2009 GCE ‘A’ Level H1 Mathematics Solution (Contributed by Hwa Chong Institution)
1)
x + 2y = 3  x = 3 – 2y
Substitute x = 3 – 2y into x2 + xy = 2:
(3 – 2y)2 + (3 – 2y)y = 2
9 – 12y + 4y2 + 3y – 2y2 = 2
2y2 – 9y + 7 = 0
(2y – 7)(y – 1) = 0
7
y=2,1
x = –4, 1
Hint to students:
Students have a wide choice of substitutions
to use:
 From equation (1): x = 3 – 2y
3–x
 From equation (1): y = 2
2 – x2
 From equation (2): y = x
All these substitutions work equally well.
2i)
y= x
(4, 2)
(0, 0)
y = x/2
(ii)
 x dx
x3/2
= 3/2 + c
2
= 3 x3/2 + c
1
 2 x dx
1
= 4 x2 + c
(iii)
Area
1
4
=
0 x – 2 x dx
2
1
4
= [ 3 x3/2 – 4 x2 ]0
16
= 3 –4
4
=3
Hint to students:
Even though the integration has to be done
manually because the question specifies
“Without using a calculator”, it is helpful to
use the GC to check that the answer obtained
is numerically correct.
3i)
Let y = ex. Then x = ln y
h(x) = f –1 (x + 2) = ln (x + 2)
1
(ii)
x = –2
(0, ln 2)
(–1, 0)
(iii)
ln (x + 2) = –x + 2
From GC, x = 0.926
4i)
Hint to students:
The y-axis is a vertical asymptote to the
curve. Do not draw the curve as if it touches
the y-axis, even though it may appear so on
the GC in some views. Zoom in to get a
better view.
(–1, 0)
(1, 0)
dy
1
=
1
+
dx
x2
dy 5
When x = 2, dx = 4 .
(ii)
Therefore gradient of normal is –
(iii)
4
.
5
3
When x = 2, y = 2 .
Therefore equation of normal is
3
4
y – 2 = – 5 (x – 2)
10y – 15 = –8(x – 2)
8x + 10y – 31 = 0
(iv)
When x = 0, y =
31
31
. So N = (0, ).
10
10
2
Hint to students:
It is helpful to draw a diagram even though it
is not asked for in the question. This will help
you to see how the area of triangle PTN can
be calculated.
N
P
T
3 5
Equation of tangent at P is y – 2 = 4 (x – 2)
i.e. 4y – 6 = 5x – 10
4y = 5x – 4
When x = 0, y = –1. So T = (0, –1).
31
41
NT = 10 – (–1) = 10
1
41
Area of triangle PTN = 2  2  10
41
= 10
5i)
dy
2
dx = 6x – 10x – 4
dy
dx
= 6x2 – 10x – 4
Hint to students:
The derivative have to be computed manually
since the questions requires exact
coordinates. However it is helpful to use the
GC to check that the answers are numerically
correct.
= 2(3x2 – 5x – 2)
= 2(3x + 1)(x – 2) = 0
1
x=–3,2
100
y = 27 , –9
1
coordinates of stationery points = (– 3 ,
100
27 ) and (2, –9).
(ii)
1 100
(– 3 , 27 ) (0, 3)
(–1, 0)
1
( , 0)
2
(3, 0)
(2, –9)
2x3 – 5x2 – 4x + 3 > 0
1
 –1 < x < 2 or x > 3
(iii)
2e3x – 5e2x – 4ex + 3 > 0
3
1
 –1 < ex < 2 or ex > 3
1
 ex < 2 or ex > 3
1
 x < ln 2 or x > ln 3.
6i)
0.2  0.7 = 0.14
(ii)
P(call for A and A is in office) +
P(call for B and B is in office) + P(call for C
and C is in office)
= 0.14 + 0.3  0.6 + 0.5  0.8
= 0.14 + 0.18 + 0.4
= 0.72
(iii) P(call is for C | researcher being
called not in office)
P(call is for C and C is not in office)
= P(researcher being called not in office)
0.5  0.2
= 1 – 0.72
0.1
= 0.28
= 0.357
7i)
P(A  B) = P(A) + P(B) – P(A  B)
17 1 2
30 = 3 + 5 – P(A  B)
1 2 17 1
P(A  B) = 3 + 5 – 30 = 6
1 2
= 
3 5
2 1
= 15  6 = P(A  B).
Hence A and B are not independent.
(ii)
P(A) P(B)
(iii)
Method 1:
A
Hint to students:
Drawing a Venn diagram is the preferred way
of solving such problem.
B
2
5
13
30
13 2 5
P(A  B) = 30 + 5 = 6
4
Method 2:
P(A  B)
= P(A) + P(A  B)
= 1 – P(A) + P(A  B)
1 1
=1–3+6
5
=6
8i)
Let X = lifetime of a component.
X ~ N(120, 182)
P(X > 144) = 0.091211 = 0.0912
(ii)
2! P(X > 144) P(X < 144)
= 2  0.091211  (1 – 0.091211)
= 0.166
Hint to students:
It is necessary to multiply by 2! since there
are 2! ways of arranging these 2 components.
H0 :  = 120
H1 :  > 120
Hint to students:
It is necessary to give the conclusion in the
context of the problem, instead of a generic
conclusion.
Since p–value = 0.058 > 0.05, we do not
reject H0 . There is insufficient evidence at
5% level to say that the mean lifetime is
longer than 120 days.
9i)
Hint to students:
Students are reminded to label and indicate
the scale on the axes. It is advisable to draw
the scatter diagram to scale and to copy what
appears on the screen of the GC as closely as
possible.
y
18
15
x
100
200
(ii)
5
r = 0.931.
The scatter diagram shows the data lying
close to a straight line. This agrees with the
value of r which is close to 1.
(iii) Regression line of y on x is
y = 0.0123x + 15.5
(iv)
Estimated weight
= 0.012329(135) + 15.4866
= 17.2 kg
(v)
Since y = 20 is outside the range of
the data values, it is unsuitable to use the
equation in part (iii) is the regression line of y
on x to estimate the amount of liquid nutrient.
10i) Let X = no. of students out of 10 who
failed the piano examination.
X ~ B(10, 0.2)
P(X = 2) = 0.302
(ii)
Let Y = no. of students out of 10 who
will be awarded distinction.
Y ~ B(10, 0.150.8) = B(10, 0.12)
P(Y < 2)
= P(Y  1)
= 0.658
(iii) Let W = no. of students out of 50 who
failed the examination.
W ~ B(50, 0.2)
Since n = 50 is large, np = 10 > 5, nq = 40 >
5, W ~ N(10, 8) approximately.
P(W  12)
Hint to students:
Some students mix up the use of nq and npq.
Also it is common for students to forget to do
continuity correction.
6
= P(W  12.5) by continuity correction
 0.812
11ai) 72  8 = 9.
We first determine the sampling interval = 72
 8 = 9.
Then choose random starting number
between 1 to 9 (inclusive), e.g. 5.
Then we sample claim no. 5, 14, 23, 32, 41,...
(ii)
Systematic sampling is a better
indication since the first 8 claims received
may all come from the same area.
5320
x = 120 + 1000 = 1044.33
1
53202
s2 = 119 [ 8 282 000 – 120 ] = 67614.7
Teaching Point:
The formula for the unbiased estimate of the
variance is found in MF15. There is no need
for students to memorise it.
(bi)
(ii)
A sample statistic T is an unbiased
estimate of a population parameter  if E(T)
= .
(iii)
H0 :  = 1000
H1 :   1000
H0 is rejected

 p–value = 0.0618 < 100
  > 6.18.
12a) Let X = mass of a plum.
X ~ N(, 2)
P(X < 22) = 30%
7
22 – 
) = 0.3

22 – 
= –0.52440

22 –  = –0.52440 –––(1)
P(X > 29) = 20%
P(X < 29) = 80%
29 – 
P(Z <
) = 0.8

29 – 
= 0.84162

29 –  = 0.84162
–––(2)
(2) – (1)  7 = 1.36602
  = 5.1244 = 5.12
  = 24.7
P(Z <
(b)
Let A, N = mass of an apple and a
nectarine respectively.
A ~ (0.15, 0.032)
N ~ (0.07, 0.022)
(i)
A1 + A2 – N1 –...– N4
~ N(20.15 – 40.07, 20.032 + 40.022)
~ N(0.02, 0.0034)
Hint to students:
There is no need to square 2 and 4 when
computing variance since we are dealing with
sums of normal variables.
P(A1 + A2 > N1 +...+ N4)
= P(A1 + A2 – N1 –...– N4 > 0)
= 0.634
(ii)
Let Y = total cost of 2 apples & 4
nectarines.
E(Y) = 920.15 + 1240.07 = 6.06
Var(Y) = 9220.032 + 12240.022 = 0.3762
P(5 < Y < 6)
= 0.419
Hint to students:
Students need to square 9 and 12 when
computing variance since we are dealing with
the multiples of normal variables.
8