) ( )44 1 44 39 2 72 25 47 3 2 1x x−= − = − = = − −
( ) ( ) ( )55 1 55 49 2 144 47 97 3 2 1x x−= − = − = = − −
OR Using G.C,
( ) ( )222 11 3 2 1x = = − −
( ) ( )333 25 3 2 1x = = − −
3
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( ) ( )444 47 3 2 1x = = − −
( ) ( )555 97 3 2 1x = = − −
From above, Conjecture: ( ) ( )3 2 1 nn
nx = − −
Let Pn be the statement that ( ) ( )3 2 1 nnnx = − − for n +∈
When n = 1,
L.H.S ( ) ( )111 3 2 1 7 R.H.Sx= = − − = =
Hence P1 is true. Assume Pk is true for some k +∈ i.e ( ) ( )3 2 1 kk
kx = − − To show that Pk+1 is true, i.e ( ) ( ) 11
1 3 2 1 kkkx +++ = − −
So ( )( ) ( )1 1
1 1 1L.H.S 9 2 kk kx x+ −+ + −= = −
( ) ( ) ( )9 2 3 2 1 kk k⎡ ⎤= − − −⎣ ⎦
( ) ( ) ( ) 19 2 3 2 1 kk k += − − −
( ) ( ) ( ) 19 2 3 2 1 kk k += − − −
( ) ( ) 16 2 1 kk += − −
( )( ) ( ) 13 2 2 1 kk += − −
( ) ( ) 113 2 1 R.H.Skk ++= − − =
Hence Pk true => Pk+1 is true. Since P1 is true and Pk=> Pk+1 is also true, by Mathematical Induction, Pn is true for all positive integer n.
4
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4 3 sinye e x x= + +
3 sinye e x x= + +
Differentiating w.r.t. x, 3 d3 1 cos
dy ye x
x= +
3e3y d2 ydx2 + 9e3y dy
dx⎛⎝⎜
⎞⎠⎟
2
+ sin x = 0 (shown)
When x = 0,
3 1sin3
ye e x x y= + + ⇒ =
3 d d 23 1 cosd d 3
y y ye xx x e= + ⇒ =
3e3y d2 ydx2 + 9e3y dy
dx⎛⎝⎜
⎞⎠⎟
2
+ sin x = 0 ⇒ 3ed2 ydx2 + 9e
23e
⎛⎝⎜
⎞⎠⎟
2
= 0 ⇒d2 ydx2 = −
43e2
Hence, the Maclaurin series of y is
y =13+
23e
x +−
43e2
⎛⎝⎜
⎞⎠⎟
2!x2 + ...
≈13+
23e
x −2
3e2 x2
5
(i) Least value of k = 1.5
(ii) let 22 3y x x= − −
⇒ 21 252( )4 8
y x= − − ⇒ 1 8 254 16
yx += ±
1 1 8 25f ( ) , 04 16
xx x− +∴ = + ≥
(iii) 22 3x x x− − = ⇒ 1 7
2x +=
5
[Turn Over
6(a)
(b) Area = 22 2r rhπ π+ (where h is the height of the cylinder)
22100 2 (21) 2 (7)r rhπ π= +
2100 14 (3 )r r hπ= + 150 3h r
rπ= −
3 2 3 2 3 3
3
2 2 150 2 3 150 33 3 3
71503
V r r h r r r r r rr
r r
π π π π π ππ
π
⎛ ⎞= + = + − = + −⎜ ⎟⎝ ⎠
= −
2d 150 7 0
dV rr
π= − =
150 (reject <0)7
r rπ
⇒ =
2
2d 14 0 Volume is max.d
V rr
π= − < ⇒
Thus, cost of hemispherical top= 2πr2(21) 1502 (21)= $9007
ππ
=
A(-1, 0) B(1, 0)
x =2
y
x
y= f’(x)
x =0
6
[Turn Over
7(a) (i) 64 3 16 0ww i iw∗ + + =
( )( ) ( )64 3 16 0x yi x yi i i x yi+ − + + + =
2 2 64 3 16 16 0x y i ix y+ + + − =
( ) ( )2 2 16 16 64 3 0x y y x i+ − + + =
Comparing coefficients,
( )16 64 3 0x + = and ( )2 2 16 0x y y+ − =
4 3x = − 2 16 48 0y y− + =
( )( )4 12 0y y− − =
y = 4 since y < 5
4 3 4w i∴ = − +
(ii) 8w = , ( ) 5arg6
w π= , thus
5 58 cos sin6 6
n n n nw iπ π⎛ ⎞= +⎜ ⎟⎝ ⎠
Since nw is real, Im ( ) 0nw = , 5 ,6n k kπ π= ∈
6 ,5kn k⇒ = ∈
For n to be an integer, 6 ,n k k= ∈ .
(b) 1 3 1z i iz− − = +
Let z x iy= + , then ( )1 3 1x iy i i x iy+ − − = + +
( )1 1 3 1x i y ix y− + − = + −
( ) ( ) ( )1 1 3 1x i y y ix− + − = − +
( ) ( ) ( )2 2 221 1 3 1x y x y− + − = + −
( )2 2 2 22 1 2 1 9 1 2x x y y x y y− + + − + = + − +
7
[Turn Over
2 28 2 8 16 7x x y y+ + − = −
( )
221 91
8 64x y⎛ ⎞+ + − =⎜ ⎟
⎝ ⎠
The locus is a circle of centre 1 ,18
⎛ ⎞−⎜ ⎟⎝ ⎠
of radius 38
units.
8 Total upward distance covered by n jumps = 1(1− 0.9n )1− 0.9
=10(1− 0.9n )
Total sliding distance after n jumps
=n2
2(0.3) + (n −1)(−0.02)⎡⎣ ⎤⎦ =0.3n − 0.01n(n −1) = −0.01n2 + 0.31n
Checking after which jump, the frog will not slide back: 0.3,0.28,0.26,...,0.02,0
0 0.3 ( 1)( 0.02) 0 16nU n n= ⇒ + − − = ⇒ =
(i) To clear the 4 m mark,
10(1− 0.9n )⎡⎣
⎤⎦− −0.01n2 + 0.31n⎡
⎣⎤⎦ ≥ 4
From GC, n ≥8.8715 Therefore, 9 jumps are needed.
(ii) After 15 jumps/slides, there will be no more sliding.
y
x 0 2
8 1
8−
48
−
318
1
58
×
8
[Turn Over
Hence, after 15 jumps/slides, distance climbed
= 10(1− 0.915)⎡⎣
⎤⎦− −0.01(15)2 + 0.31(15)⎡
⎣⎤⎦
=5.5411m
Subsequent jumps will follow GP with first term 150.9 and ratio 0.9.
Hence 0.915(1− 0.9n )1− 0.9
≥ 6.5 − 5.5411
From GC, n ≥ 5.95 Therefore, 6+15=21 jumps are needed.
(iii) Subsequent jumps will follow GP with first term 150.9 and ratio 0.9. Hence
Highest mark that the frog can reach
= 0.915
1− 0.9+ 5.5411
= 7.60 m
9(i) 1 11 , 1
4 3AB BC
−⎛ ⎞ ⎛ ⎞→ →⎜ ⎟ ⎜ ⎟= − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
⇒
712
AB BC⎛ ⎞→ → ⎜ ⎟× = ⎜ ⎟⎜ ⎟⎝ ⎠
Equation of plane π : 7 1 7 71 3 1 10 1 102 0 2 2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⋅ = ⋅ = ⇒ ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
r r
(ii) Let foot of perpendicular from D be F.
Now, 4 7
5 2OF
μμμ
+⎛ ⎞→ ⎜ ⎟= ⎜ ⎟⎜ ⎟+⎝ ⎠
Since F is on π , 4 7 7
1 105 2 2
μμμ
+⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠
1427
μ⇒ = −
9
[Turn Over
10271427
10727
OF
⎛ ⎞→ ⎜ ⎟
= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
10 9827 2714 1427 27
107 2827 27
405
DF
⎛ ⎞ ⎛ ⎞−⎛ ⎞→ ⎜ ⎟ ⎜ ⎟⎜ ⎟= − − = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠ ⎝ ⎠
392 14 627 9
DF→
⇒ = =
(iii) Equation of line l :4 00 15 1
λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
r
Thus since 4
5ON λ
λ
⎛ ⎞→ ⎜ ⎟= ⎜ ⎟⎜ ⎟+⎝ ⎠
lies on plane π ,
4 7
281 103
5 2λ λλ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⋅ = ⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠
Therefore 283
133
4ON
⎛ ⎞→ ⎜ ⎟
= −⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
Now, 1 '2
OF OD OD→ → →⎛ ⎞
= +⎜ ⎟⎝ ⎠
88272827
7927
'OD
⎛ ⎞−→ ⎜ ⎟
⇒ = −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
Thus 2827
7' 8
7ND
−⎛ ⎞→ ⎜ ⎟⇒ = ⎜ ⎟⎜ ⎟⎝ ⎠
⇒283
133
4 7' : 8
7l β
⎛ ⎞ −⎛ ⎞⎜ ⎟ ⎜ ⎟= − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠
r
10
[Turn Over
10(a) y =x2 + ax + b
x + 9= x + a − 9( )+ 81− 9a + b
x + 9
Equation of asymptotes are: ( )9y x a= + − and 9x = − When x = −9 , 12y = − So a = 6 When y = 0 , 2 6 0x x b+ + = Since curve cuts the x-axis only once, Discriminant = 62 − 4b= 0 9b∴ = For the line y kx= to have no intersection with the transformed curve ( )f x p q+ − , the point of intersection has to be shifted to the origin. Hence p = −9 and q = −12
y
x
y = x − 3
x = − 9
( )3,0−
f ( )y x=
( )9, 12− −
11
[Turn Over
(b)
x
y
( )3 , 2a a−
( )5 , 3a a
a− 2a
4x a=
0y = 0
( )5 , 3a a−
( )3 , 2a a− −
( )2 fy x=
( )1
fy
x=
13 ,2
aa
⎛ ⎞−⎜ ⎟⎝ ⎠
15 ,3
aa
⎛ ⎞⎜ ⎟⎝ ⎠
4a
2x a= x a= −
0y = x
y
0
12
[Turn Over
11(i) z5 = 32e π +2kπ( )i , k = 0,±1,±2
252 , 0, 1, 2
k iz e k
π π+⎛ ⎞⎜ ⎟⎝ ⎠= = ± ±
52i
z eπ
= ,352
ie
π
, 52i
eπ
−
, 2 ieπ ,352
ie
π−
(ii) w5 = 32i ( )55 32 32iw iw⇒ = − ⇒ = −
Let z iw= ⇒ w iz= −
Area of triangle OW1W21 22 2 sin 1.902 5
π= × × × = units2.
(iii) ( )3 3
5 5 5 52 2 2 2 2i i i i iz e z e z e z e z e
π π π ππ− −⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞
− − − − −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
( )3 3
2 25 5 5 52 4 2 4 2i i i i
z z e e z z e e zπ π π π
− −⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞− + + − + + +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
( )2 2 34cos 4 4cos 4 25 5
z z z z zπ π⎛ ⎞⎛ ⎞− + − + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
p = 4,α =π5
,β =3π5
,k = 2
End of Paper
2 R
I
1
[Turn Over
SERANGOON JUNIOR COLLEGE
2010 JC2 PRELIMINARY EXAMINATION
MATHEMATICS
Higher 2 9740/2
Solutions
Section A: Pure Mathematics [40 marks]
1
( )
180 230 70 3875045 50 10 8750300 400 12 100 3 4 12 0
x y zx y zx y z x y z
+ + =+ + =
+ = ⇒ + − = Using G.C, 100, 75 and 50x y z= = = Profit collected by outlet A ( ) ( ) ( )$100 75 $75 120 $50 20 $17500= + + =
2 (i)
(ii) x = a sin2 t, y = a cos t
ttatx cossin2
dd
= , taty sin
dd
−=
ttta
taxy sec
21
cossin2sin
dd
−=−
=
At the point P where t = 3π , 1
dd
−=xy
y
x
a
a
2
[Turn Over
Equation of normal at P ( 3 1,4 2
a a ):
y – y1 = m (x – x1)
1 1 3[ ]2 1 4
y a x a−− = −
−
Equation of normal: 1
4y x a= −
(iii) Equation of Curve: x = a sin2 t, y = a cos t
Equation of normal: 14
y x a= −
Solving, 24 cos 4 sina t a t a= −
24cos 4cos 3 0t t+ − = (2cos 3)(2cos 1) 0t t+ − = 1 3cos , cos2 2
t t= = − (rejected)
3t π= (point P)
Hence, the normal at P does not meet the curve again.
3 (i)
1/2x
• A’(‐3/2, 3)
-1/2
1
y
B’(0, -1) C’(-1/2,0)
•
•
3
[Turn Over
(ii)
(iii)
4(a)(i) 1 1ln dt 1 ln dtt t tt t+⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠∫ ∫
1ln dt ln dtt tt
= +∫ ∫
( )2ln
ln 1 dt2t
t t c= − + +∫
( )2ln
ln2t
t t t c= − + +
0
1 x
•
A’(4, 7)
2
3
y
B’(1, -1)
C’(3,1)
•
•
2x
• A’’(‐4,3)
-2
1
y
B’(1, -1)
C’’(-3,0)
B’’(-1, -1)
A’(4, 3)
C’(3,0)
•
•
• •
•
4
[Turn Over
(a)(ii) Area R 2 ln 2
1 dy x
+= ∫
( )2
1
1ln 1 dt t tt
⎛ ⎞= − +⎜ ⎟⎝ ⎠∫
2
1
11 ln dtt t tt
⎡ + ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫
( )22 22
1 1
lnln
2 2tt t t t t
⎡ ⎤⎡ ⎤= + − − +⎢ ⎥⎢ ⎥
⎢ ⎥⎣ ⎦ ⎣ ⎦
( ) ( )2ln 212 2 1 2ln 2 2 1
2 2
⎡ ⎤⎛ ⎞= + − − − − + − −⎢ ⎥⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦
( ) ( )2ln 213 2ln 2 2 1
2 2
⎡ ⎤= − − − + − −⎢ ⎥
⎢ ⎥⎣ ⎦
( )2ln 27 2ln 2
2 2= − − units2
(b)
V = ( ) ( ) ( )2 2 20 42
5 02 4 d 2 4 2 4 dx x x x x xπ π
−
⎡ ⎤ ⎡ ⎤− − − − + − − − − − + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫
= 327.25 units3
S
5− 4
y
x
5
[Turn Over
5 x is the number of infected orangutans at any time t. dxdt
∝ x 35 − x( ) ⇒ dxdt
= kx 35 − x( ), k > 0
⇒1
x 35 − x( )∫ dx = k dt∫
⇒135
1x+
135 − x
⎛⎝⎜
⎞⎠⎟∫ dx = k dt∫
⇒ ln x − ln 35 − x( )= 35kt + c
⇒ lnx
35 − x⎛⎝⎜
⎞⎠⎟= 35kt +C
35 35
35kt c ktx e Ae
x+⇒ = =
−, 0A >
35
3535
1
kt
ktAexAe
⇒ =+
When t=0, x=5, 16
A⇒ =
35635 1
6 ktxe
⎡ ⎤= −⎢ ⎥
+⎣ ⎦
As , 35t x→∞ →
Section B: Statistics [60 marks]
6 (i) As there are only 10 three-star restaurants, there is a high chance that they might not
get selected at all during random sampling.
OR The numbers of the respective types of restaurants selected might not be representative of the restaurants in the town.
(ii) Stratified sampling. Set up the 3 respective sub-groups.
- Use population proportions to determine the numbers required for each sub-group: Three-star(1), two-star(6), one-star(3).
- Randomly select the restaurants based on the respective numbers determined
x
t 0
35
5
6
[Turn Over
7(i)
(ii) P(rains on a Thursday)
1 1 1 1 2 2 4 2 1 4 1 2=5 3 3 5 3 3 5 3 3 5 3 321 745 15
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
= =
(iii) P(rains on at least 2 days given it rains on a Thursday)
P(A B)= P(A\B) = P(B)∩
1 1 1 1 2 2 4 2 1 13135 3 3 5 3 3 5 3 3 45
21 21 2145 45
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅= = =
8(i) No. of ways = 9! 12602!3!4!
=
(ii) Case 1: uni and unagi
No. of ways = ( )7! 2 2102!4!
=
Case 2: uni and ebi
No. of ways = ( )7! 2 2803!3!
=
Tuesday
15
R
Wednesday
R’
45
β
R’
R
1 β−
α
R’
R
1 α−
α
R’
R
1 α−
β
R’
R
1 β−
α
R’
R
1 α−
β
R’
R
1 β−
Thursday
7
[Turn Over
Case 3: ebi and unagi
No. of ways = ( )7! 2 4202!2!3!
=
Total no. of ways = 910 (iii) - Unagi sushi - Uni sushi
Since there are only 2 ways of arranging the uni and unagi sushi plates on the round table,
No. of ways = ( )54 2 10C =
9 (i)
Least squares regression line: y = 0.785x + 24.4
(ii) Model C. Possible reasons:
- Shape of the points follow the shape of an logarithmic graph - y increases as x increases (other 2 choices have p decreasing as x increases)
(iii) a = - 37.4, b = 26.3
The value of r increases after transformation which indicates there is a better linear correlation between ln x and y instead of just x and y
8
[Turn Over
(iv) lny a b x= + 8, 17.31when x y∴ = =
Extrapolation done in calculating y when x = 8. It might not be reliable as there may not be a linear relationship between ln x and y outside the range of data
10(i) Assuming that the IQ points of children is normally distributed. Ho: μ = 100 H1: μ >100 (test that the average IQ points of students who have participated in the DDD program has increased) 1-tailed test at 5% significance level.
Under Ho , X − μS
n~ t(n −1) .
Where s2 =n
n −1samplevariance( )= 15
1415( )2 =
337514
= 241.07143
By using GC, 1.996testt = and p-value = 0.0230.
Since p-value=0.0230<0.05, we reject Ho and conclude that here, we have sufficient evidence at 5% significance level that the IQ points of children in the DDD program has increased.
(ii) 5% significance level means there is a probability of 0.05 that we conclude that the mean IQ points have increased when in fact it did not (reject Ho when Ho is true).
(iii) Population standard deviation σ is now known and a larger sample size of 50 children is taken.
Under Ho , 2
~ 100,50
X N σ⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
by Central Limit Theorem (since n=50 is large)
At 5% significance level, Ho is not rejected when test 1.64485z < .
Thus 100 1.64485
50
xσ−
< 100 0.233x σ⇒ < +
9
[Turn Over
11(i) Let X be the r.v “number of pairs of twins born weekly”, ( )2X Po
Using G.C,when x = 1 and 2, ( ) ( )1 2 0.27067P X P X= = = = give the highest probability.
Hence the most probable numbers are 1 pair or 2 pairs of twins. (ii) In a two-week period, 1 2 ~ (4)X X Po+ 1 2( 7) 0.948866 0.949P X X+ ≤ = ≈
(iii) Let Y be the r.v “number of two-week period with more than 7 pairs of twins born in 78
two-week period” ( )78,0.051134Y B
Since 78n = is large, ( ) ( )78 0.051134 3.9884 5np = = <
So ( )3.9884Y Po approximately
( )( )
( )
less than 73 two-week periods with at most 7 pairs of twins born
6
1 50.21306
P
P Y
P Y
= ≥
= − ≤
=
0.213≈
(iv) Let n be the number of consecutive weeks ( )1 2 ... 2nX X X Po n+ + + Since 2n > 10 ( )5n >Q , so ( )1 2 ... 2 , 2nX X X N n n+ + + approximately
( ) ( )1 2 1 2.... 20 0.5 ... 20.5 0.5n nC CP X X X P X X X+ + + ≤ < ⎯⎯→ + + + < < Using G.C,when n =10, ( )1 10... 20.5 0.50997P X X+ + < =
when n =11, ( )1 11... 20.5 0.47282P X X+ + < = Thus least n is 11
(v)
50
1 2~ 2,50 50
ii
XX N= ⎛ ⎞= ⎜ ⎟
⎝ ⎠
∑ approximately, by CLT since n is large.
Thus ( )1.8 0.15865 0.159(3 s.f)P X < = =
10
[Turn Over
12
(i) Let M denote the r.v of the mass of a male hornbill. Let F denote the r.v of the mass of a female hornbill.
( ) ( )2 2N 3500,150 N 3000,M F σ
( )P 3200 0.05F > =
( )P 3200 0.95F ≤ =
3200 3000P 0.95Zσ−⎛ ⎞≤ =⎜ ⎟
⎝ ⎠
200σ
= 1.64485 ⇒σ = 121.59 = 122 (3s.f.)
(ii) M1 − M2 ~ N(0,2 ×1502 )
P M1 − M2 ≥ 100( )= P M1 − M2 ≥ 100( )+ P M1 − M2 ≤ −100( ) ( )1 22P 100M M= − ≤ −
=0.36265=0.363 (3 s.f.)
(iii) Let ( )1 2 3 4 5 1 22T F F F F F M M= + + + + − +
( )2 3 2N 5 3000 4 3500, 5 121.49 2 150T × − × × + ×
( )P 0 0.976T > ≈
(iv) Probability required
[ ] [ ]3 24! ( 3600) ( 3600)2!2!
P M P M= > ≤
=0.053967
=0.0540 (3 s.f.)
End of Paper
1
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS 9740/01 Paper 1 15 September 2010 Additional Materials: Answer paper 3 hours List of Formula (MF15)
READ THESE INSTRUCTIONS FIRST Write your Name and Civics Group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 5 printed pages and 1 blank page.
© TJC 2010 [ Turn over
2
1 The points A, B and C have position vectors given respectively by 2i +3k , 2j − k and αi − (α – 2)j + (2α – 1)k where α ∈ . Show that A, B and C are collinear. Hence or otherwise, find α such that the C divides the line segment AB in the ratio 2:1.
[3]
[3]
2 In a chemical reaction, a compound A is converted to a compound B at a rate proportional to square of the amount of A that has yet to be converted. Initially, 120 g of A is present. After 1 hour, 20 g of A remains. If x denotes the amount of A that is
converted after t hours, show that 5 1
ptxt
=+
, where p is an integer to be determined.
[6]
3 The complex number z satisfies the equation 3 2 1 0z az az+ − − = , where a∈ . (i) Verify that z = 1 is a root of the equation. (ii) Given that one of the complex roots is 2i, find a. (iii) Find the third root of the equation.
[1] [2] [3]
4 A rectangular cuboid ABCDEFGH has a constant height of 2 cm and a base length of x cm which increases with time t (measured in seconds). The cuboid, which is initially a cube, is expanding at a constant rate of 104 cm3s−1. (i) Show that 22 4AF x= + . (ii) Find the rate of increase of the distance AF when t = 3.
[1] [6]
A B
C D
E F
G H
x
x
2
3
5 The functions f, g and h are defined by 2f : 4 2x x x+ ≥ − , x∈ , ( )g : ln 0x x x− < , x∈ ,
h : 4 4x x xλ− + > , x∈ . (i) Given that the composite function ( )fh x exists, find the least value of λ .
(ii) Find ( )1g x− and state the domain of 1g− . (iii) On a clearly labeled diagram, sketch the graphs of ( )gy x= , ( )1gy x−= and
( )1g gy x−= .
[3] [2]
[2]
6 It is given that for all x∈ ,
( ) 2f 4 6x x x= + + ,
( ) 2g 2x x x= − − and
( ) 3 2h x ax bx c= + + .
(a) Solve algebraically the inequality ( )( )
f1
gxx
≤ .
Hence solve the inequality ( )( )
f e1
g e
x
x
−
−≤ .
(b) If ( ) ( )h fy x x= + passes through the points ( ) ( )1, 18 , 1, 14− − − and ( )3,30 ,
evaluate ( )h 101 .
[3] [2]
[4]
7 It is given that
1
2 12 r
n
nr
rS=
−=∑ .
(i) Express 1, 2, 3S S S and 4S in the form 3 ab
−
, where ,a b∈ . Hence obtain a
conjecture for nS in terms of n. (ii) Prove your conjecture using mathematical induction.
(iii) Hence find 10
2 12 r
N
r
r+
=
+∑ in terms of N.
[3] [4] [2]
4
8 (i) Expand ( )
521 2x −− as a series of ascending powers of x up to and including the
term in 2x . State the range of x for which the expansion is valid.
(ii) Given that ( )321 2y x −= − , show that ( ) d1 2 3
dyx yx
− = .
By repeated differentiation of this result, show that the Maclaurin’s series expansion of y in ascending powers of x, up to and including the term in 3x , is 2 31 3x px qx+ + + , where the numerical values of p and q are to be determined.
(iii) Explain briefly how the result in (i) can be used as a check on the correctness of
your answer in (ii).
[3]
[1] [4] [2]
9 (a) A geometric sequence { nx } has first term a and common ratio r. The sequence of numbers { ny } satisfies the relation 3log ( )n ny x= for n +∈ .
(i) If the product of 5x and 16x is 81, find the value of 20
31log k
kx
=∑ .
(ii) Show that { ny } is an arithmetic sequence. (b) The output of a coal mine in any year is 10% less than in the preceding year. Prove that the output of the coal mined cannot exceed ten times the output in the first year.
It is decided to close the mine when the total output exceeds nine times the output in the first year. Determine the maximum number of years the mine will be in operation, and give your answers correct to the nearest year.
[4] [2] [2] [3]
10 The plane p1 has equation 3 6x y z+ − = and the point A has position vector i + 2k. Given that the point B is the foot of perpendicular from A to p1, find the position vector of B.
Another plane p2 has equation 1 00 1 , ,1 1
λ µ λ µ = + − ∈
r . Given that p2 intersects
p1 at the line l, find the vector equation of l and show that the shortest distance from A to
l is 776
.
A plane p3 passes through the points A and B. Given that p1, p2 and p3 do not have a common point, find the equation of p3.
[5] [6]
[3]
5
11 (a) Find sec cosec dx x x∫ .
(b) By using the substitution lnu x= , find ( )1 ln ln dx xx∫ .
(c) The diagram shows the region R bounded by the circle 2 2 4x y+ = , the curve
1yx
= and the line 1x = − .
(i) The curve 1y
x= intersects the circle at a point A, and the line 1x = − cuts the
circle at the point B. Find the coordinates of the points A and B. (ii) Calculate the area of the region R. (iii) Find the volume of the solid formed when R is rotated through 4 right angles about the y –axis.
[2] [4] [2] [3]
[4]
END OF PAPER
1
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS 9740/02 Paper 2 20 September 2010 Additional Materials: Answer paper 3 hours Graph paper
List of Formula (MF15) READ THESE INSTRUCTIONS FIRST Write your Name and Civics Group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages.
© TJC 2010 [ Turn over
2
Section A: Pure Mathematics [40 marks]
1 A curve C is represented by the parametric equations
2 2x t t= − , 3 9y t t= − for 2t ≤ .
Find the equation of the tangent to the curve C which is parallel to the y-axis. [4] The figure below shows the region R bounded by the curve C and the x-axis. Show that the curve C intersects the x-axis at the point (15, 0). [1] Hence find the exact value of the area of R. [4]
2 By sketching the graphs of xy e= and 2 2y x= + on the same diagram, find the number of roots of the equation 2 2 0x xe − − = . [2] (i) Given that the root β lies in the interval (a, 0), state the largest integer value of a. [1]
A sequence of negative numbers, nx , satisfy the relation
11 ( 2)2n
nxex + = − , for .n +∈
(ii) If the sequence converges, show that it converges to β . [2]
It is given that 0nxβ < < .
(iii) By considering 1 nnx x+ − , and with the aid of the sketch in (i), show that 1 nnx x+ < . [2]
(iv) Show that 1nx β+ > . [2]
(v) Describe the behaviour of the sequence for the case when 1 0xβ < < . [1]
3
3 (a) Find the roots of the equation 3
3 i 03 iz
z +
+ = +
. [4]
(b) A complex number z satisfies i 3 5z + = .
(i) Sketch the locus of the points which represents z in an Argand diagram. [2] (ii) Find z when 3z + is a minimum. [4]
4 (a)
The diagram shows the graph of ( )fy x= with asymptotes 1x = − and 6 4y x= + .
On separate diagrams sketch the graphs of
(i) ( )fy x′= [2]
(ii) ( )2 fy x= . [2]
(b) The curve C has equation
2 2 51
x xyx
− − +=
−.
(i) Describe how the curve with equation 24y xx
= − − can be transformed to C. [3]
(ii) State the equations of all the asymptotes of C and hence sketch C. [2] (You are not required to find the x and y intercepts.)
(iii) By adding a suitable graph to the sketch in (ii), deduce the number of roots of
the equation ( ) 22 2
2
1 2 5 4 12 1
x x xx
− − − ++ + = −
. [2]
y
x
1x = −
6 4y x= +
4
1 2−
( )fy x=
4
Section B: Statistics [60 marks]
5 Through-Train Junior College (TTJC) has four levels of students: IP1, IP2, JC1 and JC2. The population breakdown for the year 2010 is as follows:
Level IP1 IP2 JC1 JC2 Number of students 78 70 692 760
Mr Koh wants to find out how students spend their time after lessons. He plans to get 10% of the student population to do a survey. His method of conducting the survey is to spend one hour each day at the canteen and ask if the students entering the canteen are willing to do a survey for him. He continues to do that until he obtains the number of surveys required. (i) State a flaw in his sampling method. [1] Mr Koh’s friend, Mr Toh, suggests he modify his sampling method to collect 40 surveys from each level instead. (ii) State the name of this sampling method. [1] Mr Koh finally decided to obtain a stratified sample across different levels of students. (iii) Describe how a sample of size 160 might be obtained. [2]
6 The time taken, in minutes, for John, a typist, to type 5000 words is normally distributed with mean 34 and standard deviation 2.1.
John attends a training course to improve his typing speed. After the course, John records the times he take to type 5000 words on 8 separate occasions, and obtains the following times (in minutes):
36.5 33.4 27.7 31.5 33.0 34.9 30.7 29.5 Assuming that the initial standard deviation has not changed, test at the 2% level of significance whether John’s typing speed has improved after the training course. [4] Explain what is meant by 2% level of significance in the context of the question. [1]
5
7 An Extreme Ironing Club consists of 11 members: 6 male and 5 female. It is also known
that 4 of the male members are local while 3 of the female members are foreigners. (a) Find the number of ways for all 11 members to stand in a row so that no two
foreigners are adjacent. [2] (b) 4 members are to be chosen to enter a competition. In how many ways can this be
done if:
(i) there are no restrictions, [1]
(ii) there must be at least 1 male member and at least 1 female member, [2]
(iii) there must be at least 1 female member and at least 1 foreign member (a female member who is a foreigner will satisfy both conditions). [3]
8 Temasek United Football Club (TUFC) is participating in a football competition. 3 points are awarded for a win, 1 point is awarded for a draw and no points are awarded for a loss. In this competition, TUFC plays three matches in the first round. Being a good team, it may be assumed that for any match in the first round, the probability that TUFC wins is
12
and the probability that TUFC loses is
15
. All the matches are played independently.
(i) Find the probability that TUFC draws a match. [1] (ii) Find the probability that after three games, TUFC has exactly three points. [3] (iii) To qualify for the second round, TUFC needs to get at least five points. Given that TUFC does not win the first game, find the probability that they will still qualify for
the second round. [4]
9 The times taken, in seconds, for two swimmers, A and B, to complete a 100-metre freestyle race are independent and normally distributed with means 48.0 and 47.2 and standard deviations 0.5 and 0.8 respectively. The two swimmers compete in a 100- metre race for which the world record is 46.9 seconds. (i) Show that the probability that at least one of the two swimmers breaks the world
record during the race is 0.363 correct to 3 significant figures. [3] (ii) Find the probability of Swimmer B beating Swimmer A. [2] (iii) If A and B are to meet 20 times for the 100m freestyle race, how many times do
you expect A to beat B? Give your answer correct to the nearest integer. [2] (iv) Find the probability that the total sum of four randomly chosen timings of A is
more than 4 times a randomly chosen timing of B. [3]
6
10 At a stall in a fun-fair, games of chance are played. The probability that a participant
wins a prize in each game is 0.05. (a) If the stall holder wants the probability of more than 5 prizes to be won on a
particular day to be less than 0.1, find the largest number of games that can be played on that day. [3]
(b) On another day, N games are played. Using a suitable approximation, find the
probability that prizes are won in at most 5% of the games if :
(i) N = 60, [3]
(ii) N = 200. [3] (c) The stall is open for 70 days and on each day, 60 games are played. Find the
probability that the average number of prizes won each day is between 2.5 and 3 inclusive. [4]
11 Mary recorded the length of time, y minutes, taken to travel to her office when leaving home x minutes after 7 am on nine selected mornings. The results are as follows.
x 0 5 10 20 25 30 40 50 60 y 20 23 29 33 35 39 40 48 51
(i) State, giving a reason, which of the least squares regression lines, x on y or y on x,
should be used to express possible relation between x and y. [1]
(ii) Find the equation of the regression line chosen in part (i) and interpret the slope of your regression line. [3]
(iii) Calculate the product moment correlation coefficient and interpret your value in
the context of this question. [2] Mary needs to arrive at her office no later than 8.30 am. The number of minutes by which Mary arrives at her office early, when leaving home x minutes after 7 am, is denoted by z.
(iv) Write z in terms of x and y. [1]
(v) Estimate, to the nearest minute, the latest time that Mary can leave home without
arriving late at office. [3] Comment on the reliability of this value. [2]
1
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS 9740/01 Paper 1 15 September 2010
Solutions
1 0 2 2 12 0 2 2 11 3 4 2
AB−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
( )2 2 1
2 0 2 2 12 1 3 2 4 2
ACα αα α αα α
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − + − = − + = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
Therefore AB is parallel to AC and since A is a common point, we have A, B and C collinear. From above, we have
22
AC ABα−=
uuur uuur
Since C divides the line segment AB in the ratio 2:1, 23
AC AB=uuur uuur
.
2 22 3α−
∴ = 23
α⇒ =
Alternative 1for last part: Since C divides the line segment AB in the ratio 2:1, we have AB:BC = 2:1
23
OA OBOC +∴ =
uuur uuuruuur
2 0
12 0 2 23
2 1 3 1
αα
α
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⇒ − = +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦
23
α⇒ =
2 ( )2120dx k x
dt= −
( ) 2120 x dx k dt−− =∫ ∫
2
( ) 1120 x kt c−− = +
When t = 0, ( ) ( )1 1120 0 0120
k c c−− = + ⇒ =
When t =1, ( ) ( )1 1 120 1120 24
k k− = + ⇒ =
1 1 1 5 1
120 24 120 120tt
x+
= + =−
6005 1
txt
=+
So 600p = .
3 (i) When z = 1,
( ) ( ) ( )3 21 1 1 1 0LHS a a RHS= + − − = =
Hence z = 1 is a root of the equation. (ii) Since 2i is a root, ( ) ( ) ( )3 22 2 2 1 0i a i a i∴ + − − = 8i 4 2 i 1 0a a− − − − = ( )4 2i 1 8ia − − = +
1 8i 31 i4 2i 2
a += = − −− −
(iii) Let z = b be the other root. ( )( )( ) 3 21 2i 1z b z z z az az− − − = + − − Comparing the constant term, we have
2 i 1b− = − 1 i2
b⇒ = −
4 (i) Clearly 2 2 2AF AB BC CF= + + 22 4x= +
3
(ii) Let 22V x=
d d4d dV xxt t
⇒ =d 26dxt x
⇒ = - - - - - - (1)
Let y AF= .
Differentiate (i) wrt x : 2
d 2d 2 4y xx x=
+- - - - - - (2)
When 3t = , 32 104(3) 320V = + = 4 10x⇒ = or 4 10x = − (NA since x > 0)
So d d d 52 26d d d 18 9y y xt x t= ⋅ = =
Rate of increase of AF at t = 3 is 269
cms−1.
Alternative solution:
dVdt
= 104
⇒V = 104t + c
When t =0, V = 8. This gives us c = 8. V = 104t + 8. Volume of cuboid = 2x2 .
From (i), let AF = y = 2x2 + 4 = V + 4
⇒
dydV
=12
V + 4( )−12 .
dydt
=dydV
⎛⎝⎜
⎞⎠⎟
dVdt
⎛⎝⎜
⎞⎠⎟=
1042 V + 4
=52
V + 4.
When t = 3, V = 104(3) + 8 = 320.
Therefore,
dydt
=52
320 + 4=
269
cm s-1.
4
5 (i) ( )fh x exist ( ) [ )h f , 2,R D⇒ ⊆ ⇒ ∞ ⊆ − ∞ ⇒λ least 2= −λ
(ii) For ( )1g x− ,
Let ( )lny x= −
y yx e x e⇒− = ⇒ = − ( )1g xx e x−⇒ = − ∈ (iii)
6 (a) ( )
( )( )
( )( )2
2
f 5 84 61 1 0 0g 2 2 1
x xx xx x x x x
++ +≤ ⇒ − ≤ ⇒ ≤
− − − +
( )( )( ) [ ]5 8 2 1 0 1, 2x x x x⇒ + − + ≤ ≠ − 8 or 1 25
x x⇒ ≤ − − < <
( )( )
f 81 or 1 25g
xx x
x
ee e
e
−− −
−≤ ⇒ ≤ − − < <
0 2xe−⇒ < <
1ln2
x⇒ >
(b) ( ) ( )h fy x x= +
( ) ( )3 2 2 3 24 6 1 4 6y ax bx c x x ax b x x c⇒ = + + + + + = + + + + + Since it passes through the points ( ) ( )1, 18 , 1, 14− − − and ( )3,30
( ) ( )( ) ( ) ( )3 21 1 1 4 1 6 18a b c− + + − + − + + = − 21 (1)a b c⇒ − + + = − −− −
( ) ( )( ) ( ) ( )3 21 1 1 4 1 6 14a b c+ + + + + = − 25 (2)a b c⇒ + + = − − − −
( ) ( )( ) ( ) ( )3 23 1 3 4 3 6 30a b c+ + + + + = 27 9 3 (3)a b c⇒ + + = − − −
Solving (1),(2) and (3) gives 2, 10, 33a b c= − = = − ( ) ( )3 2h 2 10 33 h 101 1958625x x x= − + − ⇒ = −
( )1gy x−=
( )gy x=
( )1g gy x−=
x
y
y =0
x =0
5
7 (i) 11 532 2S = = − , 2
5 734 4S = = − , 315 938 8S = = − and 4
37 11316 16S = = − .
Conjecture for 2 332n nnS += − .
(ii) Let nP be the statement “ 2 332n nnS += − ” , for all n +∈ .
When n = 1, LHS of 1P = 112S = , RHS of 1P = 1
2(1) 3 13 22+− = . 1P∴ is true.
Assume that kP is true for some k +∈ . i.e. 2 332k kkS += − .
( )1
1
1
11
1
2 1When 1, LHS of 2
2 1 12 1 2 2
2 3 2 1 32 2
r
r k
k k
k
kr
k
r
rn k S
kr
k k
+
+
+
+=
=
−= + =
+ −−= +
+ += − +
∑
∑
( )
1
1 1
4 6 2 1 32
2 1 3 3 = RHS of
2
k
k k
k k
kS
+
+ +
+ − −= −
+ += −
Hence, kP is true ⇒ 1kP + is true and since 1P is true, by mathematical induction,
nP is true for all n +∈ .
(iii) 1
1
0 1
2 1 2 12 2r r
N N
r r
r r+
+
= =
+ −=∑ ∑ ( )
1 1
2 1 3 2 5 3 32 2N N
N N+ +
+ + += − = − .
8
(i) ( ) ( ) ( )5 2 22
5 75 352 21 2 1 2 2 ... 1 52 2! 2
x x x x x−
⎛ ⎞⎛ ⎞− −⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠− = + − − + − + ≈ + +⎜ ⎟⎝ ⎠
Expansion is valid if 1 12 2
x− < <
(ii) ( ) ( ) ( ) ( ) ( )3 5 5 12 2 2
31 2 1 2 2 3 1 2 3 1 22
dyy x x x y xdx
− − − −= − ⇒ = − − − = − = −
( )1 2 3 (1)dyx ydx
⇒ − = −− −
Differentiate (1) wrt x,
6
( ) ( )2 2
2 22 1 2 3 1 2 5 (2)dy d y dy d y dyx xdx dx dx dx dx
⇒ − + − = ⇒ − = − −−
Differentiate (2) wrt x,
( ) ( )2 3 2 3 2
2 3 2 3 22 1 2 5 1 2 7 (3)d y d y d y d y d yx xdx dx dx dx dx
⇒ − + − = ⇒ − = − − −
When 2 3
2 30, 1, 3, 15, 105dy d y d yx ydx dx dx
= = = = =
Hence,
2 3 2 315 105 15 351 3 1 32! 3! 2 2
y x x x x x x≈ + + + = + + +
15 35,2 2
p q= =
(iii) From (ii), ( )3
2 3215 351 2 1 32 2
x x x x−− ≈ + + +
Diff wrt x, ( ) ( ) ( ) ( )5
223 15 351 2 2 3 2 32 2 2
x x x−⎛ ⎞− − − ≈ + +⎜ ⎟⎝ ⎠
( )5
22351 2 1 52
x x x−⇒ − ≈ + + same as (i)
Alternative 1: Now ( )5
2 3235 3151 2 1 5 ...2 8
x x x x−− = + + + +
( ) ( ) ( )3 52 21 2 1 2 1 2x x x− −⇒ − = − −
( )2 335 3151 5 ... 1 22 8
x x x x⎛ ⎞= + + + + −⎜ ⎟⎝ ⎠
2 315 351 3 ...2 2
x x x= + + + + same as (ii)
Alternative 2: ( ) ( ) ( )5 3 12 21 2 1 2 1 2x x x− − −− = − −
( )2 3 2 315 351 3 ... 1 2 4 8 ...2 2
x x x x x x⎛ ⎞= + + + + + + + +⎜ ⎟⎝ ⎠
2351 5 ...2
x x= + + + same as (i)
7
9 (a)(i) 5x ( 16x ) = 81 ⇒ ( )4 15 81ar ar = ⇒ 2 19 81a r =
20
31log k
kx
=∑ = 3 1 3 2 3 3 3 20log log log ... logx x x x+ + + + = ( )3 1 2 3 20log ...x x x x
= ( )( ) ( )( )2 19
3log ...a ar ar ar
= ( )20 1 2 ... 19
3log a r + + +
= ( )19 20
20 23log a r⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
= ( )102 19
3log a r = ( )10
3log 81 10(4) 40= =
(ii) 1 3 3 1 31
log ( ) log ( ) log nn n n n
n
xxy xx
y − −−
= − =−
1
3 2logn
n
arar
−
−= 3log r= is a constant free from n.
Hence, { ny } is an arithmetic sequence. (b) Let the amount of coal mined in the first year be a.
2 3The maximum total amount of coal mined 0.9 0.9 0.9 ....a a a a= + + + +
101 0.9a a= =−
Let n be the number of year at which the mine will be in operation.
(1 0.9 ) 91 0.9na a− >−
0.9 0.1n⇒ <
lg 0.1 21.854lg 0.9n⇒ > ≈
The mine will be in operation for at most 22 years.
8
10
The vector equation of p1 is 11 63
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟−⎝ ⎠
r ---(1)
The vector equation of the line passing thru A and perpendicular to p1 is
1 10 1 ,2 3
λ λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
r ----(2)
Sub (2) into (1):
1 1
1 62 3 3
λλλ
+⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− −⎝ ⎠⎝ ⎠
1 6 9 6λ λ λ+ + − + = 1λ⇒ =
Therefore 1 1 20 1 12 3 1
OB⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
uuur
1 0 10 1 11 1 1
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟× − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Q and origin is in p2,
Therefore p2 has Cartesian equation 0x y z− − = ---(3) And p1 has Cartesion equation 3 6x y z+ − = ---(4) Using GC to solve (3) and (4): We have x = 3+2λ, y = 3+λ, z = λ.
Therefore the equation of l is 3 23 1 ,0 1
r λ λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + ∈⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
9
Shortest distance = 2 2 2
3 1 23 0 10 2 1
2 1 1
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− ×⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
+ +
2 2 53 1 62 1 4
6 6
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟× −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠= =
( ) ( )2 225 6 4 77
66
+ − + −= =
Since p3 passes through the points A and B, therefore p3 // 113
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠
.
Since p1, p2 and p3 do not have a common point, therefore p3 // 211
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
.
An equation of p3 is 1 1 20 1 1 , ,2 3 1
r α β α β⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + + ∈⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠
.
11 (a) 1sec cosec d 2 d2sin cos
x x x xx x
=∫ ∫ 2 cosec 2 dx x= ∫
( )ln cot 2 cosec2x x c= − + +
Alternative 1: 1sec cosec d dsin cos
x x x xx x
=∫ ∫2sec d ln | tan |
tanx x x cx
= = +∫
Alternative 2: 2 2sin cossec cosec d d tan d cot d
sin cosx xx x x x x x x xx x+
= = +∫ ∫ ∫ ∫
ln | sec | ln | sin |x x c= + +
(b) Let lnu x=1du
dx x⇒ =
10
( )1 ln ln d ln dx x u ux
=∫ ∫
1ln d lnu u u x u u u cu
= − = − +∫
( ) ( )( )ln ln ln 1x x c= − +
(c) (i) Solving 2 2 4x y+ = and 1yx
= ⇒ A is ( )2 3, 2 3− + − −
or (−1.931852,−0.5176381) from GC.
Solving 2 2 4x y+ = and 1x = − ⇒ B is (−1,− 3 ) or (−1,−1.73205).
(ii) Area 1 1
2
2 3 2 3
14 d dx x xx
− −
− + − +
= − −∫ ∫
0.546= [From GC]
(iv) Volume 2 3 2 3
2 22
13
14 d (1) ( 3 1) dy y yy
− − − −
−−
= − − − −∫ ∫π π π 4.74=
[From GC]
[2]
[3]
[4]
END OF PAPER
2 3− + 2 3
1
3
− +
−
−
2
2
1
TEMASEK JUNIOR COLLEGE, SINGAPORE Preliminary Examination Higher 2
MATHEMATICS 9740/02 Paper 2 20 September 2010
Solutions
Section A: Pure Mathematics [40 marks]
1 23 9/
2 2dy dy dx tdx dt dt t
−= =
−
For tangent parallel to the y-axis, 2 2 0 1t t− = ⇒ = Equation of tangent is 1x = − When y = 0, 3 9 0 ( 3) ( 3) 0t t t t t− = ⇒ + − =
3t∴ = − , 0 or 3 [NA since 2t ≤ ]
When 3t = − ( ) ( )23 2 3 15, 0x y⇒ = − − − = =
∴ C intersects the x-axis at the point (15, 0).
Area of R 15
0
y dx= ∫
( ) ( )3
3
0
9 2 2t t t dt−
= − −∫
( )0
4 3 2
3
2 2 18 18t t t t dt−
= − − − +∫
05 4
3 2
3
2 6 95 2t t t t
−
⎡ ⎤= − − − +⎢ ⎥
⎣ ⎦= 105.3
2 2 2 0xe x− − = 2 2xe x⇒ = +
From the diagram, there are two intersection points. 2 2 0xe x∴ − − = has two roots.
2
(i) Since graph of 2 2y x= + cuts x-axis at x = − 1, ( )1, 0 , 1aβ ∈ − ∴ = − .
Given that 11 ( 2)2n
nxex + = − , 0nx < , for .n +∈
(ii) If the sequence converges to a number L, then when n →∞ , nx L→ , 1nx L+ → .
1 ( 2)22 2 0
L
L
L e
e L
⇒ = −
⇒ − − =
L β∴ = as 0nx < , for all .n +∈ (iii) If 0nxβ < < , (iv)
( )
( )( )
1
1
1 ( 2)21 2 221 2 22
< 0
as 2 2 from the above diagramHence, .
n n n
n
n
n
n n
n
n
n
n
x
x
x
x
x x e x
e x
e x
e xx x
+
+
− = − −
= − −
= − +
< +<
From (iii) and (iv), 21 30 ....x x x β> > > > > . It is a decreasing sequence which converges to β .
3 (a) Mtd1 3 3
3 i 3 i03 i 3 iz z
z z⎛ ⎞ ⎛ ⎞+ +
+ = ⇒ = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
( ) ( )644
4 3 i 2 iz eπ
⇒ = − + = −
( ) ( ) ( )2 5
3 3 334 4 4 4 4 22 2 2 2i i i kiz e e e eπ π πππ π+ − − +
⇒ = = = =
( )3 6 112
242 2 , 0, 1, 2
ki ik
z e e k
ππ
π⎛ ⎞−⎜ ⎟
⎜ ⎟⎝ ⎠
− +
⇒ = = = ± Mtd2
( )3 4
23 i 0 13 i 3 i
i kiz z e ez
π ππ +⎛ ⎞ ⎛ ⎞++ = ⇒ = − = =⎜ ⎟ ⎜ ⎟
+ +⎝ ⎠ ⎝ ⎠
( )24
6
0, 1, 23 i 2
ki
i
z z e ke
π π
π
+
⇒ = = = ± −+
( )2 34 4 4
6
,2
ki i i
i
z e e ee
π π π π
π
+± ±
⇒ = =
5 7 1112 12 12 122 , 2 , 2 , 2
i i i iz e e e e
π π π π− −
⇒ =
( ) ( )( )1
1
> as = is an function.
2 > 2 1 1 2 > 2 2 2
1 > 2 from (ii)2
>
n
n
n
n
n
n
x x
x
x
x
e e y e
e e
e e
x
x
β
β
β
β
β
β
+
+
>
⇒ ↑
⇒ − −
⇒ − −
⇒
⇒
3
Mtd3 3 4
3 i 0 1 03 i 3 iz z
z⎛ ⎞ ⎛ ⎞+
+ = ⇒ + =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
2 2 , 2 2 , 2 2 , 2 23 iz i i i i⇒ = − + − − + −+
( )( ) ( )( ) ( )( ) ( )( )3 i 2 2 , 3 i 2 2 , 3 i 2 2 , 3 i 2 2z i i i i⇒ = + − + + − − + + + −
( ) ( ) ( ) ( )6 2 6 2 , 2 6 6 2 ,z i i⎡ ⎤ ⎡ ⎤⇒ = − − + − − + − −⎣ ⎦ ⎣ ⎦
( ) ( ) ( ) ( )6 2 6 2 , 6 2 6 2i i⎡ ⎤ ⎡ ⎤− + + + − −⎣ ⎦ ⎣ ⎦ .
Or 1.93 0.518 , 0.518 1.93 , 0.518 1.93 , 1.93 0.518i i i i− + − − + − . Mtd4
( )3 3
4 33 i 3 i0 8 3 i 8 8 3 1683 i
iz z z i i ez i z
π−⎛ ⎞ + +
+ = ⇒ = − ⇒ = − + = − =⎜ ⎟+⎝ ⎠
( )3 6 112
242 2 , 0, 1, 2
ki ik
z e e k
ππ
π⎛ ⎞−⎜ ⎟
⎜ ⎟⎝ ⎠
− +
⇒ = = = ± Mtd5
( ) ( )3
4 44 2 0.52363 i 0 3 i 3 i 4
3 iiz z z e
z−⎛ ⎞ +
+ = ⇒ = − + ⇒ = ± − + = ±⎜ ⎟+⎝ ⎠
0.52364 1.93 0.518 , 1.93 0.518 ,0.518 1.93 , 0.518 1.93iz e i i i i−⇒ = ± ± = − − + + − − (b) i 3 5 3i 5z z+ = ⇒ − =
Mtd1
Let Q represent the number 3− in the Argand Diagram. Minimum of 3z + is equal to the length QR.
To find z, we find the complex number represented by R in the above Argand Diagram.
From diagram, 3tan 33
θ = = ⇒ 3πθ = .
x-coordinates of R is 1 55cos 52 2
θ ⎛ ⎞− = − = −⎜ ⎟⎝ ⎠
y-coordinates of R is 3 6 5 33 5sin 3 52 2
θ⎛ ⎞ −
− = − =⎜ ⎟⎜ ⎟⎝ ⎠
Therefore, required 5 6 5 3 i2 2
z −= − +
•P
O
3
•Q θ
R • S x
y
4
Mtd2 By similar triangle:
2 3PQ = ; 5 2 3QR = − ; 5 32
PS = ; 52
RS =
So 5 5 5 5, 3 3 3 32 2 2 2
P z i⎛ ⎞⎛ ⎞ ⎛ ⎞= − − − ⇒ = − − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ or 2.5 1.33z i= − −
Mtd3 R is the intersection of Line: 3 3y x= + and Circle: ( )2 2 23 5y x− + = .
Solving gives 5 5 5 5, 3 3 3 32 2 2 2
P z i⎛ ⎞⎛ ⎞ ⎛ ⎞= − − − ⇒ = − − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ or 2.5 1.33z i= − −
4 (a)(ii)
(b) Note: 2 2 5 23
1 1x xy x
x x− − +
= = − − +− −
Mtd1 24y xx
= − − A⎯⎯→2 24 4y x xx x
= − − − = − − +−
B⎯⎯→ ( ) ( )2 21 4 3
1 1y x x
x x= − − − + = − − +
− −
A: a reflection about y-axis B: a translation of 1 unit in the positive direction of x-axis.
Mtd2 A: a translation of 1 unit in the negative direction of x-axis. B: a reflection in y-axis. Mtd3 A: a translation of 1 unit in the positive direction of x-axis. B: a reflection in x-axis.
C: a translation of 8 units in the negative direction of y-axis.
y
x
1x = −
12−
( )2 fy x=
0
5
2 2 51
x xyx
− − +=
− ( ) ( )2
22
14 1
2x
y−
+ + =
1x =
3y x= − −
Mtd4 A: a translation of 1 unit in the positive direction of x-axis. B: a translation of 8 units in the positive direction of y-axis.
C: a reflection in x-axis.
(ii) 2 2 5 23
1 1x xy x
x x− − +
= = − − +− −
,
Asymptotes are: x = 1 and y = −x−3
(iii) Adding graph of ( ) ( )2
22
14 1
2x
y−
+ + =
to part (ii), there are 4 roots for the
equation ( ) 22 2
2
1 2 5 4 12 1
x x xx
− ⎛ ⎞− − ++ + =⎜ ⎟−⎝ ⎠
.
Section B: Statistics [60 marks]
5 (i) − The sample may not represent all levels of college’s population.
− Students who do not go to the college canteen during that one hour will be excluded from the sample.
− The sample is not representative because it excludes students who are not willing to take part in the survey.
(ii) Quota sampling. (iii) Use Stratified sampling: Draw random samples from each level as follows:
Level IP1 IP2 JC1 JC2
Number of students
78 1601600
7.8 8
×
= ≈
70 1601600
7
×
=
692 1601600
69.2 69
×
= ≈
760 1601600
76
×
=
6
6 oH : 34μ =
1H : 34μ < Level of significance: 2%
Test statistic: ~ (0, 1)/
X Nnμ
σ−
Assume oH is true, zcal = 32.15 342.1/ 8
− = −2.4917 (5 s.f.) , p –value = 0.00636 < 0.02
Since the p –value is less than the level of significance, we reject oH at 2% level of significance. There is sufficient evidence to conclude that John’s typing speed has improved after the training course. 2% level of significance means there is 2% chance that we wrongly support the claim that John has improved his typing speed after the training course.
7 Male Female
Local 4 2 Foreign 2 3
(a) Number of ways = 6! × 7P5 = 1814400
(b) (i) Number of ways = 11C4 = 330
(ii) Number of ways = No. w/o restriction − No. w/o males − No. w/o females
= 11C4 − 5C4 − 6C4 = 310
(iii) No. of ways = No. w/o restriction − No. w/o females − No. w/o foreigners
+ No. with only local males
= 11C4 − 6C4 − 6C4 + 4C4 = 301.
8(i) P(TUFC draws a match) = 1−
12−
15=
310
(ii) P(three points after 3 games)
= P(WLL) + P(DDD)
= ( )2 31 1 33
2 5 10⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= 0.087 (iii) P(qualifies for second round | TUFC did not win the first game)
( )( )
P qualifies and did not win the first gameP did not win the first game
=
7
( ) ( ) ( ) ( )P DWW +P DWD +P DDW +P LWW= 11
2−
2 2 23 1 3 1 1 1210 2 10 2 5 2
12
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠=
= 0.43
9 Let A be the timing of swimmer A for the 100m freestyle. 2~ (48.0, 0.5 )A N Let B be the timing of swimmer B for the 100m freestyle. 2~ (47.2, 0.8 )B N (i) Mtd1
Required Probability = 1− P(non among the two broke the world record) = ( ) ( )1 46.9 46.9P A P B− ≥ ≥
= ( )( )1 0.986097 0.646170 0.36281 0.363− = ≈ Mtd2
Required Probability = P(A breaks record) + P(B breaks record) – P(both break record) = ( ) ( ) ( ) ( )46.9 46.9 46.9 46.9P A P B P A P B< + < − < <
0.363≈ Mtd3
Required Probability = P(A breaks record, B don’t break record) + P(A don’t break record, B breaks record) + P(both break record)
= ( ) ( )46.9, 46.9 46.9, 46.9P A B P A B< ≥ + ≥ <
( ) ( )46.9 46.9P A P B+ < < 0.363≈
(ii) ~ (0.8,0.89)A B N− ( ) ( )0 0.801781 0.802P A B P A B> = − > = ≈ (iii) ( )20 1 0.801781 4× − ≈ Expected times out of 20 is 4 times. (iv) Let 1 2 3 4 4 ~ (3.2,11.24)W A A A A B N= + + + −
( ) ( )1 2 3 4 4 0 0.83008 0.830P A A A A B P W+ + + > = > = ≈
10 (a) Let X be the number of prizes won out of N games. ( )~ ,0.05X B N
( )5 0.1P X > < ⇒ ( )5 0.9P X ≤ ≥
From G.C., When N = 63, ( )5 0.90551 0.9P X ≤ = >
When N = 64, ( )5 0.89990 0.9P X ≤ = < Therefore, largest number of games is 63.
8
(b)(i) Let Y be the number of prizes won out of 60 games. ( )~ 60,0.05Y B Since 60 0.05 3 5× = < , ( )~ 3oY P approximately
( )( ) ( )5% 60 3 0.64723 0.647≤ = ≤ = =P Y P Y (3 s.f.)
(ii) Let S be the number of prizes won out of 200 games. ( )~ 200,0.05S B Since 200 0.05 10 5× = > , 200 0.95 190 5× = > ( )~ 10, 9.5S N approximately
( )( ) ( ) ( )5% 200 10 10.5
0.564434 0.564 (3 s.f.)
P S P S P S≤ = ≤ ≈ <
= =
(c) Mtd1:
( )~ 60,0.05Y B , E(Y) = 60 0.05 3× = , Var(Y) = 60(0.05)(0.95) = 2.85 Since sample size 70 is large, by CLT,
2.85~ 3,70
⎛ ⎞⎜ ⎟⎝ ⎠
Y N , i.e. ( )~ 3,0.04071Y N
( )2.5 3 0.49340 0.493P Y≤ ≤ = = (3 s.f.)
Mtd2: Let T be the total number of prizes won out of 70 × 60 games. ( )~ 4200, 0.05T B
( )2.5 3 175 210 0.51333 0.51370
⎛ ⎞≤ ≤ = ≤ ≤ = ≈⎜ ⎟⎝ ⎠
TP P T
11
(i) The least squares regression line of y on x should be used as x is an independent variable and y depends on x.
(ii) Equation of the regression line of y on x is y = 21.9 + 0.504x The slope of this regression line means Mary’s travelling time will increase by 0.504
minutes for each minute she leaves home after 7 am. (iii) The product moment correlation coefficient, r = 0.988 which indicates the time that Mary leave home after 7 am and the time that she takes
to travel are strongly positively linearly correlated. (iv) z = Time available – Time taken = 90 – x – y. (v) For Mary not to be late, x + y ≤ 90
⇒ x + 21.881 + 0.504x ≤ 90 ⇒ x ≤ 45.292
The latest time that Mary can leave home is 7.45am. This value is reliable as the value of r is close to 1 and x = 45 is within the given set of data. The estimate is an interpolation.
VICTORIA JUNIOR COLLEGE
Preliminary Examination Higher 2
MATHEMATICS 9740/ 01 (Paper 1)
September 2010
3 hours Additional materials: Answer paper
Graph Paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 5 printed pages
© VJC 2010 VICTORIA JUNIOR COLLEGE [Turn over
1 The equation of a curve C is 5 . Show that every line parallel to the x-axis cuts C at two distinct points. [3]
2 23x xy y
Without differentiating, explain, giving a reason, if there is any point on the curve at which the tangent is parallel to the x-axis. [1]
2 In the game of TapFarm, a player is given plots of land to grow tomatoes, pumpkins and cherries. The player can only plant a type of fruits on a plot of land each time and each plot of land produces 1 kg of fruits. The time taken from planting to harvesting the fruits, cost price and selling price for 1kg of fruits are as follows:
Time required (Hours) Cost price ($) Selling price ($) Tomatoes 3 15 35 Pumpkin 5 a 50 Cherries 16 45 100
Tommy has only 1 plot of land. He used $545 and spent 154 hours playing the game before harvesting a total of 23 kg of fruits and earned a profit of $735. Showing your working clearly, find the value of a. [4]
[You are to assume that Tommy harvests the fruits once they are ready and there is no time lapse between harvesting and planting new fruits.]
3 Find the exact value of 2 3
2 2 1
d( 1)
xx
x by using the substitution tanx . [5]
4 (i) Given that 1
2x , using the substitution 2 1,xy show algebraically that
22 8 2 1 2x x is always positive, [2]
(ii) Hence, solve the inequality
2
2
22 8 2 1 20,
x x x x
x k x m
where k and m are real and 0 < k < m. [4]
5 (a) Solve the equation 4i( 2) (1 i) 0z
giving the roots in the form iez p r , where p is a real number, r > 0 and . [4]
(b) Given that 1 i 2 is a root of the equation 3 23 3z az bz 0 ,
find the values of the real numbers a and b. [3]
2
6 (i) Express 2
( 1) ( 1)r r r in partial fractions. [2]
(ii) Hence, find 2
1
( 1) ( 1)
n
r r r r . (There is no need to express your answer as a single
algebraic fraction.) [4]
(iii) Given that 2
1
( 1) ( 1)
n
r
kr r r
, for all , find the least value of k, showing your
working clearly. [2]
2n
7 Positive odd integers, starting at 1, are grouped into sets containing 1, 3, 9, … odd integers ,
as indicated below, so that the number of odd integers in each set after the first is thrice the number of odd integers in the previous set.
{1}, {3, 5, 7}, {9, 11, 13, 15, 17, 19, 21, 23, 25}, …
Find, in terms of k,
(i) the number of terms in the kth set, [1] (ii) the number of terms in the first k sets, [2] (iii) the first integer in the kth set, [1] (iv) the last integer in the kth set, [2] (v) the sum of all integers in the first k sets. [3]
8 The planes 1 and 2 have equations . 6 r i j k and . 2 4 12 r i j k
respectively. The point A has position vector 9 7 5 i j k .
(i) Find the position vector of the foot of perpendicular from A to 2 . [3]
(ii) Find a vector equation of the line of intersection of 1 and 2 [2]
The plane has equation 3
3 3 9 b r i j k i j k i j k ,
where , are real parameters and b is a constant.
Given that have no point in common, find the value of b. [3] 1 2, and 3
3 meets and in lines and respectively. Without finding the equations of 1 2 1l 2l 1l
and , describe the relationship between and , giving a reason. [2] 2l 1l 2l
9 A sequence of real positive numbers satisfies the recurrence relation 1 2 3, , , ...u u u
1 4 4 1,n n nu u u n and 1 15u .
(i) Prove by induction that for all 24 1nu n 1 n . [4]
(ii) Determine, giving your reasons, if this is a converging or diverging sequence. [2] The sequence is modified to
1
1 14 4 1, n
n n
v nv v
and 1
1
15v .
As , nn v .
(iii) Find the value of correct to 3 decimal places. [2] (iv) Show, graphically or otherwise, that if nv , 1n nv v . [3]
[Turn over
3
10 (a) Sketch the curve given by the equation
2 2
21
9
x y
k , 0 3.k [2]
Given that m is a constant and the equation 2 2 2
21
9
x m x
k
has 2 real roots, use your sketch above to find, in terms of k, an inequality satisfied by . [3] | |m
(b) A curve C has equation 3 ( 4)
2
x xy
x
.
(i) By considering d
d
y
x, show that the gradient of C is always positive. [3]
(ii) Find the equations of the asymptotes of C. [2] (iii) Sketch C. [1]
y 11 (a)
1
2
x
1
O
y = 1
y = 0 R
e
1 e
x
xy
(i) The diagram shows the curve C with equation e
.1 e
x
xy
The region R is
bounded by C, the positive x-axis and the positive y-axis. Find the exact area of R. [3]
(ii) Denoting the answer you have obtained in part (i) by q, write down, in terms
of q, the area of R if C is scaled by factor 2 parallel to the y-axis. [1]
(iii) Sketch the graph of 2 e
1 e 2
x
xy
1
, indicating clearly the equations of the
asymptotes and the shape of the curve for points near y = 0. [3]
4
y 11 (b)
O
( , ln )A a a
B
y = ln x
x
5
x A girl intends to design a bowl by rotating a section AB of the curve completely
about the y-axis. She wants the bowl to hold 300 cmlny
3 of fluid.
Given that the diameter of the rim is twice that of the base, write down, in terms of a, the coordinates of the point B. Hence find the exact value of a. [5]
12 Verify that the general solution of the differential equation,d
( 1)(1d
y)x y x
x is
. [1] e 1xy Ax (i) Sketch three members of the family of solution curves, one for positive A, one for
negative A and one for A = 0. [3] (ii) It is given that A = 3 and k is a positive constant.
Find
03 e d
kxx x in terms of k. Hence, state the value of
03 e dxx x
. [4]
Indicate, on a clearly labeled diagram, the region whose area is given by
03 e dxx x
. [1]
(iii) A point P is conditioned to move along the curve 3 e 1xy x such that the x-coordinate of P increases at a constant rate of 2 units per second.
(a) State the range of values of x for which d
0d
y
t . [1]
(b) Find the x-coordinate of the point on the curve at which P is moving such
thatd
2dt
y . [3]
1
VICTORIA JUNIOR COLLEGE
Preliminary Examination Higher 2
MATHEMATICS 9740/ 02
Paper 2
September 2010
3 hours Additional materials: Answer paper
Graph paper List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 6 printed pages
© VJC 2010 VICTORIA JUNIOR COLLEGE [Turn over
2
Section A: Pure Mathematics [40 marks]
1 Sketch, on an Argand diagram, the locus representing the complex number z for which
4 3i 2.z [2]
(i) Given that a is the least possible value of ,z find a. [2]
(ii) The complex number p is such that
4 3i 2 and .p p a
State the exact value of arg p . [1]
(iii) Deduce the greatest value of argz
p
, giving your answer correct to 2 decimal
places. [2] 2 (i) A curve C has equation f ( ).y x
The diagram shows the curve C1 with equation and the curve C2 f ( )y x 2 with
equation fy x . The line 3x is an asymptote to C1 and C1 passes through
the points (1, 0) and (3, 0). The lines and 4y x 4y x are asymptotes to C2. C2 passes through the
points (3, 0), (3, 0) and 2
0,3
0x
. The minimum points on C2 have coordinates
and . 1, 1 1, 1
Given that f ( for ) 0x 1 and the point 7
, 92
is the only other
turning point on C, sketch C, indicating clearly the intercepts, the equations of the asymptotes and the coordinates of the turning points. [4]
(ii) Find the x-coordinates of the stationary points on the curve 2f ( )y x . [3]
2 f ( )y x fy x y
O
x = –3
–1 3 x
y
O
(–1, – )1
–3 3 y = x – 4y = – x – 4
(1, –1)
x
3
3 The plane and the line have equations l
1
. 1 4
a
r and respectively,
where
1 1
2
2 1
r 0
is a real parameter and a is a constant. (a) It is given that . Find 1a (i) the acute angle between and the plane 0,z [2] (ii) the exact perpendicular distance of the point (1, 3, 2) from . [3] (b) It is given that Find the acute angle between and 3.a l . [3]
4 (i) Obtain the expansion of 2
1
1 xup to and including the term in 4x . [2]
(ii) Given that x is small such that powers of x above 3x could be ignored, use your answer in part (i) to show that 1sin 3x x bx where b is a constant to be found.
[3] (iii) State the equation of the tangent to the curve at the origin. On a single
diagram, sketch this tangent and the graph of .
1siny 3y x bx x
You should make clear the relationship between the two graphs for points near the origin. [3]
5 The functions f, g and h are defined by
2
1f : , , 0,
g : , ,
2010h : , , 0.
x x x xx
x x x
x xx
x
(i) Sketch the graph of y = f(x). [2] (ii) Define in a similar form. [3] 1f
(iii) Use a non-calculator method to solve 1f ( ) 4.x [2] (iv) State, giving a reason, whether fg exists. [2] (v) Find 21h ( ).x [1]
[Turn over
4
Section B: Probability and Statistics [60 marks]
6 A circular disc is divided into n equal sectors which are labeled 1, 2, 3, … n –1 and a “skunk”. A game is played by spinning a pointer pivoted at the centre of the disc at most three times. If the result is a “skunk”, the game ends and the player loses all his previous winnings. If the result from a spin is k where k = 1, 2, 3, … or n – 1, then the player wins $k. He continues spinning and adding to his winnings until a maximum of three spins or a “skunk” is spun.
Show that the probability of winning $3 is 3
1
n. [1]
Find, in terms of n, the probability that the player (i) wins nothing when the game ends, [2] (ii) wins nothing when the game ends, given that he wins $3 in his first spin. [2]
[There is no need for you to simplify your answer in both cases.] 7 In a statistical investigation, a researcher wants to find out how a person’s maximum
walking speed varies with age. He selects a random sample of 12 males of certain ages and measures their individual maximum walking speeds. Their ages, t years and maximum walking speeds, w ms–1 are shown in the table below.
t 20 25 30 35 40 45 50 55 60 65 70 75
w 2.59 2.55 2.85 2.62 2.48 2.43 2.32 2.27 2.34 2.28 2.19 2.10
(i) Draw a scatter diagram for the data. [1] (ii) State, giving a reason, whether a regression line of w on t or t on w could be used
to estimate the age of a male who has a maximum walking speed of 2.65 ms–1. (There is no need to do any calculations.) [1]
The researcher decides to study the maximum walking speed of males between the age of 30 and 55 inclusive. It is given that the correlation coefficient for the six data points is . 0.965
(iii) State, giving a reason, whether the regression line stated in (ii) is suitable for this study. [1]
(iv) For this study, the variables y is defined by 1
yt
. For the variables y and w,
(a) calculate the product moment correlation coefficient and comment on its value, [2]
(b) calculate the equation of the appropriate regression line, [1] (c) determine the best estimate that you can of the maximum walking speed
when the age is 43. [1]
5
8 The distribution of the masses of Perayaan balls has a mean of 420 g and a standard deviation of 6 g. Find the probability that 50 randomly selected Perayaan balls have a combined mass which is more than 20.9 kg. [2] State, with a reason, whether it is necessary to assume that the masses of Perayaan balls follow a normal distribution. [1]
The masses of Jubalani balls follow a normal distribution with a mean of 440 g and a standard deviation of 1 g. Find the probability that the combined mass of 50 randomly selected Perayaan balls differs from 50 times the mass of one Jubalani ball by less than 900 g. [3] State the assumption that you have made in arriving at your answer. [1]
9 The number of arrivals per minute at a fast food drive-through outlet has a Poisson
distribution with mean . On a weekend evening, 0.8 . Find the probability that in a 10-minute interval, there will be at most 10 arrivals given that there are more than 5 arrivals. [3]
In view of space constraints, the management wants to control the number of arrivals during the peak period which spans over a 30-minute interval. By using a normal distribution to approximate the Poisson distribution, find, to 4 decimal places, the largest value of such that the probability of having more than 30 arrivals during the peak period is less than 0.05. [5]
10 Annabel has 7 tiles each lettered A, N, N, A, B, E, L respectively. A code-word is
formed when some tiles are picked and arranged to form a “word”.
(a) Find the number of different ways in which a 4-letter code-word can be formed (i) if the first letter is N and last letter is E, [3] (ii) if there are no restrictions. [5]
(b) Annabel picks up 4 tiles and arranges them in a random order. Find the probability that the tiles spell ANNA. [2]
11 The weekly earnings, in dollars, at two casinos are modeled by independent normal
distributions with means and standard deviations as shown in the table. Mean Earnings Standard Deviation casino 1 600 000 50 000 casino 2 700 000 75 000
(i) Find the probability that in 2 randomly chosen weeks, the total earnings at casino 2 exceed 1 500 000 dollars. [2]
(ii) Find the probability that in a 12-week period, the weekly earnings at casino 1 exceeds $650 000 in at least 3 weeks. [3]
(iii) The government imposes a weekly tax on the earnings at casino 1 and 2 at a rate of 7% and 10% respectively. Find the probability that the tax exceeds $99 000 in any given week. Hence, by using a suitable approximation, find the probability that in a year consisting of 52 weeks, the weekly tax received by the government from the two casinos exceeds $99 000 in at least 45 weeks. [6]
[Turn over
6
12 (a) Club Gunpla emailed all of its 1600 members to find out which of the four Celestial Being mechs was the most popular. The members were asked to select their favourite mech. 226 members responded and it was concluded that Gundam Exia is the most popular of the four mechs among the club members.
Explain if the sampling method used is a random one. [2]
(b) A large department store wants to find out how much its customers spend on
Gundam model kits. From a random selection of 100 transactions, the results are summarized by
3500x , 2 220 400x ,
where $x is the amount spent on Gundam model kits in a single transaction.
The distributor claimed that the mean amount a customer spent on Gundam model kits is $40. Test whether the distributor has overstated his claim at the 5% significance level. [6]
State, giving a reason, whether any assumption is needed for the test to be valid. [1]
(c) In testing the mean breaking strain of a type of fishing line, a researcher measured
the breaking strain of 80 fishing lines. He carried out a t-test at the 5% significance level and, based on the sample results, he concluded that the population mean breaking strain is significantly different from 0 kN.
If the researcher had carried out a z-test instead, determine which of the following 2 statements is correct, giving clear reasons to support your claim.
(I) The researcher would have concluded that there is significant evidence at the 5% significance level that the population mean breaking strain is different from 0 kN.
(II) It is not possible for the researcher to conclude, by using only the information given, whether there is significant evidence at the 5% significance level that the population mean breaking strain is different from
0 kN . [3]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1
1
1 A curve has equation given by y = 1)(f
+xx , where f(x) is a quadratic function. Given
that the curve passes through the points (1, 4), (−3, −12) and (−2, −14), find the equation of the curve.
Hence sketch the curve, showing clearly the coordinates of the turning points and the equations of the asymptotes. [5]
2 (i) Show that )2)(1(
)12(2)1(
)12)(12()2)(1()32)(12(
+++
=+
+−−
++++
nnnn
nnnn
nnnn . [2]
(ii) Find ∑= ++
+N
n nnnn
1 )2)(1()12( , in terms of N. [3]
(iii) Hence, find the value of ...765
11654
9543
7432
5+
××+
××+
××+
×× [2]
3 The curve C has equation bxx
ay−
= 2 , where a and b are positive constants.
(i) Find, by differentiation, the coordinates of the turning point(s) of C. [3]
(ii) Sketch the graph of C, showing clearly the coordinates of any turning point(s) and the equations of the asymptotes. [2]
(iii) Hence, find the range of values of k, where k is a constant, for which the equation ( ) 02 =−− abxxk has no real root. [2]
(iv) On separate diagrams, draw sketches of the graphs of
(a) ( ) ( )bxbbx
ay+−+
= 2 , [2]
(b) bxx
ay−
−= 22 . [2]
4 The functions f and g are defined by
( ) 83:f 2 +−xx , ℜ∈x , 3≤x ,
1:g 2 +xex , ℜ∈x .
(i) Show that 1f − exists and define 1f − , giving its rule and domain. [4]
(ii) Determine, with a reason, whether the composite function gf 1− exists or not. [2]
(iii) If α and β are real numbers such that α < β ≤ 3 and gf(α) > gf(β), show that α < 6 − β. [3]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1
2
5 The nth term of a sequence is given by un = n2n for n ≥ 1 and the sum of the first n terms is denoted by Sn.
(i) Write down the values of S1, S2, S3, S4, S5. [2]
(ii) By considering Sn – 2, find a conjecture for the general term Sn in the form of Sn = p2(p + 2) + 2 , where p is in terms of n. [2]
(iii) Prove the conjecture by mathematical induction for all positive integers n. [4]
6 (i) By using the substitution θsinax = , find, in terms of a and π ,
⌡⌠
−
a
xxa0
22 .d [4]
(ii) Find, in terms of n and e, ( ) ,d0
⌡⌠ +
nnx xenx where 0≠n . [4]
(iii) Hence, find the exact area of the region R bounded by the line x = 1 and by
the curves 21 xy −= and ( ) xexy 1+= . [2]
7 A curve C has equation 2
2
216
xxy
+= . The region R is enclosed by C and the line
x = 2. Another region S is enclosed by C and the lines x = 2 and x = k (k > 2). The volume of solid formed by region R is equal to the volume of solid formed by region S when both R and S are rotated completely about the x-axis. Find the exact value of k. [4]
8 Adam has many marbles that he wants to put in boxes.
(i) If he puts 13 marbles in the first box and for each subsequent box, he puts double the number of marbles he puts in the previous box, he will need ( )12 +k boxes for all his marbles. Given that he has 104 marbles in the kth box, find the number of marbles he has. [4]
(ii) If he puts 13 marbles in the first box and for each subsequent box, he puts 13 marbles more than what he puts in the previous box, how many boxes will he need and what is the number of marbles in the last box? [4]
9 (a) The complex numbers z and w are such that z = 1 + i p, w = 1 + i q, where p
and q are real and p is positive. Given that zw = 3 – 4i, find the exact values of p and q. [4]
(b) A complex number a is given by a = 1 + i 3 . By using De Moivre’s theorem, express 932 ....1 aaaa +++++ in the form x + i y, where x and y are exact values to be determined. [4]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1
3
10 It is given that ( )xy sin1ln += .
(i) Find xy
dd in terms of x and show that ( ) 0sin
ddcos
ddsin1 2
2
=+++ xxyx
xyx . [2]
(ii) By further differentiation of the result in (i), find the Maclaurin’s series for y, up to and including the term in x3. [3]
(iii) Deduce the Maclaurin’s series for
+ xe x
sin1ln
sin, up to and including the term
in x3. [3]
11 The rate at which a substance evaporates is proportional to the volume of the
substance which has not yet evaporated. The initial volume of the substance is A m 3 and the volume which has evaporated at time t minutes is x m 3 . Given that it takes
(2ln2) minutes for half of its initial volume to evaporate, show that
−=
− teAx 2
1
1 .
Find the additional time needed for three quarters of the substance to evaporate, giving your answer in exact form. [6]
12 The diagram shows a prism with the horizontal rectangular base PQRS. The
triangular planes APS and BQR are vertical and AB is horizontal.
Given that PQ = SR = AB = 3 units, PS = QR = 2 units and each of the planes ABQP
and ABRS is inclined at an angle θ to the horizontal, where 43tan =θ .
The point P is taken as the origin for position vectors, with unit vectors i and j parallel to PQ and PS respectively and unit vector k perpendicular to the plane PQRS.
(i) Find →PA . [1]
(ii) Find the exact value of the cosine of the angle PAR. [3]
(iii) Find a vector equation of the line AR. [1]
(iv) Show that the foot of perpendicular from P to the line AR has coordinates
−
169132 ,
169162 ,
16921 . [3]
S
A B
P Q
R
i
j k
3
2 θ
θ
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 1
4
13 The diagram shows the cross-section of a cone of radius r and height h which is inscribed in a sphere of fixed radius R. Show that
( )hRhV −= 231 2π ,
where V is the volume of the cone.
Prove that, as r and h varies, the maximum value of V is obtained when 22 2rh = . [8]
~ End of Paper ~
h
r
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2
1
Section A: Pure Mathematics [40 marks]
1 A curve has parametric equations ( )tx −= 13 , 3
1t
y = for 0≠t .
(i) Find xy
dd in terms of t and deduce that the curve is an increasing function. [2]
(ii) Find the equation of L1, the tangent to the curve at the point
− 3
1,33t
t .
Hence, find the coordinates of point P on the curve at which L1 passes through the origin O. [4]
(iii) The line L2 is another tangent to the curve which is parallel to L1. Find the equation of L2. [3]
(iv) The line L2 cuts the y-axis at Q. Find the exact area of triangle OPQ. [2]
2 (i) The equation 2x2 – x – ln(x + 1) = 0 has 2 real roots α and β , where α <β. Find the values of α and β, giving any non-exact answers correct to 3 decimal places. [1]
(ii) A sequence of positive real numbers , , , 321 xxx satisfies the recurrence
relation xn+1 = 2
)1ln( nn xx ++
for 1n ≥ .
Prove algebraically that, if the sequence converges, then it converges to β . [3]
(iii) If x1 = 2, write down the values of x2 and x3 .
Sketch the graphs of y =2
)1ln( xx ++ and y = x on the same axes.
Illustrate on your diagram how the sequence , , , 321 xxx will converge to β starting with x1 = 2. [3]
3 (a) Solve the equation 44 −=z , expressing each of the roots in the form ba i+ , where a and b are real. [4]
Hence write down the solutions of the equation 4)i 2( 4 −=−w . [2]
(b) On a single, clearly labelled diagram, sketch the loci of w and z defined by 5 1 −=− zz and 2 i1 =−+w . [2]
(i) State the minimum value of wz − . [1] (ii) Find the exact values of the modulus and argument of w for which
4
)i1arg( π−=−+w .
Hence, express w in the form yx i+ , where x and y are exact real values. [3]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2
2
4 The plane p1 has equation x + y – 2 z = 4.
(i) A plane p2 with equation 2x + ay + bz = −4 (where a and b are constants) is parallel to p1, find the values of a and b and also the exact distance between p1 and p2 . [3]
(ii) Another plane p3 contains a point with coordinates (2, 4, 1) and a line with
equation r =
−+
332
111
s , where s ∈ℜ. Show that the equation of p3 can be
expressed as 3x – y + z = 3.
Find also a vector equation of the line l where p3 meets p1. [4]
(iii) The plane p4 has equation 5x + y + λ z = µ, where λ and µ are constants. (a) If all three planes p1 , p3 and p4 intersect in the line l , find the values of λ and µ . [2]
(b) Deduce the geometrical relationship of the three planes p1, p3 and p4 if λ = −3 and µ ≠ 11. [1]
Section B: Statistics [60 marks]
5 A Residents’ Committee wants to propose improvements to the recreational facilities in Punggol. In order to find out the needs of the adults and children of both genders, a survey is to be carried out on a sample of 120 residents who are at least 5 years old. Given that of the 6525 male residents who are at least 5 years old, there are 1450 children. On the other hand, there are 7975 females who are at least 5 years old, out of which 2175 are children. Describe how the sample can be obtained using quota sampling. [1]
Name and describe another method of sampling in which each group is represented proportionately. [3]
State an advantage of quota sampling over the above method. [1]
6 A recent research study indicates that 16% of the total population are aged 60 years or
more and that 18% of the total population have a measurable hearing defect. Furthermore, 65% of those aged 60 years or more have a measurable hearing defect.
Find the probability that a randomly chosen person from the population
(i) has a measurable hearing defect, given that he is less than 60 years old, [3]
(ii) is either aged 60 years or more, or has a measurable hearing defect, or both. [2]
State, with a reason, whether or not the events ‘a person is 60 years old or more’ and ‘a person has a measurable hearing defect’ are independent. [1]
Find, correct to 3 decimal places, the probability that, if two persons are chosen at random from the population, at least one of them will be aged 60 years or more and at least one of them will have a measurable hearing defect. [4]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2
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7 A manufacturer claims that the breaking strength of a climbing rope is normally distributed with mean 1702 N and standard deviation 105 N. A random sample of 10 climbing ropes is tested and the mean breaking strength of the sample is x N. A test is then carried out at the 5% significance level to determine whether the manufacturer’s claim is valid. Given that the null hypothesis is rejected in favour of the alternative hypothesis, find the range of possible values of x . [4]
The manufacturer believes that adding a special chemical to the ropes will increase the mean breaking strength and change the standard deviation. A random sample of 10 such ropes is found to have a mean breaking strength of 1724 N and a standard deviation of 35 N. Carry out a significance test at the 5% level to decide whether this result provides sufficient evidence to confirm the manufacturer’s belief that the mean breaking strength has increased. [4]
8 An experiment carried out to see how the intensity of radiation from a particular radioactive source, I, varies with time, t. In the following table, the values of t may be considered to be exact, while the values of I are subject to experimental errors.
I 22.5 25.0 28.0 30.5 38.0 40.5 42.5 48.0 54.5 55.0 70.0 t 44.0 42.0 33.5 28.0 18.0 13.6 15.0 10.3 9.0 6.3 4.0
(i) Sketch the scatter diagram for the given data. [2]
(ii) State, with a reason, which of the following models is more appropriate to fit the data points:
A: I = a +bt2, where a > 0 and b < 0; B: I = atb, where a > 0 and b < 0. [2]
(iii) For the appropriate model, find the product-moment correlation coefficient for the transformed data. Find an estimate for a, correct to 3 decimal places. [2]
(iv) Hence, estimate the time when the value of intensity of radiation is 35.0 and comment on the reliability of the estimate found. [2]
9 After production, blocks of butter are wrapped by a machine. Each block is supposed to weigh 220 g but the blocks produced have masses which are normally distributed. A worker then packs every six dozen randomly chosen blocks of butter in a box of mass 175 g. It is known that the total mass of each box of butter follows a normal distribution with mean 16.375 kg and standard deviation 0.149 kg.
(i) Three such boxes of butter are chosen at random. Find the probability that each box of butter weighs more than its mean mass. [1]
(ii) Find the percentage of blocks of butter which weigh at least 220 g. [3]
(iii) To increase the consumers’ confidence, the company wishes to adjust the mean mass of a block of butter such that more than 75 % of the blocks will have a mass of not less than 220 g while the standard deviation remains unchanged.
Find the least value of the mean mass after the adjustment. [3]
Yishun Junior College♦ 2010 Preliminary Exam ♦H2 Maths 9740 Paper 2
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10 (a) Find the number of 8-letter code-words that can be formed using the letters A, B, C, D, E if
(i) there are no restrictions, [1]
(ii) each vowel (A, E) appears once and each consonant (B, C, D) appears twice, [1]
(iii) each letter occurs at least once and the letters appear in alphabetical order. [3]
(b) At a particular reception, 9 guests are to stand at 3 identical round cocktail tables. How many ways can this be done if there must be at least two people at each table? [3]
11 A multiple choice test consists of 20 questions, each with four possible answers, of which only one is correct. If a student randomly chooses the answer to each question, find the probability of getting at least four but less than nine correct answers. [2]
Suppose that each correct answer is awarded five marks and each incorrect answer carries a penalty of one mark, what is the expected score obtained by a student? [2]
50 students took the test. Using a suitable approximation, find the probability that the mean score is more than 12 marks. [3]
12 At a post office, the number of customers purchasing postal products in a half-hour period during peak hours follows a Poisson distribution with mean 5. During the off- peak period, the number of customers purchasing postal products in a two-hour period is an independent Poisson distribution with mean 10. The peak period for the post office is from 12 pm to 1.30 pm and the post office is open from 8 am to 5 pm.
(i) Find the probability that there are more than 15 customers from 12.30 pm to 1.30 pm. [2]
(ii) Using suitable approximations, find the probability that the total number of customers during off-peak hours is less than thrice the total number of customers during peak hours on a particular day. [5]
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