Final exam Fall 2017, questions and answers PDF

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DATE: December 9, 2017

International College of Manitoba FINAL EXAMINATION TITLE PAGE

DEPARTMENT & COURSE NO: MATH1500

TIME: 2 hours

EXAMINATION: Introduction to Calculus

EXAMINER: Various

NAME: (Print in ink) STUDENT ID: SIGNATURE: (in ink) (I understand that cheating is a serious offense) CHECK THE BOX IN FRONT OF YOUR INSTRUCTORS NAME  Maryna Shulyakova

 Yaser Maddahi (Monday morning)

 Oumar Gueye

 Yaser Maddahi (Monday afternoon)

 Liliana Menjivar

 Yaser Maddahi (Tuesday morning)

 Hamidreza Farhadi

 Yaser Maddahi (Tuesday afternoon)

 Sushil Kumar (Monday class)

 Sushil Kumar (Wednesday class)

 Liangjin Yao (Tuesday class)

 Liangjin Yao (Thursday class)

 Vladimir Nosov

INSTRUCTIONS TO STUDENTS: This is a 2 hours exam. Show all your work and justify your answers. Unjustified answers will

Question

Points

1

10

2

13

3

12

4

6

have all the pages. You may remove the blank pages if

5

4

you want, but be careful not to loosen the staples.

6

21

Answer each question on the exam paper in the space

7

7

Total:

73

receive LITTLE or NO CREDIT. No texts, notes, or other aids are permitted. Calculators, cell phones, electronic translators, and other electronic devices are not permitted. This exam has a title page, 7 pages of questions, and 2 blank pages for rough work. Please check that you

provided beneath the question. If you need more room, you may continue your work on the revised side of the page, but CLEARLY INDICATE that your work is continued.

Score

International College of Manitoba FINAL EXAMINATION

DATE: December 9, 2017

PAGE: 1 of 9 DEPARTMENT & COURSE NO: MATH1500

TIME: 2 hours

EXAMINATION: Introduction to Calculus

EXAMINER: Various

1. For each of the following limits, find the limit, if it exists. If the limit does not exist, explicitly indicate and briefly justify whether it approaches positive infinity (∞), negative infinity (−∞), or neither. (a) [3 points] lim

t→−1

√

t2 + 8 − 3 t2 + 3t + 2

Solution: √ √ √ t2 + 8 − 3 t2 + 8 + 3 t2 + 8 − 3 lim √ (1 mark) = lim t→−1 t2 + 3t + 2 t→−1 (t + 1)(t + 2) t2 + 8 + 3 t2 − 1 √ = lim t→−1 (t + 1)(t + 2)( t2 + 8 + 3) (t − 1)(t + 1) √ (1 mark) = lim t→−1 (t + 1)(t + 2)( t2 + 8 + 3) (t − 1) √ = lim t→−1 (t + 2)( t2 + 8 + 3) −2 1 = = − (1 mark) 3 (1)(6)

(b) [4 points]

lim

√

x→−∞

3x2 − 5x − 2 3x − 4

Solution: √ p √ 3x2 − 5x − 2 x2 3 − 5/x − 2/x2 lim (1 mark) = lim x→−∞ x→−∞ x(3 − 4/x) 3x p− 4 |x| 3 − 5/x − 2/x2 = lim (1 mark) x→−∞ px(3 − 4/x) −x 3 − 5/x − 2/x2 (1 mark) = lim x→−∞ x(3 − 4/x) √ 3 =− (1 mark) 3

(c) [3 points] lim

sin(θ + 2) − 3θ − 10

θ→−2 θ 2

Solution: sin(θ + 2) sin(θ + 2) = lim lim 2 (0.5 mark) θ→−2 (θ − 5)(θ + 2) θ→−2 θ − 3θ − 10 sin(θ + 2) 1 = lim (1 mark) lim θ→−2 θ→−2 θ − 5 θ+2 ) (1 mark) 7 1 = (1)(− 1

= −7 (0.5 mark)

International College of Manitoba FINAL EXAMINATION

DATE: December 9, 2017

PAGE: 2 of 9 DEPARTMENT & COURSE NO: MATH1500

TIME: 2 hours

EXAMINATION: Introduction to Calculus

EXAMINER: Various

dy of each of the following functions. DO NOT SIMPLIFY YOUR ANSWER AFTER dx YOU EVALUATE THE DERIVATIVE. Circle your final answer.

2. Find the derivative

√ (a) [3 points] y = ( x − x4 )3 − 2x csc x Solution: √ √ ′ ′ y ′ = 3( x − x4 )2 ( x − x4 ) − 2x csc x (x csc x) ln 2 (1 mark) √ 1 y ′ = 3( x − x4 )2 ( √ − 4x3 ) − 2x csc x (csc x − x. csc x. cot x) ln 2 (2 mark) 2 x

(b) [4 points] y =

R e3x√ 3

sec t dt

Solution: Let u = e3x ,

(c) [6 points] sec

√

du = 3e3x (1 mark) dx Z u√ dy d sec t dt (1 mark) = dx dx 3 Z u √ d du = sec t dt · (1 mark) du 3 dx √ = 3 sec e3x e3x (1 mark)

x − y = 3xy − y 2

Solution: √

x − y)′ = (3xy − y 2 )′ (0.5 mark, each 0.5 mark) z }| { √ √ 1 − y′ = 3xy . ln 3 . (y + xy ′ ) − 2yy ′ (3.5 marks, each 0.5 mark) sec x − y tan x − y . √ | {z } 2 x − y |{z} |{z} | {z } |{z} | {z } z }| { √ √ sec x − y tan x − y xy y.3 . ln 3 − √ 2 x−y ′ √ √ (2 marks, each 1 mark) y = sec x − y tan x − y xy − √ − x.3 ln 3 + 2y 2 x−y (sec

OR,

′

|

z

sec

{z

√

}| { √ x −√y tan x − y − y.3xy . ln 3

}

y = sec |

√

2 x −√y x −√y tan x − y + x.3xy ln 3 − 2y {z } 2 x−y

International College of Manitoba FINAL EXAMINATION

DATE: December 9, 2017

PAGE: 3 of 9 DEPARTMENT & COURSE NO: MATH1500

TIME: 2 hours

EXAMINATION: Introduction to Calculus

EXAMINER: Various

3. Evaluate the following integrals.

(a) [4 points]

Z

0

π/2 

 2ex + 3 cos x dx

Solution:  π/2  = 2ex + 3 sin x)  (1 mark, 0.5 mark for each term) 0     π/2 0 = 2e + 3 sin(π/2) − 2e + 3 sin(0) (1 mark, 0.5 mark each substitution) = 2eπ/2 + 3 − 2 − 0

(1 mark, 0.25 mark for each term)

= 2eπ/2 + 1 (1 mark)

(b) [3 points]

Z 

x + 1 sec2 x + √ dx x

Solution:

=

Z 

 sec2 x + x1/2 + x−1/2 dx (1 mark)

2 = tan x + x3/2 + 2x1/2 + C (2 marks, 0.5 mark for each term) 3

Z 3    − 2 dx (c) [5 points] x 1

Solution:

Since x − 2 = 0 =⇒ x = 2 and 2 ∈ (1, 3)

(0.5 mark), the integral should be re-written in the

following form:   (x − 2) (This line could also be written as by definition of absolue value, |x − 2| =  −(x − 2) Z 3    x − 2 dx (1 mark, 0.5 mark for each term) − x − 2 dx + 2 Z1 2 h i3 i2 h 1 1 2 = − x − 2x + x2 − 2x (1 mark, 0.5 mark for each term) 1 2 2 2 =



−

1 2





1

2



1

2



1

2



if 2 < x < 3 if 1 < x < 2

(2) + 2(2) − 2 = 1 (0.5 mark)

(2) − 2(2) − − (1) + 2(1) + (3) − 2(3) (2 marks, 2 0.5 mark for each term) 2 2

DATE: December 9, 2017

International College of Manitoba FINAL EXAMINATION PAGE: 4 of 9

DEPARTMENT & COURSE NO: MATH1500

TIME: 2 hours

EXAMINATION: Introduction to Calculus

EXAMINER: Various

4. [6 points] Find the area of the region bounded by the graph of y = x2 − 3x + 2 and below the x-axis. Solution: y = x2 − 3x + 2 = (x − 1)(x − 2) = 0. Limits of Integration: x = 1, 2 (1 mark) Z 2 (x2 − 3x + 2) dx (1 mark) 1  2 x3 3 2  (1.5 mark) − x + 2x = 3 2 1   (1)3 3  (2)3 3  − (2)2 + 2(2) − − (1)2 + 2(1) = 3 2 3 2 1 (0.5 mark) =−  6   1 1 AREA =  −  = units2 (1 mark) 6 6

5. [4 points] Find the values of a and b that make f (x) continuous at x = 1.    2ax2 − b   f (x) = −x2 + 2ax    3

if x < 1 if x > 1 if x = 1.

Solution: Show that lim− f (x) = lim+ f (x) = f (1). x→1

x→1

(1 mark)

lim f (x) = lim− 2ax2 − b = 2a − b = 3.

x→1−

x→1

(1 mark)

lim f (x) = lim −x2 + 2ax = −1 + 2a = 3. +

x→1+

x→1

a = 2 and b = 1 (1 mark).

(1 mark)

(1 mark)...