2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

1. f (2)
f (2  h)  f (2)
 lim 1M
h 0 h
(2  h) 2  1  22  1
 lim
h 0 h
h 2  4h  3  3
 lim 1M
h 0
h( h 2  4h  3  3)
h(h  4)
 lim
h 0
h( h 2  4h  3  3)
h4
 lim 1M withhold 1M if the step is skipped
h 0
h 2  4h  3  3
2
 1A
3

(4)
3
 2
2. (a) 1  
 x
2 3
2 2 2
 1  3   3     1M
x x x
6 12 8
 1  2  3 1A
x x x

(b) Note that ( x  k )5  x5  C15 kx4  C25 k 2 x3   k5 . 1M

C k  6C k  12  12
5
2
2 5
1
1M
10k  30k  0
2

k (k  3)  0
k  0 (rejected) or k  3 1A

(5)
3. (a) OP
3 1
 a b
1 3 1 3
3 1
 a b 1A
4 4

3 1  3 1  2
(b) (i)  a  b    a  b   OP 1M
4 4  4 4 
2
9 2  3  1  1 2  17 
a  2    a  b  b   
16  4  4  16  8 
9  3  1  1 17
( 2) 2  2    a  b  (2) 2 
16   
4 4 16 8
a b  2 1A

2021-DSE-MATH-EP(M2)-MS 1 1 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

a b
(ii) cos AOB  1M
a b
2

( 2)(2)
1

2

AOB  1A
4

(5)
4. (a) f ( x)
( x  5)(2)  (2 x  4)(1)
 1M
( x  5) 2
6
 1A
( x  5) 2

2x  4
(b) y
x5
When y  0 , we have
2x  4
0
x 5
x2
∴ G cuts the x-axis at A(2, 0). 1A
6
f (2)  
(2  5) 2
2

3
Slope of the normal to G at A
 2
 1    
 3
3
 1M
2
The equation of the normal to G at A is
3
y  0  ( x  2) 1M
2
3x  2 y  6  0 1A

(6)

2021-DSE-MATH-EP(M2)-MS 2 2 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

5. (a) (i) Note that
1 1 
2 2 1
3  2
 2(2)  (1)(3)   (2)( )    (1)(2)(2)   (2)(3) 1
 2  7  5
2

 (2  5)(  1)
Since (E) has a unique solution, we have
1 1 
2 2 1 0 1M
3  2
i.e. (2  5)(  1)  0
5
So, we have   and   1 . 1A
2
5 5
Thus, we have   1 , 1    or   .
2 2

(ii) Since (E) has a unique solution, we have

1 1 1
2 2 5
3  t
z 1M for Cramer’s Rule
(2  5)(  1)
2t  5(3)  2  5  (1)(2)(t )  2(3)

(2  5)(  1)
4t  3  21

(2  5)(  1) 1A

(b) When   1 , the augmented matrix of (E) is

 1 1 1 1   1 1 1 1 
   
 2 2 1 5  ~  0 4 1 3 
 3 1 2 t   0 4 1 t  3 
   
 1 1 1 1 
 
1 3 
~ 0 1 
 4 4  1M
0 0 0 t  6 

Since (E) has infinitely many solutions,
we have t  6 .
Thus, the solution set of (E) is
 3u  7 u  3  
 , , u  : u  R . 1A
 4 4  
(7)

2021-DSE-MATH-EP(M2)-MS 3 3 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

6. (a)  sin(ln x) dx
 x sin(ln x)   x d [sin(ln x)]
1
 x sin(ln x)   x cos(ln x)  dx 1M
x
 x sin(ln x)   cos(ln x) dx
 x sin(ln x)  x cos(ln x)   x d [cos(ln x )]
 x sin(ln x)  x cos(ln x)   sin(ln x) dx 1M

So, we have

 sin(ln x) dx  x sin(ln x)  x cos(ln x)   sin(ln x) dx

2 sin(ln x) dx  x sin(ln x)  x cos(ln x)  C 1 1M
1
 sin(ln x) dx  2 x[sin(ln x)  cos(ln x)]  C 1A

e 1
(b) 0
sin[ln (x  1)] dx


  sin(ln u ) du (by letting u  x  1 )

e

1
1M

e
1 
  u[sin(ln u )  cos(ln u )] (by (a)) 1M
2 1
1 1
 e (sin   cos  )  (sin 0  cos 0)
2 2

e 1
 1A
2

Alternative Solution
Let u  ln( x  1) , we have
eu  x  1 and dx  eu du .
e 1
0
sin[ln (x  1)] dx

  eu sin u du 1M
0

  sin u d (eu )
0

 [eu sin u ]0   eu cos u du
0

 0   cos u d (eu )
0

 [eu cos u ]0   eu sin u du
0

2021-DSE-MATH-EP(M2)-MS 4 4 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

So, we have
 
 e sin u du  [eu cos u ]0   eu sin u du
u
0 0

2 e u
sin u du  [eu cos u ]0
0
e 1 1
 0
sin[ln (x  1)] dx   (e cos   e0 cos 0)
2
1M

e 1
 1A
2

(7)
7. (a) f ( x)   (2  2 x) dx
 2 x  x2  C 1M
Since  passes through the point (–2, 7), we have
2(2)  (2) 2  C  7 1M
C  15
∴ The equation of  is y   x 2  2 x  15 . 1A

(b) (i) By substituting the equation of L into the equation

of , we have
x  9   x 2  2 x  15 1M
x2  x  6  0
( x  2)( x  3)  0
x  2 or x  3
By substituting x  2 into the equation of L, we
have
y  2  9  7
By substituting x  3 into the equation of L, we
have
y  3  9  12
∴ The coordinates of the required points are
(–2, 7) and (3, 12). 1A

2021-DSE-MATH-EP(M2)-MS 5 5 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

(ii) By substituting y  0 into the equation of , we
have
 x 2  2 x  15  0
x 2  2 x  15  0
( x  3)( x  5)  0
x  3 or x  5
∴ The x-intercepts of  are –3 and 5.
The required area
2
  ( x 2  2 x  15) dx   ( x  9) dx
3
1M
3 2

  ( x 2  2 x  15) dx
5

3
2 3
 x3   x2  1M
    x 2  15 x     9 x 
 3  3  2  2
5
 x3 
    x 2  15 x 
 3 3
11 95 40
  
3 2 3
129
 1A
2

(8)
8. (a) When n  1 ,
L.H.S.  sin x
1  cos 2 x 2sin 2 x
R.H.S.    sin x 1
2sin x 2sin x
∴ L.H.S. = R.H.S.
∴ The statement is true for n  1 .
m
1  cos 2mx
Assume that  sin(2k  1) x 
k 1 2sin x
for some 1M

positive integer m.
When n  m  1 ,
m 1
L.H.S.   sin(2k  1) x
k 1
m
  sin(2k  1) x  sin[2( m  1)  1] x
k 1

1  cos 2mx
  sin(2m  1) x 1M by induction assumption
2sin x
1  cos 2mx  2sin(2m  1) x sin x

2sin x
1  cos 2mx  [cos 2mx  cos(2m  2) x]
 1M
2sin x
1  cos 2(m  1) x

2sin x
 R.H.S.
∴ The statement is true for n  m  1 .

2021-DSE-MATH-EP(M2)-MS 6 6 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

By mathematical induction, we have
n
1  cos 2nx
 sin(2k  1) x 
k 1 2sin x
for all positive

integers n. 1

101 103 105 199

sin  sin  sin   sin
(b) 12 12 12 12

cos
12
       
sin [2(51)  1]     sin [2(52)  1]    
  12     12  
   
 sin [2(100)  1]   
  12  


cos
12
100
   

k  51
sin (2k  1)   
  12  


cos
12
100
    50
   
 sin (2k  1)  12     sin (2k  1)  12  
k 1 k 1
 1M

cos
12
       
1  cos  2(100)    1  cos  2(50)   
  12     12  
 
2sin 2sin
 12 12 for using (a)
 1M
cos
12
  1   1 
1    2    1  2 
     
 
2sin cos
12 12
1

 
sin 2  
 12 
2 1A
(8)
9. (a) (i) y  ax  bx 3

dy
 a  3bx 2
dx
∵ (2, 16) is a stationary point.
dy
∴ 0 1M
dx x  2
a  12b  0 (1)

2021-DSE-MATH-EP(M2)-MS 7 7 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

By substituting (2, 16) into C1 , we have,
2a  8b  16 1M
a  4b  8 (2)
By solving (1) and (2), we have
a  12 , b  1 1A

dy
(ii)  12  3x 2
dx
d2y
 6 x 1M
dx 2
d2y
 6(2)
dx 2 x  2
 12  0
∴ (2, 16) is a maximum point of C1 . 1A

d2y
(iii) Solving  0 , we have x  0 .
dx 2
By substituting x  0 into C1 , we have
y  12(0)  (0)3  0
x0 x0 x0
2
d y
+ 0 – 1M
dx 2
∴ (0, 0) is the point of inflexion of C1 . 1A
(7)
(b) (i) PQ  (12r  r )  ln(12r  r )
3 3

 12r  r 3  ln(12r  r 3 ) 1A

(ii) Let A square units be the area of △OPQ.

1
A r[12r  r 3  ln(12r  r 3 )] 1M
2
1
 [12r 2  r 4  r ln(12r  r 3 )]
2
dA 1  r (12  3r 2 )  dr
  24r  4r 3  ln(12r  r 3 )   1M
dt 2  12r  r 3  dt
24(2)  4(2)3  ln[12(2)  23 ]
dA 1 
  2[12  3(2) 2 ] 4 1M
dt r  2 2   
 12(2)  2 3

 32  2 ln16
∴ The required rate of change is 32  2ln16 1A
square units per minute.
(5)

2021-DSE-MATH-EP(M2)-MS 8 8 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

10. (a) (i) sec x  sec x tan x
2

1 1 sin x
 2
 
cos x cos x cos x
1  sin x

cos 2 x
1  sin x
 1M
1  sin 2 x
1  sin x

(1  sin x)(1  sin x)
1
 1
1  sin x

 
1
(ii) 0
4
1  sin x
dx   4 (sec 2 x  sec x tan x) dx
0

 
  4 sec 2 x dx   4 sec x tan x dx 1M
0 0
 
 [tan x]04  [sec x]04 1M
 1 2 1
 2 1A

(5)
(b) (i) Let u    x , then du  dx .

0
f (    x) cos(  x) dx

  f (  x) cos(  x) dx
0

   f (u ) cos(u ) du
0
1M


  f ( x) cos x dx 1
0

1 1
(ii) Note that  .
1  sin(  x) 1  sin x
1
Let f ( x)  .
1  sin x
 
1  cos   x 

 4  dx
04  3 
1  sin   x
 4 
 
  cos   x 
4
1
dx   4  4  dx
0  3  0  3 
1  sin   x 1  sin   x
 4   4 

2021-DSE-MATH-EP(M2)-MS 9 9 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

 
  cos   x 
4
1
dx   4  4  dx
0   0   
1  sin   x  1  sin     x 
4   4 

1 cos x
   du   4
0
dx 1M for using (b) (i)
4 1  sin u
0 1  sin x

 
1 1
4 du   4 d (1  sin x)
0 1  sin u 0 1  sin x


 2  [ln(1  sin x)]04 (by (a)(i)) 1M
 1 
 2  ln 1   1A
 2

(5)
(c) The required volume
2

 1  cos x 
   4   dx 1M
4  1  sin x 


  1 
cos x 
    4 dx   4 dx 
 1  sin x  1  sin x
 4 4 
  
  ([tan x] 4  [sec x] 4  [ln(1  sin x)] 4 ) 1M
  
4 4 4

   1   1   
  (1  1)  ( 2  2)  ln 1    ln 1   
   2  2   
  2  1 
   2  ln    1A
  2  1  

(3)
11. (a) (i) ( P  2I )( P  3I )  4I
a 2 2  a  3 2   4 0
   
 1 b  2  1 b  3  0 4 
 (a  2)(a  3)  2 2(a  2)  2(b  3)   4 0 
  
 a 3b  2 2  (b  2)(b  3)   0 4 
 a 2  5a  8 2(a  b  5)   4 0 
   1M
 a b5 b 2  5b  8   0 4 
So, we have
 a 2  5a  8  4

2(a  b  5)  0
 1M
 a b5  0
 b 2  5b  8  4

∴ a  4, b  1 1A for both correct

2021-DSE-MATH-EP(M2)-MS 10 10 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

(ii) ( P  2 I )( P  3I )  4 I
P 2  5P  6 I  4 I
5P  P 2  2 I
1  1 
P  (5I  P)    (5I  P)  P  I 1M
2  2 
1
∴ P 1 exists and P 1  (5I  P ) . 1M
2
1  5  4 0  2 
i.e. P 1   
2  0 1 5 1 
 1 
 1
 2  1A
 1 2 

 2 
Alternative Solution
4 2
det P  2 1M
1 1
1
P 1  adj P 1M
det P
1  1 2 
  
2  1 4 
 1 
 1
 2  1A
 1 2 

 2 
(6)

(b) (i) PQP 1

 1 
1
 4 2   0 1  2
   
1 1 2 3   1 2 

 2 
1 0
  1A
0 2
n
1 0
( PQP 1 ) n   
0 2
1 0 
PQ n P 1   n 
1M
0 2 
1 0 
Q n  P 1  n 
P
0 2 
 1 
 2 1  1 0  4 2 
  n  
  1 2  0 2  1 1 
 
 2 
 22 n
1  2n 
 n 1  1A
 2  2 1  2n 1 

2021-DSE-MATH-EP(M2)-MS 11 11 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

0 0  0 0 
(ii) R2    
r 0  r 0 
0 0
  1A
0 0

(iii) A  Q  PT  xI
 0 1  4 1  x 0 
   
 2 3   2 1  0 x 
4 x 0 
 
 4 4  x
When x  4 , we have
0 0
A  1M
 4 0
From (b)(ii), we have
0 0
An    for n  2 1M
0 0
100
∴  (k  1) A
k 1
k
 (1  1) A  2 A when x  4 . 1A

i.e. The claim is agreed.

(7)
12. (a) (i) OA  OB
i j k
 2 3 2 1M
2 2 3
 5i  10 j  10k 1A

(ii) The required unit vector

OA  OB

OA  OB
1
 (5i  10 j  10k ) 1M
(5)  (10) 2  (10) 2
2

1 2 2
  i  j k 1A
3 3 3
(4)
1
(b) (i) Area of △OAB  OA  OB
2
15
 1M
2
Volume of tetrahedron OABC

1
 (area of △OAB)( OC )
3

2021-DSE-MATH-EP(M2)-MS 12 12 © Pearson Education Asia Limited 2020

 2021 Mock Paper (Extended Part) - Module 2 (Marking Scheme)

Solutions Marks Remarks

So, we have
1  15  25
  ( OC ) 
3 2  2
OC  5 1M

 1 2 2 
∴ OC  5   i  j  k 
 3 3 3 
5 10 10 1A
  i  j k
3 3 3

(ii) CA  OA  OC
1 19 4
  i  j k 1M
3 3 3
CB  OB  OC for either one
11 16 1
 i  j k
3 3 3
CA  CB

i j k
1 19 4
 
3 3 3
11 16 1
3 3 3
 5i  5 j  25k 1A

(5)
(c) The acute angle between △OAB and △ABC is equal to
the acute angle between OA  OB and CA  CB . 1M accept using other vectors
Let  be the acute angle between △OAB and △ABC. perpendicular to △OAB and
△ABC
(OA  OB )  (CA  CB )
cos   1M
OA  OB CA  CB
(5i  10 j  10k )  (5i  5 j  25k )

15 (5) 2  52  (25) 2
25  50  250

15 675
15

675
1A
  54.7 (cor. to the nearest 0.1)
(3)

2021-DSE-MATH-EP(M2)-MS 13 13 © Pearson Education Asia Limited 2020