Q&A Assessment 2 MAT438 (DEC 2021)

FSKM/ MATEMATIK/UiTM Pahang



ONLINE ASSESSMENT 2/ MAT438/ DEC 2021 (ODL)

UNIVERSITI TEKNOLOGI
MARA PAHANG

MAT438
ONLINE ASSESSMENT 2
10:00 am – 11:10 am (70 minutes) - Answer time
11:10 am – 11:40 am (30 minutes) – Submission time

Name : ________________________________________________
UiTM ID : ________________________________________________
Group : ________________________________________________
Lecturer : ________________________________________________

Instructions to candidates

• Answer ALL Questions.
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not allowed to share your answers with others.
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• Upload all your answers in one file in pdf format in folder ‘ASSESSMENT 2 MAT438’

in google classroom.
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the submission.
• Save your file name as (example)

1. Ahmad Bin Husin (AS2551A)

Your name Your group

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Quiz 3 Test 2 Test 3 Assessment 2
(5 Marks) (15 Marks) (15 Marks) TOTAL

(35 Marks)

ONLINE ASSESSMENT 2/ MAT438/ DEC 2021 (ODL)

PART A : QUIZ 3 (5 marks) 2% (5 marks)

Question 1 (4 marks)
Solve the hyperbolic equation (6 marks)

2 ℎ2(2 ) + 10 = 10 ℎ(2 )

PART B : TEST 2 (15 marks) 10%

Question 2:


Find for the following ∶
a) = 2 ℎ( (2 ))

b) ( 2 ) = 2 + ℎ(3 + 1)

Question 3

Evaluate ∫ 1 ℎ (3 ) (5 marks)

0 9 + ℎ (3 )

PART C : TEST 3 (15 marks) 10%

Question 4

( )a) Differentiate y = sinh−1 ex2 −4 with respect to x . Hence, find dy when x = 2.
dx
(4 marks)

b) Solve cos 2x dx by substitution method. (5 marks)

(sin(2x))2 − 5

Question 5 (6 marks)

Solve (1− 4x ) e3xdx using integration by parts method.

END OF QUESTION PAPER
ALL THE BEST!!!

1

ONLINE ASSESSMENT 2/ MAT438/ DEC 2021 (ODL)

APPENDIX 1 (1)

TABLE OF INTEGRALS

(ax + b)n +1 + C; n  −1
 + 1) n = −1
1. (ax + b)n dx =  a(n

 1 ln | ax + b | +C;
 a

2. 1 dx = ln | x | +C
x

3. sin(ax)dx = − 1 cos(ax) + C
a

4. cos(ax) dx = 1 sin(ax) + C
a

5. sec 2 (ax)dx = 1 tan(ax) + C
a

6. sec (ax)dx = 1 ln | sec (ax) + tan(ax) | +C
a

7. sec (ax) tan(ax)dx = 1 sec (ax) + C
a

8. sinh (ax) dx = 1 cosh(ax) + C
a

9. cosh(ax)dx = 1 sinh (ax) + C
a

10. sec h2 (ax)dx = 1 tanh (ax) + C
a

11. csc h2 (ax)dx = − 1 coth(ax) + C
a

12. sec h(ax) tanh (ax)dx = − 1 sec h(ax) + C
a

13. csc h(ax)coth(ax) dx = − 1 csc h(ax) + C
a

14. 1 dx = sin−1  x  + C
a2 − x2 a

2

ONLINE ASSESSMENT 2/ MAT438/ DEC 2021 (ODL)

APPENDIX 1 (2)

15. a2 1 x2 dx = 1 tan −1  x  +C
+ a  a 

16. 1 dx = 1 sec −1  x  + Ca
x x2 − a2 a

17. 1 dx = sinh−1  x  + C =ln | x + a2 + x2 | + C
a2 + x2 a

18. 1 dx = cosh−1  x  + C =ln | x + x2 − a2 | + C, if x  a
x2 − a2 a

1 tanh −1  x  + C, if | x | a
a coth−1  a  + C, if | x | a
19. 1 dx = 1 ln x +a + C =   x 
a2 − x2 2a x −a  1

a  a 

20. 1 dx = − 1 sec h−1 x + C = − 1 ln a + a2 − x2 + C, if 0  x  a
x a2 − x2 a aa x

21. 1 dx = − 1 csc h−1 x + C = − 1 ln a + a2 + x2 + C, if x  0aaa x
x a2 + x2

TRIGONOMETRIC IDENTITIES

1. sin2 x + cos2 x = 1
2. sin 2x = 2 sin x cos x
3. cos 2x = cos2 x − sin2 x

HYPERBOLIC FUNCTIONS

1. sinh x = e x − e −x
2

2. cosh x = ex + e−x
2

3. cosh2 x − sinh2 x = 1

3



ONLINE ASSESSMENT 2/MAT438/ Solution DEC 2021 (ODL)
Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

PART A : QUIZ 3 (5 marks) 2%

Question 1 : (5 marks)
Solve the hyperbolic equation

2 ℎ2(2 ) + 10 = 10 ℎ(2 )

Solution Q1 :

2 ℎ2(2 ) + 10 = 10 ℎ(2 )

2 (1 + ℎ2(2 )) + 10 = 10 ℎ(2 ) M1

2 (1 + 2) + 10 = 10 ℎ = ℎ(2 )

2 + 2 2 + 10 = 10

2 2 − 10 + 12 = 0 M1
2 − 5 + 6 = 0

( − 2)( − 3) = 0 M1

− 2 = 0, − 3 = 0

= 2 = 3

ℎ(2 ) = 2 ℎ(2 ) = 3 M1
2 = ℎ−1(2) 2 = ℎ−1(3)

= 0.7218 # = 0.9092 # A1 5

1

ONLINE ASSESSMENT 2/MAT438/ Solution DEC 2021 (ODL)
Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

PART B : TEST 2 (15 marks) 10%

Question 2 : (4 marks)
(6 marks)

Find for the following ∶ 4
a) = 2 ℎ( (2 ))
b) ( 2 ) = 2 + ℎ(3 + 1)

Solution Q2a :
= 2 ℎ( (2 ))

= 2 = ℎ ( (2 ))
′ = 2 ∙ 2
′ = − ℎ2( (2 )) ∙ 1 ∙ 2
′ = 2 2 2

M1 ′ = − ℎ2( (2 ))


M1 M1

Differentiate using Product Rule,

= ′ + ′


= 2 2 ℎ ( (2 )) − 2 ℎ2 ( (2 )) A# 1


2

ONLINE ASSESSMENT 2/MAT438/ Solution DEC 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

Solution Q2b : ( 2 ) = 2 + ℎ(3 + 1)

= ( 2 ) = 2 + = 2 = ℎ(3 + 1)
′ = − ℎ(3 + 1) ℎ(3 + 1) ∙ (3 + 0)
= 2 + ′ = 2 ∙ (2 ′ = −3 ℎ(3 + 1) ℎ(3 + 1)
)
1 1 M1
′ = 2∙ +

′ = 2 2

′ = 2 + 1

M1

M1 M1

( 2 ) = 2 + ℎ(3 + 1)

= +

, ′ = ′ + ′

2 1 = 2 2 − 3 ℎ(3 + 1) ℎ(3 + 1)
+

2 1 M1

+ 3 ℎ(3 + 1) ℎ(3 + 1) = 2 2 −

2 + 3 ℎ(3 + 1) ℎ(3 + 1) = (2 2 − 1
)

= 2 + 3 ℎ(3 + 1) ℎ(3 + 1) # A1


2 2 − 1


6

3

ONLINE ASSESSMENT 2/MAT438/ Solution DEC 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

Question 3 :

Evaluate ∫ 1 ℎ (3 ) (5 marks)

0 9 + ℎ (3 )

Solution Q3 :

ℎ (3 ) ℎ(3 ) M1
∫ 9 + ℎ (3 ) = ∫ ∙ 3 ℎ(3 ) M1

= 9 + ℎ (3 )

1 = ℎ(3 ) ∙ 3
= ∫ ∙ 3

11 M1
= 3 ∫ = 3 ℎ(3 )

1 M1
3
= | | +

= 1 |9 + ℎ(3 )| +
3

Hence,

∫ 1 ℎ (3 ) = 1 |9 + 1
[3
0 9 + ℎ (3 ) ℎ(3 )|]

0

= 1 |9 + ℎ(3)| − 1 |9 + ℎ(0)| 5
3 3

= 0.2494 # A1

4

ONLINE ASSESSMENT 2/MAT438/ Solution DEC 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

PART C : TEST 3 (15 marks) 10%

No Marks
4
( )4a Differentiate y = sinh−1 ex2 −4 with respect to x . Hence, find dy
dx
when x = 2.

( )y = sinh−1 ex2 −4

dy = √ √1 ex2 −4 2x
dx ( )ex2 −4 2

+1

when x = 2

dy = √ √1 e22 −4 2(2) = 4 4
dx ( )e22 −4 2 2

+1

4b Solve  cos 2x 5
dx by substitution method.

(sin(2x))2 − 5

cos(2 x ) cos 2x du
sin2 (2x) −
(u )2 − 5 2 cos 2x
√ = 
dx
5

√= 1 1 du2 a2 = 5 u2 = (sin 2x )2
u2 − 2 a= 5
√u = 2cos2x
5
du = dx
√= 1  u  2cos 2x
2 cosh−1  6  + c
 

= 1 cosh−1  sin(2 x )  + c
2  5 
√
or = 1 ln sin(2x) + sin2(2x) − 5 + c 5
2

5

ONLINE ASSESSMENT 2/MAT438/ Solution DEC 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

No Marks
6
5 Solve (1− 4x ) e3xdx using integration by parts method.

 (1− 4x ) e3xdx

u = 1− 4x dv = e3xdx

√ √du = −4

dx
v = 1 e3x
3

du = −4dx

√ √udv = uv − vdu = (1− 4x ) 1 e3x − 1 e3x (−4dx )

  3 3
 √= (1− 4x ) e3x + 4 e3xdx

33

= (1− 4x ) e3x + 4  1  e3 x + c
3  3 
3 6

√= (1− 4x ) e3x + 4 + c
39

END OF ANSWER SCHEME
HAPPY MARKING

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