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Year 2013 is
going to be a
year worth
remembering!
Work hard, try hard and NO regret!!
Additional
Mathematics
4038
Six Years Series
2007 – 2012
Questions and
Full solutions
Dear Students
This is a compilation of the latest past six years Additional Mathematics O level 4038.
It is meant for you as a reference and personal use. The material is copyrighted and it is only for
your personal use.
Good luck.
Mr Ang June 2013
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 1
1. Find the set of values of the constant k for which the 1y k x intersects the curve
2 6y x x k at two distinct points.
Solution :
21 6k x x x k
2 6 2 0x k x k
Discriminant, 2
6 4 1 2k k
2 20 36k k
Let 0 to get two distinct roots,
2 20 36 0k k
18 2 0k k
2k or 18k
Alternative solution:
21 6k x x x k
2 6 2 0x k x k --------- (1)
For equation (1) to have two distinct solutions, we just need the minimum point of this “smiley”
curve to be below the x-axis, or negative.
Let 2 6 2y x k x k
Line of symmetry,
6 6
2 1 2
k kx
[Analysis]
To solve simultaneous equations that will give two distinct solutions. After combining the two
equation, the quadratic equation will have to have a positive discriminant.
k
w 2 20 36w k k
2 18
In Summary:
This is a rather routine question. Knowing the discriminant of quadratic
equation well, particularly nature of roots, will always be important. Solving
quadratic inequality occurs frequently in this paper.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 2
When 6
2
kx
,
2
min
6 66 2
2 2
k ky k k
21
6 24
k k
2
5 94
kk
For two distinct roots, min 0y
2
5 9 04
kk
2 20 36 0k k
18 2 0k k
2k or 18k
An alternative approach to solve quadratic inequality
Given a “Smiley” quadratic curve, like 2 20 36 0w k k , it is easier to consider the
complementary set of 2 20 36 0w k k .
Consider 2 20 36 0w k k ,
18 2 0k k
2 18k This is the range of values that we need to exclude!
So, the solution to 2 20 36 0k k , is
2k or 18k
Using this approach, we always turn any quadratic inequality into a “Smiley” and then with the
knowledge of as 2 0ax bx c , where 0a . The solution to this kind of inequality is always
1 2x x x ., where 2
1 2ax bx c a x x x x and 1 2x x .
k
w 2 20 36w k k
2 18
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 3
2. Given that 1 3s , express 2 2
1
s
s
in the form 3a b , where a and b are integers.
Solution :
By long division,
2 2 3
11 1
ss
s s
Given that 1 3s ,
2 2 31 3 1
1 1 3 1
s
s
33
2 3
3 2 33
2 3 2 3
2
2
6 3 33
2 3
3 6 3 3
6 2 3
[Analysis]
This is about rationalizing denominator where
1 a
aa and
2
11 a b a b
a ba b a b a b
.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 4
Alternative solution:
Given that 1 3s ,
2
2 1 3 22
1 1 3 1
s
s
2
1 2 3 3 2
2 3
6 2 3
2 3
6 2 3 2 3
2 3 2 3
2
2
2 6 2 3 3 6 2 3
2 3
12 4 3 6 3 6
4 3
6 2 3
In Summary:
Take note of the first solution. After simplifying the algebraic expression before
ratinalising the denominator can usually make the question easier to do.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 5
3. The value, V dollars, of a house is given by 0
ktV V e , where 0V dollars is the original value
of the house when built, t is the time in years since it was built and k is a constant.
Calculate
(i) the value of k if, after 10 years, the value of the house has doubled,
(ii) the value of t when the value of the house is three times its original value.
Solution :
(i)
Given that when 10t , 02V V
10
0 02k
V V e ,
102 ke
ln 2 10k
1
ln 2 0.069310
k
(ii)
When 03V V ,
0 03 ktV V e
3 kte
ln3 kt
ln 3 ln 3
15.8ln 2
10
tk
[Analysis]
A straight forward question on exponential(indices) and logarithm.
In Summary:
Answer for the part (ii) may vary a little, depending on the value used in k.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 6
4.
(i) Find the integer which satisfies the equation 3 2 11 3 0x x x .
(ii) Find, in the form 3a b , a and b are integers, the other values which satisfy the
equation.
Solution :
(i)
Let 3 2f 11 3x x x x ,
when 3x , 3 2
f 3 3 3 11 3 3 0
therefore 3x is a factor of 3 2f 11 3x x x x
By long division,
2f 3 4 1x x x x .
23 4 1 0x x x
3x or 2 4 1 0x x
Since the discriminant of 2 4 1 0x x ,
Discriminant, 2
4 4 1 1 12 0 is not a square number, the roots will
not be a rational root. So the only integer root is 3x .
Alternatively,
f 1 0 , f 1 0 , f 3 0 , so the only integer root is 3x .
(ii)
When 2 4 1 0x x , by quadratic formula,
24 4 4 1 1
2 1x
2 3x
[Analysis]
This question involves solving a polynomial equation. So, factor theorem maybe a productive
tool. In part (i), the possible integer zeros(roots) are 1, 3 . Part (ii) is to be completed with
the application of quadratic formula.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 7
Alternatively,
When 2 4 1 0x x , by completing the square method,
2 4 1 0x x
2 22 2 2 2 2 1 0x x
2
2 3 0x
2
2 3x
2 3x
2 3x
Alternative solution:
3 2 11 3 0x x x
3 2 23 4 11 3 0x x x x
2 23 4 11 3 0x x x x
2 3 4 1 3 0x x x x
23 4 1 0x x x
In Summary:
Take note that in part (i), it is sufficient to apply remainder theorem with non-
zero remainders for all possible roots to find all integer(rational) roots.
Finding roots with completing the square method, may be easier in this case.
Long division of polynomial is really a very useful technique.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 8
5.
(i) Differentiate ln sin x with respect to x .
(ii)
The diagram shows part of the curve coty x , cutting the x -axis at ,02
. The line
3y intersects the curve at P . A line is drawn from P , parallel to the y -axis, to
meet the x -axis at Q . Use your result from part (i) to find the area of the shaded
region.
Solution:
(i) Given that ln siny x ,
Let sinu x , cosdu
xdx
lny u , 1dy
du u
By chain rule,
dy dy du
dx du dx
1
cosdy
xdx u
cos
sin
dy x
dx x
cotdy
xdx
[Analysis]
Part (i) is to apply chain rule. Part (ii) is to integrate the area under the curve, which will
require the x-coordinate of the point Q to be found.
x
y coty x
3 P
Q
2
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 9
(ii)
Given that coty x , when 3y , where 02
x
3 cot x
1
3tan x
1
tan3
x
1 1tan
3x
6
x
2
2
6
6
cot ln sinxdx x
ln sin ln sin2 6
1
ln 1 ln2
ln 2 0.693 square units
In Summary:
Part (i) solution shows the full application of chain rule. Part (ii) provides an
interesting twist, make sure that you understand the range of x that is acceptable in
this question. Additional solution will need to be rejected.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 10
6. The matrices A and B are such that 2
1A= B . Given that
2 1
2 1
B= , find
(i) the matrix A ,
(ii) the value of the constant k for which 1 4k B A I , where I is the identity matrix.
Solution:
(i) Given that 2 1
2 1
B= ,
det B=2 1-2 1 4
11 11
2 24
B =
2
21
1 1 1 1 1 1 1 31 1 1
2 2 2 2 2 2 6 24 16 16
B
2
11 31
6 216
A= B
(ii) For 1 4k B A I
11 1
2 24
kk
B =
1 3 1 0 3 3 1 14 1 3
46 2 0 1 6 6 2 216 4 4
A I
1 1 1 13
2 2 2 24 4
k
Therefore , 3k
[Analysis]
Consider 1A A=I
, given that A is a 2 2 matrix, 1 0
0 1I
. 2AA=A .
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 11
Alternative solution:
(i) Given that 2
1A= B ,
1 1 BA=BB B
1BA=IB
1BBA=BB
2B A=I
Let a b
c d
A
2 1 2 1 1 0
2 1 2 1 0 1
a b
c d
2 3 1 0
6 1 0 1
a b
c d
2 3 2 3 1 0
6 6 0 1
a c b d
a c b d
1
2 3 1 2 1 3 16
6 0 6 6
16
aa c a c
a c a cc
3
2 3 1 2 1 3 16
6 0 6 2
16
bb d b d
b d b dd
1 3
1 3116 16
6 2 6 216
16 16
A
(ii) Given that 1 4k B A I ,
1 4k B B A I B
4k I AB IB
1 14k I B B B B
14k I B I B
14k I B B ------------ (1)
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 12
det B=2 1-2 1 4
11 11
2 24
B =
Substitute B and 1B into (1),
1 0 1 1 2 11
40 1 2 2 2 14
k
1 0 3 0
0 1 0 3k
1 0 1 0
30 1 0 1
k
Therefore , 3k
In Summary:
Learn to apply some matrix pre- or post multiplications can be helpful as
shown in part (ii) of the alternative solution. Beware that AB BA , in general.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 13
7. A curve has the equation 2
2
xey
x
, where 2x .
(i) Find dy
dx.
(ii) Show that the y -coordinate of the turning point is 3
2
e.
(iii) By considering the sign of dy
dx, or otherwise, determine whether the turning point is a
maximum or a minimum.
Solution:
(i) Given 2
2
xey
x
, for 2x
2 2
2
2 2 1
2
x xe x edy
dx x
2
2
2 3
2
xe xdy
dx x
(ii) When 0dy
dx ,
2
2
2 30
2
xe x
x
As 2 0xe for all real x , and 2
2 0x , therefore only
2 3 0x .
3
2x 2
[Analysis]
Apply quotient rule in part (i). The tangent at turning point is at zero gradient, 0dy
dx . The
real challenge is in part (iii).
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 14
So, when 3
2x ,
32
23
3
22
32
2
ey e
e
(iii)
x 1.5 ( 1.8) 1.5 1.5 ( 1.2)
dy
dx 0 0 0
The turning point at 3
2x is a minimum point.
Alternative solution:
(i) Given 2
2
xey
x
, for 2x
2
ln ln2
xey
x
2ln ln ln 2xy e x
ln 2 ln 2y x x
2 ln 2ln d x xd y
dx dx
1 1
22
dy
y dx x
2 3
2
dy xy
dx x
2
2
2 3
2
xe xdy
dx x
x
y
2
2
xey
x
1.2 1.8 1.5
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 15
(iii)
2 3
2
dy xy
dx x
2
2
2 3
2 3 2
2
xd
d y dy x xy
dx dx x dx
,product rule
Let 2 3
2
xu
x
,
ln ln 2 3 ln 2u x x
1 2 1
2 3 2
du
u dx x x
2 1
2 3 2
duu
dx x x
2 3 2 1
2 2 3 2
du x
dx x x x
Therefore,
2
2
2 3 2 3 2 3 2 1
2 2 2 2 3 2
d y x x xy y
dx x x x x x
2
2
2 3 2 2 2
2 2 2 3
d y x xy
dx x x x
2 2
2
2 3 2 2 2
2 2 2 2 3
xd y e x x
dx x x x x
2 2
22
2 3 2 2 2
2 22
x x xd y e
dx x xx
When 3
2x ,
23
22 0 4 0
d ye
dx
Therefore the turning point at 3
2x is a minimum.
In Summary:
Clearly, doing a 2nd
derivative on this function is not going to be any easy.
Observe the technique of taking logarithm before differentiation help to get the job
done.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 16
9. The function f : 2cos3x x is defined for 0 x .
(i) State the period f .
(ii) State the amplitude of f .
(iii) Sketch the graph of fy x .
(iv) On the diagram drawn in part (iii), sketch the graph of 2
1x
y
for 0 x .
(v) State the number of solutions, for 0 x , of the equation 2 cos3 2x x .
Solution:
(i) The period of f is 2
3
.
(ii) The amplitude of f is 2 .
(iii) Sketch the graph of fy x .
x 0 6
3
2
2
3
5
6
y 2 0 2 0 2 0 2
[Analysis]
Apply quotient rule in part (i). The tangent at turning point is at zero gradient, 0dy
dx . The
real challenge is in part (iii).
0 x
y
2
2
6
3
2
2
3
5
6
2cos3y x
21
xy
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 17
(iv) For the graph of 2
1x
y
for 0 x .
(v) State the number of solutions, for 0 x , of the equation 2 cos3 2x x .
2 cos3 2x x
2
2cos3 1x
x
The number of solutions is the number of points of intersections between the curve,
2cos3y x , and the line, 2
1x
y
,is 3.
x 0
y 1 1
In Summary:
This question may see repeated visits in this exam. A good ability to sketch curve,
trigo curves in this case, will be important.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 18
10. (a) Solve the inequality 2 3 5x .
(b) The graph of 2y x c passes through the point 1,5 .
(i) Find the possible values of the constant c .
(ii) Find, for each value of c , the x -coordinate of the point where the graph meets
the x -axis.
Solution:
(a)
When 3
2 3 02
x x , 2 3 2 3x x .
2 3 5x
2 3 5x
4x
Therefore, 3
42
x
When 3
2 3 02
x x , 2 3 2 3x x .
2 3 5x
2 3 5x
1x
Therefore, 3
12
x
Combining the two inequality, we get 1 4x
[Analysis]
This is a simple question on linear modulus. 1st, is to make a sketch of 2 3y x . 2
nd , then
place a line of 5y . We can then clearly infer from the sketch the desire solution.
Continue to sketch the V-shape curve that passes through 1,5 . It will be self-evident there
the two ‘c’ values are to be.
O x
y
3
2
5
2 3y x 3
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 19
(b) When 2y x c passes through the point 1,5 , substitute 1x and 5y into the
modulus equation.
(i) 5 2 c
When 2 0 2c c , 2 2c c .
5 2 c
3c
When 2 0 2c c , 2 2c c .
5 2 c
7c
(ii)
When 3c , 0y ,
0 2 3x
0 2 3x
3
2x
When 7c , 0y ,
0 2 7x
0 2 7x
7 1
32 2
x
In Summary:
Modulus is a key topic to understand as it will be use in ‘A’ level. It is best by
going through this solution very carefully so as to enhance your understanding.
O x
y
5
1 2
c 2
c
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 20
Alternative solution:
(a) Solve the inequality 2 3 5x .
Since 2
2 3 2 3x x ,
2
2 3 5x
2 22 3 5x
2 22 3 5 0x
2 3 5 2 3 5 0x x
2 2 2 8 0x x
1 4 0x x
1 4x
(b) When 2y x c passes through the point 1,5 , substitute 1x and 5y into the
modulus equation.
(i) 5 2 c
2
5 2 c
225 2 c
2 22 5 0c
2 5 2 5 0c c
3 7 0c c
3c or 7c
In Summary:
The alternative solution is actually easier to understand than those as shown in
textbook.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 21
12. Answer only one of the following two alternatives.
EITHER
The diagram shows a car badge ATBUCVDWA , in which ABCD is a square of side 8cm .
Points P , Q , R and S are the mid-points of AB , BC , CD and DA respectively. The arc
BUC is part of a circle with centre S . Arcs CVD , DWA and ATB have centres P , Q and
R respectively.
(i) Show that angle BSC is 0.927 radians, correct to 3 significant figures.
(ii) Find the perimeter of the car badge.
(iii) Find the area of the car badge.
Solution:
(i) 2 2 2 2 24 8BS AB AW
4 5BS
By cosine rule,
2 2 2
2 cosBC BS CS BS CS BSC
2
28 2 4 5 2 4 5 4 5 cos BSC
160 64 160cos BSC
[Analysis]
The content of this question has been moved to Mathematics syllabuses.
A
B C
D
T
U
V
W
S
P
Q
R
A
B C
D
T
U
V
W
S
P
Q
R
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 22
3cos
5BSC
1 3cos
5BSC
0.927BSC radians
(ii)
4 5 0.927BC
The perimeter 4 4 5 0.927 33.2 cm
(iii)
Area of sector SBUC 21
0.927 4 5 37.082
Area of the car badge 24 37.08 8 84.3 2cm
In Summary:
This question will now appear in General Mathematics paper.
5
3
4
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 23
OR
The table shows experimental values of two variables x and y .
It is known that x and y are related by the equation 10 xy Ab , where A and b are
constants.
(i) Using graph paper, draw the graph of lg 10y against x and use your graph to
estimate the value of A and of b .
(ii) By drawing a suitable line on your graph, solve the equation 210x xAb .
Solution:
(i)
10 xy Ab
10 xy Ab
lg 10 lg xy Ab
lg 10 lg lgy A x b
lg A is the vertical axis intercept, lgb is the gradient of the line.
x 0.5 1.0 1.5 2.0
y 15.9 19.1 23.4 30.2
x 0.5 1.0 1.5 2.0
y 15.9 19.1 23.4 30.2
10y 5.9 9.1 13.4 20.2
lg 10y 0.77 0.96 1.13 1.31
[Analysis]
Part (i) effectively tell us what we should be doing with the given curve. Copy the table and
update it very carefully.
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 24
From the graph,
lg 0.6A
3.98A
0.65
lg 0.37141.75
b
2.35b
(ii)
Given that 210x xAb ,
lg lg 2A x b x
The line to add is lg 10 2y x , from the graph, 0.35x .
x
ln 10y
O 0.5 1 1.5 2
0.5
1
1.5
1.75
0.65
0.35
lg 10 2y x lg 10 lg lgy A x b
Solutions to O Level Add Math paper 1 2007
By KL Ang, Jun 2013 Page 25
In Summary:
Be careful that it is lg not ln in this question. Students do have to expect this
type of question to appear regularly in this exam. Your answers may vary, depending
on the drawing. It is advisable to draw the curve as large as possible.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 26
1. Prove the identity 2
11 sin 1 sin
1 tanA A
A
.
Solution :
L.H.S. 2
1
1 tan A
2
1
sec A
2cos A
21 sin A
1 sin 1 sinA A
= R.H.S.
Alternative solution :
R.H.S. 1 sin 1 sinA A
21 sin A
2cos A
2
1
sec A
2
1
1 tan A
= L.H.S.
In Summary:
These identities are useful not only in proving questions about identities, but
also in solve trigo equations and some trigo applications.
[Analysis]
Recognise that 2 21 tan sec , 2 2sin cos 1 , 1
seccos
.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 27
2. A company supplies 4 garden centres – Allseed, Budwise, Croppers and Digwell – with
bags of compost, which are sold in 3 sizes – large 145 litres, medium 75 litres and small 20
litres. The number of bags of compost supplied to each garden centre in one delivery is
shown in the following table.
Over a six-month period Allseed received 5 such deliveries, Budwise 6, Croppers 8 and
Digwell 7. Write down three matrices such that matrix multiplication will give the total
amount of compost supplied over the six-month period and hence find this total.
Solution :
4,3 3,1 4,1
200 500 200 Allseed145
300 600 0 Budwise75 per delivery
0 400 300 Croppers20
0 700 0 Digwell
,
We will need to construct a delivery matrix to pre- or post- multiply the result of the above
matrix. In order to get a (1,1) matrix from (4,1), we will perform 1,4 4,1 1,1 . So we are
to pre-multiply a (1,4) delivery matrix. The three matrices are,
200 500 200
300 600 0
0 400 300
0 700 0
,
145
75
20
, 5 6 8 7
Large Medium Small
Allseed 200 500 200
Budwise 300 600 -
Croppers - 400 300
Digwell - 700 -
[Analysis]
The first challenge is to figure out what is in the question. The total compost ordered =
(frequent of delivery) x (Amounts per delivery) x (capacity of each size). Next, it is to figure
out what matrices will permit this to happen.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 28
200 500 200145
300 600 05 6 8 7 75
0 400 30020
0 700 0
70500
885005 6 8 7
36000
52500
1539000
The total amount of compost supplied over the six-month period is 1539000 litres
In Summary:
The difficulty in this question is in multiplying out the matrices correctly.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 29
3. The line 2 3 12x y meets the curve 2 4 8y x at the points P and Q . Find the length
of the line PQ .
Solution :
To solve 2
2 3 12
4 8
x y
y x
,
Let 2 12 3x y
4 24 6x y --------- (1)
Substitute (1) into 2 4 8y x ,
2 24 6y y
2 6 16 0y y
2 8 0y y
8y or 2y
When 8y , 18x .
When 2y , 3x .
2 2
18 3 8 2PQ
325
5 13 units
Alternative solution:
Given that 2 3 12x y ,
3
62
yx
3336 6
2 2 2
p qp
q p
y yyx x
2 2
2 9 9
4 4
p q q p
q p
y y y yx x
2 2
q p q pPQ x x y y
2
29
4
q p
q p
y yy y
2
13
4
q py y
13
2q py y
To solve 2
2 3 12
4 8
x y
y x
,
Let 2 12 3x y
4 24 6x y --------- (1)
Substitute (1) into 2 4 8y x ,
2 24 6y y
2 6 16 0y y
2 8 0y y
8y or 2y
[Analysis]
Points of intersections of a line and a curve, is found by solving simultaneous equations. It is
easier to replace variable x . The length 2 2
q p q pPQ x x y y .
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 30
Alternative solution:
Let 3 12 2y x --------- (1)
29 36 72y x
2
3 36 72y x --------- (2)
Substitute (1) into (2),
2
12 2 36 72x x
2144 48 4 36 72x x x
24 84 216 0x x
2 21 54 0x x
3 18 0x x
3x or 18x
When 18x , 8y .
When 3x , 2y .
13 13
2 8 10 5 132 2
PQ units
In Summary:
Students should check the answers against the given equations to guard against
mistakes. Avoid substituting fraction into equation.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 31
4. Find the coefficient of 3x in binomial expansion of
(i) 7
1 2x ,
(ii) 721 7 1 2x x .
Solution :
(i) Given that 7
1 2x ,
Each term in this expansion, 77 7
1 2 2r r r r
rT x xr r
7 0 1 2 30 1 2 3
7 7 7 71 2 2 2 2 2
0 1 2 3x x x x x
7 2 31 2 1 14 84 280x x x x
the coefficient of 3x term is 280 .
(ii)
Given that 721 7 1 2x x
2 2 31 7 1 14 84 280x x x x
2 3 2 2 31 14 84 280 7 1 14 84 280x x x x x x x
2 3 2 31 14 84 280 7 98x x x x x
2 31 14 77 182x x x
the coefficient of 3x term is 182 .
Alternative solution:
(i) Given that 7
1 2x ,
Each term in this expansion, 77 7
1 2 2r r r r
rT x xr r
When 3r , 3 3 3 3
3
7 7 6 52 8 280
3 3!T x x x
the coefficient of 3x term is 280 .
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 32
(ii)
Given that 721 7 1 2x x
7 721 2 7 1 2x x x
Expansion of the first term is as shown in part (i), 7
2r r
rT xr
Expansion of the second term, 2 27 7
7 2 7 2r rr r
rT x x xr r
When 2 3r , 1r . 1 3 3 3
1
77 2 7 14 98
1T x x x
the coefficient of 3x term is 280 98 182 .
In Summary:
A routine question, Be competent in handling the term expression, Tr . In this
way, the question can be done much quicker.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 33
5. (i) Differentiate tan 2 1x with respect to x .
(ii) Explain why the curve tan 2 1y x has no stationary points.
(iii) Find, in terms of p , the approximate change in tan 2 1x as x increases from 1 to
1 p , where p is small.
Solution:
(i)
Let tan 2 1y x .
2dsec 2 1 2
d
yx
x
2d2sec 2 1
d
yx
x
(ii) For tan 2 1y x to have stationary points, d
0d
y
x .
22sec 2 1 0x
2sec 2 1 0x
sec 2 1 0x
Since sec 2 1 1x or sec 2 1 1x , sec 2 1 0x
Therefore d
0d
y
x , tan 2 1y x has no stationary point.
(iii) Given that d
d
y y
x x
, where f fy x x x
1x , 1 1x p p , 22sec 2 1y
xx
,
22 sec 3 2.04y p p
[Analysis]
Part (i) can be done by quotient rule of sin
tancos
. Part (ii) is to show that the first
derivative cannot be zero. Part (iii) is no longer in the current examination. It is included for
completeness of this question.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 34
7. Solve the equation
(i) log 72 3 log 3x x ,
(ii) 5 253log log 10y y .
Solution:
(i) Given that log 72 3 log 3x x , where 0x and 1x
log 72 log 3 3x x
log 72 3 3x
327 8 x
33 3 2x
6x
(ii) Given that 5 253log log 10y y , where 0y
55
5
log3log 10
log 25
yy
55
log3log 10
2
yy
5 56log log 20y y
55log 20y
5log 4y
45y
625y
In Summary:
Students will be well advised to master solving logarithmic equation. Base
change formula is a common requirement. One cannot separate indices operations
from logarithm.
[Analysis]
Part (i) is to see that log 72x and log 3x share the same base. Part (ii) will need to apply the
base change formula.
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 35
9. (a) Solve the equation 22cos 5sin 1 0x x for 0 360x .
(b) Solve the equation tan 1 cot 2 0y y for 0 2y radians.
Solution:
(a) Given that 22cos 5sin 1 0x x , for 0 360x
22 1 sin 5sin 1 0x x
22sin 5sin 3 0x x
22sin 5sin 3 0x x
2sin 1 sin 3 0x x
2sin 1 0x or sin 3 0x
1
sin2
x sin 3x Rejected, 1 sin 1x
1 1sin
2x
, x is in 3
rd and 4
th Quadrants.
Principal angle, 30x
210 ,330x
(b) Given that tan 1 cot 2 0y y for 0 2y radians,
1
tan 1 2 0tan
yy
tan 1 2 0y
tan 3y
1tan 3y , y is in 2nd
and 4th
Quadrants
Principal angle, 1.249y
1.249, 2 1.249y
1.89, 5.03y
In Summary:
The solution is with principal angle method.
[Analysis]
To solve part (a), just need to change 2 2cos 1 sinx x , then it becomes a quadratic equation
look-a-like. Using 1
cottan
xx
to simplify the equation
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 36
11. Solutions to this question by accurate drawing will not be accepted.
The diagram, which is not drawn to scale, shows a triangle ABC in which the point A is
9,9 and the point B is 1, 3 . The point C lies on the perpendicular bisector of AB
and the equation of the line BC is 8 11y x . Find
(i) the equation of the perpendicular bisector of AB ,
(ii) the coordinates of C .
The point D is such that ACBD is a rhombus.
(iii) Find the coordinates of D .
(iv) Show that 2AB CD .
[Analysis]
A perpendicular bisector divides a line segment into two equal parts, passes through the
midpoint of AB . Use the gradient of AB to find the gradient of the normal. Use the equation
of the perpendicular bisector and the equation of BC to find the point C .
Lastly, the points D and E are the intersections of the two curves.
x
C
9,9 A
1, 3B O
y
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 37
Solution:
(i) Let the mid-point of AB be M ,
1 9 3 9
,2 2
M
5,3M
Gradient of AB 3 9 3
1 9 2
Gradient normal of AB 1 2
3 3
2
The equation of the perpendicular bisector of AB ,
2
3 53
y x
2 19
3 3
xy
(ii)
2 19
3 3
8 11
xy
y x
,
2 19
8 113 3
xx
2 19 24 33x x
52 26x
2x
When 2x , 5y , the coordinates of 2,5C
(iii) The mid-point of CD is 5,3M . Let ,D x y ,
2 5
, 5,32 2
x yM
2
52
x
53
2
y
8x 1y
the coordinates of 8,1D .
x
C
9,9 A
1, 3B O
y
M
D
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 38
(iv)
2 2
1 9 3 9 208AB
2 2
8 2 1 5 52CD
208 52
4 21352
AB
CD
Therefore, 2 CD AB
Alternative solution:
(iii) The mid-point of CD is 5,3M and 2,5C .
5 3,3 2 8,1D
the coordinates of 8,1D .
(iv)
tanCD CM
AB BM
3tan
2 , gradient of the line AB
tan 8 , gradient of the line BC
tan tan
tan1 tan tan
38
23
1 82
13
2
13
x
C
9,9 A
1, 3B O
y
M
D
x
2,5C
5,3M
O
y
D
2
3 2
3
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 39
1
2
1
tan2
CD CM
AB BM
Therefore, 2 CD AB
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 40
12. Answer only one of the following two alternatives.
EITHER
A particle starts from a fixed point A and travels in a straight line. The velocity, 1msv ,
of the particle, st after leaving A , is given by 1 4 9v t t .
(i) Find the acceleration of the particle when it is at instantaneous rest.
(ii) Obtain an expression, in terms of t , for the displacement, from A , of the particle st
after leaving A .
Solution:
(i) The particle is at instantaneous rest when its velocity is zero.
Let 0v , 0t
0 1 4 9t t
4 9 1t t
Since 4 9 0t and 1 0t when 0t ,
2 2
4 9 1t t
24 9 2 1t t t
2 2 8 0t t
4 2 0t t
4t or 2t Rejected as 0t
Acceleration,
d 1
1 4d 2 4 9
v
t t
d 2
1d 4 9
v
t t
When 4t ,
2d 2 31 ms
d 54 4 9
v
t
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 41
(ii) Let the displacement from A , of the particle be s t ,
ds t v t
1 4 9 dt t t
2 3
21
4 92 6
ts t t t c where c is an integration constant.
When 0t , displacement is zero, 0 0s .
3
21
0 96
c
9
02
c
9
2c
2 3
21 9
4 92 6 2
ts t t t
Alternative solution:
(i) The particle is at instantaneous rest when its velocity is zero.
Let 0v , 0t
0 1 4 9t t
4 4 4 4 9 0t t
4 9 4 4 9 5 0t t
Let 4 9 0u t ,
2 4 5 0u u
1 5 0u u
5u or 1u , Rejected as 0u .
5 4 9t
25 4 9t
4t
Let 4 9u t ,
2 4 9u t , 2 9
4
ut
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 42
2d
4d
u
t
d
2 4d
uu
t
d 2
d
u
t u
1 4 9v t t
2 9
14
uv u
d
1d 2
v u
u
d d d
d d d
v v u
t u t
d 2
1d 2
v u
t u
d 2
1d
v
t u
d 2
1d 4 9
v
t t
When 4t ,
2d 2 31 ms
d 54 4 9
v
t
(ii) Let the displacement from A , of the particle be s t , when 0t , 0 0s .
0
0 d
t
s t s v t
0
1 4 9 d
t
t t t
2 3
2
0
14 9
2 6
t
tt t
2 3 3
2 21 1
4 9 92 6 6
tt t
2 3
21 9
4 92 6 2
tt t
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 43
O
2
161
5y
x
x
y
Q
P
A
OR
The diagram shows part of the curve
2
161
5y
x
, cutting the x -axis at Q . The
tangent at the point P on the curve cuts the x -axis at A . Given that the gradient of this
tangent is 4 , calculate
(i) the coordinates of P ,
(ii) the area of the shaded region PQA .
Solution:
(i)
2
161
5y
x
3
d 162 1
d 5
y
x x
3
d 32
d 5
y
x x
At P , d
4d
y
x ,
3
324
5 x
3
5 8x
5 2x
3x
When 3x , 3y . 3,3P
O
2
161
5y
x
x
y
Q
P
A T
Solutions to O Level Add Math paper 2 2007
By KL Ang, Jun 2013 Page 44
(ii)
When 0y ,
2
160 1
5 x
2
5 16x
5 4x
5 4x
1x or 9x , Rejected since 3x
1,0Q
Equation of PA ,
3 4 3y x
4 9y x
When 0y , 9
4x .
9,0
4A
Let point 3,0T ,
the area of the shaded region PQA ,
3 3
2
91
4
161 d 4 9 d
5x x x
x
3
32
9
41
162 9
5x x x
x
2
216 16 9 93 1 2 3 9 3 2 9
5 3 5 1 4 4
81
5 3 98
9
28
7
8 square units
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 45
1.
The diagram shows a triangle ABC in which 4cmAB , 2cmBC and angle
120ABC . The line CB is extended to the point X where angle 90AXB .
(i) Find the exact length of AX .
(ii) Show that angle 1 3tan
2ACB
.
Solution :
(i)
60ABX OR By similar triangle with 30-60-Right triangle
4sin 60AX 4
23
AX
3
42
2 3 cmAX
2 3 cm
[Analysis]
First, recognize that ABX is a special 30-60-Right triangle. Part (i) demands “exact” value
which implies that NO rounding of numbers.
A
C
B
X
4cm
2cm
120
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 46
(ii)
4cos60BX OR By similar triangle with 30-60-Right triangle
1
42
4
1 2
BX
2cm 2cmBX
Therefore,
tanAX
ACBCX
2 3
2 2
3
2
1 3tan
2ACB
In Summary:
A very important point for students is to be able to handle Surds at ease. Any
rounding-off of answers with calculator will result in losing answer marks.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 47
2. Solve, for x and y , the simultaneous equations
9 27 1yx ,
8 2 16 2x
y
Solution :
Given that 9 27 1yx ,
2 33 3 1x y
2 3 03 3x y
2 3 0x y ---------- (1)
Given that 8 2 16 2x
y ,
6 8
2 2 2 2y x
6 9
2 2y x
6 9y x ---------- (2)
(1)+2x(2),
15 18y
1
15
y
Substitute 1
15
y into (1), we get
9
5x .
In Summary:
Let me highlight again here that Competency in indices operations is the
foundation in doing well in this exam. Remember to check each solution.
[Analysis]
Recognize that 29 3 and 327 3 , and also 2 3 6
38 2 2 2
and
2 4 8
416 2 2 2
. Then, apply the rules of indices.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 48
Alternative solution:
Given that 9 27 1yx ,
3 3log 9 27 log 1yx
3 3 3log 9 log 27 log 1yx
3 32 log 3 3 log 3 0x y
2 3 0x y ---------- (3)
Given that 8 2 16 2x
y ,
2 2log 8 2 log 16 2x
y
2 2 2 2log 8 log 2 log 16 log 2x
y
2 2 2 2
13 log 2 log 2 4log 2 log 2
2 2
xy
13 4
2 2
xy
93
2 2
xy --------- (4)
Solving (3) and (4) simultaneously to get the solutions.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 49
3. Given that 7 8
1 6
A , find 1A and hence solve the simultaneous equations
8 7 11 0p q ,
6 7 0p q .
Solution :
Given that
7 8
1 6
A ,
the determinant, 7 6 8 1 50
1
6 8
6 81 50 50
1 7 1 750
50 50
A
Given that 8 7 11 0p q and 6 7 0p q , we get
7 8 11q p
6 7q p
So,
7 8 11
1 6 7
q
p
6 8
1150 50
1 7 7
50 50
q
p
[Analysis]
Question on Matrix is an on-and-off sub-topic. Students can employ Matrix operations to
solve any linear simultaneous equations. One will need to know determinant to start.
Remember, “hence” in the question implies that you MUST use 1A in your solution, no mark
will be awarded for otherwise.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 50
6 11 8 7 1
50 5
61 11 7 7
550
q
p
Therefore, 1
15
p , 1
5q .
Alternative solution:
Let 1Aa b
c d
,
7 8 1 0
1 6 0 1
a b
c d
7 8 6 1 0
7 8 6 0 1
a b a b
c d c d
7 1
8 6 0
7 0
8 6 1
a b
a b
c d
c d
6 8,
50 50
1 7,
50 50
a b
c d
1
6 8
50 50
1 7
50 50
A
In Summary:
As a reminder, the second part MUST be done with the inverse matrix.
[Analysis]
Consider 1A A=I
, given that A is a 2 2 matrix, 1 0
0 1I
.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 51
4.
(i) Find 3 lnd
x xdx
.
(ii) Hence find 2 ln .x xdx
Solution :
(i)
3
2 3
2 2
ln
13 ln
3 ln
dx x
dx
x x xx
x x x
(ii)
3 2 2ln 3 lnd
x x x x xdx
3 2 2ln 3 lnd
x x dx x x x dxdx
3 2 2ln 3 lnd
x x dx x xdx x dxdx
2 3 23 ln lnd
x xdx x x dx x dxdx
3 31ln
3x x x c
[Analysis]
Part (i) can be done with product rule. Part (ii) is to be completed with the result from part (i).
The method used in part (ii) is VERY important.
In Summary:
This is an excellent question on integration. Students are best to familiar
themselves with the approach to integration, as often, integration begins with
constructing a differentiation.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 52
5.
(i) Express
8 46
5 1
x
x x
in partial fractions.
(ii) Hence, or otherwise, find the gradient of the curve
8 46
5 1
xy
x x
at the point
where 2x .
Solution(1):
(i)
8 46
5 1 5 1
x A B
x x x x
-------- (1)
(1) 5x ,
8 46
51 1
x BA x
x x
Let 5x ,
8 5 46
5 1A
1A
(1) 1x ,
8 46
15 5
x Ax B
x x
Let 1x ,
8 1 46
1 5B
9B
so,
8 46 9 1
5 1 1 5
x
x x x x
.
[Analysis]
Read careful the steps taken in part (i). This is essentially the ‘cover-up’ method. Part (ii) is to
follow from part (i) or to be taken completely as new. Gradient of a curve is dy
dx
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 53
(ii)
8 46 9 1
5 1 1 5
xy
x x x x
1 1
9 1 5y x x
2 2
9 1 1 1 5dy
x xdx
2 2
9 1
1 5
dy
dx x x
At 2x ,
2 2
9 1 8
92 1 2 5
dy
dx
Solution(2):
(i)
8 5 68 46
5 1 5 1
xx
x x x x
8 5 6
5 1 5 1
x
x x x x
8 6
1 5 1x x x
By inspection,
6 1 1
5 1 5 1x x x x
8 46 8 1 1
5 1 1 5 1
x
x x x x x
Therefore,
8 46 9 1
5 1 1 5
x
x x x x
(ii)
Given that 2
8 46 8 46
5 1 4 5
x xy
x x x x
, by quotient rule
2
22
8 4 5 8 46 2 4
4 5
x x x xdy
dx x x
2
22
8 92 224
4 5
dy x x
dx x x
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 54
At 2x ,
2
22
8 2 92 2 224 8
92 4 2 5
dy
dx
In Summary:
Part (i) solutions show that there is really no need to memorise any specific
method to decompose a fraction. However, the cover-up method is the quickest to the
solution. Part (ii) again demonstrates that the “Hence” approach is the quickest to the
solution. Since the question provide alternative approaches, students should be
reminded to apply the normal classroom practice as their response.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 55
6. A cyclist starts from rest from a point A and travels in a straight line until he comes to rest
at a point B . During the motion, his velocity, 1msv , is given by 216
2v t t , where t is
the time in seconds after leaving A . Find
(i) the time taken for the cyclist to travel from A to B ,
(ii) the distance AB ,
(iii) the acceleration of the cyclist when 8t .
Solution:
(i) At A and B , the velocity is zero.
210 6
2t t
2 12 0t t
12 0t t
0t or 12t
the time taken is 12 seconds
(ii) The area under the velocity curve yields the displacement.
12 12
2
0 0
16
2vdx t t dx
123
2
0
36
tt
[Analysis]
Make a sketch of the velocity curve to visualize the question.
0v
A B
0v
0t
v
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 56
3
2 123 12 144m
6
(iii) Acceleration is the derivative of the velocity.
Given that 216
2v t t ,
acceleration, 6dv
tdt
When 8t , 26 8 2msdv
dt
In Summary:
When integrating velocity for displacement, one has to be mindful about the
distance travelled. For distance travelled, the integration is to be carried out by taking
the absolute value of the negative area. Question may also be asked for the velocity
when the acceleration is zero, or to find the maximum velocity.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 57
7. The equation of a curve is sin
2 cos
xy
x
. Find the x -coordinate, where 0
2x
, of the
point at which the tangent to the curve is parallel to the x -axis.
Solution(1):
Given sin
2 cos
xy
x
,
2
cos 2 cos sin sin
2 cos
x x x xdy
dx x
2
2cos 1
2 cos
dy x
dx x
When 0dy
dx ,
2
2cos 10
2 cos
x
x
As 2 cos 0x for all real x ,
2cos 1 0x .
1cos
2x
1 1cos
2x
3x
for 0
2x
[Analysis]
The tangent parallel to the x -axis is at zero gradient, 0dy
dx .
In Summary:
Competent in using quotient rule and solve simple trigo equation are needed
here. Question is simple.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 58
8. (i) Show that 2sin3 sin 4sin cosx x x x .
(ii) Find all the angles between 0 and which satisfy the equation
2sin3 sin 2cosx x x .
sin sin cos cos sinA B A B A B
(i) Show that 2sin3 sin 4sin cosx x x x .
Solution(1):
From L.H.S sin3 sinx x
sin 2 sin 2x x x x
2sin 2 cosx x
2 2sin cos cosx x x
24sin cosx x
=R.H.S
Solution(2):
From L.H.S sin3 sinx x
sin 2 cos cos2 sin sinx x x x x
2 22sin cos cos cos sin sin sinx x x x x x x
2 2 22sin cos sin cos sin 1x x x x x
2 2 22sin cos sin cos cosx x x x x
24sin cosx x
=R.H.S
[Analysis]
In part (i) an inspection of the identity, suggests that multiple angles formulae or factor
formulae maybe productive. eg sin sin cos cos sinA B A B A B ;
sin sin 2sin cosA B A B A B , where 2A x and B x .
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 59
Solution(3):
From R.H.S 24sin cosx x
2 2sin cos cosx x x
2 sin 2 cosx x
sin 2 sin 2x x x x
sin3 sinx x
=L.H.S
(ii) Solve, for 0 x radians, the equation
2sin3 sin 2cosx x x .
Solution(1):
Given that 2sin3 sin 2cosx x x , for 0 x ,
2sin3 sin 2cos 0x x x
2 24sin cos 2cos 0x x x
2cos 4sin 2 0x x
2cos 0x or 4sin 2 0x
2
x
1
sin2
x
5
,6 6
x
Therefore the solutions are 5
, ,6 2 6
x
In Summary:
Take note that applying the factor formulae is most time efficient (solution(1)).
There are numerous ways to answer this question, of varying length.
[Analysis]
Students can answer this part as if it is a fresh one taking no reference from the earlier part.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 60
Solution(2):
Given that 2sin3 sin 2cosx x x , for 0 x ,
2sin3 sin 2cos 0x x x
2sin 2 cos cos2 sin sin 2cos 0x x x x x x
2 2 22sin cos cos cos sin sin sin 2cos 0x x x x x x x x
2 2 22sin cos 2cos 1 sin sin 2cos 0x x x x x x
2 24sin cos 2cos 0x x x
2cos 4sin 2 0x x
2cos 0x or 4sin 2 0x
2
x
1
sin2
x
5
,6 6
x
Therefore the solutions are 5
, ,6 2 6
x
In Summary:
Trigo questions are typically multiple approaches. Take note that solution(1) is
most efficient. The knowledge of trigo ratios of special angles and triangles are needed.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 61
9. Ann is older than her sister Betty. Their ages in years are such that twice the square of
Betty’s age subtracted from the square of Ann’s age gives a number equal to 6 times the
difference of their ages. Given also that the sum of their ages is equal to 5 times the
difference of their ages, find the age in years of each of the sisters.
Solution:
Let Ann’s age be A , and Betty’s age be B , where A B .
2 22 6A B A B ---------- (1)
5A B A B
2 3A B
3
2A B ---------- (2)
Substitute (2) into (1), 2
3 04
BB
3 04
BB
0B or 12B , so Betty is 12 years old
(Rejected)
Substitute 12B into (2), 18A , so Ann is 18 years old.
[Analysis]
This is a very odd question. 1st, is to formulate equations, then 2
nd to solve them.
In Summary:
This is a very unusual question.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 62
10. (a) Find the smallest value of the integer a for which 2 5 2ax x is positive for all values
of x .
(b) Find the smallest value of the integer b for which 25 2x bx is negative for all
values of x .
Solution(1):
(a)
The curve 2 5 2y ax x needs to be a “smiley”. So 0a .
As this curve should not touch the x -axis, there should not be any real root from the
equation 2 5 2 0ax x . Therefore, the discriminant is negative.
The discriminant, 25 4 2 0a
25
8a
The smallest value of the integer 4a .
(b)
The curve 25 2y x bx is a “frown”.
As this curve should not touch the x -axis, there should not be any real root from the
equation 25 2 0x bx . Therefore, the discriminant is negative.
The discriminant, 2 4 5 2 0b
2 40b
40 40b
The smallest value of the integer 6b .
Solution(2):
(a)
The curve 2 5 2y ax x needs to be a “smiley”. So 0a . The lowest point of this
curve should not touch the x -axis, therefore the minimum value of 2 5 2ax x is to be positive.
By completing the square,
[Analysis]
Consider the curve 2 5 2y ax x , this curve needs to “hang” above x -axis.
Consider the curve 25 2y x bx , this curve needs to “hang” below x -axis.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 63
2 2
2 5 5 25 2
2 2ax x a x
a a a
2
2
5 8 25
2 4
aa x
a a
25 8 25
2 4
aa x
a a
Therefore, the minimum value 8 25
4
a
a
is positive.
8 25
04
a
a
Given that 0a ,
8 25 0a
25
8a
The smallest value of the integer 4a .
(b)
The curve 25 2y x bx is a “frown”. The highest point of the curve should not touch
the x -axis, therefore the maximum value of 25 2x bx is to be negative.
The line of symmetry of the curve,
2 5 10
b bx
The maximum value of the curve lies on the line of symmetry.
Let 10
bx ,
2
5 210 10
b bb
2 2
220 10
b b
2
220
b
The maximum value 2
2 020
b ,
2 40b
40 40b
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 64
The smallest value of the integer 6b .
Solution(3):
(a)
The turning point of the curve 2 5 2y ax x , is 2 5dy
axdx
.
When 0dy
dx ,
5
2x
a , this is the line of symmetry.
The rest of the solution follows the method in solution(2)(b).
(b)
The turning point of the curve 25 2y x bx , is 10dy
x bdx
.
When 0dy
dx ,
10
bx , this is the line of symmetry.
The rest of the solution follows that of solution(2)(b).
In Summary:
Solution (1) is the shortest. The other solutions show how other methods can also
be employed to tackle the question. Visualising the question in quadratic curves is
important.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 65
11. (i) In the binomial expansion of
7k
xx
, where k is a positive constant, the coefficients
of 3x and x are the same. Find the value of k .
(ii) Using the value of k found in part (i), find the coefficient of 7x in the expansion of
7
21 5k
x xx
.
Solution:
7 7
7
21
k kx x
x x
The r th term of this expansion, 7 7
2
r
r r
kT x C
x
, where 0,1, ,7r
7 7 2r r
r rT C k x
For 3x term, 7 2 3r
2r
The coefficient of this 3x term,
7 2
2
2
2
7!
2!5!
21
C k
k
k
For x term, 7 2 1r
3r
The coefficient of this x term,
[Analysis]
Part (i) has a little complication with this binomial expansion of
7k
xx
, which can be
rewritten as
7
7
21
kx
x
.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 66
7 3
3
3
3
7!
3!4!
35
C k
k
k
Therefore,
2 321 35k k
Since 0k , 3 5k
3
5k
(ii)
7
21 5k
x xx
7 7
25k k
x x xx x
The expansion of the first term above, is as given in part (i), 7 7 2r r
r rT C k x
The expansion of the second term above, 7 9 25 r r
r rT C k x
For 7x term,
from the first term, 7 2 7r
0r
from the second term, 9 2 7r
1r
coefficient from 1st term, 7 0
0 1C k
coefficient from 2nd
term, 7 1
15 21C k
The coefficient of 7x term is, 1 21 20
In Summary:
Students should be well familiar with using the general term expression to
location the coefficient of the required expansion.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 67
12.
The variables x and y are connected by the equation xy kb , where k and b are constants.
Experimental values of x and y were obtained. The diagram above shows the straight line graph,
passing through the points 0,1.3 and 11,0.8 , obtained by plotting lg y against x . Estimate
(i) the value, to 2 significant figures, of k and of b .
(ii) the value of y when 8x .
Solution:
(i)
Given that 0,1.3 and 11,0.8 ,
Gradient, 1.3 0.8 1
lg0 11 22
b
1
2210b
0.90b
When 0x , lg 1.3y .
lg 1.3k 1.310k
20k
O x
lg y
0,1.3
11,0.8
[Analysis]
The curve xy kb is to be linearised as lg lg lgy k x b , in which lgb and lg k are gradient
and lg y -axis intercept respectively.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 68
(ii)
The equation of the line is, lg 1.322
xy
When 8x , 8
lg 1.322
y
lg 0.936y
0.93610y
8.63y
Alternative solution:
(i)
Let the equation of the line be Y mX C , where lg y Y , lg k C , x X
Given that 0,1.3 and 11,0.8 ,
1.3 C
0.8 11 1.3m
1
22m
Therefore,
lg 1.3k 1.310k
20k
and,
1
lg22
b
1
2210b
0.90b
(ii)
20 0.9x
y
When 8x , 8
20 0.9 8.61y
In Summary:
Be careful that it is lg not ln in this question. Students do have to expect this
type of question to appear regularly in this exam.
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 69
13.
The diagram shows a glass window, PQRST , consisting of a rectangle PQST of height h cm and
width 6x cm and an isosceles triangle QRS in which 5QR RS x cm. The perimeter of the
window is 360cm.
(i) Show that the area of the window, 2cmA , is given by 21080 36A x x .
Given that x can vary,
(ii) find the stationary value of A ,
(iii) determine whether this stationary value is a maximum or a minimum.
Solution:
(i)
Given that the perimeter is 360cm,
360 6 2 10x h x
180 8h x --------- (1)
The area,
1
6 4 62
A h x x x
26 12A xh x --------- (2)
[Analysis]
This is a typical min/max problem with a fixed perimeter to find largest area.
6 cmx
Q
P T
S
R
cmh cmh
5 cmx 5 cmx
Solutions to O Level Add Math paper 1 2008
By KL Ang, Jun 2013 Page 70
Substitute (1) into (2),
26 180 8 12A x x x
21080 36A x x
(ii)
1080 72dA
xdx
Let 0dA
dx , 0 1080 72x
15x
When 15x , the stationary value of A ,
2 21080 15 36 15 8100cmA
(iii) 2
272
d A
dx
The stationary point is a maximum point.
Alternative solution for (ii) and (iii)
21080 36A x x
236 30A x x
2 236 15 36 15A x
2
36 15 8100A x
At 15x , the value of the stationary value is 28100cm
2
36 15 8100 8100x
The stationary point is a maximum point.
In Summary:
The alternative solution is a non-calculus one.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 71
1. A man buys a new motorcycle. After t months its value $V is given by 10000 ptV e ,
where p is a constant.
(i) Find the value of the motorcycle when the man bought it.
The value of the motorcycle after 12 months is expected to be $4000. Calculate
(ii) the expected value of the motorcycle after 18 months,
(iii) the age of the motorcycle, to the nearest month, when its expected value will be
$1000.
Solution :
(i) When 0t , 0
10000p
V e
10000V
The value of the motorcycle when the man bought it, is $1000.
(ii) When 12t , 4000V 12
4000 10000p
e
2
12 ln5
p
1
ln 2 ln5 0.0763612
p
When 18t , 18
10000p
V e
1 2
18 ln12 510000V e
3 2ln
2 510000V e
3 2ln
2 510000V e
2529.82V
the expected value of the motorcycle after 18 months is, $2529.82.
(iii) When 1000V ,
1000 10000 pte
[Analysis]
In this question, the value of the motorcycle is modeled by the formula, 10000 ptV e . The
rest of the question is straight forward which involves exponential and logarithm.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 72
1 2ln
12 51
10
t
e
1 2
ln ln10 12 5
t
2
12ln10 ln5
t
30t
the age of the motorcycle, to the nearest month is 30 months.
In Summary:
Please make sure that you are comfortable with operations with indices and
logarithms. Depending on the features available with your electronic calculators, the
actual steps taken can vary. Answer for part (ii) may vary.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 73
2. The roots of the quadratic equation 22 4 3 0x x are and . Find the quadratic
equation whose roots are 2 2 and 2 2 .
Solution :
Given that 22 4 3 0x x ,
Sum of roots, 4
22
Product of roots,
3
2
For a quadratic equation whose roots are 2 2 and 2 2 ,
Sum of roots, 22 2 2 22 2 4 2 4 5
Product of roots,
2 2 22 2 2 2 33
2 2 2 4 2 2 44
Therefore, 2 335 0
4x x ,
or 24 20 33 0x x
[Analysis]
A direct application of Sum of Roots(SOR), and Product of Roots(POR).
Recall that 2 2 2 2 .
In Summary:
This type of questions used to be popular a few decades ago, is making a come
back. Do expect this trend to continue. In general, 24 20 33 0k x x where 0k
will be the required solution.
[Analysis]
A direct application of Sum of Roots(SOR), and Product of Roots(POR).
Recall that 2 2 2 2 .
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 74
3.
(i) Prove the identity tan cot 2cosec2A A A .
(ii) Find all the angles between 0 and 360 which satisfy the equation tan cot 3A A .
Solution :
(i) Given that tan cot 2cosec2A A A ,
L.H.S.= tan cotA A ,
sin cos
cos sin
A A
A A
2 2sin cos
sin cos
A A
A A
2 1
2sin cosA A
2
sin 2A
=cosec2A
=R.H.S.
(ii) Given that tan cot 3A A for 0 360A ,
2cosec2 3A
3
cosec22
A
1 3
sin 2 2A
2
sin 23
A
[Analysis]
Part (i) needs to apply 1 cos
cottan sin
AA
A A . And the left hand side is a double angle, 2A
that provide a hint of possible use of double angle formulae. Take note that
1cosec2
sin 2A
A . Part (ii) has to provide solution for 0 360A which is also for
0 2 720A .
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 75
1 22 sin
3A
for 0 2 720A
Principal angle, 2 41.81A
2 41.81 , 138.19 ,401.81 ,498.19A
20.9 , 69.1 ,200.9 ,249.1A
In Summary:
Students need to recall that 1 cos
cottan sin
AA
A A ;
1cosec
sinA
A ;
1sec
cosA
A ;
2 2sin cos 1A A ; sin 2 2sin cosA A A . The principal angle is the value returned
from your calculator.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 76
4. Solve the equation
(i) 3 32 log 3 7 log 2 3x x ,
(ii) 53log log 5 2yy .
Solution :
(i)
Given that 3 32 log 3 7 log 2 3x x ,
3 3 32log 3 log 3 7 log 2 3x x
2
3 3 3log 3 log 3 7 log 2 3x x
3 3log 9 3 7 log 2 3x x
9 3 7 2 3x x
25 60 0x
12 2
25 5
x
(ii)
Given that 53log log 5 2yy ,
5
5
13log 2
logy
y
Let 5logu y ,
1
3 2uu
23 2 1 0u u
3 1 1 0u u
1
3u or 1u
5
1log
3y 5log 1y
[Analysis]
In part (i), 1
3 7 0 23
x x and 1
2 3 0 12
x x . Together, 1
23
x .
In part (ii), 0y and 1y . Apply base change formula.
Must check that the solutions are admissible with the given equation.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 77
1
35y
5y
1
lg lg53
y
lg 0.23299y
0.2329910y
0.585y
Alternative solution:
(iii)
Given that 53log log 5 2yy ,
lg lg5
3 2lg5 lg
y
y
Let lg
lg5
yu ,
1
3 2uu
23 2 1 0u u
3 1 1 0u u
1
3u or 1u
lg 1
lg5 3
y
lg1
lg5
y
1
lg lg53
y lg lg5y
1
35y
5y
0.585y
In Summary:
Be mindful of the restrictions imposed by logarithm.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 78
5. The term containing the highest power of x in the polynomial f x is 42x . Two of the
roots of the equation f 0x are 1 and 2 . Given that 2 3 1x x is a quadratic factor
of f x , find
(i) an expression for f x in descending powers of x ,
(ii) the number of real roots of the equation f 0x , justifying your answer,
(iii) the remainder when f x is divided by 2 1x .
Solution:
(i)
2f 2 3 1 1 2x x x x x
4 3 2f 2 8 4 10 4x x x x x
(ii) Given the 2 3 1x x ,
the discriminant 2
3 4 1 1 5 0 , this quadratic factor contains 2 real roots.
The number of real roots is 4.
(iii) when f x is divided by 2 1x ,
21 1 1 1 1
f 2 3 1 1 22 2 2 2 2
1 9 1
f 12 8 8
[Analysis]
Need to construct f x . The equation, f 0x has two roots, 1 and 2 , so 1x and 2x
are factors of f x . Together with 2 3 1x x , that doesn’t contain the above two roots, then
a possible polynomial is 2 3 1 1 2x x x x . But the highest degree term is 42x , so a
factor 2 is needed to complete f x .
In Summary:
The difficulty in this question is in the construction of the correct polynomial.
Part (ii) needs not find the actual roots. Part(iii) needs to apply remainder theorem.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 79
6.
The diagram shows a circle, centre O , with diameter AB . The pointC lies on the circle.
The tangent to the circle at A meets BC extended at D . The tangent to the circle at C
meets the line AD at E .
(i) Prove that triangles AEO and CEO are congruent.
(ii) Prove that E is the mid-point of AD .
Solution:
(i) Consider AEO and CEO
90OAE OCE (tangent radius)
OA OC (radii of the circle)
OE OE (common hypotenuse)
By RHS rule, AEO is congruent to CEO .
Therefore,
AOE COE
Hence, line OE is the angle bisector of AOC .
[Analysis]
In general, be ready to apply geometrical properties of circle; congruent triangles; as well as
similar triangles. Part (i) is about applying ASA, or SAS, or SSS, or RHS rules for
congruency. Part (ii) may apply mid-point or intercept theorem.
B
E
A
C
O
D
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 80
(ii)
2AOC AOE
2AOC ABC ( s at centre of circle = 2 s at circumference)
2 2AOE ABC
AOE ABC
Consider OAE and BAD ,
90OAE BAD (common angle)
AOE ABC (corresponding angle)
OAE and BAD are similar.
AO AE
AB AD
2
radius AE
radius AD
1
2
AE
AD
Therefore, E is the mid-point of AD .
Alternative solution for (ii):
90ACB (Right s in semi-circle)
90ACB ACD (Supplementary )
AEO CEO ,
AE CE
AEC is an isosceles triangle.
EAC ECA
90EAC ADC
90ECA DCE ACD
ADC DCE
EDC is an isosceles triangle.
EC ED
AE CE ,
AE ED , hence E is the mid-point of AD .
In Summary:
Before writing any proof, one shall think through the entire proving steps. This
thought process usually required looking at the result, and then work backward until it
meets the forward process, starting from the given conditions.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 81
7. The function f is defined by f 4cos2 2x x .
(i) State the amplitude of f .
(ii) State the period of f .
The equation of a curve is 4cos2 2y x for 0 180x .
(iii) Find the coordinates of the minimum point of the curve.
(iv) Find the coordinates of the points where the curve meets the x -axis.
(v) Sketch the graph of 4cos2 2y x for 0 180x .
(vi) Sketch the graph of 4cos2 2y x for 0 180x .
Solution:
(i)
The amplitude of f is 4.
(ii)
The period of f is 360
1802
.
(iii)
4 4cos2 4x
6 4cos2 2 2x
The minimum value of y is 6 when cos2 1x
cos2 1x for 0 180x
12 cos 1x for 0 2 360x
Principal angle, 2 180x
90x
The coordinates of the minimum point of the curve is 90 , 6
[Analysis]
Part (i) and (ii) are straight forward about periodic trigo curve. The rest of the question, (iii),
(iv) and (v) will be better to start with part (v). Remember to mark clearly the intercepts and
turning points.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 82
(iv)
The x -axis is at 0y .
0 4cos2 2x for 0 180x
0 2cos2 1x
1
cos 22
x
1 12 cos
2x
for 0 2 360x
Principal angle, 2 60x
2 60 ,300x
30 ,150x
The coordinates of the points where the curve meets the x -axis are 30 ,0 , 150 ,0
(v)
The graph of 4cos2 2y x for 0 180x .
x 0 45 90 135 180
y 2 2 6 2 2
0 x
y 3
2
90
4
30 60
2
6
120 150 180
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 83
(vi)
The graph of 4cos2 2y x for 0 180x .
x 0 45 90 135 180
y 2 2 6 2 2
In Summary:
It is best to have made a simple sketch of the equation first to get a “visual”
picture of the problem. Then transfers the result of part (iii) and (iv) onto this sketch.
Finally, construct a ‘proper’ sketch with the 5-point values table.
x 0
y
90 30 60
2
120 150 180
6
4
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 84
0
3y x ax b
2,0 x
y 8.
The diagram shows part of the curve 3y x ax b , where a and b are positive constants.
The curve has a minimum point at 2,0 . Find
(i) the value of a and of b ,
(ii) the coordinates of the maximum point of the curve,
(iii) the area of the shaded region.
Solution:
(i) At 2,0 ,
3
0 2 2a b
2 8a b ---------- (1)
23dy
x adx
2
0 3 2 a
12a
Substitute 12a into (1), we get 16b .
[Analysis]
The curve passes through point 2,0 and its gradient is zero at this point too. With two
unknown a and b , these will be enough to solve part (i). The turning points occur at zero
gradient, 0dy
dx . Of course, we already know the minimum point. The area is simply the
integration of the curve from 0 to 2 .
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 85
(ii) The equation of the curve,
3 12 16y x x
23 12dy
xdx
20 3 4x
0 3 2 2x x
2x or 2x
At 2x , 3
2 12 2 16 32y
The coordinates of the maximum point of the curve is at 2,32
(iii)
The area of the shaded region 2
3
0
12 16x x dx
24
2
0
6 164
xx x
4
226 2 16 2 0
4
12 square units
In Summary:
You should read through the entire question, make a plan on how to approach
them before beginning to answer them.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 86
9.
The diagram shows a straight road OP . A runner leaves the road at O and runs 4km in a
straight line to a point A . She then turns through 90 and runs 2km in a straight line to a
point B . The angle POA is , where 0 90 , and the perpendicular distance of B
from the road OP is kmL .
(i) Show that 4sin 2cosL .
(ii) Express L in the form sinR , where 0R and 0 90 .
(iii) Find the value of for which 3L .
Solution:
(i) From the graph,
4sin 2cosL
4sin 2cosL
[Analysis]
For part (i), the challenge is to locate the value of 2cos . Study the diagram to find angle
around the length of 2 km. Part (ii) is about R-formulae. Part (iii) will need to solve a trigo
equation.
2km
P
A
B
O
4km
kmL
2km
P
A
B
O
4km
kmL
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 87
(ii)
sin sin cos cos sinR R
4sin 2cos sin cos cos sinL R R
cos 4R --------- (1)
sin 2R --------- (2)
2 2
1 2 , 2 2 2 2
sin cos 2 4R R
2 2 2sin cos 20R
2 20R
20 2 5R
2
1’
sin 2 1tan
cos 4 2
1 1tan 26.6
2
1 12 5 sin tan
2L
(iii)
When 3L ,
1 13 2 5 sin tan
2
1 1 3sin tan
2 2 5
1 11 3tan sin
2 2 5
Principal angle, 1 1tan 42.1
2
1 1tan 42.1 ,137.9
2
42.1 26.6 or 137.9 26.6
68.7 164.5 (inadmissible as 90 )
The value of for which 3L is 68.7 .
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 88
Alternative solution:
(ii)
2 24 2 20 2 5R
4cos
2 5
2sin
2 5
2 1
tan4 2
1 1tan 26.6
2
2 5 sin 26.6L
(iii)
When 3L ,
3 2 5 sin 26.6
3
sin 26.62 5
Let be the basic angle,
3sin
2 5
42.1
26.6 42.1 ,137.9
42.1 26.6 or 137.9 26.6
68.7 164.5 (inadmissible as 90 )
The value of for which 3L is 68.7 .
In Summary:
The method employed in the part (ii) of the alternative solution is more time
efficient. The part (iii) of the alternative solution uses the basic angle approach.
4
2 2 5
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 89
10. A curve is such that
2
6
2 1
dy
dx x
and 2,9P is a point on the curve. The normal to the
curve at P meets the y -axis at Q and the x -axis R .
(i) Find the coordinates of the mid-point of QR .
(ii) Find the equation of the curve.
A point ,x y moves along the curve in such a way that the x -coordinate increases at a
constant rate of 0.03 units per second.
(iii) Find the rate of change of the y -coordinate as the point passes through P .
Solution:
(i)
Let the equation of the normal be y mx q , the gradient of the curve at 2,9P ,
2
6 2
32 2 1
dy
dx
1 3
2m
dy
dx
3
9 22
q
12q
equation of the normal,
3
122
y x
When 0y , 8x , the coordinates of R is 8,0 ; the coordinates of Q is 0,12 .
The coordinates of the mid-point of QR is 8 12
, 4,62 2
.
[Analysis]
Gradient of the curve is dy
dx . The equation of the curve will be
dyy dx
dx
with a given
point. Gradient of normal, at a point is 1
gradient at the point
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 90
(ii)
The equation of the curve will be dy
y dxdx
2
6 3
2 12 1y dx c
xx
, where c is a constant
The curve passes through 2,9P ,
3
92 2 1
c
8c
The equation of the curve is 3
82 1
yx
.
(iii)
At 2,9P , 2
3
dy
dx , 0.03
dx
dt ,
By chain rule, dy dy dx
dt dx dt
2 30.02
3 100
dy
dt units per second
In Summary:
This is a very traditional question on finding the equation of a curve, given its
derivatives.
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 91
11.
The diagram shows two circles 1C and 2C . Circle 1C has its centre at the origin O .
Circle 2C passes through O and has its centre at Q . The point 8, 6P lies on both
circles and OP is a diameter of 2C .
(i) Find the equation of circle 1C .
(ii) Find the equation of circle 2C .
The line through Q perpendicular to OP meets the circle 1C at the points A and B .
(iii) Show that the x -coordinates of A and B are 3a b and 3a b respectively,
where a and b are integers to be found.
Solution:
(i) For 1C ,
228 6 10r OP
The equation of circle 1C ,
2 2 100x y --------- (1)
[Analysis]
The equation of a circle, centre at ,a b is 2 2 2x a y b r , where r is the radius of the
circle. Part (iii) is about solving simultaneous equations, a circle and a line.
8, 6P
x
y
A
B
O
1C
Q
2C
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 92
(ii)
The centre of circle 2C is the mid-point of OP , Q
8 6
, 4, 32 2
The radius of circle 2C , is half of OP , is 5
2 2
4 3 25x y
(iii)
Gradient of OP6 3
8 4
Gradient of AB1 4
3 3
4
The equation of line AB ,
4
3 43
y x
4 25
3 3y x --------- (2)
substitute (2) into (1), 2
2 4 25100
3 3x x
225 200 275 0x x 2 8 11 0x x
By quadratic formula,
28 8 4 1 11
2 1x
4 27x
4 3 3x
Therefore, 4a and 3b .
Solutions to O Level Add Math paper 2 2008
By KL Ang, Jun 2013 Page 93
Alternative solution:
10OA OB
5OQ
QA QB (perpendicular radius bisects chord)
By Pythagoras Theorem,
5 3QA
Gradient of OP6 3
8 4
Gradient of AB1 4
3 3
4
By similar triangles with the gradient of AB and AQx A ,
3 3AQx
Therefore, 4 3 3, 3 4 3A , 4 3 3, 3 4 3B
4a and 3b
8, 6P
x
y
A
B
O
1C
4, 3Q
2C
5 3
4
3
A
B
4, 3 Q Ax
5
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 94
1. The expression 3 22 3x ax bx , where a and b are constants, has a factor of 1x and
leaves a remainder of 15 when divided by 2x . Find the value of a and b .
Solution (1):
Let 3 22 3f x x ax bx
By Factor Theorem, when 1 0x , 1 0f
3 2
1 2 1 1 1 3f a b
0 5a b
5a b ------------- (1)
By Remainder Theorem, when 2 0x , 2 15f
3 2
2 2 2 2 2 3f a b
15 4 2 13a b
4 2 28a b
2 14a b ----------- (2)
(1) + (2), we get
5a b ------------- (1)
(+) 2 14a b ------------- (2)
3 9
3
a
a
Substitute 3a into (1) or (2), we get 8b
[Analysis]
First, recognize that the expression 3 22 3x ax bx is a polynomial, and there are factor
1x and remainder 15 when this polynomial is divided by 2x . So naturally, we can
connect these to the Factor and Remainder Theorems of polynomial.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 95
Solution (2):
When 3 22 3x ax bx is divided by 1x , the remainder is 0.
Therefore,
5 0a b
5a b --------------- (1)
Similarly, when 3 22 3x ax bx is divided by 2x , the remainder is 15 .
Therefore, from the long division below, we get
4 2 13 15a b
2 14a b -------------- (2)
So, (1) + (2), 3a , substitute 3a into (1) or (2), we get 8b
3 2
3 2
2
2
2 3
2 2
2 3
2 2
2 3
2 2
5
x ax bx
x x
a x bx
a x a x
a b x
a b x a b
a b
1x
22 2 2x a x a b
[Analysis]
Since the expression 3 22 3x ax bx is a polynomial, and when divided by 1x and 2x
leave a remainder 0 and 15 respectively. Then we may proceed with long division of this
polynomial. This long division can alternatively be carried out by synthetic division.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 96
Solution (3):
Let 3 22 3 1 2x ax bx x x mx n kx p
By the inspection of the highest degree term, we get 2m .
when 1x , 1 0k p
k p --------- (3)
When 2x , 2 0k p
2 15k p -------- (4)
Substitute (3) into (4), we get 5k , 5p .
3 22 3 1 2 2 5 5x ax bx x x x n x
3 22 5 2 1 2 2x ax b x x x x n
3 2
3 2
2
2
2 3
2 4
4 3
4 2 4
2 8 3
2 8 2 2 8
4 2 13
x ax bx
x x
a x bx
a x a x
a b x
a b x a b
a b
2x
22 4 2 8x a x a b
[Analysis]
Since the expression 3 22 3x ax bx is a polynomial, and when divided by 1x and 2x
leave a remainder 0 and 15 respectively. We may propose rewriting this polynomial as
1 2x x mx n kx p . When 1x , 1 0k p . When 2x , 2 15k p .
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 97
By the inspection of the constant term, we get 1n .
3 22 3 1 2 2 1 5 5x ax bx x x x x
2
2 2
3 2 2
3 2
2 2 1 5 5
2 2 2 5 5
2 2 4 4 3
2 3 8 3
x x x x
x x x x x x
x x x x x
x x x
By equating the coefficients, we get 3a , 8b .
In Summary:
Solution (1) is the most direct and efficient way to solve this question. Factor
and Remainder Theorems are frequently tested. Selecting a time efficient method is
key factor in examination performance.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 98
2. Given that ln x
yx
for 0x , find the set of values of x for which y is an increasing
function of x .
Solution :
Given that ln x
yx
for 0x , we can obtain the gradient by differentiation with quotient rule.
2
11 lnx x
dy x
dx x
2
11 ln
dyx
dx x
For the function to be increasing, the gradient 0dy
dx .
2
11 ln 0x
x
Given that 0x , 2 0x , then 2
10
x for 0x .
Therefore , 1 ln 0x
ln 1x
1x e
So the range of values of x is 0 x e .
[Analysis]
This question is most difficult as student may not understand what an increasing function is.
Let’s informally say that an increasing function is one that the value of the dependent
variable(y or f(x)) increases as the value of x increases. On a graph, this can be seen as an
upward sloping curve. If a curve is continuously upward sloping, then the gradient must be
non-negative , meaning, positive or zero.
In Summary:
This question is particularly challenging as the demand on understanding is
beyond the routine ones.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 99
3. Without using a calculator, find the values of the integers a and b for which the solution
of the equation
24 3 6x x
is 7
a b.
Solution :
24 3 6x x
8 3 3 3 2x x
8 2x x
22
2
8 1 2
8 1 8 1 2 8 1
8 1 16 2
8 1 4 2
x
x
x
x
or
7 4 2x
Therefore, 4a , 2b .
[Analysis]
Observe that there is a common factor 3 in the equation, so we shall simplify it. Next, the
answer is to be expressed as 7
a bx
. This is the same as 7x a b . Collect the
common x terms together.
2
2
2
8 1 2
2
8 1
2 8 1
8 1 8 1
16 2
8 1
4 2
7
4 2
7
x
x
x
x
x
x
In Summary:
This question tests on simplifying surds, surd conjugates and rationalizing
surds. This topic is a regular visitor in O level.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 100
4. Solve the equation
(i) lg 14 lg 2 2lg5x x ,
(ii) 2 4log log 6y y .
Solution :
(i) lg 14 lg 2 2lg5x x OR (i) lg 14 lg 2 2lg5x x
214lg lg5
2
x
x
2lg 14 lg 2 lg5x x
14
252
x
x
lg 14 lg 25 2x x
14 25 2x x
64 24x
8 3x
8
3x
2
23
x
[Analysis]
Part (i) can be done with product rule or quotient rules of logarithm. A very straight forward
question. For lg 14x , 14 0x , so 14x . Similarly, for lg 2x , 2 0x , so
2x . Therefore the equation can only admit the values of 2x .
In Summary:
As in solving any equation, students are advised to check for correctness and
admissibility of the answers to the given equation. A logarithmic function of the type
loga x requires that 0x and, 0a but also 1a . This topic is regularly tested in O
level.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 101
Solution(1) :
(ii) 2 4log log 6y y OR (ii) 2 4log log 6y y
22
2
loglog 6
log 4
yy 4
4
4
loglog 6
log 2
yy
22
2
loglog 6
2log 2
yy Since
1
24 4 4
1 1log 2 log 4 log 4
2 2 ,
Since 2log 2 1 , 44
loglog 6
1
2
yy
22
loglog 6
2
yy 4 42log log 6y y
2
3log 6
2y 43log 6y
2log 4y 4log 2y
42y 24y
16y 16y
[Analysis]
Part (ii) can be done with base change formula of logarithm. For lg y , 0y . Therefore the
equation can only admit the values of 0y .
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 102
Solution(2) :
Let 2log y n , then 2ny and 4log y m , then 24 2m my . We then get
22 2n m
2n m ---------- (1)
Given that 2 4log log 6y y , we get
6n m ------- (2)
Substitute (1) into (2), we get
3 6m
So, 2m
Hence, 2 22 16y
[Analysis]
Part (ii) can be done by exchanging the logarithm with index number where loga x n can be
written as nx a .
In Summary:
As in solving any equation, students are advised to check for correctness and
admissibility of the answers to the given equation. Remember that logarithm and
indices are the two sides of a coin. Solution(2) is a lot quicker.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 103
5. The normal to the curve 1 3tany x , at the point where the curve crosses the y-axis,
passes through the point ,3k . Find the value of k .
Solution:
Given that 1 3tany x ,
23secdy
xdx
2
3
cos
dy
dx x
At 0x , 1y ,
2
3
cos 0
dy
dx
For cos 0 1 ,
3dy
dx
The gradient of the normal will be
1 1
3 3
.
The normal also passes through the point 0,1 , so the equation of the normal line is
1
1 03
y x
13
xy
The normal passes through the point ,3k , so
3 13
k
6k
[Analysis]
It is advisable to make a sketch of the curve 1 3tany x to better visualize the problem.
“the curve crosses the y-axis” at 0x , 1y , where tan 0 0 . Here x is in radian. The
normal is perpendicular to the tangent at point 0,1 where the gradient at this point is the
gradient of the tangent.
0
1
3tany x
1 3tany x
In Summary:
In this case, the correct sketch of the curve serves as a guide to check our work.
The equation of the normal matches that in the sketch.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 104
6. A curve has the equation 22 6y x x c , where c is a constant.
(i) In the case where 20c , find the set of values of x for which 0y .
(ii) Find the value of c for which the line 2 8y x is a tangent to the curve.
Solution:
(i) In the case where 20c , find the set of values of x for which 0y .
Given that 22 6 20y x x
22 6 20 0x x
2 3 10 0x x
5 2 0x x OR
2 2
2 3 33 10 0
2 2x x
With the aid of the quadratic curve,
23 9 40
02 4 4
x
23 49
2 4x
49 3 49
4 2 4x
7 3 7
2 2 2x
7 3 3 3 7 3
2 2 2 2 2 2x
2 5x
the solution is 2 5x
(i)
[Analysis]
Part (i) is about solving quadratic inequality.
In Summary:
It is usually quicker to solve the inequality with graphical aided approach. The
completing the square method remain useful which must be master by all students. For 2x a , we then have x a or x a .
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 105
(ii) Find the value of c for which the line 2 8y x is a tangent to the curve.
Solution(1):
Given that 22 6y x x c , Given that 2 8y x ,
4 6dy
xdx
2 8y x
the gradient of the line is 2 .
Therefore, at the point of tangent,
2 4 6x
1x
When 1x , 2 1 8 6y
The point of tangent must also be on the curve.
2
6 2 1 6 1 c
10c
Solution(2):
Given that 22 6y x x c and 2 8y x ,
28 2 2 6x x x c
22 4 8 0x x c
The discriminant of this quadratic equation is to be zero.
2
4 4 2 8 0c
10c
[Analysis]
Part (ii) requires the point of tangent on the curve must have the same gradient as the line. the
coordinate of the point of tangent is to be on the line as well as on the curve.
[Analysis]
Part (ii) requires that the curve and the line touches at one point. This is equivalent to a
repeated root case with discriminant being zero.
In Summary:
By comparison, the nature of root method is much shorter than that of the
coordinate geometry approach.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 106
7. Find the coordinates of the mid-point of the straight line joining the points of intersection
of the curve 2 22 5 68x y x and the line 2 3 9y x .
Solution(1):
Given 2 3 9y x ,
2 9 3y x ------------ (1)
Given 2 22 5 68x y x ,
222 2 10 136x y x ----------- (2)
Substitute (1) into (2),
222 9 3 10 136x x x
211 44 55 0x x 2 4 5 0x x
5 1 0x x
1x or 5x
When 1x , 9 3 1
62
y
.
When 5x , 9 3 5
32
y
.
The x coordinate of midpoint 1 5
22
The y coordinate of midpoint 3 6 3 1
12 2 2
Solution(2):
Given 2 3 9y x ,
2 9 3y x ------------ (1)
[Analysis]
Similar to Question 6, we can find the points of intersection by solving a linear and an non-
linear equation. Then determines the coordinates of the intersections, before finding the mid-
point of the two coordinates.
[Analysis]
We can make use of Sum of Roots to give an alternative solution.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 107
Given 2 22 5 68x y x ,
222 2 10 136x y x ----------- (2)
Substitute (1) into (2),
222 9 3 10 136x x x
211 44 55 0x x 2 4 5 0x x
Let the two roots be 1x and 2x , then by sum of roots, we get
1 2 4x x
The x coordinate of midpoint 1 2 42
2 2
x x
The y coordinate of midpoint 1 2
2
y y
1 2
1 2
9 3 9 3
4
18 3
4
18 3 4
4
11
2
x x
x x
In Summary:
Notice that, it is easier to replace 2y into the curve rather than that of y . This
little trick makes the equation much easier to simplify.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 108
8. (i) Show that 2cos3 cos 4sin cosx x x x .
(ii) Hence, or otherwise, solve, for 0 x radians, the equation
cos3 2cos 0x x .
(i) Show that 2cos3 cos 4sin cosx x x x .
Solution(1):
From L.H.S cos3 cosx x
cos2 cos sin 2 sin cosx x x x x
2 2 2cos sin cos 2cos sin cosx x x x x x
2 2 2cos 1 cos cos 2cos sin cosx x x x x x
2 22cos 1 cos 2cos sin cosx x x x x
2 2cos 2cos 1 2sin 1x x x
2 22cos cos 1 sinx x x
22cos 2sinx x
24sin cosx x
=R.H.S
Solution(2):
From L.H.S cos3 cosx x
cos 2 cos 2x x x x
2sin 2 sinx x
2 2cos sin sinx x x
24sin cosx x
=R.H.S
[Analysis]
In part (i) an inspection of the identity, suggests that multiple angles formulae or factor
formulae maybe productive. eg cos cos cos sin sinA B A B A B ;
cos cos 2sin sinA B A B A B .
[Analysis]
In part (i) an inspection of the identity, suggests that multiple angles formulae or factor
formulae maybe productive. eg cos cos cos sin sinA B A B A B ;
cos cos 2sin sinA B A B A B .
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 109
Solution(3):
From R.H.S 24sin cosx x
22 2sin cosx x
2 cos2 1 cosx x
2cos2 cos 2cosx x x
cos2 cos cos2 cos cos cosx x x x x x
cos2 cos 1 cos2 cos cosx x x x x
2cos2 cos 2sin cos cosx x x x x
cos2 cos 2sin cos sin cosx x x x x x
cos2 cos sin 2 sin cosx x x x x
cos3 cosx x
=L.H.S
(ii) Hence, or otherwise, solve, for 0 x radians, the equation
cos3 2cos 0x x .
Solution(1):
Given that cos3 2cos 0x x , for 0 x ,
cos3 cos 3cos 0x x x
24sin cos 3cos 0x x x
2cos 4sin 3 0x x
cos 0x or 24sin 3 0x
In Summary:
Take note that applying the factor formulae is most time efficient (solution(2)).
There are numerous ways to answer this question, of varying length.
[Analysis]
In this part, “Hence” is referring to using the result shown in part (i) to answer this question;
“or otherwise” means that students can answer this part as if it is a fresh one taking no
reference from the earlier part.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 110
2
x
2 3sin
4x
3
sin2
x
3
sin2
x or 3
sin2
x
2
,3 3
x
Inadmissible
Therefore the solutions are 2
, ,3 2 3
x
Solution(2):
cos3 2cos 0x x
cos2 cos sin 2 sin 2cos 0x x x x x
2 2 2cos sin cos 2cos sin 2cos 0x x x x x x
2 22cos 1 cos 2cos sin 2cos 0x x x x x
2 2cos 2cos 1 2sin 2 0x x x
2 2cos 2cos 2sin 1 0x x x
2 2cos 2cos 2 1 cos 1 0x x x
2cos 4cos 1 0x x
cos 0x or 24cos 1 0x
2
x
2 1cos
4x
1
cos2
x
1
cos2
x or 1
cos2
x
3
x
2
3x
Therefore the solutions are 2
, ,3 2 3
x
In Summary:
Trigo questions are typically multiple approaches. Take note that solution(1) is
most efficient. The knowledge of trigo ratios of special angles and triangles are needed.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 111
9. The function f is defined, for 0x , by f 3sin 13
xx
.
(i) State the maximum and minimum values of f x .
(ii) State the amplitude of f.
(iii) State the period of f.
(iv) Find the smallest value of x such that f 0x .
(v) Sketch the graph of 3sin 13
xy
for 0 540x .
Solution:
(i) State the maximum and minimum values of f x .
1 sin 13
x
3 3sin 33
x
3 1 3sin 1 3 13
x
4 f 2x
Therefore the minimum value is 4 , and the maximum value is 2 .
(ii) State the amplitude of f.
The amplitude is 3 .
(iii) State the period of f.
The period of the function f is 360
10801
3
[Analysis]
This question tests specifically all about trigo function and its related characteristics.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 112
(vi) Find the smallest value of x such that f 0x .
3sin 1 03
x
1sin
3 3
x
19.473
x
58.4x (to the nearest .1 degree)
(vii) Sketch the graph of 3sin 13
xy
for 0 540x .
1
2
270 270
540 x
y
58.4 481.6
In Summary:
Trigo curve sketching is an important part in this paper. This is a rather easy
question.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 113
10. The mass, m mg, of a radioactive substance decreases with time, t hours. Measured
values of m and t are given in the following table.
t (hours) 2 4 6 8 10
m (mg) 48.2 41.5 35.7 30.7 26.5
It is known that m and t are related by the equation 0
ktm m e , where 0m and k are
constants.
(i) Plot ln m against t for the given data and draw a straight line graph.
(ii) Use your graph to estimate the value of 0m and of k .
(iii) Estimate the number of hours for the mass of the substance to be halved.
Solution:
t (hours) 2 4 6 8 10
m (mg) 48.2 41.5 35.7 30.7 26.5
ln m 3.88 3.73 3.58 3.42 3.28
(i) Plot ln m against t for the given data and draw a straight line graph.
See the graph on the next page. 0ln lnm m kt
(ii) Use your graph to estimate the value of 0m and of k .
From the graph, 0ln 4.05m
4.05
0 57.4m e mg
3.95 3.300.077
1.2 9.6k
0.077k
[Analysis]
This question tests specifically on “straight line graph” or linear law. The word “Estimate” is
referred to reading from the graph.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 114
(iii) Estimate the number of hours for the mass of the substance to be halved.
Given that 0
ktm m e , when 0
2
mm ,
00
2
ktmm e OR 0
0ln ln ln ln 2 4.05 0.69 3.362
mm m
1
2
kte From the graph, 9t hours
ln 2 kt
ln 2
tk
ln 2
90.077
t hours
In Summary:
Part (iii) the reading from the graph solution is preferred. In the exam, student
should use the full length of the graph paper for the vertical axis. The quality of your
solution may vary, largely dependent on the scales selected. It is advisable to make your
line as large as possible. This question is also very time-consuming, students beware.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 115
0 2 4 6 8 10 t
3
ln m
4
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 116
11. Solutions to this question by accurate drawing will not be accepted.
The diagram shows a trapezium ABCD in which AB is parallel to DC and angle 90BAD .
The point A is 0,6 and the point D is 2, 2 .
(i) Find the equation of AB .
Given that B lies on the line y x , find
(ii) the coordinates of B .
Given that the length of DC is twice the length of AB , find
(iii) the coordinates of C ,
(iv) the area of the trapezium ABCD .
Solution:
(i) Find the equation of AB .
Given that A 0,6 , AB AD , D 2, 2 ,
Gradient of AD , 2 6
42 0
ADm
Gradient of AB , 1 1 1
4 4AB
AD
mm
Equation of the line AB ,
1
6 04
y x
1
64
y x
0 x
y
0,6 A
B
C
2, 2D
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 117
(ii) the coordinates of B .
Given that B lies on the line y x , B ,B Bx x ; Point B is on the line AB .
16
4B Bx x
36
4Bx
8Bx
the coordinates of B 8,8 .
(iii) the coordinates of C ,
Given that the length of DC is twice the length of AB ,
by similar triangle, we can deduce that 2C D B Ax x x x
2 2 8 0 18Cx
2C D B Ay y y y
2 2 8 6 2Cy
the coordinates of C 18,2 .
(iv) the area of the trapezium ABCD .
the length AD , 2 2
2 0 2 6 68AD
the length AB , 2 2
8 0 8 6 68AB
the area of the trapezium ABCD 1
2AB DC AD
1
22
AB AB AD
1
32
AB AD
13 68 68
2
102 square units
In Summary:
This is a fully coordinate geometry question. There are a number of ways to
find the area of the trapezium. Pay particular attention to part(iii).
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 118
12. A curve has the equation 2 1 4 1y x x .
(i) Express dy
dx in the form
4 1
kx
x , where k is a constant.
Hence
(ii) find the rate of change of x when 2x , given that y is changing at a constant rate
of 2 units per second,
(iii) evaluate 2
0
3
4 1
xdx
x .
Solution:
(i) Express dy
dx in the form
4 1
kx
x , where k is a constant.
Given that 2 1 4 1y x x ,
1 4
2 4 1 2 12 4 1
dyx x
dx x
2 4 1 2 2 1
4 1 4 1
x xdy
dx x x
OR
1 42 4 1 2 1
24 1 4 1
kxx x
x x
12
4 1
dy x
dx x
Let 2x ,
2 1 4
2 3 33 2 3
k
2 18 6k
12k
12
4 1
dy x
dx x
[Analysis]
Part (i) is about Product rule; part (ii) is a simple rate of change. Take note that “Hence”
requires part (i) to be used in these two parts.
Solutions to O Level Add Math paper 1 2009
By KL Ang, Jun 2013 Page 119
(ii) find the rate of change of x when 2x , given that y is changing at a constant rate
of 2 units per second,
Given that 2dy
dt , we apply chain rule
dy dy dx
dt dx dt OR
1dy dy dt dy dy
dxdx dt dx dx dt
dt
OR dx dx dy
dt dy dt
1dx dy
dydt dt
dx
At 2x ,
12 28
4 2 1
dy
dx
Therefore , 2 8dx
dt
1
4
dx
dt
(iii) evaluate 2
0
3
4 1
xdx
x .
2
0
3
4 1
xdx
x
2
0
1 12
4 4 1
xdx
x
2
0
1 12
4 4 1
xdx
x
2
0
1
4
dydx
dx
2
0
1
4y
12 2 1 4 2 1 2 0 1 4 0 1
4
5 1
22 2
In Summary:
This is the only question on integration in this paper. Do pay particular
attention on part (ii). This is a rather easy question.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 120
1. A and B are acute angles such that 3
sin8
A B and 5
sin cos8
A B . Without using a
calculator, find the value of
(i) cos sinA B ,
(ii) sin A B ,
(iii) tan
tan
A
B.
Solution :
Given that 0 , 90A B , 3
sin8
A B , 5
sin cos8
A B ,
(i) cos sinA B ,
sin sin cos cos sinA B A B A B
cos sin sin cos sinA B A B A B
5 3 1
cos sin8 8 4
A B
(ii) sin A B ,
sin sin cos cos sinA B A B A B
5 1 7
sin8 4 8
A B
(iii) tan
tan
A
B.
sin 5
tan sin cos 5cos 8sin 1tan cos sin 2
cos 4
A
A A BABB A B
B
[Analysis]
This question is solely based on the multiple angle formulae of sine ratio. Since 0 , 90A B ,
the ratios of sine and cosine of angles of A and, of B are positive(in quadrant I).
In Summary:
Please make sure that you are comfortable with operations with fractions. Be
careful with the sign of each ratios.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 121
2. (i) Express 2
7
2 6x x in partial fractions.
(ii) Hence evaluate 9
23
7
2 6dx
x x .
Solution :
(i) Express 2
7
2 6x x in partial fractions.
22 6 2 3 2x x x x
2
7 7
2 6 2 3 2x x x x
Let
7
2 3 2 2 3 2
A B
x x x x
Therefore,
By covering up method, OR By equating coefficients, OR By assigning values
when 2x , 7 2 2 3A x B x 7 2 2 3A x B x
7
12 2 3
B
, 7 2 2 3A B x A B when 2x , 1B
when 3
2x ,
2 0
2 3 7
A B
A B
when 3
2x , 2A
72
32
2
A
1B , 2A
7 1 2
2 3 2 2 2 3x x x x
[Analysis]
Part (i) is a straight forward partial fractions. Part (ii) MUST use part (i) to complete the
question. The rule of integration is
'fln f
f
xdx x c
x .
[Analysis]
Part (i) is a straight forward partial fractions. Part (ii) MUST use part (i) to complete the
question. The rule of integration is
'fln f
f
xdx x c
x .
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 122
(ii) Hence evaluate 9
23
7
2 6dx
x x .
9
23
9
3
9 9
3 3
7
2 6
1 2
2 2 3
1 2
2 2 3
dxx x
dxx x
dx dxx x
9
23
9
3
9
3
7
2 6
1 2
2 2 3
ln 2 ln 2 3
dxx x
dxx x
x x
9 9
3 3ln 2 ln 2 3
ln 7 ln1 ln 21 ln 9
7 9ln
21
ln 3
1.10
x x
9
3
2ln
2 3
7 1ln ln
21 9
7 1ln
21 9
ln 3 1.10
x
x
Additional note:
If f Fx dx x , then f F Fb
ax dx b a .
So,
2
7 1 2 1 2
2 6 2 2 3 2 2 3dx dx dx dx
x x x x x x
2
ln 2 ln 2 3 ln2 3
xx x c c
x
9
23
7 9 2 3 2 7 1ln ln ln ln ln3
2 6 2 9 3 2 3 3 21 9dx c c
x x
In Summary:
This is a typical use of partial fractions with integration. More of this kind can
be expected. There are more ways to decompose a fraction than have been shown.
OR
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 123
3.
(i) Use the substitution 2xu to express the equation 28 2 15x x as an cubic
equation in u .
(ii) Show that 3u is the only real solution of this equation.
(iii) Hence solve the equation 28 2 15x x .
Solution :
(i) Use the substitution 2xu to express the equation 28 2 15x x as an cubic
equation in u .
Let 2 0xu ,
28 2 15x x
3
2 4 2 15 0x x
3 4 15 0u u
(ii) Show that 3u is the only real solution of this equation.
Let 3f 4 15u u u
3f 3 3 4 3 15 0 therefore 3u is a factor of the function f u .
2f 3 3 5u u u u by long division
Given that f 0u ,
23 3 5 0u u u
3 0u or 2 3 5 0u u
3u Discriminant 23 4 1 5 11 0 , no real root.
OR
2 23 3
5 02 2
u
23 11
02 4
u
, no real root
[Analysis]
Part (i) needs to apply indices rules for 3
3 3 38 2 2 2x
x x x u and
2 22 2 2 4 2 4x x x u . Part (ii) is to show that the quadratic factor is non-zero, or has no
real number root.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 124
(iii) Hence solve the equation 28 2 15x x .
2 3x
lg 2 lg3x
lg31.58
lg 2x
Additional Note on factorization:
(1) by synthetic div
3 24 15 3 3 5u u u u u
(2) by equating coefficients
Let 3 24 15 3u u u au bu c
3 23 3 3au b a u c b u c
therefore, 1a , 5c , 3b
3 24 15 3 3 5u u u u u
(3) by inspection and assigning value
Let 3 24 15 3 5u u u u bu
When 1u , 1 4 15 2 1 5b
3b , 3 24 15 3 3 5u u u u u
In Summary:
This question comprises of topics from indices, exponential equation, solving
polynomial equation, nature of roots in quadratics and logarithm. There are
numerous ways to factorize a polynomial. In part (iii) ln 3
1.58ln 2
x also gives the
same result.
1
3
3
9
5
3 1 0 4 15
15
0
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 125
4.
The diagram shows an isosceles triangle ABC in which AC BC . Lines are drawn from
A and B to meet BC and AC at P and Q respectively. The lines AP and BQ intersect
at X . Given that PC QC , show that
(i) AXB is an isosceles triangle,
(ii) PX QX .
Solution :
(i)
Given that PC QC , AC BC
PC QC
BC BP AC AQ
Since BC AC , we get
BP AQ
For AQB and BPA ,
AB BA , common side
BAQ ABP , ABC is isosceles
AQ BP , as shown above
AQB and BPA are congruent(SAS).
Therefore, ABQ BAP
ABX BAX
Therefore, AXB is an isosceles triangle.
[Analysis]
Essentially, this is the only pure geometrical proof question in Add Math 2009. In part (i), to
show that AXB is an isosceles, we just need to show either AX BX or XAB XBA . In
part (ii), PX QX is the same as AP AX BQ BX , with AP BQ and AX BX from
part (i).
A B
C
P Q
X
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 126
(ii)
Since AQB and BPA are congruent,
BQ AP
BX XQ AX XP
Since AXB is an isosceles,
BX AX
XQ XP
PX QX
Alternative solution:
(i)
For ACP and BCQ ,
AC BC , ABC is an isosceles
ACP BCQ , common angle
PC QC , as given
ACP and BCQ are congruent (SAS)
therefore, CAP CBQ
also AP BQ
Since CAB CBA , then
CAP PAB CBQ QBA
But CAP CBQ , therefore
PAB QBA
so XAB XBA
Therefore, AXB is an isosceles.
(ii)
Since AXB is an isosceles, then AX BX .
AP BQ ,from above
AX XP BX XQ
Therefore, PX QX
In Summary:
Before making attempt to answer a geometrical proof, students should look at
what the given conditions and what these mean to the diagram. More importantly,
look at what is required to prove and work backward to the point where it meets the
given conditions. Typically, there are multiple ways to prove a geometrical question.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 127
5. (i) Write down the first three terms in the expansion, in ascending powers of x , of
24
nx
, where n is a positive integer greater than 2 .
The first two terms in the expansion, in ascending powers of x , of 1 24
nx
x
are
2a bx , where a and b are constants.
(ii) Find the value of n .
(iii) Hence find the value of a and of b .
Solution:
(i) Given that 24
nx
,
0 1 2
1 2 0
0 1 22 2 2 2 24 4 4 4 4
n n
n n n n n n n
n
x x x x xC C C C
2 21 1 22
24 32
nnn
n n xn x
(ii) Given that 21 24
nx
x a bx
,
2 21 1 22
1 2 1 24 4 32
n nnn
n n xx n xx x
2 2 2 31 1 21 2 1 22 2
2 24 32 4 32
n nn nn n
n n x n n xn x n xx
1 2 1 24 2 2 1 2 8 2
24 32
n n n n
nn x n n n x
1 2 28 2 17 2
24 32
n n
nn x n n x
[Analysis]
This is regarding binomial theorem, 0 1 1 0
0 1
n n n n n n n
na b C a b C a b C a b .
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 128
1 2 2
28 2 17 2
24 32
n n
nn x n n x
a bx
18 2
04
nn xx
Since 12 0n , therefore
8 0n
8n
(iii) 82 256a
68 9 2144
32b
Alternative solution:
(a)
1 21
2 2 1 2 14 8 8 2! 8 8
nn n
n nn nx x x x x
n
21
2 18 128
nn n xnx
21 22
28 128
nnn
n n xn x
2 21 1 22
24 32
nnn
n n xn x
(b)
Let 2
0 1 2f 24
n
n
n
xx c c x c x c x
When 0x ,
02n c
[Analysis]
Rewrite binomial theorem, 1 2
11 1
2!
n nn n n
n nb b b ba b a a n
a a a a
.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 129
1
' 1 1
1 2
1f 2 2
4 4
n
n
n
xx n c c x nc x
When 0x ,
1
1
2
4
nnc
2 2
'' 2
2
1f 1 2 2 1
4 4
n
n
n
xx n n c n n c x
When 0x ,
2
2
1 2
32
nn nc
Therefore,
2 21 1 22
2 24 4 32
n nnn
n n xx n x
In Summary:
The difficulty in this question is in the care of indices. Alternative solution (a) is
a little trick to make the expansion much easier to handle. Alternative solution (b) uses
differentiation to expand the series.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 130
6.
The diagram shows part of the curve 1 2cosy x , meeting the x -axis at the points A
and B .
(i) Show that the x -coordinate of A is 2
3
and find the x -coordinate of B .
(ii) Find the total area of the shaded regions.
Solution:
(i)
1 2cos 0x
1cos
2x
x is in 2nd
and 3rd
quadrants.
,3 3
x
2 4
,3 3
x
Therefore, 2
3A
,
4
3B
[Analysis]
Part (i) is about solving simple trigo equation; part (ii) is integrating area under the curve.
Take note that the area below the x-axis is negative, so the two shaded regions are to be
integrated separately.
0
1 2cosy x
A B x
y
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 131
(ii) the total area of the shaded regions.
42
332
03
1 2cos 1 2cosx dx x dx
42
33203
2sin 2sinx x x x
2 2 4 4 2 2
2sin 0 2sin 0 2sin 2sin3 3 3 3 3 3
2 3 2 3 3
2 2 23 2 3 2 2
2 2
3 2 33 3
2 2
3 3 33 3
since 3 3 0
2 23 3 3
3 3
3 3 5.20 square units
In Summary:
Be mindful of the trigo ratio of special angles and triangles.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 132
7. (i) Find the coordinates of all the points at which the graph of 3 5 2y x meet the
coordinate axes.
(ii) Sketch the graph of 3 5 2y x .
(iii) Solve the equation 3 5 2x x .
Solution:
(i)
When 0y , the x-axis intercepts
3 5 2 0x
3 5 2x
When 2
13
x , When 2
13
x ,
3 5 2x 3 5 2x
7 12
3 3x 1x
When 0x , the y-axis intercept
5 2 3y
(ii)
[Analysis]
Part (i) is about y- and x- axes intercepts. Part (ii) should show the turning point and symmetry.
Part (iii) is the solution of the line y x and 3 5 2y x .
0 x
y 3
2
3 1
1 2
1
2
21 , 2
3
12
3
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 133
(iii)
3 5 2x x ,
2 3 5x x since 3 5 0x , then 2 0x . Therefore, 2x
When 2
13
x , When 2
13
x ,
2 3 5x x 2 3 5x x
2 7x 4 3x
7 13
2 2x
3
4x
Alternative solution:
Part (iii)
3 5 2x x
2 3 5x x since 3 5 0x , then 2 0x . Therefore, 2x
2
2 3 5x x
2 2
2 3 5x x
2 2
2 3 5 0x x
2 3 5 2 3 5 0x x x x
4 3 2 7 0x x
4 3 0x or 2 7 0x
3
4x
7 13
2 2x
In Summary:
Be mindful that when simplifying modulus, there is a particular range of values
of x for each of them. Any solution that does not fall in this range will need to be
rejected. All solutions MUST be tested with the given equation.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 134
8. A motorcycle is driven along a straight horizontal road. As it passes a point A the brakes
are applied and the motorcycle slows down, coming to rest at a point B . For the journey
from A to B , the distance, s metres, of the motorcycle from A , t seconds after passing
A , is given by
10400 1 16t
s e t
(i) Find an expression, in terms of t , for the velocity of the motorcycle during the
journey from A to B .
(ii) Find an expression, in terms of t , for the acceleration of the motorcycle during the
journey from A to B .
(iii) Find the velocity of the motorcycle at A .
(iv) Show that the time taken for the journey from A to B is approximately 9.163
seconds.
(v) Find the average speed of the motorcycle for the journey from A to B .
Solution:
(i) Let the velocity be V ,
101
400 1610
tds
V edt
10 m40 16s
t
V e
[Analysis]
It will be best to organize all the pieces of information into a diagram:
A B s
10s 400 1 16t
e t
0t Bt t
V 0V
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 135
(ii) Let the acceleration be a ,
101
4010
tdV
a edt
102
m4s
t
a e
(iii) At A when 0t , at the start of the journey,
0 m40 16 24s
V e
(iv) Let the time taken for the journey from A to B be Bt , where the velocity is zero.
100 40 16Bt
e
102
5
Bt
e
2ln
10 5
Bt
510ln 9.163 s
2Bt
(v) Let the distance from A to B be Bs
10400 1 16Bt
B Bs e t
2
400 1 16 9.163 93.392 93.3m5
Bs
Average speed93.393 m10.19 10.2
s9.163
In Summary:
If you are to introduce some symbols, should state them clearly before use.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 136
9.
The diagram shows a circle with centre 2, 1C and radius 5 .
(i) Given that the equation of the circle is 2 2 2 2 0x y gx fy c , find the value of
each of the constants g , f and c .
The points A and B lie on the circle such that the line AC is parallel to the x -axis and
the line AB passes through the origin 0 .
(ii) Write down the coordinates of A .
(iii) Find the equation AB .
(iv) Find the coordinates of B .
Solution:
(i) Centre 2, 1C and radius 5 ,
2g
1 1f
2 2 22 1 5 20c
2, 1C
x
y
A
B
0
[Analysis]
If you can remember that 2 22 2 2 2 2 22 2 0x y ax by a b r x a y b r where
,a b being the centre of the circle, you may quickly pick up a few points.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 137
(ii) Point A is on 1y and 5 units to the left of 2x ,
the coordinates of 3, 1A .
(iii) The line AB passes through the origin, 0,0O and 3, 1A .
Let the equation AB be y kx ,
1 3k
1
3k
equation AB : 1
3y x
(v) To find the coordinates of B .
2 2 4 2 20 0
3
x y x y
y x
2 23 4 3 2 20 0y y y y
210 10 20 0y y 2 2 0y y
2 1 0y y
1y or 2y
When 2y , 1
23
x
6x
the coordinates of 6,2B .
In Summary:
This is the only coordinate geometrical question on circle. Questions relating to
circle can be expected in the future installations in this exam. It can be set with
geometrical properties of circles.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 138
Alternative solution:
Part (i)
Equation of the circle,
2 2 22 1 5x y
2 24 4 2 1 25x x y y 2 2 4 2 20 0x y x y
Therefore,
2g
1f
20c
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 139
10. A curve is such that 2
26 6
d yx
dx . The curve passes through the point 3,10 and at this
point the gradient of the curve is 12 . Find the coordinates of the stationary point of the curve
and determine the nature of this stationary point.
Solution:
Given that 2
26 6
d yx
dx , the gradient of the curve
2
2
dy d ydx
dx dx
6 6dy
x dxdx
23 6x x c
At 3,10 , the gradient of the curve is 12 .
2
12 3 3 6 3 c
3c
Therefore, the equation of the gradient
23 6 3dy
x xdx
The stationary points are where the gradient is zero.
20 3 6 3x x 20 2 1x x
2
0 1x
0 1x
1x
There is only one stationary point, with 2
26 1 6 0
d y
dx , it has not determined the nature of
this stationary point.
[Analysis]
You remember that gradient of the curve, 2
2
dy d ydx
dx dx
.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 140
x 1 1 1
dy
dx 0 0 0
From the table, we can deduce that the stationary point is a point of inflexion.
In Summary:
This is a very traditional question on derivatives of a curve and the determination
of a stationary point.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 141
11.
The diagram shows three fixed points 0 , A and D such that 17cmOA , 31cmOD and
angle 90AOD . The lines AB and DC are perpendicular to the line OC which makes an
angle with the line OD . The angle can vary in such a way that the point B lies between
the point O and C .
(i) Show that 48cos 14sin cmAB BC CD .
(ii) Find the values of for which 49cmAB BC CD .
(iii) State the maximum value of AB BC CD and the corresponding value of .
Solution:
(i)
Given that 90AOD , COD , therefore
90AOB
As 90ABO , then
OAB
17cosAB
As 17sinOB , and 31cosOC ,
BC OC OB
31cos 17sin
O 31cm
17cm
A
B
C
D
[Analysis]
Remember point B is to be in between OC .
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 142
31sinCD
17cos 31cos 17sin 31sin
AB BC CD
48cos 14sin cm
(ii)
2 214 48 50
48cos 14sin 49
48 14 49cos sin
50 50 50
48 24tan
14 7
49sin cos cos sin
50 73.74
49
sin50
78.52 , 101.15
78.52 or 101.15
78.52 73.74 101.15 73.74
4.78 4.8 27.41 27.4
(iii)
48cos 14sin
50sin
AB BC CD
50 sin 50
For maximum value of 50AB BC CD ,
sin 1
90
90 73.74 16.26 16.3
In Summary:
One must make sure that OB OC for all solutions.
Solutions to O Level Add Math paper 2 2009
By KL Ang, Jun 2013 Page 143
Additional information:
For OB OC , we can determine the largest .
From the diagram above, we get
31tan
17
61.26
Therefore 61.26 .
17
31
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 144
1. The function f is defined by 4 3f 4x x x kx , where k is a constant.
(i) Given that 2x is a factor of f x , find the value of k .
(ii) Using the value of k found in part (i), find the remainder when f x is divided by
2x .
Solution :
(i) Given that 2x is a factor of f x , then f 2 0
4 3f 2 2 2 2 4k
0 4 2k
2k
(ii) When 2x is to divide the polynomial f x , then the remainder is f 2 .
4 3
f 2 2 2 2 2 4
f 2 24
Alternative solution:
(i) By long division,
4 3 3 24 2 4 2 2 4 4x x kx x x x k x k
Since 2x is a factor of f x , the remainder 2 4 4 0k .
2 4 4 0k
2k
[Analysis]
A very routine application of Factor & Remainder Theorems of polynomial, which states that
“If x a is a factor of a polynomial f x , then f 0a .” Likewise, “If x a is to divide a
polynomial f x , then the remainder is f a .”
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 145
(ii) When 2x is to divide the polynomial f x , we just need to apply the similar
method as in part (i) .
By Long Division,
4 3 3 22 4 3 6 14 2 24x x x x x x x
the remainder is 24 .
In Summary:
This is a rather routine question. Factors(Multiplication) and Division are two
sides of the same coin. Students MUST be able to reason this well. The question can
be quickly done with Synthetic divisions.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 146
2. (i) Show that 2
sin cos 1 sin 2x x x .
(ii) Hence find, in terms of , the value of 2
2
0
sin cosx x dx
.
Solution :
(i)
L.H.S. 2
sin cosx x
2 2sin cos 2sin cosx x x x
Since 2 2sin cos 1x x and 2sin cos sin 2x x x ,
1 sin 2x
= R.H.S.
(ii)
2
2
0
sin cosx x dx
2
0
1 sin 2x dx
2
0
cos 2
2
xx
cos cos0
2 2 2
1 1
2 2 2
12
[Analysis]
Observe that there is a double angle in the equation in part(i). So, the use of double angle
maybe productive.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 147
Alternative solution:
(i)
R.H.S. 1 sin 2x
Since 2 21 sin cosx x and sin 2 2sin cosx x x ,
2 2sin cos 2sin cosx x x x
2
sin cosx x
= L.H.S.
(ii)
Consider
2
sin cosx x dx
1 sin 2x dx
cos 2
2
xx c where c is the integration constant
Therefore,
2
2
0
sin cosx x dx
cos 2cos 2 02
02 2 2
c c
1 1
2 2 2
12
In Summary:
Part (ii) MUST be solved with the result from part (i).
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 148
3. Using a separate diagram for each part, represent on the number line the solution set of
(i) 3 2 18x x ,
(ii) 23 5 1x x .
State the set of values of x which satisfy both of these inequalities.
Solution :
(i)
3 2 18x x ,
3 x
The solution set is : 3x x
(ii)
23 5 1x x
23 14 0x x
3 7 2 0x x
2x or 1
23
x
The solution set is 1
: 2 or 23
x x x
[Analysis]
This is an interesting and unexpected question on inequalities. Part (i) is on linear inequality,
part (ii) is of quadratic inequality. Last part requires answer to be in Set notation.
In Summary:
Take note of the set notation! Other form of set notation may also be
acceptable.
3
3 x
x
2
2x
x
12
3x
12
3
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 149
4. A curve has the equation sin 2 3cosy x x .
(i) Find the gradient of the curve when 6
x
.
(ii) Given that x is increasing at a constant rate of 0.06 units per second, find the rate of
change of y when 6
x
.
Solution :
(i)
Given that sin 2 3cosy x x ,
2cos 2 3sindy
x xdx
when 6
x
,
2cos 2 3sin6 6
dy
dx
2cos 3sin3 6
.
1 12 3
2 2
12
2
(ii)
When6
x
, given that 0.06dx
dt
dy dy dx
dt dx dt
5 6
0.152 100
dy
dt units per second
[Analysis]
Gradient of a curve is dy
dx. Rate of change
dy dy dx
dt dx dt . This kind of question has been
repeated in recent years.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 150
5. (i) Sketch the curve 29y x for 5 5x .
(ii) Find the x -coordinates of the points of intersection of the curve 29y x and
the line 27y .
Solution:
(i) Given that 29y x for 5 5x ,
When 0x , 9y .
When 0y , 20 9 x ,
3x
When 5x , 16y .
(ii)
Given that 29y x , when 27y ,
227 9 x
Since the horizontal line can only be intersecting on the reflected part of the quadratic curve,
227 9 x
236 x
6x
the x -coordinates of the points of intersection are 6 and 6 .
[Analysis]
Part (i) is a quadratic modulus. Let’s begin by sketching the quadratic before applying the
modulus. Part (ii) is to solve the simultaneous equations of a line and this modulus.
In Summary:
To change from y = f(x) to y = |f(x)|, is to reflect the part of curve that are below
the x-axis to that of above the axis. The reflected parts have the equation y = - f(x).
x
y
29y x
3 3 O
9
5 5
16
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 151
6. (i) Differentiate 2xxe with respect to x .
(ii) Use your answer to part (i) to show that
1 22
0
1
4
x exe dx
.
Solution:
(i) By product rule, let 2xy xe
2 21 2x xdye x e
dx
2 22 x xxe e
(ii) From part (i),
2 22 x xdyxe e
dx
2 22 x xdydx xe dx e dx
dx
2 22 x xdyxe dx dx e dx
dx
2
222
xx e
xe dx y c where c is a integration constant
2 2
2
12 4
x xx xe e
xe dx c where 12
cc is a integration constant
1 2 2 0 02
1 1
0
0
2 4 2 4
x e e e exe dx c c
2 1
4 4
e
2 1
4
e
[Analysis]
Use product rule for part (i). Then a little trick to use part (i) for integration in part (ii).
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 152
Alternative solution:
(i) Let 2xy xe ,
2ln ln xy xe
ln ln 2y x x
1 1
2dy
y dx x
1
2dy
ydx x
2 12xdy
xedx x
2 22x xdye xe
dx
(ii) Given from part (i), 2 22x xdye xe
dx ,
2
21
2 2
xxdy e
xedx
1 1 22
0 0
1
2 2
xx dy e
xe dx dxdx
1 1
2
0 0
1 1
2 2
xdydx e dx
dx
11 2
0 0
1 1
2 4
xy e
1
2 2
0
1 11
2 4
xxe e
2 21 11
2 4e e
2 1
4
e
In Summary:
Do pay particular attention to how the integration is carried out. This
approach has been required frequently in recent exam.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 153
7. The table shows experimental values of two variables x and y , which are connected by an
equation of the form nyx k , where n and k are constants
(i) Using a scale of 1cm to 0.1unit on each axis, plot lg y against lg x and draw a
straight line graph.
(ii) Use your graph to estimate the value of k and of n .
Solution:
(i) Given that nyx k ,
lg lg lgy k n x
(ii)
From the graph, the vertical intercept
lg 1.94k
87.1k
From the graph, the gradient
0.8
0.6n
4 1
13 3
n
x 2 8 14 20
y 33.00 5.07 2.38 1.47
x 2 8 14 20
y 33.00 5.07 2.38 1.47
lg x 0.30 0.90 1.15 1.30
lg y 1.52 0.71 0.38 0.18
[Analysis]
This is a question on linear graph. The given equation, nyx k , can be expressed as
lg lg lgy k n x , where n and lg k are gradient and vertical axis intercept respectively.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 154
O lg x
lg y
0.5 1.0 1.5
0.5
1.0
1.5
In Summary:
This question is very straight forward and very time consuming. Students are
mindful of the time spent on the drawing of the graph.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 155
8. The equation of a curve is 3 23 9y x x x k , where k is a constant.
(i) Find the set of values of x for which y is decreasing.
(ii) Find the possible values of k for which the x -axis is a tangent to the curve.
Solution:
(i) Given that 3 23 9y x x x k ,
23 6 9dy
x xdx
For decreasing range of the curve, 0dy
dx ,
23 6 9 0x x
2 2 3 0x x
3 1 0x x
3 1x
The solution set is : 3 1x x .
(ii) From part (i), 23 6 9dy
x xdx
, let 0dy
dx ,
23 6 9 0x x
3 1 0x x
3x or 1x
When 3x , 0y ,
3 2
0 3 3 3 9 3 k
27k
When 1x , 0y ,
3 20 1 3 1 9 1 k
5k
[Analysis]
Part (i) tests on decreasing function where 0dy
dx . Part (ii) test on gradient being zero, 0
dy
dx
and 0y .
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 156
Alternative solution:
(ii)
From part (i), 23 6 9dy
x xdx
, let 0dy
dx ,
23 6 9 0x x
3 1 0x x
3x or 1x
When 3x , 0y ,
23 23 9 3x x x k x x p
Let 3x , 27 0k
27k
Similarly,
When 1x , 0y ,
23 23 9 1x x x k x x q
Let 1x , 5 0k --------- (2)
5k
In Summary:
This is a good question with a bit of non-routine.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 157
9. Given that the roots of 23 2 1 0x x are and , find the quadratic equation whose
roots are 2 and 2 .
Solution:
Given that 23 2 1 0x x ,
2
3
2
3
1
3
the quadratic equation whose roots are 2 and 2 ,
Sum of roots, 2 2 3 2
Product of roots, 22 22 2 2 5 2
22 1
23 3
11
9
2 112 0
9x x
29 18 11 0x x
In Summary:
Modulus is a key topic to understand as it will be use in ‘A’ level. It is best by
going through this solution very carefully so as to enhance your understanding.
[Analysis]
This question is about Sum of roots, b
a and Product of roots,
c
a , where
2 0ax bx c .
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 158
10. Without using a calculator, show that
(i) tan 75 2 3 ,
(ii) 2sec 75 4tan75 .
Solution:
(i) tan 75 tan 45 30 where tan 45 1 , 3
tan 303
tan 45 tan 30
1 tan 45 tan 30
31
3
31
3
3 3
3
3 3
3
3 3
3 3
2
3 3
3 3 3 3
12 6 3
6
2 3
[Analysis]
This is an un-usual question. For part (i), we can make use of sum of angles for tangent and the
ratios of special triangles. As for part (ii), we may try the Pythagoras identity, and then apply
the result from part (i) to finish the question.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 159
(ii) 2 2sec 75 1 tan 75
2
1 tan 75
2
1 2 3
1 7 4 3
8 4 3
4 2 3
4tan75
Alternative solution:
(i) Construct a right angled triangle of 75 , ABC .
75BAC , 15ACB , 60BAD , 15CAD
therefore, ACD is an isosceles triangle.
AD DC
Let 1AB , 2AD , then 3BD
3 2BC BD DC BD AD
2 3
tan 75 2 31
BC
AB
(ii) To show 2sec 75 4tan75 , is to show
2sec 75 4tan75 0
2sec 75 4tan75 21 tan 75 4tan75
2
1 2 3 4 2 3
1 7 4 3 8 4 3
0
In Summary:
Knowledge on surds, rationalization of denominators, ratios of special triangles,
and Pythagoras identities are a must.
75
60
A B
C
D
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 160
11. A curve is such that 2
d 82
d
y
x x .
(i) Given that the curve passes through the point 1,5 , find the equation of the curve.
(ii) Find the x -coordinates of the stationary points of the curve.
(iii) Obtain an expression for 2
2
d
d
y
x and hence, or otherwise, determine the nature of each
stationary point.
Solution:
(i) Given that 2
d 82
d
y
x x ,
2
d 8d 2 d
d
yx x
x x
let
1u
x ,
2
1du
dx x
8
2y x cx
At 1,5 , 8
5 2 11
c
15c
the equation of the curve , 8
15 2y xx
.
(ii) Let d
0d
y
x ,
2
80 2
x
2 4x
2x
[Analysis]
Using integration of 2
d 82
d
y
x x in Part (i) to get the equation of the curve that also passes
through the point 1,5 . Let d
0d
y
x , the gradient to be zero to find the stationary points.
Lastly, to determine the nature of each stationary point.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 161
(iii) Given that 2
d 82
d
y
x x ,
2
2 3
d 16
d
y
x x
When 2x ,
2
2
d2 0
d
y
x
At 2x , this is a locally maximum point
When 2x ,
2
2
d2 0
d
y
x
At 2x , this is a locally minimum point
Alternative solution:
(iii) Given that 2
d 82
d
y
x x ,
When 2x ,
At 2x , the turning point is a locally maximum.
Similarly,
When 2x ,
At 2x , the turning point is a locally minimum.
x 2 ( 1.9) 2 2 2.1
d
d
y
x 0 0 0
x 2 ( 2.1) 2 2 1.9
d
d
y
x 0 0 0
In Summary:
A routine question on first, identifying the curve; secondly, locating the turning
points; lastly, the type of turning points.
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 162
12. (i) Write down the equation of the circle with centre 3,2A and radius 5 .
This circle intersects the y -axis at points P and Q .
(ii) Find the length of PQ .
A second circle, centre B , also passes through P and Q .
(iii) State the y -coordinate of B .
Given that the x -coordinate of B is positive and that the radius of the second circle is 80 ,
find
(iv) the x -coordinate of B .
The equation of the circle, centre B , which passes through P and Q , may be written in the
form 2 2 2 2 0x y gx fy c .
(v) State the value of g and of f , and find the value of c .
Solution:
(i) Given that the circle with centre 3,2A and radius 5 , the equation of this circle is
2 2 23 2 5x y
(ii) Points P and Q are on the y -axis 0x .
2 2 20 3 2 5y
2
2 16y
2 4y
6y or 2y
the length of PQ , 6 2 8PQ units
[Analysis]
This question asks very routine questions about circle and co-ordinate geometry. Remember,
the y -axis is on the line 0x . The equation of circle is 2 2 2x a y b r where point
,a b is the centre of the circle with radius .r
Solutions to O Level Add Math paper 1 2010
By KL Ang, Jun 2013 Page 163
(iii) The y -coordinate of B is the mid-point of PQ , 6 2
22
y
(iv) Let the centre of , 2B a , the equation of the circle is
22 2
2 80x a y
2 2
2 80x a y
At point 0,6P ,
2 2
0 6 2 80a
2 64a
8a or 8a (rejected, since the x -coordinate of B is positive.)
the x -coordinate of B is 8x
(v) The equation of circle, centre B ,
2 2
8 2 80x y
2 22 8 64 2 2 4 80 0x x y y
2 2 2 8 2 2 12 0x y x y
Equate 2 2 2 2 0x y gx fy c with the above equation, we get
8g
2f
12c
In Summary:
Remember that 2 2 2 2 0x y gx fy c is the same as 2 2 2x a y b r .
Part (iii) makes use of the geometrical property of circle “ Perpendicular of radius
bisects chord” . It is advisable to make a sketch of the two circles to “visualize” the
question.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 164
1. Solve the equation 23cot 10cosec 5 for 0 360 .
Solution :
Given that 23cot 10cosec 5 ,
23 cosec 1 10cosec 5
23cosec 10cosec 8 0
cosec 4 3cosec 2 0
cosec 4 or 2
cosec3
(Rejected, since sin 1 , then cosec 1 )
1
sin4
1 1sin
4
where is in 3
rd and 4
th quadrants.
Principal angle, 14.47
180 14.47 ,360 14.47
194.5 ,345.5
Alternative solution :
For solving 1
sin4
, is in 3rd
and 4th
quadrants.
Let basic angle be ,
1sin
4
14.47
180 14.47 ,360 14.47
194.5 ,345.5
[Analysis]
The given trigo equation contains the un-common ratios of cot and cosec . It has been a
common practice to change these ratios into cos
cotsin
and
1cosec
sin
. If this
approach is taken, then the question actually would have become more difficult. Rather, it is
more productive to apply Pythagoras identity, 2 21 cot cosec .
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 165
Alternative solution:
Given that 23cot 10cosec 5 ,
2cos 1
3 10 5sin sin
where sin 0
2 23cos 10sin 5sin
2 23 1 sin 10sin 5sin 0
23 10sin 8sin 0
28sin 10sin 3 0
4sin 1 2sin 3 0
1
sin4
or 3
sin2
(Rejected, as sin 1 )
(The rest of the solutions are as shown before.)
In Summary:
Be well verse in the 3 lesser use ratios. Do not simply convert them to sine,
cosine or tangent ratios. Students MUST be very competent in solving simple trigo
equations either by principal angle or basic angle.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 166
2.
The diagram shows a triangular piece of land PQR in which angle 90PQR , 8mPQ
and 12mQR . A rectangle QUVW is to be used as the base of a greenhouse, where U ,
V and W lie on QR , RP and PQ respectively, mQU x and mQW y .
(i) Show that 2
83
xy .
(ii) Express the area, 2mA , of the base of the greenhouse in terms of x .
(iii) Given that x can vary, find the maximum value of A .
Solution :
(i)
Given that 0 8y and 0 12x ,
,V x y , 0,0Q , 0,8P , 12,0R .
Let equation of the line PR be
y mx c
When 0, 8x y ,
8c
When 12, 0x y ,
12 8 0m
2
3m
Therefore, 2
83
xy ,
[Analysis]
We will first cast this question into coordinate geometry.
12m
W
P
R
V
Q
8m
U
my
mx
W
0,8P
12,0
R
,V x y
Q U
y
x
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 167
(ii)
The area of the greenhouse,
A xy
2
83
xx
(iii)
2
83
xA x
22
83
xA x
2212
3A x x
2 2 2212 6 6
3A x x
226 36
3A x
22
6 243
A x
When 6x , the maximum value of 224mA .
Alternative solutions:
(i)
Given that 0 8y and 0 12x ,
,V x y , 0,0Q , 0,8P , 12,0R .
The gradient of PR ,
8 0 2
0 12 3m
equation of the line PR ,
2
8 03
y x
2
83
xy
(i)
Given that 0 8y and 0 12x ,
,V x y , 0,0Q , 0,8P , 12,0R .
By intercepts formula of straight line, equation
of the line PR ,
112 8
x y
2 3 24x y
3 24 2y x
2
83
xy
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 168
(i)
Given that 0 8y and 0 12x ,
,V x y , 0,0Q , 0,8P , 12,0R .
Knowing two points on the line, we have
8 0 8
0 12 0
y
x
After rearrangement,
2
83
xy
(i)
By similar triangles of RVU and RPQ , we
have
RU UV
RQ QP
12
12 8
x y
After rearrangement,
2
83
xy
(iii)
2
83
xA x
22
83
xA x
d 4
8d 3
A x
x
Let d
0d
A
x ,
40 8
3
x
6x
2
2
d 4
d 3
A
x
When 6x , 2
2
d 40
d 3
A
x , the turning point is a maximum point.
2 6
6 83
A
24A
When 6x , the maximum value of 224mA .
In Summary:
The question can be tackled by different means, of which, the coordinate
geometrical approach is most preferred. In solving for min/max value, students should
be well aware of the conditions that 0 8y and 0 12x .
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 169
3. Without using a calculator, solve, for x and y , the simultaneous equations
32 2 1x y ,
1
123 27 81x y x .
Solution :
Given that 32 2 1x y ,
32 2 1x y ,
52 2 1x y
5 02 2x y
5 0x y
5y x --------- (1)
Given that 1
123 27 81x y x
1123
8127
x
xy
412
3
33
3
x
xy
4
12 33 3x y x
412 3x y
x --------- (2)
Substitute (1) into (2),
4
12 15x xx
4
16 12xx
0x , 216 12 4 0x x
24 3 1 0x x
4 1 1 0x x
Alternative solution:
Given that 32 2 1x y ,
2 2log 32 2 log 1x y
2 2log 32 log 2 0x y
5
2 2log 2 log 2 0x y
2 25 log 2 log 2 0x y
25 log 2 0x y
5 0x y --------- (3)
Given that 1
123 27 81x y x
112
3 3
3log log 81
27
x
xy
1
12
3 3 3log 3 log 27 log 81x y x
4
12 3
3 3 3log 3 log 3 log 3 0x y x
3 3 3
412 log 3 3 log 3 log 3 0x y
x
3
412 3 log 3 0x y
x
4
12 3 0x yx
--------- (4)
3 3 4 ,
216 12 4 0x x (the rest of the
solution follows as shown on the left.)
[Analysis]
Notice that 532 2 in the first equation; 3 427 3 ,81 3 in the second equation. Naturally, we
can start by applying indices laws to simplify the equations. Alternatively, we may try
logarithmic laws.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 170
1
4x or 1x
When 1
4x ,
5
4y .
When 1x , 5y .
In Summary:
Students should check the answers against the given equations to guard against
mistakes.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 171
4. (i) Given that the constant term in the binomial expansion of
8
3
kx
x
is 7 , find the
value of the positive constant k .
(ii) Using the value of k found in part (i), show that there is no constant term in the
expansion of 8
4
31
kx x
x
.
Solution :
(i) Given that 0k ,
8
3
kx
x
8
8
41
kx
x
Each term of this expansion, 8 8 4
4
8 81
rn r r r
r
kT x k x
r rx
For the constant term,
8 4 0r
2r
Therefore, 2 0
2
8
2T k x
, since 2 7T , we have
28!7
6!2!k
2 1
4k
1
2k
Since 0k , 1
2k
[Analysis]
A very routine question that simply applies binomial theorem. A constant term is one that is
free of the factor x .
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 172
(ii)
Given that 8
4
31
kx x
x
8 8
4
3 3
k kx x x
x x
8 8
8 4 12 4
0 0
8 8r rr rk x k xr r
For the constant term, 2r , for the first expansion; 3r , for the second expansion.
The sum of the constant terms,
38 7 6 1
73! 2
8 7 1
71 8
7 7
0
Hence, there is NO constant term.
Alternative solution:
(i) Given that 0k ,
8
3
kx
x
Each term of this expansion, 8 8 4
3
8 8r
r r r
r
kT x k x
r rx
The rest will be the same as before.
In Summary:
A routine question, Be competent in handling the term expression, Tr .
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 173
5. The curve 25 xy e intersects the coordinates axes at the points A and B .
(i) Given that the line AB passes through the point with coordinates ln 5,k , find the
value of k .
(ii) In order to solve the equation ln 9x x , a graph of a suitable straight line is
drawn on the same set of axes as the graph of 25 xy e . Find the equation of this
straight line.
Solution:
(i)
When 0x , 05 4y e .
When 0y , 25 0xe ,
2 5xe
ln 5
2x
The coordinates of the points A and B are either 0,4 or ln 5
,02
.
Let the equation of the line AB be y mx c .
therefore , 4c .
At ln 5
,02
, ln 5
0 42
m
8
ln 5m
Equation of the line AB , 8
4ln 5
y x
At ln 5,k , 8
ln 5 4ln 5
k
4k
[Analysis]
It will be beneficial to sketch the curve of 25 xy e to as to visualize the problem. Part (i) is
then quite straight forward. Part (ii) may not be clear at first. The trick is to create 25 xe out
of ln 9x x . We need the e term which can be obtained by taking anit-log on the given
equation.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 174
(ii) Given that ln 9x x ,
9xe x
22
9xe x
2 9xe x
2 9xe x
25 4xe x
Therefore, the line to add in the graph of 25 xy e in order to solve the equation ln 9x x ,
is 4y x
Alternative solution:
(i)
When 0x , 05 4y e .
When 0y , 25 0xe ,
2 5xe
ln 5
2x
The coordinates of the points A and B be 0,4 and ln 5
,02
respectively.
Let ln5,C k , ln5,0D .
90AOB CDB
ln 5
2OB DB
OBA DBC (vertically opposite angle)
AOB CDB , the two triangles are congruent.
4OA DC ,
therefore 4k
In Summary:
It is troublesome to form the equation of the line, before finding k. Part (ii) is
usually flexible. Just ask how to get 25 xe from ln 9x x .
x
y
25 xy e
O
4
ln 5
2
B ln 5
k
A
C
5
D
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 175
6.
The diagram shows a point X on a circle and XY is a tangent to the circle. Points A , B
and C lie on the circle such that XA bisects angle YXB and YAC is a straight line. The
line YC and XB intersect at D .
(i) Prove that AX AB .
(ii) Prove that CD bisects angle XCB .
(iii) Prove that triangles CDX and CBA are similar.
Solution:
(i) Given that XA bisects angle YXB ,
YXA BXA
YXA XBA (alternate segment theorem)
Therefore, XBA BXA
Therefore, AXB is an isosceles triangle.
Hence, AX AB
(ii)
XCA XBA ( s in same segment)
BCA BXA ( s in same segment)
[Analysis]
Part (i) is to show that ABX is an isosceles triangle. This is to say that we need to show
ABX BXA . Part (ii) is to show that XCD BCD .
We already know by then from part (ii) that XCD BCD , we will just need to show that
CXD CAB in Part (iii).
A
Y
C
X
D
B
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 176
From part (i),
XBA BXA
Therefore, XCA BCA
XCD BCD ( D is on the line YAC )
Therefore, CD bisects angle XCB .
(iii)
Consider CDX and CBA ,
XCD BCD (from part (ii))
CXD CXB CAB ( s in same segment)
therefore, XDC ABC
So, CDX CBA
In Summary:
Before writing any proof, one shall think through the entire proving steps. This
thought process usually required looking at the result, and then work backward until it
meets the forward process, starting from the given conditions.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 177
7.
The diagram shows part of the curve 2 5y x passing through the point P and meeting the
x -axis at the point Q . The line 2x passes through P and intersects the x -axis at the point S .
Lines from Q meet 2x at the points R and T such that QR is parallel to the tangent to the
curve at P , and RS ST . Find
(i) the equation of QR ,
(ii) the area of the shaded region.
Solution:
(i) At Q , 0y .
0 2 5x
2 5 0x
5
2x
Coordinates of point 5
,02
Q
O
2 5y x
x
y
Q
P
R
T
S
2x
[Analysis]
Since the line QR is parallel to the tangent of the curve that passes through point
where , 2x its gradient can be found with dy
dx. Knowing the point Q , we can then find the
equation of QR .
Recognise that QRS is congruent to QTS , the area of the shaded region will be the area
under the curve and the area under the line QR .
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 178
Gradient of the curve, 2 5y x , by chain rule,
1
2d 1
2 5 2d 2
yx
x
d 1
d 2 5
y
x x
At P , 2x .
d 1 1
d 32 2 5
y
x
As QR is parallel to the tangent of the curve at P , the gradient of QR is 1
3.
The equation of QR ,
1 5
03 2
y x
1 5
3 6y x
(ii)
As RS ST , QS is common, 90QSR QST , by SAS, QRS is congruent to QTS .
Area of QTS area of QRS .
Area of the shaded region,
2 2
5 5
2 2
52 5d d
3 6
xx x x
2 2
5 5
2 2
1 2 53 2 5d d
3 6 6
xx x x
22 23
2
5 52 2
1 52 5
3 6 6
xx x
3
21 4 10 25 25
9 03 6 6 24 12
14 25
96 24
27
98
3
128
square units
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 179
Alternative solution:
(ii)
Gradient of 1
3
TS RSQT
QS QS
The equation of QT , 5
,02
Q
.
1 5
03 2
y x
1 5
3 6y x
Area of the shaded region,
2 2
5 5
2 2
52 5d d
3 6
xx x x
2 2
5 5
2 2
1 2 53 2 5d d
3 6 6
xx x x
2 2
5 5
2 2
1 2 53 2 5d d
3 6 6
xx x x
22 23
2
5 52 2
1 52 5
3 6 6
xx x
3
21 4 10 25 25
9 03 6 6 24 12
14 25
96 24
27
98
3
128
square units
In Summary:
The alternative solution on part (ii) uses the idea that the area enclosed by the
curve and a line. The same method can also be applied for the area enclosed by two
curves.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 180
8. Two particles, P and Q leave a point O at the same time and travel in the same direction
along the same straight line. Particle P starts with a velocity of 9m/s and moves with a
constant acceleration of 21.5m/s . Particle Q starts from rest and moves with an
acceleration of 2m/sa , where 12
ta and t seconds is the time since leaving O . Find
(i) the velocity of each particle in terms of t ,
(ii) the distance travelled by each particle in terms of t .
Hence find
(iii) the distance from O at which Q collides with P ,
(iv) the speed of each particle at the point of collision.
Solution:
(i)
For particle P , Velocity of P , V . Acceleration of P ,
d 3
d 2
V
t
1
d 3 3d d
d 2 2
VV t t t c
t where 1c is the integration constant.
When 0t , 9V .
1
39 0
2c
1 9c
3
92
V t
[Analysis]
The curve passes through point 2,0 and its gradient is zero at this point too. With two
unknown a and b , these will be enough to solve part (i). The turning points occur at zero
gradient, 0dy
dx . Of course, we already know the minimum point. The area is simply the
integration of the curve from 0 to 2 .
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 181
For particle Q , Velocity of Q , V . Acceleration of Q ,
d
1d 2
V t
t
2
2
dd 1 d
d 2 4
V t tV t t t c
t where 2c is the integration constant.
When 0t , 0V .
2
2
00 0
4c
2 0c
2
4
tV t
(ii) When 0t , 3
9 02
V t ; 2
04
tV t .
For particle P , distance travelled of P , S .
dS V t
3
9d2
t t
2
3
39
4
tt c
When 0t , 0S .
3 0c
23
94
tS t
For particle Q , distance travelled of Q , S .
dS V t
2
d4
tt t
2 3
42 12
t tc
When 0t , 0S .
4 0c
2 3
2 12
t tS
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 182
(iii)
When Q and P meet, the distance travelled must be the same.
2 3 23
92 12 4
t t tt
2 3 26 9 108t t t t
3 23 108 0t t t
2 3 108 0t t t
9 12 0t t t
0t or 9t or 12t
N.A. N.A.
When 12t , 2 312 12
2162 12
S m
(iv)
When 12t ,
for particle P ,
3
12 9 272
V m/s
for particle Q ,
212
12 484
V m/s
In Summary:
You should read through the entire question, make a plan on how to approach
them before beginning to answer them.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 183
9.
The diagram shows a triangle ABC with vertices at 0,5A , 8,14B and runs ,15C k .
Given that AB BC ,
(i) find the value of k .
A line is drawn from B to meet the x -axis at D such that AD CD .
(ii) Find the equation of BD and the coordinates of D .
(iii) Show that the area of the triangle ABC is 2
7 of the area of the quadrilateral ABCD .
Solution:
(i) Given that AB BC ,
2 2 2 2
0 8 5 14 8 15 14k
2
64 81 8 1k
2
8 144k
8 12k
[Analysis]
For part (i), the AB BC . Part (ii) is to sketch out D and BD to visualize the problem.
Recognise that the two diagonals BD and AC are perpendicular to one another. Part (iii) will
need to find the length ratio.
x
,15C k
0,5 A
8,14B
O
y
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 184
20k or 4k , Rejected since 0k
(ii) Let the diagonals BD and AC intersect at M . Quadrilateral ABCD is a kite . Therefore,
the two diagonals are perpendicular to each other.
Gradient AC , 15 5 1
20 0 2
Gradient BD , 1
21
2
The equation of BD ,
14 2 8y x
2 30y x
When 0y ,
0 2 30x
15x
the coordinates of 15,0D .
(iii) Quadrilateral ABCD is a kite where AM CM , or M is the mid-point on AC
the coordinates of 0 20 5 15
,2 2
M
.
10,10M
area of the triangle ABC 1
2AC BM
area of the quadrilateral ABCD1
2AC BD
1area of triangle 2
1area of quadrilateral
2
AC BMABC
ABCDAC BD
BM
BD
2 2
2 2
10 8 10 14
15 8 0 14
20
245
x
20,15C
0,5 A
8,14B
O
y
D
M
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 185
4
49
2
7
Alternative solution:
(iii)
area of the triangle ABC
0 20 8 01
5 15 14 52
1
0 280 40 100 1202
50 square units
area of the quadrilateral ABCD
0 15 20 8 01
5 0 15 14 52
1
225 280 40 75 1202
1
545 1952
175 square units
area of triangle 50
area of quadrilateral 175
ABC
ABCD
2
7
In Summary:
The method employed in the part (iii) of the alternative solution is an easier way
to solve this question.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 186
10. (i) Given that
2
22
3 4 20
2 1 42 1 4
x x A Bx C
x xx x
, where A , B and C are constants,
find the value of A and of B and show that 0C .
(ii) Differentiate 2ln 4x with respect to x .
(iii) Using the results from part (i) and (ii), find
2
2
3 4 20d
2 1 4
x xx
x x
.
Solution:
(i)
Given that
2
22
3 4 20
2 1 42 1 4
x x A Bx C
x xx x
,
22
2 2
4 2 13 4 20
2 1 4 2 1 4
A x Bx C xx x
x x x x
2 23 4 20 4 2 1x x A x Bx C x
2 23 4 20 2 2 4x x A B x B C x A C
Equating the coefficients,
2 3A B --------- (1)
2 4B C --------- (2)
4 20A C --------- (3)
2 2 1 , 4 5C A --------- (4)
4 3 4 , 17 85A
5A
Substitute 5A into 1 , 4B
Substitute 5A into 3 , 0C
[Analysis]
Part (i) is a partial fractions with an irreducible quadratic, 2 4x .
Apply chain rule for part (ii). Apply
'fln f
f
xdx x c
x
.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 187
(ii)
Let 2ln 4y x and 2 4u x ,
thus lny u . d 1
du
y
u .
d2
d
ux
x
By chain rule,
d d d
d d d
y y u
x u x .
d 1
2d
yx
x u
2
d 2
d 4
y x
x x
(iii)
2
2
3 4 20d
2 1 4
x xx
x x
2
5 4d
2 1 4
xx
x x
2
5 4d d
2 1 4
xx x
x x
2
5 2 2d 2 d
2 2 1 4
xx x
x x
25ln 2 1 2ln 4
2x x c where c is an integration constant
Alternative solution:
(i) Given that
2
22
3 4 20
2 1 42 1 4
x x A Bx C
x xx x
,
2 23 4 20 4 2 1x x A x Bx C x
When 1
2x , 5A . 2 23 4 20 5 4 2 1x x x Bx C x
28 4 2 1x x Bx C x
4 2 1 2 1x x Bx C x , Hence 4B , 0C .
In Summary:
A very-much guided question. There are many ways to solve the partial
fractions.
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 188
11.
The diagram shows the curves 4cosy x and 2 3siny x for 0 2x radians. The
points A and B are turning points on the curve 2 3siny x and the point C is a turning
point on the curve 4cosy x . The curves intersect at points D and E .
(i) Write down the coordinates of A , B and C .
(ii) Express the equation 4cos 2 3sinx x in the form cos x a k , where and k
are constants to be found.
(iii) Hence find, in radians, the x -coordinate of D and of E .
Solution:
(i) For points A and B ,
2 3siny x
d
3cosd
yx
x
[Analysis]
Recognise that points A , B and C are the min/max locations in each curve. We can apply
differentiation to find the turning points on each curve.
Next, is a R-formula application, where cos cos cos sin sinx a x a x a .
Lastly, the points D and E are the intersections of the two curves.
2 x
y
A
B O
C
D
E
4cosy x
2 3siny x
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 189
Let d
0d
y
x ,
0 3cos x
cos 0x
3
,2 2
x
When 2
x
, 2 3sin 2 3 52
y
When 3
2x
,
32 3sin 2 3 1
2y
coordinates of ,52
A
, 3
, 12
B
Similarly,
For point C ,
4cosy x
d
4sind
yx
x
Let d
0d
y
x ,
0 4sin x
sin 0x
0, , 2x
When 0x , 4cos 0 4y , Rejected, not point C
When x , 4cos 4y
When 2x , 4cos 0 4y , Rejected, not point C
coordinates of , 4C
(ii)
Given that 4cos 2 3sinx x ,
4cos 3sin 2x x
4 3 2
cos sin5 5 5
x x
2
cos cos sin sin5
x x
4
3 5
Solutions to O Level Add Math paper 2 2010
By KL Ang, Jun 2013 Page 190
2
cos5
x where 1 3tan
4
(iii) To find D and E , is to solve 2 3sin
4cos
y x
y x
.
4cos 2 3sinx x
From part (ii), we get
2
cos5
x where 1 3tan 0.6435
4
1 2cos
5x
angle x is in Quadrant 1st and 4
th.
Principal angle, 1.1592x
1.1592x or 2 1.1592x
1.1592x 2 1.1592x
0.516x radians 4.48x radians
the x -coordinate of D is 0.516x radians ; and of E is 4.48x radians
Alternative solution:
(i) For points A and B , given that 2 3siny x ,
1 sin 1x
1 2 3sin 5x
When sin 1x , 2
x
, 5y , ,52
A
.
When sin 1x , 3
or2 2
x
, 1y , 3
, 12
B
.
For point C ,
4cosy x
1 cos 1x
4 cos 4x
When cos 1x , x , 4y , , 4C .
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 191
1. Given thatx
yxe
for 0x , find the range of values of x for which y is a decreasing
function of x. [4]
Solution :
Given thatx
yxe
for 0x ,
2
ee
d
d
x
x
x
y xx
2
1e
d
d
x
x
x
y x
Let 0d
d
x
y,
01e
2
x
xx
Because 0e,2 xx for all value of x, therefore
01x
1x
Combine with the valid range of x, we have
10 x
Since 1x , is the turning point of the function, it is at the lowest value. So, this point is to be
included. The final range of the decreasing function is
10 x
[Analysis]
A function is decreasing only if its first derivative is negative, i.e. 0d
d
x
y. To differentiate a
function of the form v
yu
, we apply quotient rule, that is 2
d
d
d
d
d
d
v
x
vu
x
uv
x
y
. If the end
points are turning points, then it must be included in the range. Take note that the function
begins with 0x .
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 192
Alternative solution:
Given thatx
yxe
for 0x ,
2
ee
d
d
x
x
x
y xx
2
1e
d
d
x
x
x
y x
Let 0d
d
x
y,
0
1e2
x
xx
Because 0e,2 xx for all value of x, therefore
01x
1x ,is the turning point of the function.
When 1x , 01x
01e
2
x
xx
When 1x , 01x
01e
2
x
xx
1x is a minimum point.
For 0d
d
x
y, 10 x
Since 1x , is the turning point of the function, it is at the lowest value. So, this point is to be
included. The final range of the decreasing function is
10 x
In Summary:
Students are to be familiar with the monotonicity of basic functions such as
,,log,1
,1
,,,2
32 xaxxx
xxx . Each year, there will be one question on this.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 193
2.
The diagram shows part of a straight line graph drawn to represent the equation bx
ay
,
where a and b are constants. Given that the line passes through (2, 6) and has gradient 3 ,
find the value of a and of b. [4]
Solution :
Given that bx
ay
,
abyyx
abyxy , the line equation
Since the gradient is 3 ,
b3
3b
Sub. (2, 6) into the line equation,
a 236
12a
O
6,2
y
xy
[Analysis]
Observe that the coordinates (2, 6) is for (y, xy). The equation bx
ay
does not match the
line equation in the graph, so it needs to re-write first, xy in terms of y.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 194
Alternative solution:
Given that the line passes through (2, 6) and has gradient 3 ,
32
6
y
xy
636 yxy
123 yxy
123 xy
3
12
xy
Compare with bx
ay
, we get
12a and 3b
In Summary:
Be very careful with the handling of the coordinate (y, xy). There is little
difficulty with the coordinate geometry and re-writing equation.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 195
3. Without using a calculator, find the value of x6 , given that xx 22 123 . [4]
Solution :
xx 22 123
x
x
12
1239
2
16123 xx
1636 x
1662 x
166 x
46 x (Rejected 46 x as 06 x )
Alternative solution:
xx 22 123
12lg23lg2 xx
12lg12lg23lg23lg xx
3lg12lg212lg3lg x
4lg236lg x
36lg
4lg2x
6lg
4lgx
4log6x
46 x
[Analysis]
This question is asking for the value of x6 , not the value of x. The first attempt is to transform
the equation into x6 term, by applying laws of indices.
In Summary:
Log and Indices are two of a kind. These topics are a regular set in this exam.
Be very competent with Log and Indices laws.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 196
4. Solve the equation cos542cos2 for 20 , giving your answers, in radians,
correct to 2 decimal places. [5]
Solution :
Given that cos542cos2 ,
cos5422cos4 2
06cos52cos4 2
02cos3cos4 .
03cos4 or 02cos
4
3cos 2cos , Rejected as 1cos
Principal value, 418.2
418.2 or 418.2
418.22
864.3
42.2 (2 d.p.) 86.3 (2 d.p.)
Alternative solution:
Given that cos542cos2 ,
06cos52cos4 2
02cos3cos4 .
4
3cos 2cos , Rejected as 1cos
Basic angle = 722.0
722.0 or 722.0
418.2 863.3
42.2 (2 d.p.) 86.3 (2 d.p.)
[Analysis]
Changing the double angle with 1cos22cos 2 . Take note that the answer is rounded to
2 decimal places.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 197
5. The variable x and y increase in such a way that, when 2x , the rate of increase of x with
respect to time is twice the rate of increase of y with respect to time. Given that
14 xky , where k is a constant, find the value of k. [5]
Solution:
Given that 14 xky ,
4142
1
d
d2
1
xk
x
y
14
2
d
d
x
k
x
y
When 2x , t
y
t
x
d
d2
d
d
124
2
d
d
k
x
y
3
2
d
d k
x
y
t
x
x
y
t
y
d
d
d
d
d
d
t
xk
t
y
d
d
3
2
d
d
t
yk
t
y
d
d2
3
2
d
d
4
3k
[Analysis]
Rate of change t
x
x
y
t
y
d
d
d
d
d
d . Given that
t
y
t
x
d
d2
d
d , need to find the value of
x
y
d
d when
2x .
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 198
Alternative solution:
When 2x , t
y
t
x
d
d2
d
d
t
x
t
y
d
d
2
1
d
d
Therefore,
2
1
d
d
x
y
Given that 14 xky ,
4142
1
d
d2
1
xk
x
y
14
2
d
d
x
k
x
y
When 2x ,
124
2
d
d
k
x
y
3
2
d
d k
x
y
3
2
2
1 k
4
3k
In Summary:
Rate of change has been featured regular since year 2009. It is usually simple
and direct.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 199
x
x
y
6. A solid rectangular block has a square base of side x cm and a height of y cm. The total
surface area of the rectangular block is 120 cm2 and the total length of the 12 edges is 54 cm.
Show that 02092 xx and find the possible values of x and of y. [5]
Solution:
25448
112042 2
yx
xyx
From (2), we get
34272 xy
From (1), we get
120222 2 yxx
Subst (3) into the above equation,
12042722 2 xxx
0120546 2 xx
02092 xx
054 xx
4x or 5x
When 4x , 5.5y .
When 5x , 5.3y .
In Summary:
This type of question was usually set in E-math paper in the past.
[Analysis]
Sketch a cuboid to assist in formulating the two equations: surface areas and edges.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 200
7. The coefficient of x in the expansion of 6132 axx is 3.
(i) Find the value of the constant a. [4]
(ii) Using your value of a, find the coefficient of x2 in the expansion of 6132 axx .
[2]
Solution:
(i)
26
!2
56611 axaxax
221561 xaax
226156132132 xaaxxaxx
2222 1561315612 xaaxxxaax
22 18303122 xaaxa
For the x term,
3123 a
2
1a
(ii)
For the x2 term, the coefficient is
aa 1830 2
2
18
4
30
2
3
[Analysis]
This is a question can be solved with expansion of the first three terms of 61 ax or by the
general term, r + 1 term expression.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 201
Alternative solution:
(i) For 61 ax , the general term expression is rr axC6 , where 6,,2,1,0 r
rr
r
r axCxaxCaxx 66
632132 , where 6,,2,1,0 r
For the x term, the coefficient is
3320
06
1
16 aCaC
3312 a
2
1a
(ii)
For the x2 term, the coefficient is
116
2
26 32 aCaC
1263152 aa
aa 1830 2
2
18
4
30
2
3
In Summary:
This question is very straight forward. This topic is a regular visitor in this
exam.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 202
8.
The diagram shows part of the graph of 123 xy .
(i) Find the coordinates of the points A, B and C. [3]
(ii) Solve the equation xx 123 . [3]
Solution:
(i) Given that 123 xy , when 0y
1230 x
123 x
When 012 x , When 012 x ,
123 x 123 x
1x 2x
0,1C 0,2A
When 012 x , 2
1x , 3y
3,
2
1B
A
O
B
x
y
C
[Analysis]
(i) A and C are the x-axis intercepts where 0y . B is the turning point of the linear
modulus where 012 x .
(ii) Need to check for the validity of the solutions.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 203
(ii)
xx 123
123 xx
When 012 x , When 012 x ,
123 xx 123 xx
3
2x 4x
Alternative solution:
(i) Given that 123 xy , when 0y
1230 x
123 x
2123 x
22 123 x
223120 x
22420 xx
2x or 1x
0,2A 0,1C
When 012 x , 2
1x , 3y
3,
2
1B
(ii)
xx 123
123 xx
22123 xx
012322 xx
0324 xx
4x 3
2x
In Summary:
It is good to know solve linear modulus equation and the related properties in
graph.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 204
9. The equation of a curve is baxy 3 where a and b are constants. The equation of the
normal to the curve at the point where 1x is 1225 xy . Find the value of a and of b.
[6]
Solution:
Given that 1225 xy , when 1x , 2y
12 ba
5
12
5
2 xy
The tangent to the curve at 1x , 2y , is 2
5
d
d
x
y
the curve is baxy 3
23d
dax
x
y
When 1x ,
ax
y3
d
d
a32
5
6
5a
Subst, 6
5a into (1),
b6
52
6
7b
In Summary:
Very simple application of differentiation on normal and tangent.
[Analysis]
This equation 1225 xy is normal to the curve. The tangent on the curve passes through
the same point on the normal and perpendicular to this normal.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 205
10.
The diagram shows a rectangle ABCD in which 322 AB cm. Given that the area of
the rectangle is 39 cm2, find, without using a calculator,
(i) the exact value of BC in the form ba cm, where a and b are integers, [4]
(ii) the exact value of (AC)2 in the form dc cm
2, where c and d are integers. [3]
Solution:
(i) 69322 BC
69322322322 BC
3521538 BC
32355 BC
323 BC
318 BC
(ii) By Pythagoras theorem,
222323322 AC
62322
AC
24322
AC
A B
C D
322 cm
[Analysis]
This is about surd conjugates, simplifying surd. As for part (ii), we may try the Pythagoras
theorem for the right triangle ABC.
In Summary:
Simplifying surds and solving surd equation are a must.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 206
11.
The diagram shows a rhombus ABCD in which the coordinates of the points A and C are
2,2 and 2,2 respectively. Given that the point B lies on the y-axis, find
(i) the coordinates of B and of D. [6]
(ii) the area of the rhombus. [2]
Solution:
(i) Let the mid-point of the two diagonals be M,
2
82,
2
102M
5,4M
The gradient of AC
210
28
2
1
The gradient of BD 2
A 2,2
O
B
x
y
C 8,10
D
[Analysis]
The property of a rhombus is such that the two diagonals are perpendicular to each other. The
two diagonals bisect each other.
A 2,2
O
B
x
y
C 8,10
D
M
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 207
Let ByB ,0 ,
240
5
By 13By
13,0B
Let DD yxD , , 5,4M
42
0
Dx 5
2
13
Dy
8Dx 3Dy
3,8 D
(ii) the area of the rhombus,
BDAC2
1
2222 1681262
1
22 2182162
1
5862
1
120 cm2
Alternative solution:
(ii) the area of the rhombus,
238132
281002
2
1
6641301630262
1
200402
1
120
the area of the rhombus is 120 cm2
In Summary:
A simple routine question on coordinate geometry. Familiarity with
parallelogram is most useful.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 208
12. A particle P leaves a fixed point O and moves in a straight line so that, t s after leaving O, its
displacement, s m, from O is given by
ttts 1ln
Find, when 20t ,
(i) the displacement of P from O, [1]
(ii) the velocity of P, [4]
(iii) the acceleration of P. [4]
Solution:
(i) Given that the displacement ttts 1ln ,
When 20t ,
20120ln20 s
2021ln20 s
890.40s
9.40s m (3 s.f.)
(ii) As velocity, t
sv
d
d ,
11
11ln
d
d
t
tt
t
s
1
11ln
ttv
When 20t ,
21
121ln v
9969.2v
00.3v m-1
s (3 s.f.)
[Analysis]
This question asks very routine questions about derivation from displacement to velocity and
then acceleration.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 209
(iii) the acceleration t
va
d
d
21
1
1
1
d
d
ttt
v
21
2
t
ta
When 20t ,
221
22a
221
22a
04988.0a
0499.0a m-2
s (3 s.f.)
In Summary:
Be familiar with the chain rule, product rule and quotient rule of
differentiation.
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 210
13.
The diagram shows a rod OA, which is hinged at O, and a rod AB, which is hinged at A.
The rods can move in the xy-plane with origin O where the x and y axes are horizontal and
vertical respectively. The rod OA can turn about O and is inclined at an angle to the
y-axis, where 900 . The rod AB can turn about A in such a way that its inclination to
the horizontal is also . The lengths of OA and AB are 5 m and 3 m respectively.
Given that B is h m from the x-axis,
(i) find the values of the integers a and b for which
sincos bah . [2]
Using the values of a and b found in part (i),
(ii) express h in the form sinR , where 0R and 900 . [4]
Hence
(iii) state the maximum value of h and find the corresponding value of , [3]
(iv) find the value of when 17h . [2]
Solution:
A
O
B
x
y
3 m
5 m
h m
[Analysis]
Reading the statements in this question maybe a little confusing at first. Even with the aide of
the diagram, it is possible to answer part (i).
Solutions to O Level Add Math paper 1 2011
By KL Ang, Jun 2013 Page 211
(i) Refer to the diagram,
sin3cos5 h
Therefore,
5a , 3b
(ii) cossinsincossin RRR
3cos
5sin
R
R
3
5tan
036.59
3
5tan 1
222 35 R
34R
0.59sin34 h
(iii) h is maximum when 10.59sin .
10.59sin
1sin0.59 1
900.59
0.31
Therefore 34h when 0.31 .
(iv) When 17h ,
0.59sin3417
2
10.59sin
2
1sin0.59 1
Principal value, 450.59
450.59 or 451800.59
14 rejected 76
A
O
B
x
y
3 m
5 m
h m
sin3
cos5
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 212
1. The equation of a curve is kkxxy 62 2 , where k is a constant.
(i) Find the range of values of k for which the curve lies completely above the x-axis.
[4]
(ii) In the case where 2k , find the values of m for which the line 4mxy is a
tangent to the curve. [4]
Solution :
(i) The curve is to hang completely above x-axis.
That is kkxx 620 2 has no real root.
It is necessary and sufficient that the discriminant
of the equation kkxx 620 2 is negative.
06242
kk
04882 kk
0412 kk
From the sketch,
412 k
(ii) When 2k , kkxxy 62 2 has no real root.
4
62 2
mxy
kkxxy has repeat roots.
4224 2 xxmx
8220 2 xmx
The discriminant,
824202 m
22820 m
6100 mm
10m or 6m
[Analysis]
This question is on nature of roots, in relation to quadratic equation. A sketch of its graph
would have made the problem a lot clearer. Need to know how to answer quadratic
inequality. Part (ii) is about a curve tangent to a line, or a repeated root where discriminant is
zero.
x
y
O
y
k O
x
y
O 4
4 mxy
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 213
Alternative solution:
(i) Given that the quadratic curve is above x-axis, the minimum value of y must be positive.
ymin is located on the line of symmetry.
Line of symmetry 422
kkx
kk
kk
y
6
442
2
min
68
2
min kk
y
Since ymin > 0,
068
2
kk
04882 kk
0412 kk
From the sketch,
412 k
(ii) Given that the line 4 mxy is tangent to the curve kkxxy 62 2 , when 2k ,
the gradient at the point of tangent must be the same.
24d
d x
x
y , must be the same as the gradient of the line, m.
24 xm
422424 2 xxxx
042 x
2x or 2x
When 2x , 6224 m
When 2x , 10224 m
In Summary:
The topic on Quadratics remains the most important content to be mastered.
k O
x
y
O
ymin
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 214
2. The function f is given by bxax tan:f , where a and b are positive integers and
22
x .
(i) Given that 0f x when 2
x , find the smallest possible value of b. [1]
(ii) Using the value of b found in part (i) and given that the gradient of the graph of xy f
is 12 at the point where 8
x , find the value of a. [3]
(iii) Sketch the graph of xy f . [3]
Solution :
(i) Refer to the sketch, the period of bxay tan is 2
.
2
b
2b
(ii) When 8
x , 12
d
d
x
y
xax
y2sec2
d
d 2
82sec212 2
a
4sec6 2
a
4cos6 2
a 2
1
4cos
3a
(iii) as shown above.
[Analysis]
Tangent curve might not be familiar to many students. A sketch of the curve would be useful.
Part (i), the period of tan x is 180 . Part (ii) is about differentiation.
4
O
y
x
4
2
2
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 215
3. The expression baxx 32 , where a and b are constants, has a factor of 2x and leaves a
remainder of 35 when divided by 3x .
(i) Find the value of a and of b. [4]
(ii) Using the values of a and b found in part (i), solve the equation 02 3 baxx ,
expressing non-integer roots in the form 2
dc , where c and d are integers. [4]
Solution :
(i) Let baxxx 32p
By Factor Theorem, 02p ,
ba 2222p3
12160 ba
By Remainder Theorem, 353p ,
ba 3323p3
ba 35435
23190 ba
12 ,
7a
Sub. 7a into (1),
2b
272p 3 xxx
(ii) 0272 3 xx
By long division,
1422272 23 xxxxx
01422 2 xxx
2x or
2
22
22
124164
x
[Analysis]
This is a typical Factor and Remainder Theorems question. Part (ii) simply asks for
application of quadratic formula or completing square after factorised x+2.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 216
4. The roots of the quadratic equation 0542 2 xx are and .
(i) Show that 5
222
. [4]
(ii) Given that the roots of 02 baxx are 2
and 2
, find the value of a and
of b, where a and b are constants. [4]
Solution :
(i) Given that 0542 2 xx ,
22
4
2
5
12
5242
222
5
2
2
5
122
(ii)
5
184
5
2422
22
5
212524122
22
05
21
5
182 xx
[Analysis]
Recognize that this is a Sum-of-roots, Product-of-roots question, a
b and
a
c of
a quadratic equation 02 cbxax .
In Summary:
A routine question, Be competent in handling the algebraic indentity.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 217
5. (a) Solve the equation xx 273 log4log .
(b) The curve naxy , where a and n are constants, passes through the points (2, 40),
(3, 135) and (4, k). Find the values of n, a and k.
Solution:
(a) Given
xx 273 log4log
27log
log4log
3
33
xx
3
3
333
3log
log3log4log
xx
3log3
log3loglog
3
34
33
xx
3
log3loglog 34
33
xx
4
33
3 3log3
loglog
xx
4
33 3loglog3
2x
4
33 3log2
3log x
2
34
33 3loglog
x
6
33 3loglog x
63x
729x
[Analysis]
Part (a) requires base change formula. Part (b) simply find the values of the unknowns.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 218
(b)
When 2x , 40y , na240
When 3x , 135y , na3135
n
n
3
2
135
40
n
3
2
27
8
n
3
2
3
23
3
n
3
2
3
23
3n
therefore 3240 a
5a
When 4x , ky , 345k
320k
In Summary:
Application of Log and Indices laws are key elements in this paper.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 219
6.
A garden is being designed to include a semicircular pond and a lawn. The radius of the pond
is r m and the length of the lawn, which is in the shape of a rectangle with the semicircle
removed, is l m as shown in the diagram above.
(i) Given that the area of the lawn is 400 m2, express l in terms of r. [2]
(ii) Given that the perimeter of the lawn is P m, show that r
rP400
22
3
. [2]
(iii) Given that r and l can vary, find the value of r for which P has a stationary value and
determine whether this value of P is a maximum or a minimum. [5]
Solution:
(i) Area of the lawn = 400 m2,
4002
12 2 rrl
rr
l 4
1
2
400
2r m
l m
r m
POND
LAWN
[Analysis]
Part (i) is to form an equation with the area. Part (ii) is to form an equation with the perimeter,
and then replace l from part (i) to obtain the desired expression. Part (iii) is to differentiate P,
then set it to zero to locate the turning points.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 220
(ii)
rlrP 22
rrr
rP
42
40022
r
rP400
22
3
(iii)
2
4002
2
3
d
d
rr
P
When 0d
d
r
P,
2
4002
2
30
r
2
434002
r
43
8002
r
43
800
r take 0r only.
72.7r m (3 s.f.)
0800
d
d32
2
rr
P the turning point is a minimum.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 221
7.
The diagram shows two intersecting circles C1 and C2, with centres P and Q respectively.
The point R lies on both circles and the line PR is a tangent to C2. A line L passes through Q.
The point E lies on C2. The line PE is perpendicular to L and intersects C2 at A, C1 at B and
L at D.
(i) By considering triangles QAD and QED show that AD = ED. [4]
(ii) Show that PEPAADPD 22 . [3]
(iii) Hence show that 222 PBADPD . [2]
Solution:
(i) Consider QDA and QDE
EQAQ (radius of circle C2)
QDQD (common side)
90QDEQDA
Therefore, QDA QDE (RHS)
Therefore, AD = ED
A
D
L E
B
C1
R
P
Q
C2
[Analysis]
Part (i) requires proof of congruent triangles. Part (ii) applies algebraic identity and finally part
(iii) uses tangent-secant theorem.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 222
(ii)
22 ADPD
ADPDADPD
PADEPD DEAD , from part (i)
PAPE
(iii)
By Tangent-Secant Theorem,
2PRPAPE
222 PRADPD PEPAADPD 22
222 PBADPD PBPR , radius of circle C1
In Summary:
This geometrical proof is the easiest since 2009.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 223
8. (i) Express 22
12
138
xx
xx in partial fractions. [5]
(ii) Hence find
x
xx
xxd
12
1382
2
. [5]
Solution:
(i) Let
22
2
121212
138
x
C
x
B
x
A
xx
xx
CxxxBxAxx 121213822
When 0x ,
A1
CxxxBxxx 121213822
CxxxBxxx 121213822
CxxxBxx 124 2
When 1x ,
)1(33 CB
When 1x ,
)2(5 CB
(1)+(2),
B28
B4
9C
22
2
12
9
12
41
12
138
xxxxx
xx
[Analysis]
First recongise that the numerator is of degree 2 and the denominator is of degree 3. Therefore
this is a proper fraction. Again, observe that 2x + 1 is a repeated factor. Apply
'fln f
f
xdx x c
x
in part (ii)
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 224
(ii)
x
xx
xxd
12
1382
2
xxxx
d12
9
12
412
cx
xx
12
1
2
912ln2ln
In Summary:
This s a typical question that combining partial fraction with integration.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 225
9.
The diagram shows part of the curve 2
sin3x
y . The line 3
4x meets the curve at P and
the x-axis at Q. The tangent to the curve at P meets the x-axis at R. Find
(i) the length of QR.
(ii) the area of the shaded region.
Solution:
(i)
3
2sin3,
3
4 P for
2
3
3sin
3
2sin
2
33,
3
4P
x
P
Q R O
y 3
4x
2sin3
xy
[Analysis]
For part (i), PQ can be found by simply substituting 3
4x into
2sin3
xy . Then by
finding the gradient with 2
cos2
3
d
d x
x
y , then QR is found with similar triangles. Part (ii) is to
find the area under the curve from 0x to 3
4x by integration, then combines with the
area of triangle PQR.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 226
2cos
2
3
d
d x
x
y
When 3
4x ,
3
2cos
2
3
d
d
x
y
4
3
d
d
x
y for
2
1
3
2cos
Therefore,
4
3
QR
PQ
PQQR3
4
32QR units for 2
33PQ
(ii) the area of the shaded region = area under the curve from 0x to 3
4x + area of triangle
PQR.
the area of the shaded region
QRPQxx
2
1d
2sin3
3
4
0
323
4
2
1d
2sin
2
16
3
4
0
xx
3
34
2cos6
3
4
0
x
3
340cos
3
2cos6
3
341
2
16
3.163
349
unit
2 (3 s.f.)
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 227
10. Without using a calculator
(i) find the exact value of 75cos15cos , [3]
(ii) find the exact value of 75cos15cos , [2]
(iii) show that 2
375cos15cos 22 , [2]
(iv) state the acute angle such that 75cossin , [1]
(v) use the results of parts (iii) and (iv) to find the exact value of 15cos2 . [3]
Solution:
(i)
Given that 75cos15cos
30sin45sin2
30sin45sin2 for xx sinsin
2
1
2
12
2
1
2
175cos15cos
(ii)
Given that 75cos15cos
30cos45cos2
30cos45cos2 for xx coscos
2
3
2
12
[Analysis]
This question examines trigo identities and ratios of special angles.
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 228
2
3
2
3
(iii)
Given that 75cos15cos 22
75cos15cos75cos15cos
2
1
2
3
2
3
(iv)
Given that 75cossin
15sinsin
15 or 165 (rejected)
(v)
75cos2
315cos 22
275cos2
3
215sin2
3
2
315sin15cos 22
2
3115cos2 2 for xx 22 cos1sin
2
1
4
315cos2
4
2315cos2
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 229
11. A circle, centre C, has a diameter AB where A is the point (1, 1) and B is the point (7, 9).
(i) Find the coordinates of C and the radius of the circle. [4]
(ii) Find the equation of the circle. [2]
(iii) Show that the equation of the tangent to the circle at B is 5734 xy . [3]
The lowest point on the circle is D.
(iv) Explain why D lies on the x-axis. [1]
(v) Find the coordinates of the point at which the tangents to the circle at B and D intersect.
[1]
Solution:
(i)
5,42
91,
2
71CC
coordinates of 5,4C
(ii)
106436191722
AB
255422 yx
(iii)
Gradient of AB, 3
4
17
19
ABm
Gradient perpendicular of AB, 4
3ABm
[Analysis]
Part (i) makes use of midpoint formula; part (ii) first find the length AB; part (iii) is to first find
the gradient AB then forming the equation of tangent; Part (iv) is to related the centre of the
circle, its radius and x-axis; Part (v) is to set 0y .
Solutions to O Level Add Math paper 2 2011
By KL Ang, Jun 2013 Page 230
The equation of the tangent at B,
74
39 xy
5734 xy
(iv)
For point 5,4C , and radius of 5 units, the circle touches x-axis. So, the lowest point, D on
the circle is also tangent to the x-axis.
(v)
The tangent of the circle at D is the x-axis ,i.e. 0y .
When 0y ,
57304 x
19x
coordinates of the point of intersection 0,19
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 231
1. The equation of a curve is 3
12
x
xy . Given that x is changing at a constant rate of
0.4 units per second, find the rate of change of y when 2x . [4]
Solution :
Given that3
12
x
xy for 3x ,
23
1223
d
d
x
xx
x
y
23
5
d
d
xx
y
Let 2x , 2.0d
d
x
y
08.04.02.0d
d
t
y units per second
Alternative solution:
Given that3
12
x
xy for 3x ,
3
532
3
562
x
x
x
xy
3
52
xy
[Analysis]
Recognise that the question is about rate of change, i.e. t
x
x
y
t
y
d
d
d
d
d
d . Given that 4.0
d
d
t
x, to find
t
y
d
d at 2x . To differentiate a function of the form
vy
u , we apply quotient rule, that is
2
d
d
d
d
d
d
v
x
vu
x
uv
x
y
. Take note that the function implies that 3x .
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 232
23
5
d
d
xx
y
Let 2x , 2.0d
d
x
y
08.04.02.0d
d
t
y units per second
In Summary:
Simple rate of change question. The alternative solution by-passes the quotient
rule, after reducing the improper fraction to proper.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 233
2.
The diagram shows a triangle ABC in which angle A is 6
radians, angle B is a right angle,
M is the mid-point of AB and the length of AC is 8 m. Without using a calculator, find the
value of the integer k such that angle
14sin 1 k
ACM . [5]
Solution :
Since ABC is a 30°-60°-90° triangle, by similar triangles, we have
ABCABC
321 .
Given that 8CA m, then, 4BC m and 34AB m.
Since M is midpoint on AB, then 32 MBAM m.
By Pythagoras theorem on triangle MBC, we get
2843222
MC m.
[Analysis]
Observe that the triangle ABC is a 30°-60°-90°. 82
130sin
AB . Rewrite
angle
14sin 1 k
ACM as 14
sink
ACM . With right-angled triangle, sine ratio or sine rule are
likely to be productive. 0k , since k . Take note that “Without using a calculator” in the
question which implies approximated values from calculator are inadmissible.
M A B
C
8 m
6
rad
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 234
Applying Sine rule on triangle ACM, we get
AM
ACM
CM
sin6sin
AMCM
ACM
6
sin
sin
28
332
28
2
1
sin ACM
142
3
14
k
21k
21k
Alternative solution:
Since ABC is a 30°-60°-90° triangle, , we have
2
1
86sin
BC and
2
3
86cos
AB
Therefore, 4BC m and 34AB m.
Since M is midpoint on AB, then 32 MBAM m.
By Pythagoras theorem on triangle MBC, we get
2843222
MC m.
MCBACM 3
MCBACM
3sinsin
MCBMCBk
sin
3coscos
3sin
14
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 235
Since 2
3
3sin
,
2
1
3cos
,
28
32sin
MC
MBMCB ,
28
4cos
MC
BCMCB , we get
28
32
2
1
28
4
2
3
14
k
28
3
28
32
14
k
28
3
14
k
142
3
14
k
21k
21k
In Summary:
Once a student is able to recognize the special 30°-60°-90° triangle and able to
recall the trigo ratios of this triangle, the question becomes very easy.
In the alternate solution, we can make use of the trigo identity of sin (A – B) to find the
answer.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 236
3. (i) Find xxx
lnd
d 2 . [2]
(ii) Hence find xxx dln [3]
Solution :
(i) By product rule,
xxxxxx
x
1ln2ln
d
d 22
xxx ln2
(ii) From part (i), we have
xxxxxx
ln2lnd
d 2
xxxxxxxx
dln2dlnd
d 2
xxxxxxx ddln2ln2
cx
xxxxx 2dln2ln
22
cx
xxxxx 2lndln2
22
Cxxx
xxx 42
lndln
22
, where C is an arbitrary constant.
[Analysis]
This question is essentially the same as question 4 in Paper 1, 2008, (4018). Part (i) requires
product rule to differentiate. Part (ii) is to make use of the result of part (i) to integrate.
In Summary:
Very simple, direct question. Log and Exponential operations are very
important topics in this paper. These topics are a regular set in this exam. Be very
competent with Log and Indices laws.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 237
4. Find the inverse of the matrix
1753 and hence solve the simultaneous equations
.0235,0117
baba [5]
Solution :
Given that
1753M , the determinant, 38353
1753
3751
38
11M
Given that
02350117
baba ,
117253
abab
112
1753
ab
112
3751
38
1ab
1957
38
1ab
2
12
3
ab
5.1b or 5.0a
[Analysis]
There are two parts to this question, firstly is to find the inverse matrix of M, M–1
. M × M–1
= I,
where I is the identity matrix. Need to find determinant first before formulating the M–1
. Secondly
is to apply the inverse matrix to solve for the value of a and of b.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 238
Alternative solution:
Given that
1753M , let
srqp1M
M × M–1
= I
1001
1753
srqp
153 rp , 053 sq ,
07 rp , 17 sq
0535
153rp
rp
5535053sq
sq
138 p 38
5q
38
1p
38
3r
38
7r
38
3
38
738
5
38
1
M 1-
In Summary:
The first time a Matrix question is set on a 4038 paper. Students may not be
familiar with the way to find the inverse Matrix. The alternative should be a useful
one.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 239
5. In a chemical reaction, a solid substance slowly dissolves in a liquid. At the start of the
reaction, there are M grams of the substance. After t minutes, there are x grams of the
substance still to be dissolved. It is known that x and t are related by the equation ktMx e ,
where k is a constant. The table below shows measured values of x and t.
t (minutes) 1 2 3 4 5
x (grams) 6.55 5.36 4.40 3.60 2.94
(i) Plot xln against t and draw a straight line graph. [2]
(ii) Us your graph to estimate the value of M. [2]
(iii) Estimate the time taken for half of the substance to be dissolved. [2]
Solution:
(i) Given that ktMx e ,
ktMx elnln
ktMx elnlnln
ktMx lnln
[Analysis]
Wow, it is just like in an exercise. Put you practice in full benefit to pick up the 6 marks. Part (i)
hints that taking xln will be the place to start.
1.2
1.4
1.6
1.8
2.0
t
1
3 4 5
•
•
•
•
•
O 1 2
lnx
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 240
t (minutes) 1 2 3 4 5
x (grams) 6.55 5.36 4.40 3.60 2.94
lnx 1.879 1.679 1.482 1.281 1.078
(ii) From the graph, the lnx-axis intercept is 2.08.
08.2ln M
08.2eM
00.8M grams
(iii) 15
88.108.1
k
4
8.0k
2.0k
tx 2.008.2ln
2
Mx
ktMM
ln2
ln
ktMM ln2lnln
47.32.0
2lnt min
In Summary:
Direct copy from textbook.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 241
6. (i) Express tansec2sin2 as a quadratic expression in sin . [3]
(ii) Use your answer to part (i) to find, for 20 , the exact solutions of the equation
03tansec2sin2 [3]
Solution:
(i) Given that,
tansec2sin2
cos
sin
cos
1cossin22
x
cos
sin1cossin22
2sin4sin4
(ii) Given that 03tansec2sin2
03sin4sin4 2
03sin4sin4 2
01sin23sin2
2
3sin or
2
1sin
(rejected) 6
11
62
,
6
7
6
In Summary:
Be careful that the question asks for EXACT solutions. Another very direct
question.
[Analysis]
Part (i) requires a quadratic expression in sin , so apply double angle to 2sin and rewriting
sec and tan in sin and cos .
Part (ii) is to use part (i) to solve the given equation. 1sin1 and 20 .
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 242
7. In the expansion of 62 pxqx , the coefficient of 6x is 7 and there is no term in 5x .
Given that 0p , find the values of the constants p and q. [6]
Solution:
2415066
26
16
06 pxpxpxpx
4256 156 xppxx
4256615622 xppxxqxpxqx
42564256 1561562 xppxxqxppxxx
5267 630122 xpqpxqpx
For the x6 term,
qp 127 …….. (1)
For the x5 term,
pqp 6300 2
qpp 560
0p , qp 50 …….. (2)
(1) – (2), p77
1p
Subst 1p into (1),
5q
Alternative solution:
For 6px , the general term expression is rr pxr
66
, where 6,,2,1,0 r
rrrr pxr
qpxr
xpxqx
666 6622 , where 6,,2,1,0 r
[Analysis]
For all binomial expansion, extreme care is usually in order, as an expansion can be messy.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 243
rrrr pxr
qpxr
pxqx
676 6622 , where 6,,2,1,0 r
For the 6x term, the coefficient is
706
162 01
pqp
712 qp …….. (1)
For the 5x term, the coefficient is
016
262 12
pqp
06152 2 qpp
056 qpp
0p , qp 50 …….. (2)
(1) – (2), p77
1p
Subst 1p into (1),
5q
In Summary:
An usual suspect in this paper every year. Be careful in handling the
expansion.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 244
8. The function f is defined, for all values of x, by
cx
x
2cos2f ,
where c is a constant. The graph of xy f passes through the point
2,
3
2.
(i) Find the value of c. [2]
(ii) State the amplitude and period of f. [2]
(iii) Sketch the graph of xy f for 20 x . [2]
Solution:
(i) Given that cx
y
2cos2 , when
3
2x , 2y
c
3
2
2
1cos22
c
3cos22
Since 2
1
3cos
,
c
2
122
1c
[Analysis]
(i) Substitute the coordinate
2,
3
2 into the function to determine c.
(ii) Need to state Amplitude, a, which is the coefficient of the cosine and period as 2π/b
where bxacos .
(iii) Remember the domain, 20 x , end points, intercepts and shape.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 245
(ii)
Amplitude = 2 units
Period = 422
(iii)
In Summary:
A direct curve sketching question.
O
3
x
y
1.16π
–1
2π π
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 246
9. A particle moves in a straight line, so that, t seconds after leaving a fixed point O, its
velocity v ms –1
, is given by
31
122
tv . Find
(i) an expression for the acceleration of the particle in terms of t, [2]
(ii) the distance travelled by the particle before it comes to instantaneous rest. [5]
Solution:
(i)
Given that
31
122
tv ,
31122
d
d t
t
v
21
24
d
d
tt
v
(ii) When v = 0 ,
31
120
2
t
21
41
t
412t
21 t
1t or 3t (rejected)
Distance = 1
0
dtv
[Analysis]
(i) Acceleration = t
v
d
d.
(ii) Distance = tv d .
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 247
dtt
1
0
23
1
12
1
0
1
0
2d3d
1
12tt
t
1
0
1
0
2d3d
1
112 tt
t
013
0
1
1
112 t
t
01312
112
36
3 m
In Summary:
Very simple application of differentiation and integration.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 248
10. A curve has the equation cxxxy 243 23 , where c is a constant.
(i) Find the x-coordinates of the two stationary points on the curve. [4]
(ii) Hence find the value of c for which the minimum point of the curve lies on the x-axis.
[3]
Solution:
(i) Given that cxxxy 243 23 ,
2463d
d 2 xxx
y
0d
d
x
y, 24630 2 xx
820 2 xx
024 xx
4x or 2x
(ii) To determine the nature of stationary points,
66d
d2
2
xx
y
when 4x , 0d
d2
2
x
y, minimum point.
when 2x , 0d
d2
2
x
y, maximum point.
So, when 4x , 0y
c 424434023
80c
[Analysis]
(i) stationary points are points with zero gradient, i.e. 0d
d
x
y.
(ii) use the result in part (i) to find c when the minimum point touches the x-axis.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 249
Alternative solution:
(ii) A sketch of the curve cxxxy 243 23 ,
So, when 4x , 0y
c 424434023
80c
Alternative solution:
(ii) A sketch of the curve cxxxy 243 23 ,
4x is a repeated root. Let the other root be k.
kxxcxxx 223 4243
kxkxkxcxxx 168168243 2323
5k
80165 c
In Summary:
A straight forward question on stationary points.
O
c
x
y
–2 4
O
c
x
y
–2 4 k
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 250
11. Given that the roots of 052162 xx are 3 and 3 , find a quadratic equation
whose roots are and . [7]
Solution:
Let and being the roots of quadratic equation 02 cbxax , then a
b and
a
c where 0a .
1633 5233
164 523103 22
4 52103 22
4a
b 521023
2
ab 4 5210243 2
44
1
1a
c
ac
therefore, 02 cbxax
042 aaxax
0142 xx
In Summary:
Although the question is set in reverse order to the exercises, the approach in
solving it is no different from these exercises.
[Analysis]
Usually this type of question is that given and being the roots of quadratic equation
02 cbxax , then a
b and
a
c where 0a .
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 251
12. Without using a calculator, find
(i) the value of x such that x
x
27
139
3 , [4]
(ii) the values of the integers a and b such that 32
325
325
3
ba . [4]
Solution:
(i) Given that x
x
27
139
3 ,
xx 2733232
xx 3223 33
xx 346 33
xx 346
9
4x
(ii) 32
325
325
3
ba,
32
3332
325
3
ba
3232
32331
325
3
ba
34
3332361
325
3
ba
331325
3
ba
325343 ba
35386203 ba
33143 ba
14a and 3b
[Analysis]
This question asks indices laws and surds.
Solutions to O Level Add Math paper 1 2012
By KL Ang, Jun 2013 Page 252
13. A circle, whose equation is 03758622 yxyx , has centre C and radius r.
(i) Find the coordinates of C and the value of r. [3]
The point 12,9P lies on the circle.
(ii) Show that CP passes through the origin O. [2]
(iii) Find the equation of the circle for which OP is a diameter. [3]
Solution:
(i) Given the equation of circle, 03758622 yxyx
The centre of this circle, 4,3 C
The radius of this circle, 2043375 22 r units
(ii) 12,9P , 4,3 C
39
412
3
4
x
y
12
16
3
4
x
y
3
4
3
4
x
y
124123 xy
xy3
4 , equation of CP
The line CP intercept the y-axis at the origin.
(iii) The mid-point of OP,
6,
2
9
2
12,
2
9. length OP
2
156
2
9 2
2
2
2
2
2
156
2
9
yx
[Analysis]
This question asks very routine questions about circle.
In Summary:
No surprises. Routine question.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 253
1. The equation of a curve is xxy cos3sin2 for x0 .
(i) Write down expression for x
y
d
d and
2
2
d
d
x
y. [3]
(ii) Find the value of x for which the curve has a stationary point. [2]
(iii) Determine the nature of this stationary point. [2]
Solution :
(i) Given that xxy cos3sin2 ,
xxx
ysin3cos2
d
d
xxx
ycos3sin2
d
d2
2
(ii) When 0d
d
x
y,
xx sin3cos20
3
2tan x
55.2x rad or 70.5x rad (rejected)
(iii) When 55.2x rad,
55.2cos355.2sin2d
d2
2
x
y < 0, hence a maximum point.
[Analysis]
This question is on differentiation and nature of turning points.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 254
Alternative solution:
(i) Given that xxy cos3sin2 ,
xxy cos
13
3sin
13
213
sincoscossin13 xxy where 2
3tan
xy sin13 983.0 rad
xx
ycos13
d
d
xx
ysin13
d
d2
2
(ii) When 0d
d
x
y,
xcos130
2
x or
2
3 x
55.2x rad 70.5x rad (rejected)
(iii) When 55.2x rad,
2sin13
d
d2
2
x
y < 0, hence a maximum point.
In Summary:
Be very careful in handling the radian. The actual value of 2nd
derivative is not
as important, as we just need to know the sign of it only.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 255
2. (i) Given that xBxAxx coscoscos3cos , state the value of A and of B. [1]
(ii) Hence, or otherwise, solve the equation 0cos23cos xx for 1800 x . [5]
Solution :
(i)
2
3cos
2
3cos2cos3cos
xxxxxx
xxxx cos2cos2cos3cos
Hence, 2A , 2B
(ii) Given that 0cos23cos xx for 1800 x , so 36020 x
0coscos2cos2 xxx
012cos2cos xx
12cos2 x or 0cos x
2
12cos x 90x
principle angle, 1202x
240,1202x
120,60x
Alternative solution:
(i) xxxxxxx cossin2sincos2coscos3cos
xxxxx coscossin2cos2cos 2
xxx 2sin212coscos
xxcos2cos2
[Analysis]
Part (i) apply the sum of angle identity. Part (ii) is about solving a trigo equation.
In Summary:
Be very careful in handling equation in part (ii). Take note of the double angle
solution, because the range of 2x is larger.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 256
3. (i) Express 383
202 xx
in partial fractions. [4]
(ii) Find x
xxd
383
202
and hence evaluate xxx
d383
207
22
. [4]
Solution :
(i) Since 313383 2 xxxx ,
let 313383
12
x
B
x
A
xx
when 3x , 10
1
10
1
B ,by cover-up method
when 3
1x ,
10
3
33
1
1
A
3
10
1
13
10
3
383
12
xxxx
3
2
13
6
383
202
xxxx
(ii) x
xxd
383
202
x
xxd
3
2
13
6
x
xx
xd
3
12d
13
32
cxx 3ln213ln2 where c is an integrating constant
[Analysis]
Part (i) is a simple 2 parts decomposition. Part (ii) will need integration into ln function.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 257
xxx
d383
207
22
273ln213ln2 cxx
cc 32ln216ln237ln2121ln2
cc 5ln25ln210ln220ln2
2ln2
39.1 (3 s.f.)
In Summary:
Nothing complicated in this question, except the integration into ln function.
Students are reminded that exponential and log functions are VERY important.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 258
4.
Two circles C1 and C2, intersect at P and Q as shown in the diagram. The tangent to C1 at P
meets C2 at R and the tangent to C2 at P meets C1 at S. Prove that
(i) triangle PQR and SQP are similar, [3]
(ii) Show that 2QPQRQS . [2]
Points A and B lie on the circumferences of C1 and C2 respectively,
(iii) Prove that angle SAP = angle PBR. [3]
Solution :
(i) Consider PQR and SQP
PSQRPQ (Alternate Segment Theorem)
QPSQRP (Alternate Segment Theorem)
Therefore, PQR SQP (AA)
(ii) Since PQR SQP ,
PS
RP
QP
QR
SQ
PQ .
QP
QR
SQ
PQ
QRSQQPPQ 2QPQRQS
[Analysis]
Part (i) needs to find 2 congruent angles in both triangles. Part (ii) can be arranged into length ratio.
Part (iii) may need properties of cyclic quadrilaterals.
S
A
B
C1
R
P
Q
C2
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 259
(iii)
SAPQ and PBRQ are cyclic quadrilaterals.
180SQPSAP 180PQRPBR
PQRPBRSQPSAP
Since PQRSQP , from part (i) SQP PQR , therefore
PBRSAP
In Summary:
Need a keen sense of observation of the given diagram. When part (i) is solved,
the rest of the question becomes straight forward.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 260
5. The equation of a curve is xx BAy ee2 , where A and B are constants. The point P(0, 4)
lies on the curve and the gradient of the tangent to the curve at P is 1 .
(i) Find the value of A and of B. [4]
(ii) Using your values of A and B from part (i), obtain an expression for xy d and hence
evaluate xy d1
0 corrected to one decimal place. [4]
Solution:
(i) Given xx BAy ee2 ,
At P(0, 4),
BA4 --------- (1)
xx BAx
y ee2d
d 2
At P(0, 4), 1d
d
x
y
BA 21 --------- (2)
(1) + (2),
1A
3B
(ii)
xxy e3e2
xy d
xxx de3e2
xx xx de3de2
[Analysis]
Part (i) requires differentiation. Part (ii) simply find the integration and the value of the definite
integration.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 261
xx xx de3d2e
2
1 2
Cxx e3e2
1 2 where C is an arbitrary integrating constant
therefore xy d1
0
CC 0012 e3e
2
1e3e
2
1
CC 0012 e3e
2
1e3e
2
1
32
1e3e
2
1 12
1.5 (1 d.p.)
In Summary:
Application of differentiation and integration of exponential functions.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 262
6. (a) (i) Given that ux 4
3
8 log4log , express u in terms of x. [3]
(ii) Find the value of x for which 4log
1log5log
3
3
8
2
4 xxx . [3]
(b) Solve the equation 152ee yy . [3]
Solution:
(a)
(i) Given that ux 4
3
8 log4log ,
2
2
2
3
2
3
2
2log
log4
2log
log ux
2
log4log 2
2
ux
ux 222 log2log8log2
ux 2
8
2
2
2 log2loglog
ux
2
2
2 log256
log
256
2xu
(ii) Given that 4log
1log5log
3
3
8
2
4 xxx where 0x and ( 0x or 5x )
4log
3log
2log
log
2log
5log
2
2
3
2
3
2
2
2
2
2 xxx
2
3loglog
2
5log 22
2
2
xxx
3loglog5log 2
2
2
2
2 xxx
3log5
log 22
2
2
x
xx
35
2
2
x
xx
22 35 xxx
[Analysis]
(a) Part (i) requires base change formula. Part (ii) simply find the values of the log equation.
(b) to solve a routine exponential equation.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 263
052 2 xx
0x or 5.2x
(rejected)
(b) Given that 152ee yy ,
015e2e2 yy
03e5e yy
3e y or 5e y
(rejected) 61.15ln y (3 s.f.)
In Summary:
Part (a) requires application of log rules, part (b) is just another exercise.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 264
7. (i) Find the value of a and of b for which 232 2 xx is a factor of
bxxaxx 234 32 . [6]
(ii) Using the value of a and b found in part (i), solve the equation
032 234 bxxaxx . [3]
Solution:
(i) Given that 212232 2 xxxx , let bxxaxxx 234 32f .
When 2x , 02f .
ba 2223222f234
ba 280 --------- (1)
When 2
1x , 0
2
1f
.
ba
2
1
2
1
2
13
2
12
2
1f
234
ba 4
3
2
10 --------- (2)
(1) – (2),
a4
5
2
150
a 60
6a
b 6280
4b
[Analysis]
212232 2 xxxx , apply factor theorem to find a and b. Part (ii) factorizes completely to
find the roots.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 265
(ii) Given that 046632 234 xxxx
223246632 22234 kxxxxxxxx
When 1x ,
2123246632 k
133 k
0k
223246632 22234 xxxxxxx
02232 22 xxx
2x or 2
1x or 22 x
or 2x or 2x
In Summary:
Without fail, factor theorem and remainder theorem are regulars in this paper.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 266
8. (i) Find the set of values of m for which the curve 22 xxy and the line 1 mxy do
not intersect. [4]
(ii) Sketch the curve 22 xxy , giving the coordinates of the maximum point and of the
points where the curve meets the x-axis. [3]
(iii) Find the number of solutions of the equation 12 2 mxxx when
(a) 1m , (b) 2
1m [4]
Solution:
(i) Given 22 xxy and 1 mxy ,
221 xxmx
0122 xmx
Discriminant of the equation is to be negative,
0422
m
422m
222 m
40 m
The set of values 40: mm
(ii) Given 22 xxy , when 0y ,
220 xx
xx 20
0x or 2x
Line of symmetry, 12
02
x
The locally maximum value, 112 y
The coordinate of the maximum point is (1, 1)
[Analysis]
(i) Solving simultaneous equations with negative discriminant.
(ii) Sketch the graph by flipping up the negative part to the upper half.
(iii) Add lines to the sketch in part (ii) to determine the number of solutions.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 267
(a) 1m , 1 xy (b) 2
1m , 1
2
1 xy
number of solutions = 2 number of solutions = 3
In Summary:
This is a typical curve sketching question. Remember to label your curve and
lines clearly.
x
(1, 1)
2 O
y 1x
22 xxy
–1
1 xy
12
1 xy
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 268
9.
In the diagram ABC is a structure consisting of a rod AB of length 10 cm attached at B to a
rod BC of length 24 cm so that angle ABC = 90°. Small rings at A and B enable A to move
along a vertical wire Oy and B to move along a horizontal wire Ox. Angle OAB = Ɵ and can
vary. The horizontal distance of C from the vertical wire Oy is L cm.
(i) Explain clearly why cos24sin10 L . [2]
(ii) Express L in the form cosR , where R > 0 and 900 . [4]
(iii) Find the greatest possible value of L and the value of at which this occurs. [3]
(iv) Find the value of for which L = 20. [2]
Solution:
(i) sin10OB cm CBx , cos24xBC
cos24sin10 xBCOBL
(ii) sinsincoscoscos RRR
sinsincoscoscos24sin10 RRL
24cos R and 10sin R
12
5
24
10tan
6.22
12
5tan 1
[Analysis]
Part (i) is to formulate the projected horizontal length. Part (ii) is to express it into cosR .
Part (iii) is to find the maximum value of L. Part (iv) is simply solving the trigo equation.
O x
y
B
A
C
10 cm
24 cm
O x
y
B
A
C
10 cm
24 cm
Cx
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 269
22222410sincos RR
26R
cos26L where
12
5tan 1
(iii) When 1cos , L is at its maximum.
The largest 26L when 1cos
0
6.22
(iv) When 20L ,
cos2620
13
10cos
7.39
13
10cos 1
3.626.227.39
In Summary:
Question on R formula is a regular in this paper.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 270
10.
The diagram shows a trapezium ABCD in which AD is parallel to BC and angle BCD = 90°.
The vertices of the trapezium are at the points A(8, k), B(3, 6), C(1, 2) and D. The line with
equation 253 xy passes through M, the mid-point of AB, and through N, which lies on
AD.
(i) Show that k = 11. [3]
(ii) Find the coordinates of N and of D. [6]
(iii) Find the area of the triangle AMN. [2]
Solution:
(i)
2
6,
2
11
2
6,
2
38 kkM
Since M lies on the line 253 xy ,
252
113
2
6
k
63350 k
11k
A(8, k)
O
253 xy
y
x
C
(1, 2)
(3, 6) B
M
N
D
[Analysis]
Part (i), mid-point M, is on the line. Part (ii) needs to find the equation of AD using gradient BC.
And also need to find the equation of CD. Part (iii) begins with finding the area of BNA.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 271
(ii)
11,8A AD // BC, gradient of BC = gradient of AD 213
26
Equation of AD ,
8211 xy
52 xy
At point N, 52 xy intersects 253 xy ,
25352 xx
305 x
6x
7512 y
7,6N
CD BC, gradient of CD 2
1
Equation of CD ,
12
12 xy
5.22
1 xy
At point D, 52 xy intersects 5.22
1 xy ,
5.22
152 xx
5.75.2 x
3x
156 y
1,3D
(iii)
length CD 5211322 units
length AN 201178622 units
Area of triangle BNA 52052
1 square units
Area of triangle AMN = 2
1 of Area of triangle BNA 5.25
2
1 square units
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 272
11.
The diagram shows part of the curve xy 45 , meeting the x-axis at the point A and the
line 1x at the point B. The normal to the curve at B meets the x-axis at the point C. Find
(i) the coordinates of C , [6]
(ii) the area of the shaded region. [6]
Solution:
(i) When 1x , 345 y , coordinates of 3,1B
xxx
y
45
2
45
4
2
1
d
d
When 1x ,
3
2
45
2
d
d
x
y
The gradient of the tangent at B is 3
2, therefore the normal at B is
2
3 .
Equation of BC ,
12
33 xy
5.42
3 xy
x
B
A C O
y 1x
xy 45
[Analysis]
Part (i) requires differentiation to find gradient then the normal equation. Part (ii) breaks the shaded
area into two regions.
Solutions to O Level Add Math paper 2 2012
By KL Ang, Jun 2013 Page 273
When 0y ,
5.42
30 x
3x
coordinates of 0,3C
(ii) When 0y ,
x450
4
5x
coordinates of
0,
4
5A
area of the shaded region
3132
145
1
4
5
dxx
34542
3
6
11
4
5
dxx
3456
11
4
52
3
x
30456
12
3
2
3
3276
1
35.4
5.7 square units
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