2324 S6 Mathematics Extended Module 1 1st term examination Solution
Section A : 50 marks
1. Let X be a discrete random variable with probability function as shown in the following table,
where k is a positive constant:
x
1
2
4
k
P(X = x)
0.2
0.4
0.16
0.24
It is given that Var(X) = 7.0464.
(a) Find the value of k.
(b) Find Var(15 – 3X).
(c) Let Y be another discrete random variable with probability function as shown in the following
table:
y
1
2
3
4
P(Y = y)
0.1
0.4
0.3
0.2
Let G be the event that X + 2Y > 7 and H be the event that X < 3. Find P(G  H).
(7 marks)
1. (a) E(X2) = 12(0.2) + 22(0.4) + 42(0.16) + k2(0.24)
= 0.24k2 + 4.36
E(X) = 1(0.2) + 2(0.4) + 4(0.16) + k(0.24)
= 0.24k + 1.64
Var(X) = E(X2) – [E(X)]2
7.0464 = 0.24k2 + 4.36 – (0.24k + 1.64)2
0.1824k2 – 0.7872k – 5.376 = 0
k = 8 or −
70
(rejected)
19
(1M)
(1M)
(1A)
(b) Var(15 – 3X) = (–3)2Var(X)
= 9(7.0464)
= 63.4176
(1M)
(1A)
(c) P(G  H) = 0.2(0.2) + 0.4(0.3 + 0.2)
= 0.24
(1M)
(1A)
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
1
2.
In a card game, each player can choose to plays game A or game B to win a prize. The probability that a
3
player plays game B is and the probability that a player wins a prize given that the player plays game B is
4
5
4
. Given that a player does not win a prize, the probability that the player plays game A is
.
8
55
(a) Find the probability that a player plays game B and wins a prize.
(b) Given that a player plays game A, find the probability that the player wins a prize.
(c) Given that a player wins a prize, find the probability that the player plays game A.
(7 marks)
Let the probability that a player win a prize, doesn’t win a prize be P(W) and P(W),
the probability that a player plays game A and game B be P(A) and P(B).
(a)
(b)
(c)
The required probability
3 5
= 
4 8
15
=
32
(1A)
4
55
4
P(A  W) =
 P(W )
55
P(A  W) + P(A  W) = P(A)
3 4
P(A  W) = 1 − − (1 − P(W ))
4 55
15
P(B  W) =
(from (a))
32
1 4 4
15 
P(A  W) = − +  P( A  W ) + 
4 55 55 
32 
1 4 4 15
− + 
31
P(A  W) = 4 55 55 32 =
4
136
1−
55
31
31
The required probability = 136 =
.
3 34
1−
4
P( A | W ') =
The required probability
P( A  W )
=
P(W )
31
= 136
31 15
+
136 32
124
=
379
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
(1M)
(1M)
(1A)
(1A)
(1M)
(1A)
2
3.
Let a be a constant.
(a)
Expand e−ax in ascending powers of x as far as the term in x3.
(b)
Let y = (1 + ax)5e−ax. It is given that the coefficient of x3 in the expansion of y is 63.
(i) Find a.
(ii) Someone claims that the graph of y = (1 + ax)5e−ax has only one turning point.
Do you agree? Explain your answer.
(7 marks)
3. (a)
e−ax = 1 + (−ax) +
(−ax)2 (−ax)3
+
+ 
2!
3!
a 2 x 2 a 3 x3
= 1 − ax +
−
+ 
2
6
(1A)
(b) (i) y = (1 + ax)5e–ax

= (1 + 5ax + 10a 2 x 2 + 10a3 x3 + ) 1 − ax +

∵
∴

a 2 x 2 a 3 x3
−
+  
2
6

(1M)
Coefficient of x3 = 63
 a3 
 a2 
(1)  −  + (5a)   + (10a 2 )(−a) + (10a3 )(1) = 63
 6 
 2 
3
3
a 5a
− +
− 10a3 + 10a3 = 63
6
2
7a3
= 63
3
a3 = 27
a=3
(ii)
(1A)
dy
= 5(3)(1 + 3x)4e−3x + (–3)(1 + 3x)5e−3x
dx
= 3(1 + 3x)4(5 – 1 – 3x)e−3x
= 3(1 + 3x)4(4 – 3x)e−3x
dy
When
= 0,
dx
3(1 + 3x)4(4 – 3x)e−3x = 0
(1 + 3x)4(4 – 3x) = 0
x=−
4
1
or
3
3
x
x <−
dy
dx
+
(1M)
(1M)
1
3
x =−
1
3
0
4
1
− <x<
3
3
x=
+
0
The graph has only one turning point at x =
∴ The claim is agreed.
2023/24-S5-EXAM1-MATH-PAPER 1-3
4
3
x>
−
4
3
(1M)
4
.
3
(1 f.t.)
3
4.
Let A and B be two events. Denote the complementary events of A and B by A' and B'
respectively. It is given that P(A  B') = 0.56 and P(A' | B') =
(a)
(b)
2P( B )
.
3
Find P(B).
Suppose that P(A) = 0.68. Are A and B independent? Explain your answer.
(5 marks)
4.
(a)
Let P(B) = p.
P(A'  B') = P(A' | B')P(B')
(1M)
2P( B )
P(B')
3
2 p (1 − p )
=
3
=
P(A  B') + P(A'  B') = P(B')
0.56 +
2 p (1 − p )
=1–p
3
(1M)
1.68 + 2p – 2p2 = 3 – 3p
2p2 – 5p + 1.32 = 0
p = 0.3 or 2.2 (rejected)
∴ P(B) = 0.3
(b)
P(A') = 1 – 0.68
= 0.32
Noted that P(A' | B') =
2(0.3)
= 0.2  0.32.
3
∴ P(A' | B')  P(A')
∴ A' and B' are not independent.
∴ A and B are not independent.
5.
5.
(1A)
(1M)
(1A f.t.)
The weights of players in a team follows a normal distribution with a mean of  kg and a
standard deviation of  kg. It is known that 2.56% of players in the team weigh less than 55 kg
and 75.8% of players in the team weighs more than 60 kg.
(a)
Find the values of  and .
(b)
A player is selected randomly from the team. If the probability that the player selected
weighs more than  kg is less than 0.05, find the least integral value of .
(6 marks)
55 − μ 

(a) P  Z 
= 0.0256
σ 

55 − μ
= –1.95 ……..… (1)
(1M) + (1A) for either one
σ
60 − μ 

P Z 
= 0.758
σ 

60 − μ
= – 0.7 ………… (2)
σ
Solving (1) and (2),  = 62.8 and  = 4.
(1A) + (1A)
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
4
α − 62.8 

(b) P  Z 
< 0.05
4 

α − 62.8
> 1.645
4
 > 69.38
∴ The least integral value of  is 70.
6.
6.
(1M)
(1A)
Let f(x) = ln (x3 + mx2 + nx + 25) for x  0. It is known that (1, ln 29) is a turning point of the
graph of y = f(x).
(a)
Find m and n.
dy
(b)
If x = 2u, find the value of
when u = 1.
du
(6 marks)
(a) ln [13 + m(1)2 + n(1) + 25] = ln 29
(1M)
m + n = 3 ………(1)
f'(x) =
3 x 2 + 2mx + n
x 3 + mx 2 + nx + 25
∵ (1, ln 29) is a turning point.
∴
f'(1) = 0
3(1)2 + 2m(1) + n
=0
(1)3 + m(1)2 + n(1) + 25
(1M)
2m + n = –3 ………(2)
Solving (1) and (2), we have m = –6 and n = 9.
dx
(b)
= 2u ln 2
du
dy dx
dy

=
dx du
du
 3x 2 − 12 x + 9  u
= 3
 (2 ln 2)
2
 x − 6 x + 9 x + 25 
1A (both correct)
(1A)
(1M)
When u = 1, x = 21 = 2.
 3(22 ) − 12(2) + 9  1
dy
=  3
 (2 ln 2)
2
du u =1
 2 − 6(2 ) + 9(2) + 25 
=−
2 ln 2
( or −0.1540(corret to 4 d. p.))
9
2023/24-S5-EXAM1-MATH-PAPER 1-5
5
(1A)
7.
7.
Consider the curve C : y =
3x 2 − 2
.
x2 + 1
dy
.
dx
(a)
Find
(b)
If a tangent L to C passes through the point (2, 3), find the possible points of contact of
L and C.
(6 marks)
2
2
dy 6 x( x + 1) − 2 x(3x − 2)
=
(1M)
dx
( x 2 + 1)2
6 x3 + 6 x − 6 x3 + 4 x
=
( x 2 + 1)2
10 x
= 2
(1A)
( x + 1) 2
(a)
(b)
Let (a, b) be the point of contact of L and C.
3a 2 − 2
Then, b = 2
.
a +1
dy
b−3
=
dx ( a ,b ) a − 2
 3a 2 − 2 
 2
−3
a +1 
10a

=
(a 2 + 1) 2
a−2
(1M + 1M)
 3a 2 − 2  
(a 2 + 1)  2
 − 3
10a
 a +1  

=
a2 + 1
a−2
2
(3a − 2) − 3(a 2 + 1)
=
a−2
2
3a − 2 − 3a 2 − 3
=
a−2
−5
=
a−2
10a(a − 2) = −5(a2 + 1)
2a2 − 4a = −a2 − 1
3a2 − 4a + 1 = 0
(a − 1)(3a − 1) = 0
1
a = 1 or a =
3
(1M)
2
1
3  − 2
2
3
3(1) − 2 1
1
3
=− .
= . When a = , b =  2
When a = 1, b =
2
2
(1) + 1 2
3
1
  +1
3
 1
1 3
∴
The possible points of contact are  1,  and  , −  .
 2
3 2
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
6
(1A)
8.
8.
In a research, the population P (in million) of a country is modelled by P = aeb t ,
where a (a > 0)
and b are constants, t (t ≥ 0) is the number of years elapsed since the start of the research.
(a)
Express ln P as a linear function of t .
(b)
It is given that the slope and the intercept on the vertical axis of the graph of the linear
function obtained in (a) are –0.1 and 3ln 2 respectively.
(i)
Write down the values of a and b.
Hence, show that the rate of change of the population is negative for all t > 0.
(ii)
Someone claims that the population when t = 80 is less than 40% of that at the
start of the research. Do you agree? Explain your answer.
(6 marks)
b t
(a)
P = ae
ln P = ln(aeb t )
ln P = ln a + ln eb t
ln P = b t + ln a
(b) (i)
(b) (ii)
(1A)
b = −0.1
ln a = 3ln 2
ln a = ln 23
ln a = ln 8
a=8
P = 8e −0.1 t
 1 
dP
= 8(−0.1)e−0.1 t 

dt
2 t 
0.4 −0.1 t
=−
e
t
dP
< 0 for all t > 0.
dt
∴ The rate of change of the population is negative for all t > 0.
8e−0.1 80
100%
8
 40.884 171 98%
> 40%
∴ The population when t = 80 is not less than 40% of that at the start of the
research.
∴ The claim is disagreed.
2023/24-S5-EXAM1-MATH-PAPER 1-7
7
(1A)
(1A)
(1 f.t.)
(1M)
(1 f.t.)
9.
The lifetimes of the batteries produced by a factory follows a normal distribution with a mean
of  hours and a standard deviation of 160 hours.
(a)
An inspection on the lifetimes of the batteries produced by the factory is conducted to
estimate . A sample of 50 batteries produced by the factory are randomly selected and
their lifetimes are recorded below:
Lifetime (x hours)
Number of batteries
8000 < x ≤ 8100
8
8100 < x ≤ 8200
12
8200 < x ≤ 8300
15
8300 < x ≤ 8400
9
8400 < x ≤ 8500
6
Construct a 99% confidence interval for .
(4 marks)
(b)
Another sample of 14 batteries is randomly selected and their lifetimes (in hours) are
recorded as follows:
8366
8356
8382
8180
8344
8264
8098
8038
8301
8022
8330
8350
8279
8266
The two samples are then combined. A  % confidence interval for  is constructed
using the combined sample. If the lower limit of the constructed confidence interval is
8187.25, find .
(4 marks)
(c)
Suppose µ = 8322. If the lifetime of a battery is greater than 8250 hours, the battery is
classified as standard; otherwise the battery is classified as poor.
(i)
Find the probability that a randomly selected battery is standard.
(ii)
A sample of 12 batteries is randomly selected and their lifetimes are inspected
one by one. Given that at least 4 batteries are standard, find the probability
that exactly 2 poor batteries are inspected among the first 4 batteries
inspected.
(5 marks)
9.
(a)
=
The sample mean
8050(8) + 8150 (12 ) + 8250(15) + 8350(9) + 8450(6)
8 + 12 + 15 + 9 + 6
= 8236
A 99% confidence interval for 
160
160 

=  8236 − 2.575 
, 8236 + 2.575 

50
50 

= (8177.7344 hours , 8294.2656 hours) (cor. to 4 d.p.)
(b)
The mean of the combined sample
8236(50) + 8366 + 8356 + + 8266
=
50 + 14
= 8240.25
 160 
8240.25 − z 
 = 8187.25
 50 + 14 
z = 2.65
  = 0.496(2)(100)
= 99.2
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
(1A)
(1M + 1A)
(1A)
(1A)
(1A)
(1M)
(1A)
8
(c)
(i)
(ii)
Let X hours be the lifetime of a battery.
Then X ~ N(8322, 1602).
The required probability
= P(X > 8250)
8250 − 8322 

= P Z 

160


= P(Z > −0.45)
= 0.5 + 0.1736
= 0.6736
(1M)
(1A)
Let p = 0.6736
The required probability
C24 ( p 2 )(1 − p)2 [1 − C08 (1 − p)8 − C18 ( p)(1 − p)7 ]
=
(1M + 1A)
1 − C012 (1 − p)12 − C112 ( p)(1 − p)11 − C212 ( p 2 )(1 − p)10 − C312 ( p3 )(1 − p)9
= 0.2903 (cor. to 4.d.p.)
(1A)
2023/24-S5-EXAM1-MATH-PAPER 1-9
9
10.
There are 10 plants in a greenhouse. Assume that the numbers of rotten leaves on each plant in
the greenhouse follows a Poisson distribution with a mean of 2.8. The number of rotten leaves
on the plants in the greenhouse are independent. A plant with at least 4 rotten leaves is
classified as unhealthy.
(a)
Find the probability that a plant in the greenhouse is unhealthy.
(3 marks)
(b)
Find the probability that at most 3 plants in the greenhouse are unhealthy.
(3 marks)
(c)
Find the probability that exactly 3 plants in the greenhouse is unhealthy and
the total number of rotten leaves on the unhealthy plants is 13.
(2 marks)
(d)
Given that at most 3 plants in the greenhouse are unhealthy, find the probability that
the total number of rotten leaves of these plants is 13.
(4 marks)
10.
(a)
The required probability

2.82 2.83 
= = 1 − e−2.8 1 + 2.81 +
+

2!
3! 

 0.308062567
= 0.3081 (cor. to 4 d.p.)
(b)
(1A)

2.82 2.83 
Let p = 1 − e−2.8 1 + 2.81 +
+
.
2!
3! 

The required probability
= C010 (1 − p)10 + C110 ( p)(1 − p)9 + C210 ( p)2 (1 − p)8 + C310 ( p)3 (1 − p)7
 0.627983789
= 0.6280 (cor. to 4 d.p.)
(c)
(1M + 1M)
(1M + 1M)
(1A)

2.82 2.83 
+
Let p = 1 − e−2.8 1 + 2.81 +
.
2!
3! 

The required probability
2
 e −2.8 2.84   e −2.8 2.85 
= C (1 − p )  C 
 

 4!   5! 
 0.057829216
= 0.0578 (cor. to 4 d.p.)

2.82 2.83 
+
Let p = 1 − e−2.8 1 + 2.81 +
,
2!
3! 

10
3
(d)
7
3
1
(1M)
(1A)
e−2.8 2.813
and
13!
 e−2.8 2.84  e−2.8 2.89 
 e−2.8 2.85  e−2.8 2.88 
 e−2.8 2.86  e−2.8 2.87 
p2 = 2 

 + 2
 + 2



 4!  9! 
 6!  7! 
 5!   8! 
p1 =
The required probability
C10 ( p )(1 − p)9 + C210 ( p2 )(1 − p)8 + 0.057829216
 1 1
0.62798379
= 0.1029 (cor. to 4 d.p.)
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
10
(1M + 1M + 1M)
(1A)
11.
In an experiment, the radiation intensity (R units) of a chemical in a flask is given by
dR t 2 + 5
,
=
dt
2t + 1
where t (t  0) is the number of hours elapsed since the start of the experiment. The radiation
intensity is 10 units at the beginning of the experiment. It is known that the rate of change of
the radiation intensity attains its minimum value when t = T.
(a)
Find T.
(4 marks)
(b)
Find the exact value of R at t = T.
(5 marks)
(c)
The temperature (Q C) of the chemical in the flask is given by Q = ln R.
Find
dQ
at t = T.
dt
(3 marks)
1


2t + 1(2t ) − (t 2 + 5) 
 (2)
d R
2 2t + 1 

11. (a)
=
2t + 1
dt 2
2
2t (2t + 1) − (t + 5)
=
3
(2t + 1) 2
2
=
=
(1M)
3t 2 + 2t − 5
3
(2t + 1) 2
(3t + 5)(t − 1)
3
(2t + 1) 2
For
d2 R
5
= 0, t = − (rejected) or 1
2
dt
3
0t<1
t
(1M)
t=1
t>1
0
+
(1M)
2
d R
–
dt 2
dR
∴
attains its minimum at t = 1.
dt
∴T=1
(1A)
(b) When t = 1,
t2 + 5
dt
R = 10 + 
0
2t + 1
2
 −1  1 
3  u − 1 
+
5
= 10 +  
 u 2   du (Let u = 2t + 1)

1
2

2


1
3
1
(1M)
(1M + 1M)
−1
u 2 − 2u 2 + 21u 2
du
= 10 + 
1
8
3
3
3
 5
1
1  2u 2 4u 2
= 10 +
−
+ 42u 2 

8 5
3

1
 2 4
1  2(9 3) 4(3 3)

−
+ 42 3  −  − + 42  
= 10 + 
8  5
3
 
 5 3
2023/24-S5-EXAM1-MATH-PAPER 1-11
11
(1M)
78 3 − 77
15
78 3 + 73
=
(or 13.8733 (correct to 4 d.p.))
15
= 10 +
(c)
(1A)
d Q 1  dR 
= 

R  dt 
dt
(1M)
 12 + 5 
15
dQ
=


dt t =1 78 3 + 73  2(1) + 1 
(1M)
=
90
234 + 73 3
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
(1A r.t. 0.2497)
12
12.
(a)
(b)
(c)
du
p(t )
t 2 − kt + 3
= 2
Let u = 2
, where k is a positive constant. If
for some
dt (t + kt + 3)2
t + kt + 3
polynomial p(t), express p(t) in terms of k.
(2 marks)
Using integration by substitution and the identity
t2 − 3
t4 + (6 − k2)t2 + 9 = (t2 − kt + 3)(t2 + kt + 3), find  4
dt in terms of k.
t + (6 − k 2 )t 2 + 9
(4 marks)
2
A geographer studies the area of a forest. Let A(t) (in hundred km ) be the area of the
forest at time t, where t (t ≥ 0) is the number of months elapsed since the start of the
study. The geographer models the rate of change of the area of the forest by
t2 − 3
1
A' (t ) = 4
+  .
2
t + 2t + 9  2 
Find the change in area of the forest in the first two months of the study.
If the area of the forest at the start of the study is 8 hundred km2, estimate the
area of the forest after a very long time.
(7 marks)
t
(i)
(ii)
12.
(a)
du (2t − k )(t 2 + kt + 3) − (2t + k )(t 2 − kt + 3)
=
dt
(t 2 + kt + 3)2
[2t 3 + kt 2 + (6 − k 2 )t − 3k ] − [2t 3 − kt 2 + (6 − k 2 )t + 3k ]
=
(t 2 + kt + 3)2
2kt 2 − 6k
(t 2 + kt + 3)2
2k (t 2 − 3)
= 2
(t + kt + 3)2
 p(t) = 2k(t2 – 3)
(1M)
=
(1A)
2k (t 2 − 3)
t 2 − kt + 3
(b) Let u = 2
, then du = 2
dt .
(t + kt + 3)2
t + kt + 3
t2 − 3
 t 4 + (6 − k 2 )t 2 + 9dt
1
2k (t 2 − 3)
dt
=
2k  (t 2 − kt + 3)(t 2 + kt + 3)
1 t 2 + kt + 3 2k (t 2 − 3)

dt
=
2k  t 2 − kt + 3 (t 2 + kt + 3)2
1 1
=
du
2k  u
1
=
ln u + C
2k
1
t 2 − kt + 3
ln 2
+ C , where C is a constant
=
2k t + kt + 3
2023/24-S5-EXAM1-MATH-PAPER 1-13
13
(1M)
(1M)
(1A)
(1A)
(c) (i)
The required change
t
2
t2 − 3
1 
=  4
+
   dt
0 t + 2t 2 + 9
2 

(1M)
2
t

1 
 1
t 2 − 2t + 3  2  
 (by (b), k = 2)
=
ln 2
+
 2(2) t + 2t + 3 ln  1  
 

 2  0

(1M)
2
 1  t 2 − 2t + 3  1  1 t 
=  ln  2
 
−
 4  t + 2t + 3  ln 2  2   0
1  3 
3 
hundred km2
=  ln   +

 4  11  4 ln 2 
(1A: accept 0.7572 hundred km2)
1  t 2 − 2t + 3  1  1 
(ii) A(t ) = ln  2
  + C , where C is a constant.
−
4  t + 2t + 3  ln 2  2 
A(0) = 8
0
1  02 − 2(0) + 3  1  1 
ln
−
  +C =8
4  02 + 2(0) + 3  ln 2  2 
1
(1M)
C = 8+
ln 2
t
1  t 2 − 2t + 3  1  1 
1
−
 A(t ) = ln  2
  +8+

4  t + 2t + 3  ln 2  2 
ln 2
The estimated area of the forest after a very long time
 1  t 2 − 2t + 3  1  1 t
1 
−
+8+
= lim  ln  2
(1M)




t → 4
ln 2 
 t + 2t + 3  ln 2  2 


 2 3 
t
 1  1 − t + t 2  1
1
1
−
lim   + 8 +
= lim  ln 
(1M)


t → 4
t →  2 
2 3
ln 2

 1 + + 2   ln 2

t t  

1  1− 0 + 0  1
1
(0) + 8 +
= ln 
−
4  1 + 0 + 0  ln 2
ln 2
1 

2
(1A: accept 9.4427 hundred km2)
= 8 +
 hundred km
 ln 2 
t
2023/24-S6-EXAM1-MATH-EXTENDED MODULE 1
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