Exam (elaborations)

Calculus I (MAT 1320) Final Exam Questions and Solutions

13 views 0 purchase

Course
Calculus I (MAT1320)

Institution
University Of Ottawa (U Of O )

Questions and solutions to Calculus I Final Exams (Summer 2020) at the University of Ottawa.

[Show more]

Preview 5 out of 22 pages

View example

Uploaded on March 22, 2023
Number of pages 22
Written in 2020/2021
Type Exam (elaborations)
Contains Questions & answers

calculus
finals
calculus exam
solutions
2020
questions
answers
study
prep
test
material
practice
university
ottawa
mat
1320
summer
exam
of

CA$11.43

Added

Add to cart Add to wishlist

100% satisfaction guarantee
Immediately available after payment
Both online and in PDF
No strings attached

lOMoARcPSD|23478024

MAT1320 Final Exam (With Solutions)

Calculus I (University of Ottawa)

Studocu is not sponsored or endorsed by any college or university
Downloaded by Lizy Steven (lizysteven832@gmail.com)

, lOMoARcPSD|23478024

MAT1320X Solution to Final Examination Summer 2020

Solution to the Final Examination
MAT1320X, Summer 2020

Part I. Multiple-Choice Questions
3  10 = 30 points

In all questions, (A) is the right answer.

1.1. The domain of the function f (x) = 1  ln( x  e) is

(A) e < x  2e; (B) e  x  2e; (C) x > 2e; (D) x < e.

Solution. 1  ln(x  e)  0, ln(x  e)  1, 0 < x  e  e, e < x  2e.

1.2. The domain of the function f (x) = 1  ln(e  x ) is

(A) 0  x < e; (B) 0  x  e; (C) x < 0; (D) x  e.

Solution. 1  ln(e  x)  0, ln(e  x)  1, 0 < e  x  e, 0  x < e.

1.3. The domain of the function f (x) = 1  ln(e  x ) is

(A) e < x  0; (B) e  x < 0; (C) x < e; (D) x  0.

Solution. 1  ln(e + x)  0, ln(e + x)  1, 0 < e + x  e. e < x  0.

2.1. Some values of functions f (x) and g(x), and their derivatives f '(x) and g'(x) are given in the
following table:

x f (x) f '(x) g(x) g'(x)

1 3 1 2 3
2 1 2 3 5
3 2 4 1 7

Let z = h(x) = (f  g)(x), what is h(1) + h'(1)?

(A) 7; (B) 14; (C) 8; (D) 22.

Solution. h(1) = f (g(1)) = f (2) = 1. h'(1) = f '(g(1))g'(1) = f '(2)g'(1) = 2  3 = 6. h(1) + h'(1) = 7.

2.2. Some values of functions f (x) and g(x), and their derivatives f '(x) and g'(x) are given in the
following table:

1

Downloaded by Lizy Steven (lizysteven832@gmail.com)

, lOMoARcPSD|23478024

MAT1320X Solution to Final Examination Summer 2020

x f (x) f '(x) g(x) g'(x)

1 1 1 3 3
2 3 2 1 5
3 2 4 2 7

Let z = h(x) = (f  g)(x), what is h(1) + h'(1)?

(A) 14; (B) 7; (C) 8; (D) 20.

Solution. h(1) = f (g(1)) = f (3) = 2. h'(1) = f '(g(1))g'(1) = f '(3)g'(1) = 4  3 = 12. h(1) + h'(1) =
14.

2.3. Some values of functions f (x) and g(x), and their derivatives f '(x) and g'(x) are given in the
following table:

x f (x) f '(x) g(x) g'(x)

1 3 1 1 3
2 1 2 3 5
3 2 4 2 7

Let z = h(x) = (f  g)(x), what is h(2) + h'(2)?

(A) 22; (B) 14; (C) 7; (D) 8.

Solution. h(2) = f (g(2)) = f (3) = 2. h'(2) = f '(g(2))g'(2) = f '(3)g'(2) = 4  5 = 20. h(2) + h'(2) =
22.
2
e( x )
3.1. The derivative of the function f (x) = at x = 2is
x

7 4 4 4 7 4 3 4
(A) e ; (B) e ; (C) e ; (D) e .
4 3 3 4
2 2 2
2 x 2e x  e x e x (2 x 2  1) 7
Solution. By the quotient rule, f '(x) = 2
 2
. When x =2, f '(2) = e4 .
x x 4

sin 2 x 
3.2. The derivative of the function f (x) = at x = is
x 4

4(  2) 4  2 4(  2) 4  2
(A) ; (B) ; (C) ; (D) .
 2
 2
 2
2

2

Downloaded by Lizy Steven (lizysteven832@gmail.com)

, lOMoARcPSD|23478024

MAT1320X Solution to Final Examination Summer 2020

2 x sin x cos x  sin 2 x 
Solution. (B) By the quotient rule, f '(x) = 2
. When x = ,
x 4
 1 1
 
   2 2 2 4(  2)
f '    .
4  
2
2
 
4

(ln x ) 2
3.3. The derivative of the function f (x) = at x = e is
x

(A) e2; (B) e; (C) e−1; (D) e2.

ln x
2 x  (ln x ) 2
x ln x (2  ln x )
Solution. (E) Use the quotient rule. f '(x) = 2
 . When x = e, f '(e)
x x2
= e−2.

(11  3x ) e x 1
2

4.1. Let f (x) = . Then f '(1) =
( x  1) 3  2 x

9 5 7 5
(A) ; (B) ; (C)  ; (D)  .
2 2 3 4

Solution. Taking the logarithm on both sides,

2 1
ln f (x) = ln(11  3x )  ( x 2  1)  ln( x  1)  ln(3  2 x) .
3 2

Then take the derivative with respect to x on both sides:

f '( x ) 2 1 1
  2x   .
f ( x) 11  3x x 1 3  2x

 2 1 1 
f '(x) = f (x)    2x   .
 11  3x x  1 3  2x 

(11  3) e0  2 1  9
When x = 1, f (1) =  2 , f '(1) = 2    2   1  .
2 1  8 2  2

(3x  11)1/ 3 e x 1
2

4.2. Let f (x) = . Then f '(1) =
( x  3) x  5

3

Downloaded by Lizy Steven (lizysteven832@gmail.com)

, lOMoARcPSD|23478024

MAT1320X Solution to Final Examination Summer 2020

5 7 5 9
(A)  ; (B)  ; (C) ; (D)  .
4 3 2 2

Solution. Taking the logarithm on both sides,

1 1
ln f (x) = ln(3x  11)  ( x 2  1)  ln( x  3)  ln( x  5) .
3 2

Then take the derivative with respect to x on both sides:

f '( x ) 1 1 1
  2x   .
f ( x ) 3x  11 x  3 2( x  5)

 1 1 1 
f '(x) = f (x)   2x   .
 3x  11 x  3 2( x  5) 

( 3  11)1/ 3 e0 1 11 1 1 5
When x = 1, f (1) =  , f '(1) =   2      .
22 2 28 2 8 4

(2 x  5) e x 4
2

4.3. Let f (x) = . Then f '(2) =
( x  4) 3x  7

7 9 5 5
(A)  ; (B) ; (C) ; (D)  .
3 2 2 4

Solution. Taking the logarithm on both sides,

2 1
ln f (x) = ln(2 x  5)  ( x 2  4)  ln( x  4)  ln(3x  7) .
3 2

Then take the derivative with respect to x on both sides:

f '( x ) 4 1 3
  2x   .
f ( x ) 3(2 x  5) x  4 2(3x  7)

 4 1 3 
f '(x) = f (x)   2x   .
 3(2 x  5) x  4 2(3x  7) 

( 4  5) e0 1 14 1 3 2 7
When x = 2, f (2) =  , f '(1) =   4      3   .
2 1 2 23 2 2 3 3

4

Downloaded by Lizy Steven (lizysteven832@gmail.com)

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller lizysteven. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for CA$11.43. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

60904 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling