(question & solution) ASSESSMENT 1 MAT183 (TEST 30%)

CONFIDENTIAL CS/MAY 2022/MAT183

UNIVERSITI TEKNOLOGI MARA

CAWANGAN PAHANG (KAMPUS JENGKA)
ASSESSMENT 1

COURSE : CALCULUS I
COURSE CODE : MAT183
DATE : 13 MAY 2022
TIME : 3.00 PM – 4.30 PM (90 minutes)- Answering time

4.30 PM – 5.00 PM (30 minutes) – Submission time

1. This question paper consists of SIX (6) questions.

2. Answer ALL the questions and show all the necessary working solution.

3. Write your answer CLEARLY using BLUE or BLACK pen. It is your responsibility that the
copy is legible.

4. DO NOT COPY your friend's answers. NO DISCUSSION is allowed. Any similar answers
and steps of solution will be penalized.

5. Write down your FULL NAME, STUDENT ID and GROUP in all your answer sheets.

Use any mobile device app to capture your answers, then combine and convert all answers
6. into a SINGLE PDF FILE. Name your PDF file in the format <FULLNAME>_<GROUP>.

Example: ABU BIN ALI_EC1101Z

7. Any LATE submission will NOT BE EVALUATED.

8. Answer ALL questions in English.

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO

© Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

CONFIDENTIAL 2 CS/MAY 2022/MAT183

Question 1

Find the following limits.

a) lim 2x 2x 6 3
2 5x 
x 3

b) lim 3  x2  5
x2 x  2

c) lim 3x2  2
x 3
x 

(10 marks)

Question 2

The function of f x is defined as follows:

 3x , x1
 2

 x2  5  4x 1x5
f  x    x  5 , 2

 x2  2k , x  5



Find the value of k such that lim f  x  exist. Hence, check the continuity at x = 5.
x 5
(5 marks)

Question 3
Find the following limits.
lim sin 3x
x0 tan 2x

(4 marks)

© Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

CONFIDENTIAL 3 CS/MAY 2022/MAT183

Question 4

Use the definition of derivative to find f  x  for f  x   2x  3  4x .

(5 marks)

Question 5
Find dy for the following:

dx

 a) y  x2  5x  2 sin3 4x

4x  73

b) y  ex5

(9 marks)

Question 6

Find dy for ln2x  y   sin2 3x  e5y3  3xy using implicit differentiation.

dx

(7 marks)

END OF QUESTION PAPER
ALL THE BEST

© Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 1

No. SUGGESTED SOLUTION MARKS

1(a) 2 − 6 = 2( − 3)

→3 2 2 − 5 − 3 →3 ( − 3)(2 + 1)

= 2

→3 (2 + 1) 2

2
=7 #

3 − √ 2 + 5 3 − √ 2 + 5 + √ +
1(b) = ∙
− 2 − 2 + √ +
→2 →2

(3)2 − (√ 2 + 5)2
=

→2 ( − 2)(3 + √ 2 + 5)

9 − ( 2 + 5)
=

→2 ( − 2)(3 + √ 2 + 5)

9 − 2 − 5
=

→2 ( − 2)(3 + √ 2 + 5)

4 − 2
=

→2 ( − 2)(3 + √ 2 + 5)

(2 + )(2 − )
=

→2 ( − 2)(3 + √ 2 + 5)

−(2 + )( − 2)
=

→2 ( − 2)(3 + √ 2 + 5)

−(2 + )
=

→2 (3 + √ 2 + 5)

−(2 + )
=

(3 + √ 2 + 5)

−4
= (3 + 3)

−4 4
=6

2
=−3 #

1

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 1 SUGGESTED SOLUTION MARKS
No.

1(c) √ 2 (3 − 2 2)
(1 − 3 )
√3 2 − 2 =
− 3
→−∞ →−∞

| |√3 − 2
2
=
(1 − 3 )
→−∞

− √3 − 2
2
=
(1 − 3 )
→−∞

√3 − 2
(1 2
= −
− 3 )
→−∞

√3 − 2
∞
= −
(1 − −3∞)

√3 − 0 4
= − (1 + 0)

= −√3 #

2

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 2

No. SUGGESTED SOLUTION MARKS
2.
→ 5 − 2 − 4 − 5 = ( 2 − 2 ) 3
−5 2
→5+

→ 5 − ( + 1)( − 5) = 25 − 2

− 5

→ 5 −( + 1) = 25 − 2

5 + 1 = 25 − 2

6 = 25 − 2

2 = 19

19
= 2 #

( ) = 2 − 19
(5) = 52 − 19 = 25 − 19 = 6 ( )

2 − 4 − 5 ( + 1)( − 5) = → 5 −( + 1) = 6
→ 5 − − 5 = → 5 −
− 5

( 2 − 2 ) = ( 2 − 19) = 25 − 19 = 6
→5+ →5+

( ) = ( ) = 6
→5− →5+

( ) = 6 ( )

→5

( ) = (5) = 6

→5

the function is continuous at = 5

3

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 3

No. SUGGESTED SOLUTION MARKS

Find 3 4
2
→0

3 = 3

→0 2 →0 ( )

= 3

→0

= 3 ∙ )
(
→0

3 ( )
= ( ) ∙ 2
2 →0
→0

= 3 ∙ (( 1 1 ))) ∙ (0)
( 2
→0

= ( 3 ) ∙ 1
( 2 )
→0

= ( 3 )
( 2 )
→0



→0

= 3 ( 3 3 )
2 ( 2 2 )
→0



→0

3(1)
= 2(1)

3
=2 #

4

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 4 MARKS
No. SUGGESTED SOLUTION

( ) = √2 − 3 − 4
( + ℎ) = √2( + ℎ) − 3 − 4( + ℎ)

= √2 + 2ℎ − 3 − 4 − 4ℎ

′( ) = ( + ℎ) − ( )
ℎ
ℎ→0

= √2 + 2ℎ − 3 − 4 − 4ℎ − (√2 − 3 − 4 )

ℎ→0 ℎ

= √2 + 2ℎ − 3 − 4 − 4ℎ − √2 − 3 + 4
ℎ
ℎ→0

= √2 + 2ℎ − 3 − 4ℎ − √2 − 3

ℎ→0 ℎ

= √2 + 2ℎ −3 − √2 − 3 − 4ℎ
ℎ ℎ
ℎ→0 ℎ→0

= ( √2 + 2ℎ − 3 − √2 − 3 √2 + 2ℎ − 3 + √2 − 3
∙ ) − 4
ℎ→0 ℎ √2 + 2ℎ − 3 + √2 − 3 ℎ→0

= (√2 + 2ℎ − 3)2 − (√2 − 3)2 −4

ℎ→0 ℎ(√2 + 2ℎ − 3 + √2 − 3)

= (2 + 2ℎ − 3) − (2 − 3) −4

ℎ→0 ℎ(√2 + 2ℎ − 3 + √2 − 3)

= 2 + 2ℎ − 3 − 2 + 3 −4

ℎ→0 ℎ(√2 + 2ℎ − 3 + √2 − 3)

= 2ℎ −4

ℎ→0 ℎ(√2 + 2ℎ − 3 + √2 − 3)

= 2 −4

ℎ→0 √2 + 2ℎ − 3 + √2 − 3

2 5
= −4

√2 − 3 + √2 − 3

2
= −4

2√2 − 3

1
= −4#

√2 − 3

5

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 5

No. SUGGESTED SOLUTION MARKS
a)
= ( 2 − 5 + 2) = 3(4 )
′ = (2 − 5) ′ = 3 2(4 ) ∙ (4 ) ∙ 4
′ = 12 2(4 ) (4 )

= ( 2 − 5 + 2) 3(4 )

= ′ + ′


= (2 − 5) 3(4 ) + 12( 2 − 5 + 2) 2(4 ) (4 ) 4

= 2(4 )((2 − 5) (4 ) + 12( 2 − 5 + 2) (4 ))

b) = (4 − 7)3 = +5

′ = 3(4 − 7)2 ∙ 4 ′ = +5 ∙ 1

′ = 12(4 − 7)2 ′ = +5

(4 − 7)3 5
= +5
′ − ′
= 2

12 +5(4 − 7)2 − +5(4 − 7)3
= ( +5)2

+5(4 − 7)2 (12 − (4 − 7))
= ( +5)2

(4 − 7)2 (12 − 4 + 7)
= +5

(4 − 7)2 (19 − 4 )
= +5

6

ASSESSMENT 1/ MAT183/ Suggested Solution MAY 2022

Question 6 SUGGESTED SOLUTION MARKS
No. (2 + ) − 2(3 ) = 5 +3 + 3

1 = 3 =
+ ) ) ′ = 3
2 ∙ (2 + − 2 (3 ) ∙ (3 ) ∙ 3 = 5 +3 ∙ (5 + ′ + ′ ′ =


2 + 1 − 6 (3 ) (3 ) = 5 5 +3 + 3 + 3
+ +
2 2

2 1 − 5 5 +3 − 3 = 3 − 2 + 6 (3 ) (3 )
+ 2 +

1 − 5 5 +3 − 3 ) = 3 − 2 + 6 (3 ) (3 )
(2 + 2 +

3 − 2 + 6 (3 ) (3 )
2 + − 5 5 +3 − 3
=
1
2 + 7

(2 + ) (3 − 2 + 6 (3 ) (3 ))
2 +
=
(2 + ) (2 1 − 5 5 +3 − 3 )
+

3 (2 + ) − 2 + 6(2 + ) (3 ) (3 )
=
1 − 5(2 + ) 5 +3 − 3 (2 + )

7