Ad^oanced Level
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•URE MATHEMATICS
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Hung Fung Book Co., Ltd
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Advanced Level
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PURE MATHEMATICS
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Calculus anol Analytical Geometry I
K. S. Ng b.scmhon
Y. K. Kwok b sc
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Hung Fung Book Co., Ltd.
i-.D
HUNG FUNG BOOK CO., LTD.
© HUNG FUNG BOOK CO., LTD.
All rights reserved. No part of this
publication may be reproduced
without the permission of the
copyright owner.
First edition - Feb., 1993
Second edition - Sept., 2000
Third edition - Aug., 2004
Fifth published - July, 2008
ISBN 962-354-177-5
PUBLISHED BY
HUNG FUNG BOOK CO., LTD.
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Advanced Level Pure Mathematics
i
[Preface to tlhe Thard Edffion
This book follows lhe new 2006 HKAL Pure Mathematics syllabus issued by the Curriculum
Development Committee of Hong Kong and that by the Hong Kong Examinations and Assessment Authority.
In this revised edition, the core content of the HKAL Mathematics syllabus are clearly presented and
organized in logical sequences. In order to make the core content more concise and comprehensive, some
changes are found in this new edition. However, the main characteristics of the previous edition are
maintained. This book is unique in that it covers most aspects that have been deleted in the 2004 HKCEE
Additional Mathematics syllabus. The main changes are as follows:
(1) Deletion on some subject matter
From 2006 and onwards, the HKAL Pure Mathematics syllabus will be changed. In order to match
the new teaching and examination syllabus, the following topics have been deleted:
(a)
the units “The greatest coefficient and the term with the greatest absolute value” and “The
binomial series” in the chapter of Binomial Theorem:
(b)
the unit “Vectors and linear equations” in the chapter of System of Linear Equations;
(c)
the unit “improper integrals” in the chapter of Definite Integrals
(d)
the units “Arc length” and “Surfaces of revolution” in the chapter Applications of Definite
Integrals;
(e)
the whole chapters on Vectors, Three Dimensional Coordinates Geometry and Polar
Coordinates.
(2)
Addition on some new materials
Similarly, because of the change of the new HKCEE Additional Mathematics syllabus, some
examples and exercises have been added to some chapters to make this third edition more
comprehensive.
ii
Advcith ed Level Pure Maihcmaties
(a) Since the method in solving fractional inequalities and inequalities involving absolute value
have been deleted in the new Additional Mathematics syllabus, some simple examples on
these topics have been included in the chapter Inequalities.
(b) As the unit complex numbers has been deleted in the new Additional Mathematics syllabus,
the chapter on Complex Numbers has been expanded to help the students to have an easier
way to understand this new concept.
(c)
Some of trigonometric properties are not included in the new Additional Mathematics syllabus,
and so we have added some examples in the chapter Revision on Trigonometry.
(d)
In order to help the students to have a belter understanding of method of substitution in
integration, which has been deleted in the new Additional Mathematics syllabus, more examples
have been added.
Moreover, as vectors has been deleted, minor adjustment on the approach of applications in
“Coordinates transformation" in the chapter Matrices and Determinants has been done to meet the
requirement of the new syllabus.
(3) Restructuring of the sequence of some chapters
Chapter 10 on Complex Numbers in the Second Edition has been swapped to Chapter 7 in the Third
Edition. The main purpose is to organize the material in a more logical sequence.
(4) Miscellaneous exercises are included
To present a concise and comprehensive coverage of the core content of the HKAL Mathematics
syllabus, some miscellaneous exercises are included at the end of each sequence of chapters. Those
miscellaneous exercises are all related to the HKAL past examinations paper; others are some cross
topics exercises, which students would found them very useful. The core materials of those
miscellaneous exercises are divided into 2 sections: Section I and Section II. In section I, exercises
that are related to recent HKAL past examination papers are included and are classified according
to different topics. After accumulating the necessary knowledge, students can revise those exercises
according to the lime available, probably during the holidays or just before the start of the HKAL
examination. Whereas in section II, cross topics exercise are found. Those exercises are interrelated and that students could concentrate on one main aspect in this section but throughout still
show its relationship to the other aspects. This section could also serve as an overall revision of all
the main aspects highlighted in this book.
I
Preface
iii
We are indebted to the following people for their helpful comments, criticisms and advice: Mr. C.
H. Ip and Mr. Y.M. Chau. We also owe a debt of gratitude to Mr. C.C. Yiu and Mr. L. Fu who have spent
much of their precious time in proofreading the whole manuscript. Moreover, we would like to express our
deep appreciation to Mr. K.K. Lee and his colleagues of Hung Fung Book Co, Ltd; who have assisted in
publishing this book. Without their tireless, wholehearted, unique contributions, this book never would have
come about.
K. S. Ng & Y. K. Kwok
June, 2004
iv .Advanced Le vel Pun Mathematics
Preface to the Second Edffiom
The first edition of the books, which have been published for nearly ten years, is time to be revised
so as to keep abreast of the latest developments in the educational field. The topics of the present revised
textbooks, in three volumes, are written according to the latest syllabus for Hong Kong Advanced Level
Pure Mathematics issued by the Curriculum Development Committee of Hong Kong and that by the Hong
Kong Examinations Authority.
In this revised edition, the systematic approach presented in the first edition is maintained. However,
in order to provide a more comprehensive text for students working towards Hong Kong Advanced Level
Examinations, we have decided to add more exercises and examples in the new edition so that students
would consolidate the knowledge learned. In addition, more illustrations and diagrams are used to increase
students' understanding, and to stimulate their interest in studying Pure Mathematics.
There are four main changes in this new series:
(1) Restructuring of the sequence of some chapters
Chapter 7 on “Complex Numbers" in Book 1 of the first edition has been swapped to the last
chapter in the new edition. On the other hand, the application of L'Hospitafs Rule and the theory
and application of monotonic function in chapter 4 in Book 2 have all been edited to chapter 5
“Applications of Differential Calculus". As for chapter 10, the theory of polar coordinates in Book
2 has now been shifted to Appendix I of Volume 3, while the application of integral calculus on
polar coordinates has been linked to chapter 7 in the new series.
(2) Addition on some new chapters
The entirely new sections “Use of Summation Sign and Product Sign” and “Finite Series" have
been added to help explain and emphasize points. It would also help to broaden students’ scope of
knowledge.
Prefat. c
v
(3) The format of exercises in the first edition are basically kept
Revision exercises are only added to the chapter on “Mathematical Induction”. In addition, with
gratitude for the Hong Kong Examinations Authority on the permission to add the latest questions
from Advanced Level Pure Mathematics Examinations up to year 1999, students would be well
prepared to sit for the examination after being familiarize with the examination rules and format
found in these books.
(4)
Deletion on some subject matter
The theory on “Taylor’s Theorem” from chapter 4, Book 2 of the first edition has been deleted in
the second edition.
These books would have never been possible without the efforts of many people.
Our special thanks to Mr. S.H. Cheung, who have truly been our inspiration throughout the editing
process of these books. Mr. Cheung had translated these books into Chinese version and had made some
additional information on them. We have made some references to his books; as a result, these books are
more comprehensive and improvement has been made in many aspects.
Last but not least, we all owe a debt of gratitude to Mr. C.H. Ip, Miss S.L. Yau, Mr. K.W. Lam and
Mr. S.F. Tsang who have devoted much time in proofreading the whole manuscript. In addition, we would
like to express our heartfelt thanks to Mr. Chan Yiu-tung and Mr. Lee Ying-chun of Hung Fung Book Co.
Ltd., who have assisted in the preparation of the books and materialized their publication. Without their
important contribution and valuable comments, the books could not have been completed smoothly.
K. S. Ng & Y. K. Kwok
June, 2000
vi Advanced Li ve! Pure Mathematics
Preface to the Forst Edffiom)
The Hong Kong Advanced Level Pure Mathematics subject provides essential preparation in
theories and techniques in Algebra, Coordinate Geometry and Calculus for science students pursuing further
studies in Science and Engineering in tertiary' institutions. For many years, there has always been a lack of
basic textbooks written specifically for this Hong Kong Advanced Level Pure Mathematics subject. The
topics of the present textbook, in two volumes, are chosen exactly according to the new Hong Kong
Advanced Level Pure Mathematics syllabus.
The topics are organised as follows: Book 1 deals with topics in Algebra, Matrix and Solution of
Linear Systems, and Book 2 deals with Coordinate Geometry and Calculus.
This book is a complete textbook, not a study guide. Each chapter begins with an introductory
remark that motivates students’ interest in the topic. Concise and clear definitions of various mathematical
terms are listed systematically. Theorems and proofs of important results are presented succinctly and
logically. Numerous worked examples are included to illustrate the solution techniques and concepts behind
mathematical problems. A sufficient number of exercises are given and they are directly related to the
content of the chapter.
We would like to express our sincere thanks to Mr. C. K. Wong, Mr. C. H. Ip and Mr. Y. M. Chau
who have proofread the whole manuscript and made helpful comments. Also, special thanks go to Mr. Chan
Yiu-tung and Mr. Lee Ying-chun of Hung Fung Book Co. Ltd., who have assisted in the preparation of the
book and materialized its publication.
K. S. Ng & Y. K. Kwok
Feb., 1993
i
Advanced Level Pure Mathematics
Glossary of SymboQs
a gA
a is an element of the set A
(a ^,^A
n(A)
the number of elements of a finite set A
0
the empty set
(£ft)
N
the set of natural numbers {1,2, ...}
Z
the set of integers
(SK®)
Q
the set of rational numbers
R
the set of real numbers
C
the set of complex numbers
(W®)
A cz B
A is a subset of B
(A&B MTM)
u
the union of sets
(Wft)
the intersection of sets
(X®)
A'
the complement of the set A
(USA
B\A
the relative complement of a set A in another set B
(®a
[a, b]
the closed interval {a* g R : a < a* < b}
(fflUIHJ {xeR:a<x<b})
vii
viii Advanced Level Pure Mathematics
(t?. b)
the open interval (x e R : a < x < b}
(P3 Ha IW {.v g R : a <x < /;})
f: A —> B
f is a function from the domain /I to the range B
(f &&MA SIM B Wl®®)
f[A]
the direct image of the function f: A
(S’M f:
r’[B]
the inverse image of the function f: A -» B
(S’® f: A->B lWf&)
f: B-*A
the inverse function off: A —> B
B
(f: A^B iWM)
g°f
the composite function of the two functions f and g, and it is defined as g ° f(x) = g[f(x)]
(f &. g W
g o f(x) = g[f(x)J)
pn
the number of permutations of n different objects taken r at a time
(f# n
c;
(n\
Wft r
the binomial coefficient, the number of combinations of n different objects taken r at a time
the binomial coefficient
VJ
ex, exp(.r)
the exponential function with base e
In x, logrx
the logarithm of x to base e
Re z
the real part of z
(z WK)
Im z
the imaginary part of z
(z wnifS)
the modulus of z
(z w^W)
arg z
the argument of z
(zWfe’ft )
z
the complex conjugate of z
M
the inverse of the matrix M
the transpose of the matrix M
(W M fiWH W)
Glossary of Symbols ix
det M, | M |
the determinant of the square matrix M
CfrM* M iW^JA)
AB
the vector represented in magnitude and direction by the directed line segment AB
ab
i»j lit)
a
the vector a (students are advised to use a in their scripts)
(I°Jii a (^-^’If'f
a
Jfl a ))
a unit vector in the direction of the vector a
$afWW»
ij
unit vetors in the direction of the cartesian coordinate axes (students are advised to use i ,
J* in their scripts)
|a|
the magnitude of the vector a
(a M^tt)
a•b
the scalar product of the vectors a and b
(a5b
f'(x), f"(x),
the first, second,/ith derivatives of f(x) with respect to x
...,f(n)(x)
(f(x) W x M~Pg, —Pit
n IWS&)
the first, second derivatives of x with respect to t
(xWrW-P^S-PgW)
Advanced Level Pure Mathematics xi
Advanced Level Pure Mathematics
Algebra
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Statement Calculus
Sets
Mathematical Induction
Polynomials and Rational Functions
Binomial Theorem
Inequalities
Complex Numbers
Matrices and Determinants
Systems of Linear Equations
Appendix I Summation Sign and Product Sign
Appendix II Finite Series
Appendix III Revision on Trigonometry
Calculus and Analytical Geometry I
Chapter 1
Chapter 2
Functions
Limit of a Sequence
Chapter 3
Chapter 4
Limit and Continuity of a Function
Chapter 5
Derivatives
Applications of Differential Calculus
Calculus and Analytical Geometry II
Chapter 6
Indefinite Integrals
Chapter 7
Chapter 8
Definite Integrals
Applications of Definite Integrals
Chapter 9
Two Dimensional Coordinate Geometry
Advanced Level Pure Mathematics
xiii
Contents
Preface
i
Glossary of Symbols
vii
Chapter 1
Functions......................................................
.1
1.1
Introduction....................................................
1.2
Direct Images and Inverse Images................
1.3 Composition of Functions..............................
1.4
Identity Function and Inverse Function........
Injective, Surjective and Bijective Functions
1.5
1.6 Some Special Real Functions........................
1.7
Elementary Functions.....................................
Exercise 1A...............................................................
Exercise 1 B...............................................................
Exercise 1C...............................................................
Exercise 1 D..............................................................
Revision Exercise.....................................................
. 1
. 4
. 9
14
17
27
40
.. 8
14
25
47
50
Chapter 2
Limit of a Sequence.........................
55
Introduction..........................................
Convergent and Divergent Sequences
Operations on Limits of Sequences....
2.4 Sandwich Theorem for Sequences....
2.5 Monotonic Sequences.........................
2.6 The Number e......................................
Exercise 2A....................................................
Exercise 2B....................................................
Exercise 2C....................................................
Exercise 2D....................................................
Revision Exercise...........................................
55
56
60
71
78
2.1
2.2
2.3
88
68
76
85
90
90
xiv
Advanced Level Pure Mathematics
Chapter 3
Limit and Continuity of a Function
3.1
3.2
3.3
3.4
3.5
3.6
Introduction...............................
Limit of a Function...................................
Properties of Limit of a Function............
Two Important Limits..............................
Left and Right Hand Limits.....................
Continuous Functions.............................
Properties of Continuous Functions......
3.7
Exercise 3A........................................................
Exercise 3B........................................................
Exercise 3C........................................................
Exercise 3D.......................................................
Revision Exercise...............................................
Chapter 4
105
112
115
120
104
112
114
130
131
Derivatives......................
134
Introduction.................
4.2 Differentiability...........
4.3 Rules of Differentiation
4.4 Differentials.................
4.5 Higher Derivatives......
4.6 Mean Value Theorem .
Exercise 4A...........................
Exercise 4B...........................
Exercise 4C...........................
Exercise 4D..........................
Exercise 4E............................
Revision Exercise..................
134
4.1
Chapter 5
94
..94
..95
..98
Applications of Differential Calculus..............
5.1
5.2
5.3
L'Hospital's Rule..........................................................
Tangents.......................................................................
Rate of Change............................................................
5.4
5.5
5.6
Monotonic Functions..................................................
Maxima and Minima..................................................
Optimization...............................................................
5.7
Curve Sketching..........................................................
5.8
Proving Inequalities by Using Differential Calculus
Exercise 5A............................................................................
Exercise 5B............................................................................
Exercise 5C............................................................................
Exercise 5D.........................................................................
Exercise 5E................................
139
148
163
167
180
146
161
166
177
189
190
194
194
204
206
212
216
223
228
254
202
206
211
215
222
('( hi tent v
Exercise 5F..........
Exercise 5C........
Exercise 5H........
Exercise 51..........
Exercise 5J..........
Revision Exercise
XV
227
233
241
253
259
261
Miscellaneous Exercise A
275
Answers
289
Index
311
Advanced Level Pure Mathematics 1
Chapter 1
Funictaons
1.1
Introduction
When we observe some natural processes or some engineering processes, we always encounter a lot of
variables. These variables are usually not independent. For example, if 5 and t respectively denote the
distance and time after a body is released from rest under gravity, from the law of physics, 5 and t are
connected by the relation
5 = — gt2
(*)
26
where g is the acceleration due to gravity. In this circumstance, 5 is said to be a function of t and, 5 and t are
called the dependent variable and independent variable respectively.
Before proceeding to the discussion on the characteristics of some particular functions, we first define
the idea of functions from the concepts of sets.
Definition 1-1
A function {mapping} f from a set A to a set B, is a relation connecting the elements of the two sets A
and B. This relation possesses the property that for every element x in A, there exists a unique
element y in B such that y - f(x). Usually, a function is denoted by
f: A —> B
The set A is called the domain while the set B is called the range.
The element y is said to be the image of x and x is the pre-image of y.
A function may be illustrated by a diagram as shown below:
A
f
B
1
3
2
5
3
7
4
9
Fig. 1.1
2 (
As shown in Fig. 1.1, every element of the domain A has one and only one image. Moreover, the elements of
the two sets A and B are connected by the relation
f(x) = 2x + 1
Hence, f is a function from the set A = {1, 2, 3, 4} to the set B = {3, 5, 7, 9} defined by for all x eA,
f(x) = 2x + 1
A
g
1
B
A
2
1
h
2
3
2
2
4
4
3
4
B
3
5
4
6
6
(b)
(a)
Fig. 1.2
Now, consider the relations g and h represented in Fig. 1.2. In (a), the element 4 of the domain A does
not have any image, i.e. g(4) is not defined. In (b), the element 3 of the domain A of h has two different
images 4 and 6, i.e. h(3) = 4 and h(3) = 6. Hence, according to the definition, both g and h are not
functions.
A
k
1
B
3
2
4
3
4
6
Fig. 1.3
However, in Fig. 1.3, although k(l) = k(2) = 3, every element of the domain A still has a unique
image and so k is a function.
A function from R to R can also be represented by a curve on the xy-plane. For example, the graph
representing the function f: R -> R, defined by f(x) = x2, Vx 6 R, is the parabola y = x 2 on the xy-plane.
As shown in the previous discussion, a function that can be expressed by a simple equation is much
easier to understand. Moreover, the manipulation of such functions is simpler. However, some functions
cannot be represented by an equation easily. The function k in Fig. 1.3 is an example.
i A.
Functions 3
Example 1-1
Determine whether the following equations are functions from R to R.
(a) y = 2x
(b) y
_l_
x
(c) y< = x
Solution
(a) Since x is a real number which implies that 2x is also a real number, every element x
of R has an image y = 2x in R.
Further, for all xp x2e R, x, = x2 => 2xj = 2x2. That is, the image of every element of
the domain is unique.
Hence, the given equation represents a function from R to R.
(b) Since 0 e R and the reciprocal of 0 is undefined, the image of 0 does not exist and so
y
— is not a function from R to R.
x
(c) Since 4g R and note that (-2)2 = 22 = 4, the element 4 has two different images,
namely, 2 and -2, and so it is not a function.
In Fig. 1.1, f is a function from the sei /I = {1,2, 3. 4} to the set B = {3, 5, 7, 9} defined by the relation
f(x) = 2x + 1
(*)
With this relation, f can be extended to be a function from the set A, = {xeR: 0 < x < 4} to the set
= {yeR: y = 2x + 1, where xeR}. Similarly, it can also be extended to be a function from R to R.
Hence, the domain of the function is extended from the set A = {1, 2, 3, 4} to the set A, = [0, 4] and then to
the set of all real numbers R. The largest domain that can be defined by this relation is called the most
extensive possible domain of the function f. In this example, the most extensive possible domain is the set of
all real numbers R. The set of all the images of this most extensive possible domain is called the smallest set
of range.
Moreover, as in Example 1-!(/?), y = — is not a function from R to R. However, it is a function from
x
R* to R, where R* is the set of all positive real numbers. Also, it is a function from R \ {0} to R. In this
case, the most extensive set of domain is R \ {0}.
Example 1-2
State the most extensive possible domain of real numbers for each of the given relations to
be a function.
(a) y = ^x + l
(b) y =
X
X2 - I
(c) y = log io*
4
Chapter /
(a) In order that Vx +1 is a real number,
is
{xgR: x>-1}
Solution
1 > 0. Hence, the most extensive domain
(b) Since y is undefined when x = ±1, the most extensive domain for this function is the
set of all real numbers excluding ±1, i.e.
R\{1,-1}
(c) Since y is well-defined only for positive real values of x, the most extensive domain
for this function is the set of all positive real numbers.
Example 1-3
Find the smallest set of range of real numbers such that y =
x
X2 +l
is a function with the
set of all real numbers as its domain. That is, find the smallest set which contains all
images of the function.
Solution
As the domain of the function in this problem is R. for any x gR. the image of x exists
and let it be y. Then
y(x2 + I) = a*
That is,
yx2 - x + y = 0
(*)
Hence, by considering the equation (*) as a quadratic equation in x, the equation (*) must
have real roots. Such condition requires that its discriminant must be non-negative, i.e.
I2-4(y)(y) > 0
(1 —2y)(l+2y) >0
I . . 1
2~y~2
Hence, the smallest set of range is
{yeR:
1.2
Direct Images and Dnverse Images
Image and pre-image of a function f are single elements. The extension of the concept from these single
elements to sets are the direct image and inverse image. They are subsets of the range and the domain of f
respectively. Their definitions are given as follows.
Functions 5
Definition 1-2
Let f : A
B be a function from A to B and X be a subset of A, Y be a subset of B. Then the direct
image of X under f, denoted by f [X], is defined as
f [X] = {>’: y = f(x) for some xgX);
and the inverse image of Y under f, denoted by f"'[ T], is defined as
f"’[y] = {x: y = f(x) for some yG/}.
In particular, when X = A, we call f [A] the codomain of f and f [A] c B.
As shown in Fig. 1.1, if X={l,2}czA and Y = {3, 9} cz B, then the direct image of X under f is
f[XJ = {3, 5} and the inverse image of Y under f is f”*[ K] = {1,4}.
Example 1-4
Let [a, b] denote the closed interval {xgR: a <x< b} and (a, Z?) denote the open
interval {xgR : a<x<b}.
Find the direct image of the set X under the function f in the following.
(a) f : R —> R defined by f(x) = x2, Vx g R; X= [—1,2].
(b) f : Z
Z defined by f(x) = 2x, Vxg Z; X = {x : x is a positive integer less than 8).
(c)
7t) —
Vxg
(c) f
f:: (0,
(0, n)
—>> R
R defined
defined by
by f(x)
f(x) =
= cotx,
cotx, V
xg(0,
(0, 7c):,
zr); X
X=
= (-^,
(—, -y).
—).
Solution
(a) We split the set X into the non-negative and negative numbers intervals, i.e. [-1,0)
and [0, 2]. Then
and
xg[-1,0)
x2g(0, 1]
X g [0, 2]
x2 g [0, 4]
Hence, the direct image of X under f is
f[X] = (0, I] u [0,4] = [0,4].
(b) If x is a positive integer less than 8, then 2x is a positive even integer less than 16.
Hence,
f [X] = {x : x is a positive even integer less than 16}
7T
(c) V.vg(-
71
4’ 2
), 0<cotx<l. Hence, the direct image of X is
fLX] = (0, I)
Example 1-5
Find the inverse image of the set Y under the function f in the following.
(a) f : R-> R defined by f(x) = | x |, Vx g R; Y = [0, 3].
(b) f : Z —> Z defined by f(x) = 3x, Vx gZ; Y = {y : y is a positive integer less than 7}.
(c) f: [0, 2/r] —> [-1, I] defined by f(x) = cos x, Vx g [0, 2/r]; Y = [0, 1].
6
Chapii
I
Solution
(a) Since the images of .r and -v under the function f(x) = | x | are the same, the inverse
image of Y under f is [-3, 3].
(b) Since the pre-image of every element of Y must be an integer, only 3 and 6 of Y have
pre-images. Hence, the inverse image of Y under f is {1,2}. Note that some elements
in Y do not have their own pre-images.
3/r
71
(c) For 0 < a* < — or —< x < 2?r, 0 < cosx < 1.
2
2
Hence, the inverse image of Y is
[0,|]u[^,2^]
Example 1-6
Define a function f: R2 —» C by f((fl, Z?)) = a + ib, V(a, b) g R2.
Find f [XI and f ’[ FJ where X= {(fl, Z?) gR2: a = b} and T= {z g C : Re z =-Im z}.
Solution
Let z = x + iy, where x, y e R, be the image of (a, b) under f.
For (a. Z?) g X, a = b and so,
x + iy = a + ib
= fl + ia
By comparing the real parts and imaginary parts on both sides, we have
x=a
and
y=a
Eliminating a to give
x-y = 0
and hence
f [X] = {x + iy e C: x - y = 0}
Similarly, if z = x + iy, where x, y e R, is an element of T, then x = -y and so, for some
(a, b) 6 R2.
x - ix = a + ib
Hence,
x = a = -b
and so,
a+b=0
Therefore
f,[TJ = {(a, b)eR2:a + b = 0}
which represents a straight line passing through the origin.
The properties of direct images and inverse images are summarized in the following 3 theorems. The proof
of these results is beyond the scope of this text.
Functions 7
Theorem 1-1
Properties of Direct Images
Let f : A —> B be a function from A to B. For any subsets
(1)
X,cX2 => f[X,] a f[X2]
(2)
nX.nXJaftXJnftXJ
(3)
f[X,uX2] = f[X1]uf[X2]
and X2 of A we have
To verify Theorem 1-1(2) and (3), we consider f: R —> R defined by
f(x)=x2
(Vx g R)
Take X, = [-3, 1] and X2 = [-1, 2] so that
X,nX2 = [-l, 1J
and
X, u X2 = [-3, 2]
f[X,r>X2| = |0. 1]
and
f[X, uX, |
f[X, ] = [0, 9]
and
f[X2] = [0, 4]
f[XInX2]cf[X1]nf[X2]
and
f[X1uX2] = f[X,)uf[X2]
and hence,
[0,9]
On the other hand.
Therefore,
Theorem 1-2
Properties of Inverse Images
Let f : A -» B be a function from A to B. For any subsets T, and Y2 of B, we have
(2)
r,c=y2 => f-*[ y,] cz f-*[ r2]
f-1[y1nr2] = f-'[yI]nr'[}'2]
(3)
f-|[yluy2] = rl[y1]of-|[r2]
(4)
f-'[B\y,j=A\r'[yl]
(5)
f-'[y,\y2] = f-'[y,]\f-,[y2]
(1)
The readers should try to find some examples to verify these properties.
Theorem 1-3
Let f : A —> B be a function from A to B. Then
(1)
For any subset X of A, X cz f"’[ f [X] ].
(2)
For any subset Y of B, f [ f"’[ T] 1 cz Y.
8
Chapter I
To verify Theorem 1-3(1) and (2), we consider f: R —> R defined by
(Vx 6 R)
f(x) = x2
By taking X= [-1, 2], we obtain
and
f[X] = [0.4]
[-2, 2]
r’tHX]]
Hence, we have
Xcf"l[f[X]]
Similarly, by taking /=[-!. 4]. we obtain
r’[ y]
and
[-2, 2]
ar'mwo, 4]
Obviously.
ftf-’micy
Exercise 1A
1.
2.
Determine whether the following relations f from set A to set B are functions.
1x
—
Z and f(x) = 7, Vx 6;.1
A
(a)
A = Z, B
(b)
A = Z, B = Z and f(x) = 3x + 1, Vx 6/4
(c)
A = R, B = R and f(x) = x2, Vx 6/4
(d)
/4 = R. £ = R and f(x) = 101, Vx 6/4
(e)
A = R, B = R and f(x) =
(f)
A = R. B = R and f(x) =
1
, Vx 6/4
1 + x2
X
(g)
, VxgA
1 + x3
A = [-l, 1] = {xgR: -1 <x< 1}, B = [0,7Fj = {x g R: 0 < x < 7r} and f(x) = sin *x, Vx 6/4
(h)
A = (-1, 1) = {xgR: -1 <x < 1}, B = R and f(x) = l-'|x| , Vx e A
(i)
A
C, B
R and f(z) =
1
-. VzgX
M- z
Slate the most extensive possible domain of real numbers for each of the given relations to be a
function.
(a)
(c)
f(x) =
x
x2+l
f(x) = tanx
(b)
f(x) =
(d)
f(x) = cos-1(—)
x-|x|
X
(e)
f(.r) = x/(x-l)(x +2)
(f)
f(x)
| x|(x-1)
!■ unction's 9
3.
Slate the most extensive possible domain of complex numbers for the relation f(z) =
function.
l-|z|
to be a
4.
Slate the most extensive possible domain of integers for the relation f(x) = x + 1 to be a function with
2
the set of all integers as its range.
5.
Find the smallest set of range of real numbers for each of the given relations to be a function with the
set of real numbers as its domain.
(a)
f(x) = x2 - 4x + 1
(b)
(c)
f(x) = cscx
(d)
sin x
(?)
f(x)
(f)
f(x) = x-|x|
X2 - 1
x2 +1
f(x)
cot" X
6.
Let f : R —» R be a function defined by f(x) = 5x, Vx g R. Find the direct image f [Z] and the inverse
image f"’[Z].
7.
Let f: Z —> R be a function defined by f(x)
8.
Let f : C \ {0}
cos tlx, X/xeZ. Find f[Z] and f '[{I}].
C be a function defined by f(z) = —, Vz g C \ {0}.
Find f [X] and f"'[ Y], where X
z
{z eC: | z | = 1} and Y = {z g C : Re z = 1}.
9.
Construct examples to verify the results stated in Theorem 1-2.
10.
Let c be a real number not equal to 1. Define f: R2 —> C by f((x, y))
1—c
V(x, y) e R2. Find f [X]
and f"'[ T], where X = {(x, y) e R2: x = y} and Y = (z e C : | z | = 1}.
1.3
Composition of Functions
Functions can be combined to form a new function. This operation is called the composition offunctions
and it is defined as follows.
Definition 1-3
Let f : A —> B and g : B —> C be two functions. The composition of f and g, denoted by g o f, is a
function from A to C such that for all a eA, go f(^) = c with f(a) = b and g(b) = c, where b eB.
10 Chapter I
Remark 1-1
(1) From the property that
f(fl) = b
and
g(Z?) = c
it is obvious that b is an element of the codomain of f.
Hence, the composite function g ° f is defined only when the codomain of f is a
subset of the domain of g. That is,
f[A]cB
(2) For all a g A, we have
f(fl) = b
and
g(/?) = c
and so,
g o f(fl) = g(/j) = g[f(fl)] = C
(1-1)
(3) The concept of composition of functions may be illustrated in the following diagram:
b = f(a)
c= g(b)
a
A
C
= gof(a)
gof
Fig. 1.4
Example 1-7
Fig. 1.5 shows two functions f and g. Express gof in a mapping diagram. Find also the
direct image of A under gof.
A
1
f
B
B
a
2
3
4
b
C
Fig. 1.5
■
I unciions 11
Solution
With the help of mapping diagram, the composite mapping g ° f can be found as shown
in Fig. 1.6:
2
3
4
Fig. 1.6
This result is simplified in Fig. 1.7:
2
3
Fig. 1.7
In Fig. 1.7, the direct image of A under g ° f is the singleton {x}.
Remark 1-2
Note that f <> g is undefined in Example 1-7.
Example 1-8
Let f and g be two mappings both from R to R such that
f(x) = -2x
and
g(x) = x2
find g o f and fog.
Solution
By definition, both g o f and f o g are mappings from R to R. We have to find the image
of an element x of the domain under the compositions. This can be done in two ways:
Method I
By Mapping Diagram
As shown in Fig. 1.8, a typical element x is mapped to -2x under f and —2x is then mapped
to (-2x)2 = 4x2 under g.
g O f(.r) = 4.r
(Vx g R)
12 Chapter I
g
R
x2
(-2x)2=4x2/
Fig. 1.8
With similar argument, as shown in Fig. 1.9, the element x is mapped to x2 under g and x2
is then mapped to -2(x2) = -2X2 under f. That is,
(Vx g R)
f o g(x) = -2x2
g
R
f
R
x
-2x
x2
-2(x2)
Fig. 1.9
Method II
By Formula
By using the formula
g O f(x) = g[f(x)J
we have
gof(x) = g[f(x)]
= g(-2x)
= (-2x)2
= 4x2
and
fog(x) = f[g(x)]
= f(x2)
= -2(.r2)
= -2x2
(Vx g R)
Functions
Remark 1-3
13
As illustrated in Example 1-8, in general,
gof^fog
That is, composition of function is not commutative.
The idea of composition of functions can be extended to three functions. The composition of three
functions follows associative law as stated below.
Theorem 1-4
Let f: A —> B, g : B
C and h: C —> D be functions. Then
(h o g) o f = h o (g o f)
Proof
It is obvious that both (h o g) o f and h o (g o f) are functions from A to D. It suffices to
prove for any agA,
[(h o g) O fj(fl) = [h O (g o f)](fl)
Applying (1-1), we have
[(h ° g) ° f](a) = (h o g)(f(n))
= h(g(f(a)))
= h(gof(a))
= [ho(gof)](a)
Hence,
(h o g) o f
Example 1-9
h o (g o f)
Let f, g and h be mappings from R to R such that for all x g R,
f(x) = 2x,
g(x) = x2
and
(h o g) o f
h o (g o f)
h(x) = x + 1
Verify
Solution
For a typical element a g R, we have
[(h o g) o f](a) = h o g(2fl) = (la)2 + 1 = 4a2 + 1
and
[h o (g O f)](fl) = h[g o f(fl)] = h[g(2«)J = h(4fl2) = 4fl2 + 1
Hence,
(h o g) o f = h O (g o f)
h o g(x) = .v‘ + 1
14 Chapter I
Exercise 1B
1.
Let f: R —» R and g: R
R be two functions defined by
g(x) = sin.v
and
f(x) = .v2
(VxgR)
Find g o f and fog.
2.
Let f: R —> R and g: R -» R be two functions defined by
f(x) = 2x - 3
and
g(x) = x3
(Vx g R)
State the functions
3.
(a)
fof
(b) g o f
(c)
fog
(d) fogof
Let f and g be two functions from R to R defined by
f(x) = 2x,
g(x) = x + 3
and
(a)
State the functions foh and h ° g.
(b)
A function k from R to R is defined by
h(x) = x2
k(x) = 2x2 + 12x + 18
(V.v g R)
(V.v g R)
Express the function k in terms of f, g and h.
4.
Let M be the set of all 2x2 non-singular matrices. Define f: M —> R \ {0} and g:R\{0}-^R\{0}
by
(VA gM)
f(A) = det A
gW = -
(VxgR\ {0})
X
Find g o f.
Does fog exist? Why?
1.4
Bdentity Function and inverse Function
Recall the theory of matrix discussed in Advanced Level Pure Mathematics “Algebra”, we have identity
matrix and inverse of a matrix. In the theory of mapping, we can similarly define the identity function and
the inverse function of a function.
Definition 1-4
The function f: A —>A defined by
f(x) = x
is called the identity function on A and is usually denoted by iA.
(Vx gA)
Functions 15
Remark 1-4
(1) The graph representing the identity function on R is the straight line y = .v.
(2) Sometimes, the identity function may be denoted by i.
Theorem 1-5
Let f: A —> B be a function. Then
of = f
Proof
Since f is a function from A to B and iA is a function from A to A, the composite function
f o iA is a function from A to B. Similarly, iB o f is also a function from A to B.
Further, V.v eA,
by the definition of
= f(x)
identity function
and
°f(x) = 'B(fW)
= f(.r)
by the definition of
identity function
Hence,
of
f
Definition 1-5
Let f: A —> B be a function. If there exists a function g: B —> A such that
and
g o f = iA
f ° g = iB
(1-2)
g is called the inverse function of f and is usually denoted by f .
Remark 1-5
(1) Since f is a function, V.v e A, 3y 6 B such that
f(A-) = y
From the relation g ° f
i., we have
/I ’
gof(x) = z\(x)
g(fW) = GW
=> g(y) = *
i.e. the roles of image and pre-image of f and g are interchanged, and this is
illustrated in the following diagram.
16
Chapter /
A
B
B
a
a
1
2
b
b
2
3
c
c
3
A
f
1
7
g
Fig. 1.10
(2) Iff has an inverse g, we observe that every element of f has one and only one image
and that the number of elements in A is exactly equal to the number of elements in B.
Example 1-10 Let f: R \ {I} -> R \ {I} be a function defined by
f(A-) =---- 7
x-1
(VxgR\ {I})
Find the inverse function of f.
Solution
Suppose f(x) = y, where x g R \ {1}, y g R \ {1}.
If g: R \ {1} —> R \ {1} is the inverse function of f, then g(y) = f“*(y) = x.
Since f(x) = y,
x+1
y(x-l) = x-r 1
yx - x = y +1
x=
+1
y-l
That is,
r’(y)
y+i
y-l
If y is replaced by x, the inverse function of f is defined by
f-'(x) = —
x -1
Example 1-11
(VxgR\ {!})
Let f:Z—>Z be a function defined by
f(x) = 2x
(VxgZ)
Does the inverse of f exist? Why?
Solution
Suppose the inverse of f exists. Then, applying the technique as in Example 1-10, we have
r’W x
2
Functions
17
Since y is not an integer for a being an odd integer, the image of an odd integer under f
does not exist. Hence, f-1 does not exist.
Remark 1-6
Note that the range of f in Example 1-11 contains some elements which do not have
pre-images under f. Therefore, if we change the range of f to be the set Z2, where Z, is the
set of all even integers, the inverse of f exists.
From Example 1-10 and Example 1-11, the inverse of a function f may or may not exist. To clarify the
conditions under which the inverse of f exists, we have to introduce the concepts of injective, surjective and
bijective functions in the following section.
1.5
Injective, Surjective and Bijective Functions
Definition 1-6
Let f: A —> B be a function, f is said to be injective (or f is an one-to-one function) if and only if
(Vflp a2 eA)
f(fl,) = f(a,)
or equivalently,
f/i * a2
(Va,,a2eA)
f(fl,) * f(fl2)
The definition means that the pre-images of the elements of the range B. if any. must be unique. Hence,
if we want to prove that a function is not injective, we have to find two different elements in the domain A
such that their images are the same. In order to prove a function being injective, we have to show that the
pre-images of all the elements of the range B under f, if any, must be unique, i.e.
f(«2)
f^)
(Va,, a2 eA)
«i = a2
As shown in Fig. 1.11, the pre-images of all the elements of B under f are unique and so f is injective.
On the other hand, the element a of B has two different pre-images 1 and 2 under g. Therefore, g is not
injective.
B
A
a
1
g
B
a
b
b
2
3
Fig. 1.11
c
d
18 C/rnpft
/
Example 1-12 Let f: R —> R be a function defined by
(V.v g R)
f(.v) = sin.v
Is f injective? Why?
Solution
v
sin 0 = sin ^ = 0
.-.
The pre-image of 0 is not unique.
Hence, f is not injective.
In Example 1-12, if the domain of f is changed to A = [0, yj, then f(.v) = sin ,v becomes injective since
every element in the codomain [0, 1] has only one pre-image in [0, y].
Example 1-13 Let f .R—>R be a function defined by
(V.v g R)
f(.t)=.v+3
Prove that f is injective.
Solution
In order to show that f is injective, we have to check whether the statement
f(fl|) = f(fl2) => a} = aa22
(Vfl,, a2 e R)
is true. Now, Vap a2 gR,
f(fli) = f(fl2) =>
fl,+3=fl2+3
a, = a2
Hence, f is injective.
Example 1-14 Let f: R2 —> C be a function defined by
f((fl, b)) = a + bi
(V(a, b) e R)
Prove that f is injective.
Solution
V(apZ?t), (a2, £>2)gR2,
fl] + bxi = a2 + b2i
f((«p^)) = f((«2,^2))
=>
at =a2, bt =b2
(al,bl) = (a2fb2)
f is injective.
Functions 19
Definition 1-7
Let f: A —» B be a function, f is said to be surjective (or f is an onto function} if and only if VZ? g B,
gA such that f(fl) = b.
The definition means that every element of the range B has its pre-image or equivalently,
f[A] = B
For example, the function f in Fig. 1.12 is surjective but the function g is not.
A
B
f
1 2
3
b
Fig. 1.12
Example 1-15 Let f: R —> R be a function defined by
f(x)=x2
(V.r e R)
Prove that f is not surjective.
Solution
For any x g R, x1 > 0.
Hence, the pre-image of any negative element of the range R does not exist. For example,
-1 g R but the pre-image of-1 under f is purely imaginary.
That is the pre-image of-1 under f does not belong to R, and so f is not surjective.
Remark 1-7
If the range of f is changed to R* u {0}, where R+ is the set of all positive real numbers,
then f becomes surjective.
Example 1-16 Let M be the set of all 2x2 matrices. A function f:M—>R is defined by
f(A) = detA
(VAgM)
Prove that f is surjective but f is not injective.
Solution
In order to prove that f is surjective, we try to find the pre-image of an element y of the
range R.
20
Cliupie
I
Suppose the pre-image of y is
A=
a
c d
Then we have
f(A) = det A = ad - be = y
(1)
That is. we have 4 unknowns in 1 equation. As a result, we have infinitely many solutions,
and hence the pre-image of y always exists. For example, we may take
A=
(T
^0
b
Therefore, f is surjective.
Next, we want to show that f is not injective.
Note that the values of the determinants of two unequal matrices may be equal. For
example, the values of the determinants of the 2x2 matrices
y (T
0 b
A=
A=
and
f1 p
are the same. That is
f(A) = f(A,) = y
Hence, f is not injective.
Example 1-17 Let f: C \ {0} -> C be a function defined by
(VzeC\{0})
f(z) = z--
z
Prove that f is surjective.
Solution
Let w e C.
Suppose z is the pre-image of iv. Then
w = f(z)
1
W=z—
z
z2 - wz -1 = 0
vv
z=—
I
W~ 4-
2
Now, we take
z
w + V w2 4- 4
2
Next, we have to check whether the pre-image belongs to the domain C \ {0}, and this is
proved by the method of contradiction.
Functions 21
If zeC \ {0), then
vv +
z
0
w2 + 4
2
0
W = — iv2 +4
w2 = vv2 + 4
0=4
Hence, zeC\ {0}. and so f is surjective.
Definition 1 -8
Let f: A —> B be a function, f is said to be bijective (or f is an one-to-one onto function) if and only
if f is both injective and surjective.
For any bijective function, the elements in the domain A and that in the range B is an one-to-one
correspondence. That is, any element in A has one and only one image in B and any element in B also
has a unique pre-image in A.
Example 1-18 Let f: R \ {—1} —> R \ {I} be a function defined by
f(x) =
x-1
x+ 1
(VxgR\{-1})
Prove that the f is bijective.
Solution
The proof is separated into two parts: (1) to prove that f is injective; (2) to prove that I is
surjective.
(1) Vx„x2eR\{-l},
f(x,) = f(x2)
x, -1
x2 -1
(x, - l)(x2 +1) = (x2 - l)(Xj + 1)
XjX2 - x2 + X, - 1 = XjX2 - Xj + x2 - 1
2xj = 2x2
X, = x2
Hence, f is injective.
22 Chapter I
x— 1
(2) Vy g R \ {I}. we Uy to find the pre-image of y. By putting y = —-j-j-, we have
y(x4-l) = x-l
yx 4- y = x - 1
yx - x = -y - 1
x(y-l) = -y-l
-y-i
y-i
V =----- r-
(*)
.......................
Since y g R \ {1}, y * 1 and so x exists.
To complete the proof that f is surjective, we need to show that x g R \ {-1}, i.e.
x*-l.
From (*), we have
ForyeRX {I}.
x = —1---- —
y-1
-=-*0
y-1
—1
Hence, xgR\ {-!}.
From (1) and (2), f is bijective.
Theorem 1 -6
Let f: A —> B be a function. If f is bijective, then f"1 exists and (f ’) 1 = f.
Furthermore, f"1 is also bijective.
The proof is beyond the scope of this text.
Example 1-19 Let f: R\ {2} —> R\ {-3} be a function defined by
f(A-) =
Prove that f 1 exists and find f '.
3x4-4
2-x
(VxgR\ {2})
Functions 23
Solution
In order to show that f 1 exists, it is necessary to prove that f is bijective.
(1) We want to show that f is injective.
Vxi,x2gR\{2),
f(xl) = f(.r2)
2 — xj
2 — x2
=> (3xl+4)(2-x2) = (3x2+4)(2-x1)
=> 6xj - 3x,x2 + 8 - 4x2 = 6x2 - 3x,x2 + 8 - 4x(
=> 1 Oxj = 1 Ox 2
=>
*l=*2
That is, f is injective.
(2) We want to show that f is surjective.
Vy g R \ {-3}, let
3x + 4
y=~^2-x
—
Then we have
y(2-x) = 3x + 4
2y - 4 = (3 + y)x
2y —4
y+3
x
(*)
Since )>gR\{-3}, y^-3 and so x exists.
Furthermore, from (*), we have
x = 2-
y+3
Hence, 3x g R \ {2} such that f(x) = y. That is, f is surjective.
From (1) and (2), f is bijective and so f exists.
Furthermore, from (*), we have
f-*(y)=
7
y+3
(VjgR\{-3})
If y is replaced by x, we obtain
f’« =
2x —4
(VxgR\{-3))
In mathematical analysis, we always deal with functions whose domain and range are sets of real numbers,
which are called real functions. As stated in Section 1.1, a real function y = f(x) may be represented
graphically by a curve on xy-plane. Now, we are going to discuss the relationship between the graphs of the
function y = f(x) and its inverse f"1.
24 Chapter I
y
y=x
A'(y, x)
M
A(x.y)
10
Fig. 1.13
If A(x, y) is a point on the graph of y = f(x), then by the definition of inverse function, the point A'(y, x)
is a point on the graph of its inverse x = f "’(y). Since
Slope of AA' = ^Z=-l
y-x
Mid-point of AA' =
Hence, AA' is perpendicular to the straight line y = x. Also, mid-point of AA' lies on this straight line.
Hence, we have
Remark 1-8
The graphs of a real function y = f(x) and its inverse x = f-,(y) is symmetrical about the
straight line y = x. That is, the graphs of f and f-1 are mirror image to each other with
respect to the line y = x. Hence, the graph of x = f“’(y) can be obtained by reflecting the
graph of y = f(x) about the line y = x.
[Compare with the result obtained in Section 8.8 ofAdvanced Level Pure Mathematics “Algebra”.]
Fig. 1.14 shows the functions y = logiox and y = 10x, which are inverse function of each other.
y
y = 10'
y= x
y= log10x
O
Fig. 1.14
►x
Functions 25
Exercise 1C
1.
2.
Find a function f: Z —> Z such that
(a)
f is injective but not surjective,
(b)
f is surjective but not injective,
(c)
f is bijective.
Let A and B be two sets. The Cartesian product of A and
denoted by A x B, is defined as:
A x B = {(a, b): a gA and b^B}
Given that
C
f: A
are two functions. A function h: Ax B
&-B-+D
and
Cx D is defined by
h((f/, Z?)) =
g(Z?))
(V(a, b)eAxB)
Prove that
3.
(a)
if f and g are injective, then h is also injective.
(b)
if f and g are surjective, then h is also surjective.
Let A be a non-empty set and f: A —> A be a function.
It is given that f is either injective or surjective. Show that if f ° f = f, then f
4.
<r
Let f: C —» C be a function, f is said to be an isometry if and only if
|f(z,)-f(z2)| = |z1-z2|
(Vz„z2eC)
Show that all isometries are injective.
5.
Let Z+ be the set of all positive integers. Define f: Z+ —> Z+ such that
f(A)=X2
(VagZ+)
Show that f is injective but not surjective.
6.
Let f: R \ {— 1} —> R \ {— 1} be a function defined by
f(x) =
Find fof.
1-x
1-t-x
(VxgR \ {-!})
Hence find f"1.
7.
Let f: R —> R and g : R —> R be two functions defined as:
3x + 5
f(-v) = ——
Find fog.
Hence find f"1 and g"1.
,
and
( A 2x-5
g(A)
g(x) = ---
(Vx g R)
26 Chapter I
8.
Let M be the set of all 2x2 non-singular matrices.
A function f: M -> C \ {0} is defined by
(VA =
,(4)-R+W
a
b'
e M)
c
Show that f is surjective.
9.
Find the inverse of the following functions, if any.
(a)
f: R —> R such that
(V.v g R)
f(x) = 5x + 1
(b)
Let A = {(.v, y) g R2: x + y = 1}. A function f: A —> R is defined by
(V(x, y) g A)
f((x, }’)) = 2x
(c)
Let (o, b) = {xgR: a <x<b} be an open interval. A function f: (-J^y,
by
f(x) = tanx2
10.
Let f: R -> R be a function such that
f(x + y) = f(x)f(y)
(V.v, y g R)
Suppose the inverse of f exists. Show that
and
f(0) = 1
f-1(xy) = f"’(x) + f"‘(y)
(Vx, ycR)
f(xy) = f(x) + f(y)
(V.v, y g R)
f(x") = nf(x)
(V.v g R)
11. Let f: R -> R be a function such that
(a)
(b)
Show that for all integers n,
Suppose the inverse of f exists. Show that for all integers n,
[f’(x)]n = f-’Ozx)
(V.v g R)
12. Let a,, a2,..., an be n real numbers. Suppose f: R -> R is bijective. Define
g(.r) = pf(x) + q
(a)
(p, q g R, p * 0)
Show that g is bijective and
g ,(x) = f"’(p
(b)
Hence, deduce that
g
( i
-y
7
n
'X
'=1
7
) —> R is defined
Functions 27
1.6
Some Special Real Functions
There are several kinds of real functions which bear importance to mathematical analysis. They will be
discussed in the following.
A.
Even and Odd Functions
Both the even and odd functions are defined on symmetric sets of numbers. A set A is a symmetric set of
numbers when the set contains both the positive and negative values of a number, i.e.
Vx G S => -x G S
For example, if a is a positive real number, then the set {-a, a}, the open interval (-a, a), the closed interval
[-fl, and the set of real numbers R are symmetric sets of numbers. However, the open interval (-4, 2) is
not a symmetric set of numbers since -3 g (-4, 2) but 3 G (-4, 2). The even and odd functions are defined
as follows.
Definition 1-9
Let S be a symmetric set of numbers and y = f(x) be a function defined on 5. If
f(-x)
f(x)
(VxgS)
then f is said to be an even function.
Example 1-20 Obviously, the functions f(x) = |x|, f(x) = x2 and f(x)
cosx on R are even functions.
Example 1-21
Prove that the function f(x) =xsinx on R is an even function.
Solution
Vx g R, we have
f(-x) = (-x)sin(-x)
= (-x)(-sinx)
= xsinx
= f(x)
Hence, f is an even function.
r
28 Chapter I
Remark 1-9
Graph of an Even Function
Note that for any even function f, f(x) = f(-x), Vx g/1, and so the graph of an even
function is symmetrical about they-axis. Fig. 1.15 is an example of an even function.
y
y=f(x)
A(x, f(x))
>A'(-x, f(-x))/
I
I
T
I
/i
i
f
i
_______ L
L
0
Fig. 1.15
Definition 1-10
Let S be a symmetric set of numbers and y = f(x) be a function defined on S. If
f(-x) = -f(x)
(Vx g S)
then f is said to be an odd function.
Example 1-22 Obviously, the functions
f(x)=?
and
f(x) = sinx
on R are odd functions.
When n is an integer and denote an = nn + y, the function
f(x) = tan x
is also an odd functions on R \ {r?n}.
Example 1-23 Determine whether the function f(x) =x +1 is an odd function.
Solution
If f(-x) = -f(x), then
-x + 1 = -(x + l)
1=-1
which is a contradiction.
Hence, f(x) = x + I is not an odd function.
►x
Functions 29
Example 1-24 Is the function f(x) = x2 + x on R an odd function? Why?
Solution
If f(-x) = -f(x), then
(-x)2+(-x) = -(x2+x)
x2 - X = -x2 - X
2x.2 = 0
x=0
Hence, f(-x) = -f(x) occurs only when x = 0. Therefore, f is not an odd function.
Remark 1-10
Graph of an Odd Function
(1) For any odd function f, f(-x) = -f(x), Vx gA, and so it is obvious that f(0) = 0. That
is, the graph must be passing through the origin.
(2) Further, the graph contains both the points A(x, f(x)) and A'(-x, f(-x)). Then, as
shown in Fig. 1. 16, it is obvious that A, O, A' are collinear and AO = OA'. Hence, the
graph remains the same when it is rotated anticlockwise (or clockwise) through
an angle of n. [Compare with the result obtained in Section 8.8 of Advanced Level Pure
Mathematics "Algebra'.], In such circumstances, we say that the graph is symmetrical
about the origin. Fig. 1.16 shows an example of an odd function.
y
y=f(x)
1
I
7
/
’
1
1
/
/
A(-X, f(-X))
Fig. 1.16
30 Chapter 1
Theorem 1-7
Properties of Even and Odd Functions
(1)
The sum of two even functions is even.
(2)
The sum of two odd functions is odd.
(3)
The product of two even functions is even.
(4)
The product of two odd functions is even.
(5)
The product of an even function and an odd function is odd.
I
Proof
Let f and g be two functions.
(1) If both f and g are even functions. Then, for all x,
f(-x) = f(x)
and
g(-x) = g(x)
Hence
(f + g)(-x) = f(-x) + g(-x)
= f(x) + g(x)
= (f + g)(x)
The proof of the other properties is similar to that of (1) and is left as an exercise for the
readers.
Example 1-25 Since f(x) = x“ and f(x) = cosx are even functions, the function f(x) = x2 + cosx must be
an even function. Furthermore, fj(x) = x, f2(x) = tanx and f3(x) = sinx are odd functions
and so the function
F(x) = [(x2 + cosx)sinx - x] tanx
is an even function.
B.
Bounded Functions
Definition 1-11
Let f be a function defined on the set of real numbers A.
(1)
For any x 6 A, if there exists a constant M such that
f(x) < M
then f is said to be boundedfrom above on A. M is called an upper bound of f.
(2)
For any x eA, if there exists a constant M such that
f(x) > M
then f is said to be bounded from below on A. M is called an lower bound of f.
!■'unctions 31
(3)
For any x eA, if there exists a positive constant M such that
|f(x)|<AY
then f is said to be bounded on A.
Obviously, a function which is bounded from above and also bounded from below must be a
bounded function.
From the definition of bounded function, it is obvious that the graph of a bounded function lies between two
horizontal straight lines y = M and y = -M as shown in Fig. 1.17.
y
/=f(x)
-M
Fig. 1.17
Example 1-26 The function f(x) = x2 on R is bounded from below by 0 since x2 > 0. However, it is not
bounded from above.
Example 1-27 The trigonometric functions f(x) = cosx and f(x) = sinx on R are bounded functions
since
| sin .v | < 1
and
cosx|< 1
The other trigonometric functions f(x) = tan x, f(x) = sec x, f(x) = esc x and f(x) = cot x
are not bounded functions.
Example 1-28 Let f(x)
Solution
1
be function on R. Determine whether f(x) is bounded.
x2 +2x + 2
By putting y = f(x), we have
1
y = x2+2x + 2
(1)
32 Chapter /
y(r+2.v+2) = l
yr + (2y).r + (2y-l) = 0
(2)
From (1). for some real values of y, x must be real. Then, by considering (2) as a quadratic
equation in x, the discriminant of the equation (2) must be non-negative, i.e.
(2y)2-4y(2y-l) >0
4y-4y3 >0
4y(l-y)>0
0<y < 1
Hence, f is bounded from below by 0 and bounded from above by 1, and so f is a bounded
function. [Note: In this example, 1 is the least upper bound and 0 is the greatest lower bound.]
Example 1-29 Let f: R -» R be a real function such that
f(x + y) = f(x)f(y)
(Vx.yeR)
(a) Show that f is bounded from below by 0.
(b) Furthermore, if f is not identically equal to zero,
(i) find f(0).
(ii) show that
(Vx g R)
f(x) > 0
Solution
(a) VxgR,
f(x) = f(| + |)
/
\2
>0
f is bounded from below by 0
(b) (i)
f(0 + 0) = f(O)f(O)
f(O)-[f(O)]2 =0
f(O)[l-f(O)] = O
or
f(0) = 0
f(0) = I
If f(0) = 0, then VxgR,
f(x) = f(x + 0)
= f(x)f(0)
=0
which contradicts to the given condition that f is not identically equal to zero.
Function v 33
Hence, f(0)
0, and so
f(0) = 1
(ii) Now, we want to prove that
f(x) * 0
(Vx g R)
Suppose there exists x0 g R such that f(x0) = 0, then
l = f(0)
= f(x0+(-x0))
= f(x0)f(-x0)
=0
which is a contradiction.
f(x) * 0
(Vx g R)
With the result in (a), we have
f(x) > 0
(Vx g R)
Example 1-30 Let f: R —> R be a real function such that
f(x + y) = f(x) + f(y)
(Vx, y g R)
(a) Show that for all positive integers n > 1.
f(nx) = n f(x)
(Vx g R)
(b) If f is not identically equal to zero, show that f is not bounded.
Solution
(a) The result is proved by induction on n.
(1) When n = 2,
f(2x) = f(x) + f(x) = 2 f(x)
Hence, the proposition is true for n = 2.
(2) Assume the proposition is true for n = k > 2, i.e.
f(£x) = £f(x)
When n = k + 1, we have
f((Zc + l)x) = f(Ax + x)
= f(fcv) + f(x)
= £f(x) + f(x)
= (A' + l)f(x)
Hence, the proposition is also true for n = k + 1.
Therefore, by the Principle of Mathematical Induction, the proposition is true for all
positive integers n > 1.
? /
34
(b) Since f is not identically equal to zero, there exists a real number x0 such that
f(.v0) = b * 0. From the result in (a), we have
f(?LV0) = n f(.v0) = nb
Next, we want to show that f is not bounded. This is proved by contradiction.
Suppose f is bounded, i.e. there exists a positive number M such that
|f(.v)|<M
(V.v g R)
By taking a positive integer n > p-j, we have /lvogR and so,
|f(n.r0)| = |;i f(-v0)|
= «| f(A0)|
M |, |
Pi1'’1
=M
which contradicts the assumption that | f(.v) | < M (V.v g R).
Hence, f is not bounded.
c. Monotonic Functions
Now. we shall study the functions which never decrease (or increase) on a given interval.
Definition 1-12
Let f be a function defined on the set of real numbers A.
(1)
f is said to be monotonically increasing on A if and only if
a < b => f(fl) < f(Z?)
(Va.^gA)
On the other hand, f is said to be strictly increasing on A if and only if
a <b => f(a) < f(b)
(2)
(Va, b eA)
f is said to be monotonically decreasing on A if and only if
a <b => f(fl) > f(b)
(Va, be A)
On the other hand, f is said to be strictly decreasing on A if and only if
a<b =$ f(a)>f(b)
(Va, be A)
/'unctions 35
Remark 1-11
Graph of a Monotonic Function
From definition 1-12, it is obvious that the graph of an increasing function rises as x
increases while that of a decreasing function drops as x increases.
Example 1-31
By sketching the graphs of y = x and y = x2, it is obvious that y = x is an increasing
function on R while the function y = x2 is decreasing on (-«>, 0] and increasing on
[0, +oo).
Example 1-32 Prove that the function y = x3 is strictly increasing on R.
Solution
Consider
a3-b3 =(a-bXa2 +ab + b2)
1
= (a-b) a + — b \+^
2
4
(I)
If a < b, then a and b cannot be both zero, and so
« + — b |+^2>0
2
4
Since a — b < 0, from (1), we have
a3 <b3
a3-b3< 0,
Hence, a3 < b3 and the given function is strictly increasing on R.
Example 1-33 Show that the function y
Solution
— is strictly decreasing on (0, +°°).
x
For 0 < a < b, we have
_1_ _1_
a b
^>0
ab
Hence, the given function is strictly decreasing on (0, +«>).
Remark 1-12
In Example 1-33, the result also holds for a < b < 0.
y
— is strictly decreasing on (•
\ 0) and (0, +°°).
b — a > 0,
ah > 0
36 Chapter I
Example 1-34 Find an interval on which the function y = 2x2 - 8x + 1 is strictly increasing.
Solution
The given function can be rewritten as
y = 2(x-2)2-7
It is obvious that the graph is a parabola open upward and with vertex at (2, —7). Hence,
the given function is strictly increasing on [2, +©©).
Theorem 1-8
Let f: R —> R be a bijective function.
If f is strictly increasing, then f-I is also strictly increasing;
If f is strictly decreasing, then f"1 is also strictly decreasing.
Proof
We want to prove the result for strictly increasing functions first and it is to be proved by
contradiction.
Since f is a bijective function, f 1 exists.
Now, we suppose f is strictly increasing but f’1 is not strictly increasing. That is, there
exist
b2 g R such that bi < b2 and
r'tb^r^bj
Hence, we have
=> fof-'^^fof-'^)
=> ^,)>^2)
=> bj > b2
and this contradicts the assumption that b^ < b2.
Hence, f"1 is strictly increasing.
The proof for strictly decreasing function is similar and is left as an exercise for the
readers.
Example 1 -35 Obviously, y = x2 is a strictly increasing on [0, +°q). Also, its inverse function y = Vx is
also a strictly increasing on [0, +<»). Hence, Theorem 1-8 is verified.
Fitnetion* 37
Remark 1-13
D.
Note that some readers may state the result “«</?=> [(a) < [(b)" without ensuring that the
function f is strictly increasing and this will lead to some invalid results. For example, if
f(x) = -x, it should be “a < b => [(a) > [(b)" since f is strictly decreasing. Similarly, if
f(x) = x2, neither “a < b => [(a) < [(b)" nor “«</?=> [(a) > [(b)" is correct since f is
neither strictly increasing nor strictly decreasing. More properties about monotonic functions
will be discussed in Chapter 5.
Periodic Functions
A lol of natural processes occur periodically, for example, the motion of simple pendulum. This leads to the
study of periodic functions.
Definition 1-13
Let y
f(x) be a function defined on R. If there exists a positive constant T such that
f(x+T) = f(x)
(VxgR)
f(x) is called a periodic function with period T.
From Definition 1-13,
f(x — T) = f(x — T + 7) = f(x)
(Vx e R)
f(x + IT) = f(x + T + T) = f(x + T) = f(x)
(Vx e R)
Similarly,
Hence, we have
f(x + nT) = f(x)
where n is an integer, and so if T is the period of a periodic function f(x), then any multiple of T is also a
period of f(x). In general, we select the smallest positive period as the period of the periodic function.
Example 1-36 Since sin(2zr + x) = sinx (VxeR), y = sinx is a periodic function with the smallest
positive period 2ji. Similarly, the function y = cosx is also a periodic function with the
smallest positive period 2tt. On the other hand, the functions y = tanx and y = cotx are
periodic functions with the smallest positive period n.
Example 1 -37 Show that a non-constant rational function f(x) is not periodic.
Solution
A rational function is the quotient of two polynomials, i.e. f(x) =
Q(x) are two polynomials with degree less than m.
The given result is proved by contradiction.
Suppose u e R and let f(w) = c.
Since f(x) is a periodic function, then
f(u + nT) = [(u) = c
where T is the period of f(x).
P(a)
where P(x) and
Q(x)’
38 Chapter 1
Hence, we have
P(u + ?iT)
P(// + iiT) -cQ(u + nT) = 0
and so, the polynomial equation
P(x) - c Q(.v) = 0
has infinite many roots u + nT (n gZ). However, the degree of this polynomial equation
is less than m. and so the number of roots is equal to m, and so
P(.v) - c Q(.v) = 0
That is,
f(.v)
Q(.v)
This leads to a contradiction.
Therefore, a non-constant rational function f(x) cannot be periodic.
Example 1-38 (a) Let y = f(x) be a periodic function with the period T.
T.
Show that the function y = f(fi)x) is also a periodic function of period —
co
(b) Find the period of the function y = sin(ft)x + 0).
Solution
(a) Since y = f(x) is a periodic function with the period T, we have
f(x) = f(x+7)
Hence.
(
T
i{cox) = f(ft)x + T) = f co(x + —)
k
co
T.
and so, the function y = f(rox) is also a periodic function of period —
co
(b) Let f(x) = sin(x + 0). Then
f(x + 2 zr) = sin(2;r + x + 6)
= sin(x + 0)
= f(x)
and so, the period of f(x) is 2zr.
Hence, from (a), the period of the function y = sin(fi)x + 0) = f(rz)x) is —.
co
Fundions
Theorem 1-9
39
Properties on Combining Periodic Functions
Lei y = f(x) and y = g(x) be two periodic functions with the smallest positive periods 5 and T
respectively.
f
5
If —
is a rational number, then f + g, f g, fg and — are periodic functions,
T
g
Proof
S
S=m
Since — is a rational number, there exist non-zero integers m and n such that —
T
T n
Denote
T' nS = mT > 0
Then we have
f(x + nS) f(x),
g(x + mT) = g(x)
Hence,
(f+g)(x+d = f(x+r)+g(x+r)
= f(x + /zS) + g(x + nzT)
= f(x) + g(x)
Therefore, f + g is a periodic function.
The proof of the remaining results are similar and are left as an exercise for the readers.
Remark 1-14
Graph of a Periodic Function
Let y = f(x) be a periodic function with period T.
To sketch a graph of this periodic function, it is only necessary to sketch the graph within
one period [r/, a + T|. Then the other parts can be obtained by repeating the graph obtained.
This periodic property of a periodic function helps in sketching its graph. For example, the
graph of y = sin x can be obtained easily by sketching the curve for 0 < x < 2tt and then
repeating this curve to obtain the required graph. (See Fig. 1.22 in Section 1.7.)
40 Chapter I
1.7
Elementary Functions
The following are some common examples which we usually encounter in future study.
A. Basic Elementary Functions
In this part, we are going to introduce six different basic elementary functions.
I.
Constant Function
A constant function is a function of which the image of every element of the domain is the same. That is,
if c is a constant, then
y=c
is a constant function.
(1) The domain of a constant function is R.
(2)
Its graph is a straight line parallel to x-axis as shown in Fig. 1.18.
y
y=c
X
o
Fig. 1.18
II.
Power Function
Let a be a real constant. Then the function
y = .r“
(a* 0)
is called a power function.
(1)
Domain of the power function:
When a is a positive integer, the domain is (-<», +°o).
When a is a negative integer, the domain is (-©% 0) u (0, +«>).
When a is not an integer, the domain is (0, +°°) or [0, +<») according to a is negative or positive.
Functions 41
(2)
The graph of a power function is shown in the following:
y
y=x“ (a>0)
3
2
i
i
i
i
i
y=xu (a < 0)
2
' 2
1
O
y
1
■ 3
O
1
x
1
(b)
(a)
Fig. 1.19
(3)
When a > 0, the power function xa is strictly increasing for x > 0.
When a < 0, the power function xa is strictly decreasing for x > 0.
111.
Exponential Function
Let a be a real constant. Then the function
y=a
(0<n# 1)
is called an exponential function.
(1)
The domain of an exponential function is R.
(2)
The graph of an exponential function is shown in the following:
►x
O
Fig. 1.20
(3)
When 0 < a < 1, the exponential function ax is strictly decreasing.
When a > 1, the exponential function ax is strictly increasing.
42 Chapter I
(4)
In particular, when a = e, an irrational number which is approximately equal to 2.718 282 8, we write
y = exp x to denote the function y = e\ This function is an important function in mathematical
(
iy
analysis. [The irrational number e is equal to be the limit value of Inn I 1 + — I and the concept of limit will be
discussed in next chapter.]
IV.
Logarithmic Function
Let a be a real constant. Then the function
(0<at 1)
>’ = logax
is called a logarithmic function.
(1)
The domain of a logarithmic function is x > 0.
(2)
The graph of a logarithmic function is shown in the following:
y
y = log2x
y = iogi0*
\
►X
O
- Iog_j_ x
10
■ log2 X
2
Fig. 1.21
(3)
When 0 < a < 1, the logarithmic function logflx is strictly decreasing.
When a > 1, the logarithmic function Iogfla x is strictly increasing.
(4)
In particular, when a = e, the logarithmic function logex or sometimes denoted by Inx, is called the
natural logarithm of x. It is the inverse function of y = expx and is another important function in
mathematical analysis.
V.
Trigonometric Functions
The trigonometric functions are:
y = sinx,
y = cosx
y = tan x,
y = cotx
secx.
y = cscx
y
Functions 43
(1)
Domain of
y = sinx and y = cosx are R.
y = lanx and y = secx arc R \ {nn 4- —
}, where n is an integer.
2J....... —
y = colx and y = cscx are R \ {n7i}, where n is an integer.
(2)
The graph of sine function, cosine function and tangent function are shown in the following:
y
1
y = cosx
O
7T
y = sinx
2tc
K
2
-1-
2-f
3k
37T + •£•
*x
4k
Fig. 1.22
y
I
I
I
I
I
I
I
I
I
---------------- L
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
—K
O
'ZL
i2
i
2 i
i
i
i
i
h
i
i /
i /
i/
il
i|
i
i
i
i
i
i
i
I
i
x
K
i
i
i
i
i
i
i
i
i
►X
71
2
Fig. 1.23
(3)
y sinx and y = cosx are periodic functions with period 2^ while y = tanx is a periodic function
with period tt.
VI.
Inverse Circular Functions
The inverse circular functions are:
y = sin ‘x,
y = cos x
y = tan x,
y cot x
y = esc x
y = sec x,
r
44
Chapter I
(1)
Domain and principal value of inverse circular functions:
(2)
y
siiT’x
cos A
tan ‘x
Domain
-1 <X< 1
-1 <x< 1
—oo < X < +co
Principal Value
—
---- v
2 ? 2
'K
0<y<zr
71 '
7T
71
2
2
The graph of sin"’x, cos"'x, tan-lx are shown in the following:
y
y
y
y = tan-1x
y = cos-1x
y = sin“’x
K
71
2
7T
2
2
O
0
1
*x
1
*x
TV
7t_
2
2
-n -
(a)
O,
-71
(b)
(c)
Fig. 1.24
B. Elementary Functions
Definition 1-14
Elementary functions are functions obtained from some basic elementary functions through the
operations of addition, subtraction, multiplication, division and composition.
Functions 45
Example 1-39 Let
V = 1,
IV = x2
Obviously, v and iv are two basic elementary functions. Then from the definition of
elementary function, the difference between v and iv
u = v - iv = 1 — x2
is an elementary function. Furthermore, the composition of
£
y = u2
u= 1 -x2
and
is also an elementary function. Hence, the function
function obtained from
£
y = u2,
u = v — IV,
v = 1,
y = V1-a2
is an elementary
iv = x2
Example 1-40 Similar to the argument outlined in Example 1-39,
u = viv = sinx log10x
is an elementary function obtained from the product of
v = sinx,
iv = log10x
Similarly, y = iOMnt,°610x is another elementary function obtained from the composition of
y= 10",
As y
u = viv
|QSinrlogl0r can be rewrjt(en as
y = 10sinrl°Sio-'
= (ioIOSlot)sint
_ xsinr
the function y = xsin r is an elementary function.
In some cases, in order to simplify the complexity of the analysis of a function, we have to defined different
elementary functions in different intervals. This kind of function is called a. piecewise function.
Example 1-41
-x + 1
U<0)
x2
(A>0)
Let f(x) = f is a piecewise function. Its graph is as shown in the following:
y
y=f(x)
x
O
Fig. 1.25
46 Chapter I
Example 1 -42 The absolute value y = | f(.v) | is also a piecewise function. For example,
—x
(x<0)
X
(x>0)
y = |x| =
is a piecewise function and its graph is shown in the following:
I
x
O
Fig. 1.26
Example 1-43 The greatest integer function, denoted by y = [x], is the greatest integer less than or equal
to x. That is,
(n < x < n + 1, n e Z)
y= W =n
For example,
[4] = 4,
[3.1] = 3,
[-0.5] = -!, [-3.5]
-4
The greatest integer function is a piecewise function and its graph is shown in the
following:
y
y=W
2-
1-
o
■o
*x
<?■
0
-2
■o
1
-2
Fig. 1.27
2
3
Fun<
47
Exercise 1D
1.
Which of the following functions are odd functions? Which are even functions? Which are neither odd
nor even? Why?
(b)
y a + Z?sinx
y = x + cos x + e
(c)
y - x2 sin tlx
(d)
y = 2rtanx
(e)
y = (ex — e-x)cosx
(f)
y
(a)
kl
(x*0)
0
U = 0)
X
2
X
(s)
4
y - ’ sinx2
(y < X < +oo)
(-1<X<’)
2
2
2_
.x4
2.
Show that all constant functions defined on symmetric sets of numbers are even functions.
3.
Prove that
4.
(a)
the sum of two odd functions is odd.
(b)
the product of two even functions is even.
(c)
the product of two odd functions is even.
(d)
the product of an even function and an odd function is odd.
Let f be a function on R such that
f(x - y) = f(*)f(y) - f(a -
+ >’)
where a is a positive constant.
If f(0) = 1, find f(a).
Hence show that f is even.
5.
is bounded by 72 .
Show that the function f(x) =
Vx2 -t-9
(Vx, y e R)
48 Chapter I
6.
7.
Let f be a non-constant function on R such that
(a)
Show that f(0) = 1.
(b)
Show that for all integers n,
f(x + y) = f(x)f(y)
(V.v, y e R)
f(nx) = [f(x)]n
(V.v gR)
M
Show that f is positive for all real values of x.
(d)
Use the above results to show that f cannot be bounded.
Let f be a bijective mapping defined on the set of all real numbers.
Show that if f is strictly decreasing, then fis also strictly decreasing.
8.
9.
10.
Show that the function f(x) =
x + 1 is strictly decreasing on the open interval (2, +«>).
x-2
x
is a monotonic function on the closed interval [-2, 2], i.e. f(x) is
X2 +4
either increasing or decreasing on [-2, 2].
Show that the function f(.v) =
Determine which of the functions on R are periodic. In those cases which are, find the smallest
positive period.
(a) y = x+[x]
(b)
y = [x]-x
(c) y = 1 + (-1)", where n = [x]
(d) y = sec3x
(e) y = tanx
(f) y sinx + cosx
(g) y = e
(h)
(i)
11.
y=<
sinx
fl
(x = 0)
o
(x*0)
0
((2n-V)7t <x<2nn)
sinx
(2nn<x<(2n + V)7t)
Show that the function f(x) = sinx defined on R is periodic but the function g(x)
R is not periodic.
12. Let a, b be two complex numbers with b 0, ±1, ±i.
Let f be a function such that
f(z) = ^
bz + l
Suppose |f(l)| = |f(—1)| = |f(z)| = 1. Prove that b = a.
Hence show that if | a | = 1, then f is a constant function.
(VzeC\f-U)
I bJ
sinx2 defined on
Functions 49
13.
Let f(x) and g(r) be two non-constant functions on R satisfying the relation
g(r) = f(x + r) + f(x -t) - 2 f(x)
(a)
(Vx, t e R)
If x, is the absolute minimum of f(x), i.e.
f(x,) < f(x)
(Vx e R)
gW > o
(Vr g R)
then
(b)
If x2 is the absolute maximum of f(x),
f(x2)>f(x)
(Vx e R)
g(') < 0
(Vr e R)
then
(c)
14.
Deduce that f(x) cannot have both an absolute maximum and an absolute minimum.
Let f: R —> R be bijective and a{ < a2 < ••• < an, where n > 2.
(a)
Suppose f is strictly increasing. Prove that its inverse f-1 is also strictly increasing and deduce that
-
a,
(b)
VZ i=!
J
Define h(x) = pf(x) + q, where p, q e R and p * 0.
Show that
h’1(x)=r‘(^—2- )
p
and deduce that
h i
V1J
*=1
J
(5 marks)
(HKAL 1992 Paper I)
15.
Let f: [-1, 1] —> [0, tt], f(x) = arc cosx and g: R —» R, g(x) = f(cosx)
(«)
Show that g(x) is even and periodic.
(b)
Find g(x) for all x e [0, ?r].
Hence, sketch the graph of g(x) for all x e [—2/r, 2^].
(5 marks)
(HKAL 1995 Paper 1)
16.
(«)
Suppose f : R —> R is a function satisfying
f(n + x) = f(fl - x)
and
f(/? + x) = f0-x)
for all x, where a, b are constants and a> b.
Let w = 2(« - b}. Show that w is a period of f, i.e.
f(x + w) = f(x)
for all x e R.
r
50
(
(b) Suppose g : R -> R is a periodic function with period T> 0 satisfying
g(-v) = g(-v)
for all x.
Show that there exists c with 0 < c < T such that
g(c + x) = g(c - x)
for all x.
(6 marks)
(HKAL 1999 Paper II)
Revision Exercise
1.
Let f: C -» C be a non-constant function such that
+ fe) = a f(z,) + b f(Zj)
and
f(Z|42) = ffejXUz)
(Va, b gR. Vz,, z2 gC)
(a) Prove that f(0) = 0 and f(l)=l.
(b)
By finding the value of f(/), show that
ffe) z
2.
(a)
or
f(z) = z
(VzgC)
Let a, b and c be real numbers such that
c*2
a2 + b2 + c2 = 4
and
‘b. Show that zz =
•
2-c
2-c
Hence express a, b and c in terms of z and z •
Suppose z =
(b)
Let A = {(a, /?, c): a, b, cgR. c^2 and a2 + b2 + c2 = 4).
A function f: A -> C is defined by
c)) =
(')
a + ib
2-c
(V(a, b, c) g A)
Show that f is bijective.
(ii) Let B = {(a, b, c) g A : a = b}.
Sketch the direct image of B under f on the complex plane.
3.
The function f: C \ {1} —> C \ {-/} is defined by
f(z) =
(a)
Show that f is bijective.
(b)
Find and sketch the image, under f, of each of the following:
(i)
lift
(VzgC\{1})
the lower half of the imaginary axis (including the origin),
(ii) the negative real axis.
-
W + z)
l-z
Function*
4.
(a)
51
Let f: C \ {0} —> C be a function defined by
(VzgC\{0})
f(z) = z + z
By expressing z in polar form and f(z) in the form of x + fy, where x, y g R, find the image of the
circle | z | = 1 under f.
(b)
Let A
{z gC \ {0}: | z | < 1} and g: A —> C be a function defined by
(VzeA)
g(z) = f(z)
5.
(i)
Show that g is injective.
(ii)
By using the result in (a), show that g is not surjective.
Let («, Z?) denote an open interval.
Let f: A —> B and g: /? —> C be two bijective functions. Show that g ° f is also bijective.
(b)
Show that the function f: (a, £?) —> (c, d) defined by
k,
d-c,
f(x) = - ----- (x-a) + c
b-a
(Vxg(a, W
is bijective.
(c)
It is given that the function g: (0, 7i) —> (~o©, +«>) defined by
y
(Vx g (0, 71?)
cotx
is bijective.
By using the above results, construct a bijective function from (a, b) to (■
6.
', +©o).
Let 0 be a real constant. A rigid rotation is a function f: R2 —> R2 defined by
f(U, y)) = (w, v)
where
(a)
u
COS0
-sin0
x
v
sin0
cos#
y
Show that
|a| = |f(a)|
(b)
(V(x.y)eR2)
(VaeR2)
Show that f(a + b) = f(a) + f(b) for any a, b g R2.
Hence deduce that f(0) = 0.
(c)
Show that f is bijective.
Hence show that the inverse of a rigid rotation is also a rigid rotation.
52
Chapter I
7.
Let f: R -> R be a bijective strictly increasing function.
(a) Show that f1 is also a strictly increasing function.
Hence, show that if a, < a2 < ••• < an, where n > 1, then
m
J
(b) Let f: R -> R be a function defined by
f(x)=y
(i)
(Vx g R)
Show that f is a bijective strictly increasing function.
(ii) Hence, show that
,<■1^2’-^ <11
V
8.
n
Let f be a function on R such that
2f(x) cosy = f(x + y) + f(x - y)
(Vx, y g R)
Show that for all real values of x,
(a)
f(x) + f(-x) = 2a cos x, where a is a real constant.
(b)
f(zr — x) + f(-x) = 0
(c)
f(jr — x) + f(x) = 2Z? sinx, where b is a real constant.
Hence deduce that, for all real values of x,
f(x) = a cos x + b sin x
9.
A function f: R —> R is said to be additive if
f(x + y) = f(x) + f(y)
(V.v, y g R)
(a) Let f be an additive function.
(i)
Show that for all integers n, f(nx) = n f(x).
Hence deduce that f(rx) = rf(x) for any rational number r.
(ii) By using the first result of (a)(i), show that if f is also bounded on R, then f(x) is identically
equal to zero.
(b)
Suppose g is an additive function and is bounded on the interval [0, a], where a is a positive real
constant. Let h: R —> R be a function defined by
h(x) = -^x-g(x)
a
(i)
Show that h is additive and bounded on [0, a].
(ii) Show that h is a periodic function with period a.
Hence deduce that h is a bounded function.
(Hi) Prove that
g(x) = ^x
a
(VxgR)
I itnetion\
10.
Let f: R —> R and g: R —> R be two real functions. Suppose that f is a periodic function with the
smallest positive period p.
(a)
Determine whether the composite functions fog and g o f a re periodic functions. Support your
answer by giving a proof or a counter example.
(b)
Show that the function h: R —» R defined by
h(x) = f(7x)
(c)
(Vx g R)
is a periodic function with the smallest positive period P_
T
Suppose g is periodic with the smallest positive period q.
(i)
(H)
Show that if
£ is a rational number, then f + g is also a periodic function.
<7
Show that sin x + sin tlx = 0 only when x =
nn
, where n g Z.
l + (-l)”^
(Hi) Is f + g a periodic function when — is an irrational number? Support your answer.
Q
11.
53
A function f defined on R is said to be real linear if
f(ax + by) = a f(x) + b f(y)
(a)
Show that if f is a real linear function, then
f(x) = f(l)x
(b)
(Va, b, x, y e R)
(VxgR)
Suppose f is a non-constant real linear function defined for all positive real numbers such that
f(xy) = f(x) + f(y)
and
f(xfl) = a f(x)
(Vx, y > 0, a g R)
Define a function g by
(i)
Show that
(H)
Deduce that
i
g(0 = f(10')
(V/gR)
g(r) = g(l)r
(Vt gR)
f(x) = logbx
(Vx > 0)
where Z? = 10f(l0).
(c)
Suppose h is a non-constant real linear function defined for all positive real numbers such that
h(xy) = h(x)h(y)
and
h(x°) = [h(x)]a
(Vx, y > 0, a g R)
Consider the function
H(x) = log10h(x)
(i)
Show that h(x) > 0 for all x > 0.
(ii)
Use the above results to show that
h(x) = xc
where c is a real constant.
(Vx > 0)
(Vx>0)
54
Chapter /
12. Let f: R —> R be a function such that
f(.v + y) = f(x) + f(y)
for all x, y g R.
(a)
Show that
(i)
f(0) = 0.
(ii)
f(-x) = -f(x) for all x g R.
(iii) f(nx) = n f(x) for all n g Z and x g R.
(5 marks)
(b)
Show that if there exists K > 0 such that f(x) < K for all x g R, then f(x)
(c)
Suppose there exists K>0 such that f(x) < K for all x g [0, 1).
0 for all x g R.
(3 marks)
Let g(x) = f(x) - f( 1 )x for all x g R.
Show that, for all x, y g R.
(i)
g(x + y) = g(x) + g(y)
(ii)
g(x+ l) = g(x)
(iii) g(.r)< K + |f(l)|
Hence, or otherwise, show that f(x) = f(l)x for all x g R.
(7 marks)
(HKAL 1993 Paper I)
13.
A function f: C —> C is said to be real linear if
f(az, + pz2) = af(zP + Pf(z2)
for all a, p g R and zp z2 g C.
(a)
Suppose f is a real linear function. Show that
(i)
if z = 0 whenever f(z) = 0, then f is injective;
(ii) if f(i) = i f( 1) and f(z) £ 0, then f is bijective.
(4 marks)
(b)
Suppose A,//GCand
g(z) = Az + pz
for all z gC.
Show that
(i)
g is real linear;
(ii) g is injective if and only if
(8 marks)
(c)
If f is a real linear function, find a, b g C such that
f(z) = flz + bz
for all ■ eC.
(3 marks)
(HKAL 1994 Paper I)
Advanced Level Pure Mathematics
55
Chapter 2
Limit of a Sequence
2.1
Dntroductiion
Limit is the fundamental concept of calculus. In order to study calculus efficiently, we need to have a clear
concept of limit. Let us begin our study by considering the following infinite sequence:
1.1,2 4
22
±,..„
2"
or in tabular form:
1
2"
n
1
2
10
50
200
We observe that the value of
0.5
0.25
9.77 x 10"4
8.88 x IO'16
6.22 x IO’61
decreases as n increases. If we increase the value of n infinitely, (In this
the value of — will approach
case, we say that n tends to infinity and it is denoted symbolically by n —> oo.),
—
zero. In this example, zero is called the limit value of
as n tends to infinity and we write
2"
»0 as
n —> oo.
The study of limits is divided into two categories: (1) the limit of a sequence which will be discussed in
this chapter; and (2) the limit of a function which will be discussed in next chapter.
56
Chapter 2
2.2
Convergent and Divergent Sequences
A convergent sequence {.vn} is a sequence whose terms will approach a finite value a as n tends to infinity.
For example, as shown in Section 2.1. the sequence defined by .v„ = — is a convergent sequence. In order
to study the properties of convergent sequences, it is necessary to develop the following rigorous definition
about the limit of a convergent sequence.
Definition 2-1
Let {a„} be a sequence. If there exists a number a such that for any £> 0, there exists a positive
integer N such that when n > N, we have | xn - a | < £, a is called the limit value of {x,,} and we write
lim xn = a.
n—
The definition means that when n is sufficiently large, the difference between xn and a, |x(n -fl|, is
sufficiently small. Hence, we have
Remark 2-1
Geometrical Interpretation
If the sequence {.vn} converges to a, then for sufficient large value of n, xn cluster around
a.
Recall from the definition of limit that for n is sufficiently large, xn cluster around its limit value a.
Intuitively, we cannot have two distinct limits; otherwise, it is required that an infinite number of the
members of the sequence cluster around two distinct points corresponding to the two limits, and this is
impossible. Hence we have
Theorem 2-1
Uniqueness of Limit Value
The limit value of a convergent sequence is unique.
Also, for a convergent sequence, the earlier members xn may not cluster around its limit value a. But there
are only a finite number of them and all these members are bounded. The remaining members xn though
infinitely many of them, all cluster around its limit value a, and so they are bounded, too. Hence, the
following result is obvious.
Theorem 2-2
All convergent sequences are bounded. That is, if the sequence {x„} is convergent, then there exists
M > 0 such that for all n,
k |< M
1
Limit of a Sequence 57
Example 2-1
Consider the sequence defined by
When n is even, xn > 1.
1,2,3,....)
(n
n
When n is odd, xn < 1.
Also, for sufficiently large n, it is obvious that xn — 1
(-1)" is sufficiently small.
n
lim 1 + (-1T = 1
n
n—>« I
This result can also be arrived with the help of the Fig. 2.1.
x3 x5
x6 x4
X2
X
14
0
1-1
5
1
u-L
6
14 14
Fig. 2.1
When n increases, xn cluster around x = 1 and so, xn —> 1.
If lim xn
n—»«•
0, then we say that xn is infinitely small.
If {xnJ is not convergent, the sequence is said to be divergent. It is obviously that the limit value of a
divergent sequence is not finite or its limit value does not exist. In the former case, the sequence is said to
diverge to infinity.
When a sequence is divergent to infinity, the value of | xn | increases infinitely as n increases, and we
write xn —> ©©.
It should note that if xn is always positive (negative), we usually write xn —> 4-©© (-©©). However, if we
just want to discuss the magnitude of the value of xn or the sign of xn is obvious, we usually only write
xn —> ©©. For example, as n is a positive integer, n —> ©© denotes n —> 4-©©.
If xn —> +oo and there exists a positive integer N such that when n > N, xn < yn, then it is obvious that
yn -> +o°
Similarly, if xn —>
and there exists a positive integer N such that when n > N, xn > yn. then we have
Obviously, a sequence diverging to infinity must be unbounded. Therefore, as xn becomes large, i.e.
> oo, its reciprocal must be very small, i.e. it tends to zero. This leads to the following result concerning
the limit value of the reciprocal of a sequence diverging to infinity.
Theorem 2-3
Let {xnJ be a sequence. Then lim xn = ±©© if and only if lim — - 0.
n >
An
r
58
Chapin 2
Remark 2-2
From Theorem 2-3, we have
lim — = 0
n
Example 2-2
Discuss the convergence of the sequence {qn}, where q is a real number.
Solution
(I) When q > 1, we let q = 1 + /i, where h is a positive real number. Then
?"=(i+/0"
= 1+uh + • • •
>nh
//>()
lim qn =<»
If
n —> co,
then nh —> ©«
When ^<-1, we have
lim qn = lim (-l)n| q I”
n—♦°°
n—
lim qn = 00
n—
(When |<z| > 1)
1
(2) When |g|< 1, we have j—r> Land so
Id
lim
From (a)
= oo
n—
lim qn = lim
n—n—♦«
1
T
7=0
si
(3) When q = 1, we have lim^"=l.
n—
(4) When q = -1, we have
<T
1
-1
(// is even)
(n is odd)
lim qn does not exist.
The results obtained in Example 2-2 bears importance in our future studies and is summarized in the
following:
Remark 2-3
Let q be a real number. Then
0
lim q"
n—
i°°
1
not exists
(I?I<1)
(I?I>1)
(<7=1)
(<? = -!)
Limit of a Sequence 59
Example 2-3
Discuss the convergence of the sequences
(a) x„
1 . nTC
— sin —
n
2
(b) x„=[l+ (-!)"]„
0
Solution
(a) A'n=' ±1
n
(h is even)
(n is odd)
Hence, when n is sufficiently large, | xn - 01 is sufficiently small.
.. 1 . nn n
hm—sin— = 0
n->« n
2
Hence, the given sequence {xn } converges.
(b) v
2/2
(?i is even)
0
(72 is odd)
xn is unbounded (but it does not diverges to infinity) and so, it diverges.
As shown in the last case of Example 2-2, qn is the sequence {1,-1, 1,-1, 1, -1,...}. This sequence does not
approach a finite value nor tends to infinity. Such sequences are called oscillating sequences. An oscillating
sequence is said to oscillate finitely or infinitely according to whether the sequence is bounded or unbounded.
For example, the sequence {1, -1, 1,-1, 1, -1,...} oscillates finitely while as shown in Example 2-3(b), the
sequence xn = [1+ (-!)" ]/z is an example of infinitely oscillating sequence.
Hence, there are 3 possibilities for a divergent sequence.
Remark 2-4
If {x„} diverges, then
(1) xn may diverges to infinity, e.g. xn = qn, where |g|> 1;
(2) xn may oscillate finitely, e.g. yn = (-!)";
(3) xn may oscillate infinitely, e.g. zn = [1 + (-!)'']/z.
*
60 Chapter 2
2.3
Operations on Limits of Sequences
In this section, we would like to learn how to find the limit of a sequence. The following theorem
summarizes the rules of operations on the limits of two convergent sequences.
Theorem 2-4
Rules of Operations on Limits
Let {a;} and {>’„} be two convergent sequences such that lim
(1)
lim(.vn±yw) = a±Z?
71—»<»
(2)
(3)
lim — = —. provided b * 0
yn b
The proof of the validity of these operations is omitted.
Example 2-4
Find
(a) lim-^r -7/4-2
/i—
3?f+2h + 4
Vn+T
(b) lim
n->~ 27/ + 1
Solution
1
n
2
n~
1----- F —
1--4-4
772-774-2
r
(a) 11 m —=----------- = lim
3lC 4-27/4-4 n->»
3+M
n n
12
lim 1 - lim —4-lim
n->°°
n->°° f]
n—ftn
lim 3 4-lim — 4-lim4
n—>«
n—>«> fl
T
3
3
3
'-4+±
n2__ iC
7/ + 1
(b) lim
= lim n—»oo 2rt + l
2+n
=0
ff~
= a and lim yn = b. Then
ft—>oo
Limit of a Sequent
Remark 2-5
61
(1) In applying the rules of operations on limits, we have to ensure that the limits of all
the terms exist. Therefore, if noted that both the denominator and the numerator in
Example 2-4(7?) diverge to infinity, it is illogical to write
lim (/22 - n + 2)
/22 ~ /2 + 2
lim
3/i2 +2/2 + 4
/J—
lim (3//2 +2/2 + 4)
n—><»
(2) In order to find the limit of a sequence whose denominator and the numerator both
diverge to infinity, we usually divide the denominator and the numerator by suitable
/z“, where a is the highest power among the terms in the denominator and the
numerator.
Example 2-5
Solution
(2
Find lim I - +
n—>»o
72
4/2
4/2-1
Note that
lim — = 0
m->~ n
and
4
..
4/2
lim--------- = lim
„->« 4/2 _ ]
n—>« 4-1
72
4
4
1
limf—+
n—>oo
II
4/2
4/2-1
.. 2 , r
4/i
= lim — + lim -------fl
477 — 1
=0+1
=1
Example 2-6
Evaluate
2
(a)
lim (72 --^—)
n_>oo /i +
+ 1l
(b) lim (7n+T- Vn)
n—»«<>
2
Solution
72(7? + l)-722
(a) lim (//——-) = lim
n->~ /2 +n1+ 1
72 + 1
= lim -\
„_>«> n -i- 1
= lim
=1
1
l+±72
62
Chapier 2
(b)
lim (-4~n) = lim
n—n—\ljj -}■ 1 -|- yj fl
..
= hm
1
, —-—t=^
=0
Remark 2-6
(1) To evaluate the limit of the difference of two fractions which diverge infinitely, we
usually simplify the fractions and then apply the method outlined in Remark 2-5.
(2) To evaluate the limit of the difference of two surds which diverge infinitely, we
usually multiply the conjugate surd to both the denominator and the numerator and
then apply the method outlined in Remark 2-5.
Example 2-7
Find lim
n—>oo
Solution
..fl,
hm —+
n
n
n J
y(” + D
= lim -——
•+4
fl
n~
= lim4 + -!-)
n->« 2
In
— —1
2
Remark 2-7
(1) The rules of the operations on limits can only be applied to finite number of terms.
For example, as shown in Example 2-7,
( 1
2
hm I — + —+
n
n A .. 1
2
i- n r,
+ — HHim — + hm— + --- + lim — = 0
nJ «-»«n
n~ n
(2) The existence of the limit value of lim(xn±yn) does not imply the existence of the
limit value ofx„ and yn. For example, if xn = (-1)" and yn = (—l)n+l, then
=(-ir[i+(-D]
=0
and so,
lim U„±y„) = 0
n->«>
However, both xn and yn are oscillating and they are not convergent.
(3) If lim | xn | exists, {xj may not be convergent. (See Question 5 of Exercise 2A.)
4
Limit of a Sequence 63
Example 2-8
Show that the limit value lim l_2 + 3_4 + ...+(_ir,„
n—>o»
n n n n
n
Solution
When n is even, we may write ii = 2k, where k is an integer. Then
does not exist.
±_2 + 3_4 + ... + (_irlZ, = _1__ 2_ + _3__A+
n
n
n
n
n
2k 2k 2k 2k
2k
= ^{(l-2) + (3-4) + ... + [(2Z:-l)-2^}
_-k
2k
1
2
When n is odd, we may write n = 2k + 1, where k is an integer. Then
-2~+ 3
1
2A' + 1
--^ + ---+•••+(-1)"-'n n n n
n
2)1 + 1
4
2k }2k + \
+ •••
2k + l'
2k + \ 2k + \
2k + i
1
{(l-2) + (3-4)+--+[(2Z:-l)-2q} + l
2Z; + 1
-^- + 1
2&+1
j_
2
Recall that the limit value of a convergent sequence must be unique and from the result
that the sequence
l_2 + 3_4 + .„ + (_ir,n
ii
ii
we can conclude that lim
Example 2-9
n
n
n
n
n
n
n
does not exist.
n
A sequence {.vn} is defined by
*i = 1,
and
x2 = —2
2xn.2-xn.l-xn
Show that for all positive integers n,
xn
=-1“
2
Hence evaluate limx„.
zi—><»
Solution
is oscillating between -y and
The given result is to be proved by induction on n.
(1) When n = 1,
-1-
1
2
= -l-
2
= -l-(-2)
=1
= *i
0
(n= 1,2, 3,....)
,
64 Chapter 2
When n = 2,
-1-
x
X I =-l2
o
2
= -1-1
= -2
= *2
Hence, the proposition is true for n = 1 and n = 2.
(2) Assume the proposition is true for n = k and n = k + 1, k > 1, i.e.
Jt-2
= -l-
and
X
jt-i
2
When n = k + 2,
^k+2-xk+i-xk=0
^xk+2=xM+xk
k-]~
+ -1-
=
=-2-(x
X k-2~
2
k-2
2
k-2
=-2-
2 I 2.
= -2 + f-k 2.
*U2=-1 + |(-
k-1
x
k-1
2
X
-1-
k
2
Hence, the proposition is also true for n = k + 2.
By the Second Principle of Mathematical Induction, the proposition is true for all positive
integers n.
Since
-
< 1, we have
n-2
lim
n—toa
I =o
2
and so
lim xn = -1
Limit of a Sequence 65
Example 2-10 A sequence {*„} is defined by
x1 = 1
and
^n+i+^,=y
Find limx2n.
n—»«
Solution
+
*2n
1
*2n-l
X2n-l
32/1—1
1
32n-2
1
X2n-2
+
X2n-3
32n-3
X2n-3
X2n-4
32/1-4
~~X3
X2
1
32
X2n-2
+
X2
1
*1
3
Summing up, we have
'v r
^+a'1 = 3L -3
J r=l
V
x 1-
J/
3
3
X2n + 1 = ”
JJ
1-
X2n
3?
■ \2«-i
_1_
14
_i
_1_
14
_1_
3
-1
3
1
4
3
4
Since
2/1-1
j.
1
3
- — < 1, we have
3
2n-l
lim
/!—>«>
3
I
=0
and so,
3
hmx2„ =--
r
/!-><»>
4
(n = 1, 2, 3
)
66 Chapter 2
In previous section, we have discussed some divergent but bounded sequences:
a; = HA
yn
.
lift
sm —.
1+(-D"
Now. we are going to discuss some more properties on this kind of sequences.
Theorem 2-5
Let {a;} be a divergent sequence.
(1) If lim yn exists, then {.v„ ±y„} diverges.
(2) If lim y = a * 0. then both {a; vj and •! — > diverge.
n->« ■
V
Theorem 2-6
Let {a;} be a bounded sequence. That is, there exists M > 0 such that for all n, | xn | < M.
(1) If lim yn = 0. then lim xnyn = 0.
fl—>oo
fl—»»
(a\ ± yn) = °°
(2) If limy, = oo. then nlim
—
n—
Proof
We only prove (1):
kn)’n-0| = |.r„||y„|<M|y„|
where M is a positive constant.
Since lim yn = 0, for sufficiently large n, we have I y I is sufficiently small.
Hence, A/|y„| and |xnyn-0| are sufficiently small.
n—»--<>
Example 2-11 Discuss the convergence of the following sequences:
(a) xn
L
(b) K=-sin?z
n
J2„ + i
Limit of a Sequence 67
Solution
(a) Since lim——— = —^0 and nlim
[1 + (-1)" J1 does not exist, by Theorem 2-5(2),
—>oo L
«->*• 2n +1 2
the given sequence is divergent.
(b) Since I1 sin nn-»oo
I < 1 and lim — = 0, by Theorem 2-6(1), we have
1
1-
•
lim — sin/z
n—♦<»
0
Hence, the given sequence is convergent.
1 +----1 + ••• +
1
Example 2-12 Evaluate lim n + (-l)w---n->o°
I 11-3
• 3 3*5
(2/z-l)(2/z + l)
3-5
Solution
Observing that
1
(2£ —l)(2fc + l)
1(-L2 2&-1
1
)
2A' + 1
we have
-L+_l+...+___ 1___ =f------ 1----1-3
3-5
(2n-l)(2/z + l)
(2£-l)(2fc +1)
= tS2V2fc-l
1f—
1
2Z:+1
=-f(--“) +(---) + (--“)+•”
2<1 3 3 5
5 7
1
1 ■) + ( 1
+ (■
2n-l 2n + l)
2n — 3 2/z-l
’2/i + P
Hence
+_r+...+----- 1----- <1
<(-D" _L
2
1-3 3-5
(2n-l)(2n + l)
Since
lim n = 00
n—>«
by Theorem 2-6(2),
lim ?i + (-l)'n 1.3 + 3-5 + " + (2n-l)(2n + l)
/!—>»>
= oo
68
Chapjd 2
Remark 2-8
The different possibilities of the limit values on combining two sequences are summarized
in the following table.
1
2
3
4
5
6
yn
Zn
a
b
xn±yn
+oo
a
Xn±yn
+OO
Xn + yn
+oo
+OO
lim z„
n—><x>
a±b
Indeterminate
Xn + yfl
4.00
Xn-yn
4-oo
4-oo
Xn~yn
Indeterminate
7
8
9
a
b
0
Too
xnyn
xnyn
xnyn
ab
i°°
Indeterminate
10
a
b&0)
Xn
a
b
11
a(*0)
0
12
0
0
too
yn
Xn
+00
yn
Xn
Indeterminate
yn
13
Xn
+oo
a
0
yn
14
b
TOO
+oo
yn
15
±oo
Xn
TOO
Indeterminate
yn
From the above table, we can observe that the limit values in some cases cannot be
determined. These cases are known as the indeterminate forms and they can be represented
0
A
symbolically as oo — OO, 0 • oo, —
and
0
oo
oo
Exercise 2A
1.
Evaluate each of the following limits, if any.
(a)
(c)
lim
3n3-4n2+5
4«3 + 2n-l
(b)
lim
lim
2n
^n2-\
(d)
lim[- +
n
n—
(3n + l)(2n2-3)(2-n)
n3 + l
4n —
n
Limit of a Sequence 69
(?)
(g)
(i)
lim
n2
zz2+l
n(n + 2)
zz + 1
lim n(yln2 +1 -n)
n—too
3"
lim
/!—><» 9" +1
I
1+1- 1-1
(f)
lim n
(h)
lim (Vh +1 - Vh)
(j)
.. 2n+3n
lim---------n_>« 2 -3
n
n
r
i
(k)
(tn)
lim | — + e
n
lim sin/z-f-
rt—>00
2.
(I)
n—y
n3
(n)
^ln2 + 1 ,
..
n cos nn
lim
n
, 4n(-ir
n2-2
2"-1
lim —
n j-----4- j
n—>«
Find the following limits.
..
1 + 24"3 + ",, + tz
lim------------ 5---------n—»oo
(b)
fi~
1+i? +4
+...+_L
2
2
l-1+i+4+...+±
..
3
(c)
32
3"
1
1
1
lim I -7—r +r ++
zz(zz + l)
1-2 2-3
n—»°°
10
1
1-2
(d)
lim (cosh!) -An—>“>
\ n2 +1
(e)
lim V-------------------- ^“'l + 2 + 3 + -- - + £
2«3
1
(zz —l)/z
2n2 +1
zz2-l
1
n-1
zz + 1
(f)
(n is even)
lim xn , where xn
zz + 1
(n is odd)
z? — 1
3.
Find two diverging sequences {x„} and {yn} such that
(a)
{xn ± y„} is convergent;
(b)
is convergent.
4.
It is given that alt > 0 and lim an = a. Is it necessary that a > 0?
5.
It is given that lim I an I exists. Is it necessary that {a,,} is convergent?
n—>°°
n—>o«
6.
Let a be a real number. Examine the behaviour of the sequence {xn} defined by x„ -
tends to infinity. Distinguish the cases for
(i)
|«|<1
(ii)
|«| = 1
(Hi) |«|>>
an
l + fl"+1
as n
70
Chapter 2
7.
Let a and b be two real numbers. Find nlim
—-—— . Distinguish the cases for | a | > | b | and | a | < | b |.
->« d — i)
8.
A sequence {fln} of real numbers is defined by
fill I b’'
f?o = 0,
(a)
«n+1 = «n + ^_,
and
a, = -l
(//>D
Let a (< 0) and p (> 0) be the roots of the equation x2 - x - 1 = 0.
Show, by induction on n, that
an=^(a"-p")
(b)
9.
Show that lim
n—»™
f]
= p.
Let a and b be two real numbers. A sequence {xn} is defined by
=
X| = a, x2 = b and
+ Xn
*n+2
2
(H>1)
Show that for n > 1,
a
x„ = 3
-4
n-2
+ ± 2 + (421
Hence, find lim.vn.
10.
A sequence {a;,} of real numbers is defined by
*n+l
(a)
Show that for n > 1,
Xn ~Xn-l =
(b)
0/>D
2
_1_
2
n—1
(-Tl-Xo)
Find limxn.
11. The sequence {x„} is defined by
■xntl=x;-2x„+2
(a)
(n>l)
By considering yn = xn- 1, show that
(n>l)
(b)
Find the value of lim xn for the following cases:
(i)
12.
|.r,-l|<l
(ii)
|x,-l| = 1
(Hi) |x,-l|>l
Consider the sequence, {w„} in which
“i =
“„.i = 2n Using mathematical induction or otherwise, show that
n = 1, 2,... .
2urt = 2n - 1 + (-1)",
n = 1,2,... .
Hence find lim—.
n—>« fl
(4 marks)
(HKAL 1992 Paper I)
Limit of u Sequenct
13. Express
71
x+4
in partial fractions.
x2 + 3x + 2
Hence evaluate
k=2
1
k-1
Z: + 4
k2 + 3Z: + 2
(6 marks)
(HKAL J 993 Paper I)
14.
A sequence {x„} is defined by
x, = 2
1,
and
n>2
2
(a)
Write down the values of x2 - x,, x3 - x2 and x4 - xy
(b)
For n = 1, 2, 3,..., guess an expression for xn -xn^ in terms of n and prove it.
Hence find nlimx
n.
—>oo
(7 marks)
(HKAL 1996 Paper I)
15. A sequence {«„} is defined as follows:
5
(a)
Show that an =
(b)
Resolve
1
and
an+i
— = 2n + 5
an
for n = 1,2, 3,....
1__for n = 1,2, 3,....
n~ +4/1
x+2
into partial fractions.
(x2 + 4x)2
Hence or otherwise, evaluate lim
(k + 2)a2k.
(7 marks)
(HKAL 1999 Paper I)
2.4
Sandwich Theorem for Sequences
As stated in Section 2.3, before applying the rules of operations on limits, we have to ensure that the limit of
each sequence exists. However, it is often quite tedious and difficult to prove the existence of limits from the
fundamental definition. In this section, we are going to introduce a more effective way to prove the
existence of the limit of a sequence. The technique is based on the following theorem.
Theorem 2-7
Sandwich Theorem for Sequences
Let {xnJ, {y;l} and {z„} be three sequences such that
'^n — y\ — ’/>
and
Then
lim xn = lim zn = a
n—>°°
n—too
lim yn = a
/!—><»
72
Chapin 2
From the inequality xn < y„ < z„, if both {xn} and {?„} approach to a as 7i -+ ©©, {y,,} also approaches to a.
Apparently, yn is sandwiched in between xn and zn and hence the name. The trick is to find auxiliary
sequences {.v } and {- } such that lim xn = lim z„, andy„ is sandwiched in between xn and zn. Its application
n-
is shown in the following examples.
sin n
Example 2-13 Prove that lim----- = 0. (Compare with Example 2-1
n
Solution
Since -1 < sin/i < 1, we have
1 sin 71 < 1
n ~ 71n ~ n
Also.
lim (±— ) = 0
n->~
n
So. by Sandwich Theorem for Sequences,
.. sin
lim-----
n-*o<>
1
Example 2-14 Find lim A- +
n
- —»«»
k“'
Solution
71
n
0
1__
1
+ ••■ +
(2n)2/
(h + 1)2
1
Observing that the greatest term and the least term are respectively — and
and
(2n)2 ’
n
that there are n + 1 terms, we have
71 + 1
(2n)2
1
T + --- + 1 < 71 + 1
<4
+ __
(n + l):
(In)1
n2
n
By using the result that
r 71+ 1
lim ——
n-»o«
1 . 1
lim — T
+ —27 1 = 0
n-*“’ \ 71
n
71 + 1
1 rt + 1 = 0
lim----- T = 1-lim —
(2»)2 n—»<x>
n2
,•
nJ
71
we have
lim
zi—>»
-L^+...+ (2n)1 2J = 0
4+
n
(«+D2
Example 2-15 Let a be a real number greater than 1. Prove that lim —
n->°° n
ann
Solution
0.
Since a > 1, we may let a = 1 + /i, where h is a positive real number.
Then, by Binomial Theorem, we have
(1 + h)n = 1 + nh +
n(n-l)
h2
>
2
2
Since each term is
positive
I.ini it of a Sequence 73
o<4-<
a
n
2
(n-l)/?
h2
2
Since h is a positive constant, it is obvious that
i2
lim----------- 7
„_>«
_ i)/j
0
lim
=0
n—»« d
Remark 2-9
By using the inequality
(A>0, r<n)
(\ + h)n>Cnrhr
we can deduce that
rlim —
nOt
0
(a is a constant, a > 1)
n—>«■»
Note that na —> and an —> ©o as n
faster than na as n approaches infinity.
one can interpret that an tends to infinity
Example 2-16
Find lim Vn •
Solution
For n —> oo, we have n > 1, and so Vn > 1. Then we may let
n47i = \ + h
(/«>0)
H = (l + /2)n
1 + nh + ^12 lr + 2
n(n-l)
h2
>
2
Hence,
0</j2< —
n-1
and
2
0<h< —----- >0
V 71-1
and so
lim li = 0
n—>°°
liin Vh = lim (1 + /z) = 1 + lim h = 1
n—>oo
n—>oo
n—too
(as n —> 00)
74 Chapter 2
an
Example
2-17 Let a be a real number. Prove that lim —
= 0.
1
Il J
Solution
Case (i)
a=0
The result is trivial.
Case (ii) a>0
n
Let -v = — and p be a positive integer such that p + 1 > 2a. Then
" n!
-^-<1
xp
p+1
2
a
<1
XP* =
2
P^
Vl
i = £<l
«
2
By multiplying the inequalities, we have
Xn
Xp+\
XP+2
y3
XP
XP
Xp+l
Xp*
x
n-p
2
*n-l
~P
1V
l XP
0<A'n<(^J
2
As p is a fixed positive number,
rhm MY
—
2J 2
xp = 0
Hence, lim~ = 0.
Case (iii) a < 0
By setting b = -a, we have b > 0 and
lim —= lim (-l)” —
bn
Then from the result in Case (ii), lim — = 0.
fi J
Hence, with the fact that (-l)n is bounded, lim
nI
= 0.
Therefore, for any real number a, lim —- = 0.
IlI
Remark 2-10
The result in this example demonstrates that nl tends to infinity faster than the an(a> 1)
as n approaches infinity.
Limit of a Sequence 75
Example 2-18 Let A be a positive real number and {«n} be a sequence of real numbers such that
and
ax>A
iif
A2
^, = 2^+-
2i
(n>D
(a) Show that an > A for all positive integers n.
Hence show that
a„-A<|(a„.,-A)
(n>2)
(b) Find nlima
rt.
—>oo
Solution
2
(a) From the given condition that a} > A, the result is true for n = 1.
For n > 2, by using A.M. > G.M., we have
if
1
an=2I
>
i A2
—
an-l J
( A2
i
=A
Hence, an > A for all positive integers n.
Next, consider
A
11 (
2\
a. A2 1
A
“n~A = T fln-l+T--------- A
^n-l >
= if <i+A2-2fln_lA>j
2<
fln-l
J
= lf K-i-AXa^-A)
2
«n-l
1(
,
A
= -(«„_>-A) 1- —
an-\
-A)
v flr_l>A>0
0<l- — <1
an-\
(b) Repeating application of the second result in (a), we have
^p-K-2-A)
s^r^-A)
Combining with the first result in (a), we have
0<a„-A<^zr(fl|-A)
76 Chapter 2
Since lim-J-rCa.-A) = 0, lim (an -A) - 0. and so
lim an = lim (an - A + A)
n—n—♦“>
= lim(an-A) + lim A
n—n—
=0+A
=A
Exercise 2B
1.
Find the limit of each of the following, if any.
/i!
(a)
(c)
2.
(b)
lim —
n— fl
lim , ]
n->covV/i2 + l
- --- + • • • + . 1 ■■
V/12+2
ylir+n^
(d)
11
lim —
n—>«■ fj J
1__
+••• +
lim __ 1 + , 1
V/i + ft
V/i +1 V/i + 2
n—>oo
It is given that
Zcot2
4=1
kn
2/1 + 1
/i(2/i — 1)
3
and
tan 0> 6> sin 6
(When O<0<y)
Prove that
£csc 2 kn
2/z + l
4=1
/i(2/i + 2)
3
n 1
Hence, find the limit limY—.
tr k-
3.
Prove that for any positive real numbers a and b,
f
i \
lim (an + /?)"-! =0
4.
Let {an}, {bn} and {cn} be sequences of non-negative real numbers such that an + bn + cn = 1 for all
positive integers n. {SJ is a sequence defined by
\ = • - 3(«A + bnc„ + c„a„)
It is given that Sn=S".
(a)
Prove that S„ =|[(a„-Z>„)2 + (b„-c„)2 +(c„-a„)2].
Limit of a Sequence 77
(b)
Suppose at least two of <q, b} and c, are non-zero.
(i)
By considering the range of value of
find the limit value of lim Sn.
n—
(ii)
By using the above result, show that lim [3«„ - (an + bn + cn)] = 0.
1
Hence deduce that nlim
an = —
.
—>oo
Q
3‘
5.
Let// be a positive integer. It is given that (V3 -1)" = (—I)"-1 (a„ V3 ~bn) for some integers a„ and b,
(a)
Show that b,
(b)
Prove, by induction that, bn > a„ > 2"“'.
(c)
Show that
= 3a„ + b„ and a„tl = a„ + b,
'3-— <—1
2n-l •
a„ -2'-'-
Hence, find lim—.
bn
6.
A sequence {_v„} is defined by
X| > 0
7.
and
(«>1)
l-^i- V3|<a| xn - y/3 |, where 0 < k < 1.
(a)
Show that for n > 1 >
(b)
Show that {xn} is convergent and find its limit value.
A sequence {un} is defined by
zq = 4
8.
_3(l + x„)
3 + .r„
*n+i
and
Wn+1
6'0 6
+ 11
(ci)
Prove, by induction, that un > 3 for all positive integers n.
(b)
Show that for any positive integer n,
(c)
Find lim un.
Let
»„u -3
»„"3
(h>1)
<±.
10
n—>«»
3
= 2, b{ = — and an
2
2/z
2/7-1
b„
2/1 + 1 ,
—bn_} for n > 2.
2/1
2n +1 for n > 1.
(ci)
Prove that an > bn and altbn
(b)
Using (a), or otherwise, show that a2
n> 2/1 + 1 for n > 1.
Hence find lim —.
n—>“ (i
(7 marks)
(HKAL 1998 Paper II)
78
( ihiph r 2
2.5
Monotonic Sequences
In Section 1.6, we have defined what a monotonic function is. A similar definition can also be applied to
study the monotonicity of sequences and it is restated as follows:
Definition 2-2
A sequence {x„} is said to be monotonic increasing if xn ^xn+I for any positive integer n.
A sequence {x„} is said to be monotonic decreasing if x„ > xn+I for any positive integer n.
A sequence {x„} is said to be monotonic if it is either increasing or decreasing.
Example 2-19 (a) Let x„ = -. Since
1
1
X-=^T<n=X"
the sequence {x„} is monotonic decreasing.
(b) Let x„ =
" 1
. Then from the fact that
4=1 *
n+l
.
4=1 K
,
1
+l1
nn+
n
.
4=1 K
n
.
4=1 K
the sequence {x„) is monotonic increasing.
When a sequence is monotonic increasing, it either increases infinitely or approaches a finite value. If it
increases infinitely, it diverges to positive infinity. If the sequence approaches a finite value, this value must
be greater than or equal to all the terms of the sequence. That is, the sequence must be bounded from above.
This concept is important in determining whether a monotonic sequence is convergent. A similar result can
be obtained for monotonic decreasing sequence. These results are summarized in the following theorem.
Theorem 2-8
A monotonic increasing sequence is convergent if it is bounded from above.
A monotonic decreasing sequence is convergent if it is bounded from below.
The proof of this theorem is beyond the scope of this text and the application of these results is illustrated in
the following examples.
I.ifHiI of (i Sequence 19
Example 2-20 A sequence {a,,} is defined by
1
a"=4i + 32 + l + "+rTT (n> 1)
Prove that {«„} is convergent.
Solution
The proof is separated into two parts: (I) to show that it is monotonic; (2) to check
whether it is bounded from below or it is bounded from above.
(1) v
1__
^n+l
>0
3n+l +1
«n+1 > an, and so {fl„} is monotonic increasing.
(2) We want to show that {fl,,} is bounded from above. Consider
1
0" = 3TT + 32+l
+ ••• +
1
3n+l
I
4+f+'-+f
J
3*
1
|(i~)
3
3
= l(i--L)
2
3"
4
That is, it is bounded from above by y.
Hence, {a,,} is monotonic increasing and bounded from above. Therefore, {«„} is convergent.
Example 2-21
Let {an} be a sequence defined by
= 1
and
«n =
an-\ + 1
(n>2)
Show that (an} is convergent and find its limit.
Solution
From the definition of {«„}, an > 0 for all positive integers n. So {nJ is bounded from
below by 0. Also,
an
«,.+l ~an =
«n+1
~an
<0
i.e. {a,,} is monotonic decreasing.
-«n
80
Chapter 2
Hence, {«„) must be convergent and let its limit be C. Then
lim an = lim an_} = C
n—fvo
n—
Hence, we have
lim
= lim a„-i
a„-i +1
lim «„-i
lim a„_, +1
n—n—♦«>
r+i
and so, C = 0.
lim an = 0
/I—»«O
In this example, although an > 0, we have lim an = 0, and so
/!->»
an >a
lim an > a
=>
may not hold. In general, we have
Theorem 2-9
If {an} is a convergent sequence such that an < a for all n, then the limit of this sequence cannot
exceed a, i.e.
lim an < a
n—><*
Similarly, if limfln = a, \imbn=b and an<bn for all
n—»«•»
then
n—>«»
a<b
Example 2-22 A sequence {yn} is defined by
= %/2
and
y„
2 + >’„
(a) Show that {yn} is monotonic increasing.
(b) Show that {yn} is bounded from above.
(c) Find nlimy,.
—>«»
(« > 2)
Limit of a Sequence 81
Solution
(a) The result is to be proved by induction on n.
(1) When n= 1,
y2 = ^2 + V2 > V2 = >’i
Hence, the proposition is true for n = 1.
(2) Assume the proposition is true for n = k > 1, i.e.
Xw^y*
When n
and so,
yktl-yk>0
k+ 1,
y*+2 - y*+i
(2+y*+,)-(2+y*)
= y^-yl
>o
y>c+2 — y»+i
From the definition of {yj, yn > 0 for all positive integers n.
Hence,
^'(+2 - Xl+I
yL-yt.^o
(yt+2-y»+i)(yt+2+yi+i)^°
yM-yM -°
yM > yi+i
That is, the proposition is also true for n = k + 1.
Therefore, {y„} is monotonic increasing.
(b) In order to show that {yn} is bounded from above, we have to find an upper bound for
WFrom (a),
)’n ^yLi =y„ + 2
(1)
y'-y„-2<o
(y„+D(yn-2)<o
••• y. = \ -2-y, ,
V2<y„<2
>s 2
That is, {y„} is bounded from above.
(c) From the results in (a) and (b), {y„} is convergent and let its limit be L Then
lim yn = lim yn+1 = f
n—>°°
n—*°°
From (1),
(2 = C + 2
On solving, we have
£ = -l or 2
Since 42 < yn < 2, we have
lim yn = 2
n->°»
82
Chapter 2
Example 2-23 Let {«„} be a sequence defined by
and
a, = 4
an+\
_6(a2 + l)
(«>1)
a2+11
(a) Show that an > 3 for all positive integers n.
(b) Find linifln.
n—>«»
Solution
(a) The result is to be proved by induction on n.
(1) When 7i=l, a, =4>3.
Hence, the proposition is true for n
1.
(2) Assume the proposition is true for n = k> 1, i.e. ak > 3.
When n = k + 1,
«A+1“3
_6(a2 + l)
-3
«*+H
6a2+6-3a2-33
«t2 + H
3a2 -27
“ a2+11
3(at -3)(at +3)
aA2 + ll
>0
Hence, the proposition is also true for n = k + 1.
>3
Therefore, an > 3 for all positive integers n.
(b) Before finding the limit of the sequence, it is necessary to ensure that an is convergent.
From the result in (a), {flj is bounded from below by 3. Also,
6(a; + l)
~an
a; + ll
6a;+6-a,;-lla„
a;+ 11
-(an-l)(a„-2)(a„-3)
a„2 + ll
<0
••• «„>3
i.e. {flj is monotonic decreasing.
Hence, {«„} is convergent and let its limit be C Then
lim a„ = lim a„tl = (.
n—>«
n—
and so,
£
_6(f + l)
f+11
€3-6£2 +11€-6 = 0
(€ —1)(€—2)(€—3) = 0
On solving, we have
£= 1,2 or 3
83
Since an > 3, we have
lim a = 3
n—
Example 2-24 Let a and b be two positive real numbers. A sequence {xn} is defined by
ab2+x2
a +1
Xn+I =
(n>D
It is given that 0 < x, < b.
(a) Show that {xn} is monotonic increasing.
(b) Find limxn.
,1—»oo
Solution
(a) Consider
ab2 + x2 -X.1
a +1
ab2 - ax2
a+1
(1)
From the definition of {x„}, it is obvious that xn > 0 for all n, and so
b-xn
and
have the same sign.
In order to prove that {x„} is monotonic increasing, we have to show that
(n>l)
xn<b
This is done by induction on n as follows:
(1) It is given that x, < b.
Hence, the proposition is true for n = I.
(2) Assume the proposition is true for n = k > 1, i.e.
When n = k + I,
ab2 + x2
k
-b2
a +1
a +1
(•^ - b)(xk + b)
fl + l
<0
/ a. < b
*ti^2
<b
Hence, the proposition is also true for n = k + 1.
Therefore, xn < b for all positive integers n.
From (1), we can conclude that {x„} is monotonic increasing.
>0. b>0
84 Chapter 2
(b) From the results in (fl), {xj is monotonic increasing and is bounded from above by b
and so. {xn} must be convergent and let its limit be Then
lim .v = lim .r„tl = C
/i—>~
n—
and so.
f=
ab2 + f2
fl + 1
\
(a + l)f: = ab2 + C2
at2 = ab2
(2=b2
On solving, we have
t = -b or b
Since
> 0 for all n, we have
b
Example 2-25 Let a and b be two real numbers such that a > b > 0. Two sequences {<?„} and {bn} are
defined by
a+b
2
b\ =^lab
’
n -Q-l+fr,-.
""
2
b„n = Ja
n ib„ .
\ n—J n-I
(h > 2)
(a) Prove that {a„} is monotonic decreasing and {bH} is monotonic increasing.
(b) Prove that lim an
n—»°°
Solution
n—
(a) By applying the result A.M. > G.M., we have an > bn. Hence,
an~an-\
2
~an-\
b.n-1 ~ ^/i-l
2
<0
b,
That is, {aj is monotonic decreasing.
From the definition of {«„} and {bn}y both {«n} and {bn} are sequences of positive
numbers and so, bn > 0 for all integers n.
\ •'n-l
>1
That is, {bn} is monotonic increasing.
bn_i < a„
Limit of a Sequence 85
(b) Combining the results in (a), we have
ax>a2>^>an>bn>->b2^bx
That is, {«„} is a monotonic decreasing sequence bounded from below by bx and {bn}
is a monotonic increasing sequence bounded from above by a,. Hence, both {an} and
{b„} are convergent and let their limits be h and k respectively. Then
limb = k
lim an = lim a. =h,
n—»oo
and so
h+k
h, =-----2
h k
i.e.
lim an = nlim
bn
—>«
n—>°°
Exercise 2C
1.
A sequence {«„} is defined by
a,= 1
and
^4(1+0,,)
an+\
4 + «„
(n>l)
Show that {«„} is a monotonic increasing sequence.
2.
Let a and b be two positive real numbers. A sequence [xn} is defined by
Xj = b
and
X"+I = ^T/! ("sl)
Show that both sequences {xp x3, x5,...} and {x2, x4, x6,...} are monotonic decreasing.
3.
Let {an} be a sequence of real numbers. It is known that if {«n} is monotonic increasing and for all
positive integers n, an < h for some real number /i, then {«n} is convergent.
By using this result only to show that if {bn} is a monotonic decreasing sequence such that for all
positive integers
bn > k for some real number k, then {bn} is a convergent sequence.
4.
A sequence {xn} is defined by
1 +-L.
n + 1 n +2
(a)
Show that {xn} is monotonic decreasing.
(b)
Show that {xZI} is convergent.
Prove also that — < lim xZI < 1.
2
2n
(n>D
86 Chapter 2
5.
A sequence {.v„} is defined by
1 + 22 + 32 + " + n2
(a).
(b)
6.
(h>D
. the
. result.1
for zi > 1, show that {.vj is bounded from above.
By using
—<
n~ n(n-V)
Show that {x„} is convergent.
Let A-n = (l + 67)( 1 + 672 )(1 + 673 ) - (1 + 67").
(a) If 0 < 67 < 1. show that x„ is monotonic increasing.
(b)
7.
By using the inequality 1 + x < ex, show that lim x„ exists.
Let {67n} be a sequence of real numbers defined by
67, = 1
and 67n+, = sin(an) (n > 1)
By making use of the fact that sin.v <x for 0 < x < y, show that {an} is convergent and find its limit.
8.
A sequence {«„} is defined by
67, = 1
=1+-^- (n>\)
1 + «n
and
(a) Show that {an} is a monotonic increasing sequence.
(b) Show that {an} is convergent and find its limit.
9.
A sequence {«„} is defined by
67, = 3 and
= A+5 («*•)
(a) Show that {<?„} is a monotonic decreasing sequence.
(b) Show that {an} is convergent and find its limit.
10. Let M be a real number greater than 1. A sequence {67n} is defined by
67, > M and
_Af2+67n
«n+i
(n>l)
(a)
By expressing 67^, in terms of 672/I_1, show that the sequence {67,, 673, 675,...} is monotonic decreasing.
(b)
Show that the sequence {67,, 673, 675,...} is convergent and find its limit.
11. Let a be a positive real number.
(a)
Show that the equation ax2 + 2x-a = 0 has two real roots. Also show that the roots are of
opposite signs.
(b)
Let a, = a and for n > 1,67n+l be the positive real root of the equation altx2 + 2x - a,, = 0.
(i)
By expressing 67n+, in terms of an, show that the sequence {67„} is monotonic decreasing.
(ii)
Hence, show that the sequence {67n} is convergent and find its limit.
Limit of a Sequence 87
12.
Let {an} be an increasing convergent sequence of positive numbers with the limit value equal to a.
Suppose {bn} and {c„} are two sequences defined by
bi
(a)
(b)
I
1
ci = 2«i-
1 (
\
and
^ = 2(a»-i+c-i)
a„-lb„-l
Cn
(n > 2)
Show, by induction, that
(i)
both {bn} and {c„} are strictly increasing sequences,
(ii)
bn < an and cn < an for n > 1.
Show that both {bn} and [cn} are convergent.
Hence evaluate lim bn and lime,,.
/»—>»>
13.
71—
Let {<7ZI} be a sequence of positive integers.
Two sequences {/?,,} and {c„} are defined by
(a)
b} = av
b2 =
C] = 1.
c2 = a2,
+ 1,
Ck = akCk-\ + Ck-2
Show that for any positive integer n,
(-ir1
b„^n ~ Cn^\bn
(b)
where k 1,2,3,...
ck
Show that {x2, x4, x6, ...} is a strictly decreasing sequence and {xpx3, x5, ...} is a strictly
increasing sequence.
Let x. =
(i)
(ii)
14.
(&>2)
(k>2)
+ bk-2
bk =
Show that both {x2, x4, x6,...} and {xp x3, x5,...} converge to the same limit.
A sequence {«„} is defined by
0 < 67, < a2
(a)
and
an,2
an+l +“n
2
(n>l)
Show that
«n+2“«n =
(«i -«2)
2J
Hence show that the sequence {a^-il is strictly increasing while the sequence
decreasing.
15.
(b)
For any positive integers m and n, show that a!2n-l < a2m'
(c)
Prove that the two sequences
(a)
Show that if a > p > 0, then
} and {67^} converge to the same limit.
-£->
tX + 1
(b)
Let un
v _L I 11 ~fn
is strictly
p
0+1
n = 1,2,... .
Use (a) ,or otherwise, to show that
“n < W„+l
n = 1, 2,....
Hence, show that lim un exists,
n—>«
(7 marks)
(HKAL 1993 Paper II)
88 Chapter 2
2.6
The Number e
e (2.718 281 828...) is an important irrational number in calculus. In this section, we are going to investigate
n
1 + - I .The
some of the properties of this number. One of the formal definitions for e is given as /!limf
—>«•
n
following theorem states one important property of e: the interval where e lies.
Theorem 2-10
(
1V
lim | 1 + exists and is usually denoted by e. Then 2 < e < 3.
nJ
it—>°°
Proof
The proof of the existence of the limit is divided into two parts: (J)
i+’Y
monotonic increasing; and (2)
1+1
n
► is
is bounded from above.
Il J
Firstly, consider
n(zi-l)
n(n-l)---(n-r + l) .±+...
n(n-V) • —1 + ••• +--------------------------i+'Y = . +/?• —1 +--------n2
2!
, n(n-l)--(n-n + l) 1
+
n!
' nn
n
Il J
r!
1
1 11-11+-M1-1
1
= 1+1+—
2!
3!
n
1 fl-+
+ --■+—
nl
2
1
1-2 ...fl-^1
^V^ifi-i
r!
n
n
n
n
\
Il
1-2
n
n
(*)
ii
V
Note that the general term of the right hand side is
±1-11 1-1
r!
n J\
n
\
It
As n increases, each of the terms in brackets of this general term is positive and increases.
Also, the number of terms in the expansion increases with n.
So ((1+>Y
\
nJ
is monotonic increasing.
Further, from (*), with the help of the result 1 < —Ip for n > 2, we have
r! 2
i+iY < 1 + 14----- 1---- 1— *4----- r • • • 4---2!
Il J
3!
r!
/?!
<1 + 14---- 1—— + • —I----- p + • • ■ 4------ p
2 22
2
2
=2+
1-1
2
= 3—lr
2°”
<3
Limit of a Sequence 89
That is,
Since
i+iY
nJ
is bounded from above by 3.
i+nJ
>Y
is monotonic increasing and bounded from above by 3,
1+nJ>Y is
convergent.
Moreover, from (*) and (**), we have
2<e<3
In our future study, we usually deal with exponential function and logarithmic function with base e.
The logarithmic function with base equal to e is called natural logarithm, and is denoted by
■
In a = log f a
!
Example 2-26 Express the following limits in terms of e.
(a)
1
lim 14-—
n-><» \
jn
3fl
n
1
(b) lim 14------n 4-1
3n
n
Solution
1
(a) lim 1 +-L
I = lim
n—
3« '
1
3
14- —
3fl
-J
(b)
lim 14-
1
fl 4-1
lim
1+^+T
•
\ n+I
1+^tJ
e
90
Chapter 2
Exercise 2D
n
1.
By using the result e = lim | 14- — I . evaluate
n->«
(a)
2.
(b)
(c)
(e)
lim 1 4-
lim
(■4T
\
2/17
(d)
lim I 1 +
2
n+2
(f)
/I—>»
11-1
1
lim I 14- —
(a)
n
1
lim I 1 +---- -
n—><»
n
n—
n+1
2/1
n—»«> \
1
2/1-3
-3n
1
By using the result 1 — =
II
n+1
(
3
2 Y
lim I 14--4--4
/!
1
, show that lim 1 —
n—>°°
n
14- —
1
lC )
= e.
11-1
(b)
By using the result in (a), evaluate
(i)
1
lim I 1 - —
n
3/i
n—
/
7
4
2/1-3
n—><» \
(iv)
3
lim 1- —
Q \n
(iii) lim 14---4
n n J
lim 1 -
(H)
n—>«• I
4/1
4/i
5
8/r
rt
Revision Exercise
1.
2.
Let a„ =
L_ ... i------------ !---------'-ph i----I
1 + 2 + 3 + ■ • • 4- //
1+2+3
fjG’ + l)J for/z > 1.
(a)
Show that an
(b)
Hence evaluate lima,,.
n—>«
A sequence {«„} is defined by
=3
By proving that 0 < an+l - 2 <
and
2«n
(n>l)
, show that {a„} is convergent and find its limit.
Limit of a Sequence 91
3.
(a)
Prove that for any positive integer n, there exist unique positive integers an and bn such that
(1 +V3)" = an+bn^3
(b)
For an and bn determined in (a), show that for all positive integers n,
(1-V3)”
(c)
4.
Show lhat
^_V3 < 1
~2n~l
bn
and
an>bn>2n~' (n>\)
and evaluate lim—.
n—hti
Let x be a real number such that 0 < x < 1.
(a)
By setting — = 1 + y, show that
x
Hence, deduce that
1
2
(1+y)" >-»(«-Dy
2
lim xn = lim nxn = 0
n—/!—><»
(b)
Let a be a positive real number such that na < I. Show that
(! + «)"<
(c)
1
1 -na
Show that, for 0 < x < 1,
lim (1+ x")n = 1
n—>«■»
5.
It is given that for any non-zero real number x > -1, (1 + x)n > 1 + nx, where n is a positive integer
greater than 1.
Let h be a positive constant. A sequence {«„} is defined by an = hn for n > 1.
(a)
Suppose h > 1.
_i_
(i)
(ii)
Show that /z">l.
2
By putting hn = 1 + t, prove that
/z — 1
\<hn <14--^—n
(n > 2)
(Hi) Show that lim an = 1.
n—
(b)
6.
Show that for all h > 0, lim an = 1.
A sequence {«„} is defined by
0<
<3
and
«n+l =
12
l + an
(n>D
(a)
Show that {%,+1 } is monotonic increasing while {%,} is monotonic decreasing.
(b)
Show that {a^} and {a^} are both convergent.
(c)
By using the results in (b), show that {an} is convergent and find its limit.
92
Chapter 2
7.
Let a0 and Z?o be two real numbers such that 0 < Z?o < a0.
Two sequences of real numbers {«„} and {btl} are defined by
and
and
2
(a)
Show that for any positive integer n, anbt
(b)
(i)
bn =
bn_
(n> 1)
Show that for any positive integer n, an > bn.
Hence deduce that {«„) is a monotonic decreasing sequence and {bn} is a monotonic
increasing sequence.
(ii) Show that both {an} and {bn} are convergent.
(c)
(i)
LX
1
Show that for n >0, an - bn < — (o0 ~ bo) •
2nV~°
(ii) Prove that lim an = lim bn.
n—>«>
n—>«>
Hence evaluate lim an.
1 \nn
8.
For n > 1, let an = 1 + —
nJ
(a) (i)
x
n
and b = V —.
" S
r=0 H
Show that a =2 + 7).
r!
n
n
n
Hence deduce that an < bn.
(ii) Show that {bn} is monotonic increasing and bounded from above.
(b)
Show, by induction on r, that
1_
< (1 _ l)(t _ 2)... (1 _ LlI)
n
n
n
n
(2 < r < h)
Hence deduce that
9.
(c)
Show that {«„} is convergent and lim an = lim bn.
(a)
Let P and Q be two constants and P * 1. A sequence {«„} is defined by
n—/!—>“»
an = Pan-\ + Q
Guess an expression for an (n > 1) in terms of
(b)
2)
P, Q and n. Prove your assertion.
Let p and q be two non-zero constants and p*q. A sequence {bn} is defined by
bn = (P + <f>bn-l - Pqbn-2
(i)
3)
Express (bn - pbn_x) for n > 2 in terms of (b2 - pbx), q and n.
(ii) Use the result in (a) to express bn for n > 2 in terms of Z?p b2, p, q and n.
, .
(c)
.
3_ , 1 _
r
A sequence {c„} is defined by’ c"n = -c^ +‘ ~c„_
2 for n > 2.
4~'-2 —
•
Express lim cn in terms of c,, c2.
*4
Limit of a Sequence 93
10.
Suppose {«*} is a sequence of positive numbers such that a0 = a{ = 1 and
ak = ^k-i + ^k-2>
1
Let — 7- < .v < — and
3
3
(a)
For k
k = 2,3.......
S,.W= t11/ ■
k=0
0, 1,2,... , prove that
&k+l — ^k
and deduce that
at<21
Hence prove that
5„(a)<3,
72 = 0, 1,2
(6 marks)
(b)
Prove that lim S„(x) exists and equals
1
1-A-X2 ’
[Hint: Put y = -x for the cases x < 0.]
(5 marks)
(c)
Evaluate
(i)
Ml)
k=Q
k
VJ/
1 k
(H)
jt=o
5
°°
/ i vk
1=0
V —' '
(Hi)
(4 marks)
(HKAL 1992 Paper II)
11.
Let p > 0 and p * 1. {«„} is a sequence of positive numbers defined by
=2
1
(a)
Prove that lim an
n—*°°
. 1
n = 1, 2, 3,....
0 if the limit exists.
(2 marks)
(b)
(i)
If 2 = aQ < ax < a2 < • • •, show that lim an does not exist.
(ii)
If
> ak for some k > 1, show that
> an for n > k and deduce that lim an =0.
(4 marks)
(c)
(i)
If 0 < p < 1, show that lim an does not exist.
(ii)
If p > 2, show that lim an =0.
n—»«•
Il—»oo
(4 marks)
(d)
Suppose 1 < p < 2.
(i)
Prove by mathematical induction that an <
(ii)
Prove that lim an =0.
2
for n > 0.
p-1
(5 marks)
(HKAL 1995 Paper I)
94
L.'.Piccd /< vr/ Pun Muihcmatic>
Chapter 3
o
Limit and Contm^oty
of a Function
3.1
Introduction
-L may give a general idea of what is meant by continuity. The
x
1 - - graph of y = x2 in Fig. 3.1(a) is a continuous curve while that of y = — in Fig. 3.1(b) is discontinuous (or
The following graphs of y = x:2‘ and y
X
broken) at x = 0.
y
y
o
o
(a)
(b)
Fig. 3.1
Although the graph of a function may illustrate intuitively whether the function is continuous or
discontinuous, a more precise definition of continuity is required in analysis. The definition of continuity
requires the concept of the limit of a function f(x) of the continuous real variable x which will be discussed
in the following sections.
Limit and Continuity of a Function 95
3.2
Limit of a Function
A. Limit of a Function at Infinity
In Chapter 2 of this book, we have discussed the limit of a sequence, which in itself is a function of the
positive integer variable n. Therefore, any sequence xn may be represented as
x„ = f(n)
(// = 1,2,3,...)
and so, its limit value may be written as
Hni ,v„
lim f(zz)
The limit of a function fix) of the continuous real variable x can be treated similarly. When the variable
x is assumed all values corresponding to all the points on the real line, starting at some particular point of the
line and progressing continuously to the right, x is said to tend to infinity and it is denoted by x —>
The
readers should note the different meanings between the two symbols x -» °° and //—>«>. zz —> ©o means
that n progress to the right through a series of leaps while x —> °° means that x progress to the right
smoothly. Such distinction is sometimes emphasised by saying thatx tends to infinity continuously.
Similar to the concepts developed in Chapter 2, there are three possibilities for the limit of a function
when x —> 00: (1) fix) may have a finite limit value: (2) fix) may approach to infinity; and (3) fix) may
oscillate finitely or infinitely and limit value does not exist. The fundamental concept of case (1) is as
follows:
Let fix) be a function defined on R and f be a constant. If |x| is sufficiently large implies that
| f(x) - f | is sufficiently small, i.e. fix) and H. are very close, then £ is called the limit value of fix) as
V —> 00 and is denoted by
lim f(x) = t
The limit value of f(x) as x tends to positive infinity and negative infinity are represented respectively
as
and
lim f(x) = C
lim f(x) = ^
A —> +~
The readers should note that x —> 00 represents both the cases x —> +00 and x —>
where n is a positive integer, means that n —> +«>.
Example 3-1
However, n —> <»,
1
be a function of x.
Let f(x) = —— = 1x+1
x+1
As |x| is sufficiently large,
1
—---- 1 =
x+1
x+1
are very close.
lim — = lim (I x +1
=1
X->o»
is sufficiently small, i.e.
and I
96
Chapter 3
As shown in Fig. 3.2, as x —> <».•, | f(a*) -11 -4 0. That is, f(x) approaches to 1.
y
I
I
|f(x)-l|
1 1
I
i
i
i
i
/
X'
O 1
-6 -5 -4 -3 -2-1)
-► x
3
2
4
i
Fig. 3.2
B. Limit of a Function at a Point
Besides that we consider the limit value of a function as x tends to positive or negative infinity, we also
would like to consider the limit value of a function at a point. Firstly, let us consider the following example.
The function
f(A) =
x2-4
x-2
is undefined at x = 2. When x approaches to 2, we obtain the following set of values:
x
1.9
1.99
1.999
f(x)
3.9
3.99
3.999
2
~ undefined
2.001
2.01
2.1
4.001
4.01
4.1
From this table, it is obvious that as x is sufficiently close to 2, f(x) is sufficiently close to 4. This kind of
result may be represented in the form:
x2-4
■>4
x-2
or
x2-4
lim-—— = 4
x->2 x-2
In general, the limit of a function at a point may be interpreted as:
Let a, £ be two constants. If x is sufficiently close to (but not equal to) a, implies that | f(x) - ^ | is
sufficiently small, i.e. f(x) is sufficiently close to
then £ is called the limit value of f(x) as x —> a.
lim f(x) = ^
x-ia
For simple functions, the limit value can be evaluated directly by examining the values of f(x) as x
approaches a.
Limit and Continuity of a Function
Example 3-2
97
Evaluate
” 1
(a) lim
X
(b) ..lim1
x
>">0 x +1
•'->-1 X + 1
Solution
(a) As x —» 0, x + 1 —» 1.
lim *-o ,r +1
0=0
1
(b) As x —» -1, x + 1 —> 0.
When x —> -1, the denominator approaches 0 and the numerator approaches -1, and
so,
X
x+1
increases infinitely. Hence, we have
lim —A - = oo
*->-i x + 1
That is, the limit does not exist.
Example 3-3
Let f(x) =
x2-4
x-2
(a) Evaluate limf(x).
(b) Sketch the graph of f(x).
Discuss the behaviour of the graph of the points near x = 2.
Solution
(a) *:
f(x) =
x+2
undefined
(a *2)
(A = 2)
lim f(x) = lim (x + 2) = 4
(x*2)
(b) The graph of f(x) is as shown in the following:
y
4-- -
i
i
i
i
i
i
i
i
2
-2
f(x)
A
O
2
Fig. 3.3
X? -4
x-2
x
98 Chapter 5
The graph is the straight line y = x + 2 with a missing point A(2, 4).
Geometrically. “As x -> 2. the limit value of f(x) is 4.” can be interpreted as “The
graph at the left hand side and right hand side of the point A(2, 4) both approach to A
(2. 4).”.
At this state, the limit of a function al a point can only be evaluated by examining the behaviour of the
values of f(.v) as a approaches a. However, from Example 3-3(b), we can also obtain the limit of a function
by observing the graph of f(.v). Moreover, with the help of the graph, we obtain
Geometrical Interpretation of the Existence of the Limit
Remark 3-1
“As ,v -> a, the limit value of f(x) is f.” can be interpreted geometrically as “The graph al
the left hand side and right hand side of the point (a, 0 both approach to («.
Remark 3-2
3.3
The existence of lim f(.v) is independent of the existence of f(«).
x-ta
Properties of Limit of a Function
Recall from the definition lim f(x) = £ that for a* is sufficiently close to a, f(.v) is also sufficiently close to C.
As f(.v) cannot approach two different values simultaneously, we have
Theorem 3-1
Uniqueness of Limit Value
If limf(x) = /z and lim f(x) = k, then h = k.
x—x—>a
With this property, we obtain an important theorem, the Heine Theorem, with which the limit of function
and limit of sequence are related, and is stated as follows:
Theorem 3-2
Heine Theorem
Suppose lim f(x) = (. Then for all sequences {xn} with xn -> a, we have
Jim f(x„) = £
This theorem is conveniently applied to test whether the limit of a function exists. This is illustrated in the
following example.
Limit and Continuity of a bum lion 99
Example 3-4
Prove that lim sin — does not exist.
Solution
Consider the two convergent sequences {x„} and {y„}:
.t-,0
x
1
Xn
1
yn =
2nn + —
2nn---2
Obviously, both sequences converge to 0.
However,
1
sin —
sin(2n7r + y) = 1
Xn
sin — = sin(2/z/r - —)
yn
2
-1
. 1
sin —
1
sin —
yn.
That is, the two sequences
and
Xn
converge to different limits. Hence,
by Heine Theorem, lim sin— does not exist.
x->0
x
The results on limits of sequences are also valid for limit of functions. They are summarized in the following
theorems.
Theorem 3-3
If lim f(x) exists, then f(x) is bounded at the points near x = a. (The result may not hold at the point
x—»a
x = a.)
Theorem 3-4
Let limf(x) = /z and limg(x) = k.
x—>a
x—*a
If h > k, then for all the points near a, possibly except the point al x = a, we have
f(x) > g(x)
Conversely, if f(x) > g(x) for all the points near a, possibly except the point at x = a, then we have
h > k.
The operations on limits of sequences are also valid for the evaluation of the limit of a function. These
results are summarized in the following theorem.
100
Chdptei
Rules of Operations on Limits
Theorem 3-5
Let f(x) and g(x) be two functions defined on an interval containing a, possibly except a. If
lim f(x) = h and lim g(x) = k, then
(1)
lini [f(x)±g(x)] = /i±Zr
(2)
lim f(x)g(x) = hk
(3)
—- g(.r)
Remark 3-3
k
(if k * 0)
The above results on limits of functions as x -> a can also be applied to the cases as
x —>00.
In order to simplify the process in evaluating the limit of a function, we have to introduce Theorem 3-14 and
Theorem 3-15 in Section 3-7 of this chapter first.
Theorem 3-14
Let F(x) be an elementary function. If F(fl) is well-defined, then
lim F(x) = p(lim x) = F(a)
x-)a
\x—>a
/
This theorem states that for any elementary function F(x), if F(a) is well-defined, then the limit of F(x) as
x —> a can be evaluated by directly substituting x = a in F(x).
(a+3)(2a--1)
x2 + 3x - 2
Example 3-5
Evaluate lim
Solution
Since polynomials are elementary functions, we have
lim
(x + 3)(2x-l)
x2
x2+3x-2
+3x-2
[(-l) + 3][2(-l)-l]
(-1)2+3(-l)-2
2(-3)
(-4)
=3
2
Limit and Continuity of a Function 101
Theorem 3-1 5
Let lim f(x) = C. If F(x) is an elementary function such that F(0 is well-defined, then
x—>o
lim F[f(x)] = pflim f(x)l = F(£)
[x—»a
x-*a
J
This theorem helps us to evaluate the limit of composite functions.
Example 3-6
Evaluate
(a) lim
(b)
Solution
lim
10 000x4-1
x2 — 1
Vx
x—>+« V1 + 3x
In this problem, both the denominator and numerator tend to infinity and so, we cannot
apply the rules of operations on limits directly. However, by dividing the denominator
and the numerator by suitable x“, the problem can be solved.
(a) As x —> °°, — —>0, ~v—*0. Hence,
X
X
10 000 , 1
4" •>
x
x2
10 000x4-1
= lim
lim
x2 -1
_
1-4
X
10 000 4- lim -L-
lim
x_____ *-»” x
lim 1 - lim 4
X-»o»
X—X~
_04-0
1-0
=0
(b) Since Vx is an elementary function.
lim
X—>+~
Vx
V1 + 3x
lim------------ —
VX
1
= lim —
X—>+«
-4-3
X
1
J lim -4- lim 3
V ,x—Y
"V3
x—»+•»
I
102
Cliuph
3
Example 3-7
Evaluate
X* + 1
(a) rInn-----*-*-> x + l
/ i
(b) lim
x—»0
Solution
(1+x)" —1
x
In this problem, since both the denominator and the numerator tend to zero, we cannot
apply the rules of operations on limits directly. However, for rational functions, the
problem can be solved by eliminating the factor that tends to zero.
z > r x3 + l r (x + l)(x2-x + l)
(a) hm-------= hm ------- ----- -------x+l
*-*->
*->-i x + l
= lim (x2 - x + l)
=3
(b) By Binomial Theorem
(1 + x)n -1 = (1 + Cfx + C2 x2 + • • • + C>n) -1
= Cfx + CJx2 + --- + C>n
C"x+C;x2+- + C"x"
.. (l + x)"-l
hm ------ ------ = h m —-------------------- -—
’-►o
x
x-»0
x
,n-l
= lim (Cf+C2x + --- + C"x
n
= C,"
=n
Example 3-8
Evaluate
(a) lim
(b)
Solution
2-x
3-7x2+5
lim (7x2 + x - x)
In this problem, since both the denominator and the numerator tend to zero (in (a)) or the
difference of two terms which tend to infinity (in (b)), we cannot apply the rules of
operations on limits directly. However, by multiplying the conjugate surd, the limit can
be evaluated.
Limit and Continuity of a Fiinciion 103
(a) lim
(2-a)(34-7a2 4-5)
2-A' _
= lim
3-/a2 4-5
(3-Va2 4-5X34-/a2 4-5’)
= lim(2 x)(3+Ap7i)
4-x2
X-.2
lim
(2 - ,r)(3 + t/.y2 +5)
(2-x)(2 + x)
3 + Vx2+5
2 + .v
= 3 + 722+5
2+2
=3
2
= lim
(b)
lim
lim
X—>-H»
X->-t-~
= lim
= lim
A2 + A - a)(7a2 + A + A)
(V7
VA2 4-A 4-A
A
7a2 4- A 4-A
1
J14--4-1
N
A
2
Example 3-9
Solution
If lim
A3 - 6/A2 - A 4- 4
exists, find the value of a and the limit value.
A 4-1
Since the given limit exists and the denominator (a + 1) —» 0 as a —> -1, the numerator
must be divisible by (a + 1). Then, by Factor Theorem, we have
(-1)3-a(-l)2-(-1)4-4 = 0
-l-fl + l + 4 = 0
a=4
and so,
lim
A3 ~ 4A'2 - A + 4 _ lim (a-4-1)(a2 -5A4-4)
v->-1
A 4-1
A 4-1
= lim (a2 - 5a 4- 4)
= (-l)2-5(-l) + 4
= 10
104 Chapter 3
There are some properties concerning the properties of limits of bounded functions (The limits may not
exist.). They are similar to that of sequences and are summarized in the following theorems.
Theorem 3-6
Let f(x) and g(x) be two functions defined on an interval containing a, possibly except a.
Suppose f(x) is bounded at the points near a, possibly except the point at x = a. If lim g(x) = ©o and
lim h(x) = 0, then
x—>a
(1)
lim [f(x) + g(x)] = ©o
(2)
lim f(x)h(x) = 0
x—>a
Example 3-10 (a) Asx-»0, sin— <1 and lim ln|x| =
lim (sin — + In lx I) =
x-»0
x
(b) As x—>+<», |sinx|<l and lime"x=0.
lim e"xsinx = 0
X-4+»
Exercise 3A
1.
Give an example to show that the existence of lim[f(x) + g(x)] does not imply the existence of
limf(x) or limg(x).
x—x—>a
2.
Give an example to show that the existence of Tlim f(x)g(x) does not imply the existence of lim f(x)
x-»a
x—>a
or limg(x).
x-»a
3.
Evaluate the following limits.
(a)
.. 2x4-3x2 + l
lim —i
5-----6x4 — x3 — 3x
(c)
Jim
(e)
(b)
rlim —
x47-3x
------2
x +1
2x3
x2 +1
(d)
lim (3x-l)(2x + 3)
(5x + 3)(4x - 5)
r 2X-2"X
lim —------ 2 +2
(f)
r x + cosx
lim---------x->« x - 1
Limit and Continuity' of a Function 105
x—>oo
(i)
X—
(k)
3x
x-1
lim
(s)
2x
x-f-l
5x —7
'x +Vx
lim
lim (Vx 4-1 — Vx )^[x + 2
lim (7x2 4-1 -x)
(o)
lim
x—>0
lim -------—---------
(J)
lim (7x4 +1 -x2)
Vx+1
W
X—>+•»
(m)
Jx + yx + Vx
(II)
x -1
x2+3-2
(n)
lim
(P)
.. V2 4- x — 2
hm - -----------x-2
-^->2
(q)
x2-l
lim
V"i* 3x2 -x-2
(r)
.. x2—5x4-6
hm —-------------
(s)
lim
x2—3x + 2
x2 + 5x - 6
(t)
lim
x->0
(1 + x)4 -(1 + 4x)
x + x4
..
4r-4"x
(u)
lim(l + 2x)(l+3x)(l + 4x)-l
hm--------- —
x-»o 4 4-4
(v)
X
x->0
*->3 x 4-2x-15
4.
Let m and n be two positive integers. Evaluate lim
-»->o
5.
By using the Heine Theorem, show that lim cos— does not exist.
6.
Determine whether the following limits exist. If so, find the limit values.
x->0
(a)
(c)
i-
•
x
1
(b)
lim xsin —
x-»0
X
(d)
lim (x + sinx)
X—>«*>
3.4
d +_wx)
x -x
+
rhm —
1 sinx
•
X->« J
lim xtan
x
X—>«»
Two important Limits
The two limits, which are stated in Theorem 3-8 and Theorem 3-9, are important in our studies. Before
introducing the first result, we have to study Sandwich Theorem for Functions and a lemma first.
Theorem 3-7
Sandwich Theorem for Functions
Let f(x), g(x) and h(x) be three functions defined on an interval containing a, possibly except a. If
f(x) < g(x) < h(x)
and
lim f(x) = lim h(x) = £
x-»«
x-»a
then
lim g(x) = £
X—
106
Chapter
Example 3-11
Show that
x cos — <|x| for all x # 0.
x
1
Hence evaluate lim x cos —.
x-»0
Solution
X
I
For all x#0. cos— <1 and so,
x
1
cos — < | X |
X
xcos — <|x|
Hence, for all x * 0.
-|x|<xcos— <|x|
Since
lini I x I = lim (-I x I) = 0
x->0
x-»0
we have
1-
I
,r->0
x
A
hm xcos — = 0
■■■■■■
In order to prove Theorem 3-8, we have to apply the following lemma.
Lemma
If -y<x<y,then | sin x | < | x | < | tan x |.
Proof
0<x<2
The result is proved with the help of the
diagram shown in Fig. 3.4. In Fig. 3.4, the
tangent PT cuts the unit circle at T.
Case (i)
P
A
Let Z.POT = x. Then PT = tanx.
Since
Area of
AAOT
tan x
1
<
Area of
<
sector AOT
Area of
&POT
we have
x
1
T
O
0 < 7-sin x < — x < — tan x
2
2
2
That is,
| sin x | < | x | < | tan x |
Fig. 3.4
Limii and Continuity of a Function 107
Case (ii)
2
<x<0
Then 0 < -x < y, and by applying the result in Case (i), we have
|sin(-x)|<|-x|<|tan(-x)|
| - sin x | < | - x | < | - tan x |
|sinx|<|x|<|tanx|
Hence, by combining the results in Case (i) and Case (ii), the result is proved.
Theorem 3-8
..lim------sinx = lim ■
-=1
,r-»0 X
sinx
Proof
Since x —> 0, it is obvious that -y < x < y and x
0. Hence, by applying the lemma,
| sin x | < | x | < | tan x |
1<
1<
a
As
K
n <x<—
2
X
2
2 ’ sinx
1*1 J
< tan*l
| sinx |
| sinx |
X
> 0 and
1<
1
cosx
<
sinx
1
> 0. Then, we have
cosx
sinx
1
cosx
(1)
Since
2 A-
1 -cosx = 2sin2 y <
2
(1) can be rewritten as
-1 <
sinx < - cosx
x
< 1 - COSX < 7-x2
2
X
As x -> 0, and by Sandwich Theorem for Functions, we have
sinx\
x
n
lim (1 - ----- ) = 0
x—>0
108
Chapter 5
sin x
Now, as both lim 1 and lim (1------ —) exist, we have
x-»o
x
x-*0
sinx r (,
sinx
sinx x
lim------= lim 1-(1--------- )
x-»0
x-»0 '
x
-V
= lim 1 — lim (1 x->0
x->0
7
sinX)
x
= 1-0
=1
Also,
lim —
*-»o sinx
—sinx
k— = 1
hm-----x
x-»0
Example 3-12 Evaluate
(a) lim
x-»0
tan x
x
.. sin 4x
(b) lim
■’-o sin5x
Solution
(a)
lim
x->0
tan x
sinx
1
= lim----- •
x->0 j cosx
x
sinx
= lim------
v-»o
x
i-
1
hm------*-»o cosx
1
.. j sin 4x
sin4x
5x
4
(b) lim
= hm------x-»o
4x
sin5x
sin 5x 5
4
5
Example 3-13 Evaluate lim 1 - cosx
x-»0
Solution
X2
2sin2A
1 -cosx
= lim---- ——
x-»0
x->0
x2
x'-*•
lim
( sin —xV
= lim —
x-»o 4
2
x
I 2 J
2
Limit and Continuity of a Function
109
Example 3-14 Evaluate lim (x-—)secx.
X->2‘
lim (x--^)secx = lim ysec(— + y)
x->y
2
>-»o
2
Solution
= lim
>•—>o
y
COS(y + y)
Lei
zr
x----
When
/r
= -lim ——
>->o siny
= -l
In Section 2.6, we have proved that
n
lim | 1 + —
n
=e
In the section, we are going to prove a similar result, which is stated as:
Theorem 3-9
lim I 1 + —1
Proof
Case (i)
X
lim (1 + a)a = e
a-»0
X
X—
+oo
For all positive real number x, we have
[x] < x < [x] + 1
1 <1<-L
W+l A' [X]
l + [.v]+l <! + — <!+ —
[.r]
a
l+w+i r<o+i "U + wJ
xlxM
/
It is obvious that as x —> +°°, [x]
(1)
x
\
+°°. Since [x] is an integer, we have
1*1+1
lim 1+—1— II
X—
W+l
U)
= lim
[x]->+«>
lxj+l
lim
a—>+«
I+w
I
i
= lim
[iHi«
1 + —-—
W+l
__ 1_
1+
W+l
=e
111
1 + [.V]J 1 + — | = e
110
Chapter J
From (1) and by Sandwich Theorem for Functions, we have
lim fl + -' = e
.V.
•*-+- k
Case (ii) x —> -<»
Let y = -x. Then
lim h1 —1
v->+~ I
y
£_
= lim
1
= lim | 1 + ——
y-1
y-1
=e
From Case (i) and Case (ii), we have
(
1Y
lim 1 + — = e
k xJ
By putting a
1
.
—, we obtain
x
1Y
lim(l + a)a = lim 1 + —-I = e
a-»0
k x
2
Example 3-15 Evaluate lim(l-3x)x.
x-»0
Solution
Let a = -3x. Then
1
_2
lim(l-3x)x =lim (1 + a) a
x->0
a-»0
= lim (l + a)°
a-»0
= | lim(l + a)a
I a-»0
=e
Limit and Continuity of a / tuniion
Example 3-16 Evaluate lim
Solution
x2
x2 4-1
x2 -1
Method I
/
iI V
We want to transform the denominator and numerator to the form of lim 1 + — .
U—>oo
lim
1+p-
= lim
1+z
x2 4-1
x2-l
lim
X—>«
X—
l+z
= lim
and so
As x —> oo, x2 —>
= lim fl —v
lim f 1+-V
r-
—
X
—>« i
x _
x->~
x2 +1
x2 -1
lim
x->«
=e
x2
= e2
Method II
We want to express
x2+l
in the form of I + a.
x2—1
l+a=
Let
a=
x2 + l
x2-l
x2 4-1
2
-1 =
x2 -1
x2 -1
Then
2x2
lim
X—>«»
x2 4-1
x2-l
= lim
1+
X—
2
x2 -1
As x —> 00 5 a —> 0, and so
_i_
lim (14-a)a =e
a->0
Also,
..
2x2
lim
x2 4-1
x2 -1
hm -j— = 2
»-►“ x -1
= e2
2
u
111
112
Chapter 3
Similar examples have been discussed in Section 2.6. More exercises on this limit are in Exercise 3B.
Exercise 3B
i.
2.
Evaluate each of the following limits, if any.
(a)
sin5x
lim------3x
(c)
lim
(e)
lini 6* - sin 2a
’-♦o 2x + 3sin4x
A-*0
sin 3x + sin x
x2
(b)
sin4x
lim------*-*o sin7x
(d)
.. cos5x-cosx
h m------- 5-------.. 1 -2cosx + cos2x
l>m
-----------X —-------X-.0
(g)
x—>0
tan ,v - sin x
sin3 x
(h)
(i)
lim (.r-3)csc?rr
(j)
(k)
lim tan2xtan(— n- x)
rZU
4
(I)
lim
x
x->0
lim
x-»0
sinx2
x
.. sinx
h m-------
•<->*
-x
lim (1 -x)tan —
2
Evaluate each of the following limits, if any.
(a)
(c)
i- fi 1
lim 11 —
(b)
3Y
lim | 1 + —
x2)
(d)
A-
x-»«-
(e)
lim (1 + tanx)00'1
(g)
lim
3.5
r—>o°
x2+2xY
x2-l J
( 2
lim I 1+x
x—>°°
lim
X+1Y
x—
X-l J
(J)
lim
x2 -2,v- 3
x2-3x-28
(h)
lim (cosx)co,r
X
x—>0
Left and Right Hand Limits
When x is said to tend to a, it means thatx approaches a from both the left hand side and the right hand side.
However, in particular cases, the function has different behaviour on the left and right hand sides of a, and
so we have to consider x approaching a from the left hand side only or from the right hand side only. These
particular cases include the function defined differently on the left and right hand sides of a, for example, the
trigonometric function f(x) = tanx is greater than any positive real number if x approaches y from the left
hand side; while it is less than any negative real number if x approaches y from the right hand side. Another
case is that the function is only defined on one side of a. For example, the function g(x) = Jx - 2 is only
Limit and Continuity of a Function
113
defined on the right hand side of 2 and it is meaningless to consider the limit value as x approaches 2 from
the left hand side.
If .v approaches a from the left hand side, we write x -» a"; if x approaches a from the right hand side,
we write x -» a+. The fundamental concept of left and right hand limits are as follows:
Let a, C be two constants.
If x approaches a from the left hand side, i.e. x < a and x a, then £, the limit value of f(x) as x
tends to a, is said to be the left hand limit of f(x) and is denoted by
lim f(x) = £
If x approaches a from the right hand side, i.e. x > a and x —> a, then
tends to a, is said to be the right hand limit of f(x) and is denoted by
the limit value of f(x) as x
lim f(x) = £
From Section 3.2, if lim f(x) = £, then f(x) tends to f no matter x approaches a from the left hand side or
x-*a
from the right hand side. Hence, we have the following theorem.
Theorem 3-1 0
lim f(x) = lim f(x) = a.
X—>«’
lim f(x) = f
X—>rt'
x-*a
Example 3-17 Since the function f(x) = y/x-2 is only defined for x > 2. it is obviously
lim a/x - 2 = 0
lim a/x - 2
does not exist
Iim yJx-2
does not exist
Example 3-18 Show that each of the following limits does not exist.
(a)
x
lim -—x->0 x
(b) lime*
x—>0
114
Chapter 5
(a) Since
Solution
kl
x
(a>0)
-x
(A'<0)
lx I
lim j—l = lim —
x-*0-
X
-t->0-
,v
lx I
r
A*
-1
.
lim
= lim — = 1
x-»0‘ x
x-»o- x
The left and right hand limits are not equal, and so the given limit does not exist.
(b) With the substitution — = y, we have
x
2
lim ex lim ey = 0
x-»0"
_i_
lim ex
x->0'
lim ey = +°°
Hence, the given limit does not exist.
Exercise 3C
1.
2.
Evaluate each of the following limits, if any.
(a)
lim (x-[x])
(b)
lim
(c)
lim
x-»o- \-ylx
(d)
..
1-x
lim .
Determine whether the following limits exist. If so, find its limit.
(a)
(c)
lim
1
1
3 + 4'
x-2
lim
|a-2|
.t—>o
(b)
rlim —
[a]
(d)
lim A--[X]
|a-4|
3.
Vx- 1
Sketch the graph of y = f(x), where f(x)
4-x
0
x->0 X
(a #4)
(x = 4)
Limit and Continuity of a Function
3.6
115
Continuous Functions
In Section 3.1. we have a rough idea of what continuity means. In this section, we are going to give a precise
definition for continuous functions. Let us begin our discussion by considering the following example first.
x2-4
(see Fig. 3.5) is the straight line y = x + 2 with a missing
x-2
point (2, 4). Hence, the graph is broken at x = 2. In order to have a continuous graph, we have to add back
the missing point (2, 4) and so, we obtain the function
The graph of the function f(x)
x2-4
F(x) =
x-2
4
(x*2)
(x = 2)
Obviously, the graph of F(x) is continuous.
y
y
y=F(x)
4------
x2 -4
x-2
-2
o
/
4----i
i
i
i
i
i
i
i
2
f(x) =
A
/
I
'
I
I
I
I
I
I
I
x
2
-2
o
X
2
Fig. 3.5
Hence, the necessary and sufficient condition for F(x) to be continuous at x = 2 is
lim F(x) = lim F(x) = F(2)
x—>2"
x-»2*
In general, we have:
Definition 3-1
Let f(x) be a function defined on an interval containing a. f(x) is said to be continuous at x = a if
lim f(x)= lim f(x) = f(fl)
x-»a‘
x-)<f
i.e.,
lim f(x) = f(«)
x—»a
116 CM
Remark 3-4
Sometimes, the condition lim f(x) = f(r/) may be rewritten as
x—*a
lim f(ti + h) = f(fl)
/>->0
Definition 3-2
A function f(x) is called a continuous function in an open interval (a, Z?) if it is continuous at every
point in (a, b).
A function f(x) is called a continuous function on a closed interval [«, b\ if it is continuous in the
open interval (a, b) and
lim f(x) = f(Z?)
lim f(x) = f(tf),
x-*b'
j-»a*
Example 3-19 Consider the function f(x)
Since
1 +X
(-v<l)
1
U = l)
4-2x
(a>1)
lim f(x) = lim (l + x) = 2
lim f(x) = lim (4-2x) = 2
f(l) = l
lim f(x) = lim f(x)
f(l)
f(x) is discontinuous at x = 1 as shown in the following figure.
y
2i
i
i
i
i
i
i
O
y=f(x)
-► x
1
Fig. 3.6
Remark 3-5
If the left and right hand limits at x = a are equal, but not equal to f(«), then f(x) is
discontinuous at x = a. This type of discontinuity is called a removable discontinuity.
Limit and Continuity of a Function 117
Example 3-20 The graph of the greatest integer function f(x) = [x] is as follows:
y
2-
■o
1-
■o
X
O
3
2
1
Fig. 3.7
Obviously,
lim [x] = 1
x—>2'
lim [x] = 2
Hence, the greatest integer function is discontinuous at x = 2.
Remark 3-6
(1) If both left and right hand limits at the point x = a exist but they are of different
values, then f(x) is discontinuous at x = a. This type of discontinuity is called zjump
discontinuity.
(2) Removable discontinuity and jump discontinuity are called first kind discontinuity.
Example 3-21
Consider the function f(x) = —. Now
x
lim —
A->0' x
lim —
r->0- X
Hence, f(x) is discontinuous at x = 0 as shown in the following figure.
y
y=—
X
-------- ► x
O
Fig. 3.8
118
Remark 3-7
If the limit value at the point of discontinuity is equal to positive or negative infinity, then
this point of discontinuity is called an infinite discontinuity.
Example 3-22 Consider the function f(x) = cos — ,
x
A sequence {x„} is defined by:
*n =
__ 1
2nn + a
f(xn ) = cos(2n7r + a) = cos a
Obviously, as n —>
0 and f(x„) —» cos a.
x„
Hence, the limit value ofx„ is independent of ctand that of f(x„) depends on the value of a.
That is, for different value of a, f(xn) converges to different limit values, and so lim cos —
x->0
x
does not exist.
Therefore, f(x) is discontinuous at a- = 0.
Remark 3-8
(1) If the limit of f(x) at the point x = a does not exist, then f(x) is discontinuous at x = a.
In Example 3-22, the discontinuity is called a finite oscillatory discontinuity.
(2) Infinite discontinuity and finite oscillatory discontinuity are called second kind
discontinuity.
Example 3-23 Let f(x) = <
. 1
x sin —
x
a
(x*0)
(x = 0)
Find the condition that f(x) is continuous at x = 0.
Solution
Since sin— < 1, we have
A'
lim a* sin — = 0
x-»0
X
On the other hand, f(0) = a.
Hence, f(x) is continuous at a* = 0 if and only if a = 0.
Limit and Continuity of a Function 119
Example 3-24 Let f(x) = lim
x2n 4- 3
X 4n>! + t
Locate its point of discontinuity.
Hence sketch the graph of f(x).
I
Solution
When
lim x2n = lim x4fl+I = 0
2n+3
f(x) = lim —,4n+l
=3
x
+1
When
|x|>l,
lim x2n = lim x,4n+l = oo
f(x) = lim
1
3
x2n+l
4n+l
ZI->«
1+
i
----- = 0
,4n+l
X
When
x = 1,
f(l) = 2
When
x = -l,
f(-l) does not exist
0
(x<-l)
undefined
(x = -l)
f(x) = -3
(—1<X<1)
2
(x = l)
0
(x>l)
lim f(x) = 0,
lim f(x) = 3
lim f(x) = 3,
lim f(x) = 0
Hence, f(x) is discontinuous al x = ±1, and its graph is shown below:
y
?
i
i
i
i
i
i
i
i
i
i
3
y = f(x)
—9
i
;
2-
i
i
i
i
i
i
---- o----
O
Fig. 3.9
1
x
120 Chapter 3
Example 3-25 Let f(x) be a real-valued function such that
(1) f(x + y) = f(x)f(y)
(Vx,jgR)
(2) f(x) is continuous at x = 0 and f(0)
0.
Show that f(0) = 1.
Hence show that f(x) is continuous on R.
Solution
From (1), we have
f(0) = f(0)f(0)
v f(0)*0
f(0) = l
For any a g R, we have
lim f(x) = lim f(a + /i)
x—>a
A-»0
= lim f(a)f(/i)
h-»0
= f(fl)limf(/z)
A—>0
= f(«)f(O)
= f(a)
Hence, f(x) is continuous at every a g R, i.e. f(x) is continuous on R.
3.7
A.
Properties of Continuous Functions
Continuity of Elementary Functions
Theorem 3-11 Rules of Operations on Continuous Functions
Let f(x) and g(x) be continuous at x = a. Then at x = a,
(a)
f(x) ± g(x) and f(x)g(x) are also continuous at x = a\
(b)
f( y)
—— is also continuous at x = a, if g(a)
g(x)
Proof
0.
Since f(x) and g(x) are continuous at x = a,
lim f(x) = f(a),
x—
lim g(x) = g(o)
x—ta
and so,
lim f(x)g(x) = lim f(x)lim g(x)
x—x—»a
x-^a
v f(x) is
continuous at
x=0
Limit and Continuity of a Function 121
Hence, f(x)g(x) is continuous at x = a.
The proof for other properties is similar and is left for the readers.
Theorem 3-1 2
Let g(x) be continuous at x = a and f(x) be continuous at x = g(a).
Then the composite function f °g(x) = f [g(x)] is continuous at x = a.
Proof
Since g(x) is continuous at x = a, we have
lim g(x) = g(fl)
x—fa
f^limg(x)j = f[g(fl)]
Further, as f(x) is continuous at x = g(a),
lim f [g(-v)] = f [lim g(x)j = f [g(a)]
Hence, f o g(x) = f [g(x)] is continuous at x = a.
Remark 3-9
All basic elementary functions are continuous functions on their domains. These
basic elementary functions include constant function, power function, logarithmic function,
exponential function, all trigonometric functions and all inverse circular functions. For
example, the exponential function f(x) = ex is continuous on R, while the logarithmic
function f(x) = log,x is continuous in an open interval (0, <»).
Since elementary functions are functions obtained from some basic elementary functions through the
operations of addition, subtraction, multiplication, division and composition, by Theorem 3-11 and Theorem
3-12, we have
Theorem 3-13
All elementary functions are continuous on its domain.
122
Chapter 3
Example 3-26 (a) F(x) = cos71 + x2 is obtained by the composition of the following basic elementary
function:
2
U = V2,
f(w) = COS it,
Obviously. F(x) is well-defined on R. Hence, by Theorem 3-12, F(x) is continuous on
R.
(b) G(x) = (sinx) sin x = e sin x In sin x
When 0<x< —, sinx>0, Insinx is well-defined.
2
Also, G(x) is obtained by the composition of the following basic elementary function:
G(it) = e“,
u = (sinx)v
v = In vv,
vv = sinx
Hence, G(x) is continuous on (0, y).
Example 3-27 Let f(x) =«
x2-2x + l
sinx
(x>0)
(x<0)’
Determine whether f(x) is continuous at every x g R. If not, locate the point of discontinuity.
Solution
Since polynomials and sine functions are elementary functions, they are continuous on
their domain, and so f(x) must be continuous for x > 0 and x < 0. Hence, we need only to
consider the point at which the definitions on the left and right hand sides are different.
Such a point is where x = 0.
Now,
lim f(x) = lim (x2 - 2x +1)
=1
and
lim f(x)= lim sinx
x->0‘
x->0‘
=0
So, the limit value at x = 0 does not exist. Therefore, f(x) is discontinuous at x = 0.
Theorem 3-13 provides a convenient method to evaluate the limit of a continuous function. This is done just
by substituting the value of a forx. That is,
Theorem 3-14
Let F(x) be an elementary function. If F(a) is well-defined, then
lim F(x) = F^lim x) = F(a)
x->a
\x—>a
/
Limit and Continuity of a Function 123
By using this result and Theorem 3-12, we have
Theorem 3-1 5
Let limf(x)
x—>a
£. If F(x) is an elementary function such that F(£) is well-defined, then
lim F[f(x)] =
F(0
Example 3-28 Evaluate
(a) lim
x—>0
ln(l + x)
x
(b) lim tan(
x—>0
Solution
sin/tX)
x
(a) Since lim (1 + x)x = e, In it is an elementary function of u and In e is well-defined,
x—>0
we have
limine
x->0
x
£
lim ln(l + x) r
x->0
f
In lim(l + x)x
I x->0
= Ine
=1
..lim-------sino = lim (7T« sin 7tx
(b) Since
x->0
x
x-»0
TLX
and lan it is an elementary function of u and tan ft is well-defined, we have
sinzct' t
sinzzx
lim tan( ------- ) = tan hm------r—>0
X
X
= tanzr
=0
Example 3-29 Let f(x) be a continuous function defined for x > 0 and for any x, y > 0,
f(xy) = f(x) + f(y)
(a) Find f(l).
(b) Let a be a positive real number. Prove that for any rational number r,
f(flr) = rf(fl)
(*)...............................
124 Chapter 5
(c) It is known that for all real numbers x, there exists a sequence {x„} of rational
numbers such that lim x„ = x.
Show that for all x > 0
f(aJ)=xf(a)
where a is a positive real constant.
Hence show that for all x > 0
f(x) = clnx
where c is a constant.
Solution
(a) From the property of f(x), we have
f(l) = f(l) + f(l)
f(l) = 0
(b) Case (i) r is a non-negative integer
This is done by induction on r.
(1) When r = 0,
f(a°) = f(l) = 0 = 0»f(a)
Hence, it is true for r = 0.
(2) Assume the proposition is true for r = k > 0, i.e.
f(a‘) = A-f(o)
When r = £ + 1,
f(a‘+,) = f(a‘-a)
= f(a‘)+f(a)
= £f(a) + f(a)
= (A' + l)f(«)
Hence, it is also true for r = k + 1.
By the Principle of Mathematical Induction, the proposition is true for all non­
negative integers r.
Case (ii) r is a non-negative rational number
£
Suppose r = —, where p and q are two positive integers. By treating aq as a single
q
quantity, from Case (i), we have
£
£
= f(ap)
= pf(a)
£
and so,
f(a*) = £f(a)
<1
f(flr) = rf(fl)
Hence, the result is true for all non-negative rational numbers r.
Limit and Continuity of a Function 125
Case (iii) r is a negative rational number
Then -r is a positive rational number, and so
f(a"r) = -rf(a)
Hence, by using the result that
f(ar) + f(a’r) = f(ar • a~r) = f(l) = 0
we have
f(ar) = -f(a"r) = rf(a)
Hence, the result is true for all negative rational numbers r.
From Case (i), Case (ii) and Case (iii), the result is true for all rational number r.
(c) For any positive real numbers x, let {x„} be a sequence of rational numbers such that
lim x„ = x. Then,
H
lim f(ax") =
II—>00
te""')
f is continuous
=f(a~j
v power function
is continuous
= f(a')
Also, from the result in (b).
lim f(ax") =n—lim
[x f(a)lJ
>» L
zi—>oo
= f(a)lim.v„
/«—><»
= xf(a)
.* x„ is a rational
number
By comparing these two results, we have
f(ax)=xf(a)
(1)
Next, let y = ar (x > 0). Then
i.e.
Iny = xlna
In y
x = ——
Ina
and so (1) can be rewritten as
f(y) = j£2f(«)
Ina
f(a).
= -A—Iny
Ina
Now, replacing y by x, we obtain the result
f(x) = clnx
u
f(«)
•
♦
t
where c = —-— is a constant.
Ina
(x>0)
126
( 'hapter 5
B.
Properties of Continuous Functions
Properties concerning a continuous function defined on a closed interval are summarized in the following
theorems.
Theorem 3-16
If f(x) is continuous on [«, b\, then f(x) is bounded on [a, &]•
That is, for any x e [a. b], there exists a constant K > 0 such that | f(x) | < K.
Since f(x) is continuous on [«. b], f(x) must have a finite value for all x e [rz, b]. Therefore f(x) must be
bounded on [a, b]. However, the boundedness property is not valid for functions continuous on an open
interval (a, b). For example, f(x) = tanx is continuous on (0, —). However, as x —> — , tanx —> +<*>, and so
it is not bounded on (0, y). This is due to the fact that f(«) or f(b) may not be defined. In this example, f(x) is
undefined at
7T
X=—.
2
Another important property is related to the existence of the greatest and the least values of a function.
Theorem 3-1 7
If f(x) is continuous on [a, b\. then the greatest and the least values of f(x) must exist. That is, there
exist xp x2 g [a, b] such that
f(xj)<f(x)<f(x2)
f(x,) is called the least value (or absolute minimum) of f(x) on [a, Z?] while f(x2) is called the greatest
value (or absolute maximum) of f(x) on [a, £>].
This result may be arrived from the fact that as the graph is continuous on [«, b], there must be a point which
is the highest (point C) and another point which is the lowest (point A). (See Fig. 3.10.) Therefore, the
greatest and the least values must exist. It should be noted that the greatest and the least values may occur at
the end-points. More about the greatest and the least values will be discussed in Chapter 5.
y
y=f(x)
O
i
i
i
a
x2
Fig. 3.10
*X
b
Limit and Continuity of a Function
127
The next property is an important property concerning the location of real roots of an equation. This is
known as Bolzano Theorem.
Theorem 3-1 8 Bolzano Theorem
If f(x) is continuous on [«, b] and f(«)f(b) < 0, then there exists c e («, b) such that f(c) = 0.
This property is developed from the fact that when the graph of a continuous function is drawn from a point
below the x-axis (e.g. f(«) < 0) to a point above the x-axis (e.g. f(Z?) > 0), then the graph must cross the
x-axis. i.e. there exists c e(a, b) such that f(c) = 0. (See Fig. 3.11.)
y
B
y = f(x)/1
/ i
/ i
/
i
/
i
/_ £
a
c
O
x
A
Fig. 3.11
Example 3-30 The function f(x)
5cosx-3 is continuous on R. Since
f(y) = -3
f(0) = 2,
there exists at least a real number c within (0, y) such that f(c) = 0.
Example 3-31
Consider the function f(x) =
1
(a* > 0)
2
(x = 0) .
2
-1
(a < 0)
y
y=f(x)
1 ?
2
2
-► x
O
-1
Fig. 3.12
From Fig. 3.12, f is discontinuous at x = 0. Hence, although f(—1) = — 1 < 0 and
f(l) = 1 >0, we cannot find a point c between -1 and 1 such that f(c) = 0.
128 C/iapzer 3
Example 3-32 Let f(x) =.? + 2x.
(a) Show that f(x) is a strictly increasing function on R.
(h) Hence, show that the equation f(x) = 0 has a unique real root.
Solution
(a) For any xp x2 g R. if Xj < x2, then
f(.v,) - f(x2) = (x3 + 2x1) - (x2 + 2x2 )
= (xI3-x2) + 2(xI -x2)
= (Xj - x2 )(x2 + XjX2 + x2) + 2(x, - x2)
= (x, - x2 )(x2 + x,x2 + x2 + 2)
= (x,-a-,) L,+y
4—x2 +2
4 2
<0
Hence, f(x) is strictly increasing on R.
(b) Firstly, we want to show that the given equation has at least one real root.
f(-l) = -3
and
f(l) = 3
Obviously, f is continuous on [-1, 1 ].
Hence, there exists at least a real number c within (-1, 1) such that f(c) = 0.
Next, we want to show that the real root is unique.
The result is to proved by contradiction.
Suppose there exists c'tc such that f(c')
0. Then we have
f(c) = f(c') = 0
and so, f is not strictly increasing, which is a contradiction.
Therefore, the real root must be unique.
Theorem 3-19 Intermediate Value Theorem
Let f(x) be continuous on [a, b] such that f(a) * f(Z?). Then for any real number in lying between f(fl)
and f(b). there corresponds a c g (a, b) such that f(c) = in.
Proof
Consider the function g(x) = f(x) - in. Since in lies between f(a) and f(b), we have
g(fl)g(W = [f(tf) - /n][f(Z?) - in | < 0
Also, g is continuous on [«, b].
Hence, by Bolzano Theorem, there exists c g («, /?) such that
g(c) = f(c) - in = 0
That is,
f(c) = in
Limit and Continuity of a Function
129
The result of Intermediate Value Theorem is illustrated in Fig. 3.13.
y
y=f(x)
1(b)
m
f(a)
O
i
i
i
i
i
i
a
c
b
x
Fig. 3.13
Remark 3-10
By Intermediate Value Theorem, any continuous function defined on an closed interval
must be surjective.
Theorem 3-20
If f is a strictly increasing (or decreasing) function which is continuous on [#, b] with f(fl) = c and
f(b) = d, then fis also a strictly increasing (or decreasing) function which is continuous on [c, d] (or
[r/, c]).
Example 3-33 (a) The function f(.r) = ,v2 is strictly increasing and continuous on [0, 4]. Also,
f(0) = 0,
f(4) = 16
Hence, its inverse function f"’(,v) =
[0, 16].
is also strictly increasing and continuous on
(b) The function g(x) = cos.v is strictly decreasing and continuous on [0. ?r]. Also,
g(0)=l,
g(7T) = -l
Hence, its inverse function g"*(x) = cos"1* is also strictly decreasing and continuous
on [-1, 1].
130 Chapter 5
Exercise 3D
i.
Determine which of the following functions are continuous. For those functions which are not continuous,
locate the points of discontinuity.
x-1
(a)
f(.v) =
(c)
f(x) = 5X+I
(e)
2
x-2
f(x) =
2
X2 + 1
sin x
(g)
Vx
f(.v)
(x * 2)
(x > 0)
x2
f(x) =
2
3.
4.
5.
(d)
f(x) = ln|x|
(f)
f(x)
(x = 2)
COS X -1
2.
f(x) = -
(b)
f(x)
(a<0)
0
(i)
3
X 3-x
(b)
x(l + x)
-3x
(x<-4)
(-4 < x < 5)
x2-3x-20
(x > 5)
x5 + x3 + 3x
sin x
(x * 0)
0
(x = 0)
(x > 0)
(•< < 0)
Evaluate the following limits.
..
ln(l + 3x)
(a)
lim —---------- x-»0
x
(b)
lim x[ln(l + x)- In a]
(c)
lim (x +1) ln(——-)
x +1
(d)
.. ln(fl + x)-lnn
h m —---------------
Let f(x) be a function defined by f(x) = lim
,v->0
x
(«>0)
1
2 + x2n ’
(a)
1
3
Find the values of f(—), f( 1) and f(-|).
(b)
Locale the points of discontinuity and sketch the graph of f(x).
Let f(x) be a function defined by f(x) = lim a(x2"-1)
n—>°° x2”+l '
Locate the points of discontinuity and sketch the graph of f(x).
Let a be a real constant and f(x) be a function defined by f(x) = lim
,1—>oo
a(nx — n )
nx +n~x
(a)
Show that f(x) is a constant for x > 0. Show also that f(x) is also a constant for x < 0.
(b)
Find the value of a such that f(x) is a continuous function.
Limit and Continuity of a Function 131
6.
Let f(x) = .v3 - 8x + 5.
(a)
Evaluate f(0), f(l) and f(3).
(b)
Prove that the equation f(x) = 0 has exactly three real roots.
7.
Prove that the equation xex = 2 has at least one real root in [0, 1 ].
8.
Let f(x) be a continuous function defined on the interval |0, 1 ]. If 0 < f(x) < 1 for all x g [0, 1|, prove
that there is at least one real root for the equation f(x) = x in the interval [0, 1].
9.
Given that f(x) is continuous such that
f(O) = f(l)=l,
= 1,
f(.v2) = [f(.v)]2
and
f(.ry) = f(x)f(y),
(a)
Prove that for all positive integers n
[f(x)]2'
2’=f(x2’)
(b)
Evaluate lim f(x). Hence, use the result in (a) to show that f(x) = 1 for all x g [0, 1 ].
Vx.yeR
.r—>0
Revision Exercise
1.
(a)
Let f and g be two continuous real functions. Suppose x0 is a real constant such that f(x) = g(x)
for all x gR \ {x0}.
Show that f(x0) = g(x0).
(b)
Let p(x) and q(x) be two real polynomials. Suppose x0 is a real number and /z, k are two integers
such that
p(x0) * 0, q(x0) 0
and
(x - x0)* p(x) = (x - x0)* q(x)
Show that h = k and p(x) = q(x) for all x g R.
2.
Let a be a positive real number.
(a)
It is given that the function f(x) = log0x is continuous for all x > 0. Show that
lim|gjl + x)=logoe
X
A->0
Hence, or otherwise, show that
..
flr-l
lim------•t->o x
(b)
In «
By using the result of (a), show that
lim n (Vfl “ i)=inn
n->o°
132
Chapter
3.
Let f(x) be a bounded function on the interval [n, b] and, for any x,, x2 e [«, b],
f((a)
Explain why if x -> c, where x g [a. b], then c e [«, /?].
(b)
Suppose x g
(i)
b]. It is given that if c g [a, b] and x —> c. then 2x - c g [«, b].
Prove that for any x g [a. b],
f(x)-f(c)<|[f(2x-c)-f(c)]
(ii)
4.
Prove that f is continuous on the interval (a, b).
Let 0 be a real number such that 0 < 0 <
(a)
0
0
0
By using the formula sin^- = 2sin—^cos-^-, show that
sin# = 2
0
0
0
cos —
cos —COS—7
9
2
2
2
0
0
cos—nn-l
—rsin—nn-1
—r
2"
2n
Hence, find the limit value
..
0
0
0
0
hm cos—cos—COS—r ••• cos——r
2
22
23
2
(b)
Let
} and {bn} be two sequences of real numbers defined by
= cos 0.
2
(i)
Z?, = 1
b.
(”> 1)
Show that for n > 1,
.000
0
b„ = cos-cos cos-r ■■■cos^T
(ii)
5.
Evaluate limfln.
n—
Let x0 be a real constant and f be a real-valued function defined on R such that f is continuous al x0 and,
for any real numbers x and y,
f(.v + y) = f(x) + f(y)
(a)
(i)
Evaluate f(0).
fii)
Show that f is continuous at x = 0.
Hence, show that f is continuous at every real number x.
(b)
Show that for all integers n, f(/ix) = n f(x).
Hence deduce that for any rational number r, f(r) = rf(l).
(c)
Il is known that for all real numbers x, there exists a sequence {an} of rational numbers such that
lim an = x.
n—>«■»
Show that f(x) = k.x for all x e R, where k is a real constant.
Limit and Continuity of a Function 133
6.
Let f be a real-valued function defined on the set of R such that
(1)
f is a non-negative continuous function with f(l)^0;
(2)
f(x)f(y), VxjgR.
M
Show that f(x) = f(| x |) for all real values of x.
(b)
Use induction to show that for all positive integers n and for all x in R,
i
f(Vnx) = [f(x)]'1
(c)
For any non-zero integers p and q, show that
\
Q
(d)
■
It is known that for all real numbers x, there exists a sequence {an} of rational numbers such that
lim an = x.
n—»«
Show that for all x g R,
f(x) = [f(l)]
7.
Denote the open interval (0, °°) by R’. Let f: R* —> R be a continuous function such that
f(xy) = f(x) + f(y)
for all x,)’gR’.
(a)
Show that
(i)
f(D = 0,
(ii)
f(x-1) = -f(x) for all x g R+,
(Hi) f(x") = n f(x) for all n g Z and x g R+.
(6 marks)
(b)
Show that f(xr) = rf(x) for all r g Q and xgR*.
(4 marks)
(c)
Show that f(xa) = af(x) for all a g R and x g R*.
(You may use the fact that for any a g R, there exists a sequence {?•„} in Q such that lim rn = a.)
(3 marks)
(d)
If f(2) = 1, show that f(x) = log2x.
(2 marks)
(HKAL 1997 Paper II)
"7
Advanced Level Pure Mathematics
134
Chapter 4
Derivatives
4.1
Introduction
Derivative is the central concept of differential calculus. Primarily, the concept of derivative can be
developed from two different types of problems: (1) the physical problem to find the instantaneous velocity
of a panicle; (2) the geometrical problem of finding the tangent line to a curve at a given point. In this
chapter, we will illustrate the relation between the derivative and the tangent line to a curve.
A.
Derivative
When the value of a variable is changed, the change is called its increment.
Usually, the increment of the variable x is denoted by Ar, while that of y is denoted by Ay.
When the independent variable x is changed from x0 to x0 + Ar, the corresponding increment of the
dependent variable y, where y = f(x), is given by
Ay = f(x0 + Ar) - f(x0)
(4-l)
With these notation, we have
Definition 4-1
Let y = f(.r) be a function defined at x0 and all the points near x0. f(x) is said to be differentiable at x0
(or have a derivative al x0) if the limit
lim ^- = lim f(A'0+Ar)-f(x0)
a<->o Ar
Ar-»0
exists. This limit value is called the derivative of f(x) at x0 and is denoted by
or
dy
dr
(4-2)
Derivatives 135
Remark 4-1
If the notation x = x0 + Ax is used, (4-2) can be rewritten as
d^
= lim
dx . = l ^0
(4-3)
x-xn
Sometimes, in order to simplify the presentation, we use h instead of Ax, and so (4-2)
becomes
dy
dx
-lim f^o+/0-f^o)
h
(4-2')
Furthermore, if f(x) has a derivative at every point x in («, /?), then a function of the
derivative in x is obtained. In general, this function of the derivative is simply called the
derivative of f. In this case, f is said to be differentiable on («, b).
Example 4-1
Let f(x)=x2 + 1. Find f'(2).
Solution
From (4-2'), we have
r(2)=IimI(2±M2)
II
*-*0
= lim
[(2 + /2)2 + l]-[22 + l]
h-»o
h
.. 4/i + /r
= hm--------/»—>o h
= lim (4 4-/2)
h—>0
=4
Example 4-2
Solution
Let f(x) = sinx. Find f'(^).
4
f(^ + A)-f(^)
r(£) = lim_4------------ 4.
v 4 /«—>0
h
sin(—+ /z)-sin4
= lim----- —-- --------- h->0
h
n . Il
II.
2sin — cos(— + —)
= lim------ 2------h-rt
h
. h
Sin?
,n h.
= lim -^.^(-4--)
h->0
. h
sin-
2
7T
= COS —
4
I
“72
As h -> 0.
lim -JT = i
136
Chapter 4
Example 4-3
Let f(x) = x". Find f'(x).
f'(.v) = lim f(x + /l)
Solution
h-»o
a-»o
h
= lim ------ --------h
.. (r + cfx "-‘Hcy2 lr + c;x n~3h3+ ---+/in')-x"
= inn-------- 1—
h
;i-»o
+c3y/z+•••+//
= lim
fey1 + //(C"x
/i->o L 1
)]
= cy1
= ZU'”-1
Example 4-4
Let f(x) = In x. Find f '(x).
f'(x) = lim f(^ + /0-f(x)
Solution
A-»0
h
■
ln(x + /z)-lnx
= lim —------- --------h
= lim — ln(l + —)
'
h
x
h
1
= lim — In (1 + —)''
/i->0
h
’ "
I
x J
•-Inf (i + -r
r
1
= lim —
h->0 h x
-v
1
= —In lim(l + — )h
x
h-^0
v In x is continuous.
A-
1 ,
= — In e
x
= —I
X
Example 4-5
Let f(x) = ex.
(a) Find f'(0).
(b) Find f'(x).
Solution
(a)
f'(0) = lim f(0 + /1) f(0)
h
A->0
= lim-------
h->0
h
(1)
Derivatives 137
As h —> 0, eh —> 1, and so (1) is an indeterminate form.
Now, let eh = 1 + u.
Then h = ln( 1 + it).
As /z —> 0, it —» 0 , and so (1) can be rewritten as
ll
f'(0) = lim
ln(l + u)
u->0
1
lim
u—>0
— ln(l + w)
u
1
= lim--------- rM-»0
1
ln(l + w)“
In .v is continuous,
1
1
Ine
=1
/. lim < I r h)" = e
u >0
f'W = limf(* + /,)-f(x)
(b)
/,-»0
II
= lim
h-*0
ex'h-ex
h
i-
x/^-1
= lim e (—-— )
a->o
h
From (a)
= ex
Example 4-6
Let f be a real-valued function defined on R such that
f(j + y) = f(X) + f(}’)
(V.v,yeR)
Suppose f is differentiable at ,v0, where x0 g R.
............................
fW
(a) Find
the value of lim —
—.
/i->o
h
(b) Show that f is differentiable on R and express f'(x) in terms of x0.
Solution
(a) Since f is differentiable at x0, we have
U
*-»0
= lim
h
f(.V0)4-f(/l)-f(.V0)
II
h-iO
.. f(/0
= hm
h->0
and so,
h->0
II
h
138 Chapter 4
(b) For all x 6 R,
f'(a) = lim 1(x + /l)---(-^
/.-•o
h
f(x) + f(/i)-f(x)
— inn
/i->0
■
h
= rlim WO
/i-»o
h
= f'(*0)
Hence, f is differentiable on R and fz(x) = fz(x0).
B. Geometrical Interpretation of Derivatives
As shown in Fig. 4.1, P and Q are respectively the points with independent variable equal to x0 and x0 + Av:
y
y=f(x)
f(^ + Ax)
// 1
/
i
i
i
i
P.
O
i
i
xo
------- >x
x0 + Ax
Fig. 4.1
Then we have
Slope of the secant PQ = — =
PR
-° +
—LLhd
Av
As Av —> 0, the point Q moves closer and closer to P along the curve y = f(x), and the line PQ becomes
the tangent line PT of the curve at the point P. Hence,
= Jim ^(-°-+Ay) f(Ao) = Slope of the tangent PT.
dx x=*o
Hence,
Remark 4-2
f'(x0) of the function f(x) is equal to the slope of the tangent at the point (x0, f(x0)).
Derivative', 139
4.2
Differentiability
The above examples illustrate the procedure to find the derivative of a differentiable function. In this
section, we are going to examine whether a continuous function is differentiable or not.
When a function is not differentiable, the limit lim - — *
does not exists. As discussed in
a-»o
h
Chapter 3. this limit value does not exist means that it may not be found, it may tends to infinity or it
may have different left and right hand limits. These three cases will be discussed in the following.
A.
Derivatives Cannot Be Found
Example 4-7
Let f(x) = «
. —1
xsin
x
(x*0)
0
U = 0)
Show that at x = 0, f is continuous but not differentiable.
Solution
lim f(x) = lim xsin —
x->0
x-»0
x
x * 0.
=0
= f(0)
and
f is continuous at x = 0.
Also, as
r(o) = iim f(0+/,)~f(0)
A—>0
h
= lim f(/Q-f(O)
h
A—>0
/z sin — -0
h
A->0
h
= lim
i-
•
1
= lim sin —
h-,0
h
does not exist, f is not differentiable at x = 0.
B. Infinite Derivatives
Example 4-8
1
If f(x) = x3, show that at x = 0, f is continuous but not differentiable.
£
lim f(x) = lim x3 = 0 = f(0)
Solution
x—>0
•.
x->0
f is continuous at x = 0
sin — < I
x
140 Chapter 4
On the other hand.
h
h->o
i
,. zJ-o
= hm —-—
h->0
h
1
V
= hm ——
h->0 h3
= co
Hence, f is not differentiable at x = 0.
Geometrically, this result means that the slope of the tangent lends to infinity, i.e. the
tangent is a vertical line. (See Fig. 4.2.)
y
O
-*• x
Fig. 4.2
As illustrated in the above example, we have
Remark 4-3
Geometrical Interpretation of Infinite Derivative
If f is not differentiable at x = x0, but
lim f(A0+/?)-f(x0) = oo
'-o
h
(4-4)
then f is said to have an infinite derivative at x = x0 and the tangent of the graph of f at the
point (x0, f(x0)) is a vertical line.
141
C. Left Hand Derivatives and Right Hand Derivatives
For functions have different definitions at different intervals, we have to introduce the concept of left hand
derivative and right hand derivative to investigate whether they are differentiable or not.
Definition 4-2
Let f(.v) be a function continuous atx0. Denote
f-(-v0)= lim
f(x0+/Q-f(x0)
h
(4-5)
f'(x0)= lim
f(x0+/Q-f(x0)
h
(4-6)
/»->0
If the limit in (4-5) exists, then f(.v) is said to have a left hand derivative at x0.
If the limit in (4-6) exists, then f(x) is said to have a right hand derivative at x0.
As discussed in Section 3-5, if the limit of a function exists, its left and right hand limits exist and equal.
Hence, we have
Theorem 4-1
The necessary and sufficient condition for the function f(x) to be differentiable at .r0 is that the left
and right hand derivatives at x0 exist and equal. That is,
f_'(A-0) = f+'(A-0) = r
f'(.v0) = (
With the geometrical interpretation of derivative developed previously in Section 4-1, we can conclude that
if the left hand derivative C(x0) exists, then the graph of f(x) has a tangent with slope equal to f'(x0) at the
left hand side of ,v0. Similarly, if the right hand derivative f'(x0) exists, then the tangent at the right hand
side of x0 is of slope f'(x0). Hence, if the left and right hand derivatives are equal, the slopes of these two
tangents are equal, and so these two tangents become a straight line. On the other hand, if the left and right
hand derivatives are not equal, then these two tangents cannot be overlapped and a kink exists at.v0. (See
Fig. 4.3.)
y
y=f(x)
i
i
i
i
X
O
Fig. 4.3
142 Chapter 4
Example 4-9
Show that f(x) = | x | is not differentiable at x = 0.
Find also f'(x) when x * 0.
Solution
From the definition of absolute value f(x) =
x
(x>0)
-x
(x<0)
, we have
f,(0)=lim f(0 + /',)-f(0-)A—>0-
h
-h-0
..
= lim —-—
/>->o
h
= lim (-1)
h—>0‘
= -l
and
f'(0)= lim f(0 + /,) f(0)
h
.. h-0
= h m —-—
h->0*
h
= lim 1
=1
Hence, the left hand and right hand derivatives at x = 0 are unequal. Therefore. f(x) is not
differentiable at x = 0 and the graph has a kink at x = 0. (See Fig. 4.4.)
On the other hand, it is easy to show that the derivative of x is equal to 1 while that of-x is
equal to -1. Hence, when x 0, we have
I
f'(x) =
-1
U>0)
(x<0)
y
y=f(x)
0
Fig. 4.4
►x
Derivatives 143
Example 4-10
Let f(x) = ’
(a) If a = I
ax2 +b
(X<1)
x
(x>l)
and b = 0, show that f(x) is not differentiable at x = I.
(b) Find the values of a and b such that f(x) is differentiable at x = I.
Solution
(a) When a = 1 and b = 0,
f(x) =
x1
(X<1)
x
(X>1)
As h —> 0", 1 + h < 1, and so f( 1 + h) = (1 + h)2. Hence, we have
f'(l) = lim
/i—>0”
f(l + /z)-f(l)
h
- lim (l + /02-l
II
r 2h + h2
= lim--------h->oh
h—>0"
= lim (2 + /z)
A-»0~
=2
As h —» 0+, 1 + h > 1, and so f(l + h) = 1 + h. Hence, we have
f'(i)= iim f(1+/,)~f(1)A-»0*
Il
= lim (1+/Q-1
A-»0*
h
h
= ..lim —
/i-O* h
= lim 1
=1
Therefore, f'(l) * f'(l), and so f(x) is not differentiable at x = 1.
(b) Similar to the process outlined in (a).
f71) = limf(1 + /?--^
h
/»->o
+
lim
+b-\
= lim------------------h->oh
..
a—o-
(a + b - \) + 2ah + ah2
= lim------------ ;------------h
= lim (
a + b-1
+ 2a + ah)
h
In order that T(l) exists, we have
a+ b- 1 =0
and so
f'(l) = lim (2a + ah)
h->0
= 2a
(1)
144 Chapa
I
On the other hand, from (a).
f+'(l) = 1
If f(x) is differentiable at x = 1, then f'(l) = f+(l), and so
2a = 1
(2)
From (1) and (2), we have
a=b
y
2
y
(a=Vb = Q)
(a = b
i>
yr Kx)
i
i
O
0.5
1
►x
O
1
x
Fig. 4.5
D. Continuity and Differentiability
The previous examples in this section have illustrated the result that a function continuous at x0 may not
be differentiable at x0. However, we have
Theorem 4-2
If f(x) is differentiable at x0, then f(x) is continuous at x0.
Proof
Since f'(x0) exists, lim —= f'(x0) exists and so
Ax—>0 Aa
lim [f(x0 + Ax)-f(x0)l = lim Ay
Ar-»0 L
J
Ar-»0
= lim (—)Ax
Ax—»o Ar
lim —1( lim Ax)
Ar->0 Ar ) At->0
= f'(x0)(Hm Ax)
=0
(4-7)
Derivatives 145
Hence, we have
lim f(x0 + Av) = Km [f(x0 + Ar)- f(x0) + f(x0)]
= Hm [f(-v0 + Ar)- f(x0)] + f(x0)
= f(x0)
Therefore, f(x) is continuous al x0.
Example 4-11
ax
Let f(x) = - 0
sin x + b
U<0)
(x = 0) .
(x>0)
If f is differentiable al x = 0, find the values of a and b.
Solution
Consider
lim f(x)= lim ax = G
A -> O'
X -> 0'
lim f(x)= lim (sinx+ /?) = /?
x -» O’
f(0) = 0
Since f is differentiable at x = 0, f is continuous al x = 0, and so
lim f(x)= lim f(x) = f(0)
,r -> O'
a -> 0*
Hence, b = 0.
On the other hand,
f'(0)= lim
h->0'
= lim
f(0 4- /z) — f(0)
h
ah - 0
h
/i-»0'
= lim a
A-»0'
=a
f;(0)= lim W + V-Wl
+
h
a-»o*
= lim
= lim
h -♦O’
(sin/i +1?) - 0
h
sin /z
h
=1
Since f is differentiable at x
0.
f(0) = f:(0)
Hence, a = I.
I’ = 0
146
Chapter 4
Remark 4-4
The readers should have a clear concept about the difference between
(1) f(x) is well-defined at x0.
(2) the limit of f(x) at x0 exists.
(3) f(x) is continuous at x0.
(4) f(x) is differentiable at x0.
Exercise 4A
1.
Determine which of following functions are differentiable al ,r = 0.
(a)
(c)
2.
3.
4.
(b)
f(x) = x3
f(.v) =
(a>0)
X
(a<0)
(d)
f(.r)
f(.v) =
. 1
xsin —
x
0
(**0)
(.v = 0)
2x
(x>0)
0
(a<0)
Find f'(x) of the following functions. Also, locate the points at which the function is not differentiable.
(a)
[4x
f(A') = ^
2x2-6
(c)
f(x) = p + 8I
(a<3)
(a >3)
x+2
(x < -2)
(x + 2)2
(a>-2)
(b)
fU) =
(d)
f(x) = |x,22 -1
Use the definition of derivative to find the derivatives of the following functions.
(a)
f(x) = cosx
(b)
f(x) = tanx
(C)
f(x)=X2 + X
(d)
f(A)
Let f(x) = •
x3
ax + b
x
X2 + 1
(x<1)
(x> 1) '
Find the values of a and b so that f is differentiable at x = 1.
5.
6.
Let g(x) be a continuously differentiable function.
(a)
Let f(x) = (x - a)g(x). Find f'(x).
(b)
Let F(x) = | X - a |g(x). If g(<7) * 0, show that F'(«) does not exist.
Let a, b and c be three real numbers such that a <b <c.
If f(x) = |x-<7 |+|x-»|+| x-c|, find f'(x) wherever it exists and indicate where it does not.
Derivatives
7.
8.
147
Prove the following statements.
(ci)
The derivative of a differentiable even function is an odd function.
(b)
The derivative of a differentiable odd function is an even function.
(c)
The derivative of a differentiable periodic function is periodic.
Let f and g be two real-valued functions possessing the following properties:
(1)
(2)
g(x + y) = g(x)f(y) + f(x)g(y)
(Vx, y g R)
f(0) = 1, fz(0) = 0, g(0) = 0 and g'(0) = 1
Show that g'(x) = f(x) for all x g R \ {0}.
9.
Let f be a real-valued function possessing the following properties:
(1)
f(x + y) = f(x)f(y)
(2)
f(x) = 1 + xg(x), where lim g(x)
x—»0
Show that fz(x)
10.
(Vx, y g R)
1.
f(x) for all x g R.
Lei f be a real-valued function such that
(1)
f(x + y) = f(x) + f(y) + 2 xy (x + y)
(2)
f'(0)=l
(Vx,ycR)
Show that f z(x) = 1 + 2x2 for all x g R.
11.
Let f be a real-valued function such that
(Vx, a g R)
| f(A)- f(CT)|< (_V-«)2
Show that f z(x) = 0 for all x g R.
12.
Let f be a differentiable function such that for all x, y g R,
f(x + y) = f(x) + f(y) + 3xy(x + y)
(a)
Show that f'(0) = lim
(b)
Hence, or otherwise, show that for all x G R.
/.->o
h
.
f'(x) = f'(0) + 3?
and deduce that
f(.v) = f'(0)x + X3
(6 marks)
(HKAL 1992 Paper II)
13.
Let f(x) = *
• —1
x 3 sin
x
(a*0)
0
(x = 0)
(a)
Evaluate f z(x) for x * 0.
(b)
Prove that f'(0) exists.
(c)
Is f z(x) continuous at x = 0? Explain.
(6 marks)
(HKAL 1996 Paper II)
148
Chapter 4
14.
Lei f(.v)
a|.v|.
(a)
Find f'(a) for a > 0 and a < 0 respectively.
(b)
Prove that f'(0) exists.
(c)
Prove that f'(.v) is continuous al a
0.
(6 marks)
(HKAL 1997 Paper 11)
15.
Lei f(.v) = ’
4.3
A.
a2 cos —
A
if a- > 0
aa + b
if a < 0
be differentiable at a = 0. Find a and b.
(6 marks)
(HKAL 1999 Paper II)
Rules of Differentiation
Basic Formulae
The examples in Seclion 4.1 and 4.2 illustrate the method to find the derivative of a function from the first
principle. By applying this method, the derivatives of some standard functions can be found and they are
summarized in the following table.
y
dy
dv
1
c
0
2
A"
3
ex
ex
AGR
4
a'
a ' In a
a > 0, a g R
5
In a
Remark
cgR
n eR
_1_
A
6
sin a
COSA
AGR
7
COSA
-sin a
AGR
8
lan a
sec‘A
a
9
COl A
-CSC’A
A =£ 1171
10
sec a
sec a lan a
A
11
CSC A
-CSC A COlA
A * H7t
nn + —
2
H7C + —
2
149
cb:
dx
y
12
sin" x
13
cos *x
14
tan x
Remark
__ 1__
|x|<l, -^7t<y<^7t
<l-72
-I
| x | < 1, 0 < y < 7i
1
xeR, - — 7i<y<— 7t
2
2
1 + x2
Some of these standard results have been proved in previous examples and some will be proved in later
examples. The proof of the remaining results will be left as an exercise for the readers. However, in finding
the derivative of sin"'x, cos"‘x and tan"*x, the readers should note that the principal values of these functions
are adopted.
The process of finding the derivative from the first principle, in general, is lengthy. Therefore, some
rules for differentiation have to be developed to help to simplify the process. The readers may have learned
these rules in secondary school. These rules of differentiation are summarized in the following.
B.
Rules of Operations on Differentiation
Before discussing the rules of operations on differentiation, we want to introduce the following notation:
Af = f(x + Av) - f(x)
Then we have
f(x + Av) = f + Af
(4-8)
and
df(x) .. Af
= hm —
Ar—>0 /\V
dx
Rule I
Let f(x) be a differentiable function and k be a real constant. Then
df(x)
dx
Proof
A[KW]= lim ^(x+Ax)-£fW
dxL
J
Av
£(f + Af)-£f
= lim
Ar-»o
Av
/—
Af
= rlim k
Ar—>0
A.V
= A,dfU)
dx
(4-9)
150 Chapter 4
Rule II
Let f(x) and g(x) be two differentiable functions. Then
dr
Proof
dv
(4-10)
dv
— [f(x) + g(x)] = lim [f(A-^v) + g(.< + M-[f(.v) + gc<)]
dvL
J At~>0
Ax
= ljm (f + Af + g + Ag)-(f + g)
Av
Ar-+0
1- —
Af +. hm
.. —Ag
= lim
—
Ai-»o Ar Ar-»o Ar
_df(x) [ dg(x)
dr
dx
Then from Rule I. we have
^[f(-v) - g(x)] = ^j“[f(x) + (-l)g(x)]
= ^ + ^-[(-Dg(x)]
dr
dr
^df(x) dg(x)
dr
dr-
Rule 111
Product Rule
Let f(x) and g(x) be two differentiable functions. Then
£[f(x)g(x)] = f(x) dg(x) + gW df(x)
dx
Proof
dx
Since g(x) is differentiable, g(x) must be continuous, and so
lim Ag = 0
Ar-0
-?-[f(x)g(x)] = lim f(£t^)g(x + Ax)-f(x)g(x)
dr
ax—»o
Ax
= lim (f + Af)(g + Ag)-fg
Ar
Ai->o
= |im fg + fAg + gAf + AfAg-fg
Av
At->0
= lim f —-+ lim g-^- + ( lim —)(lim Ag)
Ar-»0
A.V
At->0
A.V
+8<x)^
= f(x) ^+
gW
dx
dx
A.V
Ar->0
Derivatives
Example 4-12
151
Let f(x) = |x|sinx.
(a) Show that f is differentiable at x = 0.
(b) Find fz(x) for all real values of x.
Solution
(a) Since f(x) =
-xsinx
xsinx
(x <0)
U>0) ’
f'(0) = lim -f(-Q + /i) f(°)
A->0-
h
_ jjm -Asin/i-0
h
= lim (-sin h)
h—>0~
=0
and
f'(0) = lim
f(0)
h
+
..
/zsin/2 —0
/t-,0-
h
= hm----------= lim sin h
h-rt'
=0
Then, f'(0) = f+(0) and so, f is differentiable at x = 0 and f'(0) = 0.
(b) Since both x and sin x are differentiable for all real values of x, by applying the
Product Rule, we have
d ,
x
,dsinx
dx
— (xsinx) = x(—-— ) + sin x(—) = x cos x + sin x
dx
dx
dx
Also, as xcosx + sinx = 0 when x = 0 and by combining the result in (a), the
derivative of f may be expressed as
f'(x) =
-(sinx + xcosx)
(A'<0)
sinx + xcosx
(x>0)
Rule IV Quotient Rule
Let f(,r) and g(x) be two differentiable functions and g(x) * 0. Then
f(x))
—
•
dx (g(x),
J
g(x)fz(x)-f(x)gz(x)
[gtof
(4-12)
152 Chapter 4
Proof
1 W
lim
I
+Av)
Ar-»0 ^g(x + Av)
dx^g(x)
= lim
f f + Af
Ar—>0 l^g + Ag
f(x) j 1
g(x)J Av
f
gjAv
= lim gAf~fA?-.—
A<-»o(g + Ag)g Ar
, Af fAg
hm (g-—f—)
_ A*~»° Av
Av
lim (g + Ag)g
Ar-»0
g(x)f'(x)-f(x)g'(x)
[g(A-)]2
Example 4-13 Find the derivative of cotx.
Solution
By Quotient Rule,
d cot x
dx
d (Cosx
)
dx sin x
d cos x
dsinx
sinx------- -cosx------_______ dx____________
dx
(sinx)2
sin x(—sin x) - cos x cos x
sin2 x
-(sin2 x-rcos2 x)
sin2 x
-1
=------ j---
sin" x
= - esc2 x
Example 4-14 Find the derivative of
Solution
ex tan x
for x / 0.
x
By applying the Quotient Rule and Product Rule, we have
x
d ,ex tanx
)=dLv
x
d ex tan x
-ex tanx—
dx
dx
x2
d tan x
de\ x
+ tanx—— )-e tanx
dx
dx___
2
x
x(ex sec2 x + ex tan x) - ex tan x
x2
_ e*(xsec2x + xtanx-tanx)
x2
x(ex
v g(x) is
differentiable,
as Av 0.
Ag -»()
I)cn\ iii\< '
Rule V
153
Chain Rule
Suppose the derivative A of the real-valued function y = f(«) and — of u = g(x) exist. Then the
dz/
dr
composite function y = f [g(x)] is differentiable and its derivative
dy du
dz/ dr
dr
Proof
(4-13)
= df[g(x)] dg(.r)
or
dg(.O
is given by
dx
(4-13)'
For any increment At of x, there corresponds increments Aw of u and Ay of y. Then from
the assumption that both — and
exist, we can conclude that u is a continuous
dv
dw
function in x, and so
Aw-»0
as Ax —> 0,
Hence,
Azz
..
du
dv
lim —
Ax—>0 At
Ay i- Ay dy
lim — = lim — = —
Ai/->o Aw
Ar->0 Azz
dz/
Therefore,
dy .. Ay
— = hm —
dv
Ar->0 Av
(Ay Au.
At
Ar->0 AZ/
rlim —
Ay
—
Ar—>0 Aw
Aw
lim —
Ar—>0 At
dy du
du dv
Example 4-15 Let a be a real number. Then
,alnx j
dr
dv
d(e'■alnx) d(alnx)
dv
d(alnx)
a
x
= e «In.i
= xa
a
x
= axa'}
154 Chapter 4
Example 4-16 Find the derivatives of the following functions:
(a) y = x ln( I - x)
x
(b)
Solution
dv
dr
dln(l-x) + ln(l-x)^dv
dv
= .v—!—(-l) + ln(l-.v)
(a) — = x
1-X
= ^- + ln(l-x)
x —1
-------- X
71 - x 2 •dv
(b)
-x2
dv
dx
2
1-x
dx
7i
1
-X1 -X-
= -(-2x)
271^7
1-x2
1
(1-x2)7
Example 4-17 Find the derivatives of the following functions:
(a) y = tan a 1-x
sin*
(b) y = e x
Solution
(a)
dv
= sec2 71 -x — 71 -x
dv
= sec2 71-x- __ 1_= ’7-(l-x)
27T3 x dv
sec22 71 -x
271^7
dy
dx
i
(b) ~r~e■
j
sin7
.
d . 1
sindr
x
sin—
1 d
= e x •cos—
•--- X
x dx
sin—
]
e x cos —
x
x2
Derivatives 155
Rule VI
Derivative of an Inverse Function
Let y = f(x) be a strictly increasing (or decreasing) and continuous real-valued function on an
interval (n, b). If f(x) is differentiable on («, b) and f'(x) * 0, then its inverse function x= f"‘(y) is
also differentiable and
[f‘‘(y)]' =
Proof
1
f'(x)
dv _ 1
dy ” dy
dr
and
(4-14)
For any increment Ay of y, there exists a corresponding increment Ar of x.
As discussed in Section 3.6, as y = f(x) is a strictly increasing (or decreasing) and
continuous real-valued function on an interval (a, b), f”‘(y) exists and is also a strictly
increasing (or decreasing) and continuous function.
Hence, as Ay —» 0, Av —» 0.
Moreover, since f(x) is strictly increasing (or decreasing), Ay * 0 implies that Av * 0, and
so
dv .. Av ..
1
1
— = lim — = lim ~r~ =
Ar->o Ay dy
dy A>->0
Av dv
Remark 4-5
if y
f(x) is a basic elementary function, then its inverse exists and
(1) If
^o,
dv
then
dv
dy
1
dy
dv
(4-15)
(2) If
^*0,
dy
then
dy
dv
1
dv
dy
(4-16)
- <y<•71
Example 4-18 Let y = sin *x, where -1 <x < 1 and ~—
2 'J ' 2’
It is obvious that y = sin“‘x is the inverse of x = siny. Since
dv
dy
cosy > 0
and
cos y = -^/l - sin2 y = y 1 - x“
positive root
sign is adopted
156
dy
dx
1
dx
dy
I
cosy
1
Vl-x2
Similarly, the derivative of the inverse of the other trigonometric functions are :
d
— cos
dx
d
— tan
dx
1
(When -1 < x < 1, take 0 < cos"’x < rt)
x^lZv2
1
1 + x2
(When x g R. take -y- < tan-1 x < y-)
To simplify the manipulation, we may apply these formulae directly and this is illustrated later in this
chapter.
Example 4-19 Let y
Solution
,
r-
. dV
Inx. Find -y-.
dx
When x > 0, the inverse function of y = Inx is x = ev (y g R). Then
I
dx
dy
d i
— Inx
dx
J_
ey
J_
e >0
x
Rule VII
r
*
Differentiation of Functions in Parametric Form
a
/-XL,
j-rr
li
r
-•
rr
dx
Let ::
x=f(z)
ffr) and y = g(z) be ?::z
two differentiable
drffcrcr.t-blc functions. If — = fz(/)^0 and the inverse of f(x).
dz
l = f"’(x), exists, then y can be considered as a function of x such that y = g[f“'(x)] and
dy
dy =_dz_
dx
dx
dz
.......................
(4-17)
Derivatives
Proof
157
Applying Rule VI, we have
dr
dr
_1_
dr
dr
Then, by applying Chain Rule, we obtain
dy
dr
dy , dr
dr dr
dy 1
dr ’ dx
dr
dy
__d/_
dr
dr
Example 4-20 If x = secr and y = tan r, where 0 < r < —, find —.
2
dr
Solution
dr
— = seer tanr * 0
dr
dy = sec2 r
dr
dy
dr
(0<r<|)
dy
dr
dr
dr
2
sec r
seer tanr
_X
y
dy
Sometimes, it is rather difficult to express the results in terms of x and y. Hence, we may express — results
in terms of r.
158
/ 7
c. Implicit Differentiation
In previous discussion, we have only discussed the methods in finding the derivatives of the functions
y = f(x) of which the dependent variable can be expressed in terms of x explicitly. Now, we are going to
investigate how to find the derivatives of another type of functions: the implicit function.
Let
F(x, y) = 0
(4-18)
be a function of two variables x, y. If y cannot be expressed in terms of x explicitly, then the function is
dy
called an implicit function. For example, the function x 4- y = sin at is an implicit function. To find
, we
have to
Remark 4-6
Implicit Differentiation
dy
Let f(x, y) = 0 be an implicit function. The steps in finding — of this function are
summarized as follows:
(1) Consider y in the equation f(x. y) = 0 as a function of x.
(2) Differentiate both sides with respect to x. It should be aware that the Chain Rule
should be applied to find the derivatives of the terms which is a function of y. e.g. y",
and hence we have
dy
dx
dx
and
d F(y)
dLv
d F(y) dy
dy
dx
dy in terms of x and y.
(3) Express -fdx
Such process is called implicit differentiation and the method is illustrated by the following example.
Example 4-21
dy
Find — given that y 4- xsin(x 4- y) = 2.
dLv
Solution
By differentiating both sides with respect lox, we have
dy + d '
^-4-^-[xsin(x4-y)]
=0
dx dx"
V-4-x-^-sin(x4-y)4-sin(x4-y)— = 0
dx
dx
dx
dy [ y dsin(x-f-y) d(x4-y) . ,
x
\
~ - + sin(x4-y = 0
dx
d(x4-y)
dx
—- 4- xcos(x 4- y) • (I 44- sin(x 4- y) = 0
dx
dx
[14- xcos(x 4- y)]^- 4- [xcos(x 4- y) 4- sin(x 4- y)] = 0
and so,
dy
dx
x cos(x + y) + sin(x 4- y)
14-xcos(x + y)
Derivatives 159
Example 4-22 Let y = tan 1 x-1 . Find
Solution
dx
Method I
The given equation may be rewritten as
tan y = ——x+1
Then, by differentiating both sides with respect tox, we have
2 dy
sec y —
dx
(x + l)-(x-l)
(x+1)2
..........
(1)
Since
see",2 y
2(-----x2 + 1)
1. + ftan 2 y = ,1 +, -(x-1)
----- —2 = —
(x + 1)(x + 1)2
we obtain
2(x;+l) dy =
2
(x + 1)2 * dx ~ (A'+l)2
dx
1
x2 +1
Method II
By Chain Rule,
dy =
dx
dtan
d
■ /x-1
d
lx+1
x+1 J
=— 1
x-1
1+
x+1
2
x-1
X+1
dx
d(x + l)
(x + 1) fcll-(x-l)
dx
dx______
(x+1)2
(■Y+l)2
(x + 1)2 + (x-1)
2
2(x2+l)
1
A-2 + 1
2
(x + l)~(X~l)
(x+1)2
D. Logarithmic Differentiation
When a function is a complicated expression of products or quotients involving powers, it may be advisable
to take logarithms on both sides before differentiating. This process is called logarithmic differentiation.
Let us illustrate the method with the following two examples.
160
(
Example 4-23 Let y =
Solution
Jl±A±4.Find^.
dx
Vl-x+2x2
Taking logarithms on both sides, we have
Iny
2 [ 1 n( 1 + x + x2) - ln( 1 - x + 2x2)]
2
Then differentiate both sides with respect to x to obtain
y dx
J_ [ dln(l + x + x2) e d(l+x + x2)
dLv
2 d(14-x4-x2)
l + 2x
1 _______
2\j4-x4-x2
dln(l -x + 2x2) . d(l-x + 2x2)
d(l-x4-2x
d(l-x
+ 2x2)
dx
-1 4- 4x
l-x + 2x2,
If
2-2x-3x2
2 L(1+ x + x2)(l —x + 2x2)
and so.
1C
2-2x—3x2
dLv “:2Hl + x + x!)(l-x + 2x2)
1 4- X + X2
dy
l-x + 2x2
2-2x-3x2
= —1
2 *J(l 4-x4-x")(l —X4-2x")' ?
Example 4-24 Given that y =
Solution
x3(2x-1)4
.Find
dx
Taking logarithms on both sides, we have
Iny = 31nx4-41n(2x - l)-^-ln(x4-1)
Then, by differentiating both sides with respect to x, we have
1 , dy
y dx
X
2x-l “
3 X4-1
= 9(2x - l)(x 4-1) 4- 24x(x 4-1) - x(2x - 1)
3x(2x-1)(x4-1)
= 40x24-34x-9
3x(2x-1)(x4-1)
Hence,
dy = 40x2 4-34x-9 x3(2x-1)4
dx 3x(2x- l)(x4-l)
VT+i
2
X (2x- l)’(40x2 + 34x-9)
4
3(x + l)3
Derivatives 161
Example 4-25 Find-dXw
if
dx
(a) y = a x+l, where a is a constant.
(b) y = x*4-sinx.
Solution
(a) Taking logarithms on both sides, we have
Iny = (x + l)ln a
Then, by differentiating both sides with respect tox, we have
1 dy
— • — = ln«
y dx
dy = y In a
—
dx
= a x+l In a
(b) We first evaluate the derivative ofx*.
Let z = xx. Then
In z = xlnx
1 dz
1 ,
- • — = x • — 4- In x
z dx
x
dz
(1 4- In x)z
dx
= (14- lnx)xx
dy
dx
dx* t dsinx
dx
dx
(1 4- lnx)xx 4-cosx
Exercise 4B
1.
By considering the inverse of cosx, show that
dcos"1 x
dx
2.
(-1 <X< 1)
By considering the inverse of tanx, show that
dtan"1 x
dx
3.
-1
Find
dx
if y=ln|.v + l| for a *-1.
1
1 +x2
(a e R)
■
162
Chapter 4
4.
Find the derivatives of the following functions.
I
(a)
y = 7x2 - x +1
(W
y = (x3 - ex sin x)4
(c)
y = esinx
(d)
y = tan<?J
(e)
y = csc(tan x)
(f)
y = ln(secx- tanx)
v-(^-D3
(h)
y=
x-1
x+1
■V cos x
1 + sinx
(8>
(a+2)4
I-
(i)
,y = [ln(x: +!)]■'
(j)
y
(k)
y = ln(tanes')
(i)
y = Vx’+9
(m)
y = xsinx —
(o)
y
(P)
X
vr?
sinx
(n) ' y =
sin-1 x
sin”1 71 -x2
(P)
_y = sin '(sin2x)
y = tan
2x
1 + x2
(r)
y
(s)
y = cos
1 + 2COSX
2 + cosx
(t)
2
y = 3x3 cos-1 x + (x2 + 1)( 1 -x2)2
(«)
y=e
(v)
y = (cosx)r
(nJ
y = 3xex
W
y=3 x
(y)
2
y = xx
(z)
y _ 4ln.x
y
cos ‘(cosxsinx)
tan 2
5.
6.
Find
dx
if
(a)
y = eCOSX
(b)
(c)
x2 + y2 + xcosy = 1
(d)* ln(x + y) = e
(e)
tan
(g)
— ln(x2 + y2) + tan"1 (—) = 1
2
x
Find
(a)
dx
-i
-
1
y + xy = x“ + —
y
(f)
cos
ln(ln sin x) + (sin x)lnx
1 —x
= x + y + sinx
1+y
if
x = t-cost
y = 1 - sin/
x = cos31
(b)
y = sin31
■
x = a cos i - Z?sin(
—)
b
1
x = e2' sin2 r
(c)
y = e- cos21
(d)
■
y = a sin/-bcos(—)
b
__ i
Derivatives 163
4.4
Differentials
Let y = f(x) be a differentiable function. Then by the definition of derivative,
f'(x)= lim — = —+ ct
&v-»o &x Av
Ay = f '(x)Av - a Av
where as Av —» 0, a —» 0, and so
Ay « f'(x)Ax
(4-19)
Moreover, the difference between Ay and f'(x)Ax is a Ar which approaches to zero faster than that of
Av. Hence, we have
Definition 4-3
Let y = f(x) be a differentiable function. Then f'(x)Ax is called the differentials of f(x) atx, and is
usually denoted by
dy = f'(x)Ax
(4-20)
From (4-19) and (4-20), it is obvious that the differential dy is an approximation for the increment of y, Ay.
This result can also be illustrated geometrically, as shown in Fig. 4.6.
y
y=f(x)
o/
f(x + Ax)
f(x)
O
/i
/ i
T
RjZ
Xi
i
i
dy
Ay
X
P
I
I
Fzh
x
O' Fl
X
X + AX
Fig. 4.6
As shown in this figure, as Av -> 0, the difference between Ay and dy is QR which approaches to zero
faster than that of Av. Hence, Ay = dy.
On the other hand, if y = f(x) = x, from (4-20), we have
dy = 1 • Av = Av
Also, y = x, and so
dy = d_v = Av
where dx is called the differential of x. Then by applying this result, (4-20) can be rewritten as
dy
Therefore, the derivative
f'(x)dx
(4-21)
• may be considered as the quotient of the differentials dy and dx.
164
ampler 4
From (4-20) and (4-21). we can conclude that dy and f'(x) only differ by a constant factor Ax (or dx).
Therefore, similar to the rules of operations on differentiation, we have
Rules of Operations on Differentials
Theorem 4-3
Let u = u(x) and v = v(x) be two differentiable functions and k be a constant. Then
(1)
d(u + v) = du + dv
(2)
d(Aii) = k du
(3)
d(uv)
(4)
d(-) =
v
u dv + v du
vdw — z/dv
7
Now, we are going to develop another important property on differentials.
Let y
f(u) and u = u(x) be two differentiable functions. By applying the Chain Rule, we have
c,
df du
,
Then from the definition of differentials,
dzz = u'x dx
dy = f' dx
= C-»: - dv
Hence, we have
Theorem 4-4
If the independent variable u is changed to another independent variable x, the differential of y
remains unchanged. That is
(4-22)
dy = f' dw = f' dx
Example 4-26 Find the differentials of the functions:
(a) y = eax cosbx
(b) y
Solution
(a)
v 1 - x2
dx
(-bsin bx) + aeM • cosbx
= e^tacosbx-bsinbx)
dy = e" (a cos bx - b sin £>x)dx
165
(b)
y = Vl - x 2
y3 = I-x2
3y2dy = (-2x)dx
Hom Theorem 4-4
i = -—
2xydx
,
dy
Since the differential dy is a good approximation (which is called a linear approximation) to the increment
of y and is usually written as
Ay = dy = f '(x) Ax
f(x + Av) - f(x) = f'(x)Av
we have
f(x + At) - f(x) = f z(x)Ax
(4-23)
f(x + Ar) - f(x) + f'(x)Ax
(4-23')
Rearranging,
This is a useful formula to find the approximate value of a function.
Example 4-27 Find an approximate value of sin 59'|O
Solution
Consider the function f(0) = sin 0. Then
f'(0) = cos 0
Let 0 = 60°
-.Then A0
3
-1° = -
71
AO must he
expressed in radian
180
sin59°=f(3~lfe)
7t , ,
7t y.
71
’s,n3+(“iw)e"sy
.Ji
2
71
360
« 0.857 3
From calculator, sin 59° « 0.857 2, and so the difference is quite small.
Take <3
1.7321
3.1416
166
Chapter 4
The readers may wonder why we use differentials in the above examples when a calculator can do the
job more efficiently and accurately. The answer is that our objective is to illustrate the use ol differentials.
There may be other types of problems where the concept of approximation by differentials is essential.
The following examples illustrate another important use of differentials - the use of differentials in
estimating errors that may arise from measurement errors.
Example 4-28 The radius of a spherical balloon is measured to be 12 cm, with a maximum error in
measurement of ±0.06 cm. Find the approximate maximum error in calculating the volume
of the sphere.
Solution
Let r be the measured value of the radius of the sphere and A/- be the maximum error in r.
By assuming Ar to be positive, we have
r - Ar < exact value of r < r + Ar
If V denotes the volume of the sphere, then
V
-^r3,
3
™=4xr2
dr
Hence, when we use AV to denote the error in V, Ar corresponds to the error in r, we
obtain
12
AV = 4?rr2Ar
= 4tt(1 2)2 (0.06)
= 108.6
Ar = | ±0.061
= 0.06
Take zr= 3.1416
That is, the absolute maximum error in calculating the volume
due to the error in measurement of the radius is approximately
108.6 cm3.
Exercise 4C
1.
Find the linear approximation of f(Z?) if the independent variable x is changed from a to b where
(a)
f(x) = 4x5-6x + 1, a= 1 and b = 1.01.
(b)
f(x) = 2cosx + sinx, a = 45° and b = 47°.
2.
It is given that (2, 1) is a point on the curve x2y - xy3 = 2. Find an approximate value of b if (2.1, b) is
also a point on the curve.
3.
If A = 10/zx2 and the allowable maximum error in x is I cm when x = 10 cm, find the maximum error
in A when x = 10 cm.
Derivatives 167
4.
An isosceles triangle has equal sides of length 10 cm. If the angle between the equal sides is increased
from 30° to 33°, find the change in the area of the triangle.
5.
The intensity of illumination from a source of light is inversely proportional to the square of the
distance from the source. If a student works at a desk at a certain distance from a lamp, use differentials
to find the percentage change in distance that will increase the intensity by 20%.
6.
Let S and V be respectively the surface area and volume of the sphere of radius r. Prove that
Ar . AS . A V
‘
r
7.
S ‘
? ?
V
It is given that the period T seconds of a simple pendulum of length / cm is:
T = 2rt where g = 981 cm/s2.
(a)
If the period is decreased by 0.02 second, find the change of the length of the simple pendulum
when / = 20 cm. Correct your answer to 2 decimal places.
(b)
Show that
(c)
When / = 10 cm, the time reported by this simple pendulum is correct. If the length of the
pendulum is increased by 0.05 cm, how much does the clock have to be put back each day?
4.5
d7\ d/= I<9
T ’ I
Hence find the percentage change in the period T when the length of the pendulum is extended by
0.1%.
Higher Derivatives
A. nth Derivative
As discussed in Section 4.1, when a function y = f(x) is differentiable for all x within an interval, a function
of the derivative on x is obtained and it is simply called the derivative or the first derivative. If the first
derivative of y = f(x) is also differentiable, then the function is said to have a second derivative and indeed
it is the derivative of the first derivative. The second derivative is usually denoted by
d 2y
dr”
d2Tf(x),
x y
—
dx
From the definition of second derivative,
d2y
d f dy
d.v2
dxldx
y" = (/)'
or f"(x)
168
Similarly, we have
Definition 4-4
Let n be an integer greater than 1. If a function y = f(x) has differentiable first, second,
(n - 1 )th
derivatives, then the n th derivative (or sometimes called the zzth order derivative) of f(x) exists and is
usually denoted by
dny
dx"
d" r, x
dx
y(n)
or
f(n,(x)
The nth derivative of f(x) equal to the derivative of the (zz - 1 )th derivative. That is
d"y = d d,,~ly>|
dr I dr"-1 1
dr
= (y(n',))'
•
d2v
Example 4-29 Let y = sin x. Find ^-4-
Solution
dy____}_
d-v 71-x2
d2y
dr2
d
1
dr < Vl-X2
1
--
= (-y)(l-x2) 2 (-2%)
X
d 2y
Example 4-30 If x = cos t and y = sin t, express —in terms of z.
dv
Solution
dy
dy __dz_
dr dr
dz
cosz
—sin r
= -coir
dy
Since z is a function of x, — may be considered as an implicit function of x.
Derivatives
169
Then by Chain Rule, we have
^ = -~-(-cotr)
dr2 dx
= -^-(-cotr)«ydr
dx
— —(—esc2 r) • —J—
dx
dr
2
1
= CSC* t •--------
—sin r
= - esc31
Example 4-31
Let f(x) =
x
sinx
(a<0)
(a>0)
(a) Find fz(x).
(b) Hence show that for all real values x, f(x) has a second derivative but not a third
derivative at x = 0.
Solution
(a) Since both x and sinx are differentiable, we have
(a<0)
(x>0)
1
f'W = COSA*
Obviously, the function f is continuous at x = 0; otherwise, it is meaningless to find
the derivative at x = 0. Now, the left and right hand derivatives of f at x = 0 are
computed as follows:
f'(0)= lim
f'(0)= lim
Zi—»o*
fw-f(°) = lim /? — hsin 0 = 1
h
h-»o-
Il
sin /z — sin 0 = lim sin/z ,
----- = 1
h->o‘
h
h
Since f'(0) = f'(0) = 1, f is differentiable at x = 0 and fz(0) = 1. Hence, we have
1
(A'<0)
f'W = cosx
(A->0)
(b) Similarly,
f"(x)
(x<0)
0
-sinx
(a > 0)
Note that f z(x) is also continuous at x = 0 and again consider the left and right hand
derivatives of fz(x) at x = 0, we have
f "(0) = lim
/«->o
fz(/O-fz(Q)
II
1 - cos 0
= lim
h
=0
170 Ctaty
and
.. ( fz(/i)-fz(O)
f"(0) = 1.lim
a-»oh
= lim
cos II - cos 0
h-»0‘
Il
cos/z-1
= h->0
lim
---------‘
I]
-2sin2A
= lim---------- —
A-»0’
h
. h
sin —
___
2
= lim (-sin4)*
h
h-2
2
=0
Since f"(0) = f"(0) = 0, f"(0) = 0 and so,
(x<0)
(x>0)
0
f"(x)
-sinx
Again, f zz(x) is continuous at x = 0. However,
f"z(0)
lim HAlzno) = 0
Il
f"(//)-fzz(O) .. -sin/z-sinO
-sin h
= -l
= lim
f'"(0) = lim ——---- — = lim
h->0'
h
/.-O’
h
h
Hence, f_"'(0) * f'"(0) and so, f "'(0) does not exist.
Therefore, f(x) does not have a third derivative at x = 0.
Example 4-32 Find a general formula for the n th derivative of
Solution
(a)
y = e ai (a g R)
(b)
y = sinx
(c)
y = xm (zmgR)
To deduce a general formula for the n th derivative, it is necessary to evaluate the first few
orders of derivatives to observe the pattern and then guess a formula for the n th derivative.
dy
(a) Fory = e", — = ae at
dx
d 2y
2 .
—y = n 2 e at
dx
Note that the exponent of a increases by one after each differentiation.
^ = aneax
dx"
.................................
(*)
(A formal proof for the validity of the above formula which can be achieved via
Mathematical Induction. The details are left as an exercise for the readers.)
Derivatives 171
(b) For y = sinx.
dy =
cosx = sin(x + y)
dr
d2y =_
d sln(x+
.
7C )
_
_
dr2
= cos(x + —)
= sinf (x + —) + —
I
2 2
= sin(x + -^)
Inductively,
£y = sin(x
• / +. —
an.)
dr"
(c) Fory = x"',
dy =
mx""'
dx
4=
m(»i-l)xm'2
dx2
Note that the exponent of x decreases by one after each differentiation. Hence, we
have
Case (i) tn is a non-negative integer
tn(m-V)---(tn-n + V)xm~n
d" y„,
= < in!
dx"
0
(tn > n)
(m = n)
(tn < n)
Case (ii) tn is not a non-negative integer
_dl. = tn(m -!)••• (tn -n + l)xm n
dx”'
Remark 4-7
From the part (c) of this example, if tn is a positive integer not less than n, then
111
m
dn X arxr = ar [r(r -1) • • • (r - n + l)]xr-n
dr" r=0
r=n
From the definition of nth derivative and rules of differentiation, the following result is obvious.
172
Chuph
4
Theorem 4-5
Let f(x) and g(x) be two functions which are both differentiable up to the //th order. Then
(1)
£L*fto = *^-fto
(4-24)
(2)
£^[f(.v)±g(.r)] = £rfW±^g(-r)
(4-25)
The proof is left as an exercise for the readers.
Example 4-33 Let f(x) =
Solution
x
and n be an integer greater than 1. Find f <n)(0).
1-x2
By resolving f(x) into partial fraction,
f(x) = l(-J2 1 -x
It is easy to deduce that
dx
)=
1 -x
m!
n\
1
) = (-!)"
1+X
(l + x)"+I
and
(l-x)"+l
Hence,
1 _
d" r/ x
n-r-l
(-If
(l + x)n+l
f(n)(0) = y[l (-1)"]
[°
(n is even)
(n is odd)
“ 1/7!
B.
nth Derivatives of a Product of Two Functions
Now, we are going to introduce Leibniz's Theorem for finding the n th derivative of the product of two n th
order differentiable functions. The theorem is stated as follows:
Theorem 4-6
Leibniz's Theorem
Let f(x) and g(x) be two functions with n th derivative. Then
aX
where f (0,(x) = f(x) and g <0,(x) = g(x).
r=0
to
(4-26)
-1
Derivatives 173
Proof
The result is proved by induction on n.
(1) When n = 1, by Product Rule,
^[f(x)g(x)]=f(x)g<l,(x) + f",(x)g(x)
(x)
r=0
Hence, the result is true for n = 1.
(2) Assume the result is true for n = k ( I), i.e.
-p-[f(x)g(x)] = £ G‘f(rlU)g(‘-r)
(x)
r=0
When it = k + 1,
d^1
[f(x)g(x)]
d?+*
_
d f iG‘f('’(x)g(‘-f’
=v
(x)
dr
dxlJS
k
= X-r[c‘f(r>(x)g(i'r)(x)]
r=0 a*
(x)g';'*",(x) + f(',(x)g;l“-r+,,(x)]
r=0
= Y C‘f(r+1’(x)g<l-r)(x) + Y G‘flr)(x)g(l“r+,) (X)
r=0
r=0
Jt+1
= XC*.1f<r)(x)gl':■,''■',1 (x) + XG‘fW(x)ga+,-r’ (X)
r=0
Jt+1
= Xc’-if<r,wg“+'’r) (x) + £Cf‘f('’)(x)g,‘+'-,) (X)
r=0
= C*f“+1) (x)g<0,(x) + £ct1f(r,(x)g(‘t,-r, (X)
r=l
C*f(r)(x)g(*+I"r) (x) + C‘f,0’(x)g(‘+1’ (X)
+
r=l
~ Cjt+1 T
(x)g(0,(x) + X (C‘_, + Ckr )f(f,(x)g,‘+,-r> (X)
r=l
G = G"= 1
+ C0‘+lf(0)(x)g“+l) (X)
= C‘j1'fa+,,(x)g,0)(x) + f Cr‘t,ft')(x)g,‘+1-'’
(x)
G.,+G
F=l
+ C‘+'f(0)(x)g“+,> (x)
Jt+1
= £c‘+,f‘r)(x)g(‘+,-', (x)
r=0
Hence, lhe result is also true for n = k + 1.
By Principle of Mathematical Induction, the result is true for all positive integers n.
174 Chapter 4
Example 4-34 Let y = xe at, where a is a real number. Find y<20).
Solution
By Leibniz's Theorem,
20
.r
Zi
d-v
y(2°)
d20-'
e‘u)
dx20-r
dr v
Note that for r > 2. —= 0. Hence, we have
dx
y(20)
d20 ,, +C,2o(^)(y^ea')
dLv dx
4-20fl
,9
e
at
= x«20ettl
=Az^e
Example 4-35 Let f(x) = x2sinx. Find f(n)(x) and f(n)(0), where n > 2.
Solution
Using Leibniz’s Theorem and by noting that the third and higher derivatives of x2 vanish,
we have
n
An~r
jr
sinx)
d”
d"-smx4-C.(
.
d"''
= x2 —
—x 2V
)(—
- sinx) + C"(-^-x2)(
dx"
1 dLv
d.v/"1
dx
dx
= x2 sin(x4--^)4-(n)(2x)sinf x42
x2 sin(x4--^)4-2/7xsin^x4-
2
sinx)
+ n(n-\) (2)sin x + (n-2)n
2
2!
4- n(n — 1)sin x 4-
(n-2)n
2
Hence,
(n 2
f(n,(0) = (/i)(n-l)sin| 0 +
0
-n(n-l)
n(n-\)
(n is even)
(fi = 4k 4-1, where k is a positive integer)
(n = 4k - I, where k is a positive integer)
Example 4-36 Let y = tan ‘x. Show that for n > 0,
(14-x2)y(n+2) + 2(n 4- l)xy(n+,) 4- n(n 4- l)y(n) = 0
Solution
Since y = tan ‘x,
1
14-X2
(l + x2)/ = l
y'
Derivatives 175
Differentiating both sides of this expression by (n + 1) times, by Leibniz's Theorem, we
obtain
1 71 + 1
(i+x2)^i-/+cr
v') = 0
(l+x2)y("+2> + (n + l)(2x)y',(n+D + (n + l)n (2)y(n) = 0
(1 + x2)/"*2' + 2(n + l)xy<ntl) + /z(?i + l)y(/” = 0
Example 4-37 If y = acos(lnx) + hsin(lnx), show that
x2y" + .ry' + y = 0
(1)
Hence show that for n > 0,
x2y(n+2> + (2/i + l)xy'l<"+l)+(n2 + l)y<”>=0
Solution
(2)
By Chain Rule,
y' = -flsin(lnx)« —+ bcos(lnx)- —
X
X
xy' = -a sin(ln x) + b cos(ln x)
(3)
Then by differentiating both sides of (3) with respect to x, we obtain
xy" + y' = -a cos(ln x) • — - b sin(ln x) • —
X
X
x2y" + xy' = -[a cos(ln x) + b sin(ln x)]
= -y
x2y" + xy' + y = 0
Now, by Leibniz’s Theorem, the n th derivatives of the first two terms are
d y")+c,"(-^x2)/ d"
dx
dx
dx"'1
n(n-l)
(2)y(fl)
= x2y,n+2) + n(2x)y("+11
2!
x2y'"*2> + 2Hxy("+,) + n(n-l)y(n)
and
d"
,,
d" , „„,dx., d”'1
^) = X^y+C1(^)(^ y')
= xy^+ny(n)
Hence, by differentiating both sides of (1) by n times, we obtain
[x2y(fl+2) + 2/zxy("+1) + n(7i-l)y(",]+[xy<"t,>+/ly,",]+y<” = 0
„2,,(„
x
‘.y +2> +(2„+ J);,/"*” +(,12 + 1)/” = 0
/')
176
Chapter 4
Example 4-38
If y = (cos ’a)2, show that
(1 -A'2)y"-A'y'-2 = 0
(1)
Hence show that for n > 1,
(1 - x2)y(nt2’ -(2n + l)xy(n+l> -n2y‘n) = 0
(2)
Find alsoy<n)(0) for n > 1.
-1
y' = 2(cos"‘ a) •
Solution
x2 y' + 2cos’'x = 0
(3)
Then differentiate both sides of (3) with respect to a* to obtain
V1 - x2 y" + —*—-, •.(-2x)y'
(-2x)y' +
2•
+ 22-Vl-x
-x 2 Vl-x
==0
2
(1 - a'2 )y" - xy' -2 = 0
Again, by differentiating both sides of (1) by n times, using Leibniz’s Theorem, we have
+cr
2)y(n+2)+ C1"(-2x)y(n+l) + C," (-2)y*] - [xy'"*1’ + C"y"n ]
0
-Ay<”+l)
(1 —x2)y<n+2) —2/ixy'’("+l)-n(n-l)y(n> -xy'^
1’ -wy'"
- ny ’ = 0
(1 -x2)y<,’+2)-(2n
<"+” -n2y("’ = 0
-(2n + l)xy<n+1)
Substituting x = 0 into (2), we have
y'(“+2,(0)-n2y'"’(0) = 0
y(n+2)(0) = ,i2y,")(0)
(n>\)
Then by replacing n + 2 by n, we obtain
y(”)(0) = (n-2)2/"-2>(0)
(n > 3)
By repeating the application of this result, the last term will be either y(I)(0) or y(2,(0)
according to n being odd or even. Now, from (1) and (3),
y(2)(0) = 2
Hence, if n is odd, then
and
and
y(l)(0) = . ~2
•cos~10 = -^
Vi-(0)2
y<",(0) = (n-2)2(n-4)2--l2y<,’(0)
= (n-2)2(n-4)2 •••I2 •(—^)
If n is even, then
y(n,(0) = (zi - 2)2(zz - 4)2 • -22y(2)(0)
= (z?-2)2(z?-4)2---22-2
Derivatives
Exercise 4D
1.
2.
0
(x<0)
x2
(x>0)
(a)
Find f '(x) for x
0.
(b)
Does f'(0) exist? If so, find its value.
(c)
Find f "(x) for x
(d)
Does f "(0) exist? If so, find its value.
Let f(x) = <
Let f be a function defined on R by f(x) = «
Prove that f'(0)
3.
0.
Find
(a)
x3 cos —
(x*0)
X
0
0, but f "(0) does not exist.
(x = 0)
if
dv3
x~ + y2 = 9
(b)
x2 + xy + y2 = 1
(d)
x = cos/ + sin/
y cos/-sin/
x=/
(c)
y
t
4.
If y = er sin 2x, show that y " - 2y' + 5y = 0.
5.
Let x = /"' + /
and y = t + t ’. Show that
2
m2(x2-4)f— I =/-4
v dx
Hence show that
-> dy
„
+ m‘x—-y = 0
dx
dx
6.
7.
d"y .
Let n be an integer greater than 1. Find —if
dx
(a)
y
Inx
(b) y = cosx
(c)
y = xlnx
M y = ax
(e)
y = cosx sin 3x
Let f(x) =
x3
x2-4‘
(a)
Resolve f(x) into partial fractions.
(b)
Let n be a positive integer greater than 1. Find the value of the n th derivative of f(x) at x = 0.
177
178
Chapter 4
8.
It is given that
=nl.
r=0
0.
Show that the //th derivative of (eA - 1)" is n\ at x
9.
(o)
For x * 0. show that
d e7x
—
dx
(b)
1
ex
10.
and
Let n be a positive integer. Show that, for x
d"
dx" A
d2
7
—-xex
dx2
1
ex
x3
0.
2
£
Let y = cos ’x. Show that
(1 -?)/'-r/ = 0
Hence show that
(l-x2)y(',+2)-(2n + l)xy^n+,)-n^(^,
11.
0
(/z > 0)
(l-x2)y(/,+2)-2xy(n+I) + n(n + l)y(n) =0
(n > 0)
Let y = (1 — x2)n, where n is a positive integer. Prove that
(1 - x2)y' + 2nxy = 0
Hence show that
12.
Let y = ex ln( 1 + x).
(a)
Show that
(1 +x)y"-(l + 2x)y' + xy = 0
U).
Hence find the value of y"(—
2Z*
(b)
Using (a) to show that
(l + x)y("+2) + (n-l-2x)y(n+,) + (x-2n)y(”)+ny(n’,)
0
(n > 1)
In x + Vl + x2
13.
Let y
—-—■-=—-. Show that
vi+x2
(1 + x2)y' + xy = 1
Hence show that
(l + x2)y(n+2) + (2n + 3)xy(n+,) + (/z + I)2yln)
14.
Let y(l -x2) = 1. Show that for n > 2,
(1-x2)y(n)-2nxy''(n“,)-n(n-l)y('’"2) =0
Hence evaluate y (n)(0) for n > 0.
0
(n > 0)
! >< ri\ ali < ■
15.
179
Let x = sin t and y = sin(2m + 1)/, where in is a positive integer such that 2m +1*0. Show that
(1 - r)(y')2 = (2m + 1)2(1 -/)
Hence show that for n > 0.
d-x2)y("+2)-(2n + l)xy("+,) + [(2,n + l)2-/z2]/") = 0
jA
16.
Show that for any positive integer k <n, the value of ^T[(x4-2)n(x-2)rt] is 0 at x = ±2.
17.
If y = (sin *x)2, show that
(1 -^)y"-xy' = 2
Hence show that
(l-x2)y(n+2) -(2>z4-l)xy^^+,, -n2y(n)=0
(fl>D
Find alsoy(n,(0) for n > 1.
18.
Let f(x) = xne
and g(x) = e1 f(n)(x), where n is a positive integer.
Show that g(x) is an n th degree polynomial and find its coefficients.
19. (a)
(b)
Let p(x) = ek\ where k is a constant. Find p(n)(x).
Let n be a non-negative integer.
Il is given that for any functions f(x) and g(x) with n th derivatives,
[f(x)g(x)]<n) = £arf(r,(x)g<'-r) (x)
r=0
where a/s are constants. By taking f(x) = etr and g(x) =
show that
z^n
20.
Let n be a positive integer and w(x) be a function such that 1/ z(x), w"(x),..., z/n)(x) exist.
(a)
Given that y(x) = u(x)egxy where q is a real number, expressy<w)(x) in terms of u(x), uz(x), z/"(x),
(b)
By putting zz(x) = ep\ where p is a real number, use (a) to prove the Binomial Theorem, i.e.
...,«(rt)(x).
‘
« (n\
n
\prq
(p+qY1 = X
(6 marks)
(HKAL 1993 Paper II)
21.
Let r be a real number. Define y =
X 4-1 Y
X - 1 J forx> 1.
dy _ -2ry
dx x2 -1
(a)
Show that
(b)
For n = 1,2, 3,... , show that
(x2 -1)/"+” + 2(nx + r)y'n> + (n2 -n)yt"~" = 0
where y(0> = y and y'
dx
for it >1.
(5 marks)
(HKAL 1995 Paper II)
180
Chapter 4
22.
Let f(x) = xne<L\ where a is real and n is a positive integer.
Evaluate f,2?,’(0).
(4 marks)
(HKAL 1996 Paper II)
23.
Let f(x) =
2x
x2 -1
(o)
Resolve f(x) into partial fractions.
(b)
Find f(n)(0), where n = 1,2,3,....
(5 marks)
(HKAL 1999 Paper 11)
4.6
Mean Value Theorem
In this section, we are going to introduce some important theorems in the study of the properties of a
function. The highlight is Mean Value Theorem. Before introducing this important theorem, it is necessary
to study the following definitions and theorems first.
Theorem 4-7
Fermat's Theorem
Let y = f(x) be defined and differentiable on an open interval (n, b). If f(x) attains its absolute
maximum (greatest value) or absolute minimum (least value) at x = c, where c g (a, b), then
f'(c) = 0.
Both absolute maximum and absolute minimum are called absolute extremum.
Proof
Suppose f(x) attains its absolute maximum at x = c.
Since c g(a, b), there exists sufficient small \h\ such that (c + h) e(a, b). As f(c) is the
greatest value, we have
f(c + h) < f(c)
(1)
and so,
f'(c) = lim f(c + /;) f(c) > 0
a->o
h
(1) and h < 0
f'(c) = lim f<c + /|)-1(c) < 0
v (1) and h > 0
A-»0’
h
Since f(x) is differentiable at x = c,
t'(c) = f;(c) = o
That is. f'(<?) = 0.
The proof for f(x) attaining its absolute minimum at x = c is similar and is left as an
exercise for the readers.
Derivatives
Remark 4-8
181
From Fermat’s Theorem, for any differentiable functions f(x), the tangent at the point,
where f(x) attains its absolute extremum, is a horizontal line. (See Fig. 4.7.)
y
y=f(x)
i
i
i
o
► X
c
Fig. 4.7
Example 4-39 The quadratic function f(x) = x2 - 4x + 5 is well-defined on (0, 5).
Since
f(x) = (x - 2)2 + 1
f(x) attains its least value at x = 2. (See Fig. 4.8.)
On the other hand
f'(x) = 2x-4
f'(2) exists and f'(2) = 0.
y
y=(x—2)2+1
10-
5
1-- O
x
-I----- r
1
2
3
4
Fig. 4.8
5
182
Example 4-40 Let f(x) = x3 + 3av2 + 3bx + c for all real values x.
Show that if a2 - b < 0, f(x) cannot have an absolute maximum or an absolute minimum.
Solution
The result is to be proved by contradiction.
Since f(x) is defined on an open interval, if f(x) has an absolute maximum or an absolute
minimum, by Fermat’s Theorem, there exists a real number c such that f'(c) = 0. However,
f'(x) = 3x2 + 6av + 3b = 3(x + a)2 + 3(b - a2)
and since b - a2 > 0, we have f '(x) > 0 for all real numbers x and this contradicts to the
assumption that f'(c) = 0.
Therefore, f(x) cannot have an absolute maximum or an absolute minimum.
Remark 4-9
(1) The converse of Fermat’s Theorem does not hold.
That is, f'(c) = 0 does not necessarily imply that f(x) attains its absolute extremum at
x = c. For example, the function f(x) = x3 with zero derivative at x = 0 but f(x) is a
strictly increasing function. (See Fig. 4.9.) Hence, f(0) cannot be the absolute minimum
value nor the absolute maximum value.
y
f(x) = X3
-► X
O
Fig. 4.9
(2) Fermat’s Theorem cannot be applied to functions which are not differentiable.
For example, the derivative of the function f(x) = | x | is non-zero for all x
0 but it
attains its absolute minimum at x = 0. At x = 0, f(x) is not differentiable. (See
Fig. 4.10.)
O
Fig. 4.10
x
/)cri\ fifi\ <•%
183
(3) Fermat’s Theorem cannot be applied to differentiable functions defined on closed
intervals. This is because the absolute maximum or absolute minimum may be
attained at the end-points. As a result, one of the left and right hand derivatives at c
may not exist. For example, the function f(x) = (x - 2)2 + 1 defined on [0, 5] attains
its absolute maximum at x = 5 (See Fig. 4.8.) but
f'(5) = 2(5)-4 = 6*0
Theorem 4-8
Rolle's Theorem
Let f(x) be a continuous function defined on [a, b]. If f(x) is differentiable on (a, b) and f(fl) = f(b),
then there exists c e (a, b) such that f'(c) = 0.
Proof
Since the function f(x) is continuous on [a, b], the greatest and the least values of f(x) must
exist and let it be M and in respectively.
Case (i)
M=m
Since for any x in [a, b], in < f(x) < M and so, f(x) = M. Hence, f'(x) = 0. Then we may
take c to be any value in (a, b).
Case (ii) M > m
Since f(fl) = f(b), one of the greatest and least values should not be at x = a or x = b. That
is. it should be at some point c in (a, b). Hence, by Fermat’s Theorem, f'(c) = 0.
HHEMMMI
Remark 4-10
The geometrical meaning of Rolle’s Theorem is that for any continuous function defined
on [a, b], if f(x) is differentiable on (fl, b) and f(fl) = f(£>), then there exists a point in (fl, b)
with a horizontal tangent. (See Fig. 4.11.)
y
y = f(x)
O
a
c
Fig. 4.11
b
x
184 Chapter 4
Example 4-41 The function f(.v) = (x- 2)2 + I is continuous on [0. 4]. Note that
f(0) = f(4) = 5
and so it satisfies the condition of Rolle’s Theorem.
Also, as f '(.v) = 2x - 4, f'(2) = 0 and 2 g (0. 4). Rolle’s Theorem is verified.
Theorem 4-9
Mean Value Theorem
Let f(.r) be a continuous function defined on [a, /?]. If f(x) is differentiable on («, b), then there
exists c g (a, b) such that
f'(c)
Proof
b-a
(4-27)
The proof is based on the result of Rolle’s Theorem. In order that Rolle’s Theorem can be
applied, it is necessary to define a new function g(x) such that g(fl) = g(Z?).
Suppose
g(x) = f(.r) + kx
where k is a constant.
Since g(a) = g(Z?),
f(fl) + ka = f(Z?) + kb
k = - fW-f(a)
b-a
Hence,
g(x) = f(.v)-
f(b)-f(a)
x
b-a
As g(x) satisfies the condition of Rolle’s Theorem, there exists c g(a, b) such that
g '(c) = 0, and so
fW~ f(n)
g'(x) = f'(x)b-a
g'(c) = f'(c)- fW-f(n) = 0
b-a
Therefore,
f'(c) =
f(Z?)-f(n)
b-a
The readers should note that a < b is not a necessary condition for the Mean Value Theorem.
Mean Value Theorem may be considered as the generalization of Rolle’s Theorem when f(«) * f(b),
and it is sometimes expressed as
f(b)-f(a) = f'(c)(b-a)
(4-28)
I
F
Derivatives 185
The Mean Value Theorem is important in mathematical analysis. As shown in Fig. 4.12,
Slope of 45
■
fW-f(fl)
b—a
Hence, we have
Remark 4-11
The geometrical meaning of Mean Value Theorem is that for any continuous function
defined on [«, Z?l, if f(x) is differentiable on (a, 5), then there exists a point in (a, b) with a
tangent parallel the line joining the two end-points of f(x), i.e. AB. (See Fig. 4.12.)
y
A
y=f(x)
' 1
I
I
vx
"\
.r
i
i
i
i
'
/’
Zi
1
i
0
a
c
I
I
I
I
b
x
Fig. 4.12
i
Example 4-42 By using Mean Value Theorem, show that if x < y, then
I
siny - sinx <y - x
Solution
Let f(r) = sin t. Then f(r) is continuous on [x, y] and differentiable on (x, y). Then by Mean
Value Theorem, there exists c e (x, y) such that
sin y - sin x
y-x
f'(c)
= cosc
<1
As y - x > 0, we have
siny-sinx<y-x
!
186
('htipier 4
Example 4-43 By using Mean Value Theorem, show that
e v - e a > e a(y - a)
for all real values y and a.
Solution
Let f(x) = ex.
Case (i)
y>a
Applying Mean Value Theorem on to f on (a, y), there exists c g (a, y) such that
ey-ea
c
---------= e
y-a
>ea
ey -ea > ea(y-a)
c > a and e1 is
strictly increasing
Case (ii) y = a
It is trivial that the equality holds.
Case (Hi) y<a
Applying Mean Value Theorem to f on (y, a), there exists c g (y, a) such that
ey-ea
c
--------- = e
y-a
ey -ea > ea(y-a)
.* i < (i and e ’ is
strictly increasing
.* y - a < 0
From these three cases, for all real values y and a,
ey —
- e > ea(y - a)
Example 4-44 Let a, b e R such that a <b and f(x) be a differentiable function on R such that f(«) > 0,
f(b) < 0 and f '(x) is strictly decreasing. Show that
f '(b) < 0
Solution
Applying Mean Value Theorem to f on (a, b), there exists c e (a, b) such that
f'(c) = fW-f(fl)
b-a
<0
Since f'(x) is strictly decreasing,
^fW-f(fl)
b-a
<0
I
l)(‘n\iiii\es
187
Example 4-45 Let f be a real-valued function such that f has a continuous derivative and f z(0) = 0.
Suppose an <0<bn such that as n -> °o, a„ -» 0 and bn -> 0. Define
D
Dr _f(6„)-f(a„)
Using Mean Value Theorem to show that
lim D = f'(0)
n—>o»
Solution
Applying Mean Value Theorem to f on (a„, b„), there exists c„ e (<?„, bn) such that
hn -
= f'(c„)
O„ = f'(c„)
■
As n —» oo, an —> 0, bn
0 and cn g (an, btf we have
Hence,
= lim f'(c„)
v f'(x)is
continuous
= f'(0)
In this part, we consider the application of Mean Value Theorem to the study of constant functions. As
shown in the table of standard results, we have seen that the derivative of a constant function is zero. Now,
we are going to study the converse of this result.
Theorem 4-1 0
Let f(x) be continuous on [a, b] and differentiable on (a, b). f(x) is a constant function if and only if
f'(x) = 0 for all A' e («, b).
Proof
“Only if’ part:
Since f(x) is a constant, f(x + h) = f(x) for all x, x + h e. [a, b].
fz(x)=limf(x + /,)~f(x)
h-»o
h
0
“If’ part:
Let x,, x2 6 [a, H Then by Mean Value Theorem, there exists c lying between x, and x2
such that
f(x2) - f(x,) = f '(c)(x, - x,)
Now, from the given condition that f z(x) — 0 f°r all -Y 6
^)> we have
f'(c) = 0
and so, f(x,) = f(x2) = k (say).
Since x, and x2 are two arbitrary points in [a, b], f(x) = k for all x g [a, b], i.e. f(x) is a
constant function.
188 Chapa
4
Example 4-46 Let a be a positive real number. By considering the function f(x) = In ax - Inx defined on
the set of all positive real numbers and using the result In 1 = 0, show that for any positive
real numbers a and b.
In ab In fl + \nb
Solution
By Chain Rule,
f'(.r) = — «a- —= 0
ax
x
According to Theorem 4-10, we conclude that there exists a real constant k such that
f(x) = In ax - lnx = k
Then by putting x = 1, we have
k = In a - In 1 = In a
In «x - Inx = ln«
Lastly, if we put x = Z?, we obtain the result
InflZ? = In a + In Z?
Corollary 4-1
Let f(x) and g(x) be continuous on [a, b] and differentiable on (a, b). If
f'(x) = g'(x)
for all x g (a, Z?), then for all x g [a, Z?], the difference between f(x) and g(x) is a constant, i.e.
f(x) = g(x) + constant
Proof
Since f'(x) = g'(x), we have [f(x) - g(x)]' = 0. Hence, by Theorem 4-10,
f(x) - gU) = constant
and so.
f(x) = g(x) + constant
Dt rivalix
189
Exercise 4E
1.
By using Mean Value Theorem, show that
(«)
| cos x - cos y | < | x - y |
(h)
sin px
z\p\
x
(c)
2.
< ln(I 4-x)
Let f(x) be a continuous function defined on [3, 6].
If f(x) is differentiable on (3, 6) and | f'(x)-9|<3, show that
18< f(6)-f(3)<36
3.
Let P(x) = anxn 4- a„_1xn~l + • • • +
be a polynomial with real coefficients and
-^- + ■^- + .•• + ^=0
Il +1
II
By using Mean Value Theorem, show that the equation P(x) = 0 has at least one real root between 0
and 1.
4.
By using Mean Value Theorem, show that for 0 <x< 1,
1 + x < er < 1 + ex
5.
A real-valued function f defined on an interval [a, b\ is said to be Lipschitz-continuous if there exists
a positive constant k such that
I f(x,) — f(x2) I < £ I Xj -X 2
(Vx,,X, €[«,&])
Show that if f is continuously differentiable with | f'(x) | < K for all x e [a, b], where K is a positive
constant, then f is Lipschitz-continuous.
6.
Let f be a real-valued function defined on (0, °°). If f'(f) is an increasing function, show that
f(/z) 4- f
7.
- n) < f(r) < f(/i) + f '(n 4- l)(r-/z)
(Vre(/z, n + 1))
Let a be a positive real number. The function f(x) = lnx4-ln^- is defined on the set of all positive
real numbers. By using the result In 1 = 0, show that for any positive real numbers a and b,
In— = Infl-lnb
b
8.
Let a be a positive real number. The function f(x) = |lnxr is defined on the set of all positive real
numbers. By using the result In 1 = 0, show that for any positive real numbers x and non-zero real
number r,
lnxr = rlnx
190 Chapter 4
9.
Let f be a real-valued function such that
|f(.v)-f(y)|<(.v-y)2
(V.v. y e R)
Show that f is a differentiable function.
Hence deduce that f(x) = k for all x g R, where k is a real constant.
10.
Let f(x) and g(x) be two differentiable functions defined on the interval [0, zrj such that
(1)
g(x)>0
(2)
y-f(.v) =
(3)
y-g(.v) = f(x)g(x)
dr
(4)
f(y) = 0 and g(y)=l
dr
-1
g2(-v)
\
(ci)
Find, in terms of f(x) and g(x)
(b)
Show that l + f2(x)
dv^g^x)/
1
g2U)
Revision Exercise
1.
Let f be a function defined on R. Suppose x g R. If the limit
f[(A.)=Hmnj^/o-L(A-.-/o
$
2h
exists, then f is said to be S-differentiable al x and f/x) is called the symmetric derivative of f al x.
2.
(Cl)
It is given that f is differentiable al x. Show that f is S-differenliable at x.
(b)
By considering the function g defined by g(x) = | x | for all x g R. show that the converse of the
statement of (a) is not true.
(c)
Let f and g be two continuous functions. If f and g are S-differenliable at x, show that both the
sum f + g and the product fg are S-differenliable al x.
Let f(x) and g(r) be two non-constant differentiable functions satisfying the relation
g(0 = f(x + t) + f(x - t) - 2 f(x)
(Vx, t g R)
for all real numbers x and t.
(a)
By finding the derivative of g with respect to r, show that
Hx + ^-f'Wrxf'C/O-fXO)
(Vx, h g R)
for all real numbers x and h.
(b)
Suppose that f "(0) exists. Prove that f"(x) exists for all x and that f"(x) = f"(0).
I' 191
3.
Let f be a differentiable function such that
f(x + y) = exf(y) + eyf(x)
(a)
(Vx.yeR)
Prove, by induction, that for all integers n greater than 1,
f(/tv) = 7ze(n-Utf(x)
(b)
Show that f(0) = 0. Hence express lim
(c)
Show that
in terms of f'(0).
f'(x) = f(x) + kex
(x g R)
where k is a real constant.
4.
Let nt and n be two positive integers such that in > n.
(a)
Find the n th derivative of x"’-1.
(b)
By considering x
m-1
^in
1 , show
u
•—
that
x
c'Qn - c” +£"-•••+(-i)n cnn = (-ir c;-'
5.
Let f(x) = x"( 1 - x)", where n is a positive integer.
(a)
Prove by induction on k that
(b)
Show that fa)(0) is a multiple of n!.
Hence deduce that f'A,(l) is also a multiple of nl.
6.
(a)
Let y = eXsinx. Show by induction on n that
g=2Vsin(x+^
(b)
By applying Leibniz’s Theorem and using the result in (a), show that
Cn
k sin(x +
22 sin(x +
4
7.
I
k=o
Let y = ln( 1 + x2).
(a)
Show that
(b)
(1 + x2)y' - 2x = 0
d*
Show that if k is a positive integer and z = —p[ln( l + x2) , then
(c)
(1 + a2) ^4 + 2( k +1 ).r + k(k +1 )z = 0
dx“
dx
d^ r[ln(l + x
vanishes at x = 0.
Deduce that for any positive odd integer/:, z = —
192
Chapter 4
8.
Let f be a non-constant real-valued function defined on R satisfying the following conditions:
(1)
f(x + y) = f(x)f(y)
(2)
there is a real number c such that f z(c) exists.
(V.v, y 6 R)
(a)
Find f(0). Hence show that f(x) * 0 for all real numbers x.
(b)
Show that f'(0) exists and express f'(0) in terms of f z(c) and f(c).
Hence show that for all real numbers x, f z(x) exists and
f'(-v) =
(c)
9.
-^f(x)
f(c)
By considering the derivative of the function e~L' f(x), show that f(x) = eLx for some non-zero
constant k.
Let f be a real-valued function such that for all x g R, f(x) =
Show that
(l + x2)f'(x) + xf(x)=0
Hence deduce that for n > 1,
(l + x2)f("+,) (x) + (2/t +l)xf(n,(x)4-7/2f(rt”,) (x) = 0
n-r
(b)
i
Let P„(x) = (l + x2) 2f(n)(x)
(')
(n = 0, 1,2,...)
Show that
P„tl (a) = (1 + a-2 )P„'(x) - (2m + 1 )x P„ (x)
(h = 0, 1,2,...)
Hence deduce that Pn(x) is an n th degree polynomial with leading coefficient (-1)"/?!.
(ii)
For n > 0, show that
P„tl(x) + (2m + l)x P„(x) + m2(1 + X2 )P„_, (X) = 0
and evaluate Pn(0).
(Hi) Prove by induction on r that for r= 1,2,...,
Py’(A-) = (-1)'[//(« -1)- • (m - r + I)]2 pn r(x)
where n is a positive integer.
(c)
It is given that the coefficient of xr in PJx) is
an even function for any positive integer n.
10.
^^(0)
, show that Pn(x) is either an odd function or
r!
Let f(x) be a function such that f z(x) is strictly increasing for x > 0.
(a)
Using Mean Value Theorem, or otherwise, show that
fz(£) < f(£ + l)-f(£)<f'(fc + I)
(b)
Hence, deduce that
f'(1) + f'(2) + • • • + f \n - 1) < f(n) - f( 1) < f'(2) + f'(3) + • • • + f z(n)
11.
Let f(x) be a real-valued function, x is called a fixed point of f if
f(x) = x
If f is differentiable and f z(x)
fixed point.
1 for all x, by using Mean Value Theorem, show that f has at most one
Derivative* 193
12.
Let f: R —> R be a continuously differentiable function satisfying the following conditions for all
x e R:
A.
f(x) > 0
B.
f(x + l) = f(x)
C.
f(v)f(
4
x-t-1
) = f(.r)
4
Define g(x) =-^-lnf(x) forxcR.
dx
(a) Show that for all x g R,
(i)
f'(x+ l) = f'(x)
(ii)
g(x+ l) = g(x)
(•") 4fer4 )+8( 4 ) = g(-v)
•4 \
4-
(8 marks)
(b)
Let M be a constant such that | g(x) \< M for all x 6 [0, 1 ].
(i)
Using (a), or otherwise, show that for all x eR,
Hence, deduce that for all x g R,
g(-v) = 0
(ii)
Show that for all x g R, f(x) = 1.
(7 marks)
(HKAL 1994 Paper II)
13.
Let f(x) be a differentiable function on R such that | f'(x)|<|f(x)| for all x 6 R.
(a)
Suppose a > 0 and f(«) = 0. Letx G(n, a + 1).
(i)
Using Mean Value Theorem or otherwise, show that there exists
g (a, x) such that
|f(x)|<|f(^,)|(x-«)
(ii)
Using (a)(i) or otherwise, show that for each n = 1, 2, 3
there exists
e («, x) such that
I rev) | < | f(^„) |cv (Hi) Using (a)(ii) or otherwise, show that f(x) = 0 for all x g [a, a + 1 ].
[Hint: You may use the fact that there is M > 0 such that | f(x) | < M for all x g [«, a + 1 ].]
(9 marks)
(b)
Suppose f(0) = 0.
(i)
Using (a), or otherwise, show that f(x) = 0 for all x g [0, °°).
(ii)
Show that f(x) = 0 forallxcR.
(6 marks)
(HKAL 1999 Paper II)
194
\dvtiih < d I < v<7 Pint .\hilht'imiric\
Chapter 5
AppOratoomis
Diiffeireirrtra
As discussed in Chapter 4, we have learned some theory on differential calculus. In this chapter, we are
going to discuss the applications of differential calculus. Some of these applications may also have been
discussed in secondary school.
5.1
[/Hospital's Rulle
Theoretically, if the limit values of f(x) and g(x) have been evaluated, by the rules of operations of limits, the
limits in the following forms
f(.r) ± g(.r),
f(.v)g(.v),
f(x)
gW
[KA-)]8'"
can also be found. However, the limits in the forms of
“0”
0 ’
“0-oo”
oo — oo
“0°”,
. oo0„
(5-1)
oo
cannot be determined directly from the limits values of f(x) and g(x). The seven forms in (5-1) are called the
indeterminate forms.
In Chapter 2 and 3, we have discussed the methods in evaluating some limits in indeterminant form.
Now, we are going to introduce the L’Hospital's Rule which may help us to solve the indeterminate forms
more effectively.
Applii atitms <>f Oifjcrciiiuil Calculi!'.
A.
195
L’Hospital’s Rule
Theorem 5-1
L'Hospital's Rule
Let f(x), g(x) be two differentiable functions at all the points near a, but not at a. Suppose g'(x) * 0.
If
lim f(.v) = lim g(.r) = 0
(1)
x—>a
(2)
liin
x—>a
fZ(A)
g'CO
exists or tends to infinity
i- fU) i- r'(x)
lim -7-^ = hm
then
g(x)
(5-2)
g'(-r)
A crucial step of the proof of this rule requires Cauchy Theorem and so, we will only describe the concept of
the proof briefly as follows:
Since lim f(x) = lim g(x) = 0, we have to define f(n) = g(a) = 0 so that f(x) and g(x) are continuous at
.v—>a
.r—>a
x = a. Then by applying Mean Value Theorem to f(x) and g(x), we have
f(x)
g(.r)
f(x)-f(a) ... fz(c,)
g(.r)-g(a) g'(c2)
(5-3)
where cp c2 lie between x and a.
As x —> a and cp c2 lie between x and a, it is obvious that c, —> a and c2 —> a and hence (5-2).
The author wants to emphasize that in order to complete the proof formally, we have to show that
c, = c2 and this is out of the scope of this text.
Remark 5-1
(1) L’Hospital’s Rule can also be applied to the indeterminant form
ii OO ”
OO
In this case, only the condition (1) have to be changed to
(1)'
=: oo,
oo
Iimf(x) =
Iimg(x) = oo
.v—>o
x—*a
(2) L’Hospital’s Rule can also be applied to the indeterminant form
“0
0
and
with x —> 00.
This can be done by using the substitution z = — • Then we have x —> oo implies that
z -> 0. Also, the condition that f(x) and g(x) are differentiable at x = a (In this case,
a = 0.) will be changed to the condition that f(x) and g(x) are differentiable at all
points with sufficiently large |x|. Furthermore, g'(x)
0.
1'1
196
Chapter 5
r
Example 5-1
Evaluate lim
Solution
The functions f(x) = In x and g(x) = x - I are differentiable at all points near x = I. Also
g z(x) = I * 0 and
—
x-1
I
lim Inx = lim (x-1) = 0
Hence, we obtain the indeterminate form
“0”
0
The conditions for L’Hospital’s Rule arc seen to be satisfied. Then, by applying L’Hospital’s
Rule, we have
Inx
hm----- = lim
=I
x->l _v -1
Example 5-2
t
~
In sin x
Evaluate hm--------v->o- In tan x
t
Solution
The functions f(x) = In sinx and g(x) = In tanx are differentiable at all points on the
right hand side of x = 0. Also
g z(x) = cot x sec2x
0
and
lim lnlanx =
lim In sin x =
x—»0’
.1—>0'
Hence, we obtain the indeterminate form
4* OO ”
OO
The conditions for L’Hospital’s Rule are seen to be satisfied. Then, by applying L’Hospital’s
Rule, we have
In sinx ..
cotx
hm---------= h m---------- 5—
*-*o- Intanx
cotxsec“x
= lim cos2 x
x—>0*
=1
For any elementary functions f(x) and g(x), except some special points, their derivatives are
differentiable. Hence, if they satisfy the condition that
lim f z(x) = lim gz(x) = 0 (or «>)
x—>a
(x—>°°)
x-ta
(x—»«*>)
then L’Hospital’s Rule can also be applied to their derivatives.
Sometimes, it is necessary to apply L’Hospital’s Rule more than once in order to obtain the value
of the limit. This technique is illustrated in the following example.
Applieation\ of Differential Calculus 197
Example 5-3
Evaluate lim -——.
-I (x-1)2
Solution
Note that the given limit is of the form
Then, by applying L’Hospital’s Rule, we
have
» o ••.
r x-ex
lim-------- 7 = lim--------(x-1)2 x-i 2(x-l)
= liin
(I
is transformed
into another
-e
2
•• 0 "
0
By I,'Hospital’s Rule
_1_
2
ln(l + xe2t)
x2
Example 5-4
Evaluate
Solution
Obviously, it is of the form
lim
—
. Hence, by applying L’Hospital’s Rule, we have
oo
ln(l + xe2jt)
lim
= lim
x2
e2' + 2xe2t
l+xe2j
2x
e2,(l+2x)
X—♦ +*•» 2x(l+xe2*)
= lim
= lim
\ x —»t«>
l + 2xV lim
2x J k
(*)
e2'
>
1 + xe"x .
1
= lim (— + !)• lim
.t—>+«
2x2 r x—>t«>
__
Div ide both the
denominator and
numerator by e2’
=0
Remark 5-2
In applying L’Hospital's Rule, we have to find the derivatives of the denominator and
numerator. The result obtained may be very complex. Therefore, it is suggested that the
obtained result should be simplified and separated into terms which can be evaluated
easily. For example, if we divide both the numerator and the denominator of (*) by xez\
we have
lim ln^1+;1-—= lim
X —>-r~
-+2
------ = 0
4 + 2x
c
which is a simpler solution for Example 5-4.
I
198
< . , </ ?
I
1 -cosxcos2x
x—>0
1 - cos x
Example 5-5
Evaluate lim
Solution
lim
sin x cos 2x + cos x«2 sin 2x
1 - cos x cos 2 x
= lim
.t
—
»o
A-»0
sin x
1 - cos x
~
n
r sin2x
= lim cos2x 4- lim 2cosx • lim —----x—>o
x—>o
x—>0 sinx-
, , OI. 2cos2x
1 +2 lim
x-»0
cos x
=5
Remark 5-3
.. Q -
0
is transformed
.. o ..
into another —
By L’Hospital’s Rule
I'
. .. sin2x
in the above example may also be solved as follows:
The limit lim-------sinx
sin2x -^-•2
.. sin2x
lim —----- = lim
x
—
»0
*-♦0 sinx
2x
sin x
sin 2x
x
• lim
2 lim
x—>0
*-*o sinx
2x
2
i
I'
Remark 5-4
f
cannot
cannot be
be evaluated,
evaluated, it
it does
does not
not necessarily
necessarily imply
imply that
that lim
lim -• — does
When lim
—>aa g(x)
x—>a g (x)
xx~'
not exist. What we can conclude is that L’Hospital’s Rule cannot be applied in such a
situation. For example, by L’Hospital’s Rule, the limit
lim
x + sin x
1 + cos x
= lim
x
1
cannot be determined.
However, as sinx is bounded, we have
lim
x+sinx
x
lim (1 +
sinX) _ |
x
Hence, L’Hospital’s Rule cannot be applied in this case.
1
ns oj Differential Calculus 199
B. Indeterminate Forms:
0 • °°
and “oo — oo
By applying the rules on the operations of fractions, these indeterminate forms may be transformed into the
forms
“0”
0
or
“OO”
oo
as shown in the following two examples.
71
Example 5-6
Evaluate lim (x---- )tanx.
2
Solution
2
lim (x-—)tanx = lim
22
cot X
71
X-------
71
a-»—•
2
••() • oo” js, transformed
2
“ 0 ♦’
1
= limn -CSC2 X
into
-
2
= lim (-sin2 x)
2
= -l
Example 5-7
Solution
Evaluate lim x(lnx)n, where n is a positive integer,
.x—»0*
lim x(lnx)n = lim
x—>0*
(Inxf
• oo” is transformed
x->0*
into
x
n(lnx)"'1
= lim--------x—>0*
X
X
___ 1_
By L* Hospital’s Rule
x2
= lim [ziifcT
x—»0*
_1_
X
= lim
x->o*
1
7
If n > I. transform to
— again
Appl> ing IJ Hospital's
Rule n times
= lim (-l)"?i!x
x—>0*
=0
In the above example, it is obvious that as x —» 0+, — tends to infinity is much faster than of (lnx)n.
x
1
200
Chapter 5
Remark 5-5
In the above example, “0 • ©©” may also be transformed to
I
“0”
0
j
x
lim x(ln.v)" = lim
1
x-»0*
x-»0‘
(Inx)"
fl
However, by applying L'Hospital’s Rule, we obtain
1
x
= lim
-n
1
x-»0‘
(lnx)"+l
(Inx)"
lim x(lnx)" = lim
x-»0’
x—»0*
= lim
.1-^0 •
x
x
—n
(lnx)"+l
1
which appears to be more complicated than the original form. Therefore, such a choice
should be abandoned.
Example 5-8
Evaluate lim (secx - tan x).
x—>—
Solution
1 - sin x
lim (sec x- tan x) = limn
cosx
X~*2
x~*l.
**oo - co” is transformed
•• 0 ”
,nl° 0
- COS X
..
= lim-------— stnx
= lim cotx
t->2
=0
Example 5-9
Evaluate lim
Solution
lim
x
x-1
1
In x
X
x-1
1
Inx
lim
xlnx-(x-1)
(x — l)lnx
x(—) + In x -1
= lim
x .--------t->l (x-l)(—) + lnx
x
= lim
Inx
1 , ,
1------- 1- In a*
x
x
= lim
X
-oo - co” is transformed
.. 0 into Q
By L*Hospital's Rule
Transform to —
again
Apply L'Hospital's
Rule again
2
I
Applications of Differential Calculus 201
uq0>»
C. Indeterminate Forms: “1 00
and “oon”
By finding the value of the logarithm of the given limit, these indeterminate forms may be transformed into
the form “0 •
and then into the form
“ 0 ”
77
(J
or
“ oo»
— . Then, by using the fact that logarithmic function is
00
continuous, the given limit can be found by taking exponential as the reverse process. This technique is
illustrated in the following examples.
Example 5-10 Evaluate lim cosx,cscx
x—>0
Solution
The given limit is the type “1
. Let y = cosx cscx. Then
Iny = esc x In cosx
lim Iny = lim cscxlncosx
x-»0
x-»0
cos x
= ..h mIn
-------•'->0 sinx
-sinx
= lim cosx
x—»0
cosx
=0
Since Iny is continuous,
lim Iny = ln(lim y) = 0
x-»0
x->0
lim y = lim cosxC5CX
x->0
x-»0
=1
i
Example 5-11
Evaluate lim (l-cosx)lnx.
Solution
The given limit is the type “0°”. Let y = (1 —cosx)lnx. Then
x->0
1
Iny =
In(l-cosx)
Inx
In(l-cosx)
x->0‘
Inx
sinx
1 - cosx
lim In y = lim
x—>0*
= i*™.
x
xsinx
= lim--------x-»0* 1-cosx
x cos x + sin x
= lim
x-»0*
sinx
= lim ( ——)cosx +1 |
x—>0 V sinx
J
=2
202
Since Iny is continuous.
lim hiy = Inf lim yj = 2
lim y = e2
x—>0’
Exercise 5A
1.
2.
The following limits are of the form
“0”
0
or
— . Use L’Hospital’s Rule to evaluate their limits,
oo
(b)
x + tanx
lim---------x-»o sin2x
.. ex — e~
hm-------sinx
(d)
.. tan x
hm
x-»0
x
e2x -1
(f)
lim
x-sinx
x3
(h)
lim
e~x - ex + 2x
x —sinx
lim
24 - 12x2 + x4 - 24 cos x
sin6 x
(j)
lim
x tan x
Vl-x7 -1
(k)
lim
ln(ex4-x2)
x2
(I)
.. sin2x- x2
hm —
x sin“x
('»)
.. ax -1
nm
x
(n)
..lim----------------l + lnx-xx
x-»i 1 + lnx-x
(a)
lim
(c)
(e)
lim
(g)
lim
(')
x3 - 3x2 - 2x + 6
x2-9
x—»0 ln(l + x)
1-»O
(« > 0)
The following limits are of the form “0 •
limits.
(a)
(c)
(e)
(g)
x->0
x->0
x2+3x-2
3x3 +1
- oo”. Use L’Hospital’s Rule to evaluate their
or
(b)
lim xcsc2x
lim (;r-x)tan^
(d)
lim x3e"x
lim x4 In a*
(f)
lim x
(b)
lim |---- cscx
x—>0 x
lim (
lim (
lim sin x In a
r—>0’
X
2
x-»^
x—>0’
lim x
l + ± — elnf 1 + —
X—
x
k
x
(i)
r
hm /( . 1 2—
-7>
(J)
(k)
limn (tan 5x - tan x)
(I)
sin“ x
x *2
x—>0
1Y
1+- — e
X
1
x—>0
■V
,r-l
-p-)
Inx
1
-1)
tanx X
Applications oj Ifillet- nual (Uh ulus
3.
The following limits are of the form “1
limits.
203
“0°” or “oo°”. Use L’Hospital’s Rule to evaluate their
(a)
2
lim xx
(b)
(c)
lim (tanx)cosx
(d)
—t—
1-
(x
3^2
lim I 1 + — I
x/
lim x’inx
x—>0’
2
(e)
(s)
x
lim
x->0 sinxj
2
(f)
lim x
A—>0*
lim
Wxy
X
(h)
J
2
lim (l + sinx)'
(J)
lim (cos x)lnsinx
.
x—>0
_1_
(i)
lim (1 + sin2 x)r
x—>0
I
jT
2
(
(k)
("i)
lim
• -1
sin
x
x—>0 .
X
lim x1"
sin x Ar
x )
(I)
lim
(n)
lim/r (tan x)™2'
i->0
4
i
4.
V-1V . Distinguish the cases for a > 1 and a < 1.
Let a be a positive real number. Evaluate lim
5.
Let f(x) be a continuously differentiable function such that f(0) = 0 and f '(0) * 0.
k «-l,
Show that lim x f(t) _ j
x—>0*
Verify this result by evaluating lim x,sinx
6.
7.
Evaluate the following limits.
(a)
..
(x2-l)sinx
lim —-------------- —
ln(l +sin
(c)
lim
(b)
.. x + sinx
lim----- ;—
■<-»« x - sin x
2x + sin 2x
(2x + sin x)es,njr
Let f be a real-valued function such that
f(xy) = f(x) + f(y)
(Vx, y e R)
(a)
Show that for x^O, f(x) = -f(—).
x
(b)
Hence show that if f is continuously differentiable at every x
f'(x) = fz(D
x
0, then
(VxeR, x*0)
I
204 Chaptei 5
8.
Let f be a real-valued function satisfying the condition that for any real numbers t and x,
x 4- /.
f(x)-f(/) = (x-/)f'( 2 ?
I
Suppose f has a continuous third derivative.
(a)
Show that for x * t.
4 f"(x)-f"(
)
=
—
----------- ' 2
x-t
r///z X 4~ /
9.
(b)
By finding the value of f z/,(x), show that the second derivative of f is a constant.
(a)
Evaluate lim-—
(b)
Evaluate lim (secx-tanx).
1-Vx
2
(4 marks)
(HKAL 1994 Paper II)
10.
Evaluate
(a)
r ex - 1 -sinx
hm-------- 5-------
(b)
2
3gx +2>
lim
x—>0
5
)
r->0
A"
(6 marks)
(HKAL 1998 Paper 11}
11.
(a)
Evaluate lim A s‘n A
x_>o ] _ cos v
(b)
Using (a) or otherwise, evaluate lim (1 -cosx)lnr.
1
x—»0’
(6 marks)
(HKAL 1999 Paper II)
i.
5.2
Tangents
As shown in Section 4.1, the derivative of a function f(x) at some point (x0, y0) can be visualized as the slope
of the tangent at (x0, y0). Knowing the slope, we may apply the point-slope form to show that the equation of
tangent at (x0, y0) on the graph of f(x) is given by
y-yo = (*-*o)f,(*o)
Applications of Differential Calculus
205
Example 5-12 Find all points on the curve ,Vy2 + xy = 2 where the slope of the tangent is — I.
Solution
The points with slope of tangent equals — I are those points whose derivative equal — I.
By differentiating both sides of the given equation with respect tox, we have
2 d
■>
■> d
?
d
d
x — y"+y“ — x +x—y + y —x = 0
dx
dx
dx
’ dx
2x2y^ + 2xy2+x —+ y = 0
dx
dx
x(2xy +1) -^ = ~y(2xy + 1)
dx
=_Z
dx
x
To find those points whose derivatives equal -1, we set
-^ = -1
x
y=x
Substituting this relation into the given equation, we have
x2x2 + X • X = 2
(x2)2+x2-2 = 0
(x2 + 2)(x2 -1) = 0
x2 = 1
x = ±l
Hence, the points (1.1) and (-1,-1) are with slope of tangent equal to-1.
Example 5-13 Find the point P on the graph of y - x3 such that the tangent at P cuts the x-axis at (4, 0).
Solution
Since P lies on the graph of y = x\ we may let P be (a, a3). Then the slope of the tangent
at P is equal to 3a2. Hence, the equation of tangent at P is
y-fl3 = 3a2(x-a)
3fl2x-y-2fl3 =0
(1)
It is given that the tangent passes through the point (4, 0), we have
3a2 (4)-2 a3 =0
2a2(6-a) = 0
a=0
or
a=6
When a = 0, from (1), the tangent at the point (0, 0) is y = 0, which is the x-axis. This
case should be excluded since the tangent is not cutting the x-axis.
When a = 6, a3 = 216, and so the required point is (6,216).
■
206
Chapter 5
Exercise 5B
1.
7x
Find the slope of tangent at the point (0. 0) to the curve y = —
x + 4x +1
2.
Find the equation of tangent to the curve x + 3y - y2 = 0 at the point (4, -1).
3.
Find the point(s) on the graph at which the tangent to the curve y = x3 - 3x + 1 is horizontal.
4.
Show that the tangents to the curve y = x3 - 9x2 + 30x + 1 cannot be parallel to x-axis.
5.
Find the x-coordinates of all points on the graph of y = x3 + 3x2 - 4x + 5 at which the tangent line is
parallel to the line 4x + y = 1.
6.
Find the poinl(s) on the graph of y
3
x2 -x2 at which the tangent line is perpendicular to the line
x + y = 0.
I
Let A(a, a2 + 1) be a point on the parabola y = x2 + 1. Find the equation of tangent at A to this
parabola.
7.
(b)
Let Z?(£, -b2 - 1) be a point on the parabola y = -x2 - 1. Find the equation of tangent at B to this
parabola.
(c)
Find the equation of straight line which is tangent to both parabolae y = x2 + 1 and y
-x2- 1.
Kate of Change
By the definition of derivative,
= jirn
denotes the rate of change of the dependent variable y with
respect to the independent variable x. Therefore, as functions can be used to represent the relations between
some physical quantities in natural processes or engineering processes, we may use derivative to describe
the rate of change of the processes. For example.
respect to x.
dx
is called the instantaneous rate of change of y with
If the dependent variable is the time r, then — is simply called the rate of change of y.
dr
In this section, we are going to discuss two types of problems on the rate of change of some physical
quantities with respect to the time t as follows:
(1)
Rate of Change on Explicit Functions of t
The displacement of a moving object is a function of t. Then its velocity v and acceleration a are:
ds
dr
dv d2s
a=—=
dr dr2
v=—
Applications oj Differential Cah ulus
(2)
207
Rate of Change on Implicit Functions of t
This type of problems usually involve more than one variables which are implicit functions of t. Hence
we have to differentiate both sides with respect to t. The process is similar to that in the differentiation
of implicit functions.
!
Example 5-14 An object undergoes harmonic vibration along the x-axis. Its displacement from the origin
after t s is denoted by x m, where
x = a sin r + bcosz
(a, b > 0)
(a) Let a = V3, b = 1.
, find the velocity and acceleration of this moving object.
(i) When z =
(ii) Find the maximum displacement of this moving object from the origin. Find also
its maximum velocity.
(iii) If 0 < z < 7T, find the range of z that the object moves towards the origin.
(b) Show that
Solution
(a) (i)
dx
dz
= a2+b2-x22 .
x = V3 sin t + cos t = 2 sin(z +
6
d2x
—r = -2sin(r + -)
dz
o
— = 2cos(z + —),
dz
6
7T
Hence, when Z = —, the velocity and acceleration of this moving object are
6
respectively 1 ms"1 and
(ii) From (i), the greatest values of |x| and
both equal to 2. Hence, the
maximum displacement of this moving object from the origin is 2 m and the
maximum velocity is 2 ms’.
(Hi) When the object moves towards the origin, either the object is on the left hand
side of the origin moves towards the right hand side or the object is on the right
hand side moves towards the left hand side. The first case gives x < 0 and
^>0 while the second case gives x>0 and — <0.
A'
dv
dz
2sin(f + ^)
n
-----------— = tan(z + —) < 0
2cos(r + ^)
6
0<t <7t
7T .
.2 '
3
6
Hence, for — < Z < —, the object moves towards the origin.
3
6
I
■I
208 Chapter 5
(b) Since
x = a sin/ + /?cos/
a2 +b2 sin(r + r0)
dx
dr
Go =tan
fl
a2 +b2 cos(r + r0)
we have
2
2
a2 + b 2
\dr)
chdr
i.e.
2
= a2 + b2-x2
Example 5-15 A spherical capsule is melted in water. Assuming that in the process of melting, the
capsule remains spherical.
(a) It is given that the volume of the capsule is decreased at a rale of 9/rcm3s'1. Find the
rate of change of its radius and surface area when its radius is 4 cm.
(b) If the rate of change of its volume varies directly as its surface area, prove that its
radius is decreased at a constant rate.
Solution
(a) After time r, suppose the radius, the surface area and the volume of the spherical
capsule be r cm, S cm2 and V cm3 respectively. Then
V = — 7Tr3
S
4td‘2
dS
dr
.
3
dV = 4zrr
. 2—
dr ,
—
dr
dr
dr
dr
— = 8tt/‘ —
(1)
From the given condition,
dV = -9/r
n ,
—
dr
r=4
and so,
-9zr = 4/r«42 • —
dr
(2)
dS
dr
(3)
dr
From (2) and (3),
dr
d/
64’
dS
dr
9
2
----- 71
9y
—1
Hence, when its radius is 4 cm, its radius is decreased at a rate of — cm s" while
64
its surface area is decreased at a rate of — cm2s"’.
2
Applications of Differential Calculus 209
(b) Since the rate of change of its volume varies directly as its surface area and the
capsule is melted, we have
dV
L A
— = ~k • Anr
dr
where k is a positive constant.
2
g
From (I),
-k • 4nr2 = 4/rr2 —
dr
^=-k
dr
That is, its radius is decreased at a constant rate.
Example 5-16 AB is a ladder of length 5 m leaning against a
vertical wall. Let OA = a m, OB = y m.
(a) Now, lhe foot of the ladder/1 is drawn away
from the wall at the rate of 4 ms”1. At what
rate is the top of the ladder B moving down
lhe wall when the foot of the ladder is 3 metres
from the wall?
B
5m
ym
0
A
(b) When the ladder is sliding down, find the
position that the speed of A and B are equal.
xm
O
Fig. 5.1
(c) Let Z.BAO = Q. By using the relation y = Alan 0, prove that
d0
dr
Solution
I , dy
d-r
25 dr 7 dr
(a) From Fig. 5.1,
x2 + / = 25
(1)
Performing implicit differentiation with respect to r, we obtain
o dr o dy = 0
dr
* dr
dy _ -a ch*
dr
y dr
(2)
When a = 3, y = 52 -32 =4 and — = 4. Hence, from (2)
dr
d£ = _3.4 = -3
dr
4
The negative sign means that? decreases with r. That is, the ladder is coming down at
a rate of 3 metres per second.
210
(b) When A is drawn away from the wall, x increases. On the other hand, when B moves
downwards, v decreases. Hence.
dz
and
dz
dz
dy _
dz "
x , dy
y( dz
dz
are of opposite sign. That is,
................................................. <3>
From (2) and (3),
x=y
................................................. (4)
From (1) and (4).
!
2
nc
X" + -V = 25,
X=
5
V2
When ,4 is at a distance
m from the wall, the speed of A and B are equal.
(c) From Fig. 5.1,
y = xtan 0
Differentiating both sides with respect to z. we have
d\’
dz
dA\
dz
dz
Since
tan0 = —,
x
sec0 = —
x
Substituting into (5),
/A?
kxj dz
dz
dz x
dy
dz
d0
dz
dx O4-d0
’ dz
dz
1 . dy
d.\\
25 dz J dz
Sometimes, it is more meaningful to compare the rate of change at different situations. This can be done by
finding the relative rate of change. The definition of the relative rate of change is stated as follows.
Definition 5-1
The relative rate of change of a function f(z) with respect to z is defined as
Relative rate of change of f(z) =
(5-4)
Il should be note that the relative rate of change of the function f(z) with respect to z is equal to the
logarithmic derivative of the function f(z).
Applications of Differential Calc it Iir^
211
As shown in Example 5-15,
dS
8^
„
Relative rate of change of the surface area = — = ___ d/ = 2 , dr
S
4zrr2
r d/
dV
Relative rate of change of the volume =
2 dr
a
=
4?rr~ d7 = 3 dr
1^3
3
r’df
Also, if the rate of change of radius of the spherical capsule is kept constant, then the relative rate of
change of surface area of the spherical capsule is inversely proportional to r. However, when the
relative rate of change of the volume is kept constant, the rate of change of radius is proportional to r.
Exercise 5C
1.
A sandbag is dropped from a hot-air balloon that is hovering at a height of 80 m above the ground. If
air resistance is disregarded, then the distance 5 from the ground to the sandbag after time t seconds is
given by
5 = -5/2 + 80
Find the velocity and the acceleration of the sandbag when the sandbag is 35 m above the ground.
2.
A point mass moves along the x-axis and its displacement is given by
x = 2? + pt2 + qt
(0<t< 4)
where /?, q are constants.
When t = 1. and t = 2, the direction of the motion of the point mass reverse.
3.
(a)
Find p and q.
(b)
Find the velocity and acceleration of the point mass at t = 4.
(c)
When will the velocity and acceleration of the point mass be numerically equal?
(d)
When will the velocity be the maximum? When will the acceleration be the minimum?
Boyle’s Law for confined gases states that if the temperature remains constant, then
PV=C
where P is the pressure, V is the volume and C is a constant.
Suppose that at time t (in minutes), the pressure is (20 + 2t) cm of mercury for t > 0. If the volume is
60 cm3 at t = 0, find the rate at which the volume is changing with respect to t at t = 5.
4.
The surface area of a sphere is increasing at a rate of 4 cm“s '. How fast is the volume of the sphere
increasing when the surface area is 36^ cm2.
5.
A street lamp is 6 m directly above a straight path. A man 2 m high walks away from the lamp at a
constant speed of 1 ms-1. At what rate is the length of his shadow increasing?
1
212
Chapter 5
6.
A right conical tank, with vertex at the bottom and axis of symmetry vertical, opens at the lop. The
base radius is 4 m and the height is 8 m and waler is pouring into the tank at the rale of 12 m3s"’.
7.
(a)
Find the rate of the water level rising when the water is 2 m deep.
(b)
If the volume of the water in the lank is equal to one-eighth of the capacity of the tank, find the
rate of the waler level rising.
A straight line with inclination 0 and passing through
the origin meets the parabola y = x2 at the points O and
A as shown in Fig. 5.2. Another straight line through A
and parallel to the y-axis is drawn to intersect the x-axis
at the point B. Let OA = /, the area of AOBA be S.
Given that 0 is increasing at a rate of 0.01 rads"1.
(a)
When 3 = y, find the rate of change of S.
(b)
When I = a/2 . find the rale of change of /.
y
O
B
X
Fig. 5.2
5.4
Monotonic Functions
The study of the monotonic property of a function can help us have a better understanding of the graph of
the function. Also, based on the property of increasing function and decreasing function, we can establish
inequalities which will be discussed later in this chapter.
Theorem 5-2
Let f(x) be continuous on [a. Z?] and differentiable on (a, b).
(1)
If f'(x) > 0. Vx g («, b), f(x) is strictly increasing on [fl, b].
(2)
If f'(x) < 0, Vx g (a, b), f(x) is strictly decreasing on [fl, b}.
Proof
Let a'j and x2 be two real numbers in [a, b] such that x, < x2.
From the given conditions, f(x) is continuous on [xp x2] and differentiable on (xp x2). Then
by Mean Value Theorem, there exists c g (xp x2) such that
fU.) - f(.r2) = f'(c)(X| -a,)
(1) Since x, < x2 and f z(x) > 0, Vx g (a, b), then f'(c) > 0, and so
f(x1)<f(x2)
.......................................... (1)
As X| and x2 are two arbitrary points in [a, bj, f(x) must be strictly increasing on [«, b].
-
Application* of Differential Calculus 213
(2) Similarly, as x, < x2 and f'(x) < 0, Vx e (a, £>), then f'(c) < 0, and so
f(x,) > f(x2)
............................................... (2)
Since x, and x2 are two arbitrary points in [a, b], f(x) must be strictly decreasing on
[a, b].
The converse of this theorem does not hold. For example, the function y = x3 is strictly increasing for all
real values x with f'(0) = 0. However, if the function f'(x) = 0 only occurs at some discrete points in [a, b],
then f(x) is strictly increasing in [a, b]. That is,
Corollary 5-1
Lei f(x) be a continuous function. If f'(x) > 0 (< 0) and f'(x) = 0 only occurs at some discrete points
in [a. £>], then f(x) is strictly increasing (decreasing) in [a, b].
Proof
Suppose there exists c g (a, b) such that f '(c) = 0.
Let X] and x2 be two real numbers in [fl, b] such that x, <x2.
(1) If xp x2 g [a, c] or Xp x2 g [c, /?], then from the result that f'(x) > 0 for all x in (a, c)
and (c, Z>), f(x) is strictly increasing on (fl, c) and (c, b). Hence, by using the fact that
f(x) is continuous on [fl, b], we have
f(x,) < f(x2)
(2) If Xj g [fl, c), x2 g (c, b], then Xj < c < x2, and so from (/),
f (x,) - f(x2) = [f (X,) - f (c)] + [ f(c) - f (X, )] < 0
Example 5-17 Discuss the monotonic properties of the following functions:
(a) f(x) = x3 - 3x2 + 12x - 1
(b) g(x) = ln(l-x)+ x
Solution
(a) f'(x) = 3x2 -6x + 12
= 3(x-1)2+9
>0
(Vx e R)
Hence, f(x) is strictly increasing for all real values x.
(b) The function g(x) = ln( 1 -x) + x is well-defined for x < 1.
g'W = 1 - X
x -1
214
Chapter ?
If g(x) is strictly increasing, then
-^->0
x-1
x <0 or .V > 1
Hence, when x < 0, g(x) is strictly increasing.
If g(x) is strictly decreasing, then
— <0
x-1
Hence, when 0 < x < 1, g(x) is strictly decreasing.
Example 5-18 Let f(x) = 2x4 - 12x2 - 16x - 5. Find the interval for which f(x) is strictly increasing. Find
also the interval for which f(x) is strictly decreasing.
Solution
f'(x) = 8x’-24x-16
= 8(x + l):(x-2)
The sign of f'(x) depends on the sign of x - 2.
If x > 2. f'(x) > 0 and so. f(x) is strictly increasing when x > 2.
If x < 2, f'(x) < 0. However, for all real values x, only f'(-1) = 0 and f'(2) = 0.
Obviously, f(x) is a continuous function, and so f(x) is strictly decreasing when x < 2.
One of the major application of the graph of the function f(x) is the study of the existence of real root of the
equation f(x) = 0. If f(x) is strictly increasing (or decreasing), then the graph of y = f(x) crosses the x-axis
at most at one point. Hence, we have
Corollary 5-2
If the function y = f(x) is strictly increasing or strictly decreasing on an interval [«, Z?], then the
equation f(x) = 0 has at most one real root in the interval [a, b].
2
Example 5-19 Let f(x) = cosx +
-2.
Show that for x > 0, the equation f(x) = 0 has exactly one real root.
Application1* of Differential Calculus
Solution
215
“Existence” part:
Obviously, f(x) is continuous on R.
t
2
Since f(0) = -l<0 and f(?r) =
3 > 0, by Bolzano Theorem, the equation f(x) = 0
has at least one real root on (0, n).
“Uniqueness” part:
When x > 0,
f'(x) = -sinx + x
Now, it is rather difficult to determine the sign of f'(x). In order to solve this problem, we
have to consider (f'(x))'. Then
f"(x) = -cosx + 1 > 0
(VxgR)
As f "(x) > 0 and f "(x) = 0 only occurs at x = 21171. Since f(x) is continuous on R, by
Corollary 5-1, f'(x) is strictly increasing and so,
f'(x) > f'(0) = 0
(Vx>0)
That is, f(x) is a strictly increasing for x > 0.
Hence, the given equation has at most one real root.
From the above proof, for x > 0, the equation f(x) = 0 has exactly one real root.
Exercise 5D
x3
tan x - x - — is strictly increasing for 0 < x < y.
1.
Show that the function f(x)
2.
Show that the function f(x)
3.
Determine the interval for which the function f(x) -
4.
Determine the interval in [0, 2zr] for which the function f(x) — x + 2sinx is decreasing.
5.
Let n be a positive integer and f(x) =
x
x -3 is strictly increasing for x > 1.
2x is increasing.
1 + x2
(x + n + ir*1
(x + n)n
(a)
When x > 0, show that f(x) is strictly increasing.
(b)
When x < 0, discuss the monotonic property of f(x).
[Hint: Consider the cases for n is odd and n is even.]
2
6.
Let f(x) = ex - 2 — x + y-.
Show that when x > 0, the equation f(x) = 0 has exactly one real root.
216
< 'hapu i- ?
5.5
Maxima and Minima
A. The Greatest Value and The Least Value of a Function
There are many practical problems that require the knowledge of maximum and minimum values of a
function. In Section 3.7. we have discussed the terms “absolute maximum’' and “absolute minimum” of a
function. In addition, we are going to discuss the local maximum and local minimum of a function.
For all the points near c, if
f(.v) < f(c)
then f is said to have a local maximum (relative maximum) at c. Sometimes, it is simply called the
maximum.
On the other hand, for all points near cl, if
f(.v)>fW
then f is said to have a local minimum (relative minimum) at cl. Sometimes, it is simply called the
minimum.
The values f(c) and f(<7) are called the local maximum value and the local minimum value respectively.
Both maximum and minimum are called the extremum.
The term local is used to distinguish from the absolute maximum (or minimum). In considering the
local extrema, we only localize our attention to a sufficiently small interval containing c (or cl) such that f
attains its greatest (or least) value at c (or d). Hence, one of the maximum values of f(x) may be less than the
one of its minimum values.
y
y=f(x)
D
F
E
A
C
B
f(x5)
f(x,)
O
a
x,
x2
*x
X3
X4
X5
X6
b
Fig. 5.3
As shown in Fig. 5.3, f(Xj) is one of the maximum values of f(.r) and f(x5) is one of its minimum values.
Also, the greatest and least values of f(x) may not be located at its extrema. This is because they may be
occurred at its end-points. For example, in Fig. 5.3, the greatest value is f(Z?) while the least value is f(«).
From Fermat’s Theorem, if f(c) is the extremum of a differentiable function f(x), then f '(c) = 0. Hence,
we have
I
Applications of Differential Cal< uln\
217
Theorem 5-3
Let fix) be a continuous function. The local extrema of fix) occur when f'(c) = 0 or f'(c) is not
defined.
As in Fig. 5.3, fix) attains its extremum at x,, x2, x4, x5. At these
points, we have f fix,) = f '(x5) = 0 and f'(x2), f'(x4) do not exist.
Hence, Theorem 5-3 is verified.
y
f(x) = X3
However, the converse of this property does not hold. For
example, f'(x3) = 0 and f '(x6) does not exist but they are not
extrema.
If (c, fic)) is a point on the graph of y = fix) such that
f'(c) = 0, then (c, fic)) is called a stationary point. If f'(c) = 0 or
f'(c) does not exist, then (c, fic)) is called a critical point. If
(c, fic)) is a maximum point or a minimum point, then (c, fic)) is
called a turning point or a extreme point.
As staled previously, a turning point must be a critical point
and the converse is not true. For example, consider the function
fix) = x\ Obviously, f'(0) = 0. However, as shown in Fig. 5.4,
fix) is strictly increasing and so, (0, 0) is a critical point but not a
turning point.
► X
O
Fig. 5.4
B. Sign Test for Location of Local Extrema
Theorem 5-4
Sign Test for Location of Local Extrema
Let v = fix) be a function continuous on [fl, b] and differentiable on (a, c)
such that f'(c) = 0 or f '(c) is not defined.
(c, b). Suppose c e (a, b)
(1)
If the sign of f'(x) changes from positive to negative as x increases through c, then fix) attains a
local maximum at x = c.
(2)
If the sign of f'(x) changes from negative to positive as x increases through c, then fix) attains a
local minimum at x = c.
Proof
(1) Let x c) be an arbitrary point in (a, b). If x is on the left hand side of c, then x < c
and f'(x) > 0. Hence, fix) is strictly increasing on the left of c and so, fix) < fic). On
the other hand, if x is on the right hand side of c, then x > c and f'(x) < 0. Hence, fix)
is strictly decreasing on the right of c and so, fix) < fic). As a result, fix) attains a local
maximum at x = c. (See Fig. 5.3.)
(2) The proof for the location of local minimum is similar and is left as an exercise for the
readers.
218
Example 5-20 Find the maximum and minimum points of the function f(x) = 3x4 - 4x3.
Solution
fz(x) = 12x2(x- 1)
Since f(x) is differentiable on R, i.e. there does not exist any point where f '(x) is not
defined, the local extrema occur only at stationary points where f z(x) = 0, i.e. x = 0 and
x = 1. Recall that the condition f'(x) = 0 does not necessarily give turning points. Hence,
we have to apply the sign test to confirm whether they are local extrema or not. Such a
procedure can be conveniently performed in the following table:
X
x<0
0
f'(-v)
f(x)
zzr
0<x < 1
X=0
-
■\
I
X= 1
0
+
min. pt.
indecisive
In the above table, the symbols “ 71 ” and “ \ ” represent that f(x) is strictly
increasing and strictly decreasing respectively.
Hence, x = I corresponds to a minimum point. Also, from the table, we cannot make any
conclusion on the point at x = 0. The curve at both the immediate left and right
neighbourhoods of x = 0 is decreasing and the nature at such points will be discussed later
in this chapter. Therefore, the graph has only a minimum point (1, - I) but no maximum
point.
2
Example 5-21
If f(x) = 3-(x-2)3, find the maximum and minimum points of f(x), if any.
Solution
Differentiate the given function to obtain
-2
f'W= 3(x-2)3
Note that for all real values x, f '(x) 0 but f z(x) is not defined at x = 2. Hence, the local
extremum may occur at x = 2. Now, we are going to construct a table to check for the sign
change of f z(x) in the left and right neighbourhoods of x = 2 as follows:
X
f'(x)
f(x)
A'=2
+
undefined
max. pt.
Hence, the maximum value of f(x) is f(2) = 3.
Therefore, f(x) has only a maximum point A(2, 3) but no minimum point.
Applications of Diffe rential Calculus 219
f
O
y=f(x)
x
2
Fig. 5.5
2
2
Example 5-22 Find the local extrema of the function f(x) = x3(x- I)3.
Solution
Differentiating f(x) with respect tox, we obtain
f'(x)
2
2 3112
(x-l)3 + |x3(x-l) 3
3X
i
^2(x-l)|3
xi
— +---------- T
3x3
3(x-l)3
3x-2
1
2
3x3(x-l)3
/
.
2
Hence, f (x) = 0 if x = y and f'(x) is not defined at x = 0 and x = 1. We then apply the
sign test to these critical points to determine whether they are either a minimum or a
maximum or none. The results are summarized in the following table.
I
X
x<0
x=0
f'(A')
+
undefined
f(x)
z
max. pt.
0<x< —
3
------------ r
2
3
0
min. pt.
2
5<x<l
+
x= 1
I---------------- T
undefined
+
indecisive
Z
1---------- 1
?
V4
So, the maximum and minimum values of f(x) are f(0) = 0 and f(-j) = —— respectively.
9
3
Hence, the graph of f(x) has a maximum point D(0, 0) and a minimum point A(y, - "^~)-
220 Chapter 5
C. Second Derivative Test for Location of Local Extrema
(
Second Derivative Test
Theorem 5-5
Let y = f(x) be a function with continuous first and second derivatives at x = c, and that f'(c) = 0.
(1)
If f "(c) > 0, then f(x) attains a local minimum at x = c.
(2)
If f "(c) < 0, then f(x) attains a local maximum at x = c.
Solution
(1) From the given condition that f"(c) > 0 and f'(c) = 0, we have
f (c) = hm
— ----- /]7----- — = lim
lim--- -—
- >0
h->0
h—>0
/]
h->o
From the properly of limit of functions, for sufficiently small | h |,
.. f'(c + /z)
>0
hm
—-------/i->0
/?
(*)
If h < 0, then c + h < c and from (*), f'(c + It) < 0.
If h > 0. then c + h > c and from (*), f'(c + //) > 0.
That is, the sign of f'(x) changes from negative to positive as x increases through c.
Hence, f(x) attains a local minimum at x = c.
(2) The proof is similar and is left as an exercise for the readers.
Example 5-23 Use the second derivative test to find the local extrema of the function
f(.v)=.? + 3<-24a + 3
Solution
f'(x) = 3x2 + 6x — 24
= 3(x + 4)(x-2)
f"(x) = 6(x + l)
Since
f'(-4) = 0,
f"(-4) = -18 < 0
f'(2) = 0,
f "(2) = 18 >0
Hence, f(x) attains a local maximum at x = -4 and a local minimum at x = 2.
Also, the maximum and minimum values of f(x) are f(-4) = 83 and f(2) = -25 respectively.
Applications of Differential Calculus 221
Example 5-24 Find the local extrema of f(0) = sin30cos20 on [0, —].
2
Solution
f'(0) = 3sin2 0cos3 9 - 2sin4 0cos0
=-2 sin2 0cos3 0(tan2 0--|)
Consider f'(0) = O for 0e(O, y). Then we have
tan2 9 = —,
2
9Q = tan
Also,
___ 3
__ 2n
£
(1)
3
f"(0) = -2 (sin2 0cos3 0)'(tan2 9—)-2sin2 0cos3 0-2tan0sec2 9
From (1) and by using the fact that 0 < 90 < y, we obtain
f,z(0o) = 0 + (negative term) < 0
Hence f(0) attains its maximum at 90 = tan 1
.
Then from Fig. 5.6,
sin0° = J1 ’
cos0o =
Fig. 5.6
and so, the maximum value of f(x) is f(0o) = jyrV15 .
Remark 5-6
Although the second derivative test sometimes gives us a simpler approach for testing for
local extrema when the second derivative is readily computable, it cannot be applied to
turning points whose first derivative does not exist or points whose second derivative is
zero. In such cases, we have to rely on the sign test for the first derivative as discussed
before.
222
Chapter 5
Exercise 5E
i.
Find all the critical points of the following functions. Determine whether they are turning points or not.
(a)
f(x) = 4x2 - 8x + 1
(b)
f(x) = 2? + x2 - 20x + 3
(c)
f(x)=x4-32x
(d)
f(x) = x5-2x3 + x- 12
(e)
f(-r) =
x2-25
(f)
f(x) = Va-2+x-2
(g)
f(x) = (2x-5)Vx2-4
(h)
f(x) = x2\ 2.V-5
(j)
f(.v) = |.r-3| + |2.v-9|
(i)
2.
r, x
2x - 3
Find all the turning points of the following functions and use the sign test to determine whether they
are a local maximum or a local minimum or none.
(a)
f(x) =x - tanx
(b)
f(x) = cos2x + sinx
(c)
f(x) = (x+2)’(3x-I)4
(d)
f(x) = V8-X3
(e)
f(x) = (4-x)x3
(J)
f(x) = (x —2)3Vx--T
(g)
f(x) =
(h)
x 1-sinx
f(x) =--------1 +sinx
1
3.
4.
A
1+A-2
Use the second derivative test to find the local maximum and minimum points of the following
functions, if possible.
(a)
f(x) = 3? - 3? + x + 3
(b)
f(x) = x4 - 8x3 + 22? - 24x + 10
(c)
f(x) = (?-4)2
(d)
f(-v) =
1
x2 + l
Use the second derivative test to locate local maximum and minimum points of the following functions
on the given interval.
(a)
f(x) =x — 2sinx
[0, 2?r]
(b)
f(x) = cosx - sinx
[0, 2/r]
(c)
f(x) = 2 tan x - lan2x
[0, n]
(d)
f(x) = tanx - 2secx
[-^ /r|
(e)
f(0) = 7sin0cos3 0
[0, j]
(f)
f(0) =--------- !----- 5--tan0(1 - tan“ 0)
(0, |)
Applications of Differential Calculus
5.6
223
Optimization
In practical problems, it is quite common that the absolute maximum or absolute minimum values of some
functions are required. These values are sometimes called optimal values.
In Section 5.5, we have discussed how to locate the relative extrema. However, as discussed before if
the function is defined on a closed interval, the absolute extrema may occur at the end-points Therefore ’in
order to find the absolute extrema of a function defined on a closed interval, we have to compare the relative
extrema with the end-point values, too. Such a technique is summarized as follows
Location of Optimal Values
Let f(x) be continuous on [n, b] such that f'(.r) = 0 and f'(x) docs not exist at the points
Then the greatest (least) value among
x2»...» Xn.
f(a), f(x,), f(x2), ..., f(x„), f(b)
is the greatest (least) value of f(x) on [a, b].
Corollary 5-3
Let f(.v) be continuous on [«, b]. If f(.v) has only a maximum (minimum ) point for a <x < b. Then
this maximum (minimum) value is the greatest (least) value of f(x) on [a, b\.
With the help of Fig. 5.7, the readers should have a better understanding of this corollary. Also, this result
holds when the function is continuous on an open interval.
y
y
y=f«
y=fW
0
I
I
I
I
I
I
a
*0
I
I
I
i
i
X
0
b
i
i
b
a
(b)
(a)
Fig. 5.7
i
i
X
224
Chapter 5
Example 5-25 An open box with a rectangular base is to be constructed from a rectangular piece of
cardboard 16 cm wide and 21 cm long by cutting a square from each corner and then
bending up the resulting sides. Find the size of the corner square that will produce a box
having the largest possible volume.
Solution
Let the length of the side of the square to be cut
from each comer be x cm and the volume of the
box be V cm3.
21
x
Refer to Fig. 5.8,
i
i
i
21 -2x
H6-2x
V = x(16-2x)(21-2x)
i
i
i
= 2(168.v-37x2+2x3)
v
The length and the width of the base must
be non-negative
The range of x is
i
i
i
i
i
i
i
i
16
Fig. 5.8
0 <x< 8
Obviously, at the end-points x = 0 and x = 8, the volume cannot be the greatest since
V = 0 at these points. Hence, we find the local maximum value by applying the second
derivative test. Now,
dV
ch'
2(168-74x + 6x2)
= 4(3x - 28)(x - 3)
d2V
^4
= 2(12x-74)
dx
Set
dV
dx
Also
d 2V
dx2 x=3
0, we only have x = 3 (v0<x<8, (3x-28)^0).
-76 < 0, and so at x = 3, V is a maximum.
Then from Corollary 5-3, this maximum value is the greatest value.
Hence, a 3 cm square should be cut from each corner of the cardboard in order to attain the
maximum volume.
Example 5-26 AABC is an isosceles triangle. A circle of radius 1 cm is inscribed in this triangle. Denote
the area of AABC by S cm2.
(a) When 5 is the least, show that &ABC is equilateral.
(b) If the base of AABC is not less than 4 cm, find the least value of S.
Applications of Differential Calculus 225
Solution
(a) Referring to Fig. 5.9, let Z.0CH = 0.
A
/i\
Then Z.AOG = 20, and so
HC =
I
I
1
lan0
i
i
AH = HCtan20 =
2
1 - tan2 9
O'<
1!
er \
2
s=
tan 0(1 - tan2 0)
(O<0<^)
B
H
Fig. 5.9
Consider
(o<0<4)
4
f(0) = tan 0- tan3 0
f'(0) = sec2 0-3tan2 0sec2 0
= 3 sec2 0(1-tan2 0)
f "(0) = 3(sec2 0)'(| - tan2 6) + 3 sec2 0(-2tan 6sec2 0)
When f'(0) = O, 0 = -.
6
f"(ZE)
6
o + (negative term) < 0
So, when 0 = 7-, f(0) is the greatest.
6
That is, when AABC is equilateral, S is the least.
(b) Since BC>4, HC>2 and so,
lan0 = -1- < 1
HC 2
0 < 0 < tan
2
2
_1_
--tan2 0 > -I >o
2
3
3
Hence, on [0, tan
|], m>o.
Therefore, f(0) is strictly increasing.
As a result, when 0 = tan-’ 1, f(0) is the greatest, and so S is the least and
2___
amin
112
G
tir
16
3
If the base of &ABC is not less than 4 cm, the least area is y cm2.
c
226
5
Example 5-27 A woman at a point A on the shore of a circular lake with radius 2 km wants to reach at
the point C diametrically opposite to A on the opposite side of the lake. She can row at the
rate of 2 km per hour and walk at the rate of 4 km per hour. Suppose she first rows
across to point B between A and C and then walks along the shore of the lake toward C.
If the angle BAC is 0, show that total time T (hours) of travel is given by
7=2cos 0+6
(O<0< *)
How would she minimize her travel time?
Solution
Refer to Fig. 5.10, the range of values for 0 is
O<0< —
2
Also,
A
AB = 2(2 cos 0) = 4 cos 0
BC = 2(20) = 40
Hence, the time T is given by
4cos0 40
2
4
= 2cos0 + 0
T =------------- 1------
Fig. 5.10
In order to find the least values of T. we have to find the local extrema and the values of T
al the end-points. Now,
— = -2sin0 + l
d0
When 0 = —, —
6 d0
0.
Also,
d27
-2 cos 0 < 0
d02
Hence, when 0
—, T is a maximum, and this cannot be the required value.
6
Therefore, the least value must be al the end-points.
When 0=0,
T = 2 cos 0 + 0
=2
When 0 = -,
2
T
2 cos— + —
2 2
71
2
If she walks along the lake, the travel time will be the least.
Applications of Differential Cali ulus 227
Exercise 5F
1.
A fanner has 1 000 m of fencing, which he wishes to use to fence a
rectangular field dividing it into three equal parts as shown in Fig.
5.11. Express the area of the field as a function of one of the sides
of the field. Find the dimensions and the area of the field that has
the maximum area.
Fig. 5.11
2.
A circular cylindrical metal container, open at the top, is to have a capacity of 24zrm3. The cost of the
material used for the bottom of the container is $150 per m2, and that of the material used for the
curved part is $50 per m2. If there is no waste of material, find the dimensions that will minimize the
cost of the material. Find also the minimum cost.
3.
Find the maximum volume of a right circular cylinder that can be inscribed in a cone of altitude 12 cm,
and base radius 4 cm, if the axes of the cylinder and cone coincide.
4.
A wire 30 m long is to be cut into two pieces. One of the pieces will be bent into the shape of a circle
and the other into the shape of an equilateral triangle. Where should the wire be cut so that the sum of
the areas of the circle and triangle is minimized? maximized?
5.
Fig. 5.12 shows a circle of radius b and centre O circumscribing an
isosceles triangle PQR with PQ = PR. Show that the area of triangle
PQR is the greatest when triangle PQR is equilateral.
Fig. 5.12
6.
A billboard 5 m tall is located on top of a building, with its lower edge 15 m above the level of a
viewer’s eye. How far from a point directly below the sign should a viewer stand to maximize the
angle between the lines of sight of the top and the bottom of the billboard?
5
228
Chapu
7.
P is a moving point on a quadrant of a circle with
centre at O. The radius of the circle is r. F is a fixed
point on the .v-axis such that OF = (. Let
APOF=0.
(a)
y
Z.PFQ = ty
Show that
tan0 =
rsin#
rcosG- (
a
O
(b)
When C = 2r. find the least value of 0.
(c)
When r = 2C, find the greatest value of 0.
x
Fig. 5.13
Curve Sketching
5.7
One of our major goals is to sketch the curve of a given function y = f(x). To facilitate the sketching of a
curve, besides that the location of local extrema has to be known, we also have to examine
A.
(1)
the property of concavity of the function;
(2)
the property of the function as x -» 00 and y —> «>.
Concavity and Point of Inflexion
Lei f(.r) be a funciion continuous on [a, b], For any xp x2 e (a, b), if
f(x,) + f(x2)
f(*|+*2
)>
2
2
(5-5)
then f(.v) is said to be concave downward (convex upward) on [«. b\.
On the other hand, if
2
)<
f(X|) + f(.v2)
2
(5-6)
then f(x) is said to be concave upward (convex downward) on [a, b].
y
y
y=f(x)
f(
X'+X2}
2
}
■/
f(x,) + f(x2)
2
Ar
i
i
i
i
i
i
i
i
i
i
i
i
i
i
f(x,) + f(x2)
2
x^
2
i
-► x
O
x,
x, + x2
2
1
o
X2
Fig. 5.14
x
x, + x2
2
x2
Applications t>f Differential Cult ulus 229
As shown in Fig. 5.14, M is a point representing the image of
V T V
' -2". For any convex function, M
always lies above the chord AB. and for any concave function, M always lies below AB.
Also, with the help of this diagram, we can see that the values of the slopes of tangents are increasing
for a concave upward graph while that in a concave downward graph is decreasing. Since the slope of
tangent is given by f z(x), f'(x) is increasing in a concave upward graph while that in a concave downward
graph is decreasing, and so the following result can be obtained.
Theorem 5-6
Let y = f(x) be a function with continuous first and second derivatives on (a, b).
(1)
If f "(x) < 0 for all x g (a, b), then the curve is concave downward on (a, b).
(2)
If f "(x) > 0 for all x g (a, b), then the curve is concave upward on (a, b).
Proof
(1) Let ,rpx2 g(a, b) with Xj < x2. Denote
*l+*2
*0
2
/z =
*2-X|
2
Xy-h
Ci
X, + x2
2
Xy
h
Then
x, = x0 - /z,
_1_
*2
h
J
X22 = x0 + Il
Fig. 5.15
Consider
l[f(x,)+f(x
[f(x,) + f(x22)]-f(^-)
)]-f(
= y [f(x0 + /l) - f(x0 ) + f(xo - II) - f(x0 )]
= y[f/(c,)/i-f'(c1)/i]
= |[f'(c2)-f'(c1)]/i
By Mean Value
Theorem
x, < ( | < x„ < c\
By Mean Value
Theorem
c, < c* <
= ^f"(c^c2-cl)h
fV)<0
c, - r, > 0. h > 0
<0
f(xt) + f(x2)
2
That is, f(x) is concave downward.
(2) The proof is similar and is left as an exercise for the readers.
230
5
Example 5-28 If f(x) = x3 + 3x2 - 4x - 2, determine intervals on which the graph of f is concave upward
or is concave downward.
Solution
In order to investigate the concavity of the graph, we find the second derivative f "(x) first.
Now,
f '(x) = 3x2 + 6x - 4
f "(x) = 6x + 6 = 6(x + 1)
Hence, f"(x) < 0 for x < -1 and f"(x) >0 for x > -1.
Therefore, the graph is concave downward on (-«>, -1) and concave upward on (-1, <*>).
Example 5-29 If f(x) = sinx, determine intervals on which the graph of f is concave upward or is
concave downward.
Solution
fz(x) = cosx
f"(x) = -sinx
When 2kn < x < (2k + 1 )zr. f"(x) < 0, and so the graph is concave downward.
When (2k + I )tt < x < (2k + 2)tt, f "(x) > 0. and so the graph is concave upward.
y
y = sin x
concave
downward
-2k
concave
downward
-K
concave
upward
o
t
\
► X
K
concave
upward
2k
Fig. 5.16
The study of concavity of a function provides the second derivative test to determine the local extrema of the
function. As shown in Fig. 5.17, if f'(c) = 0 and f"(c) < 0. then the graph of f(x) lies below the tangent line
at x = c, and so f(c) must be the local maximum value of f(x).
Also, in Fig. 5.17, the graph of f(x) is convex on the left hand side of the point / and is concave on its
right hand. Such point is called a point of inflexion (point of inflection). A point (c, f(c)) on the graph of f is
a point of inflexion (point of inflection) if the graph on one side of this point is concave downward and
concave upward on the other side. That is, the graph changes concavity at x = c.
Also, from Fig. 5.17, it is quite obvious that the graph of a concave f(x) lies above the tangent at any
point while that of a convex function f(x) lies below the tangent at any point. Hence, the tangent at the point
of inflexion will cut across the curve.
Applications of Differential Calculus 231
y
B
y = fW
i
i
i
i
i
i
i
i
i
i
O
i
►X
C
Fig. 5.17
Theorem 5-7
Let (c, f(c)) be a point of inflexion of the function y = f(x). Then f "(c) = 0 or f "(c) does not exist.
The readers are again warned that f "(c) = 0 does not necessarily imply that c is a point of inflexion. For
example, the function y = x4 has second derivative equals to zero at x = 0 but (0, 0) is not an inflexion
point. Indeed, (0, 0) is a local minimum point since the first derivative changes sign at x = 0. Each of these
points have to be tested to determine whether it is a point of inflexion.
The method of testing is similar to that of the location of local extrema and is illustrated in the
following examples.
Example 5-30 Find the points of inflexion of f(x) = x4 - 6x2 + 1.
Solution
f'(x) = 4x3-12x
f"(x) = 12(x-l)(x +1)
When f"(x) = 0, we have x = -l or 1.
x = —1
X
+
f(x)
-1 <x< 1
0
--------- 1—————
concave up
,
pt. of infl.
concave down
Hence, (-1,-4) and (1, -4) are points of inflexion.
x= 1
0
+
pt. of infl.
concave up
232
Chapter 5
I
Example 5-31
Let f(x) = 2 - x3. Find the points of inflexion of f(x).
Solution
rw=-|3
C"t
X
2
f (*) = —r
9x3
Note that for all x e R. f "(x)
0. Also, f"(0) is not defined.
The following table illustrates the test for concavity of the curve in the neighbourhoods of
x = 0.
x
x<0
f"(x)
concave down
f(x)
x=0
x> 0
undefined
+
pt. of infl.
concave up
Hence, (0, 2) is a point of inflexion.
2
Example 5-32 Let f(x) = x3 (5 - x). Find the points of inflexion of f(x).
Solution
f(x) = 5x3 -x3
2
5 3
10 3J —
f (x)x = yX
x
3
f''W = (y)(-|)x’’ + (-|)(|)x'5 _~10(x + l)
9x7
Note that f "(—1) = 0 and f"(0) does not exist. The test for concavity is shown in the
following table.
X
x<-l
f"W
+
f(x)
concave up
x = —I
I
-1 < x < 0
0
pt. of infl.
x=0
undefined
concave down
Hence, there is only a point of inflexion (-1, 6).
indecisive
| concave down
Applications of Differential Calculus 233
Exercise 5G
i.
Find the intervals on which the graph of f(x) is concave upward or is concave downward. Find also the
points of inflexion.
(a)
f(x) = 3x3 - 9x2 + 4x + 1
(b)
(c)
f(x) = 3x5 - 5x3
(d) f(x) = (?-l)2
(e)
f(x) = V7
(f)
f(x) = 10-Vx?
(h)
f(x) = x2(3x-5)3
f(x) = 3/ - 32? + 1
1
(g)
(i)
1
1
f(x) = 8x3 + 4x3
(j)
1 2
f(x) = 6x2 + x2
(k)
f(x) = x2 9-x2
(I)
f« = xvM-x2
B. Asymptotes
An asymptote of the curve y = f(x) is a straight line such that the distance between the curve and the straight
line tends to zero when the points on the curve approach infinity. There are three types of asymptotes,
namely, vertical asymptotes, horizontal asymptotes and oblique (or slant) asymptotes. Each type of these
asymptotes will be discussed in detail in the following.
Vertical Asymptotes
Lei y
Proof
f(x) be a function. If lim f(x) = oo, then x = c is a vertical asymptote.
x—*c
Since lim f(x) = «>, when the points on the graph of f(x) approach infinity, the distance
between these points and the line x = c is |x-c|->0.
Hence, x = c is a vertical asymptote.
234 Chapin 5
Fig. 5.18 to Fig. 5.21 show the four typical behaviours of the functions near the vertical asymptote.
(1)
lim f(.v) = +oo
(2)
lim f(x) = +oo
y
y
I
I
I
I
I
I
I
I
I
I
i
i
i
i
i
i
i
i
i
i
-► X
T
O
1 C
I
I
O
I
Fig. 5.19
Fig. 5.18
(3)
(4)
lim f(x) = -oo
lim f(x) = - oo
x—»c4
y
y
y=f(x) i
O
i
4----I C
>x
O
I
I
I
I
I
I
I
I
I
Fig. 5.20
Note that lim
y=f(x)
—► X
Fig. 5.21
Example 5-33 Find the vertical asymptote of f(x) =
Solution
I
I
X
I
I
I
I
I
I
I
I
I
I
1
x->2 X-2
1
x-2
= oo. Therefore, x = 2 is a vertical asymptote.
It is also important to examine the behaviours of f(x) when x approaches 2 either on the
left side or right side of 2.
Let x - 2 = <5. As x —> 2", <5 —» 0"; as x -4 2+, 8 —> 0+. Observe that
1
lim —-— = lim
x->2- x-2
lim-------
x-2
<5-40- 8
= lim -^ = 4-oo
o-^o’ 8
i
!
I
Applications of Differential Calculus 235
The behaviour of the curve f(x) = —— around x = 2 is shown in Fig. 5.22. It should be
noted that we need more information about the curve to complete the sketch.
y
'|y=
—
'|
x-2
ii
i\
i \
1
I
I
x
x
" - .
X
[2
P
\
i
\ i
\ i
\i
li
i
Fig. 5.22
In order to have complete information to sketch the curve, we also have to find local extremum, points of
inflexion, concavity property and symmetry of the curve, if some or all these properties exist. In latter part of
this section, we will discuss the systematic procedures to explore the necessary information for the sketching
of a given function. In this and the following examples, we take the liberty to show the sketches of the
curves, the purpose of which is to illustrate the role of asymptotes in curve sketching. The readers are
reminded that the mere knowledge of the asymptotes does not provide enough information for the sketching
of the curve.
Example 5-34 Find the vertical asymptote of f(x) =
Solution
1
(x-l)2(x+3)’
Note that the denominator is zero when x = 1 or x = -3. Consider
lim f(x)= lim f(x) =
lim f(x)
lim f(x) = +°°
Hence, x = 1 and x = -3 are two vertical asymptotes of the curve.
Fig. 5.23 shows the behaviours of the curve f(x) =
asymptotes.
1
around the vertical
(x-l)2(x+3)
236
Chaf t
y
ii
f(x) =
ii
i\
1
(x-1)2(x + 3)
•\
i
i
i
i
I
I
I
-► X
O
73
I
I
I
I
II
I
I
I
I
I
I
II
Fig. 5.23
Remark 5-7
It is easily seen that for a rational function f(x) =
g(-v)
, f(x) is undefined at the zeros of
g(x). Therefore, x = c is a vertical asymptote if and only if g(c) = 0 and h(c)
0.
When the distance between the straight line y = nix + b and the curve y = f(x) lends to zero as x tends to
infinity, the straight line y = mx + b is said to be an oblique (or slant) asymptote. In particular, when m = 0,
the asymptote becomes a horizontal straight line and is called a horizontal asymptote. Hence, we have
Horizontal and Oblique Asymptotes
y
nix + b is an asymptote of f(x) if and only if
lim [f(x) - (mx + Z?)] = 0
y
Proof
Pz
y = f(x)
a
/'0
/*
/?
f //
A
O
X
Fig. 5.24
(5-7)
Applications of Differential Calculus 237
In Fig. 5.24, the distance between the point P on the graph of f(x) and the asymptote is
equal to PH. Suppose 0 is the inclination of the asymptote. Then
PQ2 = PH2 sec2 9 = PH2(\ + tan2 0) = PW2(1 +m2)
Hence, PQ —> 0 if and only if PH —> 0.
On the other hand, PQ = | f(x)-(?HX + b)|.
Since PH —» 0 as x —»<» is the condition for y = nix + b to be an asymptote of f(x), the
necessary and sufficient condition for y = nix + b to be an asymptote of f(x) is
lim [f (x) - (nix + />)] = 0
Now, we would like to present the method for finding the constants tn and b of the oblique asymptote
y = nix + b, if such oblique asymptote exists.
Corollary 5-4
If y = nix + b is an asymptote of f(x), then
.. f(x)
tn = lim ——,
r->oo
Proof
_v
b = lim [f(x)-/nx]
(5-8)
X—>«■»
From (5-7)
lim [f(x) - (mx + /?)] = 0
lim
(*)
f(x) -tn----b = lim [f(x)-(7/Lt+ /?)]• lim — = 0
x->« L
X
x
x
Since b is a constant, lim — = 0 and so,
x
. f(x)
tn = lim----X
From (*), if tn exists, we have
b = lim [f(x)-;nx]
Remark 5-8
However, if any one of the two limits in (5-8) does not exist, then there is no oblique
asymptote.
238
X3
Example 5-35 Find all the asymptotes of y =
Solution
x2 + 1 '
Since x2 +1*0. the graph has no vertical asymptote. Now.
X3
m = lim A'~ + 1
lim
X
x2
A'2 + 1
= lim
1
1
1
x3
b = lim —--------- X
X‘ + 1
= lim —4— = lim---- Xq1— = 0
,r_>„ x +1
1 + 4x2
Hence, y = x is an oblique asymptote.
Also, it is important to know whether the curve lies above or below the asymptote when
x —> +oo and x —> -<». This can be achieved by considering the sign of
x3
-X
x2 +1
f(x)-(/nx + b)
-A'
x2+l
(1)
As x —> -oo, x < 0. Then from (1), f(x) - (ntx + b) > 0 and so the graph of f(x) lies above
the asymptote. Similarly, as x -> +oo, a > 0 and so f(x) - (mx + b) < 0. Hence, the curve
is below the asymptote. (See Fig. 5.25.)
y
X
O
Fig. 5.25
In the above example, by long division, we have
x3
X
= -X----- 5-----x2 +1
x2 + 1
Hence,
lim [f(x)-x] = lim
-x
= 0+
X2 + 1
-x
lim [f(x)-x]= lim - 2 A = 0~
X—x-X42 + 1
!■
Applications of Differential Calculus 239
From (5-7), it is obvious that y = x is an asymptote. Also, from 0* and 0", it is easy to determine whether
the curve lies above or below the asymptote.
With similar argument, we obtain a more convenient way to find the oblique asymptotes of a rational
functions as stated in the following:
Remark 5-9
Let f(x) = P(-V) be a rational function.
QU)
If deg Q(x) > deg P(x) - 1, then f(x) can be expressed as
f(x) = >nx + £? + -^^
S(x)
where deg S(x) > deg R(x). It is obvious that
lim
S(x)
nix + b is an oblique asymptote of f(x).
Hence, y
Example 5-36 Find all the asymptotes of the graph of
(a) y =
(b) y
Solution
(a)
X
x-l
x2+l
x-2
x
lim
....i------- = —°°
x->r x— 1
y
i
i
i
i
i
i
i
i
ix
lim
-----= +<*>
x-r x-l
Hence, x = 1 is a vertical asymptote.
X
^(x-D + l
x-l
x-l
= 1+—!—
x-l
lim (
lim (
X
x-l
X
x-l
x
0
-1)= lim ——— = 0"
x-»— x -1
-1)= lim —^—= 0+
*-♦+- x -1
Hence, y = 1 is a oblique asymptote. (See Fig. 5.26).
i
y=1
Fig. 5.26
240
Chapter 5
(b)
y
..
x“ + 1
lim---------
x —2
.. x2 + l
lim------x —2
x+ 2
i
i
i
Hence, x = 2 is a vertical asymptote.
O
x2 +1
,_
5
------- = X + 2 +----- —
x-2
x-2
lim
X —>-«•
lim
X—>+<»
Hence, y
Remark 5-10
x
i
i
i
X2 -F 1
i-
x2 +1
-(a + 2)
x-2
lim —— = 0f
x-2
5
A-
'/ y
li
x-2
-(a- + 2) = lim ------ = 0
x-2
x-2
Fig. 5.27
x + 2 is a oblique asymptote. (See Fig. 5.27).
Apparently, the oblique asymptotes in the above examples are found by only finding the
oblique asymptote of the graph of a function for the case x —> +<*>. However, readers are
cautioned that we also have to consider the limit x —> -<» in order to find all asymptotes.
For example, the exponential function y = ex lends to infinity as x —> +~ while it tends
to 0 as x —> -oo. As a result, y = 0 is an asymptote to the graph of y = ex. Therefore, if we
want to find all oblique asymptotes of the graph, we have to consider both cases x —>
and x —> -co.
Example 5-37 Let y = tan 1 x (- — < y < —). Find all the asymptotes of the curve.
2
2
Solution
Since tan x is defined for all x, there is no vertical asymptote.
Furthermore,
y
lim
tan 1 x
x
0
71
2
lim (tan 1 x-0)-»-y
O
lim (tan 1 x - 0) —>-7-X—/
Hence, when x —>
y
71
•
---- is a
2
horizontal asymptote.
Fig. 5.28
Similarly, when x —> +°°, y = ~- is a
horizontal asymptote.
X
Applications of Differential Calculus 241
Exercise 5H
i.
Find all the asymptotes of the graphs of the following functions.
1
3x
(a) f(x) =
(b) f(.r) =
x2-9
4-x2
(c)
f(.v) =
(e)
f(-v) =
(g)
f(-r) =
(i)
f(-v) =
(k)
(tn)
fz A
2x2
x+ 1
x3 + x2 - 6x
(d)
fW =
3x- 1
x+1
(f)
f(x) =
x3-x2 + l
x2 — 3x-4
(h)
f(x) = x + —!—
x+1
U)
f(-r)
1 +x2
x-3
x+1
y 9x~ + 4
=- -—
(I)
f(x) = x + tan x
(n)
x —2
x2+l
x-1
x-100
/x2 + 100
f(x) = xtan’ x
C. Curve Sketching
Many applications of calculus are based on the knowledge of the behaviour of the functional values f(x) as x
varies throughout a set of real numbers. Such information can be provided through the sketching of the
graph of y = f(x). The procedures for sketching a graph are summarized in the following.
Procedures for Sketching a Graph
(1)
Find the domain of f, i.e. all real numbers x such that f(x) is defined. Also, locate the points of
discontinuity of f, if any.
(2)
Investigate some special properties of f, if any, like periodic property, symmetric property.
(3)
Find the x-intercept and y-intercept of f.
(4)
Locate the turning points and points of inflexion of f. Also, examine the concavity of f.
(5)
Find the asymptotes of f, if any.
(6)
Sketch the graph with the above information.
242
Chapter 5
x3
. Sketch the graph of y = f(x).
x2 - 4
Example 5-38 Let f(x)
Solution
(I) Domain off: R\{±2).
Also, f(x) has infinite discontinuities al x = ±2 and is continuous at all other real
numbers.
(2) Since f(-x) = -f(x), the given function is an odd function. Therefore, the graph is
symmetric with respect to the origin.
Also, it is not periodic.
(3) Set x = 0, we have y = 0.
Set y = 0, we have x = 0.
Therefore, the graph intersects both the x-axis and lhe y-axis at the origin <2(0, 0).
(x2-4)(3x2)-x3(2x)
(x2-4)2
,
(4)
x2(x2-12)
(x2-4)2
and
(x2 -4)2(4x3 -24x)-x2(x2 -12)[2(x2
f"U)
4)(2x)]
(x2-4)4
8x(x2 + 12)
(x2-4)3
In order to locate lhe turning points, we have to consider the values of x for which
f'(x) = 0, and al which f'(x) is undefined. These critical points are x = 0, ±V12 and
x = ±2. The sign test of f'(x) at these points are summarized in the following table.
12
x
x = -V12
12 <x<-2
x = -2
f'W
+
0
undefined
f(x)
z
max. pt.
undefined
x
x=0
f'W
o
undefined
f(x)
indecisive
undefined
x
2 < x < V12
I
x = V12
12
f'W
0
+
fw
min. pt.
Z
x=2
Applications of Differential Calculus 243
Hence, the graph has a maximum point at A(-s[\2, -373) and a minimum point at
B(V12, 3V3).
Similarly, to locate the points of inflexion, we have to test for the points when
f (x) = 0 and at which f "(x) is undefined, i.e. x = 0 and x = ±2. The results are
summarized in the following table.
X
X = -2
f"(A)
undefined
f(A)
x= 0
+
0
concave down
undefined
—
concave up
0<x<2
x=2
x>2
undefined
+
undefined
concave up
F"(a)
f(x)
-2 < x < 0
concave down
pt. of infl.
Hence, 0(0, 0) is a point of inflexion of the graph.
(5) Since
i•
A3
i-
a3
i-
A3
lim —----- : = —oo
x->2- X - 4
,.
X3
lim -j—x->-2* x“ - 4
lim —----x-»-2- X2 -4
x = -2 and x = 2 are vertical asymptotes.
Furthermore,
X3
x2-4
4x
x2-4
lim (-^—r — x) = lim —4x = 0"
x->— x 2-4
a-*-~ x -4
X3
lim (-t—- —x) = limt
x->+~ x — 4
/.
4x
= 0+
x" -4
2
y = x is an oblique asymptote.
(6) Also, at the point of inflexion 0(0, 0), f'(.r) = 0 and so, the curve touches the jc-axis.
From the above information, the sketch of the graph of y = f(x) is shown in the
following:
244
Chapter 5
y
x=-2
x=2
ii
i
i|
x3
il
i|
it
;
11
i \
i \
i '
A
I
J___________
______ i
-21
1 z
O
V
z' I
-3^3)
B(Vl2,3^3)
I
►x
‘2
i
i
i
i i
it
|i
|i
i
i
i
Fig. 5.29
Example 5-39 Let f(x) =
Solution
1__
. Sketch the graph of y = f(x).
X2 +3
(J) Since x2 + 3 * 0 for all real values of x, the domain consists of all real numbers.
Also, there is no point of discontinuity.
(2) Since f(-x) = f(x), the given function is an even function. Therefore, the graph is
symmetric with respect to y-axis.
Also, it is not periodic.
(3) Since f(x)
0 for all real values of x, there is no x-intercept. When x = 0. y
— and
3
so, the graph passes through A(0, -j).
f'(-v)
(4)
-2x
(x2+3)2
and
(x2 +3)2(-2)-(-2x)[2(x2 +3)(2x)]
f"(x)
(x2+3)4
6(x2—1)
(x2+3)3
f'(x) exists for all values of x. In order to locate the turning points, we have to
consider the values for f'(x) = 0. This gives x = 0. The sign test of the first derivative
at this point is summarized in the following table:
Applications of Differential Calculus 245
x
x<0
x=0
f'(-v)
0
f(x)
max. pt.
0 <x
Hence, the graph has a maximum point at .4(0, |).
Similarly, to locate the points of inflexion, we have to test for the points when
1 „(x) = 0, i.e. x = ±1. The sign test of second derivative at these points are summarized
in the following table:
x = -l
X
f"(x)
+
0
f(x)
concave up
pt. of infl.
-1 <x < 1
x= 1
0
concave down
pt. of infl.
x
f"W
f(A)
concave up
Hence, B(-l, —) and 0(1 , -j) are points of inflexion of the graph.
4
(5) Obviously, the graph has no vertical asymptote.
Furthermore, as
lim
1
x2+3
lim -U-— *0+
r_>+« x +3
y = 0 is an oblique asymptote.
(6) From the above information, the sketch of the graph of y = f(x) is shown in the
following.
y
AO. 4)
8H. 1)
x
O
Fig. 5.30
246
Cluipu r 5
Example 5-40 Let f(x) = x3 - 5x3. Sketch the graph of y = f(x).
Solution
(I) The domain consists of all real numbers.
Also, there is no point of discontinuity.
(2) There is no special property for f.
(3) When x 0, y = 0; when y = 0, x = 0, 5. Hence, the graph passes through (9(0, 0)
and A(5, 0).
f'(x) = |x3
(4)
-5(|)3
= -^-(x-2)
(1)
3x3
and
-
i
?
--
„
5 2
3-5(|)(-|)x 3
f"U)
= (j)(y)x
= _12_(x + l)
(2)
9x3
In order to locate the turning points, we have to find the values for f'(x) = 0, and at
which f'(x) is undefined, i.e. x = 0 and x = 2. The sign test of the first derivative at
these points are summarized in the following table.
x
x<0
x=0
x=2
f'(x)
+
undefined
0
f(x)
Z
max. pt.
min. pl.
x
f'(x)
+
f(x)
2
Hence, the graph has a maximum point <9(0, 0) and a minimum point B(2, -3(2)3).
Similarly, to locate the points of inflexion, we have to test for the points when f "(x) = 0
and at which f"(x) is undefined, i.e. x = -1 and x = 0. The results are summarized in
the following table.
X
x<—1
f"(x)
fW
concave down
x = —1
-1 <x<0
x=0
0
+
undefined
pt. of infl.
concave up
indecisive
Applications of Differential Calculus 247
X
f"(A)
+
f(x)
concave up
Hence, C(-l, -6) is a point of inflexion of the graph.
(5) Since
1
2
lim (x3 -5x3)
2
5
lim (x3 -5x3) = +o°
X->+«o
the graph of f has no asymptote.
(6) From the above information, the sketch of the graph of y = f(x) is shown in the
following:
y
y = *3 -5X3
A/
x
O
C(-1,-6)
B(Z -3(2)h
Fig. 5.31
361 * I Sketch the graph of y = f(x).
Example 5-41
Let f(x) =
Solution
(1) Domain of f: R\{1}.
(X-1)2 '
Also, f(x) has a infinite discontinuity at x = 1 and is continuous at all other real
values of x.
(2) There is no special property for f.
(3) Set x = 0, we have y = 0.
Set y = 0, we have x = 0.
Therefore, the graph intersects both the x-axis and the y-axis at the origin 0(0, 0).
248 Chaplet 5
(4) f(x)
36|x|
"(x-1)2
36x
x > 0, x
(x-1)2
-36x
1
x<0
(x-1)2
Consider
36(x —I)2-36x[2(x-1)]
(x-1)4
_d_______
36x
dxljx-l)2
36[(x-l)-2x]
(x-1)3
-36(1 + x)
(x-1)3
f_'(0)
lim
h
A->0-
(ft-I)2
ft
-36
lim
= lim---------- l—O- (/l-l)
= -36
and
f+'(0) = lim
MT
f(0 + /?) - f (0)
h
36/z
-0
(/*-1)2
ft
= lim
h->0'
= lim
h-*0’
36
(ft-I)2
= 36
* f_'(0)
f'(x)
-36(1+x)
(x-1)3
x > 0, x # 1
36(1 + x)
. (x-1)3
x<0
Applications of Differential Calculus 249
Also,
d2
36x
dx2 (x-1)2
d | —36(14- x)
dx (x-1)3 J
(x -1)3 - (1 + x)[3(x - >/]
= -36 ------------------- £-----(x-1)6
(x-l)-3(l + x)
= -36
(x-1)4
-2x-4
= -36 •
(x-1)4
72(x + 2)
(x-1)4
and so.
72(x + 2)
(x-1)4
f"(x) =
-72(x + 2)
(x-1)4
x > 0, x * 1
x<0
In order to locate the turning points, we have to find the values for f'(x) = 0, and at
which f '(x) is undefined, i.e. x = -1 and x = 0, 1. The sign test of the first derivative
al these points are summarized in the following table.
x = —1
X
-1 <x < 0
x= 0
f'(x)
+
0
undefined
f(x)
Z
max. pt.
min. pt.
x
0 <x < 1
x=l
f'(x)
+
undefined
f(x)
x> 1
undefined
Hence, the graph has a maximum point A(-l, 9) and a minimum point 0(0, 0).
Similarly, to locate the points of inflexion, we have to test for the points when f "(x) = 0
and at which f "(x) is undefined, i.e. x = -2 and x = 0, 1. The results are summarized
in the following table.
250
-2 < a < 0
a = -2
A
0
+
concave down
pt. of infl.
A= 1
0 <A< 1
A
undefined
pl. of infl.
concave up
____________ I
f(.v)
A=0
undefined
f"(A)
—
concave up
f(A)
concave up
undefined
Hence, B(-2, 8) and (0, 0) are points of inflexion of the graph.
(5) Since
36|x|
lim
!—V
-r (a-1)-
(«-!>•
/.
a= 1
is a vertical asymptote.
Furthermore,
36|x|
.. . 36a
lim ----------------- T = 0*
lim ---- !—r = ini)
(a -1)'
~~ (.r-1)-
36|x|
lim ---- !t = lim
(A-l)-
a=0
-36a
(x-1)
2
—
0+
is an oblique asymptote.
(6) From the above information, the sketch of the graph of y
following:
f(A) is shown in the
y
i|
i
i
i
i
i
i
i
i
4(-1,9)
B(-2. 8)
O
1
1
36|x|
y ■(x-1)2
x
Fig. 5.32
Remark 5-11
In Example 5-41, (?(0, 0) is a point of inflexion as well as a minimum point. That is, a
turning point may also be a point of inflexion.
Applications oj Differential Calculus 251
Example 5-42 Sketch the graph of y2 = x3
2-x
Solution
The given function may be considered as two separate functions :
x3
y = 2-x’
y
I -----x3 first.
r
Now, we sketch the graph of y
V2-x
x3
> 0, we have 0 <x<2.
2-x
Within this domain, there is no point of discontinuity.
(1) Domain of f: For
(2) There is no special property for f.
(3) When x = 0, y = 0. Hence, the graph passes through 0(0, 0).
f'(x) = 2
(4)
(2-x)(3x2)-x3(-1)
(2-x)2
1
\2-x
- V^(3--v)
V(2-*)3
and
3__ 3 x - 77(3-x)(| 7(2-x)J(-l)
7(2-X)3
2\l~x 2
f"W =--------- (2-x)3
_3[(2-x)(l-x)+x(3-x)]
2 7^(2-x)5
3
7x(2-x)5
For 0 < x < 2, f '(x) > 0. As f(x) is continuous for 0 < x < 2, the graph is strictly
increasing and so, there is no turning point. Also, f (x) > 0 for all 0 < x < 2 and so
there is no point of inflexion. Moreover, f(x) is concave upward for 0 < x < 2.
(5) Since
lim
x->2‘
x = 2 is a vertical asymptote.
252
Chapter 5
(6) From the above information, the sketch of the graph of y =
x3
is shown in the
2-x
following:
y
i;x=2
y=
I
I
I
I
I
I
I
I
I
I
x3
2-x
O
’ X
Fig. 5.33
The graph of y = -
X3
2-x
can be obtained by reflecting the graph of y =
the x-axis. The complete graph of y2 =
x3
2-x
about
is shown in Fig. 5.34.
y
i;x=2
y
y2 = —
2-x
O
I
I
I
I
I
I
I
I
I
I
’
x
I
I
I
I
I
I
I
I
I
II
Fig. 5.34
Remark 5-12
After sketching the graph, we can examine the property of the function much more easily.
For example, the function f(x) = x4 + 4x + 6 has only a minimum point (-1, 3) and no
maximum point. Hence, the graph of y = f(x) lies above the x-axis, and so the equation
f(x) = 0 has no real root.
Applications of Differential Calculus 253
Exercise 51
1.
Sketch the graph of the following functions.
(Cl)
f« =
x2
x3 +9x
x2 +1
(c)
fW =
(e)
1
Z
f(x) = x3(x + 3)3
2.
Sketch the graph of y2
3.
Let f(x) =
(A-+1)3
(■v-1)2 ’
(d)
r/
\
x2■ + 2
f(x) =
V2x2 + 1
(f)
f(x) = xe’2
x+1
Find f'(x) and f"(x) for x
(b)
Determine the values of x such that
1.
f'(x) = 0
(ii) f'(x)>0
(Hi) f'(x)<0
f "(x) > 0
(vi) f"(x)<0
(v)
(c)
Find the local extrema and the points of inflexion of f(x).
(d)
Find the asymptotes of the graph of f(x).
(e)
Sketch the graph of f(x).
Hence, sketch the graph of y = | f(x) |.
Let f(x) =
—^- + 5
x+2
where x g R and x * 1.
(iv) f "(x) = 0
4.
f(x) = —!—
4x3.
(a)
(i)
(b)
1 000 - x 3
- , wherexeR and x0.
x
(a)
Find the range of values of x where f(x) is defined.
(b)
Find f'(x) and f "(x) for x * 0.
(c)
Show that f(x) is strictly decreasing.
(d)
Find the point of inflexion of f(x).
(e)
Find the asymptotes of the graph of f(x).
(f)
Sketch the graph of f(x).
2
(a > 0)
x2(x + 4)3
5.
Consider the function f(x) = *
2
-x2(x + 4)3
(x<0)
(a)
Find f '(x) and f "(x) for x * 0, -4.
(b)
Discuss the existence of f '(x) and f "(x) for x = 0, -4.
(c)
Determine the values of x at which the graph of f has an extreme point or a point of inflexion.
(d)
Sketch the graph of f.
254
5.8
Proving inequalities by Using Differential!
Calculus
In practical problems, we always encounter inequalities within a certain range such as:
f(x) > g(x)
(a < x < b)
F(x) = f(x) - g(x) > 0
(a<x<b)
(i)
Usually, it is transformed to
Based on the properties of increasing function and decreasing function, we can establish inequalities
and the method is outlined in the following.
Making Use of Strictly Increasing or Decreasing Functions
Want to prove that
(1)
(2)
(3)
f(x) > g(x)
Consider
F(x) = f(x) - g(x)
Try to prove that
F'(x)>0
Hence, we have
F(x) is strictly increasing on (a, b)
Try to prove that
F(x) is continuous on [«, /?]
Then we have
F(x) = f(x) - g(x) > F(«)
Try to prove that
F(fl) > 0
Then we can conclude that
f(x) > g(x)
(a<x<b)
(a<x<b)
Making Use of The Greatest and Least Values of a Function
Want to prove that
f(x) > g(x)
(1)
Consider
F(x) = f(x) - g(x)
Try to prove that
F(c) is the least value in (a, b)
Try to prove that
F(x) is continuous on [a, b]
Then we have
F(x) = f(x) - g(x) > F(c)
Try to prove that
F(c) > 0
Then we can conclude that
f(x) > g(x)
(2)
(3)
(x * c)
(a < x < /?, x * c)
*
Applications of Differential Calculus 255
Example 5-43 Show that sinx<x for all x>0.
Solution
Let f(x) = x - sinx. Then
f'(x) = 1 - cosx > 0
Further, f z(x) = 0 only occurs at some discrete points x = 2nn, where n is an integer.
Since f(x) is continuous for x > 0. f(x) is a strictly increasing function for x > 0.
Hence,
f(x) = x - sinx > f(0) = 0
(x>0)
sinx <x
(x>0)
(x+H+ir1
for x > 0.
U + n)"
Show that f(x) is a strictly increasing function.
Example 5-44 Let n be a positive integer. Define f(x) =
Hence show that
n>l
< i+->l n+l
Solution
For x > 0,
(x + n)" — (x + n + l)"u -(x + n+1)i"*1—(x + n)"
dx
dx
f'(A)
(x-H^
(x + n)" [(n + l)(x + n +1)" ] - (x+n +1)"*1 [n(x + n)"~' ]
(x + n)2"
"
(x + h)i"~' (x + n +1)" [(n + l)(x + n) - n(x + n +1)]
(x + n)2"
_ (x+h+i/'x
(x + n)nH
>0
Since f(x) is continuous for x > 0, we can conclude that f(x) is strictly increasing for x > 0,
and so
f(0) < f(D
That is,
(H+iy— < (1+n + ir*
Hn
(n + l)1
nn
(!+«)"
(1 + n + ir1
(l + n)"+l
n+l
1+1
11
1+7n
256 Chapter 5
Example 5-45 Let k be an integer greater than 1. Show that
<A* > 0)
x* + A - 1 > Ax
When does the equality hold?
Solution
Consider
f (a) = a* - kx + k - 1
f'(A) = jl(.v‘-|-l)
For 0 < x < 1, f '(x) < 0, and so f(x) is strictly decreasing for 0 < x < 1.
For x > 1, f'(x) > 0, and so f(x) is strictly increasing for x > 1.
Obviously, f(x) is continuous for x > 0.
Hence, f(l) = 0 is the least value.
As a result.
f(x)=x*-Ax + k- 1 > f(l) = 0
(a>0)
x*+ k - 1 > kx
Equality holds only when f(x) attains its least value, i.e. at x = 1.
Example 5-46 (a) Show that lnx<x-l (x>0).
When does the equality hold?
(b) Letap a2,.
an be n positive real numbers. Define
2
( "
c-= ip.
"
i
\ '=•
1=1
n
/
Using (a) to show that
G„ < A„
When does the equality hold?
Solution
(a) Let
f(x) = lnx-x + 1
r'/ x
1
1
(x>0)
1 “A
f (a ) = - - 1 =----X
X
For 0 < x < I, f '(x) > 0, and so f(x) is strictly increasing for 0 < x < 1.
For x > 1, f'(x) < 0, and so f(x) is strictly decreasing for x > 1.
Obviously, f(x) is continuous for x > 0.
Hence, f(x) attains its greatest value at x = 1, i.e. for all x > 0,
f(x)<f(l)
As a result,
f(x) = lnx-x + 1 < f(l) = o
lnx<x- 1
(a>0)
Equality holds only when f(x) attains its greatest value, i.e. at x = 1.
Applications of Differential Calculus 257
(From Fig. 5.35, the graphs of y = x - 1 and y = Inx touch at the point (1,0). Hence,
we have In x < x - 1.)
y
y = x- 1
y = In x
x
1
O
Fig. 5.35
a
(b) By putting x = y-, which is a positive real number, into the result in (a), we have
ln(-J-)<-^-l
A„
n
(z= 1,2,
n
Then by taking summation from i = 1 to n, we obtain
Zln(
1=1
A.
i=l
A
-*
\ «=1
/
4^ -z>
»=1
J
1=1
n
In
ln
nA
It is well known that In.r <0 for 0 < x < 1. Hence, we have
^<1
A"
^■<1
A
G„<A.
From the result in (a), equality holds only when x - 1, i.e.
and so,
A
A.
A
Cl | — ^2 “ ” ’
That is. the geometric mean of n positive real numbers is equal to its arithmetic mean
if and only if the n numbers are all equal.
-
258
Chapter 5
Example 5-47 Let n be a positive integer. Define
f( 0) = (1 + cos 0)n + (1 — cos 0)" (0eR)
(a) Show that for 0 g [-y, y],
f(0)<2n
(b) Explain why
f(0)<2"
for all 0 g R.
Solution
(a) f'(0) = //(! + cos 0)n"1 (- sin 0) + //(!- cos 0)""1 (sin 0)
= n sin0[(l - cos0)n"1 -(1 + COS0)”-1]
z
r-1
71 71
For 0 g---- , —),
2 2
(1 - cos 0)n_| - (1 + cos 0)'
<0
and so f'(0) = O if and only if sin 0= 0. This occurs only when 0=0.
For
2
<0<O, f'(0) > 0, and so f(0) is strictly increasing for
2
<0<O.
71
For 0<6><—, f'(0) < 0, and so f(0) is strictly decreasing for 0 < 0 < —.
2
2
Obviously, f(0) is continuous for 0 G[-y, y].
Hence, f(0) attains its greatest value at 0 = 0, i.e. for all 0G[-y,y],
f(0) < f(0)
As a result,
f(0) < (1 + cos 0)" + (1 - cos0)"
2"
(b) Since
f(7T + 0) = [1 + cos(/r + 0)]" + [1 - cos(?r + 0)]"
= (1 - cos 0)" + (1 + cos 0)”
= f(0)
f is a periodic function with period equal to ti.
TC 71
Hence, for all 0 g R, there exists 0G[-y,y]
and A'gZ such that
0 = 0 + A'7T
and so,
f(0) = f(0 + kn)
= f(0)
<2"
T
Applications of Differential Cali ulus 259
Exercise 5J
1.
Prove that --^2 <tan~‘x<x forx>0.
2.
Prove that 0<ln(l+x)-
3.
f
c/ x Inx
Let f(x) =----- .
x
2x
V3
< — for x > 0.
(a)
Show that fix) is strictly increasing on the interval (0, e).
(b)
Hence, show that if 0 < a < b < e.
ab <ba
4.
Let fix) = ln( 1 + x) - x.
By finding the greatest value of fix), prove that
ln(l +x)<x
5.
(X>-1)
Let f(x) = — for x > 0.
e
Find the greatest value of fix).
Hence show that
6.
Let t be a real number such that 0 < t < 1 and
f(x) = tx-x‘ +(!-/)
Cr>0)
By finding the least value of fix), show that
rx + (l-r)>x'
(.v>0)
Hence deduce that
(Va./?>0)
7.
Let / be a real number greater than 1. Define
f(x) = (l + x)'-l-rx
(x>-l)
By finding the least value of fix), show that
(1+x)' >l + fx
(x>-l)
and that the equality holds if and only if x = 0.
8.
(a)
Define fix) = x Inx - x + 1 for x > 0. Find the least value of fix).
(h)
Hence, or otherwise, show that when x > 0, the function g(x) -
.r
function.
l+l I is a strictly increasing
X
260
Chapter 5
9.
(a)
(b)
Show that for 0 <x < 1,
(i)
ln(l+x)<x
(ii)
-ln(l-x)>x
Let n be a positive integer greater than 1. Deduce from (a) that
ln(/i +1) - In n < — < In n - ln(/z - 1)
n
Hence show that
. , 3/z + l
ln(------- )<l + ^_ + ... + J_<ln(-^-.)
11
n n +1
3/i
n—1
(c)
Use the above results to evaluate the limit
limfl+_l_+...+±
n—>«>
10.
11
Let f be a real-valued function with f(x)
3/1.
11 4-1
A
1+X
for x > 0.
(a)
Show that f is strictly increasing for x > 0.
(b)
Let xp x2,x„ be n non-negative real numbers. Show that
n
X\X2"‘Xn
1=1
\+X}X2-‘Xn
1=1
11.
Let x and a be two real numbers such that 0 < a < 1 and 0 < x < 1.
(a)
Show that
(1 -x)a< 1 -ar
(b)
By using the result in (a), show that for any positive integers n and k,
1
/z + 1
n
n
n
kn+l -(k-l)n+}
7
Hence show that
’ + * + > +...+
Vl V2 V3
12.
(a)
For x > 0, prove that Inx < x
(b)
Prove that for r > 1,
1
Vi ooo
<150
1 where equality holds if and only if x = 1.
In — <-?r-1 r-1
II— I
Hence deduce that for n = 2, 3, 4,In n <
I
i=i b
.
(7 marks)
(1995 Paper 1)
Applications of Differential Calculus 261
13. (a)
(b)
x
is an increasing function on (-1, «>).
1+x
Using (a), or otherwise, show that for any real numbers r and 5,
Show that f(x) =
lr+d < Id , Id
I + |r + .y|
l + |r|
l+|s|
(5 marks)
(HKAL 1996 Paper I)
14. (a)
Show that for 0 < x < 1,
1 + ex > e x > 1 + x
(b)
Using mathematical induction, or otherwise, prove that for n = 0, 1,2,... and x g (0, 1 ],
^n+1
n
y
S r!
n+1
r
+—ex
— > ex > y —
(n + 1)!
" r!
Hence show that
1
1 )
1 i m | 1 + 14- — 4- • •• 4—2
nlj
.’
=*
(7 marks)
(HKAL 1996 Paper 11)
15.
By considering the function f(x) = xe~'\ or otherwise, show that if 1 <a <b, then
aeb > bea
(4 marks)
(HKAL 1997 Paper 11)
16. Suppose 0 < p < 1.
(a)
Let f(x) = xp - px + p - 1 for x > 0. Find the absolute maximum value of f(x).
(b)
Show that for all positive numbers a and /?,
apbl~p <pa + (1 -p)b
(6 marks)
(HKAL 1998 Paper I)
Revision Exercise
1.
2.
Let f(x) be a real-valued function such that f'(x) = f(x) and f(x) > 0 for all x.
(a)
Find the maximum and minimum points of f(x“), if any.
(b)
Is there any maximum and minimum points for [f(x)] ?
Let f(x) =
x3
x2 - r
where x e R and x
± 1.
(a)
Find the turning points of f(x).
(b)
Find the points of inflexion of f(x).
(c)
Find the asymptotes of the graph f(x).
(d)
Sketch the graph of f(x).
262
Chapter 5
3.
Let f(x)
(■v + 3)2
, where x g R and x
x+1
(a)
Find f'(x) and f "(x) for x * -1.
(b)
Determine the values of x such that
(i)
-1.
(ii)
f'(x) > 0
(iv) f "(x) < 0
(Hi) f"(x)>0
4.
(c)
Find the local extrema and the points of inflexion of f(x).
(d)
Find the asymptotes of the graph of f(x).
(e)
Sketch the graph of f(x).
Let f(x)
6.
4
— x — 1, where x g R and x
2 —x
(a)
Find f '(x) for x
(b)
Determine the values of x such that
(i)
5.
f'(x)<0
2.
2.
f'(x) > 0
(c)
Find the local extrema of f(x), if any.
(d)
Find the points of inflexion of f(x), if any.
(e)
Find the asymptotes of the graph of f(x).
(f)
Sketch the graph of f(x).
(ii)
f'(x)<0
b-x
The function f(x) = -^
—
. _ where a, b are two real constants and a > 0, has local extrema at x = 1
x..2 +a
and x = -1.
(a)
Determine the values of a and b.
(b)
Find the local maximum and minimum points and the points of inflexion of f(x).
(c)
Find the asymptotes of the graph of f(x).
(d)
Sketch the graph of f(x).
Let f(x) = Vx2(x + 6) .
(a)
Find f'(x) and f"(x) for x # 0, -6.
(b)
Determine the values of x such that
(i)
f'U) > 0
(Hi) f "(x) > 0
(H)
f'W<0
(iv) f "(x) < 0
(c)
Find the local extrema and the points of inflexion of f(x).
(d)
Find the asymptotes of the graph of f(x).
(e)
Sketch the graph of f(x).
Applications of Differential Calculus 263
7.
1
Let f(x) = (x-l)x3 for x g R.
(a)
Find f'(x) and f "(x) for x * 0.
(b)
Show that f'(0) does not exist.
(c)
Determine the range of values of x such that
f'(x) > 0
(ii) f'(x)<0
(iii) f"(x)>0
(iv) f"(x)<0
(i)
8.
(d)
Find the local extrema and the points of inflexion of the function.
(e)
Show that the graph of the function has no asymptotes.
(f)
Sketch the graph of the function.
Let f(x) = x2(2 - x)e x, where x g R.
(a)
Find f'(x) and f"(x).
(b)
Determine the values of x such that
(i)
f'(x) = 0
(iv) f "(x) = 0
9.
10.
(ii) f'(x)>0
(iii) f'(x)<0
f"(x)>0
(vi) f"(x) < 0
(v)
(c)
(cl)
Find the local extrema and the points of inflexion of f(x).
(e)
Sketch the graph of f(x).
Find the asymptotes of the graph of f(x).
x3
, where x g R.
Vx-1
Let f(x) =
(a)
(b)
Find the range of values of x so that f is well defined.
Find the turning points of f(x).
(c)
Find the asymptotes of the graph f(x).
(cl)
Sketch the graph of f(x).
Let f(x) =
x|x|(x-7)
, where x e R and x -1.
(a)
Find f'(x) and f "(x) for x
(b)
Discuss the existence of f'(x) and f/z(x) for x = -1,0.
(i) Find the relative maximum and minimum points and the points of inflexion.
(c)
(ii)
(cl)
-1, 0.
Find the asymptotes of the graph of f(x).
Sketch the graph of f(x).
264
Chapter 5
11.
Let f(x)
2x(x+l)
l-2.v
Define h(x) = •
(x^y) and g(x) = sin
f(-v)
(x > 0 and x
,g(-v)
(-1 <X<0)
2x
1+x2
(-1 <x < 1).
y)
Find h'(x) and h"(x), 0 for x 0.
Do h z(x) and h "(x) exist at x = 0?
(b)
Determine the values of x such that
(i)
h'(x)>0
(Hi) h"(x)>0
12.
(ii)
h'(x)<0
(iv) h"(x)<0
(c)
Find the local extrema and the points of inflexion of h(x).
(d)
Find the asymptotes of the graph of h(x).
(e)
Sketch the graph of h(x).
Let f(x) = x3 + 3<zx2 + 3Z?x + c, where a, b and c are two real constants.
(a)
(i)
Show that the graph of f(x) has a minimum point and a maximum point both lying on the
same side of y-axis if a2 > b > 0.
(ii)
Show that the graph of f(x) has a minimum point and a maximum point lying on the
opposite sides of y-axis if b < 0.
(Hi) Under what condition has the graph of f(x) a minimum point or a maximum point lying on
y-axis?
13.
(b)
Use (a), or otherwise, show that if b > 0, c < 0 and a > -jb , then the equation f(x) = 0 has a
unique positive root.
(c)
Deduce that the equation x4 + 4x3 + 2x2 - 8x + d = 0 has a unique positive root if d < 0.
(a)
Let f(z) = t3 - 12zn2Z + 16n, where tn and n are two real constants.
(i)
(ii)
Find the maximum and minimum points of the graph of f(x).
Hence show that the equation f(z) = 0 has three distinct real roots if | m |3 > | n |, one real
root if | m |3 < | n |.
14.
(b)
Use the result in (a), or otherwise, show that on the graph of y3 - 3y = x3 - 12x, there are two
points at which the tangent is parallel to x-axis, and six points at which the tangent is parallel to
y-axis.
(a)
Let f(x) = 2x3 + 9x2+ 12x + 4, where x eR.
(i)
Find the points of intersection of the graph f(x) with the x-axis.
(ii)
Find the turning points of f(x).
(Hi) Sketch the graph of f(x).
Applications oj Differential Calculus 265
(b)
Let g(x) = f[f(x)].
(i)
Show that the real roots of the equation gz(x) = 0 are —1, —2, — — and some real number cc.
5.
where -3 < a < —
2
(ii)
By using the graph of f(x), or otherwise, find the number of distinct real roots of the
equation g(x) = 0.
I
15. A right circular cone is inscribed in a sphere of radius a as shown in Fig. 5.36. Determine the base
radius of the cone to have the maximum volume.
Fig. 5.36
16.
A wall h metres high stands d metres away from a tall building. A ladder of length a metres reaches the
building from the ground outside the wall. Let 0 be the angle between the ladder and the building.
(a)
It is given that the ladder reaches the top of the wall. Show that
a = d esc 0 + h sec 0
(b)
Find the shortest ladder that will reach the wall if h = 3 and d = 3.
17. ABCD is a trapezium inscribed in a circle of radius a. AB is a diameter of the circle, and that AD
subtends an angle 3 at the centre of the circle as shown in Fig. 5.37.
A
0
B
Fig. 5.37
(a)
Express the perimeter p of the trapezium and its area 5 in terms of a and 3.
(b)
Find the greatest value of S.
(c)
Find the maximum value of p.
266
18.
A circle with radius a is inscribed in an isosceles triangle. Suppose A and a are respectively the area
and the semi-vertex angle of the isosceles triangle.
(a)
Show that
_ 2<72(1 + sinoQ2
sin 2a
Hence find the least value of A if a can vary' over all relevant values.
(b)
19.
If a is restricted to lie in the range sin
— < a < sin’1 —, find the least value of A.
5
3
A container in the form of a right circular cone is obtained by removing a sector from a circular disc of
radius a and joining the straight edges of the remaining portion. (See Fig. 5.38.) Find the maximum
capacity of the container.
Fig. 5.38
20.
(a)
Let a, b, c be three positive real constants such that ar < b < c. Suppose
f(x) = | x - a | +1 x - b 14-1.x-c|
Sketch the graph of f(x).
21.
(b)
Three men live on a straight road. They agree to meet at a point on the road so that the sum of the
distances they travel from their homes to the meeting place is least. Locate this point.
(a)
Let f(x) = (x - a)2 + b2 and g(x) = (x - c)2 + d2, where a, b, c, d are real constants and a > c.
Suppose
max(p, <?)
(if P <7)
(if p< q)
P
<7
h(x) = max[f(.r), g(x)]
f(x)
gW
(Vx e R)
(if x<x0)
(i)
Find the value of x0 such that h(x) =
(ii)
If Xj is the absolute minimum of h(x), find the value x,.
(if x>x0)
(Distinguish the cases: c < a < x0; c < x0 < a\ x0 < c < a.)
(b)
Let P, Q be two points separated by a straight line L on a plane. Two persons are at P and Q who
want to meet each other at some point on L and agree that whoever arrives there first should wait
for the other. If they start at the same time and walk at the same speed, find the point R on L so
that they can meet within the shortest time.
Applications oj Differential Calculus 267
22. Let f(x) be a differentiable function defined for x > 1 such that
r(x)=-r- 1
,
Vl + 4x2
Suppose g(n) is a function defined for u > 0 such that
f(D = 0
1
f[g(l<)] = H
M
Show that
g'(«) = -Jl + 4[g(n)]2
and
g"(«) = 4g(«)
(b)
Explain why g(z/) > 1 for all u > 0.
(c)
Let h(w) = e"-g(w) for all u > 0.
(i)
Show that h,z(«) < h'(w).
(ii)
Hence show that h(z/) cannot have a minimum for it > 0.
j
23. Let a be a positive real number greater than 1 and p, q, r and 5 be four real numbers such that
p + q= r+s
and
ft <p - q < r- s
(a)
Show that the function f(x) = ax - a~x is strictly increasing forx > 0.
(b)
By considering the values of f at -(p-q) and y(r-s), show that
ap1' - aq <ar-as
Hence deduce that if it and v are two distinct positive real numbers such that u > v, then
■
uqvp < urvs -usvr
24. A continuous real-valued function f is said to be concave downward in an open interval (a, Z?) if for any
x, y in (a, b) and for any p, q in [0, 1 ] with p + q= 1,
f(px + qy) > p f(x) + q f(y)
(a)
Show that, for any t in the interval (x, y), there exist p, q in (0, 1) with p + q = 1 such that
t = px + qy.
i
Draw a diagram to illustrate the inequality in the definition of a concave downward function.
(b)
I
It is given that f is concave downward. Show that, for any x, /, y with x < t < y,
f(r)-f(x) > f(y)-f(O
r-x
y-t
(c)
I
Let g be a function with a second derivative g„(x) 0 on (a, b). For any x, y in («, b) with x < y,
consider the function
h(r) = p g(r) + q g(y) - g(pr + qy) (x < t < y)
where p, q are fixed real numbers in [0, 1] with/) + </=!.
Show that h is monotonic increasing and hence show that g is a concave downward function.
(d)
Let a,, x2 > 0, 0<r<l andp,, p2 in [0, 1] such thatp,+p, = 1. Show that
(PiX, + p2x2)r > ptx' + p,x2
*
268
Chapter 5
25.
Let f(x) be a function with f"(x) < 0 forx > 0.
(a)
(b)
Let a > 0 and g(x) = f(x) - f(fl) - f'(n)(x - a).
(i)
Show that g(x) is strictly increasing for x < a.
(ii)
Hence show that g(x)<0 for x<a.
Use the result in (a) to show that for any integer n > 2,
f'(2) + f'(3) + -•• + f'(zz) < f(//) - f(l)
(c)
Use the above results, or otherwise, show that for any integer n > 2,
±
+±+-+±<2-l
1" 2“
n
n
26.
Let p be a real number not less than 2.
(a)
Suppose x e [0, 1]. By considering the range of values of
r i-x p-i < 1 —x
I l+x
2x
l+x
and
l+x
2x
1 —x
and
, show that
l+x
1+X
p-i
l+x
Hence show that
(1 +x)',+ (1 -*)'’< 2'~l(l +xp)
£
(b)
(c)
£
Let f(0) = (l + r2 +2/-COS0)2 + (l + r~ -2rcos0)2, where r> 0.
(i)
Show that f is a periodic function. Find the period.
(ii)
Show that f(0) < f(0) for all real values of 6.
Let Zj and z2 be two complex numbers. By using the results in (a) and (b), show that
r+iz2 r)
izi+ziMzi-zz
27.
Let
n2,an be n positive real numbers and
i
C„ =(a,a2---aj"
(a)
For x > 0, define
i
f(x) = fl, + a2 + -+an + x-(n + l)(a1a2 •••artx)n+1
Show that for any x > 0,
> f(G„)
(b)
Prove by induction on n that
G <A„
When does the equality hold?
Applications of Differential Calculus 269
28. (a)
(b)
By finding the greatest value of the function f(x) = ex - e\ or otherwise, show that for all real
numbers x,
ex<ex
(VxgR)
Let av a2
an and Z?p b2,..., bn be positive real numbers. Show that
e
iH'
i=l
ui
Hence, or otherwise, show that if Y — < n, then
i=i
bj
fl
1=1
(c)
Using the result in (b), show that for any positive numbers a i, a2,
n«.
\ «=1
29. (a)
b.
1=1
i
,l 1=1
/
Let f(x) = Inx - x + 1 for all x > 0. Find the maximum value of f(x).
Hence show that
(x>0)
Inx <x- 1
When does the equality hold?
(b)
Let ap a2, .... an and xp x2,..., xn be positive numbers such that
n, + a2 +
+ an =
+ a2x2 + ••• + arfxn = 1
Show that
Vx2
‘•<1
When does the equality hold?
(c)
Show that for any positive numbers hp b2,..., bn and yp y2,..., y„,
bp\+b2y2 + - + bnyn
b\ +b2 + -‘ + bn
30. Let f(x)
r2
r3
x2
r3
x4
'-v+ir-¥ and »w=1-"+¥-¥+¥-
(a)
Show that f(x) is strictly decreasing.
Hence deduce that the equation f(x) = 0 has only one real root.
(b)
Let a be the real root of the equation f(x) = 0.
Show that g(x) attains its absolute minimum at x = a.
Hence deduce that the equation g(x) = 0 has no real root.
I
270
Chapter 5
31.
The function f(x) is a differentiable function with
f(x) = f '(x)
(a)
and
f(0) = 1
Let a be a fixed real number. Show that for all x.
-kf(.r)f(a-A-)] = 0
dx
(b)
Deduce that
(Vx, jgR)
f(x)f(y) = f(x + y)
32.
(c)
Hence show that f(x) > 0 for all values of x and that f(x) is a strictly increasing function.
(d)
By considering the derivative of the function (?”xf(x), show that f(x) = ex.
(a)
Show that for x > 0.
x~
x — — < ln(I + x) < x
(b)
Deduce that for any positive integer n,
y-' k
1 V
✓ V i <i
L
—"T2 L
a=i n
*=I “
n r<Z,n(1 + —
n
\
i=1 —
n
Hence evaluate
lim(l+-V)(l+^r)--(l+A)
n~ n-
fj~
[Hint: You may use the result I2 +22 + ••• + /i2
33.
(a)
+ l)(2/? + l).]
Let n be a positive integer.
«
By considering
j
2sin — xcosAx, prove that
i=i
2
n
k
. 1
sin(n + —)x —sin—x
cosAx=----------- —j------—
n
For any positive integer n, define f„(x) =
(0<X<7T)
2 sin— x
2
*=i
(b)
o
A=.
sin Ax on [0,7r].
*
(i)
Prove that if f,'(cz) = 0, then by using the result in (a), show that sin/ia> 0.
(it)
It is given that for any positive integer n, fn(x) has an absolute minimum.
(1) Suppose that fm+1(x) attains its absolute minimum at some a g (0. n).
Show that fm+1(a) < 0. Hence deduce that f/n(cx) < 0.
(2) Show that if fm(x) > 0 for all x G (0, tt), then fm+1(x) attains its absolute minimum only
at x = 0 and x = n.
(Hi) Prove by induction that for any positive integer /i, flux) > 0 for all x g (0, rt).
I
♦
Applications of Differential Calculus 271
34.
Let f(x) = a x2-x 3
(a)
Find f'(x) and f"(x).
(b)
Show that f'(0) and f'(1) do not exist.
(c)
Determine the sets of values of x such that:
(2 marks)
(2 marks)
(i)
f'(x) = 0
(iv) f"(x) = 0
(ii) f'(x)>0
(Hi) f'(x)<0
f"(x)>0
(vi) f "(x) < 0
(v)
(3 marks)
(d)
Find the relative extremum point(s) and the poinl(s) of inflexion on the curve y = f(x).
(e)
Find the asymptote(s) of the curve y = f(x).
(f)
Sketch the curve y = f(x).
(3 marks)
(3 marks)
(2 marks)
(HKAL 1993 Paper II)
35.
Let a, b e R and a < b. Let f(x) be a differentiable function on R such that f(a) < 0, f(b) > 0 and f'(x)
is strictly decreasing.
(a)
Show that f '(a) > 0.
(b)
f(-Vp)
Let x0 = a and a*j = x0 f'Uo)'
(2 marks)
Show that a < x, < b, f(x,) < 0 and f '(x.) > 0.
(6 marks)
(c)
Let x0 = a and xn
Ha;,-.)
(h=1,2, 3,...).
Show that a < xn < b, f(x„) < 0 and f '(x„) > 0 (n
1,2, 3,...).
(4 marks)
(d)
Show that lim xn exists and is a zero of f(x).
n —>~
(3 marks)
(HKAL 1993 Paper II)
3’
36.
Let f(x) =
(a)
(i)
\Jx ->
x2 + r
X G R.
Evaluate f'(x) for x * 0.
Prove that f '(0) does not exist.
(ii)
Determine those values of x for which f'(x) > 0 and those values of x for which f'(x)<0.
(Hi) Find the relative extreme points of f(x).
(8 marks)
(b)
(i)
Evaluate f "(x) for x * 0. Hence determine the points of inflexion of f(x).
(ii)
Find the asymptote of the graph of f(x).
(4 marks)
(c)
Using the above results, sketch the graph of f(x).
(3 marks)
(HKAL 1994 Paper II)
5
272
Chapti
3,7*
Let a and b be positive numbers.
(a)
Prove that
a“bb > abb“
where, if the equality holds, then a = b.
(4 marks)
(b)
Using (a), or otherwise, prove that
fa+b I
a-rb
> abb“
I 2
where, if the equality holds, then a = b.
(3 marks)
(c)
Show that for 0 < .v < 1,
x'(l-x)
1
where, if the equality holds, then x = —.
2
Deduce that
a“bb>
4
a+b
2
a-rb
where, if the equality holds, then a = b.
(8 marks)
(HKAL 1995 Paper I)
38.
Let f(x) =
(a)
14 , where x — 1.
(A + l)2
(i)
Find f'(x) and f "(x) for x > 0.
(ii)
Find f'(x) and f ,z(x) for x < 0.
(Hi) Show that f'(0) does not exist.
(4 marks)
(b)
Determine the values of x for each of the following cases:
(i)
f z(x) > 0
(iii) f "(x) >0
(ii)
f'(x)<0
(iv) f "(x) < 0
(4 marks)
(c)
Find the relative extreme point(s) and point(s) of inflexion of f(x).
(d)
Find the asymptote(s) and sketch the graph of f(x).
(3 marks)
(4 marks)
(HKAL 1995 Paper II)
39.
Let f(x) =
(X-1)3
(x + 1)2 '
(a)
Find f '(x) and f "(x) for x # -1.
(b)
Determine the values of x for each of the following cases:
(2 marks)
(i)
f'(x)>0
(Hi) f"(x) > 0
(ii)
f z(x) < 0
(iv) f"(x) < 0
(3 marks)
(c)
Find the relative extreme point(s) and point(s) of inflexion of f(x).
(2 marks)
r
Applications of Differential Calculus 273
(d)
Find the asymptote(s) of f(x).
(e)
Sketch the graph of f(x).
(f)
Let g(x) = |f(x)|. Does gz(l) exist?
(2 marks)
(2 marks)
Find the asymptote(s) and sketch the graph of g(x).
(4 marks)
(HKAL 1996 Paper II)
40. Let f(x) =
(a)
(i)
J
X+ 1
Find f'(x) for x * -1,0.
Does f'(0) exist? Explain.
(ii)
Show that f"(x) =
2(2x2-8x-1)
4
for x -1, 0.
9.r’(x + l)3
(b)
(4 marks)
Determine the values of x for each of the following cases:
(i)
f'(x)>0
(ii) f'(x)<0
(Hi) f "(x) > 0
(iv) f"(x)<0
(4 marks)
(c)
Find the relative extrema and points of inflexion of f(x).
(3 marks)
(d)
Find the asymptote(s) of the graph of f(x).
(1 marks)
(e)
Sketch the graph of f(x).
(3 marks)
(HKAL 1997 Paper II)
41. Let f(A) = .r3(x+l)3.
(a)
(i)
Find f '(x) for x
(ii)
-2
Show that f"(x) = —j5—-—y for x#-l, 0.
9x5(x + 1)j
-1, 0.
(b)
Determine with reasons whether f'(-l) and f'(0) exist or not.
(c)
Determine the values of x for each of the following cases:
(2 marks)
(2 marks)
f'(x) > 0
(H) f'(x)<0
(Hi) f"(x)>0
0’v) f"(x)<0
(i)
(3 marks)
(d)
Find all relative extrema and points of inflexion of f(x).
(3 marks)
(e)
Find the asymptotes of the graph of f(x).
(2 marks)
(J)
Sketch all graph of f(x).
(3 marks)
(HKAL 1998 Paper 11)
274
Chapter 5
42.
Let /= 0.-
3J
L
and g(x) = cos x - j cos3 x, where x e /.
Let x0 e I and define xn = g(xn_j) for n = 1, 2, 3,....
(a)
Show that the equation x = g(x) has exactly one root in /.
(b)
Show that xw e / for all n.
(3 marks)
(3 marks)
(c)
3
Show that | g'(x) | < — for all x e I.
4
(d)
Let a be the root of x = g(x) mentioned in (a).
(2 marks)
o
(i)
Show that | x„ - a | < — | x„. - a | for n = 1,2, 3,... .
4
(ii)
Show that {x„} converges and lim xn = a.
n—>o°
(Hi) If
find a positive integer n such that
I x„ - a 1I < —100
—
1 "
(7 marks)
(HKAL 1998 Paper II)
43.
1
Let f (x) = xeJ for x
0.
(a)
Find lim f(x) and show that f(x) —>
(b)
Find f'(x) and f "(x) for x
as x —> 0+.
x—»0”
(3 marks)
0.
(2 marks)
(c)
Determine the values of x for each of the following cases:
(i)
f> 0
(ii)
f "(x) > 0
(3 marks)
(d)
Find all relative extrema of f(x).
(e)
Find all asymptotes of the graph of f(x).
(f)
Sketch the graph of f(x).
(2 marks)
(3 marks)
(2 marks)
(HKAL 1999 Paper II)
Advanced Level Pure Mathematics
275
Miscellaneous Exercise A
Section I
Functions
1.
(ci)
Let p and q be positive numbers. Using the fact that Inx is increasing on (0,«»), show that
(p-^)(Inp-ln<7)>0
(b)
Let a, b and c be positive numbers. Using (a) or otherwise, show that
a In ci + b In b + c In c >
a+b+c
(In a + In b +inc)
3
(6 marks)
(HKAL 2000 Paper I)
2.
(a)
If the function g : R —> R is both even and odd, show that g(x) = 0 for all x e R.
(b)
For any function f : R —> R. define
F(x) = -i[f(x) + f(-x)] and G(x) = ~ [f(x) — f(-x)]
(i)
Show that F is an even function and G is an odd function.
(u)
If f(x) = M(x) + N(x) for all x e R, where M is even and N is odd, show that M(x) = F(x)
and N(x) = G(x) for all xeR.
(6 marks)
(HKAL 2002 Paper II)
Limit of a Sequence
3.
Let a, = 1 and an+i =
4 + a; for n e N. Show that 1 < an < 2 for all n e N.
2
Hence show that {flw} is convergent and find its limit.
(6 marks)
(HKAL 2001 Paper II)
4.
(a)
Resolve
(b)
(i)
5x - 3
into partial fractions.
x(x + l)(x + 3)
Prove that
v
5k~3
*(* + l)(* + 3)
<1
for any positive integer n.
A
276 Advanced Level Pure Mathematics
(ii)
Evaluate
t
Jt=l
5Ar — 3
k(k + V)(k + 3)
(7 marks)
(HKAL 2003 Paper I)
5.
Let {*„} be a sequence of positive real numbers, where Xj = 2 and xn+I
x„ -xn + l for all n = 1,2,
3
tX
Define Sn
for all it = 1,2, 3,... .
Xi
(a)
(b)
Using mathematical induction, prove that for any positive integer n,
(i)
\ > fJ>
(U)
1
5„ = 1- *n+l “1
Using (a), or otherwise, prove that lim Sn exists.
n —» «»»
(7 marks)
(HKAL 2003 Paper I)
6.
Let
(a)
ai =1
A=1
and
an = an-! + 2b„-l
b„ =
, n = 2, 3, 4
+ Vi
Show that for any positive integer n,
(')
a„,bn>0 and a;-2b;
(ii)
(1 + V2)" =a„+bj2
(-Dn;
(4 marks)
(b)
For n = 1, 2, 3,... , define un
= ^L.
b„
+2
+1 '
(i)
Show that
(ii)
Show that u2n_i < V2 and “in >
“n+l
_3u„+4
2»„+3'
Hence show that the sequence {zq, u3, u5,...} is strictly increasing and the sequence {u2,
w4, z/6, ...} is strictly decreasing.
(Hi) Show that un*2
(iv) Show that the sequences {wp u3, u5,...} and {u2, u4, tt6, ...} converge to the same limit.
Find this limit.
(11 marks)
(HKAL 2001 Paper 1)
1.
Let {xn} be a sequence of real numbers such that x, > x2 and
3x„2-x„,-2x„ = 0
for n = 1,2, 3,....
(a)
(i)
Show that for n > 1,
2”"1
Miscellaneous Excnisc /I
(ii)
(b)
(i)
(ii)
(c)
277
Show that the sequence {xp x3, x5,...} is strictly decreasing and the sequence {x2, x4, x6,
...} is strictly increasing.
(5 marks)
For any positive integer/?, show that x^ <x2fl_1.
Show that the sequences {xp x3, x5,...} and {x2, x4, x6,...} converge to the same limit.
(6 marks)
By considering
(x„+2 - xn) or otherwise, find lim xn in terms of x, and x2.
[You may use the fact, without proof, that from (b)(ii), lim xn exists.]
n —>»
(4 marks)
(HKAL 2002 Paper I)
Limit and Continuity of a Function
8.
Let f : (0, oo) -» (0,
) be a continuous function satisfying f [f(x)] = x and f(l + x) =
f(x)
for
l + f(x)
all x.
(a)
Show that for any n G N and x g R,
f(/i + x)
fU)
1 + M f(x)
(3 marks)
(b)
Define xn = 1 +
f(l)
for any n g N. Show that
1+n f(l)
f(X") = ^2
Hence, by considering lim xn, show that f(l) = 1.
M—>«
Deduce that
=n
f(n) = - .nd
n
(c)
(5 marks)
For any q g N and q > 2, let S(q) be the statement
(i)
£ | = -2- for all p g N with 0 < p < q.”
P
Show that S(2) is true.
(ii)
Assume that S(/z) is true for 2 < It < q and h gN. Use (a) to show that
“f
f — =
I p J
P
t? + l
for 0 < p < q + 1.
Hence deduce that S(q + 1) is true.
(Hi) Use (a) to show that for any positive rational number x,
f(x) = X
(7 marks)
(HKAL 2001 Paper II)
278
□
meed Level Pun Mathemmies
Derivatives
9.
Let f and g be differentiable functions defined on R satisfying the following conditions:
A.
f '(x) = g(A') for x g R:
B.
g '(.v) = -f(x) for x g R;
C.
f(0) = 0 and g(0) = 1.
By differentiating h(x)
[f(x) - sinx]2 + (g(x) - cosx]2, or otherwise, show that
f(x) = sinx and g(x) = cosx
for xgR.
(5 marks)
(HKAL 2000 Paper II)
x2 + bx + c
10.
Let f(x) = <LinA. + 2x
.
X
(a)
If f is continuous at x
(b)
If f'(0) exists, find b.
(a- > 0)
(A < 0)
0. find c.
(6 marks)
(HKAL 2000 Paper II)
11.
Let f : R —> R be defined by
x3-12x
f(A') =
e2
e'
when x > 2
when x < 2
(ci)
Show that lim f'(x) = lim f'(x) = 0.
(b)
Is f differentiable at x = 2? Explain your answer.
(5 marks)
(HKAL 2001 Paper II)
12.
Let f: R —> R be a continuous function satisfying the following conditions:
(i)
(ii)
f(x + y) = f(x)f(y) for all x and y.
(a)
Prove that f z(x) exists and f '(x) = f(x) for every x g R.
(b)
f (v)
By considering the derivative of -A-z, show that f(x) = ex.
e
(5 marks)
(HKAL 2002 Paper II)
Miscellaneous Exercise /I
13.
Let
ax
•r
tc
if x < —,
2
ebx sinx
if x>-.
2
f(x) = ’
279
a
71
If f is continuous at —
, show that
2
bn
an
, 2
2 ~C
Furthermore, if f is differentiable at —, find the values of a and b.
2
(5 marks)
(HKAL 2002 Paper II)
14. Let f : R —> R be a non-constant function satisfying the following conditions:.
(a)
(1)
f(x + y) = f(x) + f(y) + f(x)f(y) for all x, y g R;
(2)
.. f(/i)
hm ——
h -> o h
(i)
Prove that f(0)[ 1 + f(x)J = 0 for all x G R.
(H)
Prove that f(0) = 0 and f(x)^-l for all xgR.
a, where a g R.
(5 marks)
(b)
(c)
i
V
V i
2
2
By considering f(x) = f ■? + ■? , or otherwise, prove that f(x) > -1 for all x e R.
(2 marks)
Prove that f is differentiable everywhere and that f'(x) = a[l + f(x)] for all x gR.
Deduce that a * 0.
(4 marks)
(d)
ax
- 1 for all x g R.
(4 marks)
(HKAL 2003 Paper II)
By considering the derivative of the function In [1 + f(x)J, prove that f(x) = e
Applications of Differential Calculus
15.
Show that for x > 0,
x > 1 + In x
Find the necessary and sufficient condition for the equality holds.
(5 marks)
(HKAL 2000 Paper 11)
16.
(a)
Let 0 < A < 1. Show that
xA <(1- A) + Ax-
for all x > 0.
(b)
Let a, /?, p and q be positive numbers with - + - = !. Prove that
P <1
a~pb~q <- + -•
P Q
(5 marks)
(HKAL 2001 Paper I)
280 Advanced Ltvd Pure Mathematics
17.
2
xx for all a* > 1. Find the greatest value of f(x).
(a)
Let f(x)
(b)
Using (a) or otherwise, find a positive integer in, such that
tn m >„n
for all positive integers n.
(7 marks)
(HKAL 2003 Paper II)
18.
(a)
By considering the derivative of f(x) = (1 + x)“ - 1 - ax, show that
(1 + x)“> 1 + ax
for a> 1. x>-l and x*0.
(4 marks)
(b)
Let k and m be positive integers. Show that
zn-rl
(i)
(ii)
1-1-1
I
m
Hl + 1
m +l
1
k
m
k
in
k
-1
m+l ’
m+l
k ,n - (k -1)
m+l
< km
---■
<
-
m+l
(£ + 1)
m+l 5
-k m
(6 marks)
(c)
Using (b) or otherwise, show that
1<
i
_i_
1
I2 + 22 + 32 +--- + n2
3
3
1
1
n
n2
Hence or otherwise, find
lim
n—
19.
Let f(x)
1
1
1
1
I2 +22 +32 +--- + n2
3
(5 marks)
(HKAL 2000 Paper I)
(1 + X2)2'
(a)
Find f'(x) and f"(x).
(b)
Determine the values of x for each of the following cases:
(2 marks)
(i)
f 'M > 0
(ii)
f "(x) > 0
(3 marks)
(c)
(d)
(e)
Find all the relative extreme points, points of inflexion of f(x) and asymptotes of y = f(x).
(4 marks)
Sketch the graph of f(x).
(3 marks)
Let g(x) = |f(x)|.
(i)
Does g'(0) exist? Why?
(ii)
Sketch the graph of g(x).
(3 marks)
(HKAL 2000 Paper II)
Misct'llaiicons Exercise .4
20. (a)
281
Let f be a real-valued function defined on an open interval I, and f "(x) > 0 for x e I.
(i)
Let a, b, c e I with a < c < b. Using Mean Value Theorem, or otherwise, show that
f (C) - f(6Z)
c-a
<
f (^) - f(C)
b-c
Hence show that
f(c)<|— f(a) +
b-a
b-a
(ii)
Let a, b el with a<b and A e(0, 1), show that
a < Xa + (1 - X)b < b
Hence, show that
f[Aa + (1 - X)b] < Af(a) + (1 - A)f(Z?)
(8 marks)
(b)
Let 0 < a < b. Using (a)(ii) or otherwise, show that
(i)
if p > 1 and 0 < A < 1, then
[Az? + (1 - A)Z?JP < Anp + (1 - A)/?p
(H)
if 0 < A < 1, then
Xa + (l-2.)b>a^bl~x
(7 marks)
(HKAL 2000 Paper II)
21. Let a > b > 0 and define
ax+bx V
for x>0
ab
for x = 0
2 J
f(x) =
(a)
(i)
Evaluate
lim f(x).
x—>0*
Hence show that f is continuous at x = 0.
(H)
Show that
lim f(x) = fl
(6 marks)
(b)
Let h(r) = (l + z)ln(l + /) + (1 —r)ln(l-f) for 0<r< 1 and g(x) = lnf(x) for x 0.
(i)
Show that h(r) > h(0) for 0 < t < 1.
(ii)
Show that 0 < t < 1 and
For x > 0, let t = a
ax+bx
2
ax lnaJ +bx Infr—+ ln
h(r) = 2
a
x
+b
x
ax+bx
(Hi) Show that for x > 0,
x2g'(x) =
2
ax In a1 + bx \nbx + ln
ax + bx
ax+bx
Hence deduce that f(x) is strictly increasing on [0, ~).
(9 marks)
(HKAL 2000 Paper II)
□
282
Advanced I.eve I Pure Mathematics
22.
Let f(x) = x3(6-x)3.
(a)
(i)
Find f'(x) for x s* 0,6.
Show that f'(0) and f'(6) do not exist.
-8
(Hi) Show that f"(x) = —---------y for x
x3(6-x)3
(ii)
(b)
0, 6.
(4 marks)
Determine the values of x for each of the following cases:
f'(x)>0
(ii) f'(x)<0
(iii) f "(x) > 0
(iv) f"(x)<0
(i)
(3 marks)
(c)
Find all the relative extreme points and points of inflexion of f(x).
(d)
Find all the asymptotes of the graph of f(x).
(e)
Sketch the graph of f(x).
(3 marks)
(2 marks)
(3 marks)
(HKAL 2001 Paper II)
23. (a)
Let f(x) = x3 — 3px + 1, where p g R.
(i)
Show that the equation f(x) = 0 has at least one real root.
(ii)
Using differentiation or otherwise, show that if p < 0, then the equation f(x)
and only one real root.
0 has one
(iii) If p > 0, find the range of values of p for each of the following cases:
(1)
(2)
the equation f(x) = 0 has exactly one real root,
(3)
the equation f(x) = 0 has three real roots.
the equation f(x) = 0 has exactly two distinct real roots,
(9 marks)
(b)
Let g(x) = x4 + 4x + a, where a g R.
(i)
Prove that the equation g(x) = 0 has al most two real roots.
(ii) Prove that the equation g(x) = 0 has two distinct real roots if and only if a < 3.
(6 marks)
(HKAL 2002 Paper I)
24.
8
x-1
Let
f(x) = x2-
(a)
Find f'(x) and f"(x).
(b)
Determine the range of values of x for each of the following cases:
(x*l).
(2 marks)
f'(x)>0
(ii) f'(x)<0
(iii) f"(x) >0
(iv) f"(x) < 0
(i)
(3 marks)
(c)
Find all the relative extreme point(s) and point(s) of inflexion of f(x).
(2 marks)
I
Miscellaneous Exercise /I
(d)
Find the asymptote(s) of the graph of f(x).
(e)
Sketch the graph of f(x).
283
(1 mark)
(2 marks)
(f)
Let
g(.v) = f(|.v|)
(i)
Is g(x) differentiable at x = 0? Why?
(ii)
Sketch the graph of g(x).
(|xpl).
(5 marks)
(HKAL 2002 Paper II)
25. (a)
Let g(x) be a function continuous on [a, b], differentiable in («, b), with g'(x) decreasing on
(a, b) and g(«) = g(b) = 0. Using Mean Value Theorem, show that there exists c e(a, b) such
that g is increasing on (a, c) and decreasing on (c, b).
Hence show that g(x) > 0 for all x g [«, b].
(5 marks)
(b)
Let f be a twice differentiable function and f"(x) >0 on an open interval I.
Suppose a, b, xel with a<x<b.
By considering the function
g(x) = (b - x)f(«) + (x -a)f(b) -(b- a)f(x)
or otherwise, show that
b-a
W) +
b-a
Hence, or otherwise, prove that
f(A,Xj + A^v,) < A, f(Xj) + Ao f(x2)
for all xp x2 g I, where ApAo>0 with Al + A2=l.
(5 marks)
(c)
Let Xj and x2 be positive numbers.
(i)
If ApAo^O with A] + Ao = 1, prove that
AjX! 4- A2x2 > x/'x, ^2
(ii)
If
P2 are positive numbers, prove that
Pi^+P2x2 i
P\ +P1
>XjAX2ft
(5 marks)
(HKAL 2002 Paper II)
26. Let
(a)
(x*-2).
f(x)
(i)
Find f'(x) for x^-1.
(ii)
Is f differentiable at -I ? Explain your answer.
(Hi) Find f"(x) for x*-l.
(b)
(4 marks)
Determine the range of values of x for each of the following <cases:
(i) f'(x)>0
(Hi) f"(x)>0
00 f'W<0
0'0 f"W<0
(3 marks)
284
Advanced Level Pure Mathematics
Find the relative extreme point(s) and point(s) of inflexion of f(x).
(c)
(3 marks)
Find the asymptote(s) of the graph of f(x).
(d)
(3 marks)
Sketch the graph of f(x).
(2 marks)
(HKAL 2003 Paper II)
Section II
27.
Cn
1
Prove that —— < —, where //, r are positive integers and n > r.
nr
r!
(a)
(b)
If
a2,..., an are positive real numbers and 5 = a} + a2 + ••• + an, using A.M. > G.M. and (a),
or otherwise, prove that
(l+a1)(l + fl,)---(l + fl„)<l+5 + ^ + ^ + --- + ^
(c)
Let cn = Il (1 + “)• Using (b) or otherwise, show that the sequence {c,,} is converges.
”
J
n
k=\
~
(8 marks)
(HKAL 1992 Paper I)
n
28.
For n
sin3
1, 2, 3, ... and 0gR. let sn
A =l
e_
3k
3
1
— sin 0 - — sin 30, show that
Using the identity sin3 0
n
Hence, or otherwise, evaluate
3" . (o ) i1 . n
y
= — sin —---- sin 0
4
\3 J 4
lim sn.
n—><»
(4 marks)
(HKAL 1994 Paper II)
29.
(a)
(b)
t
. tan A' — x
Evaluate lim —----x->0 x sinx
Prove that
1
< | a | for all a * 0.
x
a*sin —
1.1
-4-sin —
Hence evaluate lim-y-------- y-.
x—>0 1
1
---- sin —
X
30.
A'
(a)
, .. (ex -e x)2
Find lim---------- - —.
*-♦0 1 - cos 2x
(b)
Prove that lim x2 cos — = 0.
x-»0
(6 marks)
(HKAL 1992 Paper II)
x
(6 marks)
(HKAL 2001 Paper II)
Miscellaneous Exercise A
285
31. Evaluate
(a)
(b)
r -> 0
lim
1
sin x
..
x - COS X
1
tanx
lim -------------
v ->« x + cos x
(6 marks)
(HKAL 2003 Paper II)
32. Let f : R —» R be defined by
(a)
-1
f(x) = 0
1
Prove that f is an odd function.
(b)
Is f an injective function? Explain your answer briefly.
(c)
Sketch the graph of y = f(x + 1).
(d)
Let g(x) = f(x + 1) + f(x - 1) for all x 6 R.
when x<0
when x = 0
when x > 0
Sketch the graph of y = g(x). Write down the value(s) of x at which g(x) is discontinuous.
(6 marks)
(HKAL 2003 Paper II)
33. It is given that if {#„} is a sequence diverging to positive infinity and {bn} is an increasing sequence
implies that )inA = *.
n->~bn
=
diverging to positive infinity, then lim
n-*~b„^-b„
(a)
Let {un} be a sequence of positive real numbers.
By using this result, show that if lim un = u # 0, then lim
n->«>
(b)
It is known that lim
n—>°°
+ U1 *——
fl
1____ 1_
^n+l
2
j
3‘
By using the result in (a), show that lim (flfl*) exists and find its value.
(C)
(i)
By using the result A.M. > G.M., show that
1
1
U2
(H)
If nlim
un = u 5*0, find lim
—>oo
________1_______ in terms of u.
lf±+±+...+±
«2
Hence, show that lim
n—>°° '
-'iin = u.
“»
= u.
286
□
34.
.Advanced Level Pure Mathematics
a
b
cy
Let M = c
a
b . where a, b and c are non-negative real numbers.
<b
c
a.
(a)
Show that
det(M) = -y (a + b + c)[(tz - b)2 +(b- c)2 4- (c - a)2 ]
and
0 < det (Af) < (a + b + c)3
(4 marks)
/
(b)
Let Mn =
an
bn
Cn
Cn
an
bn
\
for any positive integer n, show that aH, bn and cn are non-negative real
Cn
numbers satisfying
an + b„ + C„
(a + b + c)n
(4 marks)
(c)
If a + b + c = 1 and at least two of a, b and c are non-zero, show that
(i)
(ii)
lim det(AT) = 0
/:->oo
lim (an - bn) = 0 and lim (an - cn) = 0
n—>oo
(iii)
35.
(a)
1
lima =23J
n->~
(7 marks)
(HKAL 1994 Paper I)
Let x > 1 and define a sequence {«n} by
al = x
+1
■
and
Cln
for n > 2
2«„-,
(i)
Show that an > 1 and an > an+l for all n.
(ii)
Show that lim an = 1.
n—»«»
(8 marks)
(b)
Let f:[l, °°) —> R be a continuous function satisfying
f(x)
f
X2 4-1^
2x J
for all x > 1.
Using (a), show that f(x) = f( 1) for all x>l.
(7 marks)
(HKAL 1996 Paper 11)
36.
(a)
Show that for any positive integer n, there exist unique positive integers an and bn such that
(l + V3)2"=an+h„73
and that
(i)
a2—3b2=22n,
(ii)
an and bn are both divisible by 2n.
(8 marks)
I
Miscellaneous Exercise A
(b)
287
For an and bH as determined in (a), show that
(1 -73)2n =an-bn^3
(2 marks)
(c)
(cl)
Using (b), or otherwise, prove that lim — = V3.
bn
(3 marks)
Using (a) and (b), or otherwise, prove that for any positive integer n, the smallest integer greater
than (1 + 73)2'’ is divisible by 2n+l.
(2 marks)
(HKAL 1997 Paper I)
37. Let
Tn.r »
k
n)
^r=Q
where Tn r is the (r+ 1 )th term in the binomial expression of
in
1+±T
nJ
1
ascending powers of —.
(a)
Show that Tn 0
T
1 n. 1 = 1 and
r!
n
n
n
for r > 2.
(2 marks)
(b)
For any fixed r, show that the sequence Tr
increasing.
Find lim Tlt
n —> oo
T
1 r+l.r*
... is bounded above and monotonic
in terms of r.
(4 marks)
(c)
Show that
T
1 ii, r-1
(i)
(")
Tnr
1
Tn.k, <
~: zJ-r+l T
(Hi) T„
(d)
>r forr> 1,
for n>k>r> I,
x'-oz: T„tk for r > 1.
(7 marks)
Show that
Tn,nk
k <
1
(r-D(r-l)!
for r > 2.
(2 marks)
(HKAL 1997 Paper I)
38. Let n = 2, 3, 4
(a)
(b)
Evaluate lim
x2n -1
x2 -1
(2 marks)
Find all the complex roots of x2" - 1 = 0.
Hence, or otherwise, show that x*71 — 1 can be factorized as
(r2 - l)(x2 - 2xcos—+ l)(x2 - 2xcos —+1) ••• (x2 -2xcos
v
n
n
n
+ 1)
(6 marks)
288
Advanced Level Pure Mathematics
(c)
n
Using (b)y or otherwise, show that
lim
x2n -1
A2 - 1
= 2 2"-2sin2sin2 — ---sin2 (n-l)n
2n
2n
2n
(4 marks)
(d)
Using (a) and (c), or otherwise, show that
2
1
. ( 7t \ . (2n\
. f(ll- 1)tt
-7= sin — sin — •••sin ------- —
Jn
\2nJ
\2nJ
k 2n
n
2
(3 marks)
(HKAL 2000 Paper I)
Advanced Level Pure Mathematics 289
Answers
Chapter 1
Exercise 1A (P. 8)
1.
7.
f“’[{1}] = {2n7t: n eZ}
8.
(a) No
(b) Yes
(c) Yes
(d) Yes
(e) Yes
(f) No
(g) No
(h) Yes
f[X] = {zeC: |z| = 1}
r'[ Y] = {.r + iy e C\ {0}: x2 + y2 - x = 0,
where x and y are real numbers}
9.
Let A = {1,2, 3, 4} and B ={2,4, 6, 8}.
Define f: A -> B such that f(x) = 2x, Vx eA.
Suppose XI = {2,4} and F2={2, 4, 6}.
(i) No
10. f[X] = {z gC: Re z = Im z}
2.
(«)R
f"‘[ /] = {(x, >’) 6 R2: x2 + y2 = (1 - c)2}
(b) The set of all negative real numbers.
(c) R \ (2n + 1)7r : n = 0, ±1, ±2,,...
2
Exercise 1B (P. 14)
(d) {x g R: x < -1 or x > 1}
(e) {x g R: x < -2 or x > 1}
(f) {x g R: x = 0 or x > 1}
1.
f o g(x) = sin2x, g o f(x) = sin x2
2.
(a) f o f(x) = 4x - 9
(b) g o f(x) = (2a - 3)3
3.
{zeC: |z|*l}
4.
The set of all odd integers.
(c) f o g(x) = 2? - 3
(</) f o g o f(x) = 2(2r - 3)3 - 3
3.
5.
(a) {x g R: x > -3}
(a) f o h(x) = 2X2; h o g(x) = (x + 3)2
(b) k = fohog
(b) {x gR: 0 <x< tc}
(c) {x g R: x < -1 or x > 1}
4.
R :----- < x <s' —
(d) x g T)
2
2.
(e) {x g R: -1 <x < 1}
(f) {xgR:x<0}
6.
f[Z] = {x: x is an integer divisible by 5};
f'[Z] = x g R: x = y, where n g Z
1
det A
fog does not exist.
g°f(A) =
Exercise 1C (P. 25)
1.
(a) f(x) = 2x
(b) f(.r) = *
(c) f(x) = x
x-1
(x > 0)
x
(x < 0)
290
Advanced Level Pure Mathematics
6.
fof=Z; p1 = f
7.
fog = /; T'
Revision Exercise (P. 50)
O* a
O’ o
=f
2.
9.
(a) f-'(x)
2(z + z)
zz +1
-2i(z-z)
b
zz + \
c=2(zz-V)
(a) a =
.v — I
5
(b) f-(A) = (|. 1-j)
(c) f"1 does not exist.
(b) (ii)
Exercise 1D (P. 47)
i.
(a) It is neither odd nor even.
(b) It is neither odd nor even.
(c) Il is an odd function.
(d) It is neither odd nor even.
(e) It is an odd function.
(J) It is an odd function.
3.
(g) Il is an even function.
4.
(b) (i) The image is a semi-circle centred at
the origin and with radius equal to 1.
(ii) The image is a line segment lying on
the imaginary axis from —1 to 1.
0.
10. (a) It is not periodic.
(b) It is periodic. Period
4.
(a) The image is a line segment lying on the
real axis from -2 to 2.
5.
(c) gof(,v) = COl
1
(c) It is periodic. Period = 2
(d) It is periodic. Period =
(e) It is periodic. Period = k
(f) It is periodic. Period = 2zr
(g) It is periodic. Period = 2zr
(h) It is not periodic.
(i) Il is periodic. Period
K
b-a
(x-fl)
10. (a) g o f is periodic.
fog is not periodic.
(c) (Hi) f + g is not periodic.
f(l)-Zf(Z)
2
f(i) + zf(Z)
b=
2
13. (c) a =
2tt
15. (b)g(x)=x
y
y = g(x)
Chapter 2
Exercise 2A (P. 68)
(b)
(c) 1
(d)4
Answers 291
(e) I
0) 1
(g)
(h)0
(i) 0
(j) -1
(k) 0
(I) 0
(m)+^
(fi) 00
(c) 1
(d) -2
(e)2
(f) 1
205
1152
Exercise 2B (P. 76)
i.
(b)j
2.
3.
15. (b)
(a)0
(b) 0
(c) 1
(d) +°°
2.
7T
6
4.
(b)(i) 0
.n+1
(a) x,=(-\Y- y„ = (-!)'
5.
(b) xn = )>„ = (-D"
4.
No
6.
(b)
5.
No
7.
(c)3
8.
(b)Q
6.
7.
9.
0
(|«|<D
_1_
lim xn = < a
(|«|>1)
j_
2
not exist
n—*<»»
(* = -!)
1
an+b''
n->oo Q
(fl = D
_ [j
(H<H)
-1
a + 2b_
3
7.
0
8.
(b)
9.
(b)
1 + V5
2
1 + V2T
2
10. (b)M
3
11. (b) (ii) 0
a; + ^
10. (b)^
3
11. (b) (i)
Exercise 2C (P. 85)
12. (b) a\ a
1
(ii) 2
(Hi) does not exist
Exercise 2D (P. 90)
3
1-
12. 1
13.
3
14. (a)
2
5
x + 2’ 6
x1 1
2’ 4’
(b) x,n -x„.,
8
_1_
2
n-1
5
’ 3
(a) e
(b) e2
£
(C) e2
(d)e-
(e) e 2
(f) e3
292
Advanced Level Pmc Mathematics
2.
(b) (i)
e 3
(g) >
(h) I
(i)
(i) Z2
(j) 0
(H) e2
3
(iii) e2
(iv) e 4
(k) ±
w-l
Revision Exercise (p. <jo)
(o) 2 + 73
1.
(W1
2
2.
2
(q) i
5
3.
(c) 73
1
(s) ~~
' '
7
(t) 0
6.
(c) 3
W9
(v) 0
7.
(c) (ii) ylaobo
9.
P"-'a, +
(a) a„
Q(I-PW~I)
1-P
4.
0
6.
(a)0
WO
(c) ~
(d) oo
(b)(i) b„=pb, i + ^''(b.,-pbt)
(") bn=p
q-p
-p"-')
Exercise 3B (P. 112)
4
5
1
5
10. (c) (i)
25
19
9
(H)
25
29
(g) |
(Hi)
600
551
d) -1
(j) ■
(k) I
(l) n
(a) e
(b) e
(c) 1
(d) e2
(e) does not exist
(f) e
(g) e2
(h)
1.
(C) -C2+-Cx
W-12
(c) oo
(h)0
zr
Chapter 3
2.
Exercise 3A (P. 104)
i.
f(x) = -, g(x) = --
2.
f(x) = -, gU) = X
3.
(a)^
(b)0
(C)
(d) —
10
(e) 1
(f) 1
X
(f) -1
X
Exercise 3C (P. 114)
A'
1.
2.
(a)0
(b)0
(c) 1
(d)Q
(a) does not exist
(b) does not exist
(c) does not exist
(d) does not exist
Answers 293
y
3.
1
o
O
4
-1-
o
5.
(b)0
6.
(a) 5, -2, 8
9.
(b) 1
x
Revision Exercise (P. 131)
4.
Exercise 3D
1.
(P. 130)
(a)
sin#
0
(b) (it)
(a) f is continuous.
5.
sin#
0
(a)(i) 0
(b) f is discontinuous at x = 0 and ±1.
(c) f is discontinuous at x = -1.
(d) f is discontinuous at x = 0.
Chapter 4
(e) f is discontinuous at x = 2.
(f) f is discontinuous at x = 5.
Exercise 4A (P. 146)
(g) f is continuous.
(h) f is discontinuous at x = 0.
i.
(i) f is continuous.
2.
(a) f is not differentiable at x = 0.
(b) f is not differentiable at x = 0.
(a) 3
(c) f is differentiable at x = 0.
(b) 1
(d) f is not differentiable at x = 0.
(c) -2
3.
a
2.
(a)
:: t
0
2 3
(b) f is discontinuous atx = ±l.
*
'4
(x<3)
4x
(x > 3)
and f is not differentiable at x = 3.
y
0-2-
(a) f'(x) =
(b) f'(x) =
o
1
(x<-2)
2(x + 2)
(x>-2)
and f is not differentiable at x = -2.
1
3'
—o-----1 O
-o
1
-3x22
(x<-2)
3x2
(x > -2)
(C) f'(x)
►X
and f is not differentiable at x = -2.
4.
f is discontinuous at x = ±1.
2x
(d) fz(x) = -2x
y
2x
(x<-D
(-1<X<1)
(X>1)
and f is not differentiable at x = ±1.
►x
3.
(a) -sinx
(b) sec2x
(c) lx+ 1
(d)
O
-X2 4- 1
(x2 + l)2
294
Advanced Level Pitre Mathematics
4.
a = 3; b = —2
5.
(a) f '(a) = g(.v) + (.v - «)g'(-v)
1
(tn) A COS A + sin A------------- y
(1-A2)2
(a < a}
-1
(a < a < b)
fz(x) =
1
(b < x < c)
3
(c < a)
and f is not differentiable at a = a, b and c.
-3
6.
1
1
A
A
(n)
(o)
(p)
13. (a) 3a2 sin —-a cos —
2a
(a- > 0)
-2a
(A- < 0)
-1
xi-V
2 sin a cos v
71 - sin4 a
2(1-a2)
1 + 6a2+a4
-cos 2 a
(r)
71 - cos2 a sin2 a
«1)
(c) It is continuous.
14. (a) f'(x) =
(sin 1 a)(cosa)71 - a2 - sin a
(sin-1 a)2 71 - a2
(s)
15. a = 0; b = 0
73
2 + cos x
9a2 \ 1 - A22 COS 1 A + A ~ 6a3
(t)
Exercise 4B (P. i6i)
7i-a2
(u)
(v) (cos a)x (In cos a - a lan a)
3.
4.
1
(iv)(ln 3e)3xex
A+1
tan—
2a-1
(a)
27a2-a+1
(b) 4(a3 - ex sin a)3 (3a2 - ex sin a - ex cos a)
W -
(z)
(d) ex(sec2<?x)
(f) -sec a
2(8-a)(2a-I)2
(g)
(a* + 2)5
(h)
1
A/(.r-l)(x+l)5
6.v[ln(x2 +1)]2
(i)
COS A- - X
(k)
(I)
5.
e'x sec2 e'x
27a tanevx
3a2
ln4 (In a - l)4lnr
(In a-)2
cot a [ ln(sin a)
(sinA),nx
In sin a
a
+(cot A')(ln A)(sin A)Inx
2a + cosy
(c)
Asiny-2y
(b)
(d)
(a + y)ex~y - 1
(a + y)ex~y +1
(e)
(2A-y)(l + y2)y2
l + 2y2 +Ay2(l + y2)
(1 + y 1 - (1 + cos a) 72(A + y) + y2 -a 2
(f)
3
4(a3 + 9)4
A
(«) ~e,cosx sinA-2Aex
A2 + 1
(J) 1 + sin a
A2
(y) (l-lnA)Ax
(c) es,nx(cosA)
(e) -csc(tan a) cot(tan a) sec2A
3 rln3 sec2 —
(8)
(1 + y)72(A + y) + y2-A2 -1 + a
y-A-
A+y
Answers 295
6.
-cos/
1 4-sin/
(b) -tan /
(c)
(c)
(~l)"-2(n-2)!
x"-1
(d) a '(In a)"
cos/(cos / - sin/)
sin/(sin / + cos/)
(e) |^4"sin(4x + ^-) + 2"sin(2x + y-)
. • ,Cltx
cos/. 4-sin(
—)
(d)
----------------' '
at
sin / 4-cos(—)
b
7.
(ci)
.r3
A-2-4
2
+ -Jx-2 X4-2
(if n is even)
0
(n!)
(b) f,fl)(0) =
(if n is odd)
Exercise 4C (P. igg)
1.
(a) -0.86
2.
1.15
3.
(b) 2.096 6
12. (a) e ■*ln2
14. /°(0) = ’
h!
(if n is even)
0
(if n is odd)
200 7T
17. y(n,(0)
2 [(/z - 2)(tz - 4) • • • 2]2
(if n is even)
0
(if ti is odd)
4.
2.27
5.
The distance has to be decreased by 10%
18. coefficient of xf is (-l)rC"n(n-l)---(r + l)
7.
(a) The length is decreased by 0.89 cm.
19. (a)knek'
(b) The period is increased by 0.05%.
(n
20. (a) £ " u(r\x)qn~reqx
(c) 216 seconds
22. C;"n!a n
Exercise 4D (P. 177)
23. (a)
1.
(a) f'(-r) =
0
(A<0)
2x
(a>0)
1 4--L
x-1 A4-1
(b) f<n,(0)
(b) f'(0) = 0
(c) f"(x) = Z
0
2
,al -y
(A>0)
(d)
6.
0
(if n is even)
Exercise 4E (P. 189)
(b) -
(-!)»-■(/,-1)!
(ci)
xn
(if n is odd)
(x<0)
(d) f "(0) does not exist.
3.
-2(n!)
6
(x + 2y)3
2
(sin/-cos/)3
in / > 2fx
10. (a)---- r—
g W
Revision Exercise (P. 190)
4.
(a) (tn - 1 )(m - 2) • • ■ (m - >i)x
8.
(a) 1
flTC K
(b) cos(x + —
2 )’
b
(b)
tn-1 —n
fz(c)
f(c)
296
Advanced I cvcl Pure Mathematics
Chapter 5
Exercise 5A (p. 202)
1.
(b) 1
3.
6.
(b) e2
Exercise 5B (P. 206)
(e) 2
(f) 0
1.
7
(S)1
(h) -2
2.
x + 5y + 1=0
(i) -L
(J) -2
3.
(1,-1), (-1,3)
5.
(0, 5), (-2, 17)
6.
(1,0)
7.
(a) v = 2ax + (1 - a2)
(k) 0
"’4
(„i)In a
(«)3
(«)0
“4
(c) 2
(d)0
(e) 0
(f) -f
(g)0
(h)0
Exercise 5C (P. 211)
<j) 2
1.
-30 ms
W oo
(I) 0
2.
(a) p = -9, q
(a) 1
(b) e2
(b) 36, 30
(c) 1
(d) 1
(e) 1
(f) not exist
(g) 1
(b) e
(d) When t = 4, the velocity is a maximum;
When t = 0, the acceleration is a
maximum.
(i) 1
(J) 1
£
(k) eb
(I) eb
(>n) 1
(n) 1
lim
x—#«>
(b) y = -2bx-(\ -b2)
(c) y = 2x, y = -2x
-10 ms'
3.
- — cm3s
3
4.
6 cm3 s
5.
— ms
2
6.
(a) — ms'
n
7.
M
12
(c) |(5±V5)
e
a_ -1 y
a —1
MO
1
a
(*<D
(«>D
(b) — ms
7T
(b) 1
(c) not exist
9.
11. (a) 2
M 1
e
4.
(b) e5
(C) 2
30
2.
3
10. (a) |
(b)0
_9_
50
100
Answers 297
Exercise 5D (P. 215)
3.
4.
5.
(g) Critical point:
(5 + 3V17 ' -15 + 3717 ^/_78 +307i7 j
8
32
[-1 ,U
47T-.
|-27T
3 ’
Turning point:
3
(5 + 3717 -15 + 3717A/_7g + 307i7 j
8
32
2253[5~.
(h) Critical point: (0,0), Gy, 49 V77’
(b) If n is odd, f is strictly increasing on
(-oo, -n - 1) and is strictly decreasing on
(—n — 1, —n) u (—n, 0).
If n is even, f is strictly increasing on
(-co, -n) and is strictly decreasing on
(j,0)
(-/i, 0).
(i) There is no critical point.
Exercise 5E (p. 222)
3
Q
1.
_ 2253f5
49 V 7
Turning point: (0, 0), Gy,
(j) Critical point: (—, y), (3,3)
(a) Critical point: (1, -3)
Turning point: (1, -3)
9 3
Turning point: Gy y)
5 -494
(b) Critical point: (-2,31), (yyy )
2.
5 -494
Turning point: (-2, 31), (-,-yy )
(a) There is no turning point.
(b) Minimum point: (2/zzr + y, 0),
(c) Critical point: (2, -48)
Turning point: (2, -48)
(2/i^-^,-2)
75-75 -12),
(d) Critical point:
9
Maximum point: (2nn + a. —),
o
9
((2n + l)zr - a, f)
(_J___
75 ’
2575
(1,-12), (-1,-12)
-12)
Turning point:
(_J___ !«_ -12),
75 ’
o
-12),
2575
(1,-12), (-1,-12)
where a = sin
4
(c) Minimum point: (y 0)
Maximum point: (-1, 256)
(d) There is no turning point.
(e) Maximum point: (1,3)
(e) There is no critical point.
(f) Minimum point: (y, -
(!.0>.
(f) Critical point: (-y,
(-2, 0)
Turning point: (-y,
)
-’p)
(g) Minimum point: (-1, - y)
—)
Maximum point: (1, y)
7T
(h) Minimum point: (2n^ + y, 0)
2
298
Advanced Level Pure Mathematics
3.
(a) Second derivative lest cannot be applied.
7.
(b) Minimum point: (1,1), (3.1)
Maximum point: (2. 2)
(c) Minimum point: (-2, 0), (2. 0)
Maximum point: (0. 16)
(J) Maximum point: (0. 1)
4.
(b) Minimum point: (~~« - x^2)
4
Maximum point: (—, V2)
4
. 1)
(c) Maximum point: (—.1)
4
(d) Minimum point: (^-, x/3)
6
6
(c) 2.03
Exercise 5G (P. 233)
1.
(a) Minimum point: (—, — - \ 3)
3 3
Maximum point:
iviaAiiuuiii
11. \(-^7-,—+ V3)
(b)
(a) Concave down: (-«». 1)
Concave up:
(L°°)
Point of inflexion: (1,-1)
(7?) Concave down:
Concave up:
(°. y)
■ 0)U(y,+~)
(•
Point of inflexion: (0. 1). (y, -
(c) Concave down:
65 509
)
27
(
V2
Concave up:
(-^’0)u(^’ + oo)
Maximum point: (—, - 43)
6
Point of inflexion: (0. 0), (—U, -
V2
3
4x/2?’
71 3 4
(e) Maximum point: (—, —)
( V2’4V2}
.
.71 343}
(f) Minimum point: (—
6,
’ ----2 )?
(d) Concave down:
(- _L -^)
x/3’VT
Concave up:
Exercise 5 F (P. 227)
(-“’-73)u(73’+oo)
1.
2.
125 mx 250 m, 31 250 (in2)
The tank is of base radius 2 m and height
6 m.
The minimum cost is $1 SOOzr.
3.
4.
256zr
cm3
9
Maximum area: the wire is bent into a circle
only.
Minimum area: the wire is cut into two
pieces of lengths
90 V3
3\l3+7T
6.
4300
and
30/r
3x/3 +7t
Point of inflexion: (- 1
4
1 4
-),
(-k,~)
vV3 9
V3'99
(e) Concave down: (0, +«>)
Concave up:
(-<», 0)
Point of inflexion: (0, 1)
(f) Concave up:
(-<», 0) u (0, +°°)
There is no point of inflexion
(5) Concave down:
Concave up:
2
•,0)U(0.y)
(
?
(j, + ~)
Point of inflexion: (y,12^0
.2
3
Answers 299
(h) Concave down:
(-00,
Exercise 5 H (p. 241)
20
3’
14
14
7
1. (a)y = O, x = ±3
Concave up:
(b) y = 0, x = ±2
(c) y = 2
,20-750 5. z 20 + 750
, + °°)
(---- 7T
14 --- ’5)U<
14
(d) y = 0, x = 0, x = 2, x = -3
Point of inflexion:
(e) y = 3, x = —1
(
20 —-750 225-20750 ,-10-3750 4
(
)3),
14
’
98
14
(/) J = x+ 1, x=l
(#)y = x+2, x = 4, x = -l
(
20+y30 225 + 20750 .-10 + 3750
(
14
98
14
(h) y = x, x = —1
(3'0)
(j) y = ±l
(k) y = ±3, x = 2
(i) y = 0, x = 3, x = -l
(i) Concave down: (0, 1)
(-00, 0) u (1, +«>)
Concave up:
Point of inflexion: (0, 0), (1, 12)
(l) y = o
(m)y = x±^-
(j) Concave down: (0, 2)
Concave up:
(2, +°°)
Point of inflexion: (2,872)
(k) Concave down:
('<) y = ±^x-i
Exercise 5I (P. 253)
27-7297
\
4
Concave up:
(—3,—
127-^/297
( \
4
,3)
)u(j27-7297
4
1.
(
y
(a)
y = x- 2
27-7297
)
4
Point of inflexion:
4
i
27 - 7297
4
27 - 7297 9 + 7297 .
4
N
4
27-7297
4
27-7297 9 + 7297 ,
4
4
V
(I) Concave down: (0,2)
Concave up:
(-2, 0)
Point of inflexion: (0, 0)
X
-4-20
y
(b)
-► X
[!
0
300
Advanc cd Level Pure Mathematics
(c)
y
3.
y= x
(a) f'(A)
f"(x)
o
(b)(i)
>6
_ (a + I)2(a--5)
(A'-I)’
_ 24(a + 1)
(A -I)4
a =-1
or 5
(ii) a < -1 or -1 < x < 1 or x > 5
(Hi) 1 < x < 5
(iv) x = -1
fv) — 1 < x < 1 or x > 1
y
(d) y = -rV2
(vi) x<-l
(c) Minimum point:
?7
(5,y)
Point of inflexion: (-1.0)
(d) x = 1, y = ,v + 5
V
(e)
7
(e)
y=f(x)
y
,y=x+2
O
X
5
■ x= 1
\ 2
-3 V
O
y
y=x + 5
y»|f(x)|
(f)
y
O
5
. ,‘x= 1
O
' y = -x - 5
X
1 75
4.
x
2.
(a) 0 <x< 10
-(A-3 4- 500)^
(b) f'U) =
f"(A)=
(1 000 - a-3)
-3 000(.v3 - 250)
7%3(1 000-a3)3
1
O
(d) x = 2503
(e)x = 0
X
Answer .s
y
(f)
A
6.
0
7.
0; x = 0
8.
WO
9.
(c) In 3
301
16. (a) 0
O
X
10
250'
Revision Exercise (P. 26D
5.
(a) f'(x) =
8x(x + 3)
2
3(x + 4)3
(x>0)
-8x(x + 3)
(x < 0 and
1.
(a) Minimum point: (0, f(0))
Maximum point does not exist.
(b) There is no turning point.
i
3(x + 4)3
2.
8(5.r2 + 30a+36)
U>0)
4
-8(5x2 +30x4-36)
4
9(x + 4)3
(x < 0 and
x * -4)
3^3
)
2
.
(-V3, -373
2 ;
Maximum point:
9(x + 4)3
f"Gv) =
(V3,
(a) Minimum point:
(b) Point of inflexion: (0, 0)
(c) y = x, x = ±1
y
(d)
(b) f'(0) exists but f '(-4) does not exist.
Both f "(0) and f "(-4) do not exist.
(c) Minimum point:
at x = -3
Maximum point:
at x = ~4
Point of inflexion: at x =
-30 ±7180
10
X
: • z°
* • /
x=0
y
(d)
-30- V180
10
---- -T-4
-30+7l80
10
-3
►X
3.
(a) f'(x) =
(x + 3)(x-l). -
O
(X + D2
’
■ '
(b)(i) x<-3 or x> 1
(ii) -3<x<-l or-l<x<l
■:
I
I
(Hi) x> -1
Exercise 5J (P. 259)
(fv)x<-l
4.
0
(c) Minimum point: (1,8)
Maximum point: (-3, 0)
5.
1
(d)y = x + 5, x = -l
8
(x + 1)3
302 Advanced Level Pine Mathematics
y
(e)
y
(d)
X = -1
y = x+ 5
1 VJ
o
6.
1
X
O
-Ji
x+4
(a) f'(x) =
x3(x + 6)3
-8
f"W = —-------- y
a-3(x + 6)3
4.
(a) f'(x) =
x(4-x)
(2-x)2
(b) (i) x < -6 or -6 < x < -4 or x > 0
(ii) —4<x<0
(b) (i) 0<x<2 or 2<x<4
(Hi) x < -6
(ii) x < 0 or x > 4
(iv) -6 < x < 0 or x > 0
(c) Minimum point: (0, 1)
Maximum point: (4, -7)
(c) Minimum point:
(0, 0)
(-4, 32b
(d) There is no point of inflexion.
Maximum point:
(e) y = -x- 1, x= 2
Point of inflexion: (-6, 0)
CD
y
(d) y = x + 2
y
(e)
|>= 2
y=x + 2
;
4
■x
o; :
'S
x
I
O
5.
(a) a = 1, b = 0
7.
(b) Minimum point:
5x-2
(a) f'(x)
; f"(x) =
2(5x + l)
3x3
Maximum point:
r-
a/3
Point of inflexion: (-V3,---- ), (0, 0),
' ' ' 4 '
(c) y = o
<*-4>
2
(c) (i) x < 0 or x > —
5
2
(ii) 0<x< —
5
(iii)
(iv)
1
5
<x<0 orx>0
_1_
5
4
9x3
Answers 303
2
3*2^
M-)
(d) Minimum point:
53
Maximum point:
(0, 0)
Point of inflexion:
(_15. _A)
p
9.
(a) x<0 or x> 1
3 377
(b) Minimum point: (—
)
2’ 2
1
(c) y = x + —, y=-X-—1
2
2
y
(d)
y
(f)
I
y=
x= 1
1
2
5
1
5
? : 1
\ • 2
8.
(a) f '(x) = x(x - 1 )(x - 4)e”x
f"(x) = -(X - 2)(x2 - 6.V + 2)e
2x(x2-2x-7)
(x+1)2
(x>0)
10. (a) f'(x) =
(b) (i) x = 0, 1 or 4
2x(x2—2x —7)
(x + 1)2
(ii) 0 <x < 1 or x> 4
(Hi) x < 0 or 1 <x < 4
2(x — l)(x2 + 4x + 7)
(x + 1)3
(iv) x = 2 or 3± 77
f"(x) =
fv) x<3-77 or 2<x<34-77
(vi) 3-77 <x <2 or x>3 + 77
i
(c) Minimum point:
Maximum point:
(0,0), (4, —32e~*)
(x < 0 and
x*-l)
(x>0)
(x<0 and
x*-l)
2(x-l)(x2+4x + 7)
(x+1)3
(b) f'(0) exists but f'(-l) does not exist.
(l.e-1)
Both f"(0) and f"(-l) do not exist.
Point of inflexion:
i
(3-V7, (-58 + 22V7)e 3-’/2), (2,0),
(c) (i) Minimum point:
Maximum point:
(3 + V7, (-58- 22V7 )e,3+77)
(14-\/8, 13-16^2)
(1-V8, -13-1672)
Point of inflexion: (0, 0), (1, -3)
(d)y = 0
(ii) x = -l
y
(e)
(d)
y
o
i+Vs
:o
►X
304
Advanced Level Pure Mathematics
11. (a) h'(x) = «
—2(2.v2-2a -1)
(1 —2a)3
(a>0 and x^—)
2
1 + a-2
(-l<x<0)
y
(Hi)
’o
12
(1-2a)3
(x>0 and -v^~)
h"(x) =
(-l<x<0)
.(l + -v2)2
h'(0) exists but h"(0) does not exist.
(b) (i)
.
1
-1 < a < — or
2
1<.< 1 +2V3
1+J3
(ii) a* >
2
(b) (ii) 4
2^2
15. ------ a
3
16. (b) 6V2
e
17. (a) 4asin—b 2a cos 0 + 2a
2
a2sin 0( 1 + cos 0)
1
(Hi) -1 < x < 0 or 0 < x < y
2
_l_
(iv)
2
(b)
(c) Maximum point: ( 1 + V3 .-2-V3)
2
... x = -y
1
3
(d)
= -x_-
18. (a) 3^3a 2
y
(e)
(c) 5a
4
19.
1+5/3
(b) 4\/2a 2
2 mr
9V3
2
O
'
►X
12. (a)(iii)b = 0 and a^ 0
y
20. (a)
O
b
a
c
13. (a) (i) If m = 0. there is no turning point.
If ni > 0,
minimum point: (2aw, —16m3+ 16/z)
maximum point: (—2m, 16/n3+ 16/z)
If m < 0,
minimum point: (-2m, 16m3 + 16/z)
maximum point: (2m, -16m3 + 16//)
(b) They meet at the home of the man who
lives between the others.
21. (a)(i)
(H)
....
(-2,
0), (-2-, 0)
0)
2
(ii) Minimum point: (-1,-1)
Maximum point: (-2, 0)
...
14. (a) (i)
xo =
(c2 + J2)-(a2 +b2)
2(c-a)
a
(c<a<x0)
*,=■ *0
(c < x0 < a)
c
(x0<c< «)
r
zb/vHY'r.s
(b) x, = —
+ b2-d2
2a
(e) y
3
(f)
y
3
*1
Fl
\d
>0
a
I*
fill
\3 ; I
' 2 V
3\ \
0
X
y
24. (a)
36. (a)(i)
f'(x)= 2(/ 2x2>-
y=((x)
f(y)
3x3(x2 + 1)2
f(0
1
a
1
(ii) x<—7= or 0<x<-?=-;
V2
V2
pf(x) + qf(y)
A
1
f(x)
1
\/2
x
O
t
x
y
(iii) Minimum point: (0, 0)
£
1 43
Maximum point: (——),
27. (b) fl, = a2 = • • = “n
i
29. (a) Maximum value: f(l) = 0
1 43
\/2 3
(b)x} =x2
2 =
“ ••• =xn= 1
32. (b) \e
(b)(i)
f"(x) =
2(14x4-23x2-l)
4
9x3(.r + l)’
2-3x
34. (a) f'(x) =
Point of inflexion:
2 1
3(l-x)3x3
(-.v0, f(-x0)) = (-1.3, 0.4),
f"(x) =------ =^-T
(x0, f(x0)) = (1.3, 0.4)
9(l-x)3x3
(c) (i)
I23 + V585
where x0=^----- —-----
2
3
(ii) y = 0
(ii) 0<x<|
y
(c)
(iii) x < 0 or |- < x < 1 or x > 1
(iv) f "(x) * 0 for all real values x.
(v) x>l
~?2
(vi) x < 0 or 0 <x< 1
(d) Minimum point:
Maximum point:
(0,0)
2
v3
Point of inflexion: (1,0)
3
J
305
306
Adviincid Level Pure Mathematics
38. (a)(i)
1-x . 2(x-2)
(A-+1)” (X+1)4
(ii)
A-l
2(2-a)
(A+l)” (X+1)4
y
(b) (i) x < -1 or 0 < x < 1
(ii) -1 <x < 0 or x > 1
(iii) x < -1 or -1 < x < 0 or x* > 2
(iv) 0 < x < 2
►X
O
(c) Minimum point:
(0, 0)
y = -x + 5
y = x- 5
Maximum point:
2
Point of inflexion: (0, 0), (2, —)
9
(d)x = -l, y 0
40. (a)(i)
f'(x) =
2-x
2
3x3(x + l)2
f'(0) does not exist.
y
(b)(i) 0<x<2
(ii) x < -1 or -1 < x < 0 or x > 2
1
7
(Hi) -1 < x < a or x > p
-► X
O
1
(a-1)2(a+5).
;
(x + 1)3
39. (a) f'(x)
(iv) x < — 1 or a < x < 0 or 0 < x < P
2
where
24(x-l)
f"U) = (x + 1)4
a=
4-3V2
2
P=
(c) Minimum point:
4.121 3
2
(b) (i) x < -5 or -1 < x < 1 or x > 1
(ii) -5<x<-l
-0.1213,
(0. 0)
(2, f(2)) - (2, 0.529 I)
(iii) x> 1
Maximum point:
(iv) x < -1 or -1 < x < 1
Point of inflexion:
(a, f(a)) = (-0.121 3, 0.278 9),
(P, f(/?)) ~ (4.121 3,0.501 9)
(c) Maximum point:
97
(-5,-y)
(d) x
Point of inflexion: (1,0)
0
-1, y
(e)
(d) x = -1, y = x - 5
y
(e)
y= x- 5
-5
;o
a O
----------- *-x
P
'.2
1
41. (a) (i)
f'(x) =
3x +1
2
1
3x3(x + l)3
(b) Both f'(0) and f'(-l) do not exist.
Answers 307
(c) (i) x<-l or --j<x<0 orx>0
Miscellaneous Exercise A (P. 275)
(ii) -l<x<-|
3.
(Hi) x < -1 or -1 < x < 0
4. (a) -1 + 4
X
x+1
(iv) x>0
3
x+3
(b)(ii) |
2
1 , -—
73 )
(-1
(d) Minimum point:
3
(-1,0)
Maximum point:
3
6.
(b) (iv) V2
7.
2 +^x
3 2
(c) hm xn =-^x,
Point of inflexion: (0, 0)
(e) y
2
2
3
10. (a) 1
(f)
(b)2
y
2_
y-
3
11. (b) f is not differentiable at x = 2
ii
2
,
13. a =—; b = —
7i
n:
15. x= 1
4
X
O
1
17. (a) ee
18. (c)
42. (d)(iii)n
2
3
1 -3r2 2 3-; f"(x) = 12x(x2 -1)
19. (a) f'(x) = 1
(1 + x2)4
(l + x2)3
14 (or above)
43. (ci)0
(b) (i)
1 i1 (b) f'(x) = (l--)e'; f"(x) = — e<
x
x
(c) (i) x < 0 or x > 1
(ii) x>0
-+<-<+
V3
(c) Minimum point:
Maximum point:
( V3’
3^
16
((_L
3V1)
V3’ 16 ’
1.
,
Point of inflexion: (-1, - —), (0, 0),
4
y
A
Asymptote:
(d)
i /
O
V3
(ii) -1 <x<0 or x> 1
(d) Minimum point: (1, e)
(e) y = x + 1
(f)
(b)3
X
y=0
308
Advanced Level Pure Mathematics
(e) (i) g'(0) does not exist
(b) (i) -1 <x< 1 or x> 1
(ii) x<-l
y
(ii)
(iii) x<\ orx>3
/
:
011
1
’T
(iv) 1 <x<3
9(x)
x
(c) Minimum point:
(-U5)
Point of inflexion: (3,5)
21. (a)(i)
\!ab
22. (a)(i)
f'(x) =
(d) a- = 1
4-x
(e)
x3(6-x)3
(b) (i)
y=f(x)
0 <x<4
(ii) x <0 or 4 < x < 6 or x > 6
oT■
(iii) x>6
-► x
3
(iv) x < 0 or 0 < x < 6
(c) Minimum point:
(0, 0)
(4. 32’)
Maximum point:
(f) (i)
g(x) is not differentiable at x = 0
y
(ii)
f x= 1
Point of inflexion: (6, 0)
y = g(x)
(d) y = -a- + 2
8 :
y
(e)
o:
-►x
3
y=f(x)
6
26. (a)(i)
23. (a)(iii)(\)
(3)
x2 +4x4-2
(x + 2)2
X > -1
X < -1, A' * -2
0<P<-V
4’
(2)
f'(x)=J
x2 + 4x + 2
(x + 2)2
(ii) f is not differentiable al -1
I
4
(x + 2)3
"■J
P>JT
(Hi) f"(X) =
43
24. (a) f'(jv) = 2x +
f"(x) = 2 —
8
U-l)2 ’
16
(x-1)3
4
U + 2)3
A' > -1
x < -1, x * -2
(b) (i)
(ii) x<-2-
2 or — 1 < x < —2 + V2
zW/AHt/v 309
(Hi) x < -2 or x > -1
33. (b)3
(iv) -2<x<-l
(c) (ii) u
(c) Minimum point:
(-3.414,5,828),
(-0.585 8,-0.171 6)
Maximum point:
(-1,0)
Point of inflexion: (-1,0)
(d) X = -2, y = a - 1, y = -x + 1
37. (b) lim
1
=-
■
rl
38. (a) n
»
kit
. .
kit
(b) x = cos— + /sin —
n
fi
where k = 0, 1,2,..., (2n ~ 1)
(e)
(-3.414, 5.828) X J
(-0.585 8. -0.171 6)
I
28. — (0-sin0)
4
29. (a)
1
3
(b) 1
30. (a) 2
31.
(b) 1
32. (b) f is not injective
y
(c)
y=f(x+ 1)
c
-1
1
O
■o
(d)
y
A
2
o-
y = g(*)
1
o
■o
o
O-
1
-2
g(x) is discontinuous at x= 1 and x = -l
r
Advanced Level Pure Mathematics
311
Index
A
180
Absolute extreme
Absolute maximum
183, 216
Absolute minimum
183. 216
126, 180,
126, 180,
Applications of differential calculus f^^65ffiffl
194
concavity El'14
curve sketching
228
228
linear approximation ^'I4igifi 165
local extrema
216
optimal value
223
point of inflexion
230, 231
rate of change >®(byi^ 206
relative rate of change
210
tangent
204
Asymptote
233
horizontal asymptote
233, 236
oblique asymptote
233, 236,
237, 239, 240
slant asymptote
233, 236
vertical asymptote 36j£233
B
Basic elementary functions ®$1WlSI®
40
constant function
40
exponential function
41
inverse circular functions SH^lSSefc 43
logarithmic function
power function
trigonometric functions
42
40
42
Bijective function
21, 22
Bolzano Theorem
127
Bounded WVr 31, 78
bounded from above #±Vr 30, 78
bounded from below WT5?- 30, 78
lower bound
30
upper bound ±5?- 30
Bounded from above
30, 78
Bounded from below
30, 78
Bounded function
30
bounded W 5? 31
bounded from above
30
bounded from below
30
lower bound
30
upper bound ±5?- 30
c
Chain Rule J®
153
Codomain
5
Composition of functions ftl'aSl® 9, 13
Concave downward T El 65 228
Concave upward _t 03 65 228
Concavity El'14 228
concave downward T El 65 228
concave upward Jt El 6tl 228
convex downward T £h 65 228
convex upward Jb dh 65 228
Constant function Itt’Sd-S® 40, 187
Continuity i®££tl4 94, 144
Continuous functions
115, 121, 126
Bolzano Theorem
127
Intermediate Value Theorem
128
properties i®?flkfill&65'l4K 120
312
.\</i <///( < </ level Pure Muihcmalies
Convergent
56, 78
Convergent sequence
56. 78. 80
Convex downward T fl fi
228
Convex upward _t fl fit) 228
Critical point EhiTr-Si 217
Curve sketching
228
D
Dependent variable J®^lii i
Derivative
148
first derivatives —167
167
higher derivatives
infinite derivatives
139
left hand derivatives
141
141
right hand derivatives
Derivative of an inverse function
Si® fit)*®®
155
Differentiability W14 139, 144
Differentiable nj{^ 134, 139
Differentials
163
errors
166, 312
rules of operations
164
Differentiation
149
differentiation of functions in parametric form
156
derivative of an inverse function
® 155
implicit differentiation
158
logarithmic differentiation
159
rules of differentiation
148
Differentiation of functions in parametric form
156
Direct image
4. 7
Discontinuity
118
finite oscillartory discontinuity W
S5 118
first kind discontinuity
infinite discontinuity
jump discontinuity MSfiWW®
117
118
117
removable discontinuity nJi|uj®fSS 116,
117
second kind discontinuity M—
118
Diverge to infinity
57
Divergent KiK 57
Divergent sequence
57
diverge to infinity
57
Domain
1
E
Elementary functions ^JJ^rlSlSS 40, 121
Errors
166
Even function
27, 28, 30
Exponential function
41
Extreme point ffittSS 217
Extremum
216
F
Fermat's Theorem 3SiWSS/ES 180, 183
Finite oscillatory discontinuity W ^^WjIrIBtSS
118
First derivative —167
First kind discontinuity
117
Function Si® 1
bijective function StMSl® 21, 22
codomain _tM 5
composition of functions ?SeJS|® 9, 13
constant function <8’®S1® 40
dependent variable )®®fi 1
direct image
4, 7
domain
1
elementary functions $Z)^fS|® 40
exponential function Ila® Si® 41
identity function ®3?S|® 14
image (£’ 1
independent variable
1
injective function I^JJ^fSl® 17
Index
inverse circular functions
inverse function
14, 24
inverse image
4, 6
logarithmic function
42
43
Inverse circular functions KH#1
principal values
44
Inverse function
14, 24
Inverse image $.(%. 4, 6
43
most extensive possible domain
SIM 3
natural logarithm 1=1
piecewise function
power function
pre-image
1
range (MM 1
real functions
J
89
45
40
Jump discontinuity
23
L
smallest set of range JVhffiM 3
surjective function
19
trigonometric functions
42
Least value Jft/hdI
Left hand derivative
Left hand limit
G
Greatest integer function
Greatest value
126, 216
46
H
Higher derivatives
167
233, 236
Horizontal asymptote
Identity function
14
Image
1
Implicit differentiation
158
Increment infi 134
Independent variable [il
Indeterminate form
1
68, 194
Infinite derivatives
Infinite discontinuity
Infinitely small
139
118
57
Injective function
Intermediate Value Theorem
17
128
117
126, 216
141
113
Leibniz’s Theorem
Limit
55, 95, 194
convergent sequence
indeterminate form T'aeS
172
56
68, 194
95, 96
limits of functions
limits of sequences
55
Limits of functions
95, 96
indeterminate form T'/eS! 194
left hand limit
113
operations on limits
100
right hand limit
113
sandwich theorem for sequences ®
105
Limits of sequences J??1] W ffi ® 55
divergent
57
indeterminate form T'zeS 68
infinitely small MUPb 57
operations on limits
60
sandwich theorem for sequences
71
Linear approximation
165
Local extrema
216
local maximum
216
local minimum
216
second derivative test —
221
sign test
217
313
314 Advanced Level Pure Miidiemuncs
g|J®
Local maximum
fj*[
216
Local maximum value
p
216
Local minimum
216
Local minimum value /linPffid'titt
Period fflM
216
Logarithmic differentiation
Periodic function
natural logarithm til
L'Hospital’s Rule
37, 38
37, 38
42
Piecewise function 5)*®1S|®
42
Point of inflexion ISSA
230. 231
Power function SS|®(
40
30
Lower bound T ??•
period iSffl
159
Logarithmic function WMlS®
37, 38
194, 198. 200
Pre-image
45
1
Principal values if61
44
M
R
Mapping
1
Range (OJt
216
Maximum
Mean Value Theorem
Minimum Sid'fit
180, 184
216
Monotonic functions
34, 212
1
Rate of change
206
Real functions
23
bounded function W^-lSlSS
30
monotonically decreasing
34
constant function
40
monotonically increasing
34
elementary functions
|2|®
even function (Sl£l®
27, 28. 30
78
Monotonic sequence
40
Monotonically decreasing
34
exponential function
Monotonically increasing
34
greatest integer function
46
inverse circular functions
43
Most extensive possible domain
M 3
logarithmic function
42
monotonic functions
34
odd function pflS®
N
41
30
37, 38
periodic function
power function $£lS®
42, 89
Natural logarithm
trigonometric functions
Relative maximum #1Wl?E
o
40
42
(61
216
Relative minimum ftl S l?Sds (61
216
210
Relative rate of change
233, 236, 237,
Oblique asymptote
239, 240
Odd function
Oscillating sequences
116,
Right hand derivative
30
Operations on limits
Optimal value
Removable discontinuity
117
•60, 100
Right hand limit
Rolle’s Theorem
223
59
Rules of differentiation
141
113
183, 184
148
Index
s
Sandwich theorem for functions iSMMiMiSizE
105
91
Sandwich theorem for sequences
91
71
Second derivative test —
221
Second kind discontinuity
Sequences
55
bounded
78
118
convergent sequence
56, 78, 80
divergent sequence
57
monotonic sequence
78
oscillating sequences
59
Sign test
217
233. 236
Slant asymptote
Smallest set of range
3
217
Stationary point
34, 36, 212,
Strictly decreasing
214
Strictly increasing
34, 36, 212, 213
Surjective function
19
Tangent WS
204
Trigonometric functions
Turning point WfaSfi
42
217
u
Upper bound _h?r
30
V
Vertical asymptote
233
315
About the consultant
Chbuiig Liu Mb (
> If% >
aJ
He was graduated from the Department of Mathematics of Guangzhou Zhongshan
University with distinctions, and completed post-graduate studies at Whuhan
University, specializing in Partial Differential Equations. In addition, he has
received the Eduacation Diploma (distinction) from the Hong Kong Chinese
University.
Prior to 1979, he served as a lecturer at the Department of Mathematics of Guangxi University
and was the Deputy Panel Chairman on the subject of Differential Equations.
Cheung Siu Ho has also taugh Mathematics, Additional Mathematics, Higher Mathematics and
Pure Mathematics at Heung To Middle school in Hong Kong for over 20 years while serving as
the Panel Chairman of Mathematics and the Prefect of Studies.
In addition to his teaching experiences, he has authored a number of published text books and
reference books as below:
(I)
Listed in the table of recommended text books of the Curriculum Development Committee
of Hong Kong
(1)
Certificate Mathematics
1986
Chiu Ming & Oxford
Cheung Siu Ho
(coordinator)
1989
(2)
(M~)®)
(3)
1997
uw
(II) Reference books
(4)
Certificate Mathematics [MC] (
) 1982
(5)
Solution to Higher Mathematics
1984
(6)
Certificate Mathematics [MC]
1985
(7)
(8)
56 & (±l§)
56 & (±^)
Chiu Ming
55
2002
WW
2005
56 -a
(±S§)
is an experienced Mathematics teacher in Hong
Kong for over 20 years. He holds a bachelor degree from Hong
Kong University and a master from University of Bristol. Over the
years, he had been frequently awarded certificates for "Outstanding
Professional Performance" from the Director of Education. Moreover,
he has been publishing a number of textbooks and reference books
in Additional Mathematics and Pure Mathematics, which are among the several acclaimed
Mathematics books in Hong Kong secondary schools.
is a Senior Lecturer in the Department of Mathematics, the Hong
Kong University of Science and Technology. He holds a bachelor degree in
Engineering from Hong Kong University, masters and doctoral
degrees in Applied Mathematics from Brown University. He has
received a number of teaching awards in his teaching career. Other
books authored by Kwok include "Applied Complex Variables"
published by Cambridge University Press and "Mathematical Models
of Financial Derivatives" published by Springer Verlag.
ISBN 962-354-177-5
Hung Fung Book Co., Ltd.
9 789623 541770