F.

6 Final Examination 2021 – 2022

Mathematics Extended Part
Module 1 (Calculus and Statistics)
Suggested Solution

Solution Marks
1(a) 2 p  (1  ap)  3 p  1
(5  a) p  0
a  5 or p  0 (rejected)
E( X )  1 2 p  4  (1  5 p)  6  3 p
4
Var( X )  12  2 p  42  (1  5 p)  62  3 p  42
 30 p
1(b) Var(2 X  a)  6E(2 X  a)
22 Var( X )  6[2E( X )  a]
4  30 p  6(2  4  5)
p  0.15
2(a) The required probability
= 1  (1  0.4)7  C17 (1  0.4)6(0.4)
= 0.841 369 6
= 0.841 4, cor. to 4 d.p.
2(b) P(David spends more than $30 in restaurant A for breakfast on a certain day)
= 0.4  0.7
= 0.28
The required probability
= 1  (1  0.28)7  C17 (1  0.28)6(0.28)
 0.626 638 293
= 0.626 6, cor. to 4 d.p.
2(c) The required probability
0.626 638 293
 0.841 369 6

= 0.744 8, cor. to 4 d.p.

3(a) P( A  B)  P( B) P( A | B)
 kP( A)  (1  0.55)
 0.45kP( A)
P( A  B)  P( A)  P( B)  P( A  B)
0.846  P( A)  kP( A)  0.45kP( A)
 (1  0.55k ) P( A)
0.846
P( A) 
1  0.55k

1
 Solution Marks
3(b)  k  0 and P( A)  0
 P( A  B)  0
 A and B are not mutually exclusive.
3(c) P( A)  1  P( A ' | B)
0.846
 1  0.55
1  0.55k
1  0.55k  1.88
k  1.6
4(a)  1
2

 
e k  1 x  

x
1 k 2
x  ...
k 2!
1 1
 1  x  2 x 2  ...
k 2k
4(b) 
x
e (1  kx )6
k

 1 
  1  x  2 x 2  ...  1  C16 ( kx )  C26 ( kx ) 2  ...
1
 k 2k 
 1 1 
  1  x  2 x 2  ...  1  6kx  15k 2 x 2  ... 
 k 2k 
 1   1 
Coefficient of x2 is (1)(15k 2 )  (1)  2     ( 6k )
 2k   k 
1
 15k 2  6
2k 2
1 529
15k 2  2
6
2k 8
120k 4  481k 2 4  0
1
∴ k  2 or k  
120
∴ k  2 (∵ k is a positive integer)
 1 
Coefficient of x is (1)( 6k )    (1)
 k 
25

2
5(a) Sample mean
= 32 cm
A 95% confidence interval for µ
 4 4 
=  32  1.96  , 32  1.96  
 20 20 
= (30.246 9 , 33.753 1), cor. to 4 d.p.

2
 Solution Marks
5(b) Let n be the sample size.
Width of a 99% confidence interval
 4 
= 2(2.575) 
 n
20.6
=
n
20.6
∴ 2
n
n  10.3
n  106.09
∴ The least sample size is 107.
6(a) 1
dv 1  43 
3

Let v  x 4 . Then  x , i.e. x 4 dx  4dv .

dx 4
3 3 1
 
 x f ( x )dx   x e dx
4 4  x4

 4  e  v dv
 4e  v  C
1

 4e  x  C
4

6(b) g (u)  e  u (u 3  3u 2  6u  3)
dg (u ) dg (u ) du

dx du dx
 1 3 
 [ e  u (u 3  3u 2  6u  3)  e  u (3u 2  6u  6)]  x 4 
4 
1   3

 e  u ( u 3  3u 2  6u  3  3u 2  6u  6)  x 4 
4 
 1 3 
3 1

 (3  x 4 )e  x  x 4 
4

4 
3  1
3 1

  x 4   e x
4

4 4
3  13

∴ h( x )  x 4 
4 4
16  3  1 4
3 1
6(c) 16 dg (u )
0 dx  0  x 4   e  x dx
dx  4 4
1 1
3 16  34  x 4 1 16 4
4 0
[ g (u )]02  x e dx  0 e  x dx
4
1 3 1
16 16 
0 e  x dx  30 x 4 e  x dx  4[ g (u )]02
4 4

 3(4)[e  x ]160  4{e 2 [23  3(22 )  6(2)  3]  e0 (3)}

4

 12e 16  12  4(35e 2  3)

4

 24  152e 2
∴ The required area is 24  152e2 .

3
 Solution Marks
7(a) dy ( x 3  3x  6)e x  (3x 2  3)e x
=
dx ( x 3  3x  6)2
( x 3  3x 2  3x  9)e x
=
( x 3  3x  6)2
7(b) dy
When = 0,
dx
x3  3x2  3x + 9 = 0
(x  3)(x2  3) = 0
x=3 or x =  3 (rejected)
x 2x<3 x=3 3<x6
dy
 0 +
dx

∴ The value of y is at the least when x = 3.

e3
∴ The least value of y is (or 0.836897371).
24
e2
When x = 2, y = or 0.923632012
8
e6
When x = 6, y = or 1.977592125
204
e6
∴ The greatest value of y is (or 1.977592125)
204
8(a) dy  1 
  
dx x 2  25 
5k ( 2 )  52
k  1
8(b) (i) By substituting P(2, n) into L: x – 25y + 73 = 0, we have

2  25n  73  0
n3

(ii) y   5 x dx
5 x
 C
 ln 5
When x = 2, y = 3.
52
3 C
 ln 5
1
C  3
25ln 5
5 x 1
∴ The equation of the curve S is y    3 .
ln 5 25ln 5

4
 Solution Marks
9(a) The required probability
1.80 e 1.8 1.81 e 1.8 1.82 e1.8
= + +
0! 1! 2!
1.8
= 4.42e
= 0.730 6, cor. to 4 d.p.
9(b) The required probability
(1.2  12)9 e  (1.2  12)
=
9!
= 0.040 9, cor. to 4 d.p.
9(c) The required probability

 1.80 e 1.8   1.24 e 1.2   1.81 e 1.8   1.23 e 1.2   1.82 e 1.8   1.22 e 1.2 
=   +  +  +
 0!   4!   1!   3!   2!   2! 

 1.83 e 1.8   1.21 e 1.2   1.84 e 1.8   1.20 e 1.2 

 3!   1!  +  4!   0! 
     

= 3.375e3
= 0.168 0, cor. to 4 d.p.
9(d) The required probability
 1.82 e 1.8   1.2 2 e 1.2 
 2!   2! 
=  
3

3.375e
 216 
= 0.345 6  or 
 625 
9(e) P(total number of breakdowns of the two machines in a month is less than 3)

 1.20 e 1.2   1.80 e 1.8 1.81 e 1.8   1.21 e 1.2 

= 4.42e1.8   + 
1!   1! 
+
 0!   0!

 1.80 e 1.8   1.22 e 1.2 

 0!   2! 
  

= 8.5e3
The required probability
 1.80 e 1.8 1.81 e 1.8   1.21 e 1.2 
 0!  1!   1! 
=  3
 
8.5e
 168 
= 0.395 3, cor. to 4 d.p.  or 
 425 

5
 Solution Marks
10(a)  105  101 
P0  Z   = 0.5  0.308 5
  
= 0.191 5
105  101
∴ = 0.5

 =8
The required probability
 95  101 105  101 
= P Z 
 8 8 
= P(0.75  Z  0.5)
= 0.273 4 + 0.191 5
= 0.464 9
10(b) The required probability

= C14 (0.464 9)(1  0.464 9)3(0.464 9)

= 0.132 5, cor. to 4 d.p.

10(c) The required probability
= C35 (0.464 9)3(1  0.464 9)2 + C45 (0.464 9)4(1  0.464 9) + (0.464 9)5
 0.434 403 398
= 0.434 4, cor. to 4 d.p.
10(d) (i) The required probability
 99  101 101  101 
= P Z 
 8 8 
= P(0.25  Z  0)
= 0.098 7

(ii) P(at least 3 participants get cash coupons and only 1 of them gets an
extra gift)
= C15C24 (0.098 7)(0.464 9  0.098 7)2(1  0.464 9)2 +
C15C34 (0.098 7)(0.464 9  0.098 7)3(1  0.464 9) +
C15 (0.098 7)(0.464 9  0.098 7)
4

 0.174 443 281

The required probability
0.174 443 281
 0.434 403 398

= 0.401 6, cor. to 4 d.p.

6
 Solution Marks
11(a) (i)
A1
1  2 1
   1ln1  2 ln 2  2(1.2 ln1.2  1.4 ln1.4  1.6ln1.6  1.8ln1.8)
2 5 
 0.638 603196
 0.6386 (cor. to 4 d.p.)

(ii)

1 f ( x )dx  1 f ( x )dx  2 f ( x )dx

4 2 4

15
16ln 2   A1  A2
4
15
A2  16ln 2   A1
4
15
 16ln 2   0.638 603196
4
 6.701 751 692
 6.7018 (cor. to 4 d.p.)
11(b) 1
f ( x )  ln x  x  
x
 ln x  1
1
f ( x ) 
x
11(c) (i)
du 1
Let u  ln x . Then,  .
dx x
dx
We have du  .
x
When x = 1, u = 0; when x = 4, u = 2ln2.
2
4 (ln x )
B  1 dx
x
 0 u 2 du
2ln 2

2ln 2
u 
3

 
 3 0
(2 ln 2) 3
 0
3
8(ln 2) 3

3

7
 Solution Marks
(ii)
1
By (b), we have f ( x )   0 for x > 0.
x
By (a)(i), we have A1  0.638 603196 .
By (a)(ii), we have A2  A1  6.063148 496 .
8(ln 2) 3
B 3
∴ 
A2  A1 6.063148 496
 0.146 469 402
 0.15
∴ Dickson’s claim is agreed.
12(a) 5
P=
1  a 5bt
5
1 + a5  bt =
P
5
 1 = a5  bt
P
5 
ln   1 = ln a5  bt
P 
5 
ln   1 = (5  bt) ln a
P 
5 
ln   1 = (b ln a)t + 5 ln a
P 
12(b) (i)
5 
When t = 0, ln   1 = ln 32.
P 
∴ 5 ln a = ln 32
a =2
5  5
When ln   1 = 0, t = .
P  4
5
∴ 0 = (b ln 2)   + 5 ln 2
4
b=4

8
 Solution Marks
(ii)
5
P=
1  2 5 4 t
dP 5(254t ln 2)( 4)
=
dt (1  254t )2
(20ln 2)(254t )
=
(1  254t )2
d 2P
dt 2
(1  254t )2 (254t ln 2)( 4)  (254t )(2)(1  254t )(254t ln 2)( 4)
= (20 ln 2)
(1  254t )4
1  254t  (254 t )(2)
= 80 (ln 2)2 (25  4t)
(1  254t )3
80(ln 2)2 (254t )(254t  1)
=
(1  254t )3

(iii)
The estimated population
 5 
= tlim  5 4t 
million
  1  2 
5
= million
1 0
= 5 million
12(c) d 2P
When = 0,
dt 2
80(ln 2) 2 (2 54t )( 2 54t  1)
=0
(1  2 54t ) 3
25  4t  1 = 0
5  4t = 0
5
t=
4
5 5 5
t 0t< t= t>
4 4 4
d 2P
+ 0 
dt 2

dP 5
∴ attains its greatest value when t = .
dt 4

9
 Solution Marks
dW dW dP
= 
dt dP dt

 3  P 3  (20 ln 2)( 25  4t )
=   ln  1M
 2  2  (1  25  4t ) 2

5
When t = ,
4
5
P= 5
5  4 
1 2 4

= 2.5
5
5  4 
dW  3  2.5 3  (20 ln 2)[ 2  4  ]
∴ 5 =
  ln 
dt  2  2   2
5  4  
t 5
4
1  2   4

 
 
 3.872 372 26
<4
∴ The rate of change of the weight of waste disposal does not exceed
dP
4 units per year when attains its greatest value.
dt
∴ The claim is not correct.

10