H2 MATHS YR 6 2011
PRELIMINARY EXAMINATION PAPER 2
SUGGESTED SOLUTIONS
Qn
1(a)
Suggested Solutions
x
y  Ae  2sin x  B

dy
 Ae x  2 cos x
dx
d2 y
 (1)
 Ae x  2sin x
 (2)
From (1),
dy
Ae x 
 2 cos x
dx
Substitute (3) into (2),
 (3)

dx
2
d2 y
 dy

   2 cos x   2sin x
 dx

dx
2
d2 y
dx 2

dy
 2  sin x  cos x 
dx
(Shown)
Alternative:
y  Ae x  2sin x  B


dy
 Ae x  2 cos x
dx
d2 y
dx
2
RHS 

 Ae x  2sin x
dy
 2  sin x  cos x 
dx

 Ae x  2 cos x  2  sin x  cos x 
 Ae x  2sin x
 LHS 
© DHS 2011
d2 y
dx 2
(Shown)
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 1 of 14
Qn
1(b)
Suggested Solutions
y  ux 
dy
du
ux
dx
dx
 (1)
Substitute (1) into d.e., we have
du 

x  u  x   4 x 2  2u 2 x 2  ux
dx 

du
x2
 4 x 2  2u 2 x 2
dx
du
 4  2u 2
dx
1
 dx   4  2 u 2 du
1
1
 dx  2  2  u 2 du
1
u
x
tan 1
c
2 2
2
1
y
x
tan 1
c
2 2
2x
Substitute x 
c

8
and y  0,

Alternative
1
u
tan 1
 2x  C
2
2
1
y
tan 1
 2x  C
2
2x
Substitute x 
C
8

8
and y  0,

4

2 
 y  2 x tan  2 2 x 

4 

© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 2 of 14
Qn
2(a)
Suggested Solutions
(1)
w * 2z  i
w  (1  2i)z  3  3i
(2)
i  w*
From (1), substitute z 
into (2)
2
(i  w*)
w  (1  2i)
 3  3i
2
2w  w *(2i  1)  4  5i
Let w  a  bi
2w  w *(2i  1)  4  5i
2(a  bi)  (a  bi)(2i  1)  4  5i
2a  2bi  2ai  a  2b  bi  4  5i
(a  2b)  (2a  3b)i  4  5i
-------------------------------------------------------------------------------Alternative
3+3i  w
From (2), substitute z 
into (1)
1  2i
 3+3i  w 
w * 2 
i
 1  2i 
w *(1  2i)  6  6i  2 w  2  i
w *(1  2i)  2 w  4  5i
Let w  a  bi
w *(1  2i)  2 w  4  5i
(a  bi)(1  2i)  2( a  bi)  4  5i
(a  2b)  (2a  3b)i  4  5i
-------------------------------------------------------------------------------Compare real parts:
(3)
a  2b  4
Compare imaginary parts: 2a  3b  5
(4)
Solve equations (3) and (4) : a  2 and b  3
w  2  3i
(3  3i)  (2  3i)
z
 1  2i
(1  2i)
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 3 of 14
Qn
2(b)
Suggested Solutions
iz  2  2 i
3
 3 
i 

4 
1
z 3  2  2i  8 2 e 
1 
 3
or 8 2 cos  
  4
1
6
8 e
z
1
or  8 6 e
1
3 
i  2 k  
3
4 
 2 k  
i
 
4
 3
1
6
or z1  8 e
 11 
i 

 12 
or
 3 
i 

4 
2 2e 

 3
  i sin  

 4
  3

  or 2 2 cos  

  4

 3
  i sin  

 4
, k  1, 0, 1
, k  1, 0, 1
1
6
, z2  8 e
 
i  
 4
1
6
, z3  8 e
 5 
i 
 12 
Conjugate on both sides of iz 3  2  2i
 iz  *   2  2 i  *
i*  z  *  2  2 i
3
Alternative
z 3  2  2i
w3  2  2i  (2  2i) *
3
w3  ( z 3 )*  ( z*)3
w  z*
i  z *  2  2i
3
Comparing i  z *  2  2i with iw3  2  2i  w  z *
3
*
1
3 
 1 i 13  2 k  34  
 i  2 k  
1
3
4 


6
6
w  z *  8 e
, k  1, 0, 1
 8 e




*
 2 k  
 1 i 2 k3  4  
 i
 
1


6
6
or  z *   8 e
  8 e  3 4  , k  1, 0, 1




1
6
or w1  8 e
© DHS 2011
 5 
i 

 12 
1
6
, w2  8 e

i 
 4
1
6
, w3  8 e
 11 
i

 12 
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 4 of 14



Qn
3(i)
Suggested Solutions
1
1
x  t  , y  2t  , t  0.
t
t
Required area
5
  y dx
0
1 
1

2t  1  2  dt

1
t  t 

2
 31.39 unit (2 d.p.)

5.1926
dx
1
 1 2
dt
t
t 2 1
 t 1
t
t 2 1
When x  5,5 
 t 2  5t  1  0
t
 t  5.1926
When x  0, 0 
(ii)
(iii)
( t  0)
( t  0)
1 dx
1
xt 
 1 2
t
dt
t
1 dy
1
y  2t  
 2 2
t
dt
t
dy dx
1
1

 1 2  2  2
dt dt
t
t
2
 2 1
t
 t  2 since t  0
1
1
x  t  , y  t  , t  0.
t
t
y
yx
y  x
y  f ( x)
x
O
y 1
y 1
x
O
1
© DHS 2011
y  1
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 5 of 14
Qn
4(a)
Suggested Solutions
3
1
1
 
(by Cover-up rule)
r (r  3) r r  3
n
1 1 1
1
1
   ... 

4 10 18
n(n  3) r 1 r  r  3


1 n
3
1 n 1
1 



 

3 r 1 r  r  3 3 r 1  r r  3 
1 1

3 1
1

2
1

3
1

4
1
4
1

5
1

6
1

7

1
1

n3
n
1
1


n2
n 1
1
1


n 1
n2
1
1


n
(n  3)


1 1 1
1
1
1 
 1   



3  2 3 n 1 n  2 n  3 
11 1  1
1
1 
  



18 3  n  1 n  2 n  3 
1
1 
 1
 As n  , 


0
 n 1 n  2 n  3 
1 1 1
1 1 1

    ...  2     ... 
2 5 9
 4 10 18


1
11

9
r 1 r  r  3 
 2
 series is convergent since sum to infinity of series exists.
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 6 of 14
Qn
4(b)
Suggested Solutions
(i)
un 1  un 
1
3
, u1 
(n  1)(n  2)
2
n
un
1
2
3
4
(ii) Conjecture: un 
5
3
7
4
9
5
2n  1
for n 
n 1
(iii)
Let Pn be the proposition un 
When n = 1,
un1
3
2
5
3
7
4
9
5

.
2n  1
for n 
n 1

.
3
2
2 1 3
 = LHS
RHS =
11 2
LHS = u1 
 P1 is true.
Assume Pk is true for some k 

, i.e. uk 
2k  1
.
k 1
We want to show that Pk+1 is also true, i.e.
2  k  1  1 2k  3
uk 1 

 k  1  1 k  2
1
(k  1)(k  2)
2k  1
1


k  1 (k  1)(k  2)
LHS = uk 1  uk 
(2k  1)(k  2)  1 2k 2  5k  3

(k  1)(k  2)
(k  1)(k  2)
(k  1)(2k  3)

(k  1)( k  2)
2k  3
=
= RHS
k 2
 Pk is true  Pk+1 is true.
Since P1 is true and Pk is true  Pk+1 is true, by mathematical induction, Pn is
true for for all n   .

© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 7 of 14
Qn
5
Suggested Solutions
X ~ B(20, p)
P( X  1)  0.8
P( X  0)  P( X  1)  0.8
(1  p) 20  20 p(1  p)19  0.8
From GC, p  0.041412  0.0414
X ~ B(20, 0.041412)
P( X  a)  0.999  P( X  a  1)  0.999
From GC,
P( X  4)  0.99888  0.999
P( X  5)  0.99988  0.999
Thus the least value of a = 6.
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 8 of 14
Qn
6(i)
Suggested Solution
No. of ways to arrange the 4 boys = 4!
No. of ways to arrange the girls such that they are separated (after the boys are
seated) = 5 P3
By (MP), total number of ways = 4! 5 P3  1440
(ii)
Alternative
 2  3  5
3!
 4!  3!10  4!  1440
3 

Consider the arrangement of <BBBB>, 3 G & 3 distinct “objects” A, B and C.
No. of ways to arrange these 7 entities at a round table = 6!
No. of ways to arrange the 4 boys in the group = 4!
Since A, B and C represent 3 identical empty seats,
6!4!
 No. of ways =
 2880
3!
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 9 of 14
Qn
7
Suggested Solution
Clear
0.175
Doesn’t clear
0.35
Clear
0.825
0.35
Clear
0.7
Doesn’t clear
0.65
Clear
0.65
Doesn’t clear
0.35
0.7
0.3
Clear
Clear
Doesn’t clear
0.65
0.3
Stage 1
Doesn’t clear
Doesn’t clear
Stage 2
Stage 3
(i)
P(game ends prematurely)   0.3 0.3  0.09
(ii)
P(clears exactly 2 stages)
  0.7  0.35 0.825   0.7  0.65  0.35    0.3 0.7  0.35 
(iii)
 0.43488  0.435 (3 d.p.)
P(clears the third stage | cleared exactly 2 stages)
 0.7  0.65 0.35   0.3 0.7  0.35 

0.43488
 0.535 (3 d.p.)
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 10 of 14
Qn
Suggested Solutions
8(a) Any one possible answer
1. Determine if there’s a relationship/trend/pattern between two variables.
2. Identify outliers or suspicious observations.
(b)(i) From GC, r = −0.930 (3 s.f.)
r = − 0.930 is close to −1  a strong negative linear correlation between x and
y.
 linear model is appropriate
(ii)
y
400
380
360
340
320
300
x
70
80
90
100
110
120
130
140
(iii) Any one possible answer
1. The scatter diagram shows that as x increases, y decreases (but at an
increasing rate)/graph is concave downwards. For a linear model, the rate of
decrease is constant.
2. Between x and y, r = − 0.930 while between x2 and y, r = − 0.952 is closer to
−1.
 the model y  a  bx 2 is better.
(iv)
From GC,
when x  150, y  446.87  0.006788(150) 2
 294 (3 s.f.)
Since city bird density x = 150 lies outside the given data range, the model may
not be valid, hence the prediction for forest bird density may not be reliable.
Any one possible answer (or any logical answer)
1. Immigration / emigration of birds from/to other countries.
2. Births/deaths/diseases
3. Not constant due to sampling variation.
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 11 of 14
Qn
9(i)
Suggested Solutions
C ~ N(68,  )
2
T ~ N(65,82 )
S ~ N(70,102 )
P(C  85)  0.05
85  68 

P Z 
 0.05
 

85  68
 1.6449

  10.335  10.3 (3 s.f.)
(ii) C ~ N(68,102 )
 102 
~ N  68,

5 

P(C  75)  0.058762  0.0588 (3 s.f.)
C
(iii)
C1  C 2 
5
 C5
Let X  0.2C  0.2T  0.6S
E( X )  0.2(68)  0.2(65)  0.6(70)  68.6
 10 2 
2
2
2
2
Var( X )  0.22 
  0.2 (8 )  0.6 (10 )  39.36
 5 
X ~ N(68.6,39.36)
P( X  80)  0.034601  0.0346 (3 s.f.)
(iv)
C or C , T and S are independent.
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 12 of 14
Qn
Suggested Solution
10(i) An unbiased estimate of population mean  ,
  x  8  8  31.333  31.3 (3 s.f.)
x
60
An unbiased estimate of population variance  2 :
2


x

8




1
2

   x  8 
  36.158  36.2 (3 s.f.) .
s2 
59 
60



(ii)
Let X represent the number of dengue cases per day with population mean .
To test H0:   30
against H1:   30
One-tail test at 5% significance level.
Since sample size n = 60 is large, by Central Limit Theorem,
 36.158 
under H0, X ~ N  30,
 approx.
60 

Using Z-test, p-value = 0.042978 = 0.0430 (3 s.f.) (from GC).
Since p-value = 0.0430 < 0.05, we reject H0 and conclude that there is sufficient
evidence at 5% level of significance that the mean number of dengue cases has
increased.
The p-value is the lowest level of significance for which the null hypothesis of
mean number of dengue cases being 30, will be rejected.
Alternative: The p-value is the probability of obtaining a test statistic at least as
extreme as 31.3, assuming that H0 is true.
(iii)
To test H0:   30
against H1:   30
One-tail test at 5% significance level.
Since sample size n = 7 is small, assuming X is normally distributed with 2
unknown,
X  30
~ t  6 .
under H0, T 
36.158 / 7
Using P(T  tc )  0.95, tc  1.943 (MF15)
To reject H0 at 5% level of significance,
tvalue  tc
x  30
 1.943
36.158 / 7
x  25.584
x  25.6 (3 s.f.)
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
Page 13 of 14
Qn
11
Suggested Solution
Let X be the number of traffic offences occurring at a particular stretch of road
between 7:30am to 9:30am on any day.
i.e. X Po( )
P( X  1)  0.95
1  P( X  0)  0.95
P( X  0)  0.05
(i)
e  0.05
  3 (nearest integer)
Let Y be the number of traffic offences occurring between 7:30am to 9:30am
during weekdays in a given week.
i.e. Y Po(15)
P(Y  17)
 1  P(Y  16)
 0.33588
(ii)
(iii)
 0.336 (3 s.f.)
Since n = 52 is large, by Central Limit Theorem,
 15 
Y N 15,  approximately
 52 
P(13  Y  16)  0.969 (3 s.f.)
Let W be the no. of weeks (out of 52 weeks) where there are at most 16 traffic
offences occurring between 7:30am to 9:30am during weekdays.
i.e. W B(52,1  0.33588)
W B(52, 0.66412)
Since n  52 is large, np  34.534  5, n(1  p)  17.466  5,
W
N  34.534,11.599  approximately
P(30  W  42)
 P(30.5  W  41.5) (c.c. applied)
(iv)
 0.861 (3 s.f.)
A Poisson distribution with mean 36 is not appropriate as the mean number of
offences occurring at a different 2-hour period of a day is unlikely to be the same
as   3 , and hence the mean number of offences occurring in a full day is
unlikely to be 12  36 as well.
© DHS 2011
9740/2011 Year 6 H2 Math Prelim Paper 2 Solutions/DHS
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