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2012 GCE ‘A’ Level H1 Maths Solution Paper 1
1
3e2 x  4  e2 x  1
4
40
u
3u 2  4u  4  0
2
u  2 or
3
2
e 2 x  or 2 (N.A.)
3
1 2
x  ln  
2 3
3u 
2(i)
Given 4 x  2 y  20  100 i.e. 2 x  y  40 ……... (i)
And
y( x  20)  3  20 x  ……………………………..(ii)
From (i), substitute y  40  2 x into (ii):
 40  2 x  x  20  60 x
 40 x  800  2 x2  40 x  60 x
 2 x2  60 x  800  0
 x2  30 x  400  0  shown 
2(ii)
From x2  30 x  400  0
 x  10 or 40 (N.A) since x  0
When x  10 , y  20 . Thus x  y  30 .
Therefore, length of HF is 30 cm.
3 (i)
Substitute y 
3 2
k into y  k 2  x 2 :
4
3 2
k  k 2  x2
4
1
 x2  k 2
4
k
k
which are the x -coordinates of
 x  or 
2
2
points of intersection of C & L .

3 (ii)
k

Area   Area of under the curve from 0 to 
2

k
3 

  Area of the rectangle with sides
and k 2 
2
4 

1
Note: Alternatively, the area
can be expressed as
k
3 2
2
2
2k2  k  x   4 k dx
 k
k  3 
 2   2  k 2  x 2  dx   k 2  
2  4 
 0
k


3 2


x
3k 3 
2

 2 k x   
 
3 0
8 


 k 3 k 3  3k 3
 2   
 2 24  4
1
 k3
6
4 (i)(a)
4 (i)(b)
4 (ii)
d
6
2ln  3x  2  =
dx
3x  2
d
8
1
2
4  2 x  1 =  8  2 x  1 
dx
 2 x  12
Note: In general, there is a
modulus for the ln in
1
dx  ln x  C
x
2
1 

2  x  x  dx
4 
1

   x   2  dx
2 
x

4

4
 x2

   ln x  2 x 
2
2
 16
 4

   ln 4  8     ln 2  4   2  ln 2
 2
 2

5(i)
y
y  2x  x2
x
–0.767 O
2
4
Points of intersection with the axes are
 0, 1 ,  0.767, 0 ,  2, 0  and  4, 0  .
5(ii)
From GC, the gradient of C at x  1.5 is
– 1.0394836 = – 1.0395 correct to 4 d.p.
5(iii)
When x  1.5 , y  0.57842712 .
Equation of tangent to C is
Note: Equation of a straight line
is y  mx  c or we can
2
y  0.57842712  1.0394836  x 1.5
 y  0.57842712  1.0394836 x  1.5592254
 y  1.0394836x  2.13765252
That is, y  1.0395x  2.1377 correct to 4 d.p.
5 (iv)
consider the form which is used
here: y  y1  m  x  x1  .
At y -axis, x  0
 coordinates of A   0, 2.1377 
At y  x
 x  1.0395x  2.1377
2.1377
x
 1.048
2.0395
 coordinates of B  1.048,1.048
Length of AB

1.048  02  1.048  2.1377 2
 1.51 correct to 3 s.f.
6(i)
6(ii)
Systematic sampling is a method of selecting
members from an ordered sampling frame in such a
way that the first member of the sample is randomly
selected out of the first k members in the ordered
sampling frame, followed by selecting every
subsequent k th member from the ordered sampling
frame for inclusion in the sample.
number of members in sampling frame
Here, k 
.
sample size
Advantage – This method is cost effective. That is, it
takes less time and effort to carry out.
Disadvantage – Not all adults will go to the
supermarket at midday due to various reasons such
as work etc. Thus the sample collected may not be
representative of the town people.
6(iii)
One possible way to carry out a more appropriate
systematic sampling is to obtain the list of all adults
in that particular town, apply systematic sampling on
it and call the adults selected in the sample to do the
survey.
7(i)
P  A  B   P  A  P  B   P  A  B 
p  p  p2 
p2  2 p 
5
9
A & B are independent.
5
 0 or 9 p 2  18 p  5  0
9
3
Note: Since the objective is to
survey on usage time on
computer, the choice of doing
survey outside the supermarket
is irrelevant.
7(ii)
1
5
or (N.A.)
3
3
1
P  A  B   p2 
9
From (i), p 
8
0.5
0.6
M
0.4
0.4
F
M
A
0.3
B
5
0.1
5
C
0.6
0.2
F
M
0.8
F
8(i)
P  A  M   0.5  0.6  0.3
8(ii)
P( F ) = 0.5  0.4  0.35  0.6  0.15  0.8  0.53
8(iii)
P( C | M ) =
P  C  M  0.15  0.2 3


PM 
1  0.53
47
9 (i)
y
150
42
x
7.5
2.0
9(ii)
r  0.98403
Since r  0.984 (correct to 3 s.f.) is close to 1 , it
suggests there is a strong negative linear correlation
between the advertised price and the age of “Pluto”
cars.
9(iii)
From GC, y  183.11584 19.21307 x
That is, y  19.21x  183.12 correct to 2 d.p.
9(iv)
When x  4 , y  106.28
Thus, the estimated advertised price is $10600.
When x  9 , y  10.23
Thus, the estimated advertised price is $1020.
4
9(v)
Since x  4 is within the data range, the estimate
obtained in (a) is reliable but x  9 is not within the
given range of data ( we are doing extrapolation),
hence the estimate obtained in (b) is unreliable.
10
Let X be the number of “Sunbrite” plants that
flower out of 12 plants.
Then X ~ B 12, 0.8 .
10(i)
P  X  10  0.28347  0.283 correct to 3 s.f.
10(ii)
P  X  8
 P  X  7   0.072555  0.0726 correct to 3 s.f.
10(iii)
Let Y be the number of “Sunbrite” plants that flower
out of 96 plants (8 trays).
Then Y ~ B  96, 0.8 .
n = 96 is large, np  76.8  5 & nq  19.2  5
 Y ~ N  76.8,15.36 approx.
Thus required probability is P Y  75
C.C.
 P Y  75.5  0.62994  0.630 correct to 3 s.f.
10(iv)
Let T be the number of gardeners who have more
than 75 of their plants flower.
Then T ~ B  3, 0.62994  .
P T  2 
 1  P T  1  0.69052  0.691 correct to 3 s.f.
11(i)
11(ii)
n  100 ,
 ( x  300)  300  60  300  299.4
x
n
100
2

( x  300)  

1 

2
2

s 
 ( x  300) 

n 1 
n


2
60   1204

1 

1240 

99 
100 
99


Let  be the mean length of string in a ball.
H0 :   300 against H1 :   300 at the 5%
significance level
X  300
Test statistic, Z 
S2
100
5
Note: In the manager’s claim
that the “average length is at
least 300 m”, we test
H1 :   300
Using z -test, from GC,
p -value  0.0427 correct to 3 s.f.
Since p -value = 0.0427 < 0.05, we reject H 0 at 5%
level of significance and conclude that there is
sufficient evidence to conclude that   300 . Hence
the manager’s claim is not valid.
11(iii)
H0 :   300 against H1 :   300 at the 10%
significance level.
k  300
Test statistic, Z 
12.1
100
For manager’s claim to be valid, we have H 0 not to
be rejected.
k  300
 1.28155
Thus Z 
12.1
100
12.1
 (1.28155)
100
 k  299.5542121
 k  300 
 least k  299.56 correct to 2 d.p.
12
Let X and Y be the masses of grapefruit of type A
and B respectively.
Then X N  0.25, 0.022  and Y N  0.35, 0.032 
12(i)
Let the total mass of 10 randomly chosen grapefruit
of type A be X1  X 2  .....  X10 .
Then X1  X 2  .....  X10
N 10  0.25,10  0.022 
 X1  X 2  .....  X10 ~ N  2.5, 0.004 
Required probability  P  X 1  X 2  ....  X 10  2.4 
 0.056923  0.0569 correct to 3 s.f.
6
12(ii)
Let M1 be the total mass of 6 randomly chosen
grapefruit of type A and
Let M 2 be the total mass of 5 randomly chosen
grapefruit of type B .
Then
M1  M 2 ~ N  6  0.25  5  0.35, 6  0.022  5  0.032 
 M1  M 2 ~ N  0.25,0.0069
Required probability is P( M 1  M 2  0.2)
 P(0.2  M 1  M 2  0.2)
 0.27361  0.274 correct to 3 s.f.
12(iii)
Note: The phrase “type A is
within 0.2 kg of … type B ”
means type A grapefruits
weigh at most 0.2 kg more than
type B and at the same time,
type A grapefruits weigh at
most 0.2 kg less than type B .
Let $ W be the price paid by Mrs Woo.
Then W  1.50  X1  X 2  X 3   2.40 Y1  Y2  Y3 
 W ~ N 1.5  3  0.25  2.4  3  0.35,1.52  3  0.022  2.42  3  0.032 
 W ~ N  3.645, 0.018252
Let $ T be the price paid by Mr Tan.
Then T  1.50  X1  X 2   X10  .
T
N 1.5 10  0.25, 1.52 10  0.022 
T
N  3.75, 0.009 
Therefore, W  T  N  3.645  3.75, 0.018252  0.009 
 W  T  N  0.105, 0.027252 
P(W  T )  P(W  T  0)  0.262372  0.262 correct to 3 s.f.
7