Calculus I (MAT 1320) Final Exam Questions and Solutions

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MAT1320 Final Exam (With Solutions)


Calculus I (University of Ottawa)




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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


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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

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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


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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


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