HCI Prelim H2 Mathematics P2 Solutions
2010 HCI H2 Mathematics Preliminary Examination Paper 2 Solution
Qn
Solutions
1
Surface area of the tin and lid
= 2 x 2  2 xy  10 x = 400
y
200  x 2  5 x
x
Volume of the container
 200  x 2  5 x 
  x2 

x


   200 x  x3  5 x 2 
dV
   200  3x 2  10 x 
dx
dV
20
0 x
or x  10 (rejected)
dx
3
20
d 2V
   6 x  10   0 when x 
2
3
dx
20
V is maximum when x 
.
3
20
55
When x 
, y
3
3
(or x = 6.67, y = 18.3).
2(i)
z 5  32  0  z 5  32 ei0 = 32ei2k
 z  2e
(ii)
2 k i
5
where k  0,  1,  2.
 2w  1 
5
The highest power in the equation 
  32 is four since the terms with w are
 w 
canceled out. Hence the equation has only four roots.
5
5
1

 2    32
w

2 k i
1
 2   z  2e 5
w
2 k i
2 k i
1
  2e 5  2  2 e 5  1
w
1
1
1
1
 
 
w1 w2 w3 w4



 2 e

2 i
5

1  e
2 i
5

1  e
4 i
5

1  e
1
4 i
5

1 

HCI Prelim H2 Mathematics P2 Solutions
2
4


 2 cos
 4
= 2  2 cos
5
5


2
4


 cos
 2  .
= 4 cos
5
5


Or use GC,
3
1
1
1
1



= 10.
w1 w2 w3 w4
S n  an 2  bn  c
U1  S1  a  b  c  100
S2  4a  2b  c  190
S10  100a  10b  c  360  100  90  550
Using GC,
a  5 , b  105 , c  0
Thus Sn  5n 2  105n
U n  Sn  Sn 1

 5n2  105n  5  n  1  105  n  1
2

 110 10n
U n  U n 1
4i
 110  10n  110  10n  10 
 10 (a constant)
Hence sequence is an AP.
0
 1  1   
 
  

OX    2   (1   )  0    2  
 t 
 0   t  
 
  

0 
0
0 
 
  

OY    0   (1   )  2    2  2  
t
 t   t  2t  
 
  

0  1       1 


 
 

XY   2  2    2    2  4 
 t  2t    t    t  3t  

 
 

OR
 1
 
AB   2 
 t 
 
 0
 
BC   2 
 2t 
 
2
HCI Prelim H2 Mathematics P2 Solutions
ii
 1    1 
  

XB  1     2    2  2 
 t   t   t 
  

0  0 
  

BY    2    2 
 2t   2t  
  

  1   0    1 

 
 

XY  XB  BY   2  2    2    2  4 
 t   t   2t    3t   t 

 
 

Suppose O, X, Y are collinear.
Then
OX  kOY
0 
1   





 2   k  2  2 
 t  
 t  2t  




1    0    1 (Out of range)
Thus O, X, Y are not collinear.
iii
0 
1    



OX OY   2   2  2 
 t    t  2t  



2
2
= (4 – 4 + t – 2t )
=0
4  t2 1
1
   0 (reject) or  
= + 2
2
4  2t 2 t  2
For all t ℝ\{0}, 0 <  < 1.
Thus XOY can be 90 when t  0 .
3
HCI Prelim H2 Mathematics P2 Solutions
iv
  1 


XY   2  4  
 t  3t  


projection vector
  1   4   4 

  
 2  4  .  1  .  1 
 t  3t    0   0 
  

17
 4
4  4  2  4  

1
17
0
 
 4
2 
  1
17  
0
5(i)
 x  6   y  2
2
 y  2
2
2
 x3
  x  3   x  6 
2
2
 9  2 x  3
(ii)
(iii)
For the equation  y  2   9  2 x  3 ,
When x  2 , y  1 .
2
4
HCI Prelim H2 Mathematics P2 Solutions
When y  7 , x  6 .
Method 1: Using
 y  2
x
 x dy
 9  2 x  3
2
2
2
 y  2  
3  y  2
1

 3

2
18
2 
9

1
R
2

7
1
2

y  2

3

9


 d y  2(6)


7
 1 

( y  2)3 
  3 y 

12


2
27  1
 

1

   21  27    3  1   12 
2

2
 10 units
Method 2: Using
 y  2
2
 y dx
 9  2 x  3
y  2  3 2 x  3
R  4(7) 
[ y  2  3 2 x  3 N.A.]
  2  3
6
2

2 x  3 dx
3 6


 
 28   2 x  (2 x  3) 2  

 2 

= { 28 – [(–12 + 27) – (–4 + 1)]}
= 10 units2
(iv)
Volume required
= vol. of cylinder – (vol. generated by curve from y = –2 to y = 1)
6
2


  (7)2 (4)  
2  3 2 x  3 dx 
2


= 196 – 92
= 327 unit3
(3 s.f.)
Use random sampling method to select a sample from each class. The number of seats from
each class would be proportional to the size of each stratum.

6

First Class
Business Class Economy Class
Any 1 of the
answers below:
4
16
60
Some
passengers have
booked a flight ticket but did not turn up or changed flight so some of the seats in the sample
may not have a passenger.
OR
5
HCI Prelim H2 Mathematics P2 Solutions
The flight is not fully booked so the chosen seat could be empty.
OR The passenger may ignore the questionnaire.
It is not appropriate to use simple random sampling as passengers from different classes may
have different opinions on the service. The number of passengers in the first class is very
small, so the passengers from the first class may not be chosen at all using the simple random
sampling method.
7(i)
No. of ways =
10!
 12600
4!3!2!
7(ii)
Case 1: The 2 blue tiles and 1 yellow tile are in the 4th row with the 4th tile being red or
green.
No. of ways
= no. of ways with B, B, Y, G in 4th row + no. of ways with B, B, Y, R in 4th row
4! 6! 4!
6!
 
 3240
= 
2! 2!2! 2! 2!2!2!
Case 2: The 2 blue tiles and 1 yellow tile are in the third row.
3!
7!
 1890
No. of ways = 
2 2!2!2!
3! 3! 4! 3! 4!
Total no. of ways = 3240 + 1890 –
–
3!
2! 2! 2! 2! 2!
= 5130 – 108 – 216 = 4806
7(iii)
No. of ways such that less than 3 yellow tiles are in the fourth row
7!
= 12600  4C3
=12600  420 =12180
4!2!
No. of ways
6!
=
× 7C4 = 2100
3!2!
2010  10(9000)
x
 9201,
10
2
 x  9000    507147

1 
2

2
s    x  9000  

9
10
9



H0 :   9000
H1 :   9000
7 last
part
8(i)
x  9000
~ t (9)
507147
9  10
p–value = 0.01265 < 0.05
Test Stat: 
Since the p –value = 0.01265 < 0.05, we reject H 0 and conclude that there is sufficient
evidence, at 5% level of significance, that the mean life span of the electronic component has
increased.
6
HCI Prelim H2 Mathematics P2 Solutions
8(ii)
H0:   9000 vs H1:   9000
252
Under H0, X ~ N(9000, 10 ) = N(9000, 62.5).
X – 9000
Test Statistic =
~ N(0, 1).
62.5
Level of significance = 1%
P(Z > 2.326347877) = 0.01
At the 1% significance level, reject H 0 if z ≥ 2.326347877.
x – 9000
≥ 2.326347877
62.5
x ≥ 9018.391395 = 9020.
z=
Assumptions: The standard deviation of the life span remains unchanged after the change in
process.
9
First
part
X ~ N(190, 576)
T  0.001( X1  ...  X 20 )  0.001(2)( X 21  ...  X 30 ) ~ N(0, 0.03456)
P( | T | 0.15)  P(  0.15  T  0.15)
= 0.580
OR
A = X1 +... + X20 – 2(X21 +... + X30) ~ N(0, 34560)
P( | A |
9(i)
0.15
)  P(  150  A  150)
0.001
= 0.580
Let Y be the r.v. denoting the mass of a randomly chosen apple from Mark's orchard.
Y ~ N( , 302 )
Since the shaded area is the same, using the symmetric property of the normal curve,
  110
9(ii)
Probability that Mark will get an apple graded as 'large' chosen at random = P (Y  150) =
7
HCI Prelim H2 Mathematics P2 Solutions
0.09121128
Let A be the r.v. denoting the number of apples graded as large out of 65 randomly chosen
apples.
A ~ B(65, 0.09121128)
P( A  5)  1  P( A  4)
 0.718
10(a)
200 1

400 2
250  300 11

(ii) P(M '  C ') 
1000
20
(i) P(A M ) =
9
1
, P  A M    P  A
20
2
A and M are not independent.
10(b) (i) No. of immigrants in the sample
= 0.2  200  250  0.3 130  300  0.05 120   225
P  A 
P(voter supports Party A given voter is an immigrant) 
0.2  450
 0.4
225
(ii) Number of immigrants supporting Party C = 0.05 120 =6
P(exactly one immigrant voter supporting Party C or
exactly one female voter supporting Party A (or both))
 P  exactly 1 immigrant voted for C 
 P  exactly 1 female voted for A   P  both 

C1 994C2  250C1 750C2  250C1 6C1 744C1
 0.434
1000
C3
6
Alternative method:
Required Probability =
6 994 993
250 750 749
6 250 744
3
3
 3!
1000 999 998
1000 999 998
1000 999 998
= 0.434
8
HCI Prelim H2 Mathematics P2 Solutions
11(i)

Let X be the r.v. denoting the number of call–ins in a week. Hence X ~ Po   .
4
We are looking for the  such that P(X ≤ 9) = 0.701
P( X  9)

From graph, the value of   32.5 (to 3 sig.fig).
The condition is that the rate of call–ins received by the centre is constant throughout a
month / the call–in occurs randomly / The call–ins occur in a month are independent of one
another
11(ii) Let Y be the r.v. denoting the number of call–ins in a week.
Y ~ Po(32.5)
Since the mean is bigger than 10, hence
Y ~ N (32.5, 32.5) approximately.
c.c
P(25  Y  40) 
 P(25.5  Y  40.5) = 0.810
11(iii Let S be the r.v. denoting the number of successful cases out of the n people in a support
)
group.
3
S ~ B(n,
)
20
Since the number of groups concerned, which is 70, is large, therefore by applying CLT,
3 
 3 
n  1  
3
20
20 
S ~ N ( n,  
) approximately.
20
70
EITHER
n
P ( S  4)
27
0.589
28
0.812
Hence minimum value of n is 28.
OR
P( S  4)  0.7
9
HCI Prelim H2 Mathematics P2 Solutions
P( S  4)  0.3
4  0.15n
P( Z 
)  0.3
0.1275n
70
4  0.15n
 0.5244
0.1275n
70
4  0.15n  (0.5244
0.15n  (0.5244
n  5.23912
0.1275
) n
70
0.1275
) n 4 0
70
or
n  5.0899(reject)
n  27.45
Least n = 28.
12(i)
Location F should be omitted as the road distance cannot be smaller than the straight line
distance, indicating that it is an incorrect data entry.
From the scatter diagram, another location that should be omitted is location H, as it is an
outlier based on the scatter diagram.
12(ii) The suitable regression line is the regression x on y:
x  0.3936554  0.81702935 y
When y = 20.0,
x  16.7 km
12(iii s
)
180
70
y
2
30
10
HCI Prelim H2 Mathematics P2 Solutions
Since the graph of s  a  b ln y is concave downwards whereas the graph of s  a  by 2 is
concave upwards, the graph of s  a  b ln y will be more suitable to describe the scatter
diagram of s and y. Hence model II is more suitable.
12(iv
)
The appropriate regression line of s on ln y is s  25.9499647  (45.24427905) ln y ,
i.e. s  25.9  (45.2) ln y (to 3 s.f.)
12(v) Since r for s and ln y is 0.992 close to 1, the linear correlation is strong between s and ln y.
Furthermore, 170 cents is within the data range of the sample. Therefore the estimation using
the line in (iv) is reliable.
Since y is the independent variable, the line found in (iv) is also suitable for the estimation.
11