anyflip ASSESSMENT 1 MAT438 NOV 2021 (Q&A)

FSKM/ MATEMATIK/UiTM Pahang



ONLINE ASSESSMENT 1/MAT438/NOV2021 (ODL)

UNIVERSITI TEKNOLOGI
MARA PAHANG

MAT438
ONLINE ASSESSMENT 1
3:00 pm – 4:00 pm (60 minutes) - Answer time
4:00 pm – 4:30 pm (30 minutes) – Submission time

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

Instructions to candidates

• Answer ALL Questions.
• You are advised to take 60 minutes to complete a written exam and 30 minutes to

save in pdf format and upload your answers.
• Write your answer CLEARLY using a BLUE or BLACK pen. It is your responsibility

that the copy is legible.
• This assessment is an Open Book Test but not an Open Discussion Test. So you

are not allowed to share your answers with others.
• Any LATE submission will NOT BE EVALUATED.
• Upload all your answers in one file in pdf format in folder ‘ASSESSMENT 1

MAT438’ in google classroom.
• It is your responsibility to make sure that your file is openable. So please recheck

after the submission.

Quiz 1 Quiz 2 Test 1 Assessment 1
(5 Marks) (5 Marks) (20 Marks) TOTAL

(30 Marks)

ONLINE ASSESSMENT 1/ MAT438/ NOV 2021 (ODL)

Quiz 1

1. Simplify sin  2 tan−1  x  using right triangle method. (5 Marks)
  3 


Quiz 2 (5 Marks)
2. By applying implicit differentiation, find dy of the following

dx

tan−1 ( x + y ) − sin(4x) = ey +3 + 3

Test 1
3. Differentiate with respect to x.

) = 3 ( −1(2 ))

−1( 2 ) (5 marks)
) = ( 2 ) (5 marks)
(5 marks)
4. Evaluate (5 Marks)

) ∫ 2(2 ) 1
2
0
4 + 2(2 )

) ∫ 2 ( ℎ ℎ )

√− 2 + 6 − 5

END OF QUESTION PAPER
ALL THE BEST!!!

ONLINE ASSESSMENT 1/ MAT438/ NOV 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 1/ MAT438/ NOV 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 1/MAT438/ Solution NOV 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

PART A (6 marks)

( −1 1 − (6 marks)
(2 ) 4)

Solution :

= −1 (21 ) 1 ඥ1 − 4 2
1
A
= 2 2
2

= 1 = 2

√1 − 4 2
= 2


4 = 1

Therefore,

( −1 (21 ) − 4 ) =
( − 4)

= −
4

1 + 4

(√1 − 4 2 ) − 1 (√1 − 4 2 − 22 )
2 2
==
(√1 − 4 2 (22 + √1 2− 4 2)
1 + 2 ) (1)

√1 − 4 2 − 2 2
=( )∙( )
2 2 + √1 − 4 2 6

√1 − 4 2 − 2
=#

√1 − 4 2 + 2

1

ONLINE ASSESSMENT 1/MAT438/ Solution NOV 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

PART B (6 marks)

Solve the hyperbolic equation (6 marks)
2 ℎ2(2 ) + 10 = 10 ℎ(2 )

Solution :

2 ℎ2(2 ) + 10 = 10 ℎ(2 )

2 (1 + ℎ2(2 )) + 10 = 10 ℎ(2 )

2 (1 + 2) + 10 = 10 ℎ = ℎ(2 )

2 + 2 2 + 10 = 10

2 2 − 10 + 12 = 0

2 − 5 + 6 = 0

( − 2)( − 3) = 0

− 2 = 0, − 3 = 0

= 2 = 3

ℎ(2 ) = 2 ℎ(2 ) = 3

2 = ℎ−1(2) 2 = ℎ−1(3)

= 0.7218 # = 0.9092 #

6

2

ONLINE ASSESSMENT 1/MAT438/ Solution NOV 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

PART C (20 marks) (5 marks)
Question 3 :
a) Differentiate = 3 ( −1(2 )) ℎ .
Solution :

= 3 ( −1(2 ))

= 3 = ( −1(2 ))
′ = 3 ∙ − 3 ∙ 3
′ = −3 3 3 ′ = 1 ∙ 1 + 1 ∙ 2
−1(2 ) (2 )2

′ = (1 + 2
4 2) −1(2 )

= ′ + ′


3 ( −1(2 )) 3 + 2 3 #
= −3 (1 + 4 2) −1(2 )

5

3

ONLINE ASSESSMENT 1/MAT438/ Solution NOV 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

Question 3 : (5 marks)
−1( 2 )

b) Differentiate = ( 2 ) ℎ .

Solution :

−1( 2 ) = ( (2 ))
= ( 2 )
′ = 1 ∙ − (2 ) ∙ 2
= −1( 2 ) (2 )
′ = 1 ∙ 2 ∙ 2
′ = −2 (2 )
ඥ1 − ( 2 )2
′ = 2 2

ඥ1 − 4

′ − ′
= 2

2 2 ( (2 )) −2 (2 ) −1( 2 ))
( √1 − 4 −

= #
2
5
( ( (2 )))

(2 2 ( (2 )) + 2 √1 − 4 (2 ) −1( 2 ) )
= √1 − 4

2

( ( (2 )))

2 2 ( (2 )) + 2 √1 − 4 (2 ) −1( 2 ) #
=
√1 − 4 2( (2 ))

4

ONLINE ASSESSMENT 1/MAT438/ Solution NOV 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

Question 4 :

a) Evaluate ∫ 2(2 ) (5 marks)
2
0
4 + 2(2 )

Solution :

2(2 ) = ∫ 2(2 )
∫ 4 + 2(2 ) (2)2
+ ( (2 ))2 = 2

2(2 ) = (2 )

= ∫ 2 + 2 ∙ 2 2(2 ) 2(2 )

= ∙ 2

11
= 2 ∫ 2 + 2 = 2 2(2 )

= 1 ∙ 1 −1 +
2 ()



= 1 ∙ 1 −1 (2 ) +
2 2 (2)

= 1 −1 (2 ) +
4 (2)

Hence,

2(2 )
+ 2(2 )
∫ 2 = 1 −1 (2 ) 2
[ ( 2 )]
0 4 4 0

= 1 −1 ( ) − 1 −1 (0) 5
4 (2) 4 (2)

=0 #

5

ONLINE ASSESSMENT 1/MAT438/ Solution NOV 2021 (ODL)

Name: ___________________________________________
ID UiTM : ___________________
Group : _________________

Question 4 :

b) ∫ 2 ( ℎ ) (5 marks)

√− 2 + 6 − 5

Solution 4b :

Using completing the squares,
− 2 + 6 − 5 = −( 2 − 6 + 5)

= −[ 2 − 6 + (−3)2− (−3)2 + 5]
= −(( − 3)2 − 4)
= 4 − ( − 3)2

Therefore,

∫ 21 = 2
= 2 ∫ = − 3
√− 2 + 6 − 5 ඥ4 − ( − 3)2
= 1
= 2∫ 1 =

ඥ(2)2 − ( − 3)2 5

= 2∫ 1

√ 2 − 2

= 2 −1 +
()


= 2 −1 −3 + #
( )
2

6