Course: Calculus I
Date: Tuesday, October 11, 2022
Instructor: Dr. Mohammed Kassim
Name:
Exam type: Midterm Exam
Duration: 1200 minutes
Semester: 1st Semester 22/23
ID:
Solution
Instructions:
1. Write your name and student ID number in the space provided above.
2. Mobiles and calculators are not allowed in this exam..
3. Answer all questions.
4. Choose the nearest number to your answer.
5. Write clearly and legibly. You may lose points for messy work.
6. Make sure that you have 16 MCQ+3 Essay Questions (7 pages + cover page).
7. For MCQ, use pen to put the code in capital letter of the correct answer, in the following table, beneath
the question number
MCQ
1
2
3
4
5
6
7
8
9
10
11
12
Answer
Written
Marks
Maximum Marks
Q1
3
Q2
2
Q3
1
Total
6
1
13
14
15
16
Total
PART 1: MULTIPLE CHOICE QUESTIONS
Choose the one alternative that best completes the statement or answers the following question (Each
question is 1.5 point):
1. Consider the function
f (x) =
If f di¤erentiable everywhere, then a + b =
8 2
< x + b;
:
x < 2;
ax + 3; x
2:
(A) 13
(B) 11
(C) 7
(D) 10
Solution: Since f is di¤erentiable everywhere, then
8
< 2x; x < 2;
f 0 (x) =
) lim+ f 0 (x) = lim f 0 (x) ) a = 4:
:
x!2
x!2
a; x 2:
(1)
Also, since f is di¤erentiable everywhere, then it must be continuous everywhere, therefore
) lim f (x) = lim f (x) ) 2a + 3 = 4 + b ) b = 2a
x!2+
x!2
1:
From (1) and (2) we have
2. If y =
p
a = 4 and b = 7 ) a + b = 11:
x
x
(A) 1 + ln
; then y 0 (2) =
p
2
p
(B) 2 2
(C) 1
(D) 1 + ln 2
Solution: We take the natural logarithm of both sides
ln y = ln
p
x
x
p
1
= x ln x = x ln x:
2
We then take derivatives of both sides with respect to x; then we solve for y 0
y0
1
y
= (1 + ln x) ) y 0 = (1 + ln x)
y
2
2
At x = 2 ) y =
p
2
2
= 2 ) y 0 (2) =
2
2
(1 + ln 2) = 1 + ln 2:
2
(2)
1
3. If y = x sin
1
(A) x sin
(B) sin
1
x+
p
1
x2 ; then
dy
=
dx
x
x
(C) 0
2x
1 x2
Solution: We will use the product rule for solving this problem
(D) sin
1
x
p
y0
=
x sin
=
sin
1
=
sin
1
1
x+
p
1
p
x+x
x+ p
x2
p
x
1
4. The equation of the tangent line to the curve y =
(A) y =
0:5x + 3
(B) y =
0:5x
(C) y =
0:5x + 2
x + x sin
1
0
1 x2
0
x + p
2 1 x2
2x
p
2 1 x2
x2
x2
1
= 1: sin
1
1
x
1
0
x2
1
= sin
x:
2
at x = 2 is
x
(D) y = 0:5x + 2
Solution: First we will …nd the derivative of the curve equation and then evaluate it at the point
(1; 2) to …nd the slope of the tangent at the point (1; 2)
y0 =
2
)m=
x2
2
2
(2)
1
and y = 1:
2
=
The equation of the tangent line is
y=
5. If f (x) =
(A)
1
(x
2
1
x+2
2
2) + 1 =
1 sin x
, then f 0 ( 2 ) =
1 + sin x
0:5
(B) 0:5
(C)
2
(D) 0
Solution: We will use the quotient rule for solving this problem
f 0 (x)
=
=
Then
(1 + sin x) (1
0
sin x)
0
(1
sin x) (1 + sin x)
2
(1 + sin x)
(1 + sin x) ( cos x) (1 sin x) (cos x)
2
(1 + sin x)
2 cos 2
f 0( ) =
2
1 + sin 2
3
2
=
0
= 0:
4
=
2 cos x
2:
(1 + sin x)
6. If y = xe2x , then y 00 (2) =
(A) 8e4
(B) 12e4
(C) 4e4
(D) 3e4
Solution: We will use the product rule for solving this problem
y0
y 00
= 2xe2x + e2x :
= 4xe2x + 2e2x + 2e2x = 4xe2x + 4e2x :
Then
y 00 (2) = 8e4 + 4e4 = 12e4 :
7. If x3
x+3
f (x)
3x2
x + 3; then limx!3 f (x) =
(A) 18
(B) 27
(C) 0
(D) 21
Solution: We will use the Sandwich Theorem for solving this problem
lim x3
x!3
x+3
27
8. Let
f (x) =
8
>
>
>
< (5
lim f (x)
lim 3x2
x+3
x!3
x!3
lim f (x)
27 ) lim f (x) = 27:
x!3
6
;
x) (4 + 2x)
>
>
>
:
x!3
x < 1;
1
p ; x > 1:
2 x
Then the number of points of discontinuity of f is
(A) 2
(B) 1
(C) 4
(D) 3
Solution: At x = 1; f is not continuous because
lim
x!1+
1
p 6= lim
2 x x!1 (5
6
1
1
) 6=
x) (4 + 2x)
2
4
1
.
2
For x > 1; f has no point of discontinuity.
For x < 1; f is not continuous at x =
Then the number of points of discontinuity of f is 2:
4
9. limx!1
(A)
p
4x2 + 4x
2x =
1
(B) 1
(C) 2
(D) 0
Solution: For this problem we have to use conjugate
p
lim
x!1
4x2
+ 4x
2x
=
=
=
=
x2
10. limx!3
lim
x!1
lim
x!1
lim
x!1
lim
x!1
p
4x2 + 4x
2x
p
p
4x2 + 4x + 2x
= lim
x!1
4x2 + 4x + 2x
4x
4x
p
= lim q
2
x!1
4x + 4x + 2x
4x2 1 + x1 + 2x
4x
4x
q
q
= lim
x!1
1
2x 1 + x + 2x
2x
1+
q
2
1+
1
x
1
x
4x2 + 4x 4x2
p
4x2 + 4x + 2x
+1
2
= 1:
= p
(1 + 0) + 1
+1
x 12
=
jx + 3j
(A) 7
(B)
7
(C) 1
(D)
1
Solution:We have
x2
Then
8 2
x
x 12
>
>
; x+3>0
>
<
x+3
x 12
=
>
jx + 3j
2
>
>
: x
lim
x2
x!3
11. limx!
2
=
x 12
; x+3<0
(x + 3)
8
(x + 3) (x
>
>
>
<
x+3
4)
>
>
(x + 3) (x 4)
>
:
=
(x + 3)
x 12
= lim x
jx + 3j
x!3
4=
1:
x 4
=
x (x + 2)
(A) 0
(B) 1
(C)
1
(D)
2
Solution: Since ( 2 + 2) is a very small negative value, then
lim
x! 2
x 4
=
x (x + 2)
5
=x
2 4
=
2 ( 0)
1:
4; x >
(x
3;
4) ; x <
3:
12. Which one of the following statements is TRUE about the curve y =
4x2 + 5x
2x2 2
(A) It has y = 2 as a H.A. and x = 2 as a V.A.
(B) It has y = 2 as a H.A. and x =
2 as V.A.
(C) It has y = 4 as a H.A. and x = 1 as a V.A.
(D) It has y = 2 as a H.A. and x =
1 as V.A.
Solution: First we …nd the H.A.
4 + x5
4
4x2 + 5x
= lim
2 = 2 = 2 ) y = 2 is a H.A:
2
1 2x
x! 1 2
2
x2
lim
x!
To …nd the V.A., we check the limits at x =
lim+
x!1
4x2 + 5x
= 1 and
2 (x2 1)
1 (the zeros of the denominator)
lim +
x! 1
13. What are the domain and range of f (x) =
x4
4x2 + 5x
=1)x=
2 (x2 1)
1 are V.A.
6x2 + 9
+4
x2
(A) Domain ( 1; 2) [ ( 2; 2) [ (2; 1) ; range ( 1; 1)
(B) Domain [0; 1); range ( 1; 1)
(C) Domain ( 1; 1) ; range [0; 1)
(D) Domain ( 1; 2) [ ( 2; 2) [ (2; 1) ; range [0; 1)
Solution: The denominator x2 + 4 is always positive and x2 + 4 6= 0 for all x. Then domain of f (x)
is R = ( 1; 1) :
f (x) =
x4
x2 3
6x2 + 9
=
x2 + 4
x2 + 4
14. Find the slope of the line 6x
0 ) The range of f (x) is [0; 1)
5y = 5
6
5
(A)
(B)
6
5
(C)
5
6
(D)
2
5
6
Solution: We have to rewrite the equation in general equation for a line y = mx + b; that is
6x
Then, the slope of the line 6x
5y = 5 ) 5y = 6x
5y = 5 is m = 65 :
6
5)y=
6
x
5
1:
15. Write an equation for the graph obtained by shifting the graph of y = (x
2 units
2
(A) y = (x
1)
(B) y = (x
1) + 1
(C) y = (x
5) + 1
(D) y = (x
5)
2
3)
1 up 2 units and right
3
2
2
2
3
Solution: Shifting up by 2 adds 2, then f (x) + 2 = (x
2
3)
1 + 2 = (x
2
3) + 1;
Shifting the function right by 2 units requires subtracting 2 from x-the parameter of f , that is f (x
2
2
2 = (x 2 3) + 1 = (x 5) + 1::
2)+
16. Determine the intervals on which the function is increasing, decreasing and constant.
(A) Increasing on ( 1; 2) ; decreasing on ( 2; 1) ; constant on (2; 1)
(B) Increasing on (2; 1) ; decreasing on ( 2; 1) ; constant on ( 2; 2)
(C) Increasing on ( 1; 2) ; decreasing on ( 1; 2) ; constant on (2; 1)
(D) Increasing on (2; 1) ; decreasing on ( 1; 2) ; constant on ( 2; 2)
Solution: Clear from the graph of the function that it is Increasing on (2; 1) ; decreasing on ( 1; 2) ;
constant on ( 2; 2) :
7
PART 2: ESSAY QUESTIONS
Write your answer in the space provided (SHOW YOUR ANSWER)
Q1) Find the tangent line to the curve
tan (xy) = xy 3 + 2y 2
8
at the point (0; 2).
Solution: Di¤erentiating implicitly with respect to x, we get
0
sec2 (xy) : (xy) = 3xy 2 y 0 + y 3 + 4yy 0
sec2 (xy) : (xy 0 + y) = 3xy 2 y 0 + y 3 + 4yy 0
At the point (0; 2) ; we have
3
sec2 (0) : (0 + 2) = 0 + (2) + 4 (2) y 0 (0) ) 2 = 8 + 8y 0 (0) ) 8y 0 (0) =
6 ) m = y 0 (0) =
The tangent line to the curve is
y=
3
(x
4
Q2) Evaluate the following limit
lim
3
x+2
4
0) + 2 =
p
p
t2
1
1 + t2
t2
t!0
Solution: To …nd this limit we have to use conjugate
p
p
p
p
1 t2
1 + t2 1 t2 + 1 + t 2
p
p
lim
t!0
t2
1 t2 + 1 + t 2
=
1 t2
1 + t2
p
p
t!0 t2
1 t2 + 1 + t 2
lim
2t2
p
1
+ 1 + t2
2
p
= lim p
t!0
1 t 2 + 1 + t2
2
p
p
=
= 1:
1 0+ 1+0
=
Q3) If f (x) =
2
and (f
x+1
g) (x) =
lim
t!0 t2
2x
; …nd g (x) :
x+3
Solution: We have
f (g (x)) = (f
g) (x) =
2x
x+3
2
2x
=
:
g (x) + 1
x+3
Solve for g (x)
2xg (x) + 2x = 2x + 6
6
3
2xg (x) = 6 ) g (x) =
=
2x
x
GOOD LUCK
8
p
t2
6
=
8
3
:
4