Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Exam Paper for Volume 1
Suggested Solutions and Marking Scheme
Suggested solutions
Marks
Remarks
Section A (30 marks)
1.
(a) (2 − x)5 = 25 + C15 (24 )(− x) + C25 (23 )(− x) 2 + C35 (22 )(− x)3 + ... 1M
= 32 − 80 x + 80 x 2 − 40 x 3 + ...
(b)
1

(2 − x) 5  3 − 
x

1A
3
2
3
 3
1
 1  1 
3
2 
3
= (32 − 80 x + 80 x − 40 x + ...) 3 + C1 (3 ) −  + C2 (3) −  +  −  
 x
 x   x  

27 9 1 

= (32 − 80 x + 80 x 2 − 40 x 3 + ...) 27 −
+ − 
x x 2 x3 

2
3
The required constant term
2.
(a)
= 32(27) + (−80)(−27) + 80(9) + (−40)(−1)
1M
= 3784
1A
d
( xe − x ) = e − x − xe − x
dx
= (1 − x)e − x
1A
xe − x = e − x −
(b)
 xe
−x
d
( xe − x )
dx
dx =  e − x dx − xe− x
1M + 1M
= − e − x (1 + x) + C
HKDSE Mathematics in Action (Extended Part)
1M for using (a)
1A
1
© Pearson Education Asia Limited 2019
Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Suggested solutions
3.
Marks




6

6
(a) ln − 0.1 = ln
− 0.1
6


y



 0.1 + aebt

bt
= ln(ae )
= ln a + bt
Remarks
1M
1A
6

(b) By (a), ln − 0.1 = ln a + bt
y

t
1
2
3
4
6

ln  − 0.1
y

1.31
1.01
0.71
0.41
1M
2
1.5


1

0.5

t
0
1
2
3
1M
4
From the graph, y-intercept  1.6
∴
∵
∴
ln a  1.6
a = 5.0 (cor. to 1 d.p.)
1A
1.31 − 0.41
= −0.3
1− 4
b = − 0.3 (cor. to 1 d.p.)
Slope 
HKDSE Mathematics in Action (Extended Part)
1A
2
© Pearson Education Asia Limited 2019
Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Suggested solutions
4.
Marks
Remarks
f ( x) = 6( x − 1)
f ( x) =  (6 x − 6)dx
1M
= 3 x 2 − 6 x + C1
f ( x) =  (3x 2 − 6 x + C1 )dx
1M
= x 3 − 3x 2 + C1 x + C2
1A
When x = 0, y = 6.
0 3 − 3(0) 2 + C1 (0) + C 2 = 6
C2 = 6
1M
When x = 2, y = 2.
2 3 − 3(2) 2 + C1 (2) + 6 = 2
C1 = 0
∴ The equation of the curve is y = x 3 − 3x 2 + 6 .
5.
for either one*
for either one*
1A
(a) When y = 0,
1M
0 = ( x − 2) 2 − 1
x − 2 = −1 or 1
x = 1 or 3
x = 1 or 9
∴ The x-intercepts of the curve are 1 and 9.
1A
(b) The area of the region bounded by the curve C and the x-axis
9
= −  [( x − 2) 2 − 1]dx
1M
1
9
= −  ( x − 4 x + 3)dx
1
9
1

8 3
= −  x 2 − x 2 + 3x
3
2
1
 9 5
= − − − 
 2 6
16
=
3
HKDSE Mathematics in Action (Extended Part)
1M
1A
3
© Pearson Education Asia Limited 2019
Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Suggested solutions
6.
(a)
Marks
y = ln( x 2 + 2 x + 5)
dy
2x + 2
= 2
dx x + 2 x + 5
2( x + 1)
= 2
x + 2x + 5
1M
Remarks
for chain rule
1A
(b) Slope of the straight line 2 x − 5 y − 3 = 0
2
=
5
2
∴ The slope of the tangent is
.
5
2( x + 1)
2
∴
=
2
x + 2x + 5 5
5x + 5 = x 2 + 2 x + 5
1M
x 2 − 3x = 0
x = 0 or 3
∴ There are two tangents to C which are parallel to the
straight line 2 x − 5 y − 3 = 0 .
∴ The claim is disagreed.
for using (a)
1M
1M
1A
Section B (30 marks)
7.
(a) ∵
∴
P attains its minimum at t =
2
.
3
dP
=0
dt t = 2
3
 2 
h 3  + k 
 3 
=0
2
  2 2  2  
3  − 4  + 4
 3  
  3 
h( 2 + k ) = 0
h = 0 (rejected) or k = − 2
HKDSE Mathematics in Action (Extended Part)
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1M
1A
© Pearson Education Asia Limited 2019
Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Suggested solutions
Marks
From the question, we have
dP
1
=
dt t = 2 2
h[3(2) − 2]
1
=
2
2
[3(2) − 4(2) + 4]
2
4h 1
=
64 2
h=8
Remarks
1M
1A
u = 3t 2 − 4t + 4
du
1A
= 6t − 4
dt
(ii) The total profit made by the company from t = 0 to t = 6
6
8(3t − 2)
=
dt
1M
0 (3t 2 − 4t + 4) 2
6
1
= 4
d (3t 2 − 4t + 4)
1M
2
0 (3t − 4t + 4) 2
(b) (i)
for using (b)(i)
6
1


= 4 − 2
 3t − 4t + 4  0
 1 1
= 4 − + 
 88 4 
21
=
22
1M
1A
Alternative Solution
Let u = 3t 2 − 4t + 4 , we have du = (6t − 4)dt .
When t = 0 , u = 3(0) 2 − 4(0) + 4 = 4 ;
when t = 6 , u = 3(6) 2 − 4(6) + 4 = 88 .
The total profit made by the company from t = 0 to t = 6
6
8(3t − 2)
=
dt
1M
0 (3t 2 − 4t + 4) 2
88 4
=  2 du
1M
4 u
88
 1
= 4− 
 u 4
 1 1
= 4 − + 
 88 4 
21
=
22
HKDSE Mathematics in Action (Extended Part)
1M
1A
5
© Pearson Education Asia Limited 2019
Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Suggested solutions
8.
(a)
Marks
Remarks
R(t ) = 3 ln(t 2 + kt + 1) + 12
R(t ) =
∵
∴
3(2t + k )
t 2 + kt + 1
1M
9
7
3[2(3) + k ] 9
=
32 + 3k + 1 7
42 + 7k = 30 + 9k
R(3) =
1M
k =6
(b) (i)
1A
R(t ) = 3 ln(t 2 + 6t + 1) + 12
∵
∴
8
=2
4
The end-points of the sub-intervals are 0, 2, 4, 6, 8.
t =
8
P =  R(t )dt
1M
0
2
 [ R(0) + 2 R(2) + 2 R(4) + 2 R(6) + R(8)]
2
= {[3 ln(1) + 12] + 2[3 ln(17) + 12] + 2[3 ln( 41) + 12]
+ 2[3 ln( 73) + 12] + [3 ln(113) + 12]}
= 175.2056 (cor. to 4 d.p.)
(ii) R(t ) =
R(t ) =
1M
1A
3(2t + 6)
t 2 + 6t + 1
3[(2)(t 2 + 6t + 1) − (2t + 6) 2 ]
(t 2 + 6t + 1) 2
1M
∴
− 3(2t 2 + 12t + 34)
(t 2 + 6t + 1) 2
 0 for 0  t  8
175.2056 million dollars is an under-estimate of
∴
the total revenue made by the company in the
coming 8 months.
1A
The total revenue must be greater than 175 million
∴
dollars.
The claim is agreed.
=
9.
1A
(a) By substituting t = 1 and C = 31.15 into C (t ) = pte − qt ,
we have
p (1)e − q (1) = 31.15
1M
for either one*
−q
pe = 31.15
ln p − q = ln 31.15 (1)
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© Pearson Education Asia Limited 2019
Module 1 Volume 1 Exam Paper - Full Solutions
Assessment Kit
Suggested solutions
Marks
By substituting t = 2 and C = 48.52 into C (t ) = pte− qt ,
we have
p (2)e − q ( 2) = 48.52
Remarks
for either one*
−2 q
2 pe = 48.52
ln p − 2q = ln 24.26 (2)
(1) – (2):
q = ln 31.15 − ln 24.26
= 0.25 (cor. to 2 sig. fig.)
1M
1A
(1)  2 – (2):
ln p = 2 ln 31.15 − ln 24.26
p = e 2 ln 31.15−ln 24.26
= 40 (cor. to 2 sig. fig.)
(b) (i)
1A
C (t ) = 40te −0.25t
C (t ) = 40e −0.25t − 10te −0.25t
= 10e −0.25t (4 − t )
Solving C (t ) = 0 , we have
1M
1M
10e −0.25t (4 − t ) = 0
t=4
The signs of C(t) are as follows:
t
t>4
0t<4
t=4
C (t )
0
+
−
1M
From the table, C(t) attains its maximum at t = 4.
When t = 4,
C (t ) = 40(4)e −0.25( 4 )
1M
 58.86
 60
∴ The concentration of poisons will not exceed 60 mg/L.
(ii) C (t ) = 10[−0.25e
−0.25t
(4 − t ) − e
−0.25t
for C(t) = 0
for testing
1A
]
= 10e −0.25t [−0.25(4 − t ) − 1]
= 10e −0.25t (0.25t − 2)
Solving C (t ) = 0 , we have
1M
10e −0.25t (0.25t − 2) = 0
t =8
The signs of C (t ) are as follows:
t
0t<8
t=8
8 < t  10
C (t )
0
−
+
∴
1M
The rate of change of the concentration of poisons
decreases for 0  t < 8, and the rate of the
concentration of poisons increases for 8 < t  10.
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1A
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