Tutorials, Past Test and Examination Booklets in Mathematics 2018
School of Mathematics
University of the Witwatersrand
CALCULUS (MATH1036)
Tutorial Questions: Semester One
and
Past Test and Examination Questions (2012-2017)
with Selected Solutions
O
THE WIT
W
A
T
IT
Y
F
RS
UNIVER
S
E
RAND
J
MATH I
This book
belongs to
O
R
HA
N N E SBU
NAME
G
STUDENT NUMBER
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
i
Contents
1
MATH1036 Tutorial Questions: Semester One
2
Calculus Tests and Examinations
23
2.1
Calculus 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
2.1.1
Calculus March 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
2.1.2
Calculus May 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
2.1.3
Calculus June 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
2.1.4
Calculus August 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.1.5
Calculus October 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
Calculus 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.2.1
Calculus March 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.2.2
Calculus April 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
2.2.3
Calculus June 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.2.4
Calculus August 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
2.2.5
Calculus October 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
Calculus 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
2.3.1
Calculus March 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
2.3.2
Calculus April 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
2.3.3
Calculus June 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
2.3.4
Calculus September 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
2.3.5
Calculus November 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
Calculus 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
2.4.1
Calculus March 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
2.4.2
Calculus April 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
2.4.3
Calculus June 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
2.4.4
Calculus September 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
2.4.5
Calculus November 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
Calculus 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
2.5.1
Calculus March 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
2.5.2
Calculus April 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
2.5.3
Calculus June 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
2.5.4
Calculus August 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
2.5.5
Calculus November 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
Calculus 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
2.6.1
Calculus March 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
2.6.2
Calculus April 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
2.6.3
Calculus June 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
2.6.4
Calculus August 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
2.6.5
Calculus November 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2
2.3
2.4
2.5
2.6
3
1
88
Calculus Solutions
92
3.1
Calculus Solutions 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
3.1.1
Calculus Solutions March 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
3.1.2
Calculus Solutions May 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
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2012–2017 Past Test and Examination Questions Booklet
3.2
3.3
3.4
3.5
3.6
Mathematics I (Major)
3.1.3
Calculus Solutions June 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
3.1.4
Calculus Solutions August 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
Calculus Solutions 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
3.2.1
Calculus Solutions April 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
3.2.2
Calculus Solutions June 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
3.2.3
3.2.4
Calculus Solutions June 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
Calculus Solutions August 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Calculus Solutions 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
3.3.1
Calculus Solutions March 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
3.3.2
Calculus Solutions April 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
3.3.3
Calculus Solutions June 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
3.3.4
Calculus Solutions September 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
3.3.5
Calculus Solutions November 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
Calculus Solutions 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
3.4.1
Calculus Solutions March 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
3.4.2
Calculus Solutions April 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
3.4.3
Calculus Solutions June 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
3.4.4
Calculus Solutions September 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
3.4.5
Calculus Solutions November 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
Calculus Solutions 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
3.5.1
Calculus Solutions March 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
3.5.2
Calculus Solutions April 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
3.5.3
Calculus Solutions June 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
3.5.4
Calculus Solutions August 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
3.5.5
Calculus Solutions November 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Calculus Solutions 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
3.6.1
Calculus Solutions March 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
3.6.2
Calculus Solutions April 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
3.6.3
Calculus Solutions June 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
3.6.4
Calculus Solutions August 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
3.6.5
Calculus Solutions November 2017 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
Mathematics I (Major)
1
2012–2017 Past Test and Examination Questions Booklet
1
MATH1036 Tutorial Questions: Semester One
In some tutorials, you will see that there are some references to questions in the textbook by Thomas. These are
extra for those students who feel they would like some more practise. Also feel free to find your own questions in
the recommended textbooks. In many textbooks, only odd numbered questions have answers provided.
Tutorial 0.1.1
For this very first tutorial, don’t worry if there is something you haven’t seen before. The important thing it that
you learn what everything means and understand the questions and answers before moving on.
The questions in this tutorial are not based on any particular section of your study guide, but are a collection of
things that new first years often do not understand fully. It will really benefit you to do this tutorial very thoroughly.
Remember: To learn you need to makes mistakes.
1. TRUE or FALSE:
(a) π = 22
7
√
2
(b) x = x
∀ x in R
(c) It is possible to draw a function which intersects its asymptote.
(d) loge ea = a = eloge a
(e) If x is negative then
(f) x − 2 =
x2 − 4
x+2
−2x
√
3|x|
is positive.
∀x ∈ R
2. What is the difference in meaning between the symbol ∀ and the symbol ∃?
2
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
3. If dxe denotes the smallest integer larger or equal to x then dπe =
4. How many numbers are there between 2 and 4?
1
1
1
, sec x :=
and cot x :=
then express cosec2 x in terms of cot2 x and express
sin x
cos x
tan x
sec2 x in terms of tan2 x. HINT: Recall that sin2 x + cos2 x = 1.
5. If cosec x :=
6. If A ⇒ B means “if A then B” or “A implies B”, then what do you think the following expressions mean:
(a) A ⇐ B
(b) A ⇔ B
7. Calculate the final mark if a student obtains the following results during the year: (See the preamble in your
study guide to see the two different ways the mark is calculated and choose the higher mark.)
Mark
13
25
15
20
62
28
March test
April test
June test
September test
November exam
Tut tests and assignments
Total
40
40
40
30
90
30
Tutorial 1.1.1
1. Thomas Exercises 11th ed. 2.1 pp. 75–76: 1, 2, 4, 7, 8 or 12th ed. 2.2 p. 73: 1, 2, 4, 7, 8.
2. If f is a function such that f (2) = 4, can you conclude anything about lim f (x)? Explain your answer and
x→2
provide a graph for f to substantiate.
3. If g is a function such that lim g(x) = 4, can you conclude anything about g(2)? Explain your answer and
x→2
provide a graph for g to substantiate.
4. Let f (x) = x − bxc. For each integer n, find lim− f (x) and lim+ f (x) if they exist.
x→n
x→n
5. (a) Estimate the value of
√
2
.
lim t t+9−3
2
t→0
Hint: Use a table for
several values of t near 0. Also, check for values of t made sufficiently small (i.e. ±0.0001; ±0.00001; ±0.000001
√
2
(b) Evaluate lim t t+9−3
.
2
t→0
Hint: You can’t apply the quotient law immediately, first rationalize the numerator.
6. Explain what it means to say that lim− f (x) = 1 and lim+ f (x) = 5.
x→2
x→2
In this situation is it possible that lim f (x) exists?. Explain.
t→1
7. Sketch the graph of the function f (x) and use it to determine the values of a for which lim f (x) exists.
x→a


 1 + x,
x2 ,
f (x) =


2 − x,

if x < −1 

if − 1 ≤ x < 1


if x ≥ 1
8. Sketch the graph of an example of a function f that satisfies all the given conditions.
lim f (x) = 1, lim+ f (x) = −1, lim− f (x) = 0, lim+ f (x) = 1, f (2) = 1, f (0) is undefined .
x→0−
x→0
x→2
x→2
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
3
Tutorial 1.1.2
1. Thomas Exercises 11th ed. 2.5 pp. 117–118: 1, 2, 6, 11, 13, 16, 21, 42 or 12th ed. 2.6 pp. 114-115: 37, 38, 42,
47, 49, 52, 57, 72.
2. If
f (x) =
( √
x − 4,
8 − 2x,
if x > 4
)
if x < 4
Determine whether lim f (x) exists.
x→4
1
3
x→1 x −1
3. Determine lim−
1
3
x→1 x −1
and lim+
from the right.
2−x
2,
x→1 (x−1)
4. Determine the limit of lim
by evaluating f (x) =
1
(x3 −1)
for values of x that approach 1 from the left and
if it exists.
5. Find limπ + 1x sec x
x→( 2 )
6. Evaluate lim− x cos ecx
x→2π
x2 −2x−8
2
x→2 x −5x+6
Q.7 Determine the limit of lim+
Q.8 Find the vertical asymptotes of the function y =
x2 +1
3x−2x2
Tutorial 1.2.1
1. Thomas Exercises 11th ed. 2.2 pp. 83–85: 1, 3, 4, 5, 10, 15, 16, 20, 28, 31, 36, 39, 42, 55, 58 or 12th ed. 2.2
pp. 74–76: 11–14, 18, 21, 22, 24, 34, 37, 42, 53, 56, 79, 82.
2. Given that lim f (x) = 2, lim g(x) = 0 and lim h(x) = −3 find the limits that exist. If the limit does not exist,
x→a
x→a
x→a
explain why.
√
a. lim[ f (x) − h(x)], b. lim( f (x))3 , c. lim 3 2 f (x) − 4h(x),
x→a
x→a
x→a
f (x)
g(x)
d. lim
, e. lim
.
x→a h(x)
x→a g(x)
3. Find the limit, if it exists.
√
√
1
1
x2 + 4 − 5
3 + x
, b. lim √
a. lim
,
x→−3 3 + x
x→1
x+3−2
cubes.
x2 − 1
√
4. Find the limit of lim √
x→1
2x + 1 − 3
c. lim
x→0
Å
ã
1
1
. d. lim
−
x→−2
x 1+x x
√
5. If you are given that lim g(x) = π1 , what can you say about g(e)?
x→e
x3 +8
x+2
Hint: Sum and difference of two
4
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
6. If you are given that lim g(x) = π1 , what can you say about lim− g(x)?
x→e
x→e
7. If you are given that f (−5) D.N.E., what can you say about lim f (x)?
x→−5
8. Evaluate the limits and justify each step by indicating the appropriate Limit Law(s).
4
−2
(a) lim 2t2t−3t+2
.
t→−2
Hint: Use√theorems 4,1,3,8 and 10.
(b) lim u4 + 3u + 6.
u→−2
Hint: Use theorems
√ 5,1,2,3,10,8 and 9.
(c) lim 1 + 3 x (2 − 6x2 + x3 )
x→8
Hint: Use theorems 4,1,2,3,8,11 and 10.
9. Evaluate the limit lim
Ä
t→2
t2 −2
t3 −3t+5
ä2
.
Hint: Use theorems 7,5,1,2,3,10,8 and 9.
10. Find lim
»
x→2
2x2 +1
3x−2 .
Hint: Use theorems 12,5,1,2,3,10,9 and 8.
11. Evaluate the limit, if it exists. lim
x→−4
12. Evaluate lim
h→0
1
−1
(x+h2 ) x2
h
√
x2 +9−5
x+4 .
.
Tutorial 1.3.1
1. Thomas Exercises 11th ed. 2.4 p. 108: 52, 58, 59, 63, 64, 65 or 12th ed. 2.6 p. 114: 14, 20, 21, 29, 30, 31.
2. Thomas Exercises 12th ed. 2.6 p. 116: 80, 82, 85.
3
3. Find the limit of lim 3/2
x→−∞ x
x3 + πx
x→∞ (x + 1)2
4. Find the limit of lim
2x
x→−∞ 3 x
5. Find the limit of lim
6. Determine whether the following function have vertical, horizontal or oblique asymptote:
a. y = 2x−3
x−2
Hint: use the one-sided limits and the infinity limit
x2
b. y = x−1
Hint: Remember to use long division.
7. Find the limit as x → −∞ of f (x) =
8. Evaluate lim
x→∞
√ x−3
4x2 +25
9. Determine the limit of lim
x→−∞
4−3x3
√
x6 +9
√
3
x−5x+3
√
3
2x+ x2 −4
Mathematics I (Major)
5
2012–2017 Past Test and Examination Questions Booklet
Tutorial 1.4.1
1. Thomas Exercises 11th ed. 2.4 p. 108–109: 12–15, 17–20, 82 or 12th ed. 2.4 pp. 91–92: 12–15, 17–20, 52.
2. For the following functions f (x), determine if the one-sided limits at the indicated point a exist, find whether
the limit at a exists, and compare with the value of f (a) if f is defined at a.
1
1
x2 − 1
a. f (x) = −
at a = 0, b. f (x) =
at a = 1, c. f (x) = bxc + b−xc at a = −3.
x |x|
|x − 1|
3. Find the limit of lim 2x+12
|x+6| , if it exists. If the limit does not exist, explain why.
x→−6
x2 +x−6
|x−2|
4. Let g(x) =
(a) Find (i) lim+ g(x) and (ii) lim− g(x).
x→2
(b) Does lim g(x) exist?
x→2
x→2
5. Let

x,



 3,
f (x) =

2 − x2 ,



x − 3,

if x < 1 


if x = 1 
if 1 < x ≤ 2 



if x > 2
Evaluate each of the following, if it exists. (a) lim− f (x); (b)lim f (x); (c) f (1); (d) lim− f (x); (e) lim+ f (x); (f)lim f (x).
x→1
6. Find the limit of
x→1
x→2
x→2
x→2
lim 2x−1
3
2
x→0.5− |2x −x |
Tutorial 1.5.1
1. Thomas Exercises 11th ed. 2.2 p. 84: 49, 50, 51a, 52a or 12th ed. 2.2 p. 75: 63, 64, 65a, 66a.
2. Thomas Exercises 11th ed. 2.4 p. 108: 22, 25, 28, 29, 31, 33, 36 or 12th ed. 2.4 p. 92: 22, 25, 28, 29, 31, 33,
35, 40.
3. Thomas Chapter 2 Additional and Advanced Exercises 11th ed. p. 143: 27–30 or 12th ed. p. 121: 27–30.
(
x if x is rational,
4. Let f (x) =
Show that lim f (x) = 0.
x→0
0 if x is irrational.
5. Use the Squeeze Theorem to show that lim x2 cos 20πx = 0. Consider f (x) = −x2 , g(x) = x2 cos 20πx and
x→0
h(x) = x2 .
6. Use the Squeeze Theorem to show that lim
x→0
√
and h(x) = x3 + x2 .
√
√
√
x3 + x2 sin πx = 0. Consider f (x) = − x3 + x2 , g(x) = x3 + x2 sin πx
7. If 4x − 9 ≤ f (x) ≤ x2 − 4x + 7 for x ≥ 0, find lim f (x).
x→4
8. Prove that lim x4 cos 2x = 0.
x→0
9. Prove that lim+
x→0
Tutorial 2.1.1
√
x [1 + sin2
2π
x
] = 0.
6
2012–2017 Past Test and Examination Questions Booklet
1. Find the values of a and b which make the function


x−1


f (x) = ax2 + b



x+1
Mathematics I (Major)
if x ≤ −2,
if − 2 < x < 1,
if x ≥ 1,
continuous at x = −2 and x = 1.
2. Find the values of a and b which make the function f continuous everywhere.

 2
x −4

if x < 2 

 x−2 ,
f (x) =
ax2 − bx + 3, if 2 ≤ x < 3




2x − a + b,
if x ≥ 3
3. Show that f is continuous on (−∞, ∞).
(
f (x) =
1 − x2 ,
√
x − 1,
if x ≤ 1
)
if x > 1
4. Show that f is continuous on (−∞, ∞).
®
f (x) =
sin x,
if x <
cos x,
if x ≥
π
4
π
4
´
5. How would you ”remove the discontinuity” of f in order to make the function continuous at 2? f (x) =
x2 −3x+2
x2 +x−6
Tutorial 2.2.1
1. Consider the function

 bxc
f (x) =
x
−1
if x , 0,
if x = 0.
Investigate continuity from the left and the right at x = 0, x = π and x = 1.
2. Use the definition of continuity and the properties of limits to show that the function is continuous at a given
point a.
√
(a) f (x) = x2 + 7 − x,√ a = 4.
3
(b) f (x) = 3x4 − 5x + x2 + 4, a = 2.
3. At which points is the function f continuous from the the right, from the left, or neither?
 2

if x < −1 

 x ,

x,
if − 1 ≤ x < 1
f (x) =


 1

if x ≥ 1
x,
4. For what value of the constant c is the function continuous on (−∞, ∞).
( 2
)
cx + 2x, if x < 2
f (x) =
x3 − cx,
if x ≥ 2
5. If f is continuous at 5 and f (5) = 2 and f (4) = 3, then lim f (4x2 − 11) = 2.
TRUE or FALSE. Give reasons.
x→2
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
7
Tutorial 2.3.1
1. The function g(x) is continuous at x = − 12 iff...
2. The function f (x) is continuous on (−3, e] iff...
3. Show that the
√ function is continuous on the given interval.
(a) f (x) = x + x − 4, (4, ∞).
x−1
, (−∞, −2).
(b) g(x) =
3x + 6
4. Why is the function continuous at every point in its domain. State the domain.
2
(a) f (x) = 2xx2−x−1
+1
(b) h(x) =
(c) g(x) =
√
3
(x−2)
x3 −2
√tan x
4−x2
5. Show that
√ the function is continuous on its domain. State the domain.
(a) f (x) = √4 x + x3 cos x.
2
(b) g(x) = xx2 −2−9 .
Tutorial 2.4.1
1. Thomas Exercises 11th ed. 2.6 pp. 129–130: 13–20, 23–33, 40, 47–50, 53–57, 59 or 12th ed. 2.5 pp. 101–103:
13–20, 23–35, 47, 55–58, 61–64, 65, 67.
2. Thomas Chapter 2 Additional and Advanced Exercises 11th ed. p. 142: 19 or 12th ed. p. 120: 19.
3. The equation x3 − 12x + 8 = 0 has three ANSWERS. Find intervals between successive integers that contain
these ANSWERS.
4. Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
(a) x4 + x − 3 = 0, (1, 2)
(b) sin x = x2 − x, (1, 2)
(c) 2 sin x = 3 − 2x, (0, 1)
5. If f is continuous on [-1,1] and f (−1) = 4 and f (1) = 3, then there exists a number r such that |r| < 1 and
f (r) = π.
6. If f (x) = x2 + 10 sin x, show that there is a number c such that f (c) = 1000.
7. Suppose f is continuous on [1,5] and the only ANSWERS of the equation f (x) = 6 are x = 1 and x = 4. If
f (2) = 8, explain why f (3) > 6.
Tutorial 2.5.1
1. Thomas Exercises 11th ed. 2.6 p. 130: 51, 52 or 12th ed. 2.5 p. 102: 59, 60.
2. Thomas Chapter 2 Practice Exercises 11th ed. p. 139: 1, 2 or 12th ed. p. 117: 1, 2. Also indicate the type of
discontinuities.
3
2
−2x
3. Does the function f (x) = x −x
have a removable discontinuity at 2? If the discontinuity is removable, find
x−2
the function g that agrees with f for x , 2 and is continuous at 2.
4. Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from
the left, or neither?
8
2012–2017 Past Test and Examination Questions Booklet
(a)
 2
x + 1,


3 − x,
f (x) =

 √ x,
(b)
5. Let
Mathematics I (Major)

if x ≤ 1 

if 1 < x ≤ 4

if x > 4 

1 + x2 ,


2 − x,
f (x) =

 (x − 2)2 ,

if x ≤ 0 

if 0 < x ≤ 2

if x > 2 
 √
−x,


3 − x,
f (x) =

 (x − 3)2 ,

if x < 0 

if 0 ≤ x < 3

if x > 3 
(a) Evaluate each limit, if it exists.
(i) lim+ f (x); (ii) lim− f (x); (iii)lim f (x); (iv) lim− f (x); (v) lim+ f (x); (vi) lim f (x).
x→0
x→0
x→0
x→3
x→3
x→3
(b) Where is f discontinuous?
(c) Sketch the graph of f
Tutorial 3.1.1
1. Thomas Exercises 11th ed. 3.1 pp. 152–153 or 12th ed. 3.2 p. 131: 1, 2, 4, 6, 12.
2. Find f 0 (a) if it exists, where
(
x3
if x ≤ 1
(a)
f (x) =
and a = 1.
cx + b if x > 1
(
3x2
if x ≤ 1
(b)
f (x) =
and a = 1.
3
2x + 1 if x > 1
3. Thomas Chapter 2 Practice Exercises 11th ed. p. 139: 13–16 or 12th ed. p. 118: 13–16. Each limit represents
the limit of some function f at some number a. State such an f and a in each case.
4. A curve has equation y = f (x).
(a) Write an expression for the slope of the secant line through the points P(3, f (3)) and Q(x, f (x)).
(b) Write an expression for the slope of the tangent line at P.
5. If f (x) = 3x2 − x3 , find f 0 (1) and use it to find an equation of the tangent line to the curve y = 3x2 − x3 at the
point (1,2).
6. Find f 0 (a).
2
(a) f (x) = 3x
√ − 4x + 1
(b) f (x) = 1 − 2x
4
(c) f (x) = √1−x
7. Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
h+3
(a) lim 2 h −8
h→0
(b) lim
1
4 −4
x− 14
(c) limπ
sin θ− 21
θ− π6
x→ 14
θ→ 6
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
9
Tutorial 3.2.1
1. Thomas Exercises 11th ed. 2.7 pp. 136–137: 5, 8, 11, 16, 17, 31, 32 or 12th ed. 3.1 p. 125: 5, 8, 11, 16, 17,
31, 33, 34.
2. Find an equation of the tangent line to the curve at the given point.
2
(a) y = 4x
√ − 3x , (2, −4)
(b) y = x, (1, 1)
(c) y = 2x+1
x+2 , (1, 1)
3. (a) Find the slope of the tangent of the curve y = 3 + 4x2 − 2x3 at the point where x = a.
(b) Find equations of the tangent lines at the points (1,5) and (2,3).
(c) Graph the curve and both tangents on common screen.
4. Find an equation of the tangent line to the graph of y = g(x) at x = 5 if g(5) = −3 and g0 (5) = 4.
5. (a) If G(x) = 4x2 − x3 , find G0 (a) and use it to find equations of the tangent lines to the curve y = 4x2 − x3 at
the points (2,8) and (3,9).
(b) Graph the curve and the tangent lines on the same screen.
6. Find an equation of the tangent to the curve y = 4 sin2 x at ( π6 , 1).
√
7. Find equations of the tangent line to the curve y = 1 + 4 sin x, at (0,1).
Tutorial 3.2.2
1. Thomas Exercises 11th ed. 2.7 pp. 137–138: 33, 39, 44 or 12th ed. 3.1 p. 126: 35, 41, 46.
2. Sketch the graph of a function g that is continuous on its domain (-5,5) and where g(0) = 1, g0 (0) = 1, g0 (−2) =
0, lim + g(x) = ∞, and lim− g(x) = 3.
x→−5
x→5
3. Sketch the graph of a function g for which g(0) = g(2) = g(4) = 0, g0 (1) = g0 (3) = 0, g0 (0) = g0 (4) = 1, g0 (2) =
−1, lim− g(x) = ∞, and lim + g(x) = −∞.
x→5
x→−1
4. Sketch the graph of a function f where the domain is (-2,2), f 0 (0) = −2, lim− f (x) = ∞, f is continuous at all
numbers in its domain except ±1 and f is odd.
x→2
√
5. Let f (x) = 3 x.
(a) If a , 0, find f 0 (a).
(b) Show that f 0 (0)√does not exist.
(c) Show that y = 3 x has a vertical tangent line at (0,0).
6. (a) If g(x) = x2/3 , show that g0 (0) does not exist.
(b) If a , 0, find g0 (a).
(c) Show that g(x) = x2/3 has a vertical tangent line at (0,0).
Tutorial 3.4.1
1. Thomas Exercises 11th ed. 3.1 pp. 153–154: 27–32 or 12th ed. 3.2 p. 132: 27–32.
2. Let f (x) = x2 .
(a) Find the values of f 0 (0), f 0 ( 21 ), f 0 (1), and f 0 (2).
(b) Deduce the values of f 0 (− 12 ), f 0 (−1), and f 0 (−2).
(c) Use the results in parts (a) and (b) to guess a formula for f 0 (x).
(d) Use the definition of the derivative to prove that your guess in part (c) is correct.
10
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
3. Find the derivative of the function using the definition of the derivative. State the domain of the function and
the domain of its derivative.
(a) f (x) = 3x − 8
(b) f (x) = 4 + 8x − 5x2
x2 −1
(c) f (x) = 2x−3
√
(d) f (x) = x + x
4. Use the definition of the derivative to find f 0 (x) and f 00 (x). f (x) = 3x2 + 2x + 1.
Check to see if your answers are reasonable.
0
0
5. Use the
√ definition of the derivative to find f (x). What are the domains of f and f .
f (x) = 6 − x.
Tutorial 3.5.1
1. Thomas Exercises 11th ed. 3.2 pp. 167–169: 9, 10, 12, 13, 15, 17, 20, 22, 43, 46, 52, 55 or 12th ed. 3.3 pp.
143–145: 9, 10, 12, 13, 15, 17, 20, 22, 57, 60, 74, 77.
2. Thomas Chapter 3 Practice Exercises 11th ed. pp. 255–257: 4, 9, 99, 101, 107, 109 or 12th ed. pp. 213–215:
4, 9, 99, 101, 107, 109.
3. Differentiate the function.
(a) g(x) = 74 x2 − 3x + 12
(b) g(x) √= x2 (1 − 2x)
(c) y = √3 x(2 + x)
(d) y = xx+x
2
x2 +4x3
√
(e) y =
x
4. Find the derivative of f (x) = (1 + 2x2 )(x − x2 ) in two ways: by using the Product rule and by performing the
multiplication first.
Do your answers agree?
5. Find the
derivative
of the function
√
4
3
F(x) = x −5xx2 + x
in two ways: by using the Quotient rule and by simplifying first. Show that your answers are equivalent. Which
method do you prefer?
6. (a) Use the Product rule twice to prove that if f, g, and h are differentiable, then ( f gh)0 = f 0 gh + f g0 h + f gh0 .
(b) Taking f = g = h in part (a), show that
d
3
2 0
dx [ f (x)] = 3[ f (x)] f (x)
(c) Use part (b) to differentiate y = (x4 + 3x3 + 17x + 82)3 .
Tutorial 3.5.2
1. Thomas Exercises 11th ed. 3.5 pp. 199–202: 1, 2, 9, 11, 14, 23, 31, 34, 67, 71, 72–74, 116 or 12th ed. 3.6 pp.
167–169: 1, 2, 9, 11, 13, 23, 31, 34, 79, 83, 84, 87, 88, 104.
2. For the following functions, find the points on the graphs where the tangents are horizontal or vertical. (a)
1
f (x) = x − ,
x
4
1
(b) g(x) = x 3 + 4x 3 ,
3
(c) h(x) = x(x + 2) 5 ,
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
11
1
(d) d(x) = xr − x r on [0, ∞], where r > 1.
3. Are there points on the curve y = x − e(−x) where the slope of the tangent is 2? If so, find them.
Hint: the first derivative of the function y with respect to x is 2.
4. Write the composite function in the form f (g(x)).
Identify the inner function u = g(x) and the outer function y = f (u). Then find the derivative dy/dx.
2 10
(a) y = (1
√− x )
(b) y = 4 + 3x
5. Find the derivative of the function.
(a) f (x) = (2x − 3)4 (x2 + x + 1)5
(r2 −1)3
(b) H(r) = (2r+1)
5
»
p
√
(c) y = x + x + x
6. Find y0 and y00 for y =
√4x .
x+1
7.Find an equation of the tangent line to the curve at the given point.
−6
(a) y = (3x
√ − 1) , (0, 1)
3
(b) y = 1 + x , (2, 3)
Tutorial 3.6.1
1. Thomas Exercises 11th ed. 3.4 pp. 186–187: 1, 2, 5, 10, 11, 24, 26 or 12th ed. 3.5 pp. 159–160: 1, 2, 6, 9, 14,
15, 18, 20.
2. Thomas Exercises 11th ed. 3.5 pp. 199–200: 3, 6, 7, 28, 43, 44, 46, 57, 69, 70 or 12th ed. 3.6 pp. 167–168: 3,
6, 7, 28, 45, 46, 48, 63, 81, 82.
3. Thomas Chapter 3 Practice Exercises 11th ed. p. 255: 20, 39 or 12th ed. pp. 213–214: 20, 39.
4. Does the curve y = sin (x − sin x) have horizontal tangents at the x − axis? Give reasons.
Hint: y0 is 0 for horizontal tangents.
5. Differentiate.
(a) f (x) = x2 sin x
(b) g(t) = 4 sec t + tan t
(c) y = 2 sec x − csc x
x
(d) y = 2−tan
x
(e) y = x2 sin x tan x
6. Find an equation of the tangent line to the curve.
(a) y = 2x sin x at the point ( π2 , π).
(b) y = tan(πx2 /4) at the point (1, 1).
7. Find the limit.
(a) lim sin 3xx2sin 5x
x→0
(b) lim csc x sin(sin x)
x→0
sin(x−1)
2
x→1 x +x−2
(c) lim
8. Write the composite function in the form f (g(x)).
Identify the inner function u = g(x) and the outer function y = f (u). Then find the derivative dy/dx.
(a) y = tan(sin
√ x)
(b) y = sin x
12
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
9. Find the derivative of the function.
(a) y = cos4 (sin3 x)
(b) y = [x + (x + sin2 x)3 ]4
Tutorial 3.7.1
1. Thomas Exercises 11th ed. 3.6 pp. 209–210: 19, 22, 25, 28, 30, 31, 35, 47, 54, 56, 59, 61 or 12th ed. 3.7 p.
174: 1, 4, 7, 10, 13, 19, 31, 38, 40, 43, 45.
Ä ä
dy
2. Find dx
if y sin 1y = 1 − xy.
3. (a) find y0 by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get y0 in terms of x.
(c) Check that your ANSWERS to parts (a) and (b0 are consistent by substituting the expression for y into the
solution for part (a).
2
(i) xy
√ + 2x√+ 3x = 4
(ii) x + y = 1
4. Find dy/dx by implicit differentiation.
(a) x3 − xy2 + y3 = 1
x2
(b) x+y
= y2 + 1
(c) x4 (x + y) = y2 (3x − y)
√
(d) xy = 1 + x2 y
(e) x sin y + y sin x = 1
5. Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
(a) y sin 2x = x cos 2y, (π/2, π/4).
(b) 2(x2 y2 )2 = 25(x2 − y2 ), (3, 1).
6. If x2 + xy + y3 = 1, find the value of y000 at the point where x = 1.
Tutorial 3.8.1
1. Thomas Exercises 11th ed. 3.2 p. 167: 24, 27, 28 or 12th ed. 3.3 p. 143: 30, 31, 40.
2. Thomas Exercises 11th ed. 3.6 p. 209: 32 or 12th ed. 3.7 p. 174: 16.
3. Thomas Practice Exercises 11th ed. pp. 255–256: 44, 73, 74 or 12th ed. p. 214: 44, 73, 74.
4. Using the Chain Rule we have
d f (x)
e
=
dx
5. Using the Chain Rule we have
d f (x)
a
=
dx
6. Differentiate
√ the function.
(a) f (x) = e
(b) f (x) = (3x2 − 5x)e x
(c) y = etan θ
2 x
e
(d) f (x) = xx2 +e
x
7. Find the limit.
3x
−3x
(a) lim ee3x −e
+e−3x
x→∞
(b) lim− e3/(2−x)
x→2
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
13
(c) lim (e−2x cos x)
x→∞
8. Find an equation of the tangent line to the curve at the given point.
(a) y = e2x cos πx, (0, 1)
x
(b) y = ex , (1, e)
Tutorial 3.8.2
1. Thomas Exercises 11th ed. 3.7 pp. 221–222: 13, 14, 21, 22, 28, 30, 33, 38, 57–60, 76, 77, 86 or 12th ed. 3.8
pp. 184–185: 13, 14, 21, 22, 28, 30, 33, 38, 57–60, 76, 77, 86.
2. Using the Chain Rule we have
d
loga f (x) =
dx
3. Differentiate.
(a) f (x) = x ln x − x
(b) f (x) = cos(ln x)
2
(c) f (x) = ln(sin
x)
√5
(d) f (x) = ln x
(e) f (x) = sin x ln(5x)
(f) y = ln(csc x − cot x)
(g) y = log2 (x log5 x)
4. Find y√0 and y00 .
(a) y = x ln x
ln x
(b) y = 1+ln
x
5. Differentiate f and find the domain of f .
x
(a) f (x) = 1−ln(x−1)
√
(b) f (x) = 2 + ln x
6. Find an equation of the tangent line to the curve at the given point.
(a) y = ln(x2 − 3x + 1), (3, 0).
(b) y = x2 ln x, (1, 0).
Tutorial 3.9.1
1. Thomas Exercises 11th ed. 3.10 pp. 250–251: 1, 2, 3, 6, 8, 9, 11, 13, 14, 21, 30, 31, 33, 38, 39, 43, 44 or 12th
ed. 3.11 pp. 210–211: 1, 2, 3, 6, 8, 9, 11, 13, 14, 21, 30, 31, 33, 38, 39, 43, 44.
2. Find the linearization L(x) of the function at a.
(a) f (x) = x3 − x2 + 3, a = −2.
(b) f (x) = sin x, a = π/6.
3. Find the differential dy of each function.
(a) y = (x2 − 3)−2
(b) y = θ2 sin
√ 2θ
(c) y = tan t
(d) y = 1x sin x
14
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
4. (a) Find
√ the differential dy and (b) evaluate dy for the given values of x and dx.
(i) y = 3 + x2 , x = 1, dx = −0.1
(ii) y = x+1
x−1 , x = 2, dx = 0.05
5. Use linear approximation (or differentials) to estimate the given number.
4
(a) (1.999)
√3
(b) 1001
Tutorial 4.1.1
1. Thomas Exercises 11th ed. 4.8 pp. 338–340: 2, 3, 7, 36, 39, 40, 55, 65, 66, 72, 73, 75, 76, 77, 83, 86 or 12th
ed. 4.8 pp. 285–286: 2, 3, 7, 36, 39, 40, 55, 65, 66, 72, 73, 75, 76, 77, 83, 86.
1
2. Integrate the function f (x) = (2+6x+9x
2
)
Hint: Complete the square.
3. Verify by differentiation that the formula is correct.
(a)
R
√
x a + bx dx =
(b)
R
x cos x2 dx = 12 x + 41 sin 2x + C
2
15b2 (3bx
− 2a)(a + bx)3/2 + C
4. Evaluate the integral.
(a)
R
(1+e x )2
ex
(b)
R
x2 x dx
(c)
R
e x cos(e x ) dx
(d)
R
(ln x)2
x
(e)
R
etan x sec2 x dx
(f)
R
sin2x
1+cos2 x
(g)
R
x
1+x4
dx
2
dx
dx
dx
5. Show that
(a)
R
1
x2 +a2
(b)
R
√ 1
a2 −x2
dx =
1
a
dx =
tan−1
1
a
sin−1
x
a
+C
x
a
+C
Tutorial 4.2.1
1. Thomas Exercises 11th ed. 5.4 p. 392: 1, 2, 14, 23, 31, 32 or 12th ed. 5.4 p. 333: 1, 2, 6, 7, 12, 16, 20, 21, 23,
24, 33, 34.
2. Evaluate the integral.
Mathematics I (Major)
(a)
R1
(b)
R4
(c)
R2
(d)
R9Ä√
(e)
R1
(f)
Re
(g)
R
0
2012–2017 Past Test and Examination Questions Booklet
15
(xe + e x ) dx
3
2 x
1
dx
e1/x
x2
4
√
0
1
dx
√1
x
x+
1+e−x
ex
x2 +x+1
x
ä
dx
dx
dx
√
3
8
√
1/ 3 1+x2
dx
3. Determine whether this statement is true or false.
(a)
R 16
(b)
R π/2
(c)
R1
2
0
dx
x
= 3 ln 2
cos x
1+sin2 x
ex
0 1+e2x
dx =
π
4
dx = arctan e −
π
4
4. What is wrong with this equation?
R1
−3 1
(a) −2 x4 dx = x−3 −2 = − 83
(b)
Rπ
π/3
sec θ tan θ dθ = sec θ
π
π/3
= −3
Tutorial 4.3.1
1. Thomas Exercises 11th ed. 5.5 pp. 402–403: 2, 7, 10, 19, 24, 34, 40, 55-57, 67 or 12th ed. 5.5 pp. 342–344:
3, 4, 5, 8, 11, 14, 21, 26, 36, 52, 67-69, 79.
2. Thomas Exercises 11th ed. 5.6 pp. 410–414: 1, 2, 4, 5, 8, 13, 23, 29, 31, 32, 37, 40, 111, 112, 115, 116 or 12th
ed. 5.6 pp. 350–353: 1, 2, 4, 5, 8, 13, 23, 29, 31, 32, 37, 40, 111, 112, 113, 115, 116.
3. The Substitution Rule gives neat formulas for functions with multiples. For example, if k is a constant, then
using the substitution u = kx will obtain:
Z
Z
ekx dx =
Z
sin kx dx =
Z
cos kx dx =
Z
sec2 kx dx =
sec kx tan kx dx =
16
2012–2017 Past Test and Examination Questions Booklet
4. Evaluate this integral by using substitution:
Z
Mathematics I (Major)
eθ cos ec eθ + 1 dθ
5. Evaluate the integral by making the given substitution.
(a)
R
cos 2x dx, u = 2x
(b)
R
x(2x2 + 3)4 dx, u = 2x2 + 3
(c)
R
√
x2 x3 + 1 dx, u = x3 + 1
(d)
R
sin2 θ cos θ dθ, u = sin θ
6. Evaluate the indefinite integral.
R
(a) (x2 + 1)(x3 + 3x)4 dx
(b)
R
cos(π/x)
x2
(c)
R
sec2 x
tan2 x
dx
dx
7. Evaluate the definite integral.
(a)
R π/6
(b)
R π/4
(c)
R 13
(d)
R2
0
sin t
cos2 t
−π/4 (x
0
0
√
3
3
dt
+ x4 t tan x) dx
1
(1+2x)2
dx
2
(x − 1)e(x−1) dx
Tutorial 4.4.1
1. Thomas Exercises 11th ed. 8.2 pp. 552–554: 1, 6, 9, 12, 14, 25, 26, 29, 30 or 12th ed. 8.1 pp. 459–460: 1, 6,
11,R13,
16, 18, 25, 26, 29, 30.
π
2. 04 sec3 x dx =
3. Evaluate the integral using integration by parts with the indicated choices of u and dv.
xe2x dx; u = x, dv = e2x dx
(a)
R
(b)
R √
x ln x dx; u = ln x, dv =
4. Evaluate the integral.
R
(a) (x2 + 2x) cos x dx
√
xdx
Mathematics I (Major)
(b)
Rπ
(c)
R
0
2012–2017 Past Test and Examination Questions Booklet
x sin x cos x dx
xe2x
(1+2x)2
dx
R
(d) (arcsin x)2 dx
5. First make a substitution and then use integration by parts to evaluate the integral.
(a)
R
cos(ln x) dx
(b)
R
x ln(1 + x) dx
6. Use integration by parts to prove the reduction formula.
R
R
(a) (ln x)n dx = x(ln x)n − n (ln x)n−1 dx
(b)
R
tan x dx =
tann−1 x
n−1
−
R
tann−2 x dx
Tutorial 5.1.1
1. Thomas Exercises 11th ed. 3.2 p. 167: 29, 31, 33, 35, 36 or 12th ed. 3.3 p. 143: 41, 45, 47, 51, 52.
2. Thomas Exercises 11th ed. 3.4 p. 186: 25 or 12th ed. 3.5 p. 160: 33.
3. Thomas Exercises 11th ed. 3.6 p. 209: 39–44 or 12th ed. 3.7 p. 174: 23–26.
4. Thomas Exercises 11th ed. 11.8 p. 794: 2, 4, 5, 7, 8 or 12th ed. 10.8 p. 606: 2, 4, 6, 7, 9, 10.
5. If f (x) = e2x , find the formula for f (n) (x).
6. Find the thousandth derivative of f (x) = xe−x .
7. Find a formula for f (n) (x) if f (x) = ln(x − 1)
Tutorial 5.2.1
1. Thomas Exercises 11th ed. 3.7, p. 221: 3, 4, 7, 9 or 12th ed. 3.8 p. 184: 3, 4, 7, 9.
√
2. Let f (x) = 1 − x2 , 0 ≤ x ≤ 1.
(a) Find f −1 . How is it related to f ?
(b) Identify the grah of f and explain your answer to part (a).
3. Suppose f −1 is the inverse function of a differentiable function f and f (4) = 5, f 0 (4) = 23 .
Find ( f −1 )0 (5).
Rx √
4. If f (x) = 3 1 + t3 dt, find ( f −1 )0 (0).
5. Differentiate the function.
3
(a) y = eax
2 x
e
(b) f (x) = xx2 +e
x
(c) f (x) = (3x2 − 5x)e x
6. Differentiate the function.
(a) y = xsin x
(b) y = (cos x) x
(c) g(x) = x4 4 x
17
18
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
7. Let f (x) = arctan(1/x) if x , 0 and f (0) = 0.
(a) Is f continuous at 0?
(b) Is f differentiable at 0?
Tutorial 5.3.1
1. Thomas Exercises 11th ed. 3.8 pp. 230–231: 49, 50, 61, 62, 64, 65, 67, 68 or 12th ed. 3.9 p. 191: 21, 22, 33,
34, 36, 37, 39, 40.
2. Thomas Exercises 11th ed. 5.5 p. 403: 49, 50, 53, 54 or 12th ed. 5.5 p. 343: 61, 62, 65, 66.
3. Thomas Exercises 11th ed. 8.6 p. 584: 23, 24, 31, 32 or 12th ed. 8.5 p. 485: 17–20. For each integral, use
appropriate methods to find the antiderivates.
4. Using the Chain Rule,
5. Using the Chain Rule,
d
arcsin f (x) =
dx
d
arctan f (x) =
dx
6. Find the
√ derivative of the function. Simplify where possible.
(a) y = tan−1
√
(b) y = x sin−1 x + 1 − x2
(c) y = sin−1 (2x + 1)
(d) F(x) = √
x sec−1 (x3 )
(e) g(x) = x2 − 1 sec−1 x
(f) y = cos−1 (e2x )
cos x
(g) y = arccos b+a
a+b cos x , 0 ≤ x ≤ π, a > b > 0.
7. Find the derivative of the function f (x) = arcsin(e x ). Find the domains of the function and its derivative.
8. Find y0 if tan−1 (x2 y) = x + xy2 .
9. Find an equation of the tangent line to the curve y = 3 arccos(x/2) at the point (1, π).
√
10. Find f 0 (x) if f (x) = 1 − x2 arcsin x. Check that your answer is reasonable by comparing the graphs of f and
f 0.
11. Evaluate the integral.
R 1/2 −1 x
√
(a) 0 sin
dx
1−x2
R
dx
√
(b)
−1
2
1−x sin
x
Tutorial 5.4.1
1. Thomas Exercises 11th ed. 3.7 pp. 221–222: 41, 42, 49, 51–54, 89–96 or 12th ed. 3.8 pp. 184–185: 41, 42,
49, 51–54, 89–96.
2. Use logarithmic differentiation to find the derivative of the function.
5
4
6
(a) y = (2x
» + 1) (X − 3)
x−1
(b) y =
4
√ x +1
2
(c) y = xe x −x (x + 1)2/3
sin x
(d) y = x
(e) y = (ln x)cos x
3. Differentiate f and find the domain of f .
x
(a) f (x) = 1−ln(x−1)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
19
(b) f (x) = ln(x2 − 2x)
4. Find an equation of the tangent line to the curve y = ln(x2 − 3x + 1), at the given point (3,0).
5. Let f (x) = cx + ln(cos x). For what values of c is f 0 (π/4) = 6?
Tutorial 5.5.1
1. Thomas Exercises 14.3 11th ed. pp. 978–980: 1, 7, 11, 17, 19, 23, 25, 29, 41, 47, 63, 66, 68, 69, 74 or 12th ed.
pp. 790-792: 1, 7, 11, 17, 19, 23, 25, 29, 41, 47, 73, 76, 79, 80, 81, 86.
2. Find the indicated partial derivative.
(a) f (x, y) = y sin−1 (xy); fy (1, 21 )
√
1− x2 +y2 +z2
(b) f (x, y, z) = ln √ 2 2 2 ; fy (1, 2, 2)
1+
x +y +z
3. Use implicit differentiation to find ∂z/∂x and ∂z/∂y.
(a) x2 + 2y2 + 3z2 = 1
(b) yz + x ln y = z2
4. Find all second partial derivatives.
(a) f (x, y) = x4 y − 2x3 y2
y
(b) z = 2x+3y
(c) f (x, y) = ln(ax + by)
5. Find the indicated partial derivative.
(a) f (x, y) = x4 y2 − x3 y; f xxx , f xyx
(b) f (x, y) = sin(2x + 5y); fyxy
2
(c) f (x, y, z) = e xyz ; f xyz
Tutorial 6.1.1
1. Thomas Exercises 11th ed. 4.6 pp. 323-324: 3, 6, 13, 16, 19, 20, 21, 27, 28, 29, 32, 33, 34, 37, 45, 47, 50, 51,
52, 57, 59, 61, 62, 63 or 12th ed. 4.5 p. 261–262: 3, 6, 13, 16, 19, 20, 21, 27, 28, 29, 32, 33, 34, 37, 45, 51, 54,
55, 56, 67, 69, 75, 76, 77.
2. Find the limit. Use the l’Hospital’s Rule where appropriate. If there is a more elementary method, consider
using it. If the l’Hospital’s Rule does not apply, explain why.
(a) lim xx−3
2 −9
x→3
(b) lim
x→4
x2 −2x−8
x−4
e2x −1
x−(x+1)
x→0 cos
√
√
1+2x− 1−4x
(d) lim
x
x→0
x
(e) lim 3x3
x
x→0 −1
(f) lim+ cosln(ex ln(x−a)
x −ea )
x→a
(c) lim
(g) lim sin 5x csc 3x
x→0
(h) lim+ (tan 2x) x
x→0
3. Use the l’Hospital’s Rule to help sketch the curve.
(a) y = 1x + ln x
(b) y = e x /x
20
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
4. What happens if you try to use l’Hospital’s Rule to find the limit? Evaluate the limit using another method.
(a) lim √ x2
(b)
x +1
sec x
x→(π/2)− tan x
x→∞
lim
Tutorial 6.2.1
1. Thomas Exercises 11th ed. 3.3 pp. 177-178: 1, 4, 6 or 12th ed. 3.4 p. 152: 1, 4, 6.
2. Thomas Exercises 11th ed. 3.9 pp. 236–240: 3, 8, 10, 17, 19, 20, 21, 23, 30 or 12th ed. 3.10 pp. 198–200: 11,
13, 18, 20, 27, 29, 30, 31, 33, 39.
√
3. Suppose y = 2x + 1, where x and y are function of t.
(a) If dx/dt = 3, find dy/dt when x = 4.
(b) If dy/dt = 5, find dx/dt when x = 12.
4. Suppose 4x2 + 9y2 = 36, where x and y are function
√ of t.
(a) If dy/dt = 31 , find dx/dt when x = 2 and y = 23 5.
√
(b) If dx/dt = 3, find dy/dt when x = −2 and y = 23 5.
5. If x2 + y2 + z2 = 9, dx/dt = 5 and dy/dt = 4, find dz/dt when (x, y, z) = (2, 2, 1).
6. (a) If A is the area of a circle with radius r and the circle expands as time passes, find dA/dt in terms of dr/dt.
(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases
at a constant rate of 1 m/s, how fast the area of the spill increasing when the radius is 30 m?
7. The area of a triangle with sides of lengths a and b and contained angle θ is
1
ab sin θ
2
(a) If a = 2cm, b = 3cm, and θ increases at a rate of 0.2 rad/min, how fast is the area increasing when θ = π/3?
(b) If a = 2cm, b increases at a rate of 1.5 cm/min, and θ increases at a rate of 0.2 rad/min, how fast is the area
increasing when b = 3cm and θ = π/3?
(c) If a increases at a rate of 2.5 cm/min=2 cm, b increases at a rate of 1.5 cm/min, and θ increases at a rate of 0.2
rad/min, how fast is the area increasing when a = 2cm, b = 3cm and θ = π/3?
A=
Tutorial 6.3.1
1. Thomas Exercises 11th ed. 4.1 pp. 272–273: 1, 5, 6, 7, 10, 14, 17, 20, 29, 33, 41, 45, 46, 48, 49, 52, 53 or 12th
ed. 4.1 pp. 227–229: 1, 5, 6, 7, 10, 14, 18, 23, 26, 35, 39, 49, 52, 55, 59, 60, 62, 63, 66, 67.
2. (a) Sketch the graph of a function on [-1,2] that has an absolute maximun but no local maximum.
(b) Sketch the graph of a function on [-1,2] that has a local maximun but no absolute maximum.
3. Sketch the graph of f by hand and use your sketch to find the absolute and local maximun and minimum values
of f .
(a) f (x) = 12 (3x − 1), x ≤ 3
(b) f (x) = 2 − 31 x, x ≥ −2
(c) f (x) = 1/x, 1 > x > 3
(d) f (x) = (
sin x, 0 < x ≤ π/2
)
2x + 1, if 0 ≤ x < 1
(e) f (x) =
4 − 2x, if 1 ≤ x ≤ 3
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
21
4. Find the critical numbers of the function.
(a) f (x) = 5x2 + 4x
(b) f (x) = 2x3 + x2 + 2x
5. A formula for the derivative of a function is given: f 0 (x) = 1 +
How many critical numbers does f have?
210 sin x
x2 −6x+10 .
6. Find the absolute maximum and absolute minimum values of f on the given interval.
(a) f (x) = 3x2 − 12x + 5, [0, 3]
x
, [0, 3]
(b) f (x) = x2 −x+1
(c) f (t) = 2 cos t + sin 2t, [0, π/2]
Tutorial 6.4.1
1. Thomas Exercises 11th ed. 4.2 pp. 282–284: 1, 3, 4, 5, 7–10, 12, 13, 15, 16, 19, 22, 51, 56, 57, 58 or 12th ed.
4.2 pp. 236–237: 1, 3, 5, 6, 11, 12, 15, 16, 18, 19, 21, 22, 25, 28, 57, 66, 67.
2. Show that the functions y(x) = Ce x , C ∈ R, are the only functions which satisfy y0 = y. Hint: Consider the
function y(x)e−x .
3. Verify that the function satisfies the three hypotheses of the Rolle’s Theorem on the given interval. Then find
all numbers c that satisfy the conclusion of the Rolle’s theorem.
(a) f (x) = 2x2 − 4x + 5, [−1, 3]
(b) f (x) = sin(π/2), [π/2, 3π/2]
4. Let f (x) = 1 − x2/3 . Show that f (−1) = f (1) but there is no number c in (-1, 1) such that f 0 (c) = 0. Why does
this not contradict Rolle’s Theorem?
5. Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all
numbers c that satisfy the conclusion of the Mean Value Theorem.
2
(a) f (x) = 2x
√3 − 3x + 1, [0, 2]
(b) f (x) = x, [0, 1]
6. Let f (x) = (x − 3)−2 . Show that there is no value c in (1, 4) such that f (4) − f (1) = f 0 (c)(4 − 1) = 0. Why does
this not contradict Mean Value Theorem?
7. Use the Mean Value Theorem to prove the inequality
| sin a − sin b| ≤ |a − b|
for all a and b
Tutorial 6.5.1
1. Thomas Exercises 11th ed. 4.3 pp. 289–290: 1, 5, 8, 9, 16, 17, 21, 23, 47, 48 or 12th ed. 4.3 pp. 241–242: 1,
5, 12, 19, 26, 27, 33, 35, 67, 68.
2. Thomas Exercises 11th ed. 4.4 pp. 298–301: 1, 2, 3, 5, 67, 68, 75, 77, 79, 83, 84 or 12th ed. 4.4 pp. 251–254:
1, 2, 3, 5, 103, 104, 111, 113, 115, 117, 118.
3. (a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f .
22
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
(c) Find the intervals of concavity and the inflection points.
(i) f (x) = x3 − 3x2 − 9x + 4
2
(ii) f (x) = x2x+3
(iii) f (x) = sin x + cos x, 0 ≤ x ≤ 2π
4. Sketch the graph of a function that satisfies all the given conditions.
(a) f 0 (1) = f 0 (−1) = 0, f 0 (x) < 0 if |x| < 1,
f 0 (x) > 0 if 1 < |x| < 2, f 0 (x) = −1 if |x| > 2,
f 00 (x) < 0 if − 2 < x < 0, infection point (0,1)
(b) f 0 (x) > 0 for all x , 1, vertical asymptote x = 1,
f 00 (x) > 0 if x < 1 or x > 3, f 00 (x) < 0 if 1 < x < 3
(c) f 0 (0) = f 0 (4) = 0, f 0 (x) = 1ifx < −1,
f 0 (x) > 0 if 0 < x < 2,
f 0 (x) < 0 if − 1 < x < 0 or 2 < x < 4 or x > 4,
lim− f (x) = ∞, lim+ f (x) = −∞
x→2
00
x→2
f (x) > 0 if −1 < x < 2 or 2 < x < 4,
f 00 (x) < 0 if x > 4
Tutorial 6.5.2
1.Thomas Exercises 11th ed. 4.4 p. 299: 21, 25, 27, 29, 30, 33, 39, 41, 42 or 12th ed. 4.4 p. 252–253: 23, 35, 37,
41, 49, 53, 55, 58, 85, 87, 88, 97.
2. Find the horizontal and vertical asymptotes of each curve.
(a) y = 5+4x
x+3
2
+x−1
(b) y = 2xx2 +x−2
(c) y = √ x−9
2
4x +3x+2
3. Find the limits as x → ∞ and as x → −∞. Use this information, together with the intercepts, to give a rough
sketch of the curve.
(a) y = 2x3 − x4
(b) y = x3 (x + 2)2 (x + 1)
(c) y = x4 − x6
4. Find the horizontal asymptotes of each curve and use them, together with concavity and intervals of increase
and decrease, to sketch of the curve.
2
(a) y = 1+2x
1+x2
(b) y = 1−x
1+x
(c) y = x2x+1
5. Sketch the curve.
(a) y = x3 + 6x2 + 9
2
−4
(b) y = xx2 −2x
sin x
(c) y = 2+cos x
Tutorial 6.6.1
1. Thomas Exercises 11th ed. 4.5 pp. 309–315: 4, 7, 8, 11, 16, 18, 19, 22, 24, 33, 43, 44, 52a, 59a or 12th ed. 4.6
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
23
pp. 268–273: 4, 7, 8, 11, 16, 18, 19, 22, 24, 39, 51, 52, 60a, 67a.
2. Thomas Chapter 4 Additional and Advanced Exercises 11th ed. pp. 348–350: 15, 17, 18, 26 or 12th ed. pp.
294–295: 15, 17, 18, 26.
3. What is the maximum vertical distance between the line y = x + 2 and the parabola y = x2 for −1 ≤ x ≤ 2?
4. What is the minimum vertical distance between the parabolas y = x2 + 1 and y = x − x2 ?
5. Find the point on the line y = 2x + 3 that is closest to the origin.
√
6. Find the point on the curve y = x that is closest to the point (3,0).
Tutorial 7.0.1
1. Thomas Exercises 7.3 (12th ed.) or 7.4 (11th ed.): 1, 3, 5, 6, 7, 9–12, 14, 15, 17, 21, 23, 25, 27, 31, 37, 40,
41, 45, 55, 61, 62, 67, 69 and 71.
2. Prove the following formulas:
ä
Ä
√
(a) sinh−1 x = ln x + x2 + 1 ,
ä
Ä
√
(b) cosh−1 x = ln x + x2 − 1 ,
Å
ã
1+x
1
−1
(c) tanh x = ln
.
2
1−x
2
Calculus Tests and Examinations
2.1
Calculus 2012
2.1.1
Calculus March 2012
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–6 are multiple choice questions. In each of these questions, circle the letter that corresponds to the correct answer.
Question 1
[2]
y
5
3
1
2
3
x
The graph above represents a function f . Which of the following is a true statement?
The limit of the above function:
A.
Exists for all x ∈ (1, 2).
24
2012–2017 Past Test and Examination Questions Booklet
B.
Exists for all x ∈ [1, 2].
C.
Exists for all x ∈ (1, 3).
D.
Does not exist at x = 3.
E.
Exists at x = 2.
Question 2
[2]
We are given that
√
…
3 − 2x2 ≤ f (x) ≤
x2
3−
for all x ∈ (−1, 1). Then
2
A.
lim f (x) does not exist because the one-sided limits are not equal.
B.
The left-hand limit exists but the right-hand limit does not exist.
C.
The limit lim f (x) exists.
D.
lim
E.
There is not enough information to determine whether or not lim f (x) exists.
x→0
x→0
x→0
√
3 − 2x2 < lim f (x) < lim
x→0
x→0
3−
x2
.
2
x→0
Question 3
√
3x2 − 12
Let f (x) =
. Which of the following statements is true?
x−2
A.
B.
C.
D.
E.
Mathematics I (Major)
[2]
lim f (x) = ∞.
x→2−
lim f (x) = 0.
x→2
lim f (x) = lim+ f (x).
x→2−
x→2
lim f (x) = ∞.
x→2+
√
lim f (x) = 2 3.
x→2
Question 4
[2]
Let g(x) = tan x. The function g:
A.
is continuous on (0, 2π).
B.
is continuous on all intervals of the form
C.
is undefined at any integer multiple of
D.
has no asymptotes.
E.
has a horizontal asymptote at y =
Question 5
x→a
lim f (x) = −2.
x→a
ã
2n − 1 2n + 1
π,
π .
2
2
π
.
2
π
.
2
[2]
Suppose that lim | f (x)| = 2. Then
A.
Å
Mathematics I (Major)
B.
lim f (x) = 2 or lim f (x) = −2.
C.
lim f (x) does not exist.
D.
there is not enough information to determine whether or not lim f (x) exists.
E.
lim f (x) exists but we do not know its value.
x→a
25
2012–2017 Past Test and Examination Questions Booklet
x→a
x→a
x→a
x→a
Question 6
[2]
y
2
3
6
x
9
The graph above represents the function f . Which of the following statements is false?
A.
The function is defined at x = 3 and x = 3 is a vertical asymptote.
B.
The function is undefined at x = 9 and lim f (x) does not exist.
C.
D.
(E)
x→9
lim f (x) = ∞.
x→∞
lim f (x) = −∞.
x→−∞
The function is continuous on (3, 9).
Total Section A: [12] marks
Section B
In this section you are expected to show all your working to earn the marks allocated.
Question 1
Evaluate the following limits if they exist:
√
√
x2 + 4 − x + 4
(a) lim
x→1
2x2 − 2
√
9x2 − 4
(b) lim
x→−∞ 3 − 2x
 2
x −4



√
if − 1 < x < 1

 4 − x2
(c) lim f (x) if f (x) =

x→−1

2


x − 4
if |x| > 1 and x , 2.
x−2
13]
(5)
(4)
(4)
26
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Question 2
[8]
(a) State the Sandwich Theorem.
(4)
1
(b) Use the Sandwich Theorem to prove that lim x2 sin 2 = 0.
x→0
x
Question 3
[7]
(a) Define the continuity of a function at a point x = c.
(2)
(4)
(b) Identify and classify any discontinuities of the function defined by
g(x) =
2.1.2
x3 − 4x2 + 4x
.
x2 − x
(5)
Calculus May 2012
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–6 are multiple choice questions. Circle the correct answer of your choice.
Question 1
Å ã0
f
(3) =
If f and g are differentiable functions with f (3) = 4, f 0 (3) = −6, g(3) = −1, g0 (3) = 9, then
g
(A)
−30
(B)
−15
(C)
−
(D)
2
3
(E)
42
2
3
Question 2
(A)
− tan x
(B)
cot2 x
(C)
− cot2 x
(D)
− cosec2 x
(E)
sec2 x
Question 3
(A)
1
(B)
ex
x
(C)
(D)
e x ln x − xe x
Å
ã
1
x
e ln x +
x
(2)
d
cot x =
dx
(2)
d x
e ln x =
dx
Mathematics I (Major)
(E)
1 x
xe
27
2012–2017 Past Test and Examination Questions Booklet
ln x
(2)
Question 4
Z
1
(A)
3 ln 3 − 6
(B)
−2.7
(C)
−1.7
(D)
3 ln 3 − 4
(E)
3 ln 3
3
(ln x − x) dx =
(2)
Question 5
0
0
Let f be a differentiable function with f (0) = 1, f (0) = 2, f (1) = 3, f (1) = 4. Then
(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
Z
0
1
f 0 (x) f (x) dx =
(2)
Question 6
√
|1 − x| x
On the interval [0, ∞), the function f (x) =
has vertical tangents at:
x+1
(A)
x = 0 only.
(B)
x = 1 only.
(C)
x = 0 and x = 1 only.
(D)
x = −1, x = 0 and x = 1.
(E)
f has no vertical tangents.
(2)
Total Section A: [12] marks
Section B
Question 1: 7 Marks
(a) State the definition of the derivative of f (x) at x = a.
[2]
(b) Prove that a differentiable function is continuous.
[5]
Question 2: 6 marks
Find the derivatives of the following functions: You do not need to simplify your answers.
2
2
(a) g(x) = ecos x
x3 − 2 x
(b) f (x) =
1 + ln x
[3]
[3]
28
2012–2017 Past Test and Examination Questions Booklet
Question 3: 8 marks
(a) Use implicit differentiation to find
Mathematics I (Major)
dy
if
dx
xy + sin y + 1 = e x .
[3]
(b) Using (a) above, find the equation of the normal line to
xy + sin y + 1 = e x
at
(0, π) .
[4]
Question 4: 8 marks
Evaluate
(a)
Z
sin(5x) cos3 (5x) dx
[4]
(b)
Z
1
4
xe x dx.
[4]
2.1.3
Calculus June 2012
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–5 are multiple choice questions. In each of these questions, circle the letter that corresponds to the correct answer.
Question 1
[2]
1
3
Suppose that f (x) = x − x.
A.
B.
The given function satisfies the conditions of Rolle’s Theorem on [−1, 1] and
1
f 0 (c) = 0 if c = √ .
3 3
The given function does not satisfy the conditions of Rolle’s Theorem on [−1, 1] but f
c=
1
√ ∈ (−1, 1).
3 3
0
Å
1
√
3 3
ã
= 0 and
C.
There is no c ∈ (−1, 1) for which f 0 (c) = 0.
D.
Rolle’s Theorem is not applicable to f on [−1, 1] because its graph does not have a horizontal tangent in
[−1, 1].
E.
None of the above
Question 2
[2]
A differentiable function g, defined on R, has a relative minimum at x = 2. Choose the correct statement from the
following:
A.
g0 is positive on (−∞, ∞).
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
B.
g0 is positive on (−∞, 2).
C.
g0 is negative on (−∞, ∞).
D.
g0 is negative on (−∞, 2).
E.
None of the above
Question 3
Å
ã
1 x
lim 1 +
=
x→∞
2x
A.
√
B.
1
.
2
C.
0.
D.
∞.
E.
None of the above.
[2]
e.
Question 4
[3]
Z
A.
B.
dx
√
=
(arccos x)2 1 − x2
arcsin x + C.
√
− 1 − x2 + C.
C.
1
+ C.
arccos x
D.
cosec x + C.
E.
None of the above.
Question 5
[3]
πi
Suppose that g(x) = f (sin x) is a one-to-one function on 0, . Let f
2
Å ã
1
1
f0
= √ .
2
3
Then (g−1 )0 (3) =
h
A.
B.
29
Å ã
1
= 3 and
2
π
.
6
√
3.
C.
1
.
3
D.
2.
E.
None of the above.
Total Section A: [12] marks
Section B
30
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
In this section you are expected to show all your working to earn the marks allocated.
Question 1
[3]
x
If f (x) = 2 , find f
(n)
(x), the n-th derivative of f .
Question 2
[10]
x
Let f (x) = 1 + arcsin √
1 + x2
(a) What is the domain of f ?
− arctan x.
(1)
0
(b) Show that f = 0.
(6)
(c) What conclusion can you draw about f from (b)?
(1)
(d) Show that f (x) = 1 for all x ∈ dom f .
Question 3
Z
dx
1
x
(a) Prove that
= arctan + C.
2
2
a +x
a
a
Z 3
dx
(b) Hence evaluate √
.
2
3 3+x
Question 4
(2)
[7]
(4)
(3)
[8]
cos x
Use l’Hospital’s Rule to evaluate the left hand limit limπ − (tan x)
x→ 2
2.1.4
.
Calculus August 2012
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–5 are multiple choice questions. Each of these questions has exactly one correct answer.
Circle the letter of your choice.
Question 1
[2]
2
The average value of the function defined by f (x) = 3x − 2x over the interval [2, 6] is:
A.
44
B.
4
C.
(3 × 62 − 2 × 6) + (3 × 22 − 2 × 2)
2
D.
(3 × 62 − 2 × 6) − (3 × 22 − 2 × 2)
E.
None of the above
Question 2
[2]
y
(y − k)2 = f (x)
x = u(y)
d
(y − k)2 = g(x)
k
x = v(y)
c
a
b
x
Mathematics I (Major)
31
2012–2017 Past Test and Examination Questions Booklet
The area A of th region bounded by the parabolas
(y − k)2 = f (x) [or x = u(y)] and (y − k)2 = g(x) [or x = v(y)] is given by
Rb
A.
a [ f (x) − g(x)] dx
Ra
Rb
B.
0 [ f (x) − g(x)] dx + a [ f (x) − g(x)] dx
Ra
Rb
C.
0 f (x) dx + a g(x) dx
Rd
D.
c [v(y) − u(y)] dx
Rd
E.
c [u(y) − v(y)] dx
Question 3
The value of
A.
B.
C.
D.
E.
Z
sin x
cos x
[2]
d
dx
Z
sin x
cos x
3
u
du is
2 + u2
d u3
du
dx 2 + u2
sin3 x cos x cos3 x sin x
+
2 + cos2 x
2 + sin2 x
sin3 x
cos3 x
−
2 + sin2 x 2 + cos2 x
ï 2
òsin x
3u (2 + u2 ) − u3 (2u)
(2 + u2 )2
cos x
None of the above.
Question 4
[2]
Which of the following statements is false?
A.
If a < b < c and f is continuous, then
Z b
a
B.
Z
a
c
f (x) dx +
Z
c
b
f (x) dx
lim cosh x = ∞
x→−∞
C.
R1
D.
cosh2 x = 1 + sinh2 x.
d
sech x = sech x tanh x
dx
E.
f (x) dx =
−1
tan3 x dx = 0
Total Section A: [8] marks
Section B
In this section you must answer each question as fully as possible to earn full marks.
Question 1
[4]
Determine the area of the region bounded by
y=
Question 2
(1 + ln x)3
, y = 0, x = 1 and x = e.
x
[6]
32
2012–2017 Past Test and Examination Questions Booklet
y
Mathematics I (Major)
y = x3
x
2
The region bounded by y = x3 , y = 0 and x = 2 is revolved about the line x = 2. What is the value of the solid so
generated?
Question 3
[6]
The Fundamental Theorem of Calculus Part 1 (FTC 1) states:
If f is continuous on [a, b], then the function defined by
Z
g(x) =
a
x
f (t) dt,
a ≤ x ≤ b,
is continuous on [a, b], is differentiable on (a, b) and g0 (x) = f (x).
Fill in the gaps in the following proof of this theorem.
Proof
g(x + h) − g(x)
=
h
=
1
=
h
x+h
Z
x
f (t) dt.
Assume, for now, that h > 0. By the Mean Value Theorem for Integrals, there exists a c in [x, x + h] such that
x+h
Z
x
f (t) dt =
Since x ≤ c ≤ x + h, we have that c → x as h → 0 from which it follows that f (c) → f (x).
This conclusion follows since
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
33
This means that limh→0+ f (c) = f (x). Hence
g0+ (x) =
=
= f (x)
By a similar argument for h < 0, it can be shown that g0− (x) = f (x).
Rx
d
Therefore g0 (x) = dx
a f (t) dt = f (x).
Question 4
Z 4
Evaluate
0
2.1.5
[8]
x dx
.
x2 − 4x + 8
Calculus October 2012
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–9 are multiple choice questions. Each of these questions has exactly one correct answer.
Circle the letter of your choice.
Question 1
[2]
Which of the following substitutions is most appropriate to evaluate
Z
x4
√
dx ?
x2 − 1
√
A.
x= u
B.
x = sin u
C.
x = tan u
D.
x = sec u
E.
x = arctan u
Question 2
[3]
x5 + 1
The rational function 2
has a partial fraction expansion of the form
(x − 1)2
A.
f (x) = Gx2 + F x +
B.
f (x) = E +
C.
f (x) = E +
D.
f (x) = F x +
A
B
+
with A, B , 0
2
(x − 1)
(x + 1)2
A
B
+
with A, B, E , 0
(x − 1)2 (x + 1)2
A
C
B
D
+
+
+
with A, B, E , 0
(x − 1)2 x − 1 (x + 1)2 x + 1
A
C
B
D
+
+
+
with A, B, F , 0
2
2
(x − 1)
x − 1 (x + 1)
x+1
34
E.
2012–2017 Past Test and Examination Questions Booklet
f (x) = Gx2 + F x + E +
Mathematics I (Major)
A
C
B
D
+
+
+
with A, B, G , 0
(x − 1)2 x − 1 (x + 1)2 x + 1
Question 4
[3]
The series
∞
X
5
(−1)n n+1
3
n=0
A.
is a convergent geometric series with
∞
X
5
5 1
(−1)n n+1 =
3
3 1+
n=0
B.
is a convergent geometric series with
∞
X
5 1
5
(−1)n n+1 =
3
3 1−
n=0
C.
is a convergent geometric series with
∞
X
1
5
(−1)n n+1 = 5
3
1−
n=0
D.
converges but does not converge absolutely
E.
diverges
1
3
1
3
1
3
Question 5
[3]
1
1
= 1 + x + x2 + x3 + . . . , it follows that
has the power series expansion
Using the geometric series
1−x
1 + x3
A.
1 + x3 + x9 + x27 + . . .
B.
1 + x3 + x6 + x9 + . . .
C.
1 − x3 + x6 − x9 + . . .
D.
1 − x3 − x6 − x9 + . . .
E.
1 − x3 + x9 − x27 + . . .
Question 6
[2]
Which of the following is a linear differential equation?
A.
y0 = xy2 + 2x
B.
y0 = xyy0 + 2x
R
y0 + xy = y(x) dx + x3
√
y0 + xy = x3 + x
√
y0 + xy = y3 + y
C.
D.
E.
For the next three questions, consider the function
f (x) =
You may use that
f 0 (x) = 1 −
Question 7
The function f
3
(x − 2)2
x2 − 1
.
x−2
and
f 00 (x) =
6
(x − 2)3
[4]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
A.
has no x- and y-intercepts
B.
has x-intercepts at x = −1 and x = 1 and an oblique asymptote y = x + 2
C.
has as natural domain the set (−∞, −1) ∪ (−1, 1) ∪ (1, 2) ∪ (2, ∞)
√
√
has critical points at x = 2 − 3 and x = 2 + 3, and has a points of inflection at x = 2
D.
E.
35
is odd, concave down on (−∞, 2) and concave up on (2, ∞)
Question 8
[3]
The function f
A.
B.
C.
D.
E.
√
√
√
√
is decreasing on (−∞, 2− 3), increasing on (2− 3, 2), decreasing on (2, 2+ 3), increasing on (2+ 3, ∞)
√
√
√
√
is decreasing on (−∞, 2− 3), increasing on (2− 3, 2), increasing on (2, 2+ 3), decreasing on (2+ 3, ∞)
√
√
√
√
is decreasing on (−∞, 2− 3), decreasing on (2− 3, 2), increasing on (2, 2+ 3), increasing on (2+ 3, ∞)
√
√
√
√
is increasing on (−∞, 2− 3), decreasing on (2− 3, 2), decreasing on (2, 2+ 3), increasing on (2+ 3, ∞)
√
√
√
√
is increasing on (−∞, 2− 3), decreasing on (2− 3, 2), increasing on (2, 2+ 3), decreasing on (2+ 3, ∞)
Question 9
At x = 2 −
√
[2]
3, f has a
A.
maximum
B.
minimum
C.
vertical asymptote
D.
point of inflection
E.
none of the above
Total Section A: [24] marks
Section B
In this section you must answer each question as fully as possible to earn full marks.
Question 1
[8]
Use your answers to MCQ questions 7, 8 and 9 to sketch the graph of the function
f (x) =
showing, if any,
• intercepts
• maxima and minima
• points of inflection
• asymptotes
x2 − 1
,
x−2
36
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
You do not need to justify your results.
Question 2
[7]
2
Find the area of the largest rectangle that can be inscribed between the curve y = 12 − x and the x-axis.
Question 3
Evaluate
Z
dx
√
(a)
2
x x2 − 4
Z √ 2
x +1
dx
(b)
x4
Question 4
Z
x4 + 2
Evaluate
dx.
(x2 − 2x + 1)(x2 + 2x)
Question 5
[12]
(6)
(6)
[12]
[12]
(a) Determine if the following series converge
∞
∞
X
X
1
.
(i)
33n 72−n ,
(ii)
n(ln n)2
n=1
n=2
(b) Show that the series
(7)
∞
X
n=2
2
n(n + 1)
converges and find its limit.
(5)
Question 6
[8]
Prove that if the series
∞
X
an converges, then lim an = 0.
n→∞
n=1
Question 7
[7]
0
2
2
(a) Find the general explicit solution of the differential equation y = (x + 4)(y + 1).
(5)
(b) Find the particular solution of the differential equation in (a) which satisfies the initial condition y(0) = 1. (2)
2.2
Calculus 2013
2.2.1
Calculus March 2013
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–6 are multiple choice questions. Each of these questions has exactly one correct answer.
Circle the letter of your choice.
Question 1
[2]
Let f and g be real-valued functions such that lim f (x) = L and lim g(x) = M exist. Which of the following
x→a
x→a
statements is false?
(A)
lim f (x) + g(x) = L + M.
x→a
√3
If L ≤ 0, then lim
(C)
If L = 0 and M = 0, lim
(D)
lim k f (x) = kL, where k is any constant.
x→a
f (x) =
√3
(B)
f (x)
x→a g(x)
x→
L.
does not exist.
Mathematics I (Major)
(E)
2012–2017 Past Test and Examination Questions Booklet
37
n
If n ∈ Z and L , 0, lim f (x) , 0.
x→a
Question 2
[2]
Which of the following statements is false?
(A)
(B)
(C)
(D)
(E)
lim (e x − e−x ) = ∞.
x→∞
lim (e x − e−x ) = −∞.
x→−∞
lim 3 x = ∞.
x→∞
lim π−x = −∞.
x→−∞
lim e−x = ∞.
x→−∞
Question 3
Let L be a non-negative number. Suppose that lim f (x) + L = 0. Then:
[2]
x→a
(A)
(B)
lim f (x) = −L and lim+ f (x) = L.
x→a
x→a−
lim f (x) = −L and lim− f (x) = L.
x→a+
x→a
(C)
There is not enough information to say that lim f (x) exists.
(D)
lim f (x) = L.
(E)
lim f (x) = −L.
x→a
x→a
x→a
Question 4


 sin(πx)
Let f (x) = 0

 2
x
[2]
if x > 0
if x = 0
if x < 0
(A)
lim f (x) does not exist.
(B)
f is continuous from the left at 0 but not from the right at 0.
(C)
f is continuous from the right at 0 but not from the left at 0.
(D)
f is continuous at 0.
(E)
None of the above.
x→0
Question 5
The graph representing the function f can be seen below.
(A)
f is defined at x = 2 and x = 2 is a vertical asymptote.
(B)
The line y = 3 is horizontal asymptote.
(C)
f is continuous on (2, 8).
(D)
(E)
lim f (x) = 3.
x→∞
f is continuous on the right of x = 8.
[2]
38
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Section B
Question 1
[17]
Evaluate the following limits if they exist:
2x3 −54
2
x→3 x −9
(a) lim
(3)
3x cos x
√
x+4−2
(3)
(c) lim 1−cos(θ)
θ2
(3)
(b) lim
x→0
θ→0
(d) lim x→2
√
√
x2 +5− 2x+5
3x2 −12
(4)
√
9x2 + 5
(e) lim
x→−∞ 1 − 2x
(4)
Question 2
[4]
(a) State the squeeze Theorem.
(2)
(b) You are given that for all x ∈ (−1, 1), the function f (x) satisfies
Show that lim f (x) = 1.
sin x
x
≤ f (x) ≤ 1 + x2 .
x→0
Question 3
(2)
[9]
Recall that for a real number a, bac is the greatest integer less or equal to a, while dae is the smallest integer greater
or equal to a.
Let n be a non-negative integer and f the function defined as follows:


 n − dxe
f (x) = 0


n − bxc
if x < n
if x = 0
if x > 0
1. Evaluate:
(a) lim− f (x),
(3)
(b) lim+ f (x).
(3)
x→n
x→n
2. Is the function f continuous at the point n? Justify your answer.
2.2.2
(3)
Calculus April 2013
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–6 are multiple choice questions. In each of these questions, circle the letter that corresponds to the correct answer.
Question 1
[2]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
39
y
5
3
1
2
3
x
The graph above represents a function f . Which of the following is a true statement? The limit of the above
function:
A.
Exists for all x ∈ (1, 2).
B.
Exists for all x ∈ [1, 2].
C.
Exists for all x ∈ (1, 3).
D.
Does not exist at x = 3.
E.
Exists at x = 2.
Question 2
[2]
We are given that
√
…
3 − 2x2 ≤ f (x) ≤
3−
2
x
for all x ∈ (−1, 1). Then
2
A.
lim f (x) does not exist because the one-sided limits are not equal.
B.
The left-hand limit exists but the right-hand limit does not exist.
C.
The limit lim f (x) exists.
D.
lim
E.
There is not enough information to determine whether or not lim f (x) exists.
x→0
x→0
x→0
√
3 − 2x2 < lim f (x) < lim
x→0
x→0
3−
x2
.
2
x→0
Question 3
√
3x2 − 12
Let f (x) =
. Which of the following statements is true?
x−2
A.
B.
C.
D.
E.
lim f (x) = ∞.
x→2−
lim f (x) = 0.
x→2
lim f (x) = lim+ f (x).
x→2−
x→2
lim f (x) = ∞.
x→2+
√
lim f (x) = 2 3.
x→2
[2]
40
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Question 4
[2]
Let g(x) = tan x. The function g:
A.
is continuous on (0, 2π).
B.
is continuous on all intervals of the form
C.
is undefined at any integer multiple of
D.
has no asymptotes.
E.
has a horizontal asymptote at y =
Å
ã
2n − 1 2n + 1
π,
π .
2
2
π
.
2
π
.
2
Question 5
[2]
Suppose that lim | f (x)| = 2. Then
x→a
A.
lim f (x) = −2.
B.
lim f (x) = 2 or lim f (x) = −2.
C.
lim f (x) does not exist.
D.
there is not enough information to determine whether or not lim f (x) exists.
E.
lim f (x) exists but we do not know its value.
x→a
x→a
x→a
x→a
x→a
x→a
Question 6
[2]
y
2
3
6
9
The graph above represents the function f . Which of the following statements is false?
A.
The function is defined at x = 3 and x = 3 is a vertical asymptote.
B.
The function is undefined at x = 9 and lim f (x) does not exist.
C.
D.
x→9
lim f (x) = ∞.
x→∞
lim f (x) = −∞.
x→−∞
x
Mathematics I (Major)
(e)
2012–2017 Past Test and Examination Questions Booklet
41
The function is continuous on (3, 9).
Section B
Question 1
[7]
(a) State the definition of the derivative of f (x) at x = a
(2)
(b) State and prove the Quotient Rule for derivatives
(5)
Question 2
[7]
1) Give the definition of an anti-derivative.
(1)
2) Find the derivatives of the following functions: You do not need to simplify your answers.
(a) f (x) =
5x
1
ex
3
(b) g(x) = e x +ln 9 .
.
(3)
ex +
(3)
Question 3
[8]
(a) Use implicit differentiation to find
dy
if e xy + y2 cos x + e x = 6.
dx
(3)
(b) Find the slope of the tangent line to e xy + y2 cos x + e x = 6 at (0, 2).
(2)
(c) Find the equation of the normal line to e xy + y2 cos x + e x = 6 at (0, 2).
(3)
Question 4
[6]
(a) Give the linear approximation of a differentiable function f at a.
√
(b) Find the linear appriximation of the function f (x) = 3 x at x = 1.
√3
(c) Use (b) above to approximate the value of 1.0006.
2.2.3
(2)
(2)
(2)
Calculus June 2013
Section A: Multiple choice questions
[10]
Instructions:
Section A, Questions 1–5 are multiple choice questions. In each of these questions, circle the letter that corresponds to the correct answer.
Question 1
You are given that f (x) and F(x) are functions which satisfy
Z
f (x)dx = F(x) + c, where c is constant.
Then it follows by definition (or otherwise) that
A.
B.
d2 F(x)
= f (x).
dx2
Rb
a f (x)dx = F(a) − F(b).
[2]
42
2012–2017 Past Test and Examination Questions Booklet
C.
d R
f (x)dx) = F(x).
dx
D.
d
[F(x)] = f (x).
dx
E.
None of the above.
Mathematics I (Major)
Question 2
[2]
Let a be a positive constant. Then
A.
d x
[a ] = xa x−1 .
dx
B.
d a
[x ] = xa ln x.
dx
C.
R
xa dx =
xa+1
+ c.
a+1
D.
R
a x dx =
a x+1
+ c.
x+1
E.
None of the above.
Question 3
R
arcsin xdx =
A.
1
√
+ c.
1 − x2
B.
− arccos x + c.
C.
arccos x + c.
Z
1
√
dx.
1 − x2
D.
E.
[2]
None of the above.
Question 4
[2]
Let f (x) = ln x be defined on the interval [1, 2]. Then
A.
f (x) is not continuous on [1, 2].
B.
f (x) is not differentiable on (1, 2).
C.
There exists c in (1, 2) such that f 0 (c) = 0.
D.
There exists c in (1, 2) such that
E.
None of the above.
1
f (2) − f (1)
=
.
c
2−1
Question 5
[2]
n
You are given that f (x) = x , where n is a positive integer and x is a positive real number. Then
A.
f 0 (2) = n2n−2 .
B.
f (2) (x) = n(n − 1)xn−1 .
C.
f (3) (1) = n(n − 1)(n − 2)(n − 3).
Mathematics I (Major)
D.
f (n) (1) = 0.
E.
f (n) (x) = n!
2012–2017 Past Test and Examination Questions Booklet
43
Section B
Question 1
[10]
(a) Use the Substitution Rule to evaluate
(5)
Z
π 1
8−2
− 12
tan4 (2x + 1) sec2 (2x + 1)dx.
(b) Evaluate
(5)
Z
(x − 1)2 cos xdx.
Question 2
[10]
2
x
1
(x − 1) 3 .
3
Let f (x) =
(a) Is the function f (x) continuous on the closed interval [0, 2]? Explain.
(1)
(b) Find the absolute maximum and minimum values of f (x) on the interval [0, 2].
(5)
(c) If c is a critical number of f (x), then the point with coordinates (c, f (c)) is called a critical point.
By drawing up a sign table of the first derivative, classify all the critical and end points of f (x) defined in
the interval [0, 2] as local maxima, minima or vertical tangents.
(4)
Question 3
[10]
(a) Find the derivative of (tan x)cos x .
(5)
1
2x
(b) Use l’Hôpital’s rule to evaluate lim (1 + 3x) .
x→0
2.2.4
(5)
Calculus August 2013
Section A: Multiple choice questions
[8]
Instructions:
Section A, Questions 1–4 are multiple choice questions. In each of these questions, circle the letter that corresponds to the correct answer.
Question 1
[2]
b−a
Let a = x0 < x1 < x2 < · · · < xn−1 < xn = b be a partition of [a, b], and ∆x =
, where a = −2 and b = 3. Let
n
ck be any sample point in the subinterval [xk−1 , xk ], where k = 1, 2 · · · , n. Then
A.
B.
lim
n→∞
lim
n→∞
n
X
3c2k ∆x = 34.
k=1
n
X
k=1
3c2k ∆x = 35.
44
C.
D.
E.
2012–2017 Past Test and Examination Questions Booklet
n
X
lim
n→∞
n→∞
3c2k ∆x = 36.
k=1
n
X
lim
Mathematics I (Major)
3c2k ∆x = 37.
k=1
None of the above.
Question 2
[2]
Suppose that u, v and w are integrable functions and
Z 8
Z 10
w(x)dx = 6.
w(x)dx = 2 and
Z
2
10
u(x)dx = −3,
Z
8
10
u(x)dx = 5,
Z
8
10
v(x)dx = 4,
−2
−2
Which of the following is false?
8
A.
Z
B.
Z
C.
Z
D.
Z
E.
None of the above.
2
u(x)dx = −8.
10
w(x)dx = −4.
8
10
8
10
w(x)dx = 4.
î
8
ó
− 6u(x) + 5v(x) dx = −10.
Question 3
[2]
2
The average value of the function f (x) = 6x − 2x over the interval [1, 3] is:
A.
f (3) − f (1)
= 24.
2
B.
f 0 (3) − f 0 (1)
= 12.
2
C.
44.
D.
22.
E.
None of the above.
Question 4
[2]
Which of the followings is false?
Z a
Z a
A.
f (x)dx = 2
f (x)dx if f is even.
0
−a
B.
Z
a
−a
C.
Z
a
f (x)dx = −2
Z
0
a
f (x)dx if f is even.
f (x)dx = 0 if f is odd.
−a
D.
Z
a
b
f (x)dx =
Z
a
c
f (x)dx −
Z
b
c
f (x)dx if f is even.
Mathematics I (Major)
E.
45
2012–2017 Past Test and Examination Questions Booklet
None of the above.
Section B
Question 1
[7]
(a) Solve cos 2x = sin 2x for x ∈ [0, π].
(2)
(b) Find the area of the region bounded above and below by the graphs y = sin 2x and y = cos 2x, where
x ∈ [0, π].
Question 2
(5)
[5]
(a) Simplify ln xn , where n ∈ Z and x > 0.
(0.5)
(b) Find the derivative of f (x) = 1 + ln x.
(c) Determine the area of the region bounded by
y=
(0.5)
(4)
(3 + ln x3 )2
,
x
y = 0, x = 1, x = e.
Question 3
[5]
(a) State the Fundamental Theorem of Calculus 1, (FTC1) with all conditions.
(2)
(b) Evaluate the following derivatives:
(3)
d
dx
Z
sec x
tan x
1
dt,
1 + t2
Question 4
[4]
2
Find the volume of the solid generated by revolving the region between the parabola x = y + 2
and the line x = 5 about the line x = 5.
y
x = y2 + 2
1
2
3
4
x
5
x=5
2.2.5
Calculus October 2013
Section A: Multiple choice questions
[24]
Instructions: Section A, Questions 1–8 are multiple choice questions. In each of these questions, circle the letter
that corresponds to the correct answer.
46
2012–2017 Past Test and Examination Questions Booklet
Question 1
Mathematics I (Major)
[3]
Which of the following is false.
A.
The general solution of the differential equation 2x(1 + y) − y
x2 − y + ln |1 + y| = k.
dy
= 0 is:
dx
B.
The above differential equation is linear.
C.
The differential equation
D.
The differential equation ye xy dx + xe xy dy = 0 is exact.
ã
Å 2
x2
y2
y
+
dy − dx = 0 is homogeneous of degree 1.
The differential equation
x
y
x
E.
x
y = 2ydx is separable.
d
Question 2
dy
The general solution of the differential equation xy − x + ln x = 0 is:
dx
[3]
x (ln x)2
−
+ c.
2
2
A.
y2 =
B.
y2
= x − (ln x)2 + c.
2
C.
y2
ln x
= x−
+ c.
2
2
D.
y2 = 2x − (ln x)2 + c.
E.
y= x−
(ln x)2
+ c.
2
Question 3
[3]
Let p be a non-negative integer. Which of the following is false?
A.
If
∞
X
1
1
converges, then lim p = 0.
p
n→∞ n
n
n=1
B.
If
∞
X
1
diverges, then 0 ≤ p ≤ 1.
p
n
n=1
C.
If lim
D.
If p > 1, then
E.
∞
X 1
1
=
0,
then
converges.
n→∞ n p
np
n=1
∞ Å
X
1
n=1
n
−
∞
X
1
converges.
p
n
n=1
1
n+1
ã
converges.
Question 4
1
The power series representing the function
is:
(x + 1)2
A.
1 − x + x2 − x3 + · · ·
[3]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
B.
1 − x2 + x4 + −x6 + · · ·
C.
−1 + 2x − 3x2 + 4x3 + · · ·
D.
1 + 2x + 3x2 + 4x3 + · · ·
E.
1 − 2x + 3x2 − 4x3 + · · ·
Question 5
The following improper integral
A.
diverges for p > 1
B.
converges for p < 1
C.
converges for p = 1
D.
is equal to
E.
is equal to
Z
∞
1
[3]
1
dx
xp
1
if p > 1
p−1
1
if p < 1
p−1
Question 6
The improper integral
Z
∞
3
A.
1
.
2
B.
1.
C.
2.
D.
3.
E.
None of the above.
[3]
1
dx is equal to
(x − 1)2
Question 7
The solution of the following integral
A.
cos(e x )
B.
− cos(e x )
C.
cos(e x ) + c
D.
− cos(e x ) + e x + c
E.
− cos(e x ) + c
[3]
Z
e x sin(e x )dx is:
Question 8
Which of the following substitutions is appropriate for the following integral, where |x| < 2?
Z
x4
√
dx.
4 − x2
A.
u = x4 .
47
[3]
48
B.
C.
2012–2017 Past Test and Examination Questions Booklet
π
<θ<
2
π
x = 2 sin θ, − < θ <
2
x = 4 sin θ, −
Mathematics I (Major)
π
.
2
π
.
2
D.
x = 2 sec θ, 0 < θ <
π
3π
or , π < θ <
.
2
2
E.
x = 2 csc θ, 0 < θ <
3π
π
or , π < θ <
.
2
2
Section B
[66]
Question 1
[5]
1) Give the general solution of the differential equation:
(3)
dy x − 2x2
=
.
dx
e2y
2) Give the particular solution of the above differential equation, where the initial condition is y(0) = 2.
Question 2
(2)
[11]
1) Let F1 (θ) = ln | sec θ| + c. Show that the derivative of F1 (θ) is
F10 (θ)
= tan θ
2) Evaluate
Z
a) F2 (θ) = sec2 θ tan θdθ.
Z
b) F3 (θ) = sec2 θ tan3 θdθ.
R
3) Evaluate F(θ) = tan5 θdθ using your answers in questions a) and b) above.
Question 3
(3)
(1)
(1)
(6)
[14]
Let q(x) be the polynomial function
q(x) = x3 − 3x + 2.
1) Evaluate q(1).
(1)
2) Use the above result to write q(x) as a product of linear factors.
(3)
3) Use the result in 2) to find the partial fractions decomposition of the rational
function
(4)
f (x) =
x3
9x
·
− 3x + 2
4)
a) Find the integral
(3)
Z
x3
9x
dx.
− 3x + 2
b) Evaluate and simplify your answer
(3)
Z
2
4
9x
dx.
x3 − 3x + 2
Mathematics I (Major)
49
2012–2017 Past Test and Examination Questions Booklet
Question 4
[18]
1)
a) When is a series said to be conditionally convergent?
cos(nπ) (−1)n
=
, where n is a positive integer.
b) Show that
n
n
∞
X
cos(nπ)
c) Is the series
conditionally convergent? Justify your anwser.
n
n=1
ã
∞ Å
X
5
1
2) Find the sum of the series
− n .
n(n + 1) 3
n=1
(2)
(1)
(4)
(11)
Question 5
1
1) Find the power series representation of the function f (x) =
and its radius of convergence.
1+x
1
2) Give with justification the power series representation of the function g(x) =
.
(1 + x)3
[10]
Question 6
(4)
(6)
[8]
1
1) Evaluate
dx.
x2 − x
2) Using your answer from Question 6-1), determine whether the following integral is
convergent or divergent
Z ∞
1
dx.
2
x −x
1
Z
2.3
Calculus 2014
2.3.1
Calculus March 2014
(4)
(4)
Question 1
[3]
Let the domain of a function f be (−5, −1) and the range of f be (1, 5) \ {2}. You are given that lim − f (x) = 2 =
x→−2
lim + f (x). Here are some true or false statements about f .
x→−2
i.
ii.
iii.
Since 2 is not in the range of f , lim f (x) does not exist.
x→−2
lim f (x) = 2 implies f (−2) = 2.
x→−2
lim f (x) = 2.
x→−2
iv.
f (−2) = 2.
A.
Only i is correct.
B.
Only ii is correct.
C.
Only iii is correct.
D.
Only iv is correct.
E.
iii and iv are correct.
Question 2
[3]
Let functions f and g be given by f (x) = e and g(x) = π defined on the real line. lim ( f (x) + g(x)) is given by:
x→π
A.
e x because e and π are irrational numbers.
50
2012–2017 Past Test and Examination Questions Booklet
B.
π because f + g approaches π.
C.
e + π because this is lim π + lim e.
D.
Nothing because the limit does not exist.
E.
None of the answers above.
x→π
Mathematics I (Major)
x→π
Question 3
[3]
y
5
2
1
2
3
x
The graph above represents a function f (x). f (2) = 0 (i.e x = 2 is a zero of the function). The limit from the right
of f i.e lim+ f (x) = f (2) and lim− f (x) = 2 . Here are some true or false statements about f .
x→2
x→2
i.
lim f (x) = f (2).
ii.
lim f (x) = 2.
iii.
lim f (x) does not exist.
x→2
x→2
x→3
Which of the following is true?
A.
Only i is correct.
B.
Only ii is correct.
C.
Only iii is correct.
D.
All of the statements are correct.
E.
None of the statements is correct.
Question 4
[2]
A student wrote f (x) = loge ((x − 2)2 /c) = − ln c + 2 loge |(x − 2)|, where c is a positive constant and x , 2. What
makes the statement incorrect?
A.
− ln c makes the statement incorrect and should be loge c.
B.
The entire simplification is incorrect.
C.
2 in front of 2 loge |(x − 2)| makes the statement incorrect.
D.
Nothing makes the statement incorrect hence the statement is correct.
E.
2 loge |(x − 2)| makes the statement incorrect. It should have simplified to loge (x2 /22 ) then the statement
would be correct.
Mathematics I (Major)
51
2012–2017 Past Test and Examination Questions Booklet
Question 5
[2]
Given the graph below
y
y = f (x) = bxc
1
−2
1
−1
2
x
−2
Here are some true/false statement/s;
i.
f (x) is not continuous at x = 1 because lim f (x) = 0.
ii.
f (x) is continuous at x = 1 because f (1) = 1.
iii.
f (x) is not continuous at x = 1 because f (x) is not defined at x = 1.
iv.
f (x) is not continuous at x = 1 because lim− f (x) = 0 , f (1) = lim+ f (x).
A.
Only i is correct.
B.
Only ii is correct.
C.
Only iii is correct.
D.
Only iv is correct.
E.
All of the above are not true.
x→1
x→1
x→1
Total Section A: [13] marks
Section B
In this section you are expected to show all your working to earn the marks allocated.
Question 1
[4]
1
Let h(x) = csc x =
. The function sin x has zeros or roots at x = nπ where n ∈ Z. Write down your answer
sin x
next to the letters A, B, C and D in the space provided below.
A.
B.
C.
D.
lim h(x) =
x→2π−
lim h(x) =
x→2π+
lim h(x) =
x→2π
Asymptote/s for h(x) is/are given by x =
52
2012–2017 Past Test and Examination Questions Booklet
Question 2
Mathematics I (Major)
[2]
We are given that some function h(x) is defined at x = a and lim h(x) exists. In order to conclude that h is a
x→a
continuous at a, state other condition/s that must be fulfilled for h to be continuous?
Question 3
[12]
(a) Find a simpler expression for the following i. and ii. below:
x
i) ln(e(e ) )
(4)
ii) e(ln πx − ln 2) for x > 0.
x3 + 1
0
(b) Directly substituting x = −1 into g(x) =
we get g(−1) = . Use limit laws and factorisation to evaluate
x+1
0
lim g(x).
(4)
x→−1
(c) It can be shown that the inequalities
(4)
1+
x2
x sin x
<
<1
6
2 − 2 cos x
hold for all values of x close to zero. What, if anything, does this tell you about
x sin x
?
x→0 2 − 2 cos x
lim
Give reasons for your answer.
Question 4
[9]
1) For what value(s) of a is
f (x) =
x −a


a + 1
if
x≥0



if
x<0
x2 + a
continuous everywhere?
(4)
2) Determine if the one-sided limits at x = 0 exist (i.e limit from the left of x = 0 and limit from right of x = 0).
What do you conclude about the limit and continuity of f (x) at x = 0.
f (x) =
1
1
− .
x |x|
(5)
2.3.2
Calculus April 2014
Section A: Multiple choice questions
Question 7
[3]
We are given that f (x) is differentiable at x = a. Here are some statements:
i.
f 0 (a) = lim
ii.
f 0 (a) = lim
iii.
f+0 (a) = lim+
x→a
h→0
x→a
f (x) − f (a)
.
x−a
f (a + h) − f (a)
.
h
f (x) − f (a)
.
x−a
Which of the following is correct?
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
A.
(i) is true and (ii) and (iii) are false.
B.
(ii) is true and (i) and (iii) are false.
C.
(i) and (ii) are true and (iii) is false.
D.
All the above are true.
E.
Two statements must be false and one true.
Question 8
53
[3]
Which of the following is incorrect?
A.
If f (x) is differentiable at x = a, then it is continuous at x = a.
B.
If f (x) is NOT continuous at x = a, then it is NOT differentiable at x = a.
Z
xn+1
For all constants n, xn dx =
+ C.
n+1
C.
f (x) − f (a)
f (x) − f (a)
= lim−
.
x→a
x−a
x−a
D.
f (x) is differentiable at x = a, implies lim+
E.
f (x) is differentiable at x = a implies f (a) exists (i.e is defined).
x→a
Question 9
[3]
Use the chain rule to decide which of the following is correct.
A.
d f (g(x)) = f 0 (x).g0 (x).
dx
B.
f 0 (x)
d ln f (x) =
.
dx
f (x)
C.
d 4
(x + a x )−1 = −(x4 + a x )−2 .(4x3 + a x ).
dx
D.
d 2
1
(sin (ln(x)) = 2 sin(ln(x)). .
dx
x
E.
d 2
(sin (2x) = 2 sin(2x). cos(2x).
dx
Question 10
Which of the following is true?
A.
B.
d d d f (x).g(x) =
f (x) .
g(x) .
dx
dx
dx
d f (x)
f (x)g0 (x) − f 0 (x)g(x)
=
.
dx g(x)
[g(x)]2
C.
d a f (x) − bg(x) = a f 0 (x) − bg0 (x) where (a and b are constants).
dx
D.
2
d f (x) = 2 f (x).
dx
E.
d 2
1
ln x = 2 .
dx
x
[3]
54
2012–2017 Past Test and Examination Questions Booklet
Question 11
Mathematics I (Major)
[3]
You are given y as a function of u i.e y = f (u), and u is implicitly a function of x. Which of the following is
incorrect?
A.
B.
C.
D.
E.
du
d d f (u) =
f (u) . .
dx
dx
dx
î
ó
d u2
2
e = eu .2u.
dx
ïZ
ò
d
f (x)dx = f (x).
dx
du
d
[sin(π − u)] = − cos(π − u) .
dx
dx
d ln u du
.
e
=
dx
dx
Question 12
[3]
For differentiable functions f (x) and g(x), which of the following is false?
Z
Z ïZ
ïZ
ò
ò
A.
f (x)g(x)dx =
f (x)dx g(x) −
f (x)dx g0 (x)dx.
B.
C.
D.
E.
Z
b
a
d
dx
Z
f (g(x))g0 (x)dx =
Z
g(b)
f (u)du.
g(a)
ïZ
ò
f (x)dx = f (x).
d f (x) dx = f (x) + C.
dx
Z
ïZ
ò
2
[ f (x)]3
f (x) dx =
f (x)dx .
3
Question 13
[3]
Which of the following is true?
Z
A.
tan x dx = sec2 x + C.
B.
Z
sec x tan x dx = tan x + C.
C.
Z
cos x dx = − sin x + C.
D.
Z
sin x dx = − cos x + C.
E.
Z
x2 dx =
ïZ
ò2
x dx .
Question 14
You are given that
A.
Z
[3]
d
dx
[F(x)] = f (x). Which of the following is true.
F(x) dx = f (x) + C.
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
B.
Z
f (x) dx = F(x) + C.
C.
Z
f 0 (x)
dx = F(x) + C.
f (x)
D.
Z
f 0 (x)
dx = ln |F(x)| + C.
f (x)
E.
None of the above.
55
Section B
Full working must be shown.
Question 1
[10]
f (x) =


2x3 + 1


if
x<1



if
x ≥ 1.
3x2
a) Which of the two formulae above is used to find f (1)?
(1)
b) Find f+0 (1) and f−0 (1).
(6)
c) Use (b) to determine whether f 0 (1) exists.
0
(1)
d) If f (1) exists, what is it?
(1)
Question 2
[6]
Use integration by parts to find
Z
e x sin x dx.
(6)
2.3.3
Calculus June 2014
Section A: Multiple choice questions
[12]
Question 5
Z
arctan x dx =
A.
x arctan x + ln(1 + x2 ) + c.
B.
x arctan x − ln(1 + x2 ) + c.
√
x arctan x + ln 1 + x2 + c.
√
x arctan x − ln 1 + x2 + c.
C.
D.
E.
1
+ c.
1 + x2
[3]
56
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Question 6
Z
x+4
dx =
x2 + 4
A.
B.
C.
D.
E.
[3]
x
1
ln(x2 + 4) + arctan
+ c.
2
2
x
1
+ c.
ln(x2 + 4) + 2 arctan
2
2
x
1
1
+ c.
ln(x2 + 4) + arctan
2
2
2
x
ln(x2 + 4) + 2 arctan
+ c.
2
None of the above.
Question 7
(sin x) − x
=
lim
x→0
x3
A.
∞ or −∞ .
B.
1.
C.
1
− .
6
D.
1
.
6
E.
Does not exists.
[3]
Question 8
[3]
Let f (x) = (x − 3) ln(x − 1) be defined on the interval [2, 3]. Which of the following is false?
A.
f (x) = 0 for some x ∈ [2, 3].
B.
f (x) is continuous on [2, 3].
C.
f (x) is differentiable on (2, 3).
D.
There exists c in (2, 3) such that f 0 (c) = 0.
E.
f 0 (x) , 0 for all x ∈ (2, 3).
Section B
In this section you are expected to show all your working to earn the marks allocated.
Question 1
ó
1
dî
arcsin x = √
(a) Prove that
.
dx
1 − x2
ó
dî
(b) Find
arcsin ln x , where e−1 < x < e.
dx
[7]
(5)
(2)
Question 2
In this question you do not need to simplify your answers.
Use the method of logarithmic differentiation to find y0 , where:
[10]
Mathematics I (Major)
√5
2012–2017 Past Test and Examination Questions Booklet
6
xe x (x4 + 10)100 .
sin x cos x
.
(b) y =
e−x x x
(a) y =
57
(5)
(5)
Question 3
[11]
(a) State the Mean Value Theorem.
(4)
ex + 1
(b) Does the function f (x) = x
satisfy the Mean Value Theorem in the interval [−1, 2]? Justify your answer.
e −1
(2)
(c) Use l’Hôpital’s rule to evaluate lim+ (sin x) x
(5)
x→0
2.3.4
Calculus September 2014
Section A: Multiple choice questions
Question 6
n
X
[3]
5
approximates the area between a curve and the positive x-axis on some interval
n
i=1
125
[a, b]. We are given that the maximum value of g in the ith subinterval is g(xi∗ ) = i3 3 . The function and the
n
interval specified in the sum are:
The sum given by
g(xi∗ )
A.
g(x) = 5x3 on [0, 53 ].
B.
g(x) = 53 x on [0, 5]
C.
g(x) = x3 on [0, 5]
D.
g(x) = x3 on [0, 53 ]
E.
g(x) = (5x)3 on [0, 5]
Question 7
[3]
The average value of a function f on an interval [a, b] is given by fave =
the average value of f (t) =
A.
B.
C.
D.
E.
1
− tan t on [0, 2] is
2
1
b−a
Z
a
b
f (x) dx. Using this information,
1
(sec 2 tan 2 − 1).
2
1
sec2 2 − 0 .
2
1
(sec 2 tan 2 − 0).
2
ä
p
1Ä
1 − ln | sec 2| − 0 .
2
p
1
− ln | sec 2|.
2
Question 8
[3]
The area formed by a continuous, non-negative function f on an interval [a, b] and the x-axis is approximated by
partitioning this interval into n subintervals [x j−1 , x j ]. Points in this subinterval a j , b j and c j where chosen such
that f (a j ), f (b j ) and f (c j ) are the minimum, midpoint and maximum value of the function in each subinterval.
The most accurate calculation of the specified area is given by
58
A.
2012–2017 Past Test and Examination Questions Booklet
n
X
f (a j )
b−a
.
n
f (b j )
b−a
.
n
f (c j )
b−a
.
n
j=1
B.
n
X
j=1
C.
n
X
j=1
b
Mathematics I (Major)
D.
Z
E.
An option giving a limit tending to infinity is missing hence a correct answer is missing from the options
given.
a
f (x) dx.
Question 9
[3]
The area formulae of simple geometric figures can be used to evaluate integrals of functions. Let
f (x) =
The integral
Z
0
2


|2x|

if
x<1



if
1 ≤ x ≤ 2.
2
f (x) dx is given by one of the following areas
A.
Two triangles both of base length of 1 and a height of 2.
B.
Two rectangles both of base length of 1 and a height of 2.
C.
A rectangle with an area of 2 and a triangle with an area of 1.
D.
A rectangle with an area of 1 and a triangle with an area of 2.
E.
A rectangle with base length of 1 and a triangle with an area of 1.
Question 10
Use the expression of
[3]
d
dx
Z
A.
2x
.
1 + x2
B.
x
2x
−
.
1 + (2x)2 1 + x2
C.
x
2x
−
.
1 + x2 1 + x2
D.
2
1
−
.
2
1 + 4x
1 + x2
E.
x
2x
+
.
1 + x2 1 + x2
v(x)
u(x)
f (t) dt to select the option giving
Section B
d
dx
2x
Z
x
dt
.
1 + t2
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
59
Full workings must be shown.
Question 1
[9]
a) Evaluate the following integral
Z
dx
.
2
2x − 8x + 10
b) Find the area formed by the region bounded above by x2 + 4y2 = 4 and below by the x-axis.
(4)
Question 2
[6]
Find the volume of the solid generated by revolving the region bounded by y =
about the line x = 4
2.3.5
√
(5)
x and the lines y = 2 and x = 0
(6)
Calculus November 2014
Section A: Multiple choice questions
[24]
Question 1
[3]
Which of the following is false?
A.
∞
X
(−1)n
n2
n=1
B.
∞
X
sin n
n2
n=1
C.
∞
X
n=1
D.
converges absolutely.
1
converges.
n ln n
∞
X
cos(nπ)
n
n=1
E.
converges absolutely.
∞
X
n=1
converges.
1
converges.
n(n + 1)
Question 2
A.
x(x x−1 ).
B.
x + ln x x .
C.
(x + ln x) x .
D.
x x + x ln x.
E.
x x (1 + ln x).
Question 3
[3] The derivative of y = x x is
e x + e−x
e x − e−x
1 + cosh 2x
Recall that cosh x =
and sinh x =
· The function f (x) =
can be represented by:
2
2
2
A.
cosh2 x.
B.
sinh2 x.
[3]
60
2012–2017 Past Test and Examination Questions Booklet
C.
1 − sinh2 x.
D.
sinh 2x.
E.
2 cosh 2x.
Question 4
[3]
Using integration by parts twice, the solution of the integral
A.
B.
C.
Z
1
1
sin(ln x) + cos(ln x) + c.
x
2
E.
None of the above.
Question 5
The following improper integral
[3]
Z
0
A.
diverges for p > −1.
B.
converges for p < −1.
C.
converges for p = −1.
D.
is equal to
1
if p > −1.
p+1
E.
is equal to
1
if p < −1.
p+1
1
x p dx
Question 6
B.
C.
D.
E.
sin(ln x) dx is:
1
sin(ln x) + cos(ln x) + c.
x
ä
1 Ä
x sin(ln x) − cos(ln x) + c.
2
1
1
sin(ln x) − cos(ln x) + c.
2
x
D.
A.
Mathematics I (Major)
[3]
1
1
+
·
2(1 − sin x) 2(1 + sin x)
1
1
+
·
1 − sin x 1 + sin x
1
1
−
·
1 − sin x 1 + sin x
1
1
−
·
2(1 − sin x) 2(1 + sin x)
−1
1
+
·
2(1 − sin x) 2(1 + sin x)
Question 7
[3]
1
The power series 1 − 2x + 4x − 8x + · · · + (−1) 2 x + · · · which converges for |x| < √ is the representation
2
of
2
4
6
n n 2n
Mathematics I (Major)
A.
Ä
ä2
1 + x2 .
B.
Ä
ä2
1 + x−2 .
C.
Ä
ä−1
1 + 2x2 .
D.
Ä
ä− 21
.
1 + 2x2
E.
Ä
ä− 12
2 + x2
.
2012–2017 Past Test and Examination Questions Booklet
Question 8
[3]
x
1
dy
= 2
e y is
The differential equation
2
dx x + y
A.
separable.
B.
linear.
C.
exact.
D.
homogeneous.
E.
none of the above.
61
Section B
[66]
Question 1
[12]
1) Show that the ordinary differential equation (ODE)
(4x + y4 )dx + (4xy3 − e3y )dy = 0
(E)
is exact.
(4)
2) Give the general solution of the ODE
(6)
(E).
Å
ã
1
3) Find the particular solution satisfying y − √
= 0.
3
Question 2
(2)
[13]
k
1) Let k be a real number not equal to −1, i. e. k ∈ R \ {−1}. Evaluate cos x sin x dx.
Z
cos5 x
√3
2) Evaluate
dx. If you wish to, you may use your answer in Question 2(1) .
sin x
Z π
2 cos5 x
√3
3) Calculate
dx.
sin x
0
R
Question 3
4
4x + 1
−
·
x + 1 x2 − x + 1
1) Find the following integrals: (Complete the square, if necessary)
Z
4
a)
dx.
x+1
Z
4x − 2
b)
dx.
x2 − x + 1
Z
3
c)
dx.
2
x −x+1
(3)
(8)
(2)
[14]
Let f be the rational function defined by f (x) = 3 +
(1)
(3)
(4)
62
2012–2017 Past Test and Examination Questions Booklet
2) Use your answers in the preceding question to find
Z
Mathematics I (Major)
3(x3 − 3x + 2)
dx.
x3 + 1
(6)
Question 4
[13]
1-a) When is a series
∞
X
an said to be absolutely convergent?
(2)
n=1
b) Is the series
∞
X
sin(2n + 1) π2
convergent? Justify your answer.
n
n=1
(3)
c) Is the series
∞
X
sin(2n + 1) π2
absolutely convergent? Justify your answer.
n
n=1
(3)
Question 5
[14]
1) For which values of x does the series
∞
X
(x − 5)n
n=1
n
converge?
(6)
2) Give with justification the power series representation of the function g(x) =
2.4
Calculus 2015
2.4.1
Calculus March 2015
Question 1
-
x2
about the point 0.
(1 + x)3
4 marks
Fill in the answers to the limits in the table:
Expressions
1: lim sinθ θ
Answers
θ→0
1
x
x→−∞ 5
lim e x
x→1
2: lim
3:
4: lim |y|
y→0
Question 2
-
3 marks
Figure 1: This is a(n)
discontinuity.
Figure 2: This is a(n)
discontinuity.
(8)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Figure 3: This is a(n)
Question 3
-
63
discontinuity.
4 marks
(a) Write down the definition of continuity of the function g(x) at x = a.
(2 marks)
(b) Write down the definition of the absolute value function (|x|) in mathematical notation (i.e. not in English
words).
(2 marks)
Question 4
-
9 marks
Show that (a) lim |x| = 0
(3 marks)
x→0
(b) Hence use the sandwich (squeeze) theorem to prove that lim(x2 − 1) = −1.
x→0
SHOW ALL WORKING.
Question 5
-
(6 marks)
6 marks
(a) The greatest integer function (also called the floor function) is defined as
bxc = greatest integer less than or equal to x.
Give a sketch of this function for −2 < x < 4 on the axes provided.
(3 marks)
(b) Evaluate the following limit algebraically (if possible). SHOW ALL WORKING.
(3 marks)
limbxc =
x→2
Question 6
-
6 marks
Find any horizontal asymptotes of the function f (x) =
Question 7
-
√
x2 +4x+4
3x2 −5x+1 .
SHOW ALL WORKING.
(6 marks)
8 marks
Write down the intervals over which the function f (x) is continuous if


x3
if x ≤ 0,


f (x) = 5x
if 0 < x < 2,


√
x + 3 if x ≥ 2.
Express your answer in interval notation and SHOW ALL WORKING.
Calculus Total 40 Marks
(8 marks)
64
2.4.2
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Calculus April 2015
Section A: Multiple choice questions
Instructions:
Circle the correct letter for your choice of answer in this section. There is only one correct answer to each question.
Question 1
[3]
The derivative of ln 32x equals
A.
2x
ln 3
B.
2x.3
C.
ln 2 ln 3
D.
ln 9
√
ln 3
E.
Question 2
Z
sec kx tan kx dx equals
A.
1
sec kx + C
k
B.
k sec kx + C
C.
1
sec kx tan kx + C
k
D.
sec kx + C
E.
None of the above
[3]
Question 3
Z 1
4(2x + 1)3 dx equals
[3]
0
A.
1
(2x + 1)4 + C
2
B.
1
(2x + 1)2 + C
2
C.
40
D.
20
E.
None of the above
Total Section A: [9] marks
Section B
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
65
FULL WORKINGS MUST BE SHOWN.
Question 1
[8]
If f and g are two differentiable functions in x, then
a) State the product rule i.e. the derivative of the product of f and g.
b) Prove the product rule i.e. the rule that gives the derivative of the product of f and g.
(2)
(6)
Question 2
[7]
Find the derivative of the following functions;
a)
sin2 (3x + 1).
(3)
b)
5x − 7
.
x3 − 3x
Do not simplify your answer.
Question 3
(4)
[6]
2 x
Let g(x) = x e and h(x) = sin x.
a) Find
g0 (x).
(3)
b) Find
d
(h ◦ g) .
dx
(3)
Question 4
Use linear approximation to estimate the value of
√3
[6]
8.01.
Question 5
Find
[5]
Z
cos(3x) sin5 (3x) dx.
Total Section B: [25] marks + [2] bonus marks
2.4.3
Calculus June 2015
Section A: Multiple choice questions
Instructions:
Section A, Questions 1–4 are multiple choice questions. In each of these questions, circle the letter that corresponds to the correct answer. There is only one correct answer to each question.
Question 1
Z π
x sin x dx =
0
A.
B.
0
π
2
[3]
66
2012–2017 Past Test and Examination Questions Booklet
C.
π
D.
π2
2
E.
π2
Question 2
Mathematics I (Major)
[2]
Let f be an invertible function which is differentiable. Which of the following is true?
A.
If f ( f −1 (a)) , 0, then ( f −1 )0 (a) =
1
.
f 0 ( f −1 (a))
B.
If f 0 ( f −1 (a)) , 0, then ( f −1 )0 (a) =
1
.
f 0 ( f −1 (a))
C.
If f ( f −1 (a)) , 0, then ( f −1 )0 (a) = − f 0 ( f −1 (a)).
D.
If f ( f −1 (a)) , 0, then ( f −1 )0 (a) = −
E.
If f 0 ( f −1 (a)) , 0, then ( f −1 )0 (a) = −
1
f 0 ( f −1 (a))
.
1
.
f 0 ( f −1 (a))
Question 3
[2]
What is the following theorem called?
Therorem. Let f : [a, b] → R be continuous and differentiable on (a, b). Then there is a c ∈ (a, b) such that
f (b) − f (a) = f 0 (c)(b − a).
A.
Fermat’s Theorem,
B.
Rolle’s Theorem,
C.
Taylor’s Theorem,
D.
Intermediate Value Theorem,
E.
Mean Value Theorem.
Question 4
[3]
Which of the following functions does not have a maximum?
A.
sin x on R.
B.
1
on R.
1 + x2
C.
x5 − 7x4 on [0, 5].
D.
1 − x2 on [−1, 1).
E.
7 − x on (1, 3].
Total Section A: [10] marks
Section B
Mathematics I (Major)
67
2012–2017 Past Test and Examination Questions Booklet
In this section you are expected to show all your working to earn the marks allocated.
Question 1
[5]
Using logarithmic differentiation, find the derivative of
f (x) =
(x2 − 4)9 √ x
e .
(x − 1)2
Do not simplify your answer.
Question 2
[10]
e x sin x − x
(a) Find lim
x→0
x2
(b) Find lim+ (sin x) x
(5)
(5)
x→0
Question 3
[6]
Let R be a rectangle where the lengths of the sides change with constant speed of 2 cm/sec whereas the perimeter
of R remains constant. Originally, R is a square whose area is 1 m2 . What is the rate of change of the area of R
when the shorter sides are 80 cm long?
Question 4
[15]
Let
f (x) =
Then
x2 − 3x − 4
.
x2 − 4
f 0 (x) = 3
and
x2 + 4
(x2 − 4)2
f 00 (x) = −6x
x2 + 12
.
(x2 − 4)3
(a) Find the x-intercepts and the y-intercepts of f .
(3)
(b) Find the horizontal and vertical asymptotes of f .
(3)
(c) Find the local minima and maxima of f and determine where f is increasing and where f is decreasing.
(3)
(d) Find the points of inflection of f and determine where f is concave up and where f is concave down.
(3)
(e) Use your answers in parts (a)–(d) to decide which sketch below is the graph of the function f .
(3)
A.
B.
y
x
y
x
68
2012–2017 Past Test and Examination Questions Booklet
C.
D.
y
y
x
E.
Mathematics I (Major)
x
None of the above.
2.4.4
Calculus September 2015
Section A: Multiple choice questions
Instructions:
Section A, questions 1 – 4 are multiple choice questions. Circle the letter corresponding to your choice of answer.
There is only one correct answer to each question.
Question 1
[2]
Which of the following equations/identities is NOT true:
A.
cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y).
B.
cosh(x) + sinh(x) = e x .
C.
tanh(x) =
D.
cosh(2x) = cosh2 (x) + sinh2 (x).
E.
cosh2 (x) = 1 + sinh2 (x).
e x + e−x
.
e x − e−x
Question 2
1
The average value of f (t) = − tan t on [0, 2] is
2
A.
1
(sec 2 tan 2 − 1).
2
B.
1
sec2 2 − 0 .
2
C.
1
(sec 2 tan 2 − 0).
2
D.
ä
p
1Ä
1 − ln | sec 2| − 0 .
2
E.
p
1
− ln | sec 2|.
2
[3]
Mathematics I (Major)
Question 3
The sum given by
n
X
2012–2017 Past Test and Examination Questions Booklet
4
g(xi∗ )
i=1
n
69
[3]
approximates the area between a curve and the positive x-axis on some interval
4
[a, b]. We are given that the minimum value of g in the ith subinterval is g(xi∗ ) = 2 − i . The function and the
n
interval specified in the sum are:
A.
g(x) = x on [2, 4].
B.
g(x) = 2 − x on [2, 4].
C.
g(x) = x − 2 on [0, 4].
D.
g(x) = 2 − x on [0, 4].
E.
g(x) = 2 − x on [−2, 2].
Question 4
[2]
The area formulae of simple geometric figures can be used to evaluate integrals of functions. If f (x) = |x| then
Z 1
integral
f (x) dx is given by one of the following areas
−2
A.
Two triangles both of base length of 1 and a height of 2.
B.
Two rectangles both of base length of 1 and a height of 2.
C.
A rectangle with an area of 2 and a triangle with an area of 1.
D.
A rectangle with an area of 1 and a triangle with an area of 2.
E.
1
Two triangles with areas 2 and .
2
Total Section A: [10] marks
Section B
Full workings must be shown.
Leave answers in exact form.
Question 1
[5]
You are asked to fence off a rectangular field with area 2400m2 and partition the field into two equal parts. What
is the smallest length of fence needed?
Question 2 Å ã
x2
dy
If y = 5 cosh
, find
at the point (2, 0). Simplify your answer.
4
dx
[4]
Question 3
Find the area bounded by y = 2x2 and y = x4 − 2x2 .
[5]
Question 4
[6]
Find the volume of the solid generated by revolving the region in the first quadrant bounded above by the curve
y = x2 , below by the x-axis and on the right by the line x = 1, about the line x = −1.
Total Section B: [20] marks
70
2.4.5
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Calculus November 2015
Section A: MULTIPLE CHOICE
Question 1
-
14 marks
Circle the correct answer in the following questions. Only one answer may be circled in each question.
(a) What is the most suitable substitution to use to solve the integral
A) t = 9 sec θ.
R
√ 1
dt?
t2 −81
B) t = 9 sin θ.
C) t = 81 sec θ.
D) t = 81 sin θ.
E) None of the above.
(2 marks)
Rx
(b) The derivative with respect to x of
π
7
2
et cosec(t)dt is
π 2
2
A) e x cosec(x) − e( 7 ) cosec( π7 ).
2
B) e x cosec(x).
2
C) 2xe x ln |cosec(x) − cot(x)|.
2
D) 2xe x ln |cosec(x) − cot(x)| −
E)
3
ex
x2
ln |cosec(x) − cot(x)| −
2π ( π7 )2
7 e
π 3
e( 7 )
( π7 )2
ln |cosec( π7 ) − cot( π7 )|.
ln |cosec( π7 ) − cot( π7 )|.
(2 marks)
(c) Odd One Out: Which of the following expressions does not represent the same sequence as the others?
©∞
¶
.
A) 2(2i−1)
i
2
B)
C)
D)
i=1
3 5 7
{1, 2 , 4 , 6 , . . . }.
¶
©∞
2n+1
.
2n
n=0
¶ m+ 1 ©∞
2
2m−1
m=0
.
E) None of the above.
(3 marks)
(d) Which test for convergence would you use on the series
∞
P
n=1
(− cos(nπ))n n!
?
n3n
A) The Alternating Series Test.
B) The Ratio Test.
C) The Integral test.
D) The Comparison Test.
E) The Root Test.
(2 marks)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
71
(e) If the power series of f at a is given by
f (x) =
∞
X
n=0
cn (x − a)n
for |x − a| < R, then
f (a)
n! .
f n (a)
a! .
f (n) (a)
n! (z
f (a) (n)
a! .
f (n) (a)
n! .
A) cn =
B) cn =
C) cn =
D) cn =
E) cn =
− a)n .
(2 marks)
(f) The partial fraction decomposition of
A)
B)
C)
D)
11
x−3 −
11
x−3 +
1 11x
10 x−3
1 11
10 x−3
x+3
x2 +1 .
x−3
x2 +1 .
1
1
− x−1
+ x+1
.
1
1
+ 10 (x+1)2 .
x2 +2
x3 −3x2 +x−3
is
E) None of the above.
(3 marks)
Section B: SHOW ALL WORKING!!!
Question 2
Find
d
dx
Ä R x2
x 2
Question 3
Evaluate
R
6 marks
1
9+t2 dt
-
R2
1
1 t+t2 dt
Question 4
Find
-
-
ä
.
(6 marks)
8 marks
by completing the square and using trigonometric substitution. Show all working.
5 marks
sin2 θ cos2 θdθ. Show all your working.
Question 5
-
(8 marks)
(5 marks)
18 marks
(a) Tick any integrals in the table that are improper:
Integral
R −∞ 2
1: 0 e x dx
R0
2
2: −∞ ew dw
R 1000000000 et2
3: R0
t dt
e
4: 0 sec α dα
(4 marks)
Improper?
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2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
R 3+e
(b) Determine whether the integral R 3 ln(x − 3)dx is convergent or divergent. If convergent, give the value the
integral converges to. HINT 1: ln xdx = x ln x − x HINT 2: l’Hôpital’s rule may be useful to you.
(8 marks)
(c) The following questions make use of the function f (x) =
i) Find partial fractions of f (x).
4
x3 −2x2 −3x .
(3 marks)
ii) If you integrated your answer in Part i) from x = 0 to x = 10, indicate which of your resulting integrals
would be improper. Do NOT evaluate the integrals!
(3 marks)
Question 6
-
6 marks
(a) Find the limit of the sequence
¶
−1
(−1)n
(b) Determine whether the sequence
Question 7
-
¶
©
(if it exists).
(2 marks)
©
(4 marks)
−3n
n2 −1
is increasing/decreasing or neither.
9 marks
(a) What is an n-th partial sum?
(b) Determine whether the series
(c) Find the sum of the series
Question 8
-
P
n≥2
(2 marks)
P
n≥1
√1
n3
converges or diverges. Give a reason for your answer.
2
n2 −1 .
(4 marks)
9 marks
(a) Prove that if the series
∞
P
n=1
an converges, then lim an = 0.
(6 marks)
n→∞
(b) What is the contrapositive of the theorem above?
Question 9
-
(3 marks)
4 marks
Write down the 6th degree Maclaurin polynomial of f (x) = 2 sin(3x).
Question 10
(3 marks)
-
(4 marks)
9 marks
(a) Show that cos x is the integrating factor of the linear differential equation
dy
cos x dx
= 1 + y sin x (where − π2 < x < π2 ).
(4 marks)
(b) Solve the differential equation from Part (a) explicitly.
(3 marks)
(c) Find the particular solution of the above differential equation if y =
2
9
when x = 0.
(2 marks)
Mathematics I (Major)
Question 11
-
2012–2017 Past Test and Examination Questions Booklet
73
1 (BONUS MARK) marks
In your course books, what does CFSSG stand for?
Calculus Total 90 (+ 1 BONUS) Marks
2.5
Calculus 2016
2.5.1
Calculus March 2016
Question 1
-
9 marks
(a) Complete the sentence in words or mathematical symbols:
For the limit lim f (x) to exist...
(3 marks)
x→a
(b) Fill in the answers to the limits in the table:
Expressions
1
1: lim(1 + x) x
Answers
x→0
2: lim + cosec θ
θ→−π
3: lim 29−x
x→∞
4: lim− byc
y→2
(4 marks)
(c) Write down a limit which shows that a function h(x) has a vertical asymptote at x = 1.5.
Question 2
-
(2 marks)
8 marks
Evaluate (and SHOW ALL WORKING)
(a) lim √
x→−2
x2 − 4
.
x+3−1
(4 marks)
74
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
x2 − 4
.
x→∞ 3x2 (x + 1) + 10(x + 1)
(b) lim
Question 3
-
(a) Prove that lim
θ→0
(4 marks)
10 marks
1 − cos θ
sin θ
= 0. You may assume that lim
= 1.
θ→0 θ
θ
(5 marks)
(b) Use the Sandwich (Squeeze) Theorem to determine lim sinx2 x . Clearly define your functions f (x), g(x) and
x→−∞
h(x) and explain your working.
(5 marks)
Question 4
-
13 marks
(a) A function f is continuous at the point a if and only if ...
(b) Indicate with ‘Y’ or ‘N’ whether the functions in the following table are continuous at x = −1:
(2 marks)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Function
1
1: f (x) = x−1
2: g(x) = |x + 1|
3: h(x) = sec(x + 1)
4: j(x) = b2xc
75
Y/N
(4 marks)
(c) Find the value(s) of a which make(s) the function g(x) continuous if
(
|x − 2| + 1
if x ≤ 1
g(x) =
2
−x + ax − 3 if x > 1.
Show all your working.
(d) Draw a function with a removable discontinuity at x = 2 on the Cartesian plane below.
(5 marks)
(2 marks)
(Calculus Total 40)
2.5.2
Calculus April 2016
Question 5
-
13 marks
d
sin 4x2 + 3x .
dx
Å 2
ã
(t + 1)5
(b) Let y = ln √
.
1−t
dy
Find .
dt
(a) Evaluate
( 2 marks)
76
2012–2017 Past Test and Examination Questions Booklet
Hint: You may use logarithm rules.
Question 6
-
Mathematics I (Major)
( 4 marks)
13 marks
Let f be a function. Complete the following in words or mathematical symbols:
(a)
(i) Left-hand derivative of f at x = a,
f−0 (a) =
( 1 mark)
(ii) Right-hand derivative of f at x = a,
f+0 (a) =
( 1 mark)
(iii) f is differentiable at a provided that · · ·
(b) Consider
f (x) =
( 2 marks)

3x2
2x + 1
if x ≤ 1,
if x > 1.
Verify whether f 0 (x) exists at x = 1.
1
(c) Find an equation of the straight line having slope that is tangent to the curve
4
√
y = x.
( 6 marks)
( 4 marks)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
2
(d) Show that the graph of f (x) = x 3 has a vertical tangent at x = 0.
Question 7
-
77
( 4 marks)
13 marks
(a) Consider the equation 2xy + π sin y = 2e.
(i) Find
dy
.
dx
(ii) Find the value of
π
dy
.
at 1,
dx
2
( 4 marks)
√3
√
(b) Find the linearization of the function f (x) = 3 x at x near 8 and use this linearization to approximate 8.12
.
( 6 marks)
Question 8
-
(a) Find
Z
13 marks
4 + x2
√ dx.
x
Simplify your answer.
(b) Evaluate
2.5.3
Z
0
π
4
( 3 marks)
tan2 x dx.
( 3 marks)
Calculus June 2016
Section A: Multiple choice questions
Question 1
Z π
(a)
esin x cos x dx =
0
A.
−e
B.
−1
C.
0
D.
1
E.
e
(b) The function f (x) = x3 + x5 − 2 is invertible.
Which of the following is true?
A.
f −1 is not differentiable at 0
B.
( f −1 )0 (0) = 0
[11]
(3)
(3)
78
2012–2017 Past Test and Examination Questions Booklet
C.
( f −1 )0 (0) = 8
D.
( f −1 )0 (0) = −8
E.
( f −1 )0 (0) =
(c) Evaluate
Z
Mathematics I (Major)
1
8
2
x3 e x dx
(5)
Section B
Question 2
[10]
Find the limits
e x − sin x − 1
(a) lim
x→0
x2 cos x
(b) limπ (sin x)sec x
(5)
(5)
x→ 2
Question 3
[10]
Thandi rides on her bicycle at a constant speed of 4 meters per second (14.4 km/h) towards a water tower, which
has a diameter of 20 meters. How fast does her horizontal angle of sight of the water tower increase when Thandi
is 30 meters away from the water tower?
(a) In the sketch of the aerial view below, label the quantities which you will use for your solution.
(2)
(b) If 2θ is Thandi’s horizontal angle of sight of the water tower, find a relation between θ and Thandi’s distance
from the water tower.
(2)
(c) Solve the problem.
(6)
Question 4
[8]
(a) What is the following theorem called?
(2)
Theorem. If f has a local maximum or minimum at an interior point c of dom( f ) and if f is differentiable at c,
then f 0 (c) = 0.
A.
Fermat’s Theorem,
B.
Intermediate Value Theorem,
C.
Mean Value Theorem,
D.
Rolle’s Theorem,
E.
Taylor’s Theorem.
(b) Prove the theorem in case of a local maximum.
Question 5
(6)
[6]
2
Find the global maximum and the global minimum of 2x − 3x 3 on [−1, 8].
Mathematics I (Major)
2.5.4
79
2012–2017 Past Test and Examination Questions Booklet
Calculus August 2016
Full workings must be shown.
Leave answers in exact form.
Question 1
[5]
1
What is the lower sum approximation of the area bounded by g(x) = between x = 1 and x = 5 using 4 partitions?
x
Question 2
What are the dimensions of the rectangle with the largest area that be inscribed in the ellipse below?
[6]
y axis
3
(x,y)
5
−5
x axis
−3
Question 3
Å
ã
1
1+x
Prove the following formula tanh x = ln
.
2
1−x
Question 4
−1
[5]
[5]
Prove that if f is continuous on a closed interval [a, b], then there exist a c in the closed interval such that
Zb
a
f (x) dx = f (c)(b − a).
Question 5
Using areas of known geometries and other properties of integrals evaluate the integral
Z2
−2
Question 6
[3]
(2 − |x|) dx.
[10]
80
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
a. What is the total area of the region bounded by x = 4y2 − 2y and x = 12y2 ?
√
b. What is the volume of a solid generated by rotating h(x) = x + 4 about the line y = 2 on [1, 4]?
(5)
(5)
Total :[30] marks
Bonus : [4] marks
2.5.5
Calculus November 2016
Section A
[10]
[MCQ1] Let y =
Z
0
√
x
sin (t2 ) dt. Then
(A)
1
√ sin (x2 ).
2 x
(B)
1
√ sin (x).
2 x
dy
=
dx
1
(C) − √ sin (x2 ).
2 x
1
(D) − √ sin (x).
2 x
(E) 0.
( 2 marks)
[MCQ2] Given that the integral
Z
∞
1



1
dx =

xp

1
p−1
if p > 1,
if p ≤ 1.
Z
Which of the following statements is true for the integral
(A)
The integral converges if p > −1.
(B)
The integral diverges if p ≤ −1.
(C)
The integral is equal to
−1
p+1
(D)
The integral is equal to
1
−p+1
(E)
The integral is equal to
1
p−1
diverges
∞
1
1
dx.
x−p
if p < −1.
if p < −1.
if p ≥ −1.
( 2 marks)
Mathematics I (Major)
[MCQ3] If cos(x) =
cos(−3x2 ) =
∞
X
(A)
∞
X
(B)
(−1)n
∞
X
(−1)n+1
∞
X
(D)
(−3x2 )2n+1
.
(2n + 1)!
32n x2n
.
(2n)!
(−1)n
(−3)2n x4n
.
2(n!)
n=0
(E)
x2n
then
(2n)!
(−1)n
n=0
81
(3x2 )2n
.
(2n)!
n=0
(C)
(−1)n
n=0
n=0
∞
X
2012–2017 Past Test and Examination Questions Booklet
None of the above.
( 3 marks)
[MCQ4] Consider the differential equation y0 − xy = 1 − y0 x2 .
It may be written in the form y0 + yP(x) = Q(x).
Which of the following statements is true?
(A)
The differential equation is variable separable.
−x
.
Q(x) = 2
x +1
(B)
1
.
x2 + 1
(C)
P(x) =
(D)
The differential equation is linear.
(E)
Its integrating factor is IF = e
R
1
dx
x2 +1
.
( 3 marks)
Section B
Question 1
-
13 marks
Evaluate the following integrals.
(a)
Z
(b)
Z
3x2 − x + 1
dx.
x(x2 + 1)
√
x
3 − 2x − x2
dx.
( 8 marks)
( 10 marks)
[18]
82
2012–2017 Past Test and Examination Questions Booklet
Question 2
Let f (x) =
-
Mathematics I (Major)
13 marks
2x2
4x
4 + 12x2
, with f 0 (x) =
and f 00 (x) =
.
2
2
2
1−x
(1 − x )
(1 − x2 )3
(a) Find the domain of f .
(b) Find the x and y intercepts (if any).
(2 marks)
(1 mark)
(c) Describe the behaviour of f as x tends to ∞ and −∞.
(2 marks)
(d) Describe the behaviour of f as x tends to 1+ , 1− , −1+ and −1− .
(2 marks)
(e) Find the equation(s) of the asymptote(s) (if any).
(2 marks)
(f) Find the coordinates of the local extrema.
(2 marks)
(g) Find intervals of increase and decrease of f .
(4 marks)
(h) Determine local maxima and local minima.
(1 mark)
(i) Determine concavity of the curve.
(3 marks)
2x2
(j) Sketch the curve of f (x) =
1 − x2
indicating point(s) of intersection (if any), asymptotes and local extrema.
(4 marks)
[23]
Question 3
-
13 marks
Determine whether the following improper integrals converge or diverge. Show all your working.
Z ∞
4
(a)
dx.
x
0
Z ∞
4
(b)
dx.
2
x +2
0
(3)
(5)
[8]
Question 4
-
13 marks
(a) Determine whether the series
∞ Ä
X
4n − 3 än
n=1
3n − 4
(b) For which value of x does the series
∞
X
n=1
converge or diverge.
(5)
(x − 3)n
converge?
n!
(7)
(−1)n
[12]
Mathematics I (Major)
Question 5
-
2012–2017 Past Test and Examination Questions Booklet
83
13 marks
Consider the differential equation (x2 − y2 )dx + x(x + 2y)dy = 0.
(a) Show that the differential equation is homogeneous.
(2)
(b) Find the general solution of the differential equation.
(15)
(c) Find the particular solution for x = 1 and y = −1.
(2)
[19]
2.6
Calculus 2017
2.6.1
Calculus March 2017
2.6.2
Calculus April 2017
Section A
[8 marks]
Instructions:
For each multiple choice question, select the letter that corresponds to the correct answer.
[A1] The normal to the curve y = 3 tan x + x2 + 1 at the point with coordinates (0, 1) is
(A) y = −3x + 1.
(B) y = 3x − 1.
1
(C) y = x − 1.
3
1
(D) y = − x + 1.
3
1
(E) y = x + 1.
3
[A2] If y = log10 (sin x) ; 0 < x < π, then
2 marks
dy
=
dx
1
.
sin x
(A)
ln 10
(B)
1
.
ln 10 sin x
(C)
1
cot x.
ln 10
(D)
1
.
ln 10 cot x
(E)
1
.
ln 10 cos x
[A3] Let f and g be differentiable functions. Which of the following is equal to
d
( f (x + g(x)))?
dx
2 marks
84
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
(A) f 0 (x) + f 0 (g(x))g0 (x).
(B) f 0 (x + g(x))g0 (x).
(C) f 0 (x + g(x))(x + g0 (x)).
(D) f 0 (x) + g0 (x)(1 + g0 (x)).
(E) f 0 (x + g(x))(1 + g0 (x)).
[A4] Which of the following is equal to
−
(B)
ln |7x| +
(D)
(E)
Z Å
1
+ eπ
7x
ã
dx?
2
+ C.
7x2
(A)
(C)
2 marks
1
eπ+1 x + C.
π+1
ln |x|
+ π eπ x + C.
7
ln |x|
+ eπ x + C.
7
ln
x
+ eπ x + C.
7
2 marks
[32 marks]
Section B
Instructions:
In this section you are expected to show all your working to earn the marks allocated.
Simplify your answers.
Question 1
-
13 marks
(a) Let y = xr for r ∈ R.
dy
= rxr−1 .
dx
Prove that
d
(b) Find
dx
Ç
Question 2
-
5
√
x
4 marks
√ å
+x−4 x
.
x
3 marks
13 marks
2
(a) Consider the equation x4 + cos y + e x y = x2 + y2 .
Find
dy
. Show all your working.
dx
(b) Find the differential dy if y = πt tπ .
6 marks
2 marks
Mathematics I (Major)
Question 3
-
85
2012–2017 Past Test and Examination Questions Booklet
13 marks
(a) Find the equation of the tangent line to the curve of
x
at x = 1.
f (x) =
x+1
4 marks
(b) For what value(s) of a does the tangent line to the curve pass through
the point (−5, a ) ?
3 marks
(c) Use linearization to approximate the value of f (1.2).
Question 4
Evaluate
Z
2
−1
2.6.3
-
2 marks
13 marks
x−1
√
dx.
x+2
8 marks
Calculus June 2017
Question 1
-
4 marks
TRUE or FALSE:
Statement
True/False
If s0 (t) and r0 (t) are continuous then
R
s0 (t)r(t)dt = r(t)s(t) −
R
s(t)r0 (t)dt.
If f is continuous on an interval I, then f −1 exists and is also continuous on I.
The function g has an absolute minimum at c if g(c) ≥ g(x) ∀x in the domain of g.
r0 (s) = 0 ⇒ the point (s, r(s)) is a local extremum of the function r(s).
Question 2
-
4 marks
(a) Write in the block how many times you need to use integration by parts in order to integrate
DO NOT do the integration!
(b) Evaluate, showing all working:
0
-
1
3
…
1
dt.
e − t2
7 marks
(a) Show all working to find
d x
dx x .
(b) Show all working to find the slant asymptote of h(t) =
Question 4
-
eu cos u du?
(1 mark)
(3 marks)
Z
Question 3
R
4 marks
2t3 −t2 −t+1
.
t2 −1
(3 marks)
4 marks
(a) Find the 3rd degree Taylor polynomial of sin x at x = 0. SHOW ALL YOUR WORKING.
3 marks
(b) Differentiate your answer above to write down the 3rd degree Taylor polynomial of cos x at x = 0. (1 mark)
86
2012–2017 Past Test and Examination Questions Booklet
Question 5
-
Mathematics I (Major)
9 marks
(a) Prove the following Theorem:
5 marks
If f 0 (x) = g0 (x) for all x in an interval (a, b), then f − g is a constant on (a, b), that is, f (x) = g(x) + C,
where C is a constant.
(b) State l’Hôpital’s Rule in full. Hint: Start with “Let f and g be differentiable functions...”.
Question 6
-
6 marks
Sketch a graph on the axes provided which satisfies the following criteria:
(a) f (x) is continuous for all positive real numbers.
(b) f (x) < 0 for x ∈ (0, 2) and f (x) > 0 for x ∈ (2, ∞)
(c) f 0 (x) > 0 for x < 5 and f 0 (x) < 0 for x > 5
(d) f 0 (x) = 0 when x = 5
(e) f 00 (x) < 0 for 0 < x < 7 and f 00 (x) > 0 for x > 7
(f) lim+ = −∞
x→0
4 marks
Mathematics I (Major)
Question 7
-
87
2012–2017 Past Test and Examination Questions Booklet
6 marks
You have 40cm of wire to form a square and a circle. How much of the wire should be used for the square and
how much for the circle to enclose the maximum total area?
Calculus Total 40
2.6.4
Calculus August 2017
Question 1
-
6 marks
TRUE or FALSE:
Statement
True/False
The derivative of cosh x is − sinh x.
For a decreasing function, the right-most endpoint will give the height for the
lower Riemann sum.
d
dx
R x2
Question 2
−5
e2t dt = 2xe2x
-
2
7 marks
(a) State and prove the Fundamental Theorem of Calculus 2.
Question 3
-
(7 marks)
7 marks
(a) Find the area of the region bounded by the graphs y = x2 and y = c in terms of the positive constant c.
marks)
(4
(b) Write down an expression for the area enclosed by x = cos y, x = 12 , y = 0 and y = 2π. DO NOT evaluate
your integral.
(3 marks)
Question 4
-
7 marks
(a) Write down an expression for the volume of the solid generated by rotating the region bounded by y = 1x + 2
and y = −x + 10 about the line y = 0. DO NOT evaluate your integral.
(5 marks)
(b) Write down an expression for the volume of the solid generated by rotating the region bounded by y = 1x + 2
and y = −x + 10 about the line y = 10. DO NOT evaluate your integral.
(2 marks)
Question 5
-
3 marks
Find the cross-sectional area of a pyramid with a height of h and a rectangular base with dimensions w and b,
sliced perpendicular to the y-axis.
Calculus Total 30 Marks
88
2.6.5
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Calculus November 2017
Multiple choice questions
Section A:
[18 marks]
Instructions:
Questions MCQ1–MCQ6 are multiple choice questions.
In each of these questions, select the letter that corresponds to the correct answer.
(MCQ1) Which of the following is the appropriate partial fraction decomposition of
(A)
(B)
(C)
(D)
(E)
t4 + 9
:
+ 9)
t2 (t2
A Bt + C
t4 + 9
≡1+ 2 + 2
+ 9)
t
t +9
t2 (t2
At + B
C
D
t4 + 9
≡
+
+
+ 9)
t2
t+3 t−3
t2 (t2
t4 + 9
A B Ct + D
≡1+ + 2 + 2
+ 9)
t
t
t +9
t2 (t2
t4 + 9
A B
C
D
≡ + 2+
+
+ 9)
t
t
t+3 t−3
t2 (t2
t4 + 9
A B Ct + D
≡1+ + 2 + 2
.
+ 9)
t
t
t +9
t2 (t2
[3]
(MCQ2) Which of the following statements is NOT TRUE?
Z 3
1
3
(A)
dx = .
2
(x
−
4)
4
0
Z 2
Z 3
Z 3
1
1
1
dx
=
dx
+
dx.
(B)
2
2
(x
−
4)
(x
−
4)
(x
−
4)2
0
2
0
Z 3
1
(C)
dx has an interior infinite discontinuity.
(x
−
4)2
0
Z 3
1
(D)
dx converges.
(x
−
4)2
0
(E) The integrand
1
is defined on (0 , 3) .
(x − 4)2
[3]
Mathematics I (Major)
(MCQ3) Consider the series
2012–2017 Past Test and Examination Questions Booklet
89
∞
X
cos (nπ)
.
n
Which of the following statements is TRUE?
n=1
(A) The series is not conditionally convergent.
(B) The series does not converge.
(C) The series is absolutely convergent.
(D) The sequence
(E)
∞
X
cos (nπ)
n=1
n
ß
=
cos (nπ)
n
™
diverges.
∞
X
1
(−1)n .
n
n=1
[3]
(MCQ4) Which of the following statements is TRUE about the series
∞
X
ln x + (ln x)2 + (ln x)3 + · · · =
(ln x)n .
n=1
(A) The series converges for
1
< x < e.
e
(B) The series converges for
1
≤ x ≤ e.
e
(C) The series does not converge for any value of x.
(D)
an+1
1
= 1 + → 1 as n → ∞.
an
n
(E) The series is convergent.
[3]
(MCQ5) Let g(x, y) = x2 y + cos y + y sin x.
Which of the following statements is TRUE?
(A) g x = 2xy + cos y + ycos x.
(B) gy = 2x − sin y + sin x.
(C) g x x = 2y − sin y + sin x.
(D) g x y = gy x .
(E) gy y = 2x − cos y.
[3]
90
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
(MCQ6) Consider the differential equation (2xy − yn ) dx + x2 dy = 0.
Which of the following statements is NOT TRUE?
(A) For n = 1 the differential equation is variable separable.
(B) For n = 2 the differential equation is homogeneous of degree 2.
(C) For n = 0 the differential equation is exact.
(D) For n = 0 the differential equation is variable separable.
(E) For n = 1 the differential equation is first order linear.
[3]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Section B
91
[72 marks]
In this section you are expected to SHOW ALL YOUR WORKING on the provided space to earn the
allocated marks.
Question 1
-
3 marks
(a) Find the partial fraction decomposition of
(b) Evaluate
Question 2
Z
1
.
z (1 + z)
[3]
ln z
dz.
(1 + z)2
-
[7]
3 marks
Evaluate the following integrals:
(a)
t
√
dt.
4
t −1
Z
Hint : You may use the fact that
(b)
Z
1
2
Z
sec θ dθ = ln |sec θ + tan θ| + C.
2s − 2
√
ds.
−s2 + 2s + 3
Question 3
(a) Evaluate
Z
[5]
3 marks
1
dx.
(1 + x2 )arctan x
(b) Hence or otherwise determine whether the integral
[4]
Z
is convergent or divergent.
Question 4
-
[9]
0
1
1
dx
(1 + x2 )arctan x
[4]
3 marks
Consider the telescoping series
ã
∞ Å
X
1
2
1
−
+
.
r+2 r+1 r
r=1
ã
n Å
X
1
2
1
(a) Find the partial sum S n =
−
+
.
r+2 r+1 r
r=1
ã
∞ Å
X
1
2
1
(b) Hence or otherwise show that the series
−
+
r+2 r+1 r
r=1
converges and find its sum.
[5]
[3]
92
2012–2017 Past Test and Examination Questions Booklet
Question 5
-
Mathematics I (Major)
3 marks
(a) Determine, giving reasons, whether the sequence {an } =
is convergent or divergent.
ß
n3 − 1 + n2 sin n
1 + 3n3
™
If it is convergent, find the value to which it converges.
[4]
(b) Determine whether the following series converge or diverge
(i)
∞
X
n3 − 1 + n2 sin n
1 + 3n3
n=1
(ii)
∞
X
(−1)n
n=1
Question 6
Given that
-
Å
.
[2]
n3 − 1 + n2 sin n
1 + 3n3
ãn
.
[4]
3 marks
∞
X
1
= 1 + x + x2 + x3 + · · · =
xn .
1−x
n=0
(a) Write down the power series of
1
.
1 + x2
[4]
(b) Hence or otherwise find a power series representation of f (x) = arctan x.
(c) Write
π
as a series.
4
Question 7
-
[4]
[2]
3 marks
2
2
x
x
dx y2 + y2 e y + 2x2 e y
.
Consider the differential equation
=
2
x
dy
2x y e y
(a) Using the substitution x = vy or otherwise, find the general solution of the
differential equation.
(b) Find the particular solution of the differential equation given that y = 1 when x = 0.
3
Calculus Solutions
3.1
Calculus Solutions 2012
3.1.1
Calculus Solutions March 2012
Section B
[10]
[2]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Question 1
(a)
lim
x→1
√
Ä√
äÄ√
ä
√
√
√
x2 + 4 − x + 4
x2 + 4 + x + 4
x2 + 4 − x + 4
ä
= lim
√
Ä√
x→1
2x2 − 2
2 x2 − 1
x2 + 4 + x + 4
x2 + 4 − (x + 4)
Ä√
ä
√
x→1 2 (x − 1) (x + 1)
x2 + 4 + x + 4
= lim
= lim
x→1
= lim
x→1
=
x (x − 1)
Ä√
ä
√
2 (x − 1) (x + 1)
x2 + 4 + x + 4
2 (x + 1)
Ä√
x
x2
+4+
√
1
ä=
Ä√
√ ä
x+4
2(1 + 1)
5+ 5
1
√
8 5
(b)
lim
x→−∞
√
9x2 − 4
=
3 − 2x
=
=
=
(c)
…
4
|x| 9 − 2
x
ã
Å
lim
x→−∞
3
−2
x
x
…
4
(−x) 9 − 2
x
ã
Å
lim
x→−∞
3
−2
x
x
…
4
− 9− 2
x
lim
3
x→−∞
−2
x
−3 3
=
−2 2
√
−3
x2 − 4
lim + f (x) = lim + √
= √ =− 3
2
x→−1
x→−1
3
4−x
x2 − 4
lim f (x) = lim −
x→−1−
x→−1 x − 2
(x − 2)(x + 2)
= lim −
= lim − (x + 2)
x→−1
x→−1
x−2
=1
Since lim + f (x) , lim − f (x), lim f (x) does not exist.
x→−1
x→−1
x→−1
Question 2
(b)
1
≤1
x2
1
⇒ − x2 ≤ x2 sin 2 ≤ x2
x
− 1 ≤ sin
But
lim(−x2 ) = lim x2 = 0.
x→0
x→0
93
94
2012–2017 Past Test and Examination Questions Booklet
Thus by the Sandwich Theorem,
lim x2 sin
x→0
Mathematics I (Major)
1
= 0.
x2
Alternatively:
1
≤1
x2
1
⇒0 ≤ x2 sin 2 ≤ x2
x
0 ≤ sin
But
lim 0 = lim x2 = 0.
x→0
x→0
Thus by the Sandwich Theorem,
lim x2 sin
x→0
1
1
= 0 ⇒ lim x2 sin 2 = 0.
2
x→0
x
x
Question 3
(b)
g(x) =
x3 − 4x2 + 4x x(x2 − 4x + 4)
=
x2 − x
x(x − 1)
Hence the rational function g is defined for all real numbers except 0 and 1, and these are the only points of
discontinuity.
For all x in the domain of g:
g(x) =
(x − 2)2
x−1
(x − 2)2
4
=
= −4
x→0 x − 1
−1
lim g(x) = lim
x→0
Thus g has a removable discontinuity at x = 0.
lim− g(x) = lim−
x→1
x→1
Thus g has an infinite discontinuity at x = 1.
3.1.2
(x − 2)2
= −∞
x−1
Calculus Solutions May 2012
Section B
Question 2
(a)
2
2
g0 (x) = ecos x (−2 cos x2 sin x2 )(2x)
î
ó
2 2
= −4x cos x2 sin x2 ecos x
(b)
f 0 (x) =
Question 3
(3x2 − 2 x ln 2)(1 + ln x) − (x3 − 2 x )
(1 + ln x)2
Å ã
1
x
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
(a) Implicit differentiation gives
y + xy0 + cos yy0 = e x
⇒ (x + cos y)y0 = e x − y
ex − y
⇒ y0 =
x + cos y
(b) By (a), substituting x = 0 and y = π, the slope of the tangent to the curve is
mt =
Then the slope of the normal is
e0 − π
1−π
=
=π−1
0 + cos π
−1
mn = −
so that the equation of the normal line is
1
1
1
=−
=
mt
π−1 1−π
1
x
1−π
y = π + mn (x − 0) = π +
Question 4
(a) Substitute u = cos(5x), and therefore
du = −5 sin(5x) dx
Then
Z
Z
1
u3 du
5
1
= − u4 + C
20
1
= − cos4 (5x) + C
20
sin(5x) cos3 (5x) dx = −
(b) Integration by parts with u = x and dv = e x dx.
Then du = dx and v = e x , and
Z
1
4
x
xe dx =
Z
4
1
u, dv
4
= uv −
Z
1
= xe x
4
1
−
4
1
Z
1
v, du
4
e x dx
4
= 4e − e − e x
4
4
1
4
= 4e − e − (e − e)
= 3e4
3.1.3
Calculus Solutions June 2012
Section B
In this section you are expected to show all your working to earn the marks allocated.
Question 1
95
96
2012–2017 Past Test and Examination Questions Booklet
We know that
f 0 (x) = 2 x ln 2
Hence
f (n) (x) = 2 x (ln 2)n
Question 2
(a) The domain of f is R.
(b)
1
f 0 (x) =
1−
= √
=
√
x2
1 + x2
√
2x
(1) 1 + x2 − x √
2 1 + x2 − 1
2
1+x
1 + x2
1 + x2 − x2
1
√
−
1 + x2 − x2 (1 + x2 ) 1 + x2 1 + x2
1 + x2
1
1
−
1 + x2 1 + x2
=0
(c) f is constant
(d) From (c) and
f (0) = 1 + 0 − arctan 0 = 1
we conclude that f (x) = 1 for all x in the domain of f .
Question 3
(a) Let u =
Then
1
x
. Then du = dx.
a
a
Z
dx
=
a2 + x2
a
du
a2 + a2 u2
Z
a
1
= 2
du
a
1 + u2
1
= arctan u + C
a
x
1
= arctan + C
a
a
Z
(b)
Z
3
√
3
dx
1
x 3
√
√
=
arctan
3 + x2
3
3 √3
Å
ã
1
3
= √ arctan √ − arctan 1
3
3
1 π π
= √
−
3 3 4
π
√
=
12 3
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
97
Question 4
Assume y = limπ − (tan x)cos x exists. Then
x→ 2
ln y = ln limπ − (tan x)cos x
x→ 2
= limπ − ln (tan x)cos x
x→ 2
= limπ − cos x ln(tan x) (”0 · ∞”)
x→ 2
Å ã
ln(tan x)
0
= limπ −
1
x→ 2
0
cos x
sec2 x
tan x
= limπ −
1
x→ 2
− 2 (− sin x)
cos x
1
= limπ −
x→ 2 sin x tan x
cos x
= limπ −
x→ 2 sin2 x
=0
Therefore limπ − (tan x)cos x exists and equals
x→ 2
y = eln y = e0 = 1
3.1.4
Calculus Solutions August 2012
Section B
In this section you must answer each question as fully as possible to earn full marks.
Question 1
The area is
A=
A=
1
2
1
(1 + ln x)3
dx
x
dx
Then x = 1 ⇒ u = 1 and x = e ⇒ u = 2 give
x
Using the substitution u = 1 + ln x we get du =
Z
e
Z
u3 du =
u4
4
2
1
=
16 1 15
− =
4
4
4
Alternatively, calculate
Z
(1 + ln x)3
dx =
x
Z
u3 du =
u4
1
+ C = (1 ln x)4 + C
4
4
Then
A=
1
(1 + ln x)4
4
e
1
=
(1 + ln e)4 (1 + ln 1)4 16 1 15
−
=
− =
4
4
4
4
4
Question 2 From the sketch it is seen that the region lies between y = 0 and y = 23 = 8, and using the disc
98
2012–2017 Past Test and Examination Questions Booklet
1
method, R = 2 − y 3 , so that the volume is
Z
V=π
8
0
Z
=π
0
8
1
(2 − y 3 )2 dy
1
2
(4 − 4y 3 + y 3 ) dy
4
3 5
= π[4y − 3y 3 + y 3 ]80
5
3
= π[32 − 3 · 16 + · 32]
5
96
= π[ − 16]
5
16π
=
5
Question 4
Z
0
4
x dx
1
=
x2 − 4x + 8 2
1
=
2
=
=
=
=
3.2
Calculus Solutions 2013
3.2.1
Calculus Solutions April 2013
Z
4
0
Z
0
4
2x − 4 + 4
dx
x2 − 4x + 8
2x − 4
dx + 2
x2 − 4x + 8
Z
0
4
dx
x2 − 4x + 8
Z
Z 4
1 4 2x − 4
dx
dx + 2
2
2
2 0 x − 4x + 8
0 x − 2) + 4
4
1
2
x−2 4
ln(x2 − 4x + 8) + arctan
0
2
2
2 0
1
(ln 8 − ln 8) + arctan 1 − arctan(−1)
2
π
2
Section B
Question 2
2)
(a)
f 0 (x) =
(e x + e−x )5 x ln 5 − 5 x (e x − e−x )
Å
ã
1 2
ex + x
e
(b)
g0 (x) = 3x2 e x
3
+ln 9
= 3x2 eln 9 e x
3
= 3x2 × 9e x
= 27x2 e x
Question 3
(a)
3
3
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
d xy
d
e + y2 cos x + e x =
[6]
dx
dx
ï
ò
d
dy
⇔ e xy . [xy] + 2y (cos x) − y2 sin x + e x = 0
dx
dx
ï
ò ï
ò
dy
dy
xy
2
x
⇔e . y+x
+ 2y (cos x) − y sin x + e = 0
dx
dx
dy
dy
⇔ ye xy + xe xy
+ 2y (cos x) − y2 sin x + e x = 0
dx
dx
dy xy
⇔
xe + 2y cos x = y2 sin x − e x − ye xy
dx
Hence,
dy y2 sin x − e x − ye xy
=
.
dx
xe xy + 2y cos x
(b)
22 sin 0 − e0 − 2e0×2 0 − 1 − 2
3
dy
|(0,2) =
=
=− .
dx
0e0×2 + 2 × 2 cos 0
0+4
4
(c) Let mnormal be the gradient of the normal to the curve at (0, 2). Then
mnormal .
dy
3
|(0,2) = −1 ⇔ − .mnormal = −1
dx
4
⇔ mnormal =
4
.
3
Hence the equation of the normal line to the curve at (0, 2) is
y−2=
4
4
(x − 0) ⇔ y = x + 2.
3
3
Question 4
1 1
1
√ .
Thus f 0 (1) = .
3 3 x2
3
Hence the linear approximation of the function f (x) at x = 1 is:
(b) f (1) = 1.
f 0 (x) =
L(x) = f (1) + f 0 (1)(x − 1)
1
= 1 + (x − 1).
3
(c) It follows from (b) that
1
L(1.0006) = 1 + (1.0006 − 1)
3
0.0006
=1+
= 1 + 0.0002
3
= 1.0002.
Hence
√3
3.2.2
Calculus Solutions June 2013
1.0006 ≈ 1.0002.
Section B
99
(3.1)
100
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Question 2
(b) The function f (x) is differentiable on the open intervals (0, 1) and (1, 2) and is not differentiable at x = 1.
Hence for x ∈ (0, 1) ∪ (1, 2) we have:
1
2
2
1
x(x − 1) 3 + x2 (x − 1)− 3
3 ñ
9
ô
√3
1
x
= x 6 x−1+ p
3
9
(x − 1)2
1
x
(6(x − 1) + x)
= p
9 3 (x − 1)2
1
x
= p
(7x − 6)
9 3 (x − 1)2
f 0 (x) =
∴ f 0 (x) = 0 ⇔ x = 0 or x = 67 .
6
12
2
Now, f (0) = 0 and f ( ) = − √3 . Also f (1) = 0 and f (2) = .
7
3
49 7
Hence the absolute minimum and absolute maximum values of f (x) on the interval [0, 2] are
12
6
2
f ( ) = − √3 and f (2) = respectively.
7
3
49 7
3.2.3
Calculus Solutions June 2013
Multiple Choice: D;C;E;D;E
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
101
102
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
103
104
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
3.2.4
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions August 2013
Multiple Choice: B;C;D;B
105
106
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
3.3
Calculus Solutions 2014
3.3.1
Calculus Solutions March 2014
107
108
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
109
110
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
111
112
3.3.2
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions April 2014
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
113
114
2012–2017 Past Test and Examination Questions Booklet
3.3.3
Mathematics I (Major)
Calculus Solutions June 2014
Multiple choice: Q5) D, Q6) B, Q7) C, Q8) E
MATH1036/7
CALCULUS
Page 3 of 5
June 2014
Section B
In this section you are expected to show all your working to earn the marks allocated.
Question 1
[7]
i
dh
1
.
(a) Prove that
arcsin x = √
dx
1 − x2
Solution
✓✏
✓✏
Let y = arcsin x,
✒✑
✓✏
0.5
−1 < x < 1.
(5)
✒✑
0.5 Then sin y = sin(arcsin x) = x
✓✏
✒✑
0.5
d(sin y)
= 1. 0.5
✒✑
dx
q
✓✏
✓✏
dy
1
dy
=1⇔
=
. 0.5 However cos y = ± 1 − sin2 y. 0.5
Thus cos y
✒✑
dx
dx cos y ✒✑
q
✓✏
✓✏
√
π
π
2
2
Since −1 < x < 1, then − < y < . 0.5 Thus cos y = 1 − sin y = 1 − x . 0.5
✒✑
2
2 ✒✑
h
i
d
1
dy
1
and it follows that
· 1❥
= √
Hence
arcsin x = √
2
2
dx
dx
1−x
1−x
and
(b) Find
Solution
i
dh
arcsin ln x , where e−1 < x < e.
dx
d(ln x)
✓✏
i ✓✏
dh
1
arcsin ln x
0.5 = p dx
0.5 = p
✒✑
✒✑
dx
1 − (ln x)2
x 1 − (ln x)2
(2)
1❥ .
Question 2
In this question you do not need to simplify your answers.
Use the method of logarithmic differentiation to find y′ , where:
√ 6
(a) y = 5 xe x (x4 + 10)100 .
[10]
(5)
Solution
ln y = ln
√5
x6
4
100
xe (x + 10)
✓✏ 1
6
0.5 = ln x + ln e x + 100 ln(x4 + 10) 1❥
✒✑ 5
✓✏
1
6
4
= ln x + x + 100 ln(x + 10). 0.5
✒✑
5
It follows that
"
# ✓✏ ′ ✓✏
y
1
100 × 4x3
d 1
d(ln y)
6
4
0.5 ⇔
0.5 =
=
ln x + x + 100 ln(x + 10)
+ 6x5 + 4
.
✒✑ y ✒✑ 5x
dx
dx 5
x + 10
1❥
Mathematics I (Major)
115
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CALCULUS
Page 4 of 5
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Hence
!
!
100 × 4x3
100 × 4x3 √5 x6 4
1
1
5
5
y =
xe (x + 10)100 .
×y=
+ 6x + 4
+ 6x + 4
5x
x + 10
5x
x + 10
′
1❥
sin x cos x
.
e−x x x
Solution
(5)
(b) y =
sin x cos x
ln y = ln
e−x x x
! ✓✏
✓✏
−x
x
0.5 = ln sin x + ln cos x − ln e − ln x 0.5
✒✑
= ln sin x + ln cos x + x − x ln x.
It follows that
✒✑
✓✏
✒✑
0.5
✓✏
y′ ✓✏ d 0.5 =
ln sin x + ln cos x + x − x ln x 0.5
✒✑
y ✒✑ dx
✓✏
✓✏
✓✏
′
sin x
cos x
y
0.5 −
0.5 + 1 − ln x − 1 · 0.5
⇔ =
✒✑
y
sin x ✒✑ cos x ✒✑
Hence
!
!
cos x sin x
cos x sin x
sin x cos x
y =
−
− ln x y =
−
− ln x
sin x cos x
sin x cos x
e−x x x
′
Question 3
1❥
[11]
(a) State the Mean Value Theorem.
(4)
The Mean Value Theorem’s statement
Let f be a function that satisfies the following conditions:
1) f is continuoous on the closed interval [a, b],
2) f is differntiable on the open interval (a, b).
Then there is a number c ∈ (a, b)
f ′ (c) =
or equivalently
Note Give
1❥
1❥
such that
1❥
1❥
f (b) − f (a)
,
b−a
1❥
f (b) − f (a) = f ′ (c)(b − a).
to any of the above two equations.
ex + 1
satisfy the Mean Value Theorem in the interval
(b) Does the function f (x) = x
e −1
[−1, 2]? Justify your answer.
(2)
116
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MATH1036/7
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Page 5 of 5
Mathematics I (Major)
June 2014
Solution
ex + 1
1❥
does not satisfy the Mean Value Theorem in the interval [−1, 2].
ex − 1
ex + 1
Because f (x) = x
is not defined at x = 0, so it is not defined on the interval [−1, 3]
e −1
and therefore it is not continuous on the interval [−1, 2].
1❥
No f (x) =
1❥ if the student writes that the function is either not defined or not continNote Give ✓✏
uous, but 0.5 if the student only says that the function is not differentiable.
✒✑
(c) Use l’Hôpital’s rule to evaluate lim+ (sin x) x
(5)
x→0
Solution
Let y = (sin x) x . Then ln y = ln(sin x) x
✓✏
✒✑
0.5
= x ln sin x
✓✏
✓✏
ln sin x
· 0.5
✒✑
✒✑
1
x
0.5
=
✓✏
ln sin x
∞
= − · 0.5 Hence arrording to l’Hôpital rule,
✒✑
1
x→0
x→0
∞
x
cos x
✓✏
ln sin x
= lim+ sin x · 0.5
lim+ ln y = lim+
✒✑
1
−1
x→0
x→0
x→0
2
x
x


 ✓✏

x  1 
Thus lim+ ln y = lim+ −x cos x
= lim+ −x cos x 
 · 0.5
x→0
x→0
x→0
sin x
 sin x  ✒✑
x
✓✏
✓✏
sin x
= 1, 0.5
Since lim+ −x cos x = 0 0.5 and lim+
✒✑ x→0
✒✑
x→0
x




✓✏
 1 


0.5
then lim+ ln y = lim+ −x cos x 
 = 0· ✒✑
x→0
x→0
 sin x 
x
✓✏
Hence lim+ y = 1 and lim+ (sin x) x = 1. 0.5
And lim+ ln y = lim+
x→0
x→0
✒✑
Total Section B: [28] marks
Mathematics I (Major)
3.3.4
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions September 2014
Multiple choice: C;E;D;C;D
117
118
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Mathematics I (Major)
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119
120
3.3.5
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Calculus Solutions November 2014
Multiple choice: Q1) C, Q2) E, Q3) A, Q4) B, Q5) D, Q6) A, Q7) C, Q8) D
MATH1036
CALCULUS
Page 5 of 10
November 2014
Section B
[66]
In this section you are expected to show all your working to earn the marks allocated.
Question 1
[12]
1) Show that the ordinary differential equation (ODE)
(4x + y4 )dx + (4xy3 − e3y )dy = 0
(E)
is exact.
(4)
Solution
∂(4x + y4 )
= 4y3
∂y
1❥ and
∂(4xy3 − e3y )
= 4y3 .
∂y
∂(4x + y4 ) ∂(4xy3 − e3y )
=
= 4y3 . 1❥
∂y
∂y
Hence the differential equation (E) is exact.
So
2) Give the general solution of the ODE
Solution
Let f (x, y) =
Z
1❥
1❥
(E).
(6)
(4x + y4 )dx.
Then f (x, y) = 2x2 + xy4 + g(y). 1❥
∂ f (x, y)
= 4xy3 + g′ (y) = 4xy3 − e3y . 1❥
∂y
Z
1
′
3y
❥
So g (y) = −e
1 and g(y) =
−e3y dy = − e3y .
3
1❥
1
Hence the general solution of the differential equation (E) is 2x2 + xy4 − e3y = C. 2❥
3
!
1
3) Find the particular solution satisfying y − √ = 0.
(2)
3
Solution
!2
✓✏
1
2 1 1 ✓✏
1
1
We have 2 − √
− √ × 0 − = C. 0.5 So C = − = . 0.5 It follows
✒✑
3
3 3 3 ✒✑
3
3
!
1
1
that the particular solution satisfying y − √ = 0 is 2x2 + xy4 − e3y = C. 1❥
3
3
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CALCULUS
Page 6 of 10
Question 2
1) Let k be a real number not equal to −1, i. e. k ∈ R \ {−1}. Evaluate
November 2014
R
[13]
k
cos x sin x dx. (3)
Solution
Z
1
1❥sink+1 x 1❥ + C. 1❥
cos x sink x dx =
k+1
Z
5
cos x
2) Evaluate
dx. If you wish to, you may use your answer in Question 2(1). (8)
√3
sin x
Solution
✓✏
✒✑ ✓✏
cos5 x = (cos2 x)2 0.5
✒✑
= (1 − sin2 x)2 cos x 0.5
= (1 − 2 sin2 x + sin4 x) cos x. 1❥
Z
1
cos5 x
= (1 − 2 sin2 x + sin4 x) cos x × sin− 3 x 1❥
√3
sin x
5
11
1
= (sin− 3 x − 2 sin 3 x + sin 3 x) cos x 1❥
cos5 x
dx =
√3
sin x
3) Calculate
Solution
Z
π
2
0
Z
1
5
11
✓✏
✒✑
✓✏
cos x(sin− 3 x − 2 sin 3 x + sin 3 x) dx
0.5
✓✏
✓✏
2
14
8
1
2
1
3
3
3
sin x 0.5 −
sin x 0.5 + C
=
sin x 0.5 +
✒✑5
✒✑11
✒✑
1
− +1
+1
+1
3
3
3
✓✏
✓✏
✓✏ ✓✏
2
8
14
3
3
3
sin 3 x 0.5 + C. 0.5
= sin 3 x 0.5 − sin 3 x 0.5 +
✒✑4
✒✑14
✒✑ ✒✑
2
cos5 x
dx.
√3
sin x
Z
π
2
0
"
# π2 ✓✏
2
8
14
3
3
3
cos5 x
dx =
0.5
sin 3 x − sin 3 x +
sin 3 x
√3
✒✑
2
4
14
sin x
0
3 ✓✏
3 3
0.5
= − +
2 4 14 ✒✑
6−3
3
=
+
4
14
3
3
= +
4 14
21 + 6 27 ❥
=
1
=
28
28
(2)
122
2012–2017 Past Test and Examination Questions Booklet
MATH1036
CALCULUS
Page 7 of 10
Mathematics I (Major)
November 2014
Question 3
[14]
4
4x + 1
−
·
x + 1 x2 − x + 1
1) Find the following integrals: (Complete the square, if necessary)
Z
4
a)
dx.
x+1
Solution
Z
4
dx = 4 ln |x + 1| + c1 . 1❥
x+1
Z
4x − 2
b)
dx.
x2 − x + 1
Solution
Z
Z
✓✏
✓✏
2x − 1
4x − 2
2
❥
❥
dx = 2
dx 1 = 2 0.5 ln |x − x + 1| 1 + c2 . 0.5
✒✑
✒✑
x2 − x + 1
x2 − x + 1
Z
3
dx.
c)
2
x −x+1
Solution
Z
Z
Z
✓✏
dx
dx
3
❥
1 =3
0.5
dx = 3
x2 − x + 1
(x − 12 )2 − 14 + 1
(x − 12 )2 + 34 ✒✑
Z
✓✏
dx
=3
0.5
√
(x − 12 )2 + ( 23 )2 ✒✑



 1
 x − 12 
= 3  √ arctan  √  + c3 1❥
3
3
2
2
!
√
2x − 1
= 2 3 arctan √
+ c3 . 1❥
3
Z
3(x3 − 3x + 2)
2) Use your answers in the preceding question to find
dx.
x3 + 1
Solution
Let f be the rational function defined by f (x) = 3 +
4
4x + 1
− 2
x+1 x −x+1
3(x3 + 1) + 4(x2 − x + 1) − (4x + 1)(x + 1) ✓✏
0.5
✒✑
x3 + 1
3x3 + 3 + 4x2 − 4x + 4 − (4x2 + 4x + x + 1) ✓✏
0.5
✒✑
x3 + 1
3x3 + 4x2 − 4x + 7 − 4x2 − 5x − 1 ✓✏
0.5
✒✑
x3 + 1
3x3 − 9x + 6 3(x3 − 3x + 2) ✓✏
=
0.5
✒✑
x3 + 1
x3 + 1
f (x) = 3 +
=
=
=
=
(1)
(3)
(4)
(6)
Mathematics I (Major)
MATH1036
Z
CALCULUS
Page 8 of 10
November 2014
!
✓✏
4x + 1
4
− 2
3+
dx
0.5
✒✑
x+1 x −x+1
Z
4x − 2 + 3
dx 1❥
= 3x + 4 ln |x + 1| −
x2 − x + 1
3(3x3 − 3x + 2)
dx =
x3 + 1
But
Z
123
2012–2017 Past Test and Examination Questions Booklet
Z
Z
✓✏
2(2x − 1)
dx
0.5
dx
+
3
dx
✒✑
x2 − x + 1
x2 − x + 1
!
√
2x − 1
2
= 2 ln |x − x + 1| + 2 3 arctan √
+ K. 1❥
3
4x − 2 + 3
dx =
x2 − x + 1
Z
Hence
!
Z
√
3(x3 − 3x + 2)
2x − 1
2
+ C. 1❥
dx = 3x + 4 ln |x + 1| − 2 ln |x − x + 1| − 2 3 arctan √
x3 + 1
3
Question 4
[13]
1-a) When is a series
∞
X
an said to be absolutely convergent?
(2)
n=1
Solution
A series is said to be absolutely convergent if
∞
X
n=1
b) Is the series
Solution
(2)
∞
X
sin(2n + 1) π2
convergent? Justify your answer.
n
n=1
π
sin(2n + 1) = (−1)n , 1❥ where n = 1, 2, · · · .
2
The series
|an | converges.
∞
X
(−1)n
So
(3)
∞
X
sin(2n + 1) π
2
n=1
n
=
· converges according to the alternating series test.
n
n=1
∞
X
sin(2n + 1) π2
series
converges. 1❥
n
n=1
∞
X
(−1)n
n=1
n
·
1❥ Hence the
∞
X
sin(2n + 1) π2
absolutely convergent? Justify your anwser.
c) Is the series
n
n=1
(3)
Solution
∞
∞
∞
∞
X
X
X
X
sin(2n + 1) π2
(−1)n
1 ❥
1
=
The series
· 1
diverges. 1❥ Therefore
=
n
n
n
n
n=1
n=1
n=1
n=1
∞
X
sin(2n + 1) π2
does not converge absolutely. 1❥
the series
n
n=1
124
2012–2017 Past Test and Examination Questions Booklet
MATH1036
CALCULUS
Page 9 of 10
Mathematics I (Major)
November 2014
!n
∞
X
3n − 4
2) Prove that the series
converges.
4n
−
3
n=1
(5)
Solution
lim
n→∞
s
n
3n − 4
4n − 3
!n
= lim
n→∞
r
n
3n − 4
4n − 3
n
1❥ = lim
n→∞
= lim
n→∞
3n − 4 ❥
1
4n − 3
3−
4−
4
n
3
n
=
3 ❥
1 < 1. 1❥
4
!n
∞
X
3n − 4
converges (absolutely). 1❥
Hence the series
4n
−
3
n=1
Question 5
1) For which values of x does the series
∞
X
n=1
[14]
(x − 5)n
converge?
n
(6)
Solution
(x − 5)n+1
1
n
n
an+1
=
|x − 5|.
×
|x − 5| 1❥ =
=
n
an
n+1
(x − 5)
n+1
1 + 1n
lim
n→∞
an+1
1
= lim
|x − 5| = |x − 5|. 1❥
n→∞ 1 + 1
an
n
The series
∞
X
(x − 5)n
n
n=1
converges if |x − 5| < 1. 1❥
|x − 5| < 1 ⇔ −1 < x − 5 < 1 ⇔ 4 < x < 6.
∞
X
(−1)n
(−1)n
If x = 4, an =
and
converges according to the alternating series test. 1❥
n
n
n=1
If x = 6, an =
X1
(1)n 1
= and the series
diverges. 1❥
n
n
n
n=1
Hence the series
∞
∞
X
(x − 5)n
n=1
n
converges if and only if x ∈ [4, 6).
1❥
Mathematics I (Major)
MATH1036
CALCULUS
Page 10 of 10
November 2014
2) Give with justification the power series representation of the function g(x) =
about the point 0.
Solution
∞
∞
X
X
1
1
n
=
=
(−x) =
(−1)n xn .
1 + x 1 − (−x) n=0
0
The function f (x) =
125
2012–2017 Past Test and Examination Questions Booklet
x2
(1 + x)3
(8)
1❥
∞
X
(−1)n xn is continuous and differentiable in the interval (−1, 1), 1❥
∞
✓✏
X
1
n
n−1
=
(−1)
nx
.
1.5
✒✑
(1 + x)2 n=0
n=0
and f ′ (x) = −
∞
✓✏
X
2
n
n−2
f (x) =
=
(−1)
n(n
−
1)x
.
1.5
✒✑
(1 + x)3 n=0
′′
1X
1
=
(−1)n n(n − 1)xn−2 , 1♠
So
3
(1 + x)
2 n=0
∞
and
X
x2
1X
n
n−2
2 ♠ 1
=
(−1)
n(n
−
1)x
×
x
1
=
(−1)n n(n − 1)xn .
(1 + x)3 2 n=0
2 n=0
Hence g(x) =
∞
∞
1X
(−1)n n(n − 1)xn . 1♠
2 n=0
∞
∞
1X
Or g(x) =
(−1)n n(n − 1)xn .
2 n=2
Total Section B: [66] marks
126
2012–2017 Past Test and Examination Questions Booklet
3.4
Calculus Solutions 2015
3.4.1
Calculus Solutions March 2015
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
127
128
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
129
130
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Mathematics I (Major)
3.4.2
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions April 2015
131
132
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Mathematics I (Major)
Mathematics I (Major)
3.4.3
2012–2017 Past Test and Examination Questions Booklet
133
Calculus Solutions June 2015
Section A: Multiple choice questions
Question 1: C;
Question 2: B;
Question 3: E;
Question 4: E;
Section B
(Note: X is one mark, (X) is 1/2 mark.)
Question 1
[5]
Let
y=
Then
(x2 − 4)9 √ x
e
(x − 1)2
(X)
Å 2
ã
√
(x − 4)9 √ x X
(X)
ln y = ln
e
= 9 ln(x2 − 4) + x − 2 ln(x − 1)
2
(x − 1)
Implicit differentiation gives
y0
9(2x)
2
1
= 2
+ √ −
y
x −4 2 x x−1
Therefore
Å
ã
(x2 − 4)9 √ x
18)
1
2
0
0
f (x) = y =
e
+ √ −
(x − 1)2
x2 − 4 2 x x − 1
Question 2
XX
X
[10]
(a)
Å ã
0
e x sin x − x
lim
2
x→0
x
0
e x cos x + e x sin x − 1
l’H
= lim
x→0
2x
x
e
(−
sin
x)
+ e x cos x + e x cos x + e x sin x
l’H
= lim
x→0
2
2
= =1
2
X
Å ã
0
0
XX
X
X
(b)
lim (sin x) x is a limit of the form (00 ).
(X)
Put y = lim+ (sin x) x .
(X)
x→0+
x→0
Then
ln y = lim x ln(sin x)
x→0+
= lim
x→0+
ln(sin x)
1
x
− cos x
l’H
= lim sin x
−1
x→0
x2
x
x cos x
x→0 sin x
=0
= lim
(0 · ∞)
∞
∞
X
(X)
X
(X)
(X)
134
2012–2017 Past Test and Examination Questions Booklet
Hence
Mathematics I (Major)
lim x ln(sin x) = y = eln y = e0 = 1.
x→0+
Question 3
(X)
[6]
Let x and y be the length and the width of the rectangle. The following sketch may be used:
(X)
y
x
We know that the perimeter is constant,
p = 2x + 2y
(X)
so that
dx dy
+
= 0.
dt
dt
Since the side lengths change with constant speed 2cm/sec, and assuming that x increases, we find
dx
dy
= 2 and
= −2.
dt
dt
(X)
X
In cm2 , the area of the rectangle is
A(x) = xy.
(X)
Then
dx
dy
dA
=y +x
= 2y − 2x.
dt
dt
dt
Originally we have that x = y are 1 m, and therefore x + y = 200. Hence, when y = 80, then x = 120, so that
dA
= 2(80) − 2(120) = −80.
dt
X
X
(X)
Therefore the area of R decreases at a rate of 80 cm2 /sec when the shorter sides are 80 cm long.
(X)
Question 4
[15]
(a) We have f (x) = 0 if and only if
0 = x2 − 3x − 4 = (x − 4)(x + 1)
X
Hence the x-intercepts are at x = 4 and x = −1 [alternatively: The x-intercepts are (4, 0) and (−1, 0)]
Since f (0) = 1, the y-intercept is at 1 [alternatively: The y-intercept is (0, 1)]
X
X
(b)
3
4
1− − 2
x2 − 3x − 4
x x =1
= lim
lim
4
x→±∞
x→±∞
x2 − 4
1− 2
x
Hence f f has one horizontal asymptote y = 1.
X
2
f is undefined when x − 4 = 0, i. e., when x = ±2. Then
f (x) → ∞ as x → −2− or x → 2−
f (x) → −∞ as x → −2 or x → 2
+
(c) Since
f 0 (x) = 3
x2 + 4
(x2 − 4)2
+
X
X
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
135
one sees that f 0 (x) > 0 for all x in the domain of f .
X
Hence f increases on (−∞, −2), −2, 2), and (2, ∞),
X
and f has no global minima and maxima.
X
(d)
Since
f 00 (x) = −6x
x2 + 12
(x2 − 4)3
it follows that f 00 is positive on (−∞, −2) and on (0, 2), and f 00 is negative on (−2, 0) and on (2, ∞). [Students may
use a second derivative table.]
X
Hence f is convex up on (−∞, −2) and on (0, 2), and f is convex down on (−2, 0) and on (2, ∞).
Since convexity changes at 0, f has a point of inflection at x = 0 [or (0, 1) is a point of inflection]
X
X
(e) B
3.4.4
Calculus Solutions September 2015
Section A: Multiple choice answers: C;E;D;E
Section B
Question 1
[5]
You are asked to fence off a rectangular field with area 2400m2 and partition the field into two equal parts. What
is the smallest length of fence needed?
Solution
y
x
Let x and y be the length and width of the field, respectively.
2400
m.X
Area of the field is given by A = xy = 2400m2 =⇒ y =
x
The length of fence required and partition the field is given by the perimeter in meters as
2400
P = 3y + 2x = 3
x
Å
ã
+ 2x.X
The perimeter is function of x. To minimise the length of fence required we differentiate P(x) and find critical
d
values at
P(x) = 0.
dx
Å
ã
d
1
P(x) = 7200 − 2 + 2 = 0X ⇔ x2 = 3600.
dx
x
Thus x = ±60X but it is a length x > 0 hence x = 60. Since 0 ≤ x ≤ 2400
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2012–2017 Past Test and Examination Questions Booklet
P(0)
Mathematics I (Major)
is undefined
P(60) = 120
P(2400) = 4803
The smallest length of fence occurs at the minimum value of x = 60m and y = 40m giving total length of
P = 240mX.
Question 2 Å ã
x2
dy
If y = 5 cosh
, find
at the point (2, 0). Simplify your answer.
4
dx
Solution
d
5 cosh
dx
x2
4
Å
ã
= 5 sinh
Å
x2
4
ã
[4]
2x
.XX
4
At the point (2, 0),
Å 2ã
dy
2 2.2
= 5 sinh
X
dx
4
4
= 5 sinh(1)X
Question 3
Find the area bounded by y = 2x2 and y = x4 − 2x2 .
[5]
Solution
We find the bounds of the enclosed area by equating the two functions,
2x2 =x4 − 2x2 X
0 =x2 (4 − x2 ) ⇔ x = 0 or x = −2 or x = 2
Area =
Z
2
2x2 − (x4 − 2x2 ) dxXX
−2
=2
2
Z
0
4x2 − x4 dx
ò2
x3 x5
=2 4 −
X
3
5 0
ï 3
ò
2
25
=2 4 −
− (0 − 0)
3
5
27
128
=
=
X
15
15
ï
Question 4
[6]
Find the volume of the solid generated by revolving the region in the first quadrant bounded above by the curve
y = x2 , below by the x-axis and on the right by the line x = 1, about the line x = −1.
Solution
We find the bounds of the enclosure by equating functions. In the first quadrant,
x2 = 1 ⇔ x = 1 =⇒ y = 1X
Or
√
y = 1 ⇔ y = 1.X
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Volume =
1
0
¶
√
2 ©
π [1 − (−1)]2 −
y − (−1)
dyXXX
Z
1
Z
=π
0
√
[4] − y + 2 y + 1 dy
p ô1
y2
y3
−4
X
= π 3y −
2
3 0
ñ
ô
√
12
13
= π 3(1) −
−4
− (0 − 0 − 0)
2
3
7
= π X.
6
ñ
Total Section B: [20] marks
137
138
3.4.5
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions November 2015
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
139
140
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Mathematics I (Major)
Mathematics I (Major)
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141
142
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Mathematics I (Major)
Mathematics I (Major)
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143
144
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Mathematics I (Major)
Mathematics I (Major)
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145
146
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Mathematics I (Major)
Mathematics I (Major)
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3.5
Calculus Solutions 2016
3.5.1
Calculus Solutions March 2016
Question 8
-
147
9 marks
(a) Complete the sentence in words or mathematical symbols:
For the limit lim f (x) to exist...
(3 marks) Both the left-hand limit lim− f (x) and the right-hand limit
x→a
x→a
lim f (x) of the function exist and are equal to each other.
x→a+
or
lim f (x) = L ⇔ lim− f (x) = L = lim+ f (x).
x→a
x→a
x→a
(b) Fill in the answers to the limits in the table:
Expressions
1
1: lim(1 + x) x
Answers
e
2: lim + cosec θ
−∞ or D.N.E.
x→0
θ→−π
3: lim 29
x→∞
−x
0
4: lim− byc
1
y→2
(4 marks)
(c) Write down a limit which shows that a function h(x) has a vertical asymptote at x = 1.5.
(2 marks)
lim h(x) = ∞
x→1.5−
or
lim h(x) = −∞
x→1.5+
or
lim h(x) = ∞
x→1.5
etc. (They can write D.N.E. too, but it is not sufficient on its own, they must show that at least one of the
limits is infinite.)
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Question 9
-
Mathematics I (Major)
8 marks
Evaluate (and SHOW ALL WORKING)
(a) lim √
x→−2
x2 − 4
.
x+3−1
(4 marks)
√
(x2 − 4)( x + 3 + 1)
x2 − 4
√
= lim √
lim √
x→−2
x + 3 − 1 x→−2 ( x + 3 − 1)( x + 3 + 1)
√
(x2 − 4)( x + 3 + 1)
= lim
x→−2
x+3−1
√
2
(x − 4)( x + 3 + 1)
= lim
x→−2
x+2
√
(x + 2)(x − 2)( x + 3 + 1)
= lim
x→−2
x+2
√
= lim (x − 2)( x + 3 + 1)
x→−2
√
= (−2 − 2)( −2 + 3 + 1)
= (−4)(2)
= −8
(b) lim
x→∞ 3x2 (x
x2 − 4
.
+ 1) + 10(x + 1)
x2 − 4
x→∞ 3x2 (x + 1) + 10(x + 1)
(x2 − 4)/x3
= lim
x→∞ (3x3 + 3x2 + 10x + 10)/x3
(divide top and bottom by x to the highest exponent in the denominator)
lim
1/x − 4/x3
x→∞ 3 + 3/x + 10/x2 + 10/x3
0
=
since all terms except 3 go to zero in the limit.
3
= 0.
= lim
(4 marks)
Mathematics I (Major)
Question 10
-
(a) Prove that lim
θ→0
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2012–2017 Past Test and Examination Questions Booklet
10 marks
1 − cos θ
sin θ
= 0. You may assume that lim
= 1.
θ→0 θ
θ
(5 marks)
Å
ãÅ
ã
1 − cos θ
1 + cos θ
1 − cos θ
= lim
lim
θ→0
θ→0
θ
θ
1 + cos θ
2
1 − cos θ
= lim
θ→0 θ(1 + cos θ)
sin2 θ
= lim
θ→0 θ(1 + cos θ)
Å
ãÅ
ã
sin θ
sin θ
= lim
θ→0
θ
1 + cos θ
Å
ãÅ
ã
sin θ
sin θ
= lim
lim
θ→0 θ
θ→0 1 + cos θ
Å
ã lim sin θ
sin θ
θ→0
= lim
θ→0 θ
lim(1 + cos θ)
θ→0
sin θ
θ→0 θ
Å
= lim
lim sin θ
ã
θ→0
1 + lim cos θ
θ→0
0
= (1)
1+1
= 0.
(b) Use the Sandwich (Squeeze) Theorem to determine lim sinx2 x . Clearly define your functions f (x), g(x) and
x→−∞
h(x) and explain your working.
(5 marks)
Since
−1 ≤ sin x ≤ 1
we have (recall x2 is always positive or zero, and we do not need to consider what happens at zero)
−
Let
f (x) := −
Now,
1
;
x2
1
sin x
1
≤ 2 ≤ 2
x2
x
x
g(x) :=
sin x
;
x2
h(x) :=
lim f (x) = lim h(x) = 0
x→−∞
x→−∞
and thus by the Sandwich Theorem
sin x
= 0.
x→−∞ x2
lim g(x) = lim
x→−∞
1
.
x2
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2012–2017 Past Test and Examination Questions Booklet
Question 11
-
Mathematics I (Major)
13 marks
(a) A function f is continuous at the point a if and only if ...
(2 marks)
f (a) = lim f (x).
x→a
(b) Indicate with ‘Y’ or ‘N’ whether the functions in the following table are continuous at x = −1:
Function
1
1: f (x) = x−1
2: g(x) = |x + 1|
3: h(x) = sec(x + 1)
4: j(x) = b2xc
Y/N
Y
Y
Y
N
(4 marks)
(c) Find the value(s) of a which make(s) the function g(x) continuous if
(
|x − 2| + 1
if x ≤ 1
g(x) =
−x2 + ax − 3 if x > 1.
Show all your working.
(5 marks) The value of the function at 1 is
g(1) = |1 − 2| + 1 = 2.
For the left limit we have
lim g(x) = lim− |x − 2| + 1 = 2
x→1−
x→1
For the right limit we have
lim g(x) = lim+ −x2 + ax − 3 = −1 + a − 3 = a − 4
x→1+
x→1
For continuity, we require that all of these three quantities are the same, i.e. a − 4 = 2, so a = 6.
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
(d) Draw a function with a removable discontinuity at x = 2 on the Cartesian plane below.
151
(2 marks)
Accept anything which is suitable. I.e., lim f (x) = L but f (2) , L or is not defined.
x→2
Please ask if you are unsure.
(Calculus Total 40)
152
3.5.2
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions April 2016
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
153
154
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Mathematics I (Major)
Mathematics I (Major)
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155
156
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Mathematics I (Major)
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157
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3.5.3
Mathematics I (Major)
Calculus Solutions June 2016
Question 1
Z π
esin x cos x dx =
(a)
[11]
(3)
0
A.
−e
B.
−1
C.m 0
D.
1
E.
e
Partial Credit:
Let u = sin x. Then du = cos x dx, X
and xZ= 0 ⇒ u = 0, x = πZ ⇒ u = 0. (X).
π
∴
0
esin x cos x dx =
0
0
eu du
(X)
= 0.
(b) The function f (x) = x3 + x5 − 2 is invertible.
(3)
Which of the following is true?
A.
f −1 is not differentiable at 0
B.
( f −1 )0 (0) = 0
C.
( f −1 )0 (0) = 8
D.
−1 0
( f ) (0) = −8
1
E.m ( f −1 )0 (0) =
8
Z
2
(c) Evaluate x3 e x dx
Partial Credit:
By inspection, f (1) = 0, so that f −1 (0) = 1. X
Then f 0 (x) = 3x2 + 5x4 (X), so that f 0 (1) = 8
1
1
1
∴ ( f −1 )0 (0) = 0 −1
= 0
= . (X)
f ( f (0))
f (1) 8
(X).
(5)
Answer:
First use integration by substitution.
Let w = x2 . Then dw = 2x dx. (X)
It follows that
Z
Z
1
2
x e dx =
x2 e x 2x dx (X)
2
Z
1
wew dw X
=
2
3 x2
Next use integration by parts.
Put u = w and dv = ew dw. (X).
Then du = dw and v = ew . (X).
Then
Z
wew dw =
Z
u dv
Z
= uv − v du
Z
= wew − ew dw X
= wew − ew + C1
= (w − 1)ew + C1
(X)
Resubstituting w gives
Z
2
Z
1
wew dw
2
1
= (w − 1)ew + C
2
1
2
= (x2 − 1)e x + C
2
x3 e x dx =
(X)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Question 2
159
[10]
Find the limits
e x − sin x − 1
(a) lim
x→0
x2 cos x
Answer:
(5)
e x − sin x − 1
x→0
x2 cos x
lim
l’H
“ 0”
X
0
e x − cos x
x→0 2x cos x − x2 sin x
= lim
l’H
“ 0”
X
e x + sin x
x→0 2 cos x − 4x sin x − x2 cos x
= lim
=
1
2
X
0
X
X
(b) limπ (sin x)sec x
(5)
x→ 2
Answer:
Let
y = limπ (sin x)sec x
(X)
x→ 2
Then
ln y = limπ ln(sin x)sec x
(X)
x→ 2
= limπ sec x ln(sin x)
“0 · ∞”
x→ 2
ln(sin x)
cos x
= limπ
x→ 2
l’H
= limπ
x→ 2
=0
cos x
sin x
− sin x
“ 0”
0
X
X
X
(X)
∴
y = e0 = 1
(X)
Question 3
[10]
Thandi rides on her bicycle at a constant speed of 4 meters per second (14.4 km/h) towards a water tower, which
has a diameter of 20 meters. How fast does her horizontal angle of sight of the water tower increase when Thandi
is 30 meters away from the water tower?
(a) In the sketch of the aerial view below, label the quantities which you will use for your solution.
Answer: These quantities are inserted in red, where x denotes Thandi’s distance from the water tower.
(2)
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2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
10
10
θ
x
(b) If 2θ is Thandi’s horizontal angle of sight of the water tower, find a relation between θ and Thandi’s distance
from the water tower.
(2)
Answer: Due to the right angle,
sin θ =
10
.
x + 10
(c) Solve the problem.
(6)
Answer: Differentiating the equation in (b) with respect to time t one gets
cos θ
dx
10
dθ
=−
2
dt
(x + 10) dθ
When x = 30, then
sin θ =
so that
cos θ =
p
2
1 − sin θ =
10 1
=
40 4
…
X
X
1
1−
=
16
√
15
4
X(X)
Therefore
10 4
1
dθ
√
= − 2 √ (−4) =
X
dt
40
15
10 15
1
√ radians per second when Thandi is 30
Thandi’s horizontal angle of sight of the water tower increases by
10 15
meters away from the water tower. X(X)
Question 4
[8]
(a) What is the following theorem called?
(2)
Theorem. If f has a local maximum or minimum at an interior point c of dom( f ) and if f is differentiable at c,
then f 0 (c) = 0.
A.n Fermat’s Theorem,
B.
Intermediate Value Theorem,
C.
Mean Value Theorem,
D.
Rolle’s Theorem,
E.
Taylor’s Theorem.
(b) Prove the theorem in case of a local maximum.
Answer:
(Let f have a local maximum at c.)
Let a, b be such that the interval (a, b) belongs to the domain of f , a < c < b and
f (x) ≤ f (c) for x ∈ (a, b), i. e.,
f (x) − f (c) ≤ 0. XX
(6)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Hence
f (x) − f (c)
≥0
x−c
if x ∈ (a, c) X
f (x) − f (c)
≤0
x−c
Since f is differentiable at c, it follows that
if x ∈ (c, b). X
and
f−0 (c) ≥ 0
and
0
f−0 (c)
⇒ f (c) =
=
f+0 (c) ≤ 0
f+0 (c)
161
X
= 0. X
Question 5
[6]
2
3
Find the global maximum and the global minimum of 2x − 3x on [−1, 8].
Answer:
2
Let f (x) = 2x − 3x 3 . Then
Å ã
1
2 −1
f (x) = 2 − 3
x 3 = 2 − 2x− 3
3
0
So
1
and f is not differentiable at 0.
f 0 (x) = 0 ⇔ x− 3 = 1 ⇔ x = 1
(X)
Hence the critical numbers are 0 and 1.
X
X
(X)
Evaluate f at the critical numbers:
f (0) = 0,
f (1) = −1,
(X)
f (−1) = −5,
(X)
(X)
f (8) = 16 − 3 · 4 = 4,
(X)
Comparing the values, if follows that
2
the global maximum of 2x − 3x 3 on [−1, 8] is 4
(X)
2
the global minimum of 2x − 3x 3 on [−1, 8] is −5
3.5.4
(X)
Calculus Solutions August 2016
Full workings must be shown.
Leave answers in exact form.
Question 1
[5]
1
between x = 1 and x = 5 using 4 partitions?
x
For the ith subinterval denoted [xi−1 , xi ] of [1, 5], minimum values of the function g(x) occur at xi .
What is the lower sum approximation of the area bounded by g(x) =
5−1
=1
4
xi = 1 + i∆x = 1 + i
X
∆x = xi − xi−1 =
L(4) =
4
X
g(xi )∆x
X
X
i=1
1
.1 +
2
77
=
60
=
1
1
1
.1 + .1 + .1
3
4
5
X
X
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2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Question 2
What are the dimensions of the rectangle with the largest area that be inscribed in the ellipse below?
y axis
[6]
3
(x,y)
5
−5
x axis
−3
Please follow student’s solution if workings are in y and A(y).
The ellipse is given by
x2 y2
+
=1
52 32
y=±
3√
25 − x2
5
X
Area = 4xy
(X)
√
12
A(x) =
x 25 − x2
X
5 Å
ã
12 √
−2x
0
2
A (x) =
25 − x + x √
5
2 25 − x2
Å
2
2ã
12 (25 − x ) − x
√
=
5
25 − x2
12 25 − 2x2
√
=
= 0X
X
5 25 − x2
5
Implying x = ± √ . The bounds of x are 0 ≤ x ≤ 5 (X)with A(0) = 0 = A(5)
2
5
3
Thus maximum occurs at x = √ and y = √
(X).
2
2
Question 3
Å
ã
1
1+x
Prove the following formula tanh x = ln
.
2
1−x
−1
X
(X) .
[5]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
163
Let
Å
ã
1
1+x
ln
2
1−x
Å
ã
1+x
= ln
1−x
y=
(X)
then
Ç
ãå
1+x
tanh y = tanh ln
X
1−x
»
Ä
ä
1+x
sinh ln
1−x
=
X
Ä »
ä
1+x
cosh ln
1−x
√ 1−x
√
ln ( 1+x
)
( 1+x )
1−x − eln
e
√ 1+x
√ 1−x
X
=
ln ( 1−x )
e»
+»
eln ( 1+x )
1+x
1−x
1−x −
1+x
= »
(X)
» 1−x 1+x
1−x +
1+x
√
√
1 + x − (1 − x)
1−x 1+x
√
·
= √
1 − x 1 + x 1 + x + (1 − x)
2x
=
2
=x
(X).
Å
If
x = tanh y then
−1
tanh x = y
(X)
OR
Question 4
[5]
Prove that is f is continuous on a closed interval [a, b], then there exist a c in the closed interval such that
Zb
a
f (x) dx = f (c)(b − a).
Suppose f is continuous on [a, b]. By Extreme Value Theorem f attains its minimum and maximum on [a, b], say
at y and z both in [a, b]X. Thus,
f (y) ≤ f (x) ≤ f (z)X
f (y)(b − a) ≤
Zb
a
f (x) dx ≤ f (z)(b − a)X.
Since (b − a) is a constant g(x) = f (x)(b − a)X is a continuous function and
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2012–2017 Past Test and Examination Questions Booklet
Zb
g(y)(≤
a
If we let N =
Zb
a
Mathematics I (Major)
f (x) dx ≤ g(z).
f (x) dx, by the Intermediate Value Theorem there must be a c ∈ (y, z) such that g(c) = NX i.e
g(c) = N
f (c)(b − a) =
Zb
f (x) dxX.
a
Question 5
Using areas of known geometries and other properties of integrals evaluate the integral
Z2
−2
[3]
(2 − |x|) dx.
Give half a mark for correct diagram if answer is wrong.
Z2
−2
(2 − |x|) dx = 2
Z2
0
(2 − |x|) dx
1
·2·2
2
=4
X
=2·
X
X
Question 6
[10]
a. What is the total area of the region bounded by x = 4y2 − 2y and x = 12y2 ?
2
2
4y − 2y = 12y
0 = 8y2 + 2y
0 = 2y(4y + 1)
1
y = 0 (X) or y = −
(X)
4
Area =
Z0
− 41
Z0
=
− 14
[4y2 − 2y − 12y2 ] dy
[−8y2 − 2y] dy
ï
0
8
X
= − y3 − y2
3
− 41
Å
ã
8 1
1
=− −
−
3 (−4)3 (−4)2
1
=
X
48
XX
(5)
Mathematics I (Major)
165
2012–2017 Past Test and Examination Questions Booklet
b. What is the volume of a solid generated by rotating h(x) =
Volume = π
Z4
1
Z4
=π
√
x + 4 about the line y = 2 on [1, 4]?
√
[ x + 4 − 2]2 dx
√
x + 4 x + 4 dx
(5)
X
X
1
4
x2 8 3
X
+ x 2 + 4x
2
3
1
64
1 8
= π[8 +
+ 16 − − − 4]
3
2 3
229
=π
X
6
ï
=π
X
Total :[30] marks
Bonus : [4] marks
166
3.5.5
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions November 2016
Mathematics I (Major)
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
167
168
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Mathematics I (Major)
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169
170
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Mathematics I (Major)
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171
172
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173
174
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Mathematics I (Major)
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175
176
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Mathematics I (Major)
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177
178
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Mathematics I (Major)
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3.6
Calculus Solutions 2017
3.6.1
Calculus Solutions March 2017
Question 1
[8]
A step function of x which is the least integer greater than or equal to x also called the ceiling function of x is
usually written dxe. The graph of f (x) below is a graphical representation of the ceiling function. Use the graph
below to answer questions that follow.
f (x) = dxe
2
1
−3
−2
−1
1
2
3
−1
−2
a.
a. What is lim − f (x)?
−2X
.
(1)
b.
b. What is lim + f (x)?
−1X
.
(1)
c.
c. What is lim f (x)?
d.
d. For any n ∈ Z is f (x) continuous at x = n? Respond by writing Yes or No
then give a reason to support your answer.
x→−2
x→−2
x→−2
does not existX
(1)
.
NoX
lim f (x) , lim+ f (x)X
x→n−
E.
(2)
x→n
Circle the letter above the graph that represents of the function g(x) = x − dxe.
(3)
AXXX
B
2
1
−2
−1
1
−1
2
−1
180
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
C
D
2
−2
−1
2
1
2
−2
−1
Question 2
a.
−1
1
2
−1
[6]
1
lim(1 + x) x =
x→0
eX
(1)
.
1
b.
Compute lim+
x→0
2e x
1
1 + ex
, using only methods covered in class so far.
1
= lim+
x→0
=
ex
2
1
x
− 1x
e (e + 1)
lim+ 2
( lim+ e− x + lim+ 1)
x→0
=
X
x→0
1
X
x→0
2
− 1x
lim e
x→0+
(5)
+1
X
∵ lim+
x→0
1
= ∞X
x
=2X.
Question 3
[6]
2
What are the asymptotes of h(x) = π − 2 ?
x
Show workings on how you got to your answer using limits.
lim π −
x→∞
2
= πXX
x2
OR
2
= πXX
x2
∴ y = π is a horizontal asymptote.X
2
lim π − 2 = −∞XX
x→0+
x
OR
2
lim− π − 2 = −∞XX
x→0
x
∴ x = 0 is a vertical asymptote.X
lim π −
x→−∞
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
Question 4
a. If 2 − x2 ≤ g(x) ≤ 2 cos(x) for all x, what is lim g(x)?
181
[6]
(4)
x→0
lim 2 − x2 = 2X
x→0
lim 2 cos(x) = 2X
x→0
∴ By Sandwich or Squeeze or Pinch TheoremX
lim g(x) = 2X
x→0
b. Give the full statement of the theorem that you used to answer part a. of this question.
If f (x) ≤ g(x) ≤ h(x) when x is near a and lim f (x) = L = lim h(x)X, then lim g(x) = L.X
x→a
Question √
5
Find lim
x→1
x2
√
x→a
(2)
x→a
√
+4− 5
, if it exists.
x+3−2
[5]
√
√
√
x2 + 4 − 5
x2 + 4 + 5
√
· √
√ X
x→1
x+3−2
x2 + 4 + 5
x2 + 4 − 5
äÄ√
= lim Ä √
√ äX
x→1
x+3−2
x2 + 4 + 5
= lim
√
√
x2 − 1
x+3+2
äÄ√
= lim Ä √
X
√ ä· √
x→1
x+3+2
x+3−2
x2 + 4 + 5
√
(x2 − 1)( x + 3 + 2)
Ä
= lim
√
√ äX
x→1 (x + 3 − 4)
x2 + 4 + 5
√
(x + 1)( x + 3 + 2)
= lim Ä √
√ ä
x→1
x2 + 4 + 5
4
=√ X
5
Question 6
√
a. Detemine if the function g(x) = |x2 − 3| is continuous at x = 3.
If it is discontinuous, state the type of the discontinuity.
lim
√
x→ 3−
lim
√
x→ 3+
√
[9]
(3)
2
|x2 − 3| = lim
√ −(x − 3) = 0, 1/2
x→ 3−
2
2
|x − 3| = lim
√ (x − 3) = 0, 1/2
√
x→ 3−
g( 3) = |( 3)2 − 3| = 01/2
√
lim
√ g(x) = g( 3)X
x→ 3
∴ g(x) is continuous at x =
√
√
b. Compute g0− ( 3) and g0+ ( 3) from first principles.
√
3. 1/2
(5)
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2012–2017 Past Test and Examination Questions Booklet
g0− (
√
Mathematics I (Major)
√
g(x) − g( 3)
√
X
3) = lim
√
x− 3
x→ 3−
−(x2 − 3)
√ 1/2
= lim
√
3
x→ 3− x −
√
√
−(x − 3)(x + 3)
√
= lim
1/2
√
x− 3
x→ 3−
√
√
= lim
3) = −2 3 1/2
√ −(x +
x→ 3−
g0+ (
√
√
g(x) − g( 3)
√
X
3) = lim
√
x− 3
x→ 3+
(x2 − 3)
√ 1/2
= lim
√
3
x→ 3− x −
√
√
(x − 3)(x + 3)
√
= lim
1/2
√
x− 3
x→ 3+
√
√
3) = 2 3 1/2
= lim
√ (x +
x→ 3+
√
c. Is g0 (x) continuous at x = 3?
If it is discontinuous, state the type of the discontinuity.
NO 1/2. Jump discontunuity. 1/2
3.6.2
(1)
Calculus Solutions April 2017
[8 marks]
Section A
Instructions:
For each multiple choice question, select the letter that corresponds to the correct answer.
[A1] The normal to the curve y = 3 tan x + x2 + 1 at the point with coordinates (0, 1) is
(A) y = −3x + 1.
(B) y = 3x − 1.
1
(C) y = x − 1.
3
1
(D) y = − x + 1.X
3
1
(E) y = x + 1.
3
[A2] If y = log10 (sin x) ; 0 < x < π, then
(A)
(B)
(C)
(D)
1
.
sin x
1
.
ln 10 sin x
1
cot x.X
ln 10
1
.
ln 10 cot x
ln 10
2 marks
dy
=
dx
Mathematics I (Major)
(E)
183
2012–2017 Past Test and Examination Questions Booklet
1
.
ln 10 cos x
2 marks
[A3] Let f and g be differentiable functions. Which of the following is equal to
d
( f (x + g(x)))?
dx
(A) f 0 (x) + f 0 (g(x))g0 (x).
(B) f 0 (x + g(x))g0 (x).
(C) f 0 (x + g(x))(x + g0 (x)).
(D) f 0 (x) + g0 (x)(1 + g0 (x)).
(E) f 0 (x + g(x))(1 + g0 (x)).X
2 marks
[A4] Which of the following is equal to
−
(B)
ln |7x| +
(D)
(E)
1
+ eπ
7x
ã
dx?
2
+ C.
7x2
(A)
(C)
Z Å
1
eπ+1 x + C.
π+1
ln |x|
+ π eπ x + C.
7
ln |x|
+ eπ x + C.X
7
x
ln
+ eπ x + C.
7
2 marks
[32 marks]
Section B
Instructions:
In this section you are expected to show all your working to earn the marks allocated.
Simplify your answers.
Question 1
-
13 marks
(a) Let y = xr for r ∈ R.
dy
Prove that
= rxr−1 .
dx
Solution
r
y = xr = eln x = erln x .
dy
r
Then
= erln x = xr r x−1 = r xr−1 .
dx
x
Ç √
√ å
d 5 x+x−4 x
(b) Find
.
dx
x
Solution
√x
√ d
dx
5
+x−4 x
x
4 marks
3 marks
=
d
dx
√
=5
=5
√
Ä
x
x
5
√
x
1
x−1 + 1 − 4 x− 2
ln 5 2 √1 x x−1 + 5
ln 5
1
3
2x 2
−5
√
√
x 1
x2
x
ä
3
(−1)x−2 + 2 x− 2
+
2
3
x2
184
2012–2017 Past Test and Examination Questions Booklet
Question 2
-
Mathematics I (Major)
13 marks
2
(a) Consider the equation x4 + cos y + e x y = x2 + y2 .
dy
Find
. Show all your working.
dx
Solution
6 marks
2
x4 + cos y + e x y = x2 + y2
2
⇔ 4x3 − siny y0 + e x y (2xy + x2 y) = 2x + 2y y0
2
2
⇔ 4x3 − siny y0 + 2xye x y + x2 e x y y0 = 2x + 2yy0
2
2
⇔ 4x3 + 2xye x y − 2x = (siny − x2 e x y + 2y)y0
⇔ y0 =
2
4x3 +2xye x y −2x
sin y−x2 e x2 y +2y
(b) Find the differential dy if y = πt tπ .
Solution
dy = (πt lnπ tπ + πt πtπ−1 )dt.
Question 3
-
2 marks
13 marks
(a) Find the equation of the tangent line to the curve of
x
at x = 1.
f (x) =
x+1
Solution
x+1−x
1
f 0 (x) = (x+1)
= (x+1)
2
2
f (1) =
1
2
,
f 0 (1) =
1
4
= f (1) + f 0 (1)(x − 1)
y
4 marks
=
1
2
+ 41 (x − 1)
=
1
4
+ 14 x
(b) For what value(s) of a does the tangent line to the curve pass through
the point (−5, a ) ?
Solution
a = 14 + 41 (−5) = −1
3 marks
(c) Use linearization to approximate the value of f (1.2).
Solution
22 11
f (1.2) ≈ 14 + 41 (1.2) = 14 + 41 12
10 = 40 = 20
Question 4
Evaluate
Z
2
−1
Solution
√
u = x + 2,
-
2 marks
13 marks
x−1
√
dx.
x+2
u2 = x + 2,
8 marks
x = u2 − 2,
2u du = dx,
x = −1
u = 1,
x=2
u = 2,
Z 2
Z 2 2
Z 2
ï 3
ò2
ï
ò
x−1
8
1
u −2−1
u
2
√
dx =
2u du = 2
− 3u = 2
−6− +3
(u − 3) du = 2
u
3
3
3
x+2
−1
1
1
1
=-4 3
Mathematics I (Major)
3.6.3
185
2012–2017 Past Test and Examination Questions Booklet
Calculus Solutions June 2017
Question 1
-
4 marks
TRUE or FALSE:
Statement
R 0
R
s (t)r(t)dt = r(t)s(t) − s(t)r0 (t)dt
True/False
If f is continuous on an interval I, then f −1 exists and is also continuous on I.
F
The function g has an absolute minimum at c if g(c) ≥ g(x) ∀x in the domain of g.
F
r0 (s) = 0 ⇒ the point (s, r(s)) is a local extremum of the function r(s).
F
Question 2
-
T
4 marks
(a)
in the block how many times you need to use integration by parts in order to integrate
R Write
eu cos u du? DO NOT do the integration!
(1 mark)
2 (or any even number)
(b) Evaluate, showing all working:
(3 marks)
Z
0
Solution
Z
1
3
0
…
1
3
…
1
dt.
e − t2
1
dt = arcsin
e − t2
Å
t
√
Ç
1
3
= arcsin
= arcsin
Question 3
-
ã
e
√
Å
1
3
0
å
e
− arcsin
Å
0
√
e
ã
ã
1
√ .
3 e
7 marks
(a) Show all working to find
d x
dx x .
4 marks
Solution
Let y := x x and take logs of both sides:
ln y = ln x x = x ln x.
Then differentiate implicitly with respect to x
y0
1
= ln x + x = ln x + 1.
y
x
186
2012–2017 Past Test and Examination Questions Booklet
Thus, since y = x x we get
Mathematics I (Major)
d x
x = y0 = x x (ln x + 1).
dx
(b) Show all working to find the slant asymptote of h(t) =
2t3 −t2 −t+1
.
t2 −1
(3 marks)
Solution
Since
t
2t3 − t2 − t + 1 (t2 − 1)(2t − 1) − t
=
= (2t − 1) − 2
.
t2 − 1
t2 − 1
t −1
The slant asymptote of h(t) is y = 2t − 1.
h(t) =
Question 4
-
4 marks
(a) Find the 3rd degree Taylor polynomial of sin x at x = 0. SHOW ALL YOUR WORKING.
3 marks
Solution
Let f (x) = sin x.
Then f 0 (x) = cos x.
And f 00 (x) = − sin x.
And f 000 (x) = − cos x.
Hence, we have
f (0) = 0
f 0 (0) = 1
f 00 (x) = 0
f 000 (0) = −1,
so that
sin x ≈ P3 (x) = x −
x3
.
3!
(b) Differentiate your answer above to write down the 3rd degree Taylor polynomial of cos x at x = 0.
mark)
cos x ≈ P03 (x) = 1 −
Question 5
-
(1
x2
.
2!
9 marks
(a) Prove the following Theorem:
5 marks
If f 0 (x) = g0 (x) for all x in an interval (a, b), then f − g is a constant on (a, b), that is, f (x) = g(x) + C,
where C is a constant.
Proof
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
187
Let F(x) = f (x) − g(x).
Then F 0 (x) = f 0 (x) − g0 (x) = 0 for all x ∈ (a, b).
By the above theorem, F is constant on (a, b), i. e., f − g is constant on (a, b).
Thus there is C ∈ R such that ( f − g)(x) = C for all x ∈ (a, b),
i. e., f (x) − g(x) = C or f (x) = g(x) + C for all x ∈ (a, b).
(b) State l’Hôpital’s Rule in full. Hint: Start with “Let f and g be differentiable functions...”.
Statement:
Let f and g be differentiable functions where g0 , 0 near x = a (except possibly at a).
If we have that
lim f (x) = 0
and
lim g(x) = 0
lim f (x) = ±∞
and
lim g(x) = ±∞.
x→a
x→a
or that
x→a
x→a
(In other words, we have an indeterminate form of the type
lim
x→a
0
0
or
∞
∞ ).
f (x)
f 0 (x)
= lim 0
g(x) x→a g (x)
if the limit on the right hand side exists (or is ∞ or −∞).
Question 6
-
6 marks
Sketch a graph on the axes provided which satisfies the following criteria:
(a) f (x) is continuous for all positive real numbers.
(b) f (x) < 0 for x ∈ (0, 2) and f (x) > 0 for x ∈ (2, ∞)
(c) f 0 (x) > 0 for x < 5 and f 0 (x) < 0 for x > 5
(d) f 0 (x) = 0 when x = 5
(e) f 00 (x) < 0 for 0 < x < 7 and f 00 (x) > 0 for x > 7
(f) lim+ = −∞
x→0
Then
4 marks
188
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
y
6
f (x)
-x
−4−3−2−1 1 2 3 4 5 6 7 8 9 101112
?
Question 7
-
6 marks
You have 40cm of wire to form a square and a circle. How much of the wire should be used for the square and
how much for the circle to enclose the maximum total area?
Solution: Let s be the length of the side of the square. Let r be the radius of the circle.
Then the perimeter of the two shapes is 40 = 4s + 2πr. This means that r = 20−2s
π .
The area of the two shapes together (the optimisation function) is
A(s) = s2 + πr2 = s2 + π
Å
20 − 2s
π
ã2
1 2
(s π + (20 − 2s)2 )
π
1
= (s2 (4 + π) − 80s + 400)
π
=
and differentiating gives
A0 (s) =
1
(2s(4 + π) − 80).
π
For critical points, we need to find where A0 (s) does not exist (nowhere) and where A0 (s) = 0, which occurs when
40
1
(2s(4 + π) − 80) = 0 ⇒ s =
π
π+4
Other possible maxima occur at the end points (i.e., where s = 0 or s = 10)
Comparing the area for these three values gives:
A(0) =
A
Å
40
π+4
ã
ã2
ô
ã
40
+ 400
π+4
ï 2
ò
ï
ò
1 40
40 · 80
1 400(π + 4) − 1600
=
−
+ 400 =
π π+4
π+4
π
π+4
400
=
.
π+4
=
1
π
ñÅ
400
π
A(10) =
40
π+4
(4 + π) − 80
Å
1
(400 + π(100) − 800 + 400) = 100.
π
Mathematics I (Major)
189
2012–2017 Past Test and Examination Questions Booklet
400
Thus (since π+4
< 100 < 400
π ) the maximal area is obtained when all of the wire is used for the circle (and the
length of the side of the square is 0).
Alternative solution: The perimeter of the two shapes is 40 = 4s + 2πr. This means that s =
The area of the two shapes (the optimisation function) is
A(r) = s2 + πr2 =
Å
40 − 2πr
4
ã2
40−2πr
4 .
+ πr2
and differentiating gives
ã
Å
ã
40 − 2πr
π
20 − πr
A (r) = 2
(− ) + 2πr = −π
+ 2πr
4
2
2
Å
0
.
For critical points, we need to find where A0 (r) does not exist (nowhere) and where A0 (r) = 0, which occurs when
20
π
− (20 − πr) + 2πr = 0 ⇒ r =
2
π+4
Other possible maxima occur at the end points (i.e. where r = 0 or r =
40
2π )
Comparing the area for these three values gives:
A(0) = 100cm2
20
A
π+4
Å
A
Thus (since the largest area is
Å
400
π ),
20
π
Ç
ã
=
Ç
ã
=
20
40 − 2π π+4
4
40 − 2π 20
π
4
å2
Å
+π
å2
Å
+π
20
π+4
20
π
ã2
ã2
=
=
400
4+π
400 2
cm
π
the maximal area is obtained when all of the wire is used for the circle.
Calculus Total 40
3.6.4
Calculus Solutions August 2017
Question 1
-
TRUE or FALSE:
6 marks
190
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Statement
True/False
The derivative of cosh x is − sinh x.
F
For a decreasing function, the right-most endpoint will give the height for the
lower Riemann sum.
d
dx
R x2
Question 2
−5
e2t dt = 2xe2x
-
T
2
T
7 marks
(a) State and prove the Fundamental Theorem of Calculus 2. (7 marks)
Statement of Fundamental Theorem of Calculus 2:
If f is continuous on the closed interval [a, b] and F is an antiderivative of f on the interval [a, b], then
Z
a
b
f (x) dx = F(b) − F(a).
Proof of Fundamental Theorem of Calculus 2:
Partition the interval [a, b] into n subintervals with endpoints
a = x0 < x1 < x2 < · · · < xn−1 < xn = b
of length ∆x = xi − xi−1 .
By pairwise subtraction and addition of like terms, we can write as a telescoping series
F(b) − F(a) = F(xn ) − F(xn−1 ) + F(xn−1 ) − · · · − F(x1 ) + F(x1 ) − F(x0 )
n
X
=
[F(xi ) − F(xi−1 )].
i=1
Since F is an antiderivative, F is continuous on [xi−1 , xi ] and differentiable on (xi−1 , xi ) for each i. By the
Mean Value Theorem, applied to F on these subintervals, for each i = 1, . . . , n there exist ci ∈ (xi−1 , xi )
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
191
such that
F 0 (ci ) =
F(xi ) − F(xi−1 )
xi − xi−1
=
F(xi ) − F(xi−1 )
∆x
F(xi ) − F(xi−1 )
∵ F is an antiderivative of f, F 0 (ci ) = f (ci )
∆x
⇒ f (ci ) ∆x = F(xi ) − F(xi−1 )
n
n
X
X
⇒
f (ci ) ∆x =
(F(xi ) − F(xi−1 )) = F(b) − F(a)
⇒ f (ci ) =
i=1
⇒ lim
n→∞
⇒
Question 3
-
i=1
n
X
i=1
Z
a
b
f (ci ) ∆x = F(b) − F(a)
f (x) dx = F(b) − F(a).
7 marks
(a) Find the area of the region bounded by the graphs y = x2 and y = c in terms of the positive constant c. (4
marks)
√
c
x3
(c − x ) dx = cx −
√
3
− c
Z
2
ò √c
√
− c
√
√
√
√
( c)3 î
(− c)3 ó
− −c c−
=c c−
3
3
√ ã
Å
√
c c
=2 c c−
3
√
c c
.
=4
3
Or use the fact that it is an even function.
Or integrate with respect to y.
(b) Write down an expression for the area enclosed by x = cos y, x = 12 , y = 0 and y = 2π. DO NOT
evaluate your integral.
(3 marks)
Points of intersection: cos y =
Z
π
3
0
Question 4
-
1
2
=⇒ y =
1
(cos x − ) dx −
2
π
3
or
5π
3
Z
π
3
5π
3
1
(cos x − ) dx +
2
Z
2π
5π
3
1
(cos x − ) dx.
2
7 marks
(a) Write down an expression for the volume of the solid generated by rotating the region bounded by
y = 1x + 2 and y = −x + 10 about the line y = 0. DO NOT evaluate your integral.
(5 marks)
192
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
Points of intersection:
√
1
+ 2 = −x + 10 =⇒ x2 − 8x + 1 = 0 =⇒ x = 4 ± 15.
x
Thus the volume is given by:
Z
π
√
4+ 15
√
4− 15
Å
ã
1
(−x + 10)2 − ( + 2)2 dx.
x
(b) Write down an expression for the volume of the solid generated by rotating the region bounded by
(2 marks)
y = 1x + 2 and y = −x + 10 about the line y = 10. DO NOT evaluate your integral.
The volume in this case is:
Z
π
Question 5
-
√
4+ 15
√
4− 15
Å
ã
1
(10 − ( + 2))2 − (10 − (−x + 10))2 dx.
x
3 marks
Find the cross-sectional area of a pyramid with a height of h and a rectangular base with dimensions w and b,
sliced perpendicular to the y-axis. Note that the y-axis is perpendicular to the base of the pyramid which is
positioned at y = 0.
Cross-sectional area:
bw(h − y)2
h2
(You can give 2 marks if they assume it is square (i.e.,
b2 (h−y)2
h2
or
w2 (h−y)2
.)
h2
Mathematics I (Major)
193
2012–2017 Past Test and Examination Questions Booklet
Calculus Total 30 Marks
3.6.5
Calculus Solutions November 2017
Section A:
Multiple choice questions
[18 marks]
Instructions:
Questions MCQ1–MCQ6 are multiple choice questions.
In each of these questions, select the letter that corresponds to the correct answer.
(MCQ1) Which of the following is the appropriate partial fraction decomposition of
t4 + 9
t2 (t2 + 9)
t4 + 9
(B) 2 2
t (t + 9)
t4 + 9
X(C) 2 2
t (t + 9)
t4 + 9
(D) 2 2
t (t + 9)
(A)
t4 + 9
:
+ 9)
t2 (t2
A Bt + C
+ 2
t2
t +9
C
At + B
D
+
≡
+
t2
t+3 t−3
A B Ct + D
≡1+ + 2 + 2
t
t
t +9
A B
C
D
≡ + 2+
+
t
t
t+3 t−3
≡1+
(MCQ2) Which of the following statements is NOT TRUE?
Z 3
1
3
(A)
dx = .
2
4
0 (x − 4)
Z 3
Z 2
Z 3
1
1
1
(B)
dx =
dx +
dx.
2
2
(x
−
4)
(x
−
4)
(x
−
4)2
0
0
2
Z 3
1
X(C)
dx has an interior infinite discontinuity.
2
0 (x − 4)
Z 3
1
(D)
dx converges.
(x
−
4)2
0
1
(E) The integrand
is defined on (0 , 3) .
(x − 4)2
[3]
∞
X
cos (nπ)
.
n
Which of the following statements is TRUE?
(MCQ3) Consider the series
n=1
(A) The series is not conditionally convergent.
(B) The series does not converge.
(C) The series is absolutely convergent.
ß
™
cos (nπ)
(D) The sequence
diverges.
n
∞
∞
X
X
1
cos (nπ)
X(E)
=
(−1)n .
n
n
n=1
n=1
(MCQ4) Which of the following statements is TRUE about the series
∞
X
ln x + (ln x)2 + (ln x)3 + · · · =
(ln x)n .
n=1
[3]
194
2012–2017 Past Test and Examination Questions Booklet
Mathematics I (Major)
1
< x < e.
e
1
(B) The series converges for ≤ x ≤ e.
e
(C) The series does not converge for any value of x.
X(A) The series converges for
1
an+1
= 1 + → 1 as n → ∞.
an
n
(E) The series is convergent.
(D)
[3]
(MCQ5) Let g(x, y) = x2 y + cos y + y sin x.
Which of the following statements is TRUE?
(A) g x = 2xy + cos y + ycos x.
(B) gy = 2x − sin y + sin x.
(C) g x x = 2y − sin y + sin x.
X(D) g x y = gy x .
(E) gy y = 2x − cos y.
[3]
(MCQ6) Consider the differential equation (2xy − yn ) dx + x2 dy = 0.
Which of the following statements is NOT TRUE?
(A) For n = 1 the differential equation is variable separable.
(B) For n = 2 the differential equation is homogeneous of degree 2.
(C) For n = 0 the differential equation is exact.
X(D) For n = 0 the differential equation is variable separable.
(E) For n = 1 the differential equation is first order linear.
Section B
[3]
[72 marks]
In this section you are expected to SHOW ALL YOUR WORKING on the provided space to earn the
allocated marks.
X = 1 mark, (X) = 12 mark
Question 1
-
3 marks
(a) Find the partial fraction decomposition of
1
.
z (1 + z)
[3]
solution
1
A
B
≡
+
X
z+1
z 1+z
1 = A(1 + z) + B z.
When z = 0, A = 1 when z = −1, B = −1 X.
1
1
1
Thus
= −
X.
z+1
z 1+z
(b) Evaluate
Z
ln z
dz.
(1 + z)2
solution
Let u = ln z, du =
1
z
dz X
[7]
dv =
1
1+z2
1
dz, v = − 1+z
X
Mathematics I (Major)
ln z
(1+z)2
R
Question 2
dz =
2012–2017 Past Test and Examination Questions Booklet
R
ln z
1 1
− 1+z
− (− 1+z
z ) dz X X
=
ln z
− 1+z
+
=
R
ln z
− 1+z
+ ( 1z −
=
ln z
− 1+z
+ ln|z| − ln|1 + z| + C X X
-
195
1
(1+z)z
R
dz
1
1+z ) dz
X
3 marks
Evaluate the following integrals:
(a)
t
√
dt.
t4 − 1
Z
Hint : You may use the fact that
Z
sec θ dθ = ln |sec θ + tan θ| + C.
solution
2
Let
2t dt = sec θ tan θ dθ X X,
R t t = sec θ X. Then
R sec
1
θ tan θ
√
dt
=
dθ X X
4
2
tan θ
√
t4 − 1 =
√
[9]
sec2 θ − 1 = tanθ X.
t −1
(b)
=
1
2
=
1
2 ln|secθ
=
1
2 ln
R
secθ dθ X
+ tanθ| + C
√
t2 +
t4 − 1 + C X X
2
2s − 2
√
ds.
−s2 + 2s + 3
1
solution
u
= −s2 + 2s + 3
Z
[5]
du =
(−2s + 2)ds X
du =
−(2s − 2)ds
R
√ 2s−2
−s2 +2s+3
Thus
ds
√1
u
du X
=
−
=
R 1
− u− 2 du
=
−2u 2 + C X
=
√
−2 −s2 + 2s + 3 + C
R
1
196
2012–2017 Past Test and Examination Questions Booklet
R2
√ 2s−2
1
−s2 +2s+3
Question 3
ds
-
(a) Evaluate
Z
solution
=
√
− −s2 + 2s + 3 |21
=
−2
=
−2
=
√
4−2 3X
î√
−4 + 4 + 3 −
Ä√
√
ó
−1 + 2 + 3 X
ä
3−2
3 marks
1
dx.
(1 + x2 )arctan x
u = arctan x,
R
1
(1+x2 )arctan x
du =
dx
=
1
dx X,
1
+
R 1 x2
u du X
[4]
(1 + x2 )du = dx
=
ln | u | +C X
=
ln | arctan x | +C X
(b) Hence or otherwise determine whether the integral
Z
is convergent or divergent.
solution
R1
1
0 (1+x2 )arctan x
dx
=
=
=
=
=
Thus
Z
0
Question 4
1
Mathematics I (Major)
R1
1
lim
2
t→0+ t (1+x )arctan x
0
1
1
(1 +
x2 )arctan
x
dx
[4]
dx X X
lim ln |arctanx|1t
t→0+
lim ln(1) − ln | arctant |
t→0+
lim −ln | arctant | X
t→0+
∞X
1
dx is divergent X.
(1 + x2 )arctanx
-
3 marks
Consider the telescoping series
ã
∞ Å
X
1
2
1
−
+
.
r+2 r+1 r
r=1
ã
n Å
X
2
1
1
(a) Find the partial sum S n =
−
+
.
r+2 r+1 r
r=1
solution
[5]
Mathematics I (Major)
Sn
=
=
=
=
n
P
r=1
n
P
r=1
n
P
r=1
1
2
1
r+2
2012–2017 Past Test and Examination Questions Booklet
n
P
−2
r=1
1
r
−1−
1
r
−
−
1
n+1
3
2
+
+
1
2
1
r+1
1
n+1
+
1
n+1
1
n+2
+
+
n
P
r=1
1
r
+
1
n+2
1
n+2
−2
X
−2
n
P
r=1
ï
1
r
n
P
r=1
1
r
−1+
+2−
2
n+1
1
n+1
+
ò
n
P
r=1
+
n
P
r=1
1
r
XXX
1
r
X
(b) Hence or otherwise show that the series
ã
∞ Å
X
1
2
1
−
+
r+2 r+1 r
r=1
converges and find its sum.
solution
ã
Å
ã
∞ Å
X
2
1
1
1
1
1
1
−
+
= lim
−
+
X = X.
n→∞ 2
r
+
2
r
+
1
r
n
+
1
n
+
2
2
r=1
ã
∞ Å
X
1
2
1
1
Therefore
−
+
converges to X.
r+2 r+1 r
2
r=1
Question 5
-
is convergent or divergent.
If it is convergent, find the value to which it converges.
solution
n3 −1+n2 sin n
1+3n3
n→∞
= lim
n→∞
1−
n→∞
=
Since lim =
[3]
3 marks
(a) Determine, giving reasons, whether the sequence {an } =
lim
197
1
3
1
n3
1
n3
+ sinn n
+3
ß
n3 − 1 + n2 sin n
1 + 3n3
™
[4]
XX
(X)
1
(X) then {an } is convergent X.
3
(b) Determine whether the following series converge or diverge
(i)
∞
X
n3 − 1 + n2 sin n
.
1 + 3n3
solution
∞
X
1
n3 − 1 + n2 sin n
Since lim = , 0 X from part (a), therefore
is divergent X.
n→∞
3
1 + 3n3
n=1
n=1
[2]
198
2012–2017 Past Test and Examination Questions Booklet
(ii)
∞
X
(−1)n
n=1
Å
n3 − 1 + n2 sin n
1 + 3n3
ãn
.
Mathematics I (Major)
[4]
solution
By Root Test:
ã
Å 3
1
n − 1 + n2 sin n n
n3 − 1 + n2 sin n
n
n
→ by part (a) X.
(−1)
X
=
1 + 3n3
1 + 3n3
3
Since 31 < 1 X, by root test, the series converges X.
Question 6
Given that
-
3 marks
∞
X
1
= 1 + x + x2 + x3 + · · · =
xn .
1−x
n=0
1
.
(a) Write down the power series of
1 + x2
solution
1
1
= 1−(−x
2) X
1+x2
= 1 − x2 + x4 − x6 + · · ·
=
∞
P
[4]
XX
(−1)n x2n X
n=0
(b) Hence or otherwise find a power series representation of f (x) = arctan x.
solution
R 1
arctan x + C = 1+x
2 dx X
=
R P∞
=C+
n=0 (−1)
∞
P
n=0
n
(−1)n
x2n dx X
x2n+1
2n+1
XX
π
(c) Write as a series.
4
solution
Putting c = 0, x = 1,
∞
X
1
π
(−1)n
= arctan 1 X =
X.
4
2n + 1
n=0
Question 7
-
[4]
3 marks
2
2
x
x
dx y2 + y2 e y + 2x2 e y
Consider the differential equation
=
.
2
x
dy
y
2x y e
(a) Using the substitution x = vy or otherwise, find the general solution of the
[2]
Mathematics I (Major)
2012–2017 Past Test and Examination Questions Booklet
differential equation.
solution
dx
dv
x = vy;
=v+y
XX
dy
dy
x 2
x 2
y2 +y2 e( y ) +2x2 e( y )
dx
=
2
x
dy
2xye( y )
2
y2 +y2 ev +2v2 y2 ev
2vy2 ev2
=
1+ev +2v2 ev
2vev2
dv
y dy
=
1+ev +2v2 ev −2v2 ev
2vev2
dv
y dy
=
1+ev
2vev2
2
2
2
2
2
[10]
2
=
dv
v + y dy
199
XX
X
2
XX
Z
2
2vev
1
dv
=
dy X
y
1 + e v2
2
ln 1 + ev = ln |y| + C X
2
x
= ln |y| + C X
ln 1 + e y
Z
(b) Find the particular solution of the differential equation given that y = 1 when x = 0.
solution
y = 1, x = 0
ln 1 + e0 = ln(1) + C. Thus C = ln(2) (X).
The particular
solution of the differential equation
ln 1 + e
x
y
2
2
x
= ln |y| + ln2 (X)
ln 1 + e y
= ln |2y|
2
x
1 + e y = 2y X
[2]