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  1:
x!2
x!2+
From (1) and (2) we have
2. If y =
p x
x ; then y 0 (2) =
(A) 1 + ln
p
(B) 2 2
a = 4 and b = 7 ) a + b = 11:
p 
2
(C) 1
(D) 1 + ln 2
Solution: We take the natural logarithm of both sides
ln y = ln
p x
p
1
x = 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 = 2 ) y 0 (2) = (1 + ln 2) = 1 + ln 2:
2
2
(2)
3. If y = x sin1 x +
p
1  x2 ; then
dy
=
dx
(A) x sin1 x
(B) sin1 x
(C) 0
2x
1  x2
Solution: We will use the product rule for solving this problem
(D) sin1 x  p
y0
=
=
=

0
p

0
x sin1 x + 1  x2 = 1: sin1 x + x sin1 x +


1
2x
sin1 x + x p
 p
1  x2
2 1  x2
x
x
sin1 x + p
p
= sin1 x:
2
2
1x
1x
4. The equation of the tangent line to the curve y =

0
1  x2
p
2 1  x2
2
at x = 2 is
x
(A) y = 0:5x + 3
(B) y = 0:5x
(C) y = 0:5x + 2
(D) y = 0:5x + 2
Solution: First we will Önd the derivative of the curve equation and then evaluate it at the point
(1; 2) to Önd the slope of the tangent at the point (1; 2)
y0 = 
2
2
1
) m =  2 =  and y = 1:
x2
2
(2)
The equation of the tangent line is
1
1
y =  (x  2) + 1 =  x + 2
2
2
5. If f (x) =
1  sin x
, then f 0 ( 2 ) =
1 + sin x
(A) 0:5
(B) 0:5
(C) 2
(D) 0
Solution: We will use the quotient rule for solving this problem
f 0 (x)
=
=
Then
0
0
(1 + sin x) (1  sin x)  (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

0
f 0( ) = 
2 = = 0:

2
4
1 + sin 2
3
=
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  lim f (x)  lim 3x2  x + 3
x!3
x!3
27 
8. Let
f (x) =
x!3
lim f (x)  27 ) lim f (x) = 27:
x!3
8
6
>
;
>
>
< (5  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
6
1
1
p 6= lim
) 6=

2
4
2 x x!1 (5  x) (4 + 2x)
1
For x < 1; f is not continuous at x =  .
2
For x > 1; f has no point of discontinuity.
Then the number of points of discontinuity of f is 2:
4
9. limx!1
p

4x2 + 4x  2x =
(A) 1
(B) 1
(C) 2
(D) 0
Solution: For this problem we have to use conjugate
lim
x!1
p
4x2

+ 4x  2x
 2

p
 p4x2 + 4x + 2x
4x + 4x  4x2
2
 = lim p

= lim
4x + 4x  2x p
x!1
x!1
4x2 + 4x + 2x
4x2 + 4x + 2x
4x
4x
 = lim q
= lim p



2
x!1
x!1
4x + 4x + 2x
4x2 1 + 1 + 2x
x
=
=
10. limx!3
lim 
x!1
x2  x  12
=
jx + 3j
4x
q
 = lim
q



x!1
2x 1 + x1 + 2x
2x
1 + x1 + 1
lim q
x!1
4x
2
1+
1
x

2
 = 1:
 = p
(1 + 0) + 1
+1
(A) 7
(B) 7
(C) 1
(D) 1
Solution:We have
8 2
x  x  12
>
>
; x+3>0
>
<
x+3
Then
8
(x + 3) (x  4)
>
>
= x  4; x > 3;
>
<
x+3
x2  x  12
=
=
>
>
jx + 3j
2
>
>
(x + 3) (x  4)
x

x

12
>
>
:
:
=  (x  4) ; x < 3:
; x+3<0
 (x + 3)
 (x + 3)
lim
x!3
11. limx!2
x2  x  12
= lim x  4 = 1:
jx + 3j
x!3
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
2  4
=
= 1:
x (x + 2)
2 (0)
5
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 Önd the H.A.
4 + x5
4x2 + 5x
4
= lim
= = 2 ) y = 2 is a H.A:
2
x!1 2x  2
x!1 2  22
2
x
lim
To Önd the V.A., we check the limits at x = 1 (the zeros of the denominator)
lim+
x!1
4x2 + 5x
= 1 and
2 (x2  1)
lim +
x!1
13. What are the domain and range of f (x) =
4x2 + 5x
= 1 ) x = 1 are V.A.
2 (x2  1)
x4  6x2 + 9
x2 + 4
(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) :
 2
2
x 3
x4  6x2 + 9
f (x) =
=
 0 ) The range of f (x) is [0; 1)
x2 + 4
x2 + 4
14. Find the slope of the line 6x  5y = 5
(A) 
(B)
6
5
(C)
5
6
(D) 
6
5
5
6
Solution: We have to rewrite the equation in general equation for a line y = mx + b; that is
6x  5y = 5 ) 5y = 6x  5 ) y =
Then, the slope of the line 6x  5y = 5 is m = 65 :
6
6
x  1:
5
2
15. Write an equation for the graph obtained by shifting the graph of y = (x  3)  1 up 2 units and right
2 units
2
(A) y = (x  1)  3
2
(B) y = (x  1) + 1
2
(C) y = (x  5) + 1
2
(D) y = (x  5)  3
2
2
Solution: Shifting up by 2 adds 2, then f (x) + 2 = (x  3)  1 + 2 = (x  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
2 = (x  2  3) + 1 = (x  5) + 1::
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
3
3
y =  (x  0) + 2 =  x + 2
4
4
Q2) Evaluate the following limit
lim
t!0
p
1  t2 
t2
Solution: To Önd this limit we have to use conjugate
p
p
p
p
1  t 2  1 + t2 1  t2 + 1 + t 2
p
p
lim
t!0
t2
1  t2 + 1 + t 2
p
1 + t2

 

1  t2  1 + t 2
p

= lim p
t!0 t2
1  t2 + 1 + t 2
2t2
p

1  t2 + 1 + t 2
2
p

= lim p
t!0
1  t 2 + 1 + t2
2
 = 1:
p
= p
10+ 1+0
=
Q3) If f (x) =
lim
t!0 t2
2
2x
and (f  g) (x) =
; Önd g (x) :
x+1
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
6
3
= :
8
4