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Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 1
1. Find the value of k for which the coefficient of x3 in the expansion of 6532 xkx
is 860. [5]
Solution :
Given that 6532 xkx , we have
rr
rkx
rkx
5
0
552
52
rr
r
rxk
r
5
0
52
5
When 3r , the coefficient of the x3 term,
332335402
!3
3452
3
5kkk
rr
rx
rx
6
0
663
63
When 3r , the coefficient of the x3 term,
5403!3
4563
3
6336
Therefore,
86040540 3 k
83 k
2k
[Analysis]
The question is looking for the x3 term in the expansion of 52 kx and 63 x .
We should use the term expression, rr
rkx
rkx
5
0
552
52 and r
r
rx
rx
6
0
663
63 .
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 2
Alternative solution:
Let 32
210
6586032 xxaxaaxkxy
2
21
54860323625
d
dxxaaxkxk
x
y
xaxkxkx
y86062330220
d
d2
432
2
2
86063120260d
d 323
3
3
xkxkx
y
Let 0x ,
86063120260323 k
433233 k
274323 k
83 k
2k
In Summary:
Binomial Theorem is a simple topic that will continue to feature in the future
examination. Each year, there will be one question on this.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 3
2. The acute angles A and B are such that 8tan BA and 3tan B . Without using a
calculator, find the exact value of Asin . [5]
Solution :
Given that 8tan BA ,
8tantan1
tantan
BA
BA
8tan31
3tan
A
A
For 3
1tan A ,
AA tan2483tan
5
1tan A
Therefore, 26
26
26
1sin A
Alternative solution:
From the ratios of tan A and tan (A+B) , we know that 90,,0 BABA
1013 22 PT
6518 22 PR
TSRPQR ~ ,
565
1 ST
65
5ST
26
26
10
65
5
sin PT
STA
[Analysis]
90,0 BA and 1800 BA .
To find the exact value of trigo ratio of Asin , we know that A is an acute angle, in quadrant I , its
value must be positive. Since 8tan BA , we can deduce that angle BA is in quadrant I
too. Therefore 900 BA .
5
1
A
5
1
A 3 B
P Q
R
S
T
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 4
In Summary:
Exact value refers to non-rounding of the answer. Students should be well
aware of doing mathematics without calculator.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 5
3. A particle moves along the curve 2
12
xy in such a way that the y-coordinate of the
particle is increasing at a constant rate of 0.03 units per second. Find the y-coordinate of
the particle at the instant that the x-coordinate of the particle is increasing at 0.12 units per
second. [5]
Solution :
Given that 2
12
xy
32d
d xx
y ------ (1)
12.0d
d03.0
x
y
4
1
d
d
x
y ------ (2)
(1) = (2),
4
12 3 x
33 2 x
2x
4
31
2
12
2y
[Analysis]
Rate of change t
x
x
y
t
y
d
d
d
d
d
d . Given that 03.0
d
d
t
y, when 12.0
d
d
t
x, we have
12.0d
d03.0
x
y , the connected rate of change. Take note that the question is asking for
the y-coordinate .
In Summary:
Typically question on rate of change is straight forward. The only issue is the
finding of y-coordinate, instead of the usual value of x.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 6
4. Express 2
22
2
xx
x as the sum of 3 partial fractions. [6]
Solution :
Let 22
222
2
x
C
x
B
x
A
xx
x.
22222 CxxBxAxx
When 2x ,
2224 C
4C
When 0x ,
222
B
2B
When 1x ,
CBA 1132
3A
2
423
2
222
2
xxxxx
x
Alternative solution:
Let 22
222
2
x
C
x
B
x
A
xx
x.
22222 CxxBxAxx
BxABxCAxx 2244 22
2B
3A
4C
2
423
2
222
2
xxxxx
x
[Analysis]
The numerator is of degree 2, the denominator is of degree 3. This is a proper fraction.
In Summary:
Make sure the given fraction is a proper fraction. Must also check the
correctness of the decompositions with any value of x.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 7
5. An experiment in Physics to find the focal length, f m, of a lens, requires the student to
place an object at a distance, u m, from the lens and to record the distance, v m, at which
the image is seen on the other side of the lens. The table below shows some results.
It is known that u, v and f are related by the equation fvu
111 . It is believed that an
error was made in recording one of the values of v.
(i) Plot v
1against
u
1 and hence determine which value of v, in the table above, is the
incorrect recording. [2]
(ii) Draw the straight line graph and use it to estimate a value of v to replace the incorrect
recording of v found in part (i). [2]
(iii) Estimate the value of f. [2]
Solution:
(i) From the graph, it shows that v = 0.263
is probably the incorrect recording.
(ii) From the graph, the estimate value of 30.41
v. v is 0.233
(iii) The v
1-axis intercept is 8.2.
u 0.150 0.200 0.250 0.300
v 0.603 0.299 0.263 0.201
u
1 6.67 5.00 4.00 3.33
u 0.150 0.200 0.250 0.300
v 0.603 0.299 0.263 0.201
v
1 1.66 3.34 3.80 4.98
[Analysis]
We were told that one of the data of v is wrong. By comparing the u and v, we suspect that the
first point, 0.603 might be the one. Then find, graphically the suitable value of v.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 8
ufv
111
Therefore, 2.81
f. Hence f = 0.122
In Summary:
Graph plotting is a very time consuming activity. It is usually best to be done as
the last question of the paper.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 10
6. (i) Prove that
tantanseccosec1
1
. [4]
(ii) Find, in radians, the acute angle for which
cot3tanseccosec1
1
. [2]
Solution:
(i)
tansectanseccosec1
tansec
tanseccosec1
1
22 tanseccosec1
tansec
cosec1
tansec
sin
11
cos
sin
cos
1
sin
sin1cos
sin1
tan
cos
sin
(ii) Given that
cot3tanseccosec1
1
.
tan
3tan
3tan2
3tan
Since 2
0
, 3tan 3
[Analysis]
Part (i) has numerous ways to do. Part (ii) is for 2
0
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 11
7.
The diagram shows a trapezium OABC in which OA is parallel to CB and O is the origin.
The side AB is parallel to the x-axis and the diagonal AC is parallel to the y-axis. The side
OA has equation xy 2 and the side OC has equation xy2
1 . The x-coordinate of A is h.
(i) Express the coordinates of A, B and C in terms of h. [5]
(ii) In the case where h = 4, find the area of the trapezium OABC. [2]
Solution:
(i) hhA 2, ,
2,h
hC ,
Equation of BC, hxh
y
2
2
When hy 2 ,
hxh
h
2
22
xh 22
7
hx4
7
[Analysis]
The first challenge is the find the x-coordinate of C. There are numerous ways to find the area.
Without knowing (i), we can still solve for area.
A
O
xy 2
x
y B
C xy2
1
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 12
hhB 2,
4
7
(ii)
When 4h ,
8,4A , 8,7B , 2,4C
the area of OABC
02880
04740
2
1
325614322
1
88462
1
21
the area of OABC is 21 square units
In Summary:
This question is very straight forward. Finding area of a quadrilateral is a
regular affair in this exam.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 13
8. It is given that f(x) is such that f (x) xx 2cos4sin . Given also that 02
f
, show
that 14cos3f4''f xxx . [7]
Solution:
Let xy f , given that xxx
y2cos4sin
d
d and when
2
x , 0y .
xxxxx
yy d2cos4sind
d
d
Cxx
y 2
2sin
4
4cos where C is a constant.
When 2
x , 0y .
C2
sin
4
2cos0
C
4
10
4
1C
2
2sin
4
4cos
4
1 xxy
xxx
y2sin24cos4
d
d2
2
2
2sin
4
4cos
4
142sin24cos44
d
d2
2 xxxxy
x
y
xxxxyx
y2sin24cos12sin24cos44
d
d2
2
14cos34d
d2
2
xyx
y
[Analysis]
This is the only non-routine question in this paper. Firstly, xx
fd
d'f . Secondly
xx
fd
d''f
2
2
. Let xy f , xxx
y2cos4sin
d
d . 14cos34
d
d2
2
xyx
y.
Without knowing (i), we can still solve for area.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 14
In Summary:
The best part of this question is that if you make any mistake along the way, you
will know it in the end and let you work backward to correct the error.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 15
9. The equation of a curve is 552 axaxy where a is a constant.
(i) In the case where a = 2, find the set of values of x for which the curve lies completely
above the line y = 9. [3]
(ii) In the case where a = 4, show that the line 2 xy is a tangent to the curve. [2]
(iii) Find the other value of a for which the line 2 xy is tangent to the curve. [3]
Solution:
(i) Given that 552 axaxy , when 2a , 352 2 xxy
For 9y , we get
9352 2 xx
01252 2 xx
0432 xx
From the graph, 4x or 2
3x
2
3or4: xxx
(ii) When 4a , the curve is 154 2 xxy . The intersection of the line and the curve is
2154 2 xxx
0144 2 xx
the discriminant of the quadratic equation,
014442
Hence the line 2 xy is tangent to the curve.
[Analysis]
(i) solve 9352 2 xx . (ii) To show 154 2 xxy and 2 xy has a repeated root.
(iii) To find a for 552 axaxy tangent to 2 xy .
4 O x
2
3
| |
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 16
(iii) Let 552 2 axaxx
0342 axax
Set the discriminant to zero,
03442
aa
041216 2 aa
034 2 aa
014 aa
4a or 1a
1a is the other value.
In Summary:
(i) needs answer in Set notation. (ii) is to show discriminant is zero. (iii) is to let
discriminant to be zero.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 17
10.
A gardener uses 130 m of fencing to enclose a plot of the shape shown above. The shape
consists of two equilateral triangles of side x m and a rectangle with sides x m and l m.
(i) Show that the area of the plot is 2
34130 2xx m2. [3]
(ii) Given that x can vary, find the value of x for which the area of the plot is stationary.
[4]
(iii) Explain why this value of x gives the gardener the largest area possible. [1]
Solution:
(i) The perimeter, 13042 xL
xL 265 ---------- (3)
The area, xxLxA2
3
2
2
3xLxA ---------- (4)
Substitute (3) into (4),
2
2
3265 xxxA
2
34130 2xxA
[Analysis]
(i) This is about surd conjugates, simplifying surd. (ii) is to find the turning point and (iii) is
to determine the nature of this turning point.
L m
x m
60°
x m
60°
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 18
(ii)
2
342130
d
d x
x
A
xx
A3465
d
d
Let 0d
d
x
A,
x34650
34
65
x
3434
3465
x
22 34
3465
x
13
3465 x
345 x
(iii) Since 034 , the coefficient of the x2 term is negative, hence equation
2
34130 2xxA
has a maximum turning point at 345 x .
In Summary:
Simplifying surds and Properties of quadratic curve.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 19
11. The point P lies on the curve xxy ln . The tangent to the curve at P is parallel to the line
32 xy .
(i) Find the coordinates of P. [5]
The normal to the curve xxy ln at P meets the line 32 xy at point Q.
(ii) Show that the x-coordinate of Q is 2ek , where k is a constant to be found. [3]
Solution:
(i) The gradient of point P is 2.
xxy ln
The gradient of the curve,
xxx
x
y 1ln
d
d
xx
yln1
d
d
Therefore,
xln12
1ln x
ex
ey
e,eP
(ii) The gradient of the normal at P is 2
1 .
The equation of the normal at P is,
e2
1e xy
e2
3
2
1 xy
Let e2
3
2
132 xx
e2
33
2
5x
e365
1x
[Analysis]
Gradient of the curve at the point of tangent is the same as the gradient of the line. Forming
equation of the normal to find the intersection.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 20
e25
3x
In Summary:
A simple routine question on coordinate geometry. The handling of e may give
problem to some students.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 21
12. A curve has the equation 4322 xy .
(i) Explain why the lowest point on the curve has coordinates
4,
2
3. [1]
(ii) Find the coordinates of the points at which the curve intersects the x-axis. [2]
(iii) Sketch the graph of 4322 xy . [3]
(iv) Using your graph, state the number of solutions to each of the following equations.
(a) 74322
x [1]
(b) 34322
x [1]
(c) 024322
x [1]
Solution:
(i) Given that 4322 xy ,
0322x
44322
x
4y , y has a minimum value of 4 .
When 2
3x , 4y .
(ii) When 0y ,
43202 x
4322x
232 x
322 x
2
5x or
2
1x
[Analysis]
Sketching Quadratic curve and modulus quadratic curve.
Solutions to O Level Add Math paper 1 2014
By KL Ang, Jan 2015 Page 22
(iii) the graph of 4322 xy
(iv)
(a) 74322
x , 2 solutions.
(b) 34322
x , 4 solutions.
(c) 24322
x , no solution.
In Summary:
Questions on Curve sketching will continue to appear in the future papers.
4
O x
2
3
| |
5
|
2
1
2
5
y
7y
3y
2y