ANGLO-CHINESE JUNIOR COLLEGE M A T H E M A T I C S DEPARTMENT

MATHEMATICS Higher 2 9 7 4 0 / 0 1

Paper 1 17 August 2011

JC 2 PRELIMINARY EXAMINATION

Time allowed: 3 hours

Additional Materials: List o f 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 o f 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 o f marks is given in brackets [ ] at the end of each question or part question. At the end o f the examination, fasten all your work securely together.

This document consists o f 5 printed pages.

^ n g l n - C h i n c M 3 « « i o r College [Turn Over

C - 4

1 Without using the graphic calculator, find the range of values of x which satisfy the 1 2

inequality > , [3] x-] x+2

x2 +2 2 Given that (x +1) is a factor of x1 +1, express as partial fractions.

r + 1 Hence or otherwise, find the coefficient of in the expansion o f (l-x + x2} in

ascending powers of x. [5]

(a) Find (i) J * t a n ( x 2 ) d x , [2]

00 f 2 - , d x . |4]

(b) Evaluate, exactly,

(i) x s in - ' ( x J )dx , [4]

(ii) J o x |* - f c | dx where 0 < e < l . 13]

The equation of the plane p} is given by x~ 6y + 2z = 5,

(i) Show that the line /, with equation r = 1

, 1 . • M 1

h

lies on the plane p%. (2)

(ii) Find the Cartesian equation of the plane p2 which is parallel to plane px and contains

the point A with coordinates (1,-1,1). (2]

(iii) Line / 2 contains point A and the point B with coordinates (9,1,1). Show that the sine

of the acute angle between l2 and p, is • 12]

Hence, or otherwise, find the exact perpendicular distance between pt and p2. [2J

5 Given that y = fi» where tan"1 y = 2 tan"1 x + — for -0.4 <, x < 0.4, show that 4

= 2 ( 1 + / ) . | 2 1

By further differentiation of this result, find the Maclaurin's series for y up to and including the term i n * 3 . [4J Denote the above Maclaurin's series for y by g(x). Find the range of values of x for which the value o f g(x) differs from fl» by less than 0.5,(2]

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Anglo-Chlnae Junior College H2 Mathematics 9740:2011 JC 2 Preliminary Examination Paper I

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6 (I) Find the roots o f the equation z 5 + 243 = 0 in exponential form. [31 (II) zs +243 can be expressed as [ z + c ] [ z 2 -(acos 0)Z + a ] [ V - (acos3#)z + 6J, where

a,b,c and 9 are real. Find the exact values of a,b,c and 0. [5]

7 An isosceles triangle has fixed base of length b cm. The other 2 equal sides of the triangle are each decreasing at the constant rate of 3 cm per second. How fast is the area changing when the triangle is equilateral? Leave your answer in terms of b. |5|

8 The region R is bounded by the curve v = - — r , the line x = — , the .r-axis and the ( l + 4 ; r 2 f 2

y-axis, as shown in the diagram below.

(1) Using the substitution x = - tan / , find the exact area of R, [6]

(II) Find the volume of the solid formed when R is rotated completely about the y-axis.|3]

Sketch the curve given by the equation y2+ax2 =5 for .tSO and y > 0 , where a is a positive constant. | 1 | The functions f and g are defined by

f :xr->j5-ax2 , xeR, 0<,x<,^

g:x H>l + e", x e R . x ^ O .

Show that f" 1 exists and define f" 1 in a similar form. 141

Given that f\x) = x, f o r a l l x e R , O J S J T S J - ,

(I) show that a = 1 without evaluating f 2 (x) , |2| (II) using the result in (I), show that fg exists and find its corresponding range. |3|

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Anglo-Chlncst Junior College H2 Mathematics 9740: 2911 JC 2 Prtiimimry Examination Paper I

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10 Prove by Mathematical Induction that

j T / * = i '-—- , f o r a l l / i g Z * . [41 ~, ( 1 - x )

Hence evaluate

3 ( 3 ) 2 + 4 ( 3 ) 3 + 5 ( 3 ) 4 + + 15(3)' 4 . |3]

11 Detectives arrive at a crime scene at 12 p.m. and found an unfinished cup of coffee at 45 ° C . In order to estimate what time the coffee was brewed, a fresh cup of coffee was made from the coffee machine in the same room, and its temperature was found to be 110"C. After leaving the new cup of coffee in the room for 5 minutes, the temperature dropped to 8 0 ° C .

It is known that the rate at which the temperature of the coffee falls is proportional to the amount by which its temperature exceeds that o f the room. Given that the crime scene is an air-conditioned room with temperature controlled at 25 "C, at what time was the unfinished cup of coffee brewed? Leave your answer to the nearest minute. [6]

12 A curve with equation y = f(x) is also defined by the parametric equations

x = \-e~!, v = l + / 2 , / e R . (i) The point P on the curve has parameter p. Given that the tangent to the curve at P

passes through the point (1,0), find the value of p. [4] (II) Given that the graph of y = f(x) has an inflexion point at t = - 1 , sketch, on separate

diagrams, the graphs of y = f(x) and y = f ' ( x ) , showing clearly the exact

coordinates of the turning point(s) and asymptote(s), i f any. [4]

13 (a) Jerry's brother gave him some game cards for his birthday. He then starts to collect cards for a total period of 3n weeks, counting his birthday gift as the first week of collection. Subsequently, the number of cards he buys each week is d more than the number he bought the previous week. I f P is the number of cards collected in the first n weeks and Q the number of cards collected in the last n weeks, show that

Q-P=2n2d (4|

(b) A fund is established with a single deposit of $2500 at the beginning of 2011 to provide an annual bursary of $150. The fund earns interest at 3.5% per annum, paid at the end of each year. I f the first bursary is awarded at the end of 2011 after interest is earned, show that at the end of n years, the amount (in dollars) remaining in the fund is

d " 0 0 ( , . „ 3 5 ) % 2 ° 5 ° ° . 7 v ; 7

When is the last year that the bursary can be awarded? [6]

- End of Paper -

Anglo-Chinese Junior College H2 Mathematics 9740: 2011 JC 2 Preliminary Examination Paper I

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ANGLO-CHIMESE JUNIOR COLLEGE M A T H E M A T I C S D E P A R T M E N T

MATHEMATICS Higher 2 9 7 4 0 / 0 2

Paper 2 18 August 2011

JC 2 PRELIMINARY E X A M I N A T I O N 1

Time allowed: 3 hours

Additional Materials: List o f 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 o f 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 stales 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 o f 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.

jAnglo-uTlj inPM J u n i o r (fiollegr [Turn Over

Section A: Pure Mathematics [40 marks]

1 The graph of y~f(x) passes through the points (1,2), (-1.3) and (2 ,2) . Given that

f '(x) = lax1 + 2bx + 2, where a and b are constants, find f (x). [4]

1 P O I f - B + £ , whereP and Q are constants, find/5and Q\\\

r ( r + l)(r + 2) r ( r + l) ( r + l)(r + 2) Hence find

(ii) — + — — + _ — + [3] 1-2-3 2-3-4 3-4-5 1 J

3 (a) A complex number z is represented by the point P on an Argand diagram. Given

that z = ( l + i ) ( / - 2 ) + - ~ — T where / is real, / ( l + i )

(i) find the imaginary part of z, [2] (ii) find the Cartesian equation of the locus of point P, [1] (iii) sketch the locus o f point P in an Argand diagram, stating clearly any

intercepts and asymptotes exactly. , [2J

(b) Sketch in an Argand diagram the locus of the point A which represents the z - 5 + 2i

complex number z such that >1 , z

3 Hence find the least value o f | z -10 + 4i| in the form -yfa where a is an integer

to be determined. [5]

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Anglo-Chinese Junior College H2 Mathematics 9740:2011 JC 2 Preliminary Examination Paper 2

Page! of 6

f4> a > f 1 ^ ( 2 )

4 The lines /, and / } have equations r = 3 ¥1 2 and r = - 1 + s k

v 4 y

respectively,

where s and / are real parameters.

(i) Given that /, and / 2 are perpendicular, show that the value of it is - 3 . [2J

(ii) Find the vector equation o f the line / } which passes through the point A with

coordinates (4,3,4) and is perpendicular to both /, and l2. [2]

(iii) Show that the point B with coordinates (2,2,5) lies on line l}. Hence find the

perpendicular distance from B to line /, exactly. [3] (iv) The planes a, and n2 are such that rtx contains /, and / 2 , and /r 2 contains /, and

is perpendicular to l 7 . Given that the plane n3 has equation r = 20, find two

conditions which relate a, b and c such that the three planes do not intersect, supporting your conditions with reasons. |3 |

x The curves C, and C 2 have equations 9y2 = (x+kj2 - 9 and —+y2=\,

Ic where k is a real constant such that k > 3. (a) Describe a sequence o f transformations which transforms the graph of x2 - y 2 = 1

to the graph of C,. [2] (b) (i) On the same diagram, sketch the graphs o f C, and C 2 , stating clearly the

coordinates o f any points o f intersection with the axes and the equations of any asymptotes. |5 |

(ii) Find the range of values of the positive constant a such that the equation x2 {x+kf-9 , - r + - = 1 a2 9

has two real roots. |2 |

Section B : Statistics [60 marks]

6 How many 6-digit numbers can be formed using the digits 0, 1,J, 2, 2 and 3 exactly once i f (i) there are no restrictions? [3] (11) the number is odd and less than 300,000? |3] [Note: '0 ' cannot be the first digit o f any number]

Anglo-Chinese Junior Cotltge H2 Mathematics 9740: 2011JC2 Preliminary Examination Paper 2

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7 A survey shows that 20% of JC students sleep before 11 p.m., 70% of them sleep between 11 p.m. and 1 a.m., and the remaining steep after 1 a.m. The probability that a student is late for school the next day is 2%, 5% and 10% i f he/she slept before 11 p.m., between 11 p.m. and 1 a.m., and after 1 a.m. the night before, respectively. (i) Show that the probability that a randomly chosen student is late for school is

0.049. (2] (ii) Find the probability that a randomly chosen student slept before 11 p.m. the night

before, given that he/she is not late for school. [3]

There are five school days in one week. (Iii) What is the probability that a randomly chosen student is on time for school every

day in a given week? |2 | (iv) Given that out of n weeks, the probability that a student is on time for school every

day in at most eight of the weeks is at most 0.14, find the least value of n. [4]

in a glass factory, it is found that 20% of glass panels produced by machine A are more than 3 mm thick. Given that the thickness of glass panels produced by machine A follows a normal distribution with mean 2.56 mm and standard deviation <T mm, show that <j = 0.523. |2] (i) For the manufacture of a certain type of windscreen, two of the glass panels

produced by machine A are used to form a double panel. Find the probability that the thickness of a double panel is between 4 mm and 6 mm. [2]

(ii) One hundred samples, each consisting o f 5 glass panels produced by machine A are tested for their thickness. In how many of these samples would you expect to find the mean thickness to be greater than 3 mm? [3]

(iii) The thickness o f glass panets produced by machine B for a certain type of shower screen has a normal distribution with mean 5.9 mm and standard deviation 0.35 mm. Find the probability that the average thickness o f 2 glass panels produced by machine A and 3 glass panels produced by machine B is at least 4,2 mm. |3]

9 The mean number of packets o f cereal sold in a supermarket per day is 124.5. Following a television advertisement, the mean daily sales over a period o f n days, is found to be 129.3 with variance 172. (i) When n = 12, test, at the 10% level o f significance, whether the mean number of

packets of cereal sold per day has increased. [5] (ii) I f it is given instead that n is large and there is sufficient evidence at the.

1% level o f significance that the mean number of packets o f cereal sold per day has changed, find the range of values of n. [3]

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Anglo-Chinese Junior College H2 Mathematics 9740:2011 JC 2 Preliminary Examination Paper 2

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10 The number of chandeliers sold in a shop in one week has a Poisson distribution with mean 3.2. (a) Show that the probability that at least 7 chandeliers are sold in a week is 0.0446.|2| (b) A random sample of 150 weeks is taken. Using a suitable approximation, find the

probability that at least 7 chandeliers are sold per week in no more than 4 o f the weeks. 14]

(c) For two randomly chosen weeks, find the probability that one chandelier is sold in one of the weeks given that a total of three chandeliers are sold. [3|

(d) Calculate the probability that the mean number o f chandeliers sold in 100 randomly chosen weeks exceeds 3. |3|

11 (a) 1000 invited guests are expected to visit a private exhibition on a particular Sunday. The organizer wishes to find out i f there is a correlation between the age of a visitor and the time he/she spent at the exhibition. With the knowledge that visitor traffic is highest at around noon, the organizer decides to choose the first 50 visitors entering the exhibition from noon to record their age, as well as the duration that they spent there. (i) Give one reason why this is not a good sampling method. [1] (ii) Describe how the sample could be chosen using systematic sampling and

explain why systematic sampling is a better sampling method in this context. [2]

(b) Eight guests at the exhibition are chosen at random. The following table records x, the time spent (to the nearest minute) andy, the age of these visitors.

x (min) 79 45 301 483 250 175 282 15 v (years) 62 73 28 19 41 50 a 85

(i) Given that the least squares regression line o fxony is x = 521.86 —6.5254y, show that a = 32, to the nearest whole number. (2)

(ii) Calculate the corresponding estimated value of x when y = 85, and comment on the suitability of the linear model. [2]

(iii) Draw the scatter diagram for these values, labelling the axes clearly. [2] (iv) Calculate the linear product moment correlation coefficient between x and y,

and comment on its value in the context of the question. (2) (v) Use a regression line to give the best estimate that you can of the age o f a

visitor who spent 430 minutes on the exhibition, leaving your answer to the nearest whole number. Comment on the reliability of this estimate. J2J

.( 'hinej* Junior Coilefc H2 Mathematics 9740: 2011 JC2 Prelininary Examination Paper 2

Pax* 6 of t

Anglo-Chinese Junior College H2 Mathematics 9740

2011 JC 2 Preliminary Exam Solutions Paperl

- U - L . JC — I x + 2

- x + 4 (x-\)(x + 2)

: = > ( x - 4 ) ( x - l ) ( x + 2 ) < 0

=> x<-2 or 1 < x £ 4

x 3 + l = (x + l ) ( x 2 - x + l )

x 2 + 2 1 _ 1

(x + l ) ( x 2 - x + l ) ~ x + l x2-x + l

\ - i x 2 + 2 I

x 3 + l x + \

= ( x 2 + 2 ) ( l + x J ) ~ ' - ( l + x)

. ( 1 + ( . 0 W + t f i l ( v ) + H K ^ ) ( , . ) t . . . ]

= (x 2 + 2 ) ( l - x 3 + x 6 - x ' + ... + ( - l ) x 3 " +...)

- ( l - x + x 2 - x 3 + . . . + ( - ! ) " x"+. . . )

Hence Coefficient o f x 3 " is

2 ( - l ) " - ( - 0 J ' - ( - ! ) '

OR 2 ( - l ) " + ( - l ) - '

OR 2 ( - l ) - - ( - l ) -

OR { -1 i f n is odd; 1 i f n is even} 3(a) 0)

(ii)

1 t - 2 x s i n ( x 2 ) xtan x 2 dx = - - ( T-^dx

] 1 ' ; 2 J cos (x ; )

= - - l n | c o s ( V | + c

1 r 2 r + 1 - 1

x + x + 3 2 J x + x + 3 dx

= i r _ 2 i ± i _ d x _ i r _ L _ _ 2 J x ! + x + 3 2 J ( ^ + i ) 2 + i i

dx

3(b) (i)

(i>)

1 , I 2 , i 1 _ , ( 2 x + l ) = - l n x + x + 3— j— tan - — r - ^ + c

2 1 TiT 7:

f ^ x s i n - ' ( x 2 ) d x = ^ s m ' ( x 2 ) - p - J L = rdX

^ s m ' ( x 2

2 v ; 2

JL 24 + 4 2

1

Since 0 < / > < l ,

J ox|jr — = £ - x ( x - 6 ) £ & + J i x ( x - 6 ) £ i f

6 +

0 I V _bxl\

3 2 -

6 +

0 _ 3 2 _

" 3 + 3 ~ 2 4(1)

= 2 - 6 + 4 = 0. Therefore /, is parallel to px

= 9 - 6 + 2 = 5. Therefore (9,1,1) lies in />,

Therefore /, lies in px.

(9 + 2X\

Alternatively, 1 + /1

2 + 4/1,

• 9 + 2 A - 6 - 6 , l + 2 + 4^ = 5 (ii) f 1 ] ( I ] f 1 1

-6 = - 1 - 6

, 2 , . 1 ; . 2 ,

Therefore r. -6 = 1 + 6 + 2 = 9

The Cartesian equation is x - 6 y + 2z = 9

(iii)

which is parallel to

Let 9 be the acute angle between /, and p, fA\ 1 •N l

- 6

v 2 ;

= Vl6 + W l + 36 + 4 sin 0

4 - 6 | = v47V4l sin#

Therefore, sin 9 -

Hence mtd

2V697

/17V41 697

Hence mtd 1 Let h be the perpendicular distance between the planes

— = sin 6 AB

Therefore A = V41

Otherwise mtd 2( find length of projection on normal)

f 1 N

6

h = BA, V41

4

Otherwise mtd 3 ( find perpendicular distance fr O to each plane) _5 9 I

h =

4 V4T

tan"' y = 2tan~'x+ — 4

Differentiate with respect to x, I dy _ 2

1 + y 2 dx~ l + x2

• ( ! • * £ - » ( « • • )

Differentiate with respect to x,

d l y . ^ ^ = 4 V ^ dx dx v ' dx1 dx ' dx

* ( l + * ' ) $ + ( 2 * - 4 , ) £ . 0

Differentiate with respect to x,

s > ( l + ^ ) £ f + 2 x ^ + ( 2 x - 4 j , ) f y + ( 2 - 4 ^ ) ^ . 0 dx1 dx1 dx dx

. ( l + x » ) ^ + ( 4 x - 4 y ) ^ f + 2 ^ - 4 f ^ Y - 0 ^ ; d x 3 " d r 1 </x U * J

Whenx = 0, y = tan — = 1 4

1 dy _ 2 ify_4

1 + 1 1 + 0 dx

( l + 0 ) ^ + (0-4) (4) = 0 = > - ^ = 16 dr 2

( 1 + 0 ) ^ + ( ° " 4 ) ( 4 ) + 2 ( 4 ) " 4 ( 1 6 ) = ° ̂ ^ = 1 2 0

y = tan 2tan ' x + — 4

, , 16 2 120 3

= l + 4x + — x 2 + — x 3

2! 3!

= l + 4x + 8x 2 + 20x 3

Sketch y = | / ( x ) - g ( x ) | and y = 0.5

For \/(x) - g(x)\ 0.5, - 0.376 < x < 0.252

6(1) z 5 = -243

=3. z 3 = 243*'* - 2 4 3 / " r t f c ' \Z

^ z = 3 ^ 5 J

z=3e 5 , 3 * 5 ,3e , / r,3e 5,3<> 5

6(H) z s +243

= ( - ( " 3 ) ) z - i e 5 z -3c" ' 5 - 3 « 5 1 z -3e 5

r - 3 ; « s + « 5 + 9 : - 3 z U s + « 5 +9 = ( Z + 3)

= (z + 3 ) ^ 2 - 3 z ( ^ 2 c o S | j + 9 p 2 - 3 z ^ 2 c o s ^ J + 9

= (z + 3)[V - [ 6 c o s ^ j z + 9jjV - ^ 6 C O S y j z + 9

Therefore a = 6, b = 9, c = 3 , 0 = — 5

Let x be the length of each of the other 2 sides of the triangle.

Area, A = — b x height

oL4 I u2 \

ax 4 ^ 4

44 44 dx Now, — = — x —

a7 dx dl

' ( t o )

W h e n , = 6. f = -3, | > J ( - 3 >

V4

- A 0 2

8(0 1 dx 1 , x = - t a n / =5 — = —sec I

2 dl 2 A i

Area/?= f 2 ! T d r Jo l + 4x 2

= p L J o ( l+tan 1 / ) U

= - f 3 c o s 2 t d/ 2 Jo

W?HC0S2t 9 Jo ?

-sec 2 / Id/

sin 2/ 73 = — + — units

. 12 16

8(ii) Volume = ffjj —

7 T '

NO

dy + /r 16

3 £ 16

= 0.589 units 3

k y y 2 + a r 2 = 5

0 £ Since any horizontal l iney = k, 0<k<\fs cuts the graph o f y = f (x) once and only once, f is

a one-one function. Hence f e x i s t s .

T a

y2=5-ax2

Let y = 7 5 - a x 2 , 0 < x £ J -

l 5 - y 2

x = , | , since 0

• r ' : m o < x ^ V 5

(i)

f ( x ) = x f o r a l l x e R , 0 £ * : £ , / -

Givcn: f 2 ( x ) = x

=>f (x) = f - ' W

= > V 5 - a x ' • ,1-

< 5 1 =>5 = - , a = -o o

.". a = 1 (shown)

Since R , = ( l , 2 ] c D ( = [0 ,V5 ] , .\fg exists.

Method 1

5 x 2

(ii)

Method 2

2 , y=g(x)

x £ 0 y - i

0

[ 0 , « ) ^ ( t , 2 ] A [ ! , 2 ) R „ = [ l , 2 )

10 Let P, be 2_,rx • 1 — j for all » e Z *

Testing P,: LHS = ( l ) (x ) = x

R H f ; _ x - 2 x 2 + * 3 _ * ( * 2 - 2 x - H )

( 1 - x ) 2 ( x 2 - 2 x + l )

LHS = RHS Hence P, is true

Assuming that Pk is true for some t e Z ' ;

. A r x-(k + l)xk,, + kxkf2

le > rx = ' : is ( i - x ) -

To test P/,+1 is true ie to test" £ r x r _x-(k + 2)xtt2 + (k + i)xk,i

( 1 - x ) 2

L H S = 2)iar* =

_ x - ( * + l)x**'+ib-*- 2 + (* + l ) x " ' ( l - x ) 2

( l - x ) 2

_ x - ( / t + l )x* t ' + fc r '* 2 + ( l - 2 x + x 2 ) (A:- ( - l )x*"

=

_ x + fcr"2-2(* + l ) x , < 2 + ( x - M ) x ' » 3

( l - x ) J

_ x + ( y - 2 / : - 2 ) x ' * 2 + ( * : + r ) x t o

( 1 - x ) 2

_ x-(k + 2)xU2+(k + \)xk+>

( 1 - x ) 2

Hence, Pk+/ is also true.

Thus, by the Principle o f Mathematical Induction, Pt true and PA is true = > P w is true. Hence

P„ is true for all n e Z '

3 ( 3 ) , + 4 ( 3 ) J + 5 ( 3 ) 4 + f - l5 (3 ) ' 4

l ( 3-16(3" i) + 15(31 7)

( 1 - 3 ) 2

104 029 569

M - 3 ( 3 3 ) + 2 ( 3 ' ' ) N

( 1 - 3 ) '

ALTERNATIVE

3 ( 3 ) 2 + 4 ( 3 ) 3 + 5 ( 3 ) 4 + + I 5 ( 3 ) M

= £ ( / - +1)3'

14 14

3-15(3 , i ) + 14(3") 3 ) 3 2 ( 3 I 3 - 1 )

( 1 - 3 ) ' 3 - 1

= 104 029 569

11

=> l n | e ? - 2 5 | « t o + C

When/ = 0, ^ = 110=> I n | l l 0 - 2 5 | = C

Whenr = 5, t? = 80 => ln |80-25 | = 5A + C

, 5 85

When 0 = 45,

ln|45 - 25| = Q I n ^ jt + In 85 = . t = 16.&2 rain * 17 rain

The estimated time when coffee was brewed is 11.43 A M 12(0 . dx .,

x = l-<? — = e . dx .,

x = l-<? — = e

y = l + / 2 = > ^ = 2, dt

dy

dx dx e -dt

When/ = p , — = 2pep, point Pis ( l - e ' ^ . l + p 2 ) , ax

Equation o f tangent at P is y-(\ p2) = 2pep[x-{l-e~p)]

y = 2pe"x-2pep(\-e-p) + (\ p2) Since tangent passes through (1,0),

0 = 2pep (\)-2pep +2p + l + p2

=> p2+2p + l = 0

= > P — 1

12(H) y = f'(x)

\i 1

(0,1)

0 x=\

13(a)

13 (b)

P = ^2a + {n-X)d]

(2n + \)th te rm= l " term o f last n terms = a+{2nd)

Q = ̂ {2[a+2nd] + (n-\)d)

Q-P = ̂ [2a + 4nd + (n-\)d]-^[2a + {n-\)d] =2n2d

ALTERNATIVE:

S„=l[2a + (n-\)d] = P

^ - y [ 2 a + ( 2 i i - l ) r f ]

S , . = ^ [ 2 a + ( 3 « - l ) < ]

= 3 o „ + ^ L ( 3 „ _ 1 ) a r _ ^ . [ 2 a + ( 2 w - l ) j ] - ^ [ 2 « + ( « - l ) o ' ]

= 2nld W h c n w - 1 : Fund has $(1.035)(2500)- 150

« = 2 : $(1.035)|"(1.035)2500-150]-150 =(1.035r2500-(1.035)150- 150

at rath year: (1.035)"2500 - (1.035)""' 150 - (1.035)""' 150...-150 =( 1.035)"2500 - (150)[ 1 +(1.035)+( 1.035) J+...+(I.035y']

= (1.035)"2500- 150[ ' • ° 3 5 " ~ 1

( 0.035

( 1 . 0 3 5 / 2 5 0 0 - ^ ( 1 . 0 3 5 ) " + . 3 0 O ! ) : )

(1.035)

12500

0.035 x ' 7 2 5 o o - 2 2 2 2 £ ] + » o o o

1.035)" 30000

M 2 2 £ . l ^ ( l . 0 3 5 ) - i o

( i .03s ) - s i5222

v 7 12500 n < 25.44 n = 25 years Last year is 2035.

Anglo-Chinese Junior College If2 Mathematics 9749

2011 JC 2 Preliminary Exam Solutions Paper 2 / • ' (x) = 3 a x 2 + 2 & r + 2

=> f(x) = axi + bx2+2x+c

Curve passes through (1,2) => a f ft+c = 0 (1)

Curve passes through (-1,3) =s>-a + b + c=5 (2)

Curve passes through (2,2) => 8a + 4£ + c = -2 (3)

5 31 = _ 8

Q ~ ' l ' = 6 ' C 3

,t \ 3 31 2 „ 8 •••f(x) = —xi +—x2 +2x —

J w 2 6 3 20) 1 1 1 1

r(r + l ) ( r + 2) 2 [ r(r +1) (r + l)(r + 2)

2 2

Sum of the series = — 1

^ 2 [ r ( r + l) (r + l)(r + 2)

- L . J j + f J - . - L ,.. 4.5 5 . 6 J {5.6 6.7

1

2(11)

(«-!)/» n(n + \)J {r,(n + l) ( « + ! ) ( « + 2 )

: I l

_i_ j i _ 2 '20 40

S. =•

2 [20 (n + \)(n + 2) As n -> co,

I

1 M2

2 0 " i + 2 + 2 n n

1 _ J l _ 1.2.3 + 2.3.4 + 3.4.5 i J _ _ L y 1

6 + 24 + 60 + ^ r ( / - + l ) ( r + 2)

__9_ J _ _ ^

~ 24 + 40 4 3(a) 0)

z = ( l + « ) ( r - 2 ) +

= ( l + t ) ( / - 2 ) +

1 - /

f ( l + J )

l - i 1 - i

= ( l + , ) ( / - 2 ) ~ /

Im(z ) = / - 2 - -

Minimum value o f 12 — (10 — 4/)J

= M B + B C or OC- O M or C M

3^29 units

4(0 /, is perpendicular to l2

f\N

= 0

=>2 + 2* + 4 = 0 =>-t = -3

(ii) PI f 2 1 2 X -3

, 4 ;

( U P f 2 1 7 // I

_7J

Equation of the line l} is r = f4 > f 2 '

3 1

(iii)

Let

(4 + 2A\

2 -, 4 - A ,

= - 1

= - 1 s - 1

Therefore A lies on ! 3

h S(2,2,5)

-4(4,3,4)

Therefore, perpendicular distance from B to /, is ^ ( 4 - 2 ) 2 + ( 3 - 2 ) z + ( 4 - 5 )

= 7(3 units (iv) For the three planes to not intersect at all, plane

ny should not contain /,

(4,3,4) cannot be a pt on plane xy,

therefore *20 => 4a + 3 £ + 4 c * 20

We also need /, to be parallel to /r 3,

therefore = 0=>a + 2o + 4c = 0

5(a) C,: 9 y 2 = ( x + A r ) J - 9

3 2

Sequence of transformations: EITHER (1) Scaling parallel to the x-axis by factor 3

(2) Translation in the negative x-axis direction by k units

OR (1) Translation in the negative x-axis direction by — units

(2) Scaling parallel to thex-axis by factor 3 (b)

(0 Intercepts: ( 3 -x .O) (-3-k,0) Intercepts: (±k,Q) , ( 0 , ± l )

Asymptotes: y = ± ( * + * )

3 x + k -x-k

i.e. y = — r ~ , y - — - —

x + k

£ , (x + A ) 2 - 9 _ t y = 3

<J2 9

fe±QL.,.,.4_„ (x + £ ) 2 (x + &) 2

Sketch the graph of C,: ^ — ~ - y 2 = I (i.e. / = i—Z—l _ j ) o n the same axes as graph of

For equation (*) to have 4 real roots, the two graphs intersect at 4 points. Since a is a positive constant, a > k + 3,

6

(0 Case

51 First digit is T o r ' 2 ' : ^ - 2 = 120

51 First digit is '3 ' : — - = 30

2!2! Total number o f 6-digit numbers is ISO.

(ii) 4 ' First and last digits are ' 1 1 : — = 12

2! 4 '

First digit IS '1 and last digit is '3 ' ; — = 12 41

First digit is ' 2 ' and last digit is ' 3 ' : — = 12

First digit is ' 2 ' and last digit is ' 1': 4! = 24 Total number o f 6-digit numbers is 12 + 12 + 12 + 24 = 60.

70) 0.2 x 0.02 + 0.7 x 0.05 + 0.1 x 0.1 = 0.049

(ii) P(slept before 11p.m. and not late for school) / P(not late for school)

- 0 - 2 * 0 - 9 8 =0.20610 1-0.049

(iii) (1 - 0.049)5 = 0.77786 = 0.778 (3 s.f.) (iv) Let X be a random variable denoting the number o f weeks such that the student is on time for

school every day in that week. X-B(n, 0.77786)

Oivcn, P ( * S 8 ) < 0 . 1 4 From G.C., « = 14.

8 ?{X > 3) = 0.2

P ( * < 3 ) = 0.8

A'-N(2.56, a2), 2 = X " 2 5 6

0 44

— = 0.8416212335 a

0- = 0.5228 = 0.523 (1) W = X, + X, ~ N(5.12,0.54663968)

P(4 < A' < 6) = 0.818 (ii)

* ~ N ( 2 V ' 5 2 2 ! | ! ]

P(X > 3) - 0.029923

Expected No. = 0.029923 x 100 = 2.99

NOTE: For students, who used the rounded off value o f a = 0.523 , we shall accept p = 0.0300 and expected no.= 3.00.

(iii) y~N(5.9 ,0 .35 J )

S = Xl+X2 + Yl+YtfYi _ N ( 4 5 6 4 i 0.0365655872)

P(5 £ 4.2) = 0.972

NOTE: For students, who use the rounded off value of a = 0.523, we shall accept variance= 0.03658232 and ?(SZ 4.2) = 0.971

9(0 H „ : // = 124.5 . .„ To test at 10% significance level

H, : fj> 124.5

s2 = —(172) = 187.63636

i r degree of freedom = 12

129 3-124 5 Test Statistic : t = , • . • r = 1.2138

187.63636 ) 12 . .

p-value = 0.1251 > 0.10

We have insufficient evidence at the 10% significance level to conclude that the mean number of packets of cereal sold per day has increased.

(ii) H 0 : = 124.5 To test at 1% significance level.

H , : p* 124.5

Under H 0 , X - N - ^ ( 1 7 2 )

124.5, I approx by CLT.

120 1-194 5 / Test stat, z = i f l i ^ L t l i =0.365997 V ^ T

Since H 0 is rejected, p-value<0.01

0 . 3 6 5 9 9 7 7 ^ 1 > 2.57583

n > 50.5 (or * £ 5 1 )

10(a) * ~ P 0 ( 3 . 2 )

P( X > 7) = 1 - P(X <, 6) = 0.0446191

(b) W~fl (150 , 0.0446191)

Since tip = 6.3692868 > 5

and nq = 143.307135 > 5

W - N(6.692865, 6.394235387)

P(fV < 4) a ?(tV < 4.5) = 0.193

(c) Y~P0(6A)

?(X = l).P(AT = 2)2 p(y = 3)

= 0.750

(d) 3 2

S ~ N(3.2, y ^ ) approximately by CLT.

P ( S > 3 ) = 0.868

11(a)

(0 These 50 visitors w i l l not be a good representation o f all the visitors as morning and late afternoon visitors are unrepresented.

(ii) Let the sample size be 50.

1000

I t = 20

Select a random number between 1 and 20, say 18. Pick the 18 t h, 38 , h , 58 , h, 78 , h 998 , n

visitor. This method is better because both the morning and the afternoon visitors are represented.

1Kb) (i)

x = 203.75 (from calculator or computation)

when x = 203.75, x = 521.86 -6.5254y

y = (521.86 - 203.75)/6.5254 = 48.74950195

using y = '£y/n

o = 8 y - ( 6 2 + 73 + 28 + 19 + 4 l + 50 + 85) = 8x48.74950195-358 =31.669 « 32 (shown)

(ii) When y = 85, x = 521.86 - 6.5254 x 85 = -32.799 = -32.8 (3s.f.) This linear model is not suitable as the age of a person cannot be negative.

(iii)

(iv)

Possible scatter diagrams,

Time Spent /min •

483- • o

15>

19 a m ^ Age /years

85 Or

Age /years

85-

19 •

r = -0.95909 J-5_

| h. Time Spent 483 /rnin

There is high negative correlation between the age of the visitors and the time they spent at the exhibition.

(v) Use the xony line to predict the age of the visitor. When x = 430, y = (521.86 - 430) /6.5254 = 14.077 « 14 This estimate is reliable as 430 falls within the given data range of x.