PURE MATHEMATICS Unit 2
FOR CAPE® EXAMINATIONS
DIPCHAND BAHALL
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Contents
INTRODUCTION
ix
MODULE 1 COMPLEX NUMBERS AND CALCULUS II
CHAPTER 1
COMPLEX NUMBERS
2
Complex numbers as an extension to the real numbers
3
Powers of i
4
Algebra of complex numbers
Addition of complex numbers
5
Subtraction of complex numbers
5
Multiplication of a complex number by a real number
5
Multiplication of complex numbers
5
Equality of complex numbers
6
Conjugate of a complex number
7
Division of complex numbers
8
Square root of a complex number
9
Roots of a polynomial
CHAPTER 2
5
11
Quadratic equations
11
Other polynomials
13
The Argand diagram
15
Addition and subtraction on the Argand diagram
15
Multiplication by i
16
Modulus (length) of a complex number
16
Argument of a complex number
17
Trigonometric or polar form of a complex number
19
Exponential form of a complex number
21
De Moivre’s theorem
22
Locus of a complex number
27
Circles
27
Perpendicular bisector of a line segment
28
Half-line
29
Straight line
30
Inequalities
31
Intersecting loci
33
Cartesian form of loci
35
DIFFERENTIATION
41
Standard differentials
42
Differentiation of ln x
Differentiation of e
x
42
43
iii
Chain rule (function of a function rule)
Differentiating exponential functions of the form y =
43
ax
Differentiating logarithms of the form y = loga x
47
Differentiation of combinations of functions
50
Differentiation of combinations involving
trigonometric functions
51
Tangents and normals
54
Gradients of tangents and normals
54
Equations of tangents and normals
56
Implicit differentiation
58
Differentiation of inverse trigonometric functions
62
Differentiation of y = sin−1x
62
tan−1x
63
Differentiation of y =
Second derivatives
65
Parametric differentiation
67
First derivative of parametric equations
67
Second derivative of parametric equations
70
Partial derivatives
CHAPTER 3
CHAPTER 4
72
First order partial derivatives
72
Second order partial derivatives
73
Applications of partial derivatives
74
PARTIAL FRACTIONS
84
Rational fractions
85
Proper fractions: Unrepeated linear factors
85
Proper fractions: Repeated linear factors
88
Proper fractions: Unrepeated quadratic factors
91
Proper fractions: Repeated quadratic factors
93
Improper fractions
94
INTEGRATION
99
Integration by recognition
100
When the numerator is the differential of the denominator
104
∫
The form ∫f ′(x)ef(x)dx
The form f ′(x)[ f(x)]n dx, n ≠ −1
iv
46
105
106
Integration by substitution
108
Integration by parts
112
Integration using partial fractions
116
Integration of trigonometric functions
121
Integrating
sin2 x
and
cos2 x
Integrating
sin3 x
123
and
cos3 x
123
Integrating powers of tan x
125
Integrating products of sines and cosines
125
Finding integrals using the standard forms
x
1
_______
dx = sin−1 ( __
∫________
a) + c
2
2
√a − x
∫
CHAPTER 5
x
1
−1 __
1 dx = __
and ______
a tan ( a ) + c
a2 + x2
REDUCTION FORMULAE
∫
Reduction formula for ∫cosn x dx
Reduction formula for ∫tann x dx
Reduction formula for sinn x dx
CHAPTER 6
126
136
137
138
139
Other reduction formulae
139
TRAPEZOIDAL RULE (TRAPEZIUM RULE)
145
The area under a curve
146
MODULE 1 TESTS
153
MODULE 2 SEQUENCES, SERIES AND APPROXIMATIONS
CHAPTER 7
CHAPTER 8
SEQUENCES
156
Types of sequence
157
Convergent sequences
157
Divergent sequences
157
Oscillating sequences
157
Periodic sequences
158
Alternating sequences
158
The terms of a sequence
158
Finding the general term of a sequence by identifying
a pattern
160
A sequence defined as a recurrence relation
161
Convergence of a sequence
162
SERIES
167
Writing a series in sigma notation (Σ)
168
Sum of a series
169
Sum of a series in terms of n
Method of differences
176
Convergence of a series
180
Tests for convergence of a series
CHAPTER 9
172
181
PRINCIPLE OF MATHEMATICAL INDUCTION (PMI):
SEQUENCES AND SERIES
185
PMI and sequences
186
PMI and series
190
v
CHAPTER 10
BINOMIAL THEOREM
196
Pascal’s triangle
197
Factorial notation
197
Combinations
199
General formula for nCr
Binomial theorem for any positive integer n
The term independent of x in an expansion
CHAPTER 11
CHAPTER 13
200
203
Extension of the binomial expansion
205
Approximations and the binomial expansion
208
Partial fractions and the binomial expansion
209
ARITHMETIC AND GEOMETRIC PROGRESSIONS
215
Arithmetic progressions
216
Sum of the first n terms of an AP
218
Proving that a sequence is an AP
220
Geometric progressions
CHAPTER 12
199
224
Sum of the first n terms of a GP (Sn)
227
Sum to infinity
229
Proving that a sequence is a GP
230
Convergence of a geometric series
231
NUMERICAL TECHNIQUES
240
The intermediate value theorem (IMVT)
241
Finding the roots of an equation
241
Graphical solution of equations
242
Interval bisection
242
Linear interpolation
243
Newton–Raphson method for finding the roots of an
equation
247
POWER SERIES
255
Power series and functions
256
Taylor expansion
256
The Maclaurin expansion
259
Maclaurin expansions of some common functions
MODULE 2 TESTS
265
267
MODULE 3 COUNTING, MATRICES AND
DIFFERENTIAL EQUATIONS
CHAPTER 14
vi
PERMUTATIONS AND COMBINATIONS
270
The counting principles
271
Multiplication rule
271
Addition rule
271
Permutations
Permutations of n distinct objects
272
Permutation of r out of n distinct objects
274
Permutations with repeated objects
275
Permutations with restrictions
277
Permutations with restrictions and repetition
281
Combinations
Combinations with repetition
CHAPTER 15
285
287
PROBABILITY
294
Sample space and sample points
295
Events: mutually exclusive; equally likely
296
Probability
296
Rules of probability
298
Conditional probability
CHAPTER 16
272
299
Tree diagrams
302
Probability and permutations
307
Probability and combinations
311
MATRICES
321
Matrices: elements and order
322
Square matrices
322
Equal matrices
323
Zero matrix
323
Addition and subtraction of matrices
323
Multiplication of a matrix by a scalar
324
Properties of matrix addition
324
Matrix multiplication
325
Properties of matrix multiplication
328
Identity matrix
328
Multiplication of square matrices
329
Transpose of a matrix
Properties of the transpose of a matrix
Determinant of a square matrix
330
330
331
Determinant of a 2 × 2 matrix
331
Determinant of a 3 × 3 matrix
331
Properties of determinants
332
Singular and non-singular matrices
334
Solving equations using determinants (Cramer’s rule)
335
Using Cramer’s rule to solve three equations in three
unknowns
Inverse of a matrix
Inverse of a 2 × 2 matrix
337
339
339
vii
Cofactors of a 3 × 3 matrix
339
Inverse of a 3 × 3 matrix
341
Properties of inverses
343
Systems of linear equations
343
Row reduction to echelon form
346
Finding the inverse of a matrix by row reduction
348
Solving simultaneous equations using row reduction
352
Systems of linear equations with two unknowns
355
Intersecting lines
355
Parallel lines
356
Lines that coincide
357
Systems of linear equations with three unknowns
358
Unique solution
358
No solutions
361
Infinite set of solutions
362
Solution of linear equations in three unknowns: geometrical
interpretation
365
CHAPTER 17
Applications of matrices
367
DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELLING
380
First order linear differential equations
381
Practical applications
Second order differential equations
388
When the roots of the AQE are real and equal
388
When the roots of the AQE are real and distinct
389
When the roots of the AQE are complex
390
Non-homogeneous second order differential equations
392
When f(x) is a polynomial of degree n
393
When f(x) is a trigonometric function
398
When f(x) is an exponential function
403
Equations reducible to a recognisable form
405
Mathematical modelling
417
MODULE 3 TESTS
421
UNIT 2—MULTIPLE CHOICE TESTS
424
INDEX
441
Answers are available online at
www.macmillan-caribbean.com/resources
viii
385
Introduction
These two volumes provide students with an understanding of pure mathematics
at the CAPE level taken from both a theoretical and an application aspect and
encourage the learning of mathematics. They provide the medium through
which a student can find problems applied to different disciplines. The concepts
are developed step by step; they start from the basics (for those who did not do
additional mathematics) and move to the more advanced content areas, thereby
satisfying the needs of the syllabus. Examination questions all seem to have answers
that are considered ‘nice’ whole numbers or small fractions that are easy to work
with; not all real-world problems have such answers and these books have avoided
that to some extent. Expect any kind of numbers for your answers; there are no
strange or weird numbers.
The objectives are outlined at the beginning of each chapter, followed by the
keywords and terms that a student should be familiar with for a better understanding
of the subject. Every student should have a section of their work book for the
language of the subject. I have met many students who do not understand terms such
as ‘root’ and ‘factor’. A dictionary developed in class from topic to topic may assist the
students in understanding the terms involved. Each objective is fulfilled throughout
the chapters with examples clearly explained. Mathematical modelling is a concept
that is developed throughout, with each chapter containing the relevant modelling
questions.
The exercises at the end of each section are graded in difficulty and have adequate
problems so that a student can move on once they feel comfortable with the concepts.
Additionally, review exercises give the student a feel for solving problems that are
varied in content. There are three multiple choice papers at the end of each Unit,
and at the end of each module there are tests based on that module. For additional
practice the student can go to the relevant past papers and solve the problems given.
After going through the questions in each chapter, a student should be able to do past
paper questions from different examining boards for further practice.
A checklist at the end of each chapter enables the student to note easily what is
understood and to what extent. A student can identify areas that need work with
proper use of this checklist. Furthermore, each chapter is summarised as far as
possible as a diagram. Students can use this to revise the content that was covered in
the chapter.
The text provides all the material that is needed for the CAPE syllabus so that
teachers will not have to search for additional material. Both new and experienced
teachers will benefit from the text since it goes through the syllabus chapter by
chapter and objective to objective. All objectives in the syllabus are dealt with
in detail and both students and teachers can work through the text comfortably
knowing that the content of the syllabus will be covered.
ix
1
Complex Numbers and
Calculus II
1
M O DUL E 1
CHAPTER 1
Complex Numbers
At the end of this chapter you should be able to:
■ express complex numbers in the form a + ib, a, b, ∊ ℝ
■ calculate the square root of a complex number
■ carry out the algebra of complex numbers (add, subtract, multiply, divide)
■ calculate the complex roots of a polynomial
■ find the modulus of a complex number
■ find the argument of a complex number
■ understand the properties of modulus and argument
■ interpret the modulus and argument of complex numbers
■ represent complex numbers on an Argand diagram (including sums,
differences, products and quotients)
■ identify and sketch loci on an Argand diagram for ∣ z − c ∣ = r, c ∊ ℂ and r ∊ ℝ
■ identify and sketch loci on an Argand diagram for ∣ z − a ∣ = ∣ z − b ∣, a, b ∊ ℂ
■ identify and sketch loci on an Argand diagram for arg (z − a) = θ, a ∊ ℂ and θ
in radians
■ identify and sketch loci on an Argand diagram for z = a + λb, a, b ∊ ℂ and λ ∊ℝ
■ convert a locus to Cartesian form.
■ use de Moivre’s theorem for n ∊ ℤ+
■ establish that eiθ = cos θ + i sin θ
KEYWORDS/TERMS
OBUVSBMOVNCFSTtJOUFHFSTtSBUJPOBMOVNCFSTt
SFBMOVNCFSTtJNBHJOBSZOVNCFSTtDPNQMFY
OVNCFStDPOKVHBUFtDPNQMFYSPPUTt"SHBOE
EJBHSBNt$BSUFTJBOGPSNtNPEVMVTtBSHVNFOUt
QPMBSGPSNtFYQPOFOUJBMGPSNtMPDVT
2
MODULE 1tCHAPTER 1
Complex numbers as an extension to
the real numbers
The number system began with the set ℕ of natural numbers which are used for
counting. The set of natural numbers is ℕ = {1, 2, 3, 4, 5, . . .}. This system was later
extended to the set of integers ℤ = {0, ±1, ±2, ±3, . . .}.
With the set of natural numbers and integers only, division is not always possible. If
1 which is not possible if x ∊ ℤ.
we try to solve the equation 2x − 1 = 0, we get x = __
2
The set of integers can be extended to the set of rational numbers ℚ where
m
ℚ = {__
n : m ∊ ℤ, n ∊ ℤ, n ≠ 0}. With this extension all equations of the form
ax + b = 0 (b ≠ 0) have solutions.
The set of rational numbers ℚ is not enough to solve equations such as x2 = 2 and
hence the number system is extended further to the set of real numbers ℝ. Each extension of the number system allows us to solve equations that are otherwise unsolvable.
One of the properties of a real number is that its square cannot be negative, e.g. there
is no real number which satisfies the equation x2 = −1. To allow us to solve equations of this form, the set of imaginary numbers was introduced. The properties of
the imaginary numbers are similar to those of the real numbers. The imaginary unit,
denoted by i, is such that i2 = −1.
With the introduction of i a new number system was developed called the complex
number system. The set of complex numbers is denoted by ℂ and, in general, a
complex number in Cartesian form can be written as z = x + iy where x, y ∊ ℝ. x is
called the real part of the complex number z, and is denoted by Re(z), and y is the
imaginary part, denoted by Im(z). Note that the imaginary part of z does not include i.
The set of real numbers is a subset of the set of complex numbers, i.e. every real
number is a complex number (with imaginary part zero), and ℝ ⊂ ℂ. By working
with the set of complex numbers we can now solve all quadratics with real coefficients. For example an equation in the form x2 + x + 1 = 0 can be solved as follows:
___
_____
__ ___
__
−1 ± √−3 = ____________
−1 ± √3 √−1 = _________
−1 ± √3 i, since i 2 = −1
−1 ± √1 − 4 = __________
x = ____________
2
2
2
2
2
This gives the two solutions of the quadratic equation x + x + 1 = 0.
Complex numbers are widely used in fields such as applied mathematics, quantum
physics and engineering.
EXAMPLE 1
Identify the real and imaginary parts of the complex numbers.
(a) 2 + 3i
(b) 4 − 2i
(c) 4x + 5yi
(d) 3x2 + (6x + y)i
(e) cos θ + i (2 sin θ) (f) (4 cos θ)i − (3 sin θ) i + 2 cos2 θ
SOLUTION
(a) Let z = 2 + 3i
Re(z) = 2
Im(z) = 3
The imaginary part is the coefficient of i.
(b) z = 4 − 2i
Re(z) = 4
Im(z) = −2
3
M O DUL E 1
Remember
Re (x + iy) = x,
Im (x + iy) = y
(c) z = 4x + 5yi
Re(z) = 4x
Im(z) = 5y
(d) z = 3x2 + (6x + y)i
Re(z) = 3x2
Im(z) = 6x + y
(e) z = cos θ + i(2 sin θ)
Re(z) = cos θ
Im(z) = 2 sin θ
(f) z = (4 cos θ)i − (3 sin θ)i + 2 cos2 θ
Rearranging
z = 2 cos2 θ + i(4 cos θ − 3 sin θ)
Re(z) = 2 cos2 θ
Im(z) = 4 cos θ − 3 sin θ
Powers of i
___
Since i = √−1 we get i 2 = −1
Notice we are
back to i.
So i 2 × i = −1 × i
⇒ i3 = −i
i × i3 = i × −i
⇒ i4 = −i2
but i2 = −1
⇒ i4 = −(−1) ⇒ i4 = 1
i × i4 = i × 1
⇒ i5 = i
Thus all powers of i can be written as 1, i, −i or −1.
For example
i10 = (i2)5 = (−1)5 = −1
i21 = i20 × i = (i2)10 × i = (−1)10 × i = i
EXAMPLE 2
SOLUTION
Identify the real and imaginary parts of the following complex numbers.
(a) 2 + 3i2 − 4i
(b) 5i + 3i3 − 4i2 + 2
(c) 7i4 + i2 − 8i3 + 8
(d) xi + yi2 + yxi2 + y3i
(a) Recall i2 = −1
So
2 + 3i2 − 4i = 2 + 3(−1) − 4i
= 2 − 3 − 4i
= −1 − 4i
Re(−1 − 4i) = −1, Im(−1 − 4i) = −4
(b) Rearrange as 2 + 5i − 4i2 + 3i3
= 2 + 5i − 4(−1) + 3(−i)
= 2 + 4 + 5i − 3i
= 6 + 2i
Re(6 + 2i) = 6, Im(6 + 2i) = 2
4
Since i2 = −1, i3 = i2 × i = −1 × i = −i
MODULE 1tCHAPTER 1
(c) Rearrange as 8 + i2 − 8i3 + 7i4
= 8 − 1 − 8(−i) + 7(1)
since i3 = −i, i4 = (−i)i = −i2 = 1
= 14 + 8i
Re(14 + 8i) = 14, Im(14 + 8i) = 8
(d) (x + y3)i + (y + yx)i2 = (x + y3)i + (y + yx)(−1)
= (x + y3)i − (y + yx)
Re((x + y3)i − (y + yx)) = −y − yx, Im((x + y3)i − (y + yx)) = x + y3
Try these 1.1
(a) Identify the real and imaginary parts of the following.
(i) 5 + 4i
(ii) 4 + 7i
(iv) 7x2 + y + i (3x − 2y)
(iii) 5x + i (3xy)
(v) 7i2 − 4i
(b) Identify the real and imaginary parts of the following.
(ii) (cos θ)i + sin θ
4xi + 3yi − 2x
(i)
(iii) 4 sin θ − (3 cos θ)i
(v)
(iv) 8 cos2 θ + 7 cos θ + i sin3 θ − i sin4 θ
8 cos2 θ i2 + 7 sin3 θ i3 + 4i4 cos 2θ + 7 sin θ
Algebra of complex numbers
Let
Add real to real
and imaginary to
imaginary.
z1 = a + bi
and
z2 = c + di
Addition of complex numbers
z1 + z2 = a + bi + c + di = (a + c) + (bi + di)
= (a + c) + (b + d)i
For example
2 + 3i + 4 + 5i = 2 + 4 + 3i + 5i
= 6 + 8i
Subtraction of complex numbers
Subtract real
from real and
imaginary from
imaginary.
z1 − z2 = (a + bi) − (c + di) = (a − c) + bi − di
= a − c + (b − d)i
Multiplication of a complex number by a real number
λz1 = λ(a + bi) = λa + λbi, λ ∊ ℝ
Multiplication of complex numbers
z1 z2 = (a + bi) (c + di)
= ac + adi + bci + bdi2 (expanding the brackets)
= ac + adi + bci − bd
(i2 = −1)
= ac − bd + i (ad + bc)
5
M O DUL E 1
EXAMPLE 3
Given that z1 = 3 + 2i and z2 = 2 + 4i, find z1 z2.
SOLUTION
z1 z2 = (3 + 2i) (2 + 4i)
= 6 + 12i + 4i + 8i2
(expanding the brackets)
= 6 − 8 + 16i
(substituting i2 = −1)
= −2 + 16i
Hence z1 z2 = −2 + 16i
Equality of complex numbers
z1 = z2 ⇒ a + bi = c + di
⇒ a = c, b = d
Two complex numbers are equal if and only if the real parts and the imaginary parts
are equal.
EXAMPLE 4
Given that z1 = 2 + 3i and z2 = 2x − yi, find the value of x and the value of y for
which z1 = z2.
SOLUTION
Since z1 = z2
2 + 3i = 2x − yi
Equating real and imaginary parts
2x = 2 ⇒ x = 1
−y = 3 ⇒ y = −3
Hence x = 1, y = −3
EXAMPLE 5
Given that z1 = 3 + 2i, z2 = 1 − i and z3 = 4 + 6i, find
(a) z1 + z2 (b) z1 − 2z2 (c) z1 z2
SOLUTION
(d) z1 z3 (e) z2 z3
(a) z1 + z2 = 3 + 2i + 1 − i = (3 + 1) + (2i − i)
=4+i
(b) z1 − 2z2 = 3 + 2i − 2(1 − i)
(multiply z2 × 2)
= 1 + 4i
(c) z1 z2 = (3 + 2i) (1 − i)
= 3 − 3i + 2i − 2i2
(expanding the brackets)
= 3 + 2 − 3i + 2i
(i2 = −1)
=5−i
(d) z1 z3 = (3 + 2i) (4 + 6i)
= 12 + 18i + 8i + 12i2
= 12 − 12 + 18i + 8i
= 26i
6
MODULE 1tCHAPTER 1
(e) z2 z3 = (1 − i)(4 + 6i)
= 4 + 6i − 4i − 6i2
(i2 = −1 ⇒ −6i2 = −6(−1) = 6)
= 4 + 6 + 6i − 4i
= 10 + 2i
Conjugate of a complex number
_
If z = a + bi, then the conjugate of z, denoted by z* or z , is defined by z* = a − bi.
Here are some complex numbers and their conjugates:
Note
We change
the sign of the
imaginary part.
Complex number
Conjugate
z = −2 + 3i
z = −2 − 3i
z = −4 − 5i
z = −4 + 5i
z = 2x + 3yi
z = 2x − 3yi
_
_
_
Properties of the conjugate
Let z = a + bi, z* = a − bi
(i) z z* = a2 + b2
Proof: z z* = (a + bi) (a − bi)
= a2 − abi + abi − b2i2
= a2 − b2 (−1)
(since i2 = −1)
= a2 + b2
A complex number multiplied by its conjugate is the real part squared plus the
imaginary part squared.
(ii) z + z* = 2Re(z)
Proof: z + z* = a + bi + a − bi
= 2a
= 2Re(z)
A complex number plus its conjugate is twice the real part of the number.
(iii) z − z* = 2Im(z)i
Proof: z − z* = a + bi − (a − bi)
= a − a + bi + bi
= 2bi
= 2Im(z)i
A complex number minus its conjugate is twice the imaginary part multiplied by i.
EXAMPLE 6
Given that z1 = 3 + 4i, z2 = 2 − 3i, find
(a) z1 + z1*
(b) z1 − z1*
(e) z1* z2
(f) z1 z2
(c) z1 z1*
(d) z2 z2*
7
M O DUL E 1
SOLUTION
(a) Since z1 = 3 + 4i, z1* = 3 − 4i
z1 + z1* = 3 + 4i + 3 − 4i = 6
(b) z1 − z1* = 3 + 4i − (3 − 4i)
= 4i + 4i = 8i
(c) z1 z1* = (3 + 4i) (3 − 4i)
= 9 − 12i + 12i − 16i2
= 9 + 16 (since i2 = −1)
= 25
(d) z2 z2* = (2 − 3i) (2 + 3i) = 22 − 9i2 = 13 (since i2 = −1)
(e) z1* z2 = (3 − 4i) (2 − 3i)
= 6 − 9i − 8i + 12i2
= 6 − 12 − 17i
(since i2 = −1)
= −6 − 17i
(f) z1 z2 = (3 + 4i) (2 − 3i)
= 6 − 9i + 8i − 12i2
= 18 − i
(i2 = −1, so −12i2 = −12(−1) = 12)
Division of complex numbers
When dividing, we make the denominator real by multiplying the numerator and
denominator by the conjugate of the denominator.
For example let z1 = a + bi, z2 = c + di
z1 ______
__
= a + bi
z2
Note
Do not learn this
as a formula.
8
c + di
Multiplying the numerator and denominator by the conjugate of the denominator
we have
z1 ______
c − di
a + bi _____
__
(the conjugate of c + di is c − di)
z2 = c + di × c − di
ac − adi + bci − bdi2
= __________________
c2 + d2
ac + bd − i(ad − bc)
(i2 = −1)
= __________________
c2 + d2
i (ad − bc)
ac + bd − _________
(the real and imaginary parts are separated)
= _______
2
2
c +d
c2 + d2
EXAMPLE 7
z1
Given that z1 = 2 + 4i and z2 = 1 − i, find __
z.
SOLUTION
z1 ______
__
= 2 + 4i
2
z2
1−i
Multiplying the numerator and denominator by the conjugate of the denominator,
(1 + i), we have
MODULE 1tCHAPTER 1
Remember
The conjugate of
x + iy is x − iy.
z1 ______
1+i
2 + 4i × _____
__
z =
2
1−i
1+i
2 + 2i + 4i + 4i2 (i2 = −1)
= ______________
1+1
−2 + 6i
2
−
4
+ 6i = _______
= _________
2
2
= −1 + 3i
EXAMPLE 8
SOLUTION
1 + 2i, find the real and imaginary part of
If z = ______
2−i
(a) z2
(b) z − __1z
1 + 2i
(a) z = ______
2−i
To write z in the form x + iy multiply the numerator and denominator by the
conjugate of the denominator (2 + i):
1 + 2i × _____
2+i
z = ______
2−i
2+i
2 − 2 + 5i = __
2 + i + 4i + 2i2 = _________
5i = i
= ______________
5
5
5
2
2
z = i = −1 = −1 + 0i
So Re (z2) = −1 Im (z2) = 0
(b) z − __1z = i − __1i
i2 − 1 = ___
−2
= _____
i
i
−2i = ____
−2i
−2 × _i = ____
= ___
−1
i
i
i2
= 2i
1 = 0 Im z − __
1 =2
So Re z − __
z
z
(
)
(
)
Square root of a complex number
_______
Remember
Two complex
numbers are
equal if and only
if their real parts
are equal and
their imaginary
parts are equal.
To express the number √(a + bi) in the form x + yi where x and y are real,
_______
let √(a + bi) = x + iy
Squaring both sides ⇒ a + bi = (x + iy)2
a + bi = x2 + i(2xy) + i2y2
a + bi = x2 − y2 + i(2xy)
(i 2 = −1)
Equating real and imaginary parts
⇒ x2 − y2 = a
[1]
2xy = b
[2]
To find x and y solve the simultaneous equations.
9
M O DUL E 1
________
EXAMPLE 9
Express √(5 + 12i) in the form x + iy.
SOLUTION
Let √(5 + 12i) = x + iy
________
5 + 12i = (x + iy)2
5 + 12i = x2 + i(2xy) + i2y2
5 + 12i = x2 − y2 + i(2xy)
Equating real and imaginary parts, we have
x2 − y2 = 5
[1]
2xy = 12
[2]
6
12 = __
y = ___
2x x
Substitute into [1]
6 2=5
x2 − __
x
36 = 5
2
x − ___
x2
36 × x2 = 5 × x2
2
× x ⇒ x2 × x2 − ___
x2
x4 − 5x2 − 36 = 0
From [2]
( )
Let a = x2
a2 − 5a − 36 = 0
So (a − 9) (a + 4) = 0
a = 9 or a = −4
So x2 = 9 or x2 = −4
Since x ∊ℝ, x2 = 9
__
x = ±√9
Note
________
Recall √(5 + 12i)
≡ x + iy
Try this 1.2
x = 3 or −3
6=2
When x = 3, y = __
3
6 = −2
When x = −3, y = −__
3
________
So √(5 + 12i) = 3 + 2i or −3 − 2i
_______
Express √(3 − 4i) in the form x + iy.
EXERCISE 1A
1
Let z1 = 2 + 4i and z2 = 3 + 5i. Express the following in the form x + iy.
z1
(a) z1 + z2
(b) z1 − z2
(c) z1 z2
(d) __
z2
In questions 2–5, evaluate the expressions in a + bi form, given that
z1 = 3 + i,
10
z2 = 4 − 3i, z3 = −1 + 2i
and
z4 = −2 − 5i.
2
(a) z1 − z2
(b) z2 + z3 − z4
(c) z1* z2
3
(a) z1 + z2
(b) z3 z4
z*
(c) __3
z*4
MODULE 1tCHAPTER 1
z*1 + z*2
(c) ______
z*3 z*4
z1 z2
(c) ______
z1 + z2
4
(a) z1 z2 z3
(b) z2 z3 + z1 z4
5
z1
(a) __
z2
z1 + z2
(b) ______
z3 z4
6
Evaluate
7
4
(a) i12
(b) i15
(c) i21
(d) __
i8
1 + 3i evaluate in the form a + ib
If z = ______
1 − 2i
1
(a) z2
(b) z − __
z
8
Find a complex number z such that z2 = −5 + 12i.
9
The complex numbers u, v and w are related by the equation
1 __
1
1 __
__
u=v+w
z1 + z2
(d) ______
z3 + z4
5
(e) ___
i20
Given that v = 1 − 2i and w = 3 + i, find the complex number u in the form
x + yi where x, y ∊ ℝ.
6 + 8i , find the value of x for which Re(z) = Im(z).
2 − i − ______
10 If z = _____
1+i
x+i
11 Find the square root of (a) 3 + 4i (b) 24 − 10i
Remember
A polynomial is an
expression of the
form anxn + an − 1
xn – 1 + … + a0
where a0, a1, …
an − 1, an are constants and n ∊ 𝕎.
The roots of an
equation y = f(x)
are the values of x
for which f(x) = 0.
Roots of a polynomial
The remainder theorem and the factor theorem hold for complex numbers as well as
real numbers. If a polynomial equation has complex roots, then
_
(i) if all coefficients are real, the roots occur in conjugate pairs α, α
(ii) a quadratic factor of the polynomial is z2 − 2 Re(α)z + |α|2
where α is one root of the equation.
_
_
If α and α are roots, then the factorised polynomial (z − α) (z − α)
_
_
= z2 − αz − α z + α α
_
_
= z2 − z(α + α) + α α
= z2 − 2 Re (α) z +|α|2
Quadratic equations
E X A M P L E 10
Solve the equation z2 + z + 1 = 0.
SOLUTION
z2 + z + 1 = 0
Using the quadratic formula, we have
____________
___
−1 ± √(1)2 − 4(1)(1) __________
= −1 ± √−3
z = __________________
2
2(1)
__ ___
__
−1 ± √3 i
−1 ± √3 √−1 = _________
= _____________
2
2
__
__
3
√
√3 i
1 − ___
1
__
Hence z = − + ___i or z = −__
2
2
2
2
11
M O DUL E 1
E X A M P L E 11
Solve the equation z2 − 2z + 2 = 0.
SOLUTION
Using the quadratic formula, we have
______________
___
−(−2) ± √(−2)2 − 4(1)(2) ________
z = _______________________
= 2 ± √−4
2
2(1)
__ ___
−2 ± √4 √−1 = _______
−2 ± 2i = −1 ± i
= _____________
2
2
Hence z = −1 + i or z = −1 − i
Note that in the above examples the coefficients of the quadratic equations are real
and the roots are conjugates of each other.
E X A M P L E 12
Solve the equation z2 + iz + 2 = 0.
SOLUTION
Using the quadratic formula with a = 1, b = i, c = 2, we have
____________
___
−(i) ± √(i) − 4(1)(2) _________
= −i ± √−9
z = ___________________
2
2
2(1)
__ ___
−i ± √9 √−1 = _______
−i ± 3i = __
2i , ____
−4i
= ____________
2
2
2 2
∴ z = i or z = −2i
In this case the
coefficient of z
is not real and
hence the roots
will not occur in
conjugate pairs.
E X A M P L E 13
Solve the equation z2 − (3 + 5i)z + 2 + 5i = 0.
SOLUTION
Using the quadratic formula with a = 1, b = −(3 + 5i), c = 2 + 5i, we have
_______________________
(−(3 + 5i))2
= 9 + 30i +25i2
= 9 − 25 +30i
= −16 + 30i
−16 + 30i
− 8 − 20i
= −24 + 10i
(3 + 5i) ± √(−(3 + 5i))2 − 4(1)(2 + 5i)
z = _________________________________
2(1)
_________
3 + 5i ± √−24 + 10i
= __________________
2_________
We next write √−24 + 10i in the form x + iy where x, y ∊ ℝ.
_________
√−24 + 10i = x + iy
Squaring both sides
−24 + 10i = (x + iy)2
−24 + 10i = x2 + i2y2 + 2ixy
−24 + 10i = x2 − y2 + 2xyi
(i2 = −1)
Equating real and imaginary parts, we have
12
Note
x2 − y2 = −24
[1]
2xy = 10
5
y = __
x
[2]
MODULE 1tCHAPTER 1
Substituting into [1]
5 2 = −24
x2 − ( __
x)
25 = −24
x2 − ___
x2
Multiplying by x2
x4 − 25 = −24x2
x4 + 24x2 − 25 = 0
(x2 + 25)(x2 − 1) = 0
∴ x2 = −25 or x2 = 1
Since x is real
x2 = 1
∴ x = ±1
5=5
When x = 1, y = __
1
5 = −5
When x = −1, y = ___
−1
_________
√−24 + 10i = 1 + 5i
_________
or √−24 + 10i = −1 − 5i
Since
_________
3
+
5i
±
−24 + 10i
√
__________________
z=
2
3 + 5i ± (1 + 5i) 3 + 5i + 1 + 5i
or
z = _______________ = _____________
2
2
4 + 10i or z = __
2
z = _______
2
2
Hence z = 2 + 5i or z = 1
3 + 5i − 1 − 5i
_____________
2
Other polynomials
E X A M P L E 14
Show that 1 + 2i is a factor of f(z) = z3 − z2 + 3z + 5. Hence solve the equation
f(z) = 0.
SOLUTION
Substituting z = 1 + 2i, we have
z3 − z2 + 3z + 5 = (1 + 2i)3 − (1 + 2i)2 + 3(1 + 2i) + 5
= 1 + 3(2i) + 3(2i)2 + (2i)3 − (1 + 4i + 4i2) + 3 + 6i + 5
= 1 + 6i + 12i2 + 8i3 − 1 − 4i − 4i2 + 3 + 6i + 5
= 1 + 6i − 12 − 8i − 1 − 4i + 4 + 3 + 6i + 5
= 13 − 13 + 12i − 12i = 0
By the factor theorem 1 + 2i is a factor of z3 − z2 + 3z + 5.
Since all the coefficients of the polynomial are real, complex roots occur in conjugate
pairs.
Since 1 + 2i is a root then 1 − 2i is also a root of the equation.
13
M O DUL E 1
A quadratic factor is
(z − (1 + 2i))(z − (1 − 2i)) = z2 − (1 + 2i)z − (1 − 2i)z + (1 + 2i)(1 − 2i)
= z2 − z − 2iz − z + 2iz + 1 − 2i + 2i − 4i2
= z2 − 2z + 5 since i2 = −1
Now z3 − z2 + 3z + 5 = (z2 − 2z + 5)(az + b)
Equating coefficients of z3 gives a = 1
Equating the constants: 5 = 5b, b = 1
∴ z3 − z2 + 3z + 5 = (z2 − 2z + 5)(z + 1) = 0
Therefore, when z + 1 = 0, z = −1
The roots of the equation are 1 + 2i, 1 − 2i, −1.
EXERCISE 1B
In questions 1–5, solve the quadratic equations.
1
z2 + 16 = 0
2
z2 − 8z + 17 = 0
3
z2 − 4z + 5 = 0
4
z2 − 6z + 13 = 0
5
z2 − 10z + 31 = 0
In questions 6–9, write each expression as a product of linear factors.
6
z2 + 1
8
z2 − 6z + 25
7
z2 − 2z + 2
9
z4 − z2 − 2z + 2
__
10 Express √2i in the form a + bi where a, b ∊ ℝ. Hence solve the equation
z2 − (3 + 5i)z − 4 + 7i = 0.
11 Given that u2 = −60 + 32i, express u in the form x + iy where x, y ∊ ℝ.
Hence solve z2 − (3 − 2i)z + 5 − 5i = 0.
12 Show that 4 + 2i is a root of the equation 3z3 − 23z2 + 52z + 20 = 0.
Hence solve the equation.
13 Show that 1 + i is a root of the equation 4z3 − 7z2 + 6z + 2 = 0, and find the
other values of z satisfying the equation.
14 Given that 3 − 2i is a root of the equation z3 − 8z2 + 25z − 26 = 0, write down
a quadratic factor of f (z) = z3 − 8z2 + 25z − 26. Hence solve the equation
f (z) = 0.
15 Solve the equation z3 − 5z2 + 8z − 6 = 0.
14
MODULE 1tCHAPTER 1
The Argand diagram
A complex number can be represented in a rectangular or Cartesian axis. An
Argand diagram is a plot of complex numbers as points in the complex plane.
The horizontal axis represents the real axis and the vertical axis represents the
imaginary
___›axis. Let z = x + yi. A point P can be used to represent the number
z where OP = xy .
Im (z)
( )
The diagram is called the Argand diagram,
named after the Swiss accountant and amateur
mathematician Jean Argand (1768–1822).
Show the following numbers on an Argand diagram.
(a) 3 + i
Note
All the complex
numbers start at
the origin.
(b) 1 − i
(c) −1 − i
When showing a complex number
on an Argand diagram the real
part is plotted along the horizontal
axis and the imaginary part along
the vertical axis. For the complex
number 3 + i, we move 3 units
to the right from the origin and
1 unit upward from the origin.
The components of the complex
number are (3, 1) and the complex
number 3 + i is represented by the vector
(d) −1 + i
(e) 4 + 2i
4
Im (z)
3
(4, 2) (e)
2
(d) (–1, 1)
4 + 2i
1
(3, 1) (a)
3+i
–2
–1
(c) (–1, –1)
( 31 ).
1
–1
2
Re (z)
3
4
(1, –1) (b)
–2
Addition and subtraction on the Argand diagram
We add and subtract complex numbers z1 and
z2 on the Argand diagram in the same way that
we add and subtract vectors. We complete a
parallelogram as shown. The leading diagonal
is z1 + z2 and the other diagonal is z2 − z1.
z1
Im (z)
z2
SOLUTION
Re (z)
x
0
+
E X A M P L E 15
y
z1
A complex
number written
as z = x + yi is in
Cartesian form.
P
Complex numbers can be treated as vectors in
the xy plane. The complex number z can be
represented on the xy plane as shown.
z2
z2
z1
z2
–
z1
Re (z)
15
M O DUL E 1
In this diagram the sum of 2 + 3i and 4 + i is
the vector 6 + 4i.
Im (z)
4
3
2 + 3i
2
6 + 4i
2 + 3i
1
4+i
1
2
Re (z)
3
Multiplication by i
4
5
6
Im (z)
Let P = z. The complex number iz can be found
by rotating OP anticlockwise through 90 °
about the origin.
z = 2 + 3i
P
3
2
–3 + 2i
1
–2
O
2 + 3i
Re (z)
–3
iz = 2i + 3i2
–1
1
2
3
= −3 + 2i
Note
The modulus of a
complex number
is the distance
from the origin
to the end point,
that is, the length
of the line OP.
E X A M P L E 16
Modulus (length) of a complex number
P
Consider a complex number, z = x + yi, in the
Cartesian form.
r
______
Let r = √ +
r is called the modulus of z and
is denoted by | z |.
x2
y
y2 ;
______
Re (z)
O
x
So r = | z | = | x + yi | = √x2 + y2
Find the modulus of the following.
(a) 2 + i
SOLUTION
Im (z)
(b) 2 − i
_______
(c) 3 + 4i
(d) cos θ + i sin θ
__
(a) | 2 + i | = √22 + 12 = √5
__________
__
(b) | 2 − i | = √22 + (−1)2 = √5
_______
___
(c) | 3 + 4i | = √32 + 42 = √25 = 5
____________
__
(d) ∣ cos θ + i sin θ ∣ = √cos2 θ + sin2 θ = √1 = 1
Notes
(i) |z| = |z*|
(ii) z z* = |z|2
Proof:
Let z = x + iy
z z* = (x + iy)(x − iy)
= x2 + y2
______
= (√x2 + y2 )2
= |z|2
16
(cos2 θ + sin2 θ = 1)
MODULE 1tCHAPTER 1
Argument of a complex number
Let z = x + iy. From the triangle:
y
tan θ = __
x
( )
y
so θ = tan−1 __
x
The angle θ is called the argument of z and is denoted by
arg(z).
Im (z)
r
Special care must be taken when finding the argument
of a complex number. The argument depends on which
quadrant the complex number lies in.
y
θ
Re (z)
x
Notes
(i)
(ii)
(iii)
(iv)
The argument is measured in radians.
The argument is measured from the positive real axis.
The principal argument of z is such that −π < θ ≤ π.
If the complex number is in the
y
1st or 4th quadrant: θ = tan−1 __x
y
2nd quadrant: θ = π + tan−1 __x
y
3rd quadrant: θ = −π + tan−1 __x
( )
( )
( )
E X A M P L E 17
Find the modulus and argument of each of the following complex numbers.
(a) 1 + i
SOLUTION
(b) 1 − i
_______
(c) −1 + i
(d) −1 − i
__
(a) | 1 + i | = √12 + 12 = √2
Im (z)
1
1
θ
Re (z)
1
( )
π (1 + i is in the first quadrant)
1 = __
arg (1 + i) = tan−1 __
4
1
___________
(b) | 1 − i | = √(1)2 + (−1)2
2
__
= √2
Im (z)
1
Re (z)
–2
θ
–1
–1
1
2
(1, –1)
–2
π (1 − i is in the fourth quadrant)
arg (1 − i) = tan−1 (−1) = − __
4
17
M O DUL E 1
__________
__
(c) | −1 + i | = √(−1)2 + 12 = √2
2
(–1, 1)
Im (z)
1
θ
–2
Re (z)
–1
1
2
–1
–2
π = ___
3π
arg (−1 + i) = π + tan−1(−1) = π + ( −__
4)
4
_____________
__
(d) | −1 − i | = √(−1)2 + (−1)2 = √2
2
Im (z)
1
Re (z)
–2
θ
–1
(–1, –1)
1
2
–1
–2
3π
−1 = −π + __
arg (−1 − i) = −π + tan−1 ___
( π4 ) = −___
4
−1
( )
E X A M P L E 18
Find the modulus and argument of −2 − i.
SOLUTION
_____________
| −2 − i | = √(−2)2 + (−1)2
_____
__
2
= √4 + 1 = √5
The complex number −2 − i lies in the 3rd
quadrant, therefore its argument should be
negative.
−1 = 0.464
Basic angle tan−1 ___
−2
1
Re (z)
( )
–2
(–2, –1)
18
θ
–1
1
–1
–2
arg (−2 − i) = −π + 0.464 = −2.678 radians
Try these 1.3
Im (z)
Find the modulus and argument of the following.
__
__
__
(a) (i) 5 + i
(ii) √3 − i
(iii) −√3 − i
(iv) −√3 + i
(b) (i) 3 + 4i
(ii) 2 − 4i
(iii) −2 + 5i
(iv) −4 − 7i
2
MODULE 1tCHAPTER 1
Trigonometric or polar form of a complex number
Note
To represent a
complex number
in polar form,
you must find
the modulus and
argument of the
number.
E X A M P L E 19
)
Im (z)
r
z = r cos θ + i r sin θ
= r (cos θ + i sin θ),
y
θ
Re (z)
x
−π < θ ≤ π
z = r (cos θ + i sin θ) is the trigonometric or
polar form of the complex number.
Determine the modulus and argument of the complex number z = 3 + i and express
z in polar form.
z=3+i
Im (z)
θ = tan _____
Re (z)
where Im (z) = 1
and Re (z) = 3
(
y
sin θ = __r ⇒ y = r sin θ
x ⇒ x = r cos θ
cos θ = __
r
Substitute into z = x + iy
_______
___
r = | z | = √32 + 12 = √10
( )
θ = arg (3 + i) = tan−1 __1 = 0.322
3
Using z = r (cos θ + i sin θ) gives the polar form of z as
___
z = √10 (cos 0.322 + i sin 0.322)
E X A M P L E 20
Determine the modulus and argument of the complex number z = −2 + i, and
express z in polar form.
SOLUTION
z = −2 + i
__________
__
r = | z | = √(−2)2 + 12 = √5
θ=π+
Im (z)
tan−1 _____
Re (z)
Im (−2 + i) = 1
Re (−2 + i) = −2
(
)
( )
1
θ = arg (z) = π + tan−1 ___
–2
= π − 0.464
= 2.678 radians
So
E X A M P L E 21
__
z = √5 (cos 2.678 + i sin 2.678)
Write the following in the form r (cos θ + i sin θ), −π < θ ≤ π, giving θ either as a
multiple of π or in radians to 3 significant figures.
__
SOLUTION
(−2 + i lies in the 2nd quadrant)
(a) 1 + i √3
(b) 1 − i
(c) −1 − i
(d) −3√3 − 3i
__
_________
__
__
_____
__
(a) r = | 1 + i √3 | = √12 + (√3 )2 = √1 + 3 = √4 = 2
__
θ = arg (1 + i√3 ) =
So
__
__
tan−1
(1)
√3 = __
π
___
3
π + i sin __
π
1 + i √3 = 2 ( cos __
3
3)
__
(1 + i√3 lies in the 1st quadrant)
19
M O DUL E 1
__________
__
(b) r = | 1 − i | = √12 + (−1)2 = √2
( )
π
θ = arg (1 − i) = tan−1 −__11 = − __
4
So
(1 − i lies in the 4th quadrant)
__
π + i sin −__
1 − i = √2 cos ( −__
( π4 )
4)
[
_____________
]
_____
__
(c) r = | −1 − i | = √(−1)2 + (−1)2 = √1 + 1 = √2
( )
π
−1 = −π + __
θ = arg (−1 − i) = −π + tan−1 ___
4
−1
3π
= − ___
4
__
3π
3π
___
So −1 − i = √2 cos −___
4 + i sin − 4
[ (
)
)]
(
_______________
__
__
(3rd quadrant)
___
______
(d) r = | −3√3 −3i | = √(−3 √3 )2 + (−3)2 = √27 + 9 = √36 = 6
__
−3__ − π = __
π−π
θ = arg (−3√3 − 3i) = tan−1 ______
6
−3√3
5π
= −___
6
__
5π
5π
___
So −3√3 − 3i = 6 cos − ___
6 + i sin − 6
(
[ (
)
)
(
(3rd quadrant)
)]
E X A M P L E 22
Find the modulus and argument of the expression z = 1 + cos θ + i sin θ.
SOLUTION
r = | z | = √(1 + cos θ)2 + sin2 θ
_________________
________________________
= √1 + 2 cos θ + cos2 θ + sin2 θ
__________
= √2 + 2 cos θ
(cos2 θ + sin2 θ = 1)
___________
= √2 (1 + cos θ)
____________
(
θ
= 2 × 2 cos2 __
2
θ
__
= 2 cos
2
√
)
( 2 cos
2
(
sin θ
arg (z) = tan−1 ________
1 + cos θ
= tan−1
(
(
So
20
)
2 sin ( ) cos ( )
2
2
_____________
θ
__
2
cos2
)
[ ()
θ
__
(2)
θ
__
)
θ = __
θ
= tan−1 tan __
2
2
θ cos __
θ + i sin __
θ
z = 2 cos __
2
2
2
( )]
θ = 1 + cos θ
__
2
)
( sin θ = 2 sin __θ2 cos __θ2 )
MODULE 1tCHAPTER 1
Exponential form of a complex number
In Module 2 you will see that ex, sin θ and cos θ can be written as polynomials as shown:
θ3 + __
θ5 − __
θ7 + . . .
sin θ = θ − __
3! 5! 7!
θ2 + __
θ4 − __
θ6 + . . .
cos θ = 1 − __
2! 4! 6!
x3 + . . .
x2 + __
x
e = 1 + x + __
2! 3!
Replacing x by iθ gives
(iθ)2 (iθ)3 (iθ)4 (iθ)5
eiθ = 1 + iθ + ____ + ____ + ____ + ____ + . . .
2!
3!
4!
5!
2
3
4
5
i
i
θ
θ
θ
θ
= 1 + iθ − __ − ___ + __ + ___ − . . .
2!
3!
4!
5!
2
4
θ + __
θ − . . . + i θ − __
θ3 + __
θ5 − . . .
= 1 − __
2! 4!
3! 5!
= cos θ + i sin θ
(
So
) (
)
eiθ = cos θ + i sin θ
Any complex number can thus be expressed in the form reiθ. This is called the
exponential form of a complex number.
E X A M P L E 23
Write the number 1 + i in exponential form.
SOLUTION
r = | 1 + i | = √12 + 12 = √2
π
θ = arg (1 + i) = tan−1 __1 = __
4
1
__ __
πi
(using reiθ)
So 1 + i = √2 e 4
E X A M P L E 24
Find the modulus and argument of z = √3 + i.
Write z in polar form and in exponential form.
SOLUTION
r = √(√3 )2 + 12 = √4 = 2
π
1__ = __
θ = arg (z) = tan−1 ___
6
√3
__
π
π
__
So √3 + i = 2 ( cos + i sin __
6
6)
_
πi
__
√3 + i = 2e 6
E X A M P L E 25
SOLUTION
_______
__
( )
__
_________
__
__
( )
1 + i, write z in the form
Given that z = _____
1−i
(a) x + iy
(b) r (cos θ + i sin θ)
Polar form:
r(cos θ + i sin θ)
Exponential form: reiθ
(c) reiθ
1+i
1 + i × _____
(a) z = _____
1−i 1+i
2i = i
1 + i + i + i = __
= ____________
2
__1 + 1
(b) z = √12 = 1
π
1 = __
arg (z) = tan−1 __
0
2
π
π
__
So z = 1 ( cos + i sin __
2
2)
πi
__
(c) z = e 2
2
( )
21
M O DUL E 1
E X A M P L E 26
Find the modulus and the argument, in radians to 3 d.p., of the following.
__
3+i
−2 + 3i
__
(a) (−2 + 3i)(1 + √3 i)
(b) _____
(c) _______
2−i
1 + i √3
Hence write each number in polar form and in exponential form.
SOLUTION
(a) (−2 + 3i)(1 + √3 i)
__
__
__
= −2 − 2√3 i + 3i + 3√3 i2
__
__
__
__
= −2 − 3√3 + i (3 − 2√3 )
_______________________
__
__
___
r = −2 − 3√3 + i (3 − 2√3 ) = √(−2 − 3√3 )2 + (3 − 2√3 )2 = √52
__
(
)
__
__
3 − 2√3__ − π = −3.077
θ = arg (−2 − 3√3 + i (3 − 2√3 )) = tan−1 _________
−2 − 3√3
__
___
(polar form)
(−2 + 3i) (1 + √3 i) = √52 [ cos (−3.077 + i sin (−3.077) ]
___
= √52 e−3.077i
(exponential form)
3 + i × _____
2 + i = ______________
6 + 3i + 2i + i2 = _________
6 − 1 + 5i = 1 + i
3 + i = _____
(b) _____
5
2−i 2−i 2+i
22 − i2
__
r = 1 + i = √2
π
arg (1 + i) = tan−1 (1) = __
4
__
3
+
i
π
π
_____ = √2 cos __ + i sin __
( 4
4)
2−i
(polar form)
__ __
πi
= √2 e 4
(exponential form)
−2 + 3i
−2 + 3i
__ = _______
__
(c) _______
×
1 + i √3
1 + i √3
__
__
__
1 − i √3__ = _____________________
−2 + i 2√3 + 3i − i23√3
________
1 − i √3
1+3
__
__
__
__
−2 + 3√3 + i (2√3 + 3) −2 + 3√3
2√3 + 3
+ i _______
= _____________________ = _________
4
4
4
_______________________
__
__
−2 + 3√3
2√3 + 3
2√3 + 3 ∣ = _________
∣ _________
+ i _______
)
4
4
4
√( −2 +4 3√3 ) + ( _______
__
(
__
__
2
__
)
(
__
2√3 + 3
_______
2
= 1.803 (3 d.p.)
)
2√3 + 3 = tan−1 _________
−2 + 3√3 + i _______
4 __ = 1.112 radians (3 d.p.)
arg _________
4
4
−2 + 3√3
_________
4
−2 + 3i
_______
__ = 1.803 (cos 1.112 + i sin 1.112)
1 + i√3
= 1.803 e1.112i
(polar form)
(exponential form)
De Moivre’s theorem
Let z = r (cos θ + i sin θ)
z2 = [r (cos θ + i sin θ)]2 = r2 (cos θ + i sin θ)2
= r2 (cos2 θ − sin2 θ + i 2 sin θ cos θ)
= r2 (cos 2θ + i sin 2θ)
22
(since cos 2θ = cos2 θ − sin2 θ and sin 2 θ = 2 sin θ cos θ)
MODULE 1tCHAPTER 1
z3 = r2 (cos 2θ + i sin 2θ) × r (cos θ + i sin θ)
= r3 (cos 2θ cos θ + i sin θ cos 2θ + i sin 2θ cos θ + i2 sin 2θ sin θ)
= r3 [cos 2θ cos θ − sin 2θ sin θ + i (sin θ cos 2θ + sin 2θ cos θ)]
= r3 [cos (2θ + θ) + i sin (2θ + θ)]
= r3 (cos 3θ + i sin 3θ)
So we have z = r (cos θ + i sin θ)
z2 = r2 (cos 2θ + i sin 2θ)
z3 = r3 (cos 3θ + i sin 3θ)
and, extending this result, we get
zn = rn (cos nθ + i sin nθ)
cos (A + B) = cos A cos B − sin A sin B
so cos 2θ cos θ − sin 2θ sin θ = cos (2θ + θ)
= cos 3θ
sin (A + B) = sin A cos B + cos A sin B
so sin 2θ cos θ + cos 2θ sin θ = sin (2θ + θ)
= sin 3θ
de Moivre’s theorem states that for any real number n
(cos θ + i sin θ)n = cos nθ + i sin nθ
E X A M P L E 27
Prove de Moivre’s theorem for any positive integer n.
SOLUTION
We require to prove that (RTP)
(cos θ + i sin θ)n = cos nθ + i sin nθ, n ∊ ℤ+
Proof: Let Pn be the statement (cos θ + i sin θ)n = cos nθ + i sin nθ.
Since (cos θ + i sin θ)1 = cos θ + i sin θ, P1 is true.
Assume that Pn is true for n = k, i.e.
(cos θ + i sin θ)k = cos k θ + i sin k θ.
RTP: true for n = k + 1, i.e. Pk+1 is true.
Proof: (cos θ + i sin θ)k+1
Recall the
principle of
mathematical
induction
(PMI): prove the
statement true
for n = 1; assume
the statement
true for n = k
and prove the
statement true
for n = k + 1.
Hence, by PMI,
it is true for all
n ∊ ℤ+.
= (cos θ + i sin θ)k (cos θ + i sin θ)
(rules of indices)
= (cos kθ + i sin kθ) (cos θ + i sin θ)
= cos kθ cos θ + i sin θ cos kθ + i sin kθ cos θ + i2 sin kθ sin θ
= (cos kθ cos θ − sin k θ sin θ) + i (sin θ cos kθ + cos θ sin kθ)
= cos (k + 1) θ + i sin (k + 1)θ
cos (A + B) = cos A cos B − sin A sin B
cos (kθ + θ) = cos kθ cos θ − sin kθ sin θ
Hence, by PMI, (cos θ + i sin θ)n = cos nθ + i sin nθ,
n ∊ ℤ+
E X A M P L E 28
Express cos 3θ and sin 3θ in terms of cos θ and sin θ only.
SOLUTION
By de Moivre’s theorem
(cos θ + i sin θ)3 = cos 3θ + i sin 3θ
Using the binomial expansion (or Pascal’s triangle) to expand (cos θ + i sin θ)3
cos3 θ + 3 cos2 θ (i sin θ) + 3 cos θ (i sin θ)2 + (i sin θ)3 = cos 3θ + i sin 3θ
Binomial
expansion is
covered in
Module 2. Pascal’s
triangle can be
used for the
expansions
involved at this
stage.
Pascal ’s t r i a n g l e
1
1
1
1
1
2
3
1
3
1
23
M O DUL E 1
{
{
cos3 θ + i 3 cos2 θ sin θ + i2 3 cos θ sin2 θ + i3 sin3 θ = cos 3θ + i sin 3θ
cos3 θ − 3 cos θ sin2 θ + i 3 cos2 θ sin θ − i sin3 θ = cos 3θ + i sin 3θ
cos3 θ − 3 cos θ sin2 θ + i (3 cos2 θ sin θ − sin3 θ) = cos 3θ + i sin 3θ
SFBMQBSU
SFBMQBSU
Equating real and imaginary parts
cos 3θ = cos3 θ − 3 cos θ sin2 θ = cos3 θ − 3 cos θ (1 − cos2 θ)
= cos3 θ − 3 cos θ + 3 cos3 θ
= 4 cos3 θ − 3 cos θ
sin 3θ = 3 cos2 θ sin θ − sin3 θ = 3 (1 − sin2 θ) sin θ − sin3 θ
= 3 sin θ − 3 sin3 θ − sin3 θ
= 3 sin θ − 4 sin3 θ
E X A M P L E 29
Using de Moivre’s theorem, express cos 5θ in terms of cos θ only.
SOLUTION
By de Moivre’s theorem
Pa s c a l ’s t r i a n g l e
1
cos 5θ + i sin 5θ = (cos θ + i sin θ)5
Using Pascal’s triangle, expand (cos θ + i sin
___
i = √−1
i2 = −1
i3 = −i
i4 = 1
i5 = i
1
θ)5:
cos 5θ + i sin 5θ
1
1
1
2
3
1
3
1
1 4 6 4 1
1 5 10 10 5 1
= cos5 θ + i 5 cos4 θ sin θ + i2 10 cos3 θ sin2 θ + i3 10 cos2 θ sin3 θ
+ i4 5 cos θ sin4 θ + i5 sin5 θ
= cos5 θ + i 5 cos4 θ sin θ − 10 cos3 θ sin2 θ − i 10 cos2 θ sin3 θ + 5 cos θ sin4 θ + i sin5 θ
= cos5 θ − 10 cos3 θ sin2 θ + 5 cos θ sin4 θ + i (5 cos4 θ sin θ − 10 cos2 θ sin3 θ + sin5 θ)
Equating real parts
cos 5θ = cos5 θ − 10 cos3 θ sin2 θ + 5 cos θ sin4 θ
Substituting sin2 θ = 1 − cos2 θ, we have
cos 5θ = cos5 θ − 10 cos3 θ (1 − cos2 θ) + 5 cos θ (1 − cos2 θ)2
= cos5 θ − 10 cos3 θ + 10 cos5 θ + 5 cos θ (1 − 2 cos2 θ + cos4 θ)
= cos5 θ − 10 cos3 θ + 10 cos5 θ + 5 cos θ − 10 cos3 θ + 5 cos5 θ
= 16 cos5 θ − 20 cos3 θ + 5 cos θ
E X A M P L E 30
Express tan 4θ in terms of tan θ using de Moivre’s theorem.
SOLUTION
By de Moivre’s theorem cos 4θ + i sin 4θ = (cos θ + i sin θ)4
Using Pascal’s triangle, expand (cos θ + i sin θ)4:
cos 4θ + i sin 4θ
= cos4 θ + i 4 cos3 θ sin θ − 6 cos2 θ sin2 θ − i 4 cos θ sin3 θ + sin4 θ
= cos4 θ − 6 cos2 θ sin2 θ + sin4 θ + i (4 cos3 θ sin θ − 4 cos θ sin3 θ)
24
MODULE 1tCHAPTER 1
Equating real and imaginary parts
cos 4θ = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ
sin 4θ = 4 cos3 θ sin θ − 4 cos θ sin3 θ
sin 4θ = __________________________
4 cos3 θ sin θ − 4 cos θ sin3 θ
tan 4θ = ______
cos 4θ cos4 θ − 6 cos2 θ sin2 θ + sin4 θ
Divide numerator and denominator by cos4 θ
4 cos θ sin θ − 4 cos θ sin θ
________________________
3
3
θ
tan 4θ = __________________________
cos4 θ −6 cos2 θ sin2 θ + sin4 θ
_________________________
cos4 θ
cos4
4 sin θ
4 sin θ − ______
_____
3
tan 4θ =
cos θ
4 tan θ − 4 tan3 θ
_____________________
= _________________
cos3 θ
cos4 θ − ______
6 sin2 θ + _____
sin4 θ
_____
4
2
cos θ
cos4 θ
cos θ
E X A M P L E 31
π + i sin __
π 8.
Evaluate ( cos __
4
4)
SOLUTION
( cos __π4 + i sin __π4 ) 8
8π
8π + i sin ___
= cos ___
4
4
= cos 2π + i sin 2π
=1
E X A M P L E 32
SOLUTION
sin (−θ) = −sin θ
cos (−θ) = cos θ
1 − 6 tan2 θ + tan4 θ
(using de Moivre’s theorem)
(cos 2π = 1, sin 2π = 0)
π − i sin __
π 8.
Evaluate ( cos __
6
6)
( cos __π6 − i sin __π6 ) 8
8
π + i sin −__
= [ cos ( −__
( π6 ) ]
6)
(
)
(
8π
8π
___
= cos −___
6 + i sin − 6
__
1 ___
√3 i
+
= −__
2
2
(write in the form cos θ + i sin θ)
)
(using de Moivre’s theorem)
E X A M P L E 33
Find the value of (1 + i)10.
SOLUTION
We first write 1 + i in polar form and then use de Moivre’s theorem.
__
|1 + i| = √2
π
arg (1 + i) = tan−1 (1) = __
4
__
π + i sin __
π
1 + i = √2 ( cos __
4
4)
__
π + i sin __
π 10
(1 + i)10 = √2 ( cos __
4
4)
__
10π
10π + i sin ____
= (√2 )10 cos ____
4
4
= 25 ( 0 + i ) = 32i
[
]
(
)
(using de Moivre’s theorem)
25
M O DUL E 1
EXERCISE 1C
1
Find the modulus of each of the following.
(a) 2 + 5i
(b) 3 + 7i
(c) −1 − 4i
(d) −1 + 2i
(e) cos θ + i(2 sin θ)
2
Find the argument of each of the following.
(a) 2 + 4i
3
(c) −1 + 2i
(d) −4 − 2i
Write each of the following in exponential and polar form.
__
(a) 2 − √3i
4
(b) 3 − i
__
(b) 2i − √3
(c) 1 − i
Using de Moivre’s theorem, find the value of each of the following.
2π
2π + i sin ___
π + i sin __
π 9
(a) ( cos __
(b) 2 cos ___
5
5
3
3)
8
6
π
π
π
π
___
___
__
__
+ i sin )
(c) ( cos
(d) ( cos + i sin )
18
18
2
2
Write each of the following in the form x + iy.
[ (
5
)]
10
__
(a) (1 + i)20
(b) (3 − √3 i)12
__
(c) (−√3 + i)9
(d) (1 − i)5
6
π + i sin __
π .
Evaluate ( cos __
6
6)
7
Use de Moivre’s theorem to express sin 4θ and cos 4θ in terms of sin θ and cos θ.
−3
Hence show that cos 4θ = 8 cos4 θ − 8 cos2 θ + 1 and
sin 4θ = 4 sin θ(2 cos3 θ − cos θ).
8
Find cos 7θ in terms of cos θ.
9
Use de Moivre’s theorem to express sin 3θ and cos 3θ in terms of sin θ and cos θ.
Hence find tan 3θ in terms of tan θ.
10 Use de Moivre’s theorem to find tan 5θ in terms of tan θ.
11 Use de Moivre’s theorem to prove that if θ is not a multiple of π,
sin 5θ = 16 cos4 θ − 12 cos2 θ + 1.
_____
sin θ
12 Use de Moivre’s theorem to prove that
cos 3θ + i sin 3θ = cos 2θ − i sin 2θ.
______________
cos 5θ + i sin 5θ
4 tan θ − 4 tan θ .
13 Show that tan 4θ = _________________
2
4
3
1 − 6 tan θ + tan θ
Use your result to solve the equation
t = tan θ
nπ .
giving your answer in the form tan ___
16
t4 + 4t3 − 6t2 − 4t + 1 = 0,
26
MODULE 1tCHAPTER 1
14 Use de Moivre’s theorem to simplify the following expressions.
(a) (cos 3θ + i sin 3θ) (cos θ + i sin θ)5
(b) (cos 2θ + i sin 2θ) (cos θ + i sin θ)7
cos θ − i sin θ
(c) ______________
cos 4θ − i sin 4θ
Locus of a complex number
Let z = x + iy and P(x, y) ≡ z. The locus of the point P is a set of points in the
complex plane which satisfies a given condition. There are four standard forms
for the locus of a complex number in the Argand diagram. When identifying a
locus, there is a variable complex number z, a fixed complex number and a condition
placed on the variable number. The condition is generally the modulus or argument.
Circles
When using the
equation, make
sure that the
coefficient of z
is 1 and the
complex number
c is governed by
the negative sign.
Let c ∊ ℂ and r ∊ ℝ. The locus of z satisfying the condition
r
|z − c| = r
c
is a circle with centre c and radius r.
Re (z)
0
__
E X A M P L E 34
Describe and sketch the locus of z where |z − 1 − i| = √2
SOLUTION
|z − 1 − i| = √2
Write the equation in the form
|z − c| = r where
the negative sign
governs the fixed
number 1 + i.
__
Im (z)
__
⇒ |z − (1 + i)| = √2
The locus of __
z is a circle centre (1, 1)
and radius √2
√2
(1, 1)
Re (z)
0
E X A M P L E 35
Describe and sketch the locus of z where |z − 2 + i| = 1.
SOLUTION
|z − 2 + i| = 1
Is the locus in the
standard form?
Im (z)
⇒ |z − (2 − i)| = 1
So z lies on a circle with centre
at (2, −1) and radius 1 unit.
Im (z)
Re (z)
0
(2, –1)
27
M O DUL E 1
E X A M P L E 36
SOLUTION
Describe and sketch the locus of z where
|z + 2 + 4i| = 2.
Im (z)
|z − (−2 − 4i)| = 2
Re (z)
z lies on a circle with centre (−2, −4) and
radius 2.
(–2, –4)
Perpendicular bisector of a line segment
Let a, b ∊ ℂ. The locus of z satisfying the condition
Im (z)
|z − a| = |z − b|
b
is the perpendicular bisector of the line joining a to b.
locus of z
a
Re (z)
E X A M P L E 37
Describe and sketch the locus of z satisfying the condition |z − 1 + i| = |z − 1 − i|.
SOLUTION
|z − 1 + i| = |z − 1 − i|
Im (z)
⇒ |z − (1 − i)| = |z − (1 + i)|
The locus of z is the perpendicular
bisector of the line joining (1, −1)
to (1, 1). This is the real axis.
locus of z
2
1
(1, 1)
Re (z)
1
–1
–2
28
(1, –1)
MODULE 1tCHAPTER 1
E X A M P L E 38
Describe and sketch the locus of z satisfying the condition |z − 2 − 2i| = |z + i|.
SOLUTION
Write the equation in the form
locus of z
4
|z − (2 + 2i)| = |z − (−i)|
Im (z)
3
(2, 2)
The locus of z is the perpendicular bisector
of the line joining (2, 2) to (0, −1).
2
1
–4
–3
–2
(1, 12)
0
–1
–1
1
2
Re (z)
3
4
(0, –1)
–2
E X A M P L E 39
Describe and sketch the locus of z satisfying the condition |z + 1 + 2i| = |z − 1 + 3i|.
SOLUTION
The locus of z is the perpendicular bisector
of the line joining (−1, −2) to (1, −3).
2
Im (z)
1
Re (z)
We write the
equation in the
form
| z − a | = | z − b |:
| z − (−1−2i) | =
| z − (1 − 3i) |
–3
–2
0
–1
1
2
–1
(–1, –2)
–2
–3
3
4
locus of z
(0, – 52 )
(1, –3)
–4
Half-line
Can you identify
why it is a halfline and why
the end point is
excluded?
Let a ∊ ℂ, and the angle θ be measured in radians.
The locus of z satisfying the condition
Im (z)
locus of z
arg (z − a) = θ
is a half-line starting at a
(but excluding a) and making
an angle of θ radians with the
positive real axis.
θ
a
Re (z)
0
29
M O DUL E 1
E X A M P L E 40
SOLUTION
Write the equation in the form
arg (z − a) = θ.
E X A M P L E 41
SOLUTION
Write the equation in the form
arg (z − a) = θ.
E X A M P L E 42
SOLUTION
π.
Describe and sketch the locus of z where arg (z − 1 − i) = __
4
π
arg (z − 1 − i) = __
4
π
⇒ (z − (1 + i)) = __
4
The locus of z is a half-line starting at (1, 1)
[excluding (1, 1)] and making an angle
π radians with the real axis.
of __
4
Im (z)
locus of z
π
4
(1, 1)
3π .
Describe and sketch the locus of z where arg (z + 1 + i) = ___
4
3π
arg (z + 1 + i) = ___
4
3π
⇒ arg (z − (−1 − i)) = ___
4
The locus of z is a half-line starting at
(−1, −1) [excluding (−1, −1)] and
3π radians with
making an angle of ___
4
the positive real axis.
Im (z)
locus of z
Re (z)
3π 0
4
(–1, –1)
π.
Describe and sketch the locus of z where arg (z + 2 + 3i) = __
6
Writing the equation in the form
arg (z − a) = θ we have
π.
arg (z − (−2 − 3i) = __
6
The locus of z is a half-line starting
at (−2, −3) [excluding (−2, −3)] and
π radians with the
making an angle of __
6
positive real axis.
Im (z)
Re (z)
0
–2
locus of z
π
6
(–2, –3)
Straight line
Let z = a + λb, λ ∊ ℝ, a, b ∊ ℂ.
The locus of z is a line passing through a and parallel to b.
30
Re (z)
0
–3
MODULE 1tCHAPTER 1
E X A M P L E 43
SOLUTION
Describe and sketch the locus of z
where z = (1 + i) + λ(2 − i), λ ∊ ℝ.
The locus of z is a line passing through
(1, 1) and parallel to (2 − i).
4
Im (z)
3
locus of z
2
1 (1, 1)
Re (z)
–4
–3
–2
0
–1
1
2
–1
3
4
(2, –1)
–2
E X A M P L E 44
Describe and sketch the locus of z where
z = −1 + 2i + λ(3 + 4i), λ ∊ ℝ.
Im (z)
4
SOLUTION
(3, 4)
3
The locus of z is a line passing through
(−1, 2) and parallel to 3 + 4i.
(–1, 2)
locus of z
2
1
Re (z)
–4
–3
–2
–1
1
2
3
4
–1
–2
E X A M P L E 45
SOLUTION
Sketch the locus of z where
z = 3 + 2i + λ(−1 − 3i), λ ∊ ℝ.
4
Im (z)
3
The locus of z is a line passing through
(3, 2) and parallel to −1 − 3i.
2
(3, 2)
1
Re (z)
–4
–3
–2
–1
1
2
3
4
–1
–2
(–1, –3)
locus of z
–3
–4
Inequalities
When identifying the region represented by an inequality, we first draw the region
bounded by the boundary line. Then we can shade the appropriate region satisfied by
the inequality.
31
M O DUL E 1
E X A M P L E 46
Indicate on an Argand diagram the set of points satisfying the conditions
π.
|z − 1 − i| ≤ 2 and arg (z − 1 − i) ≤ __
4
SOLUTION
First we ignore the inequality and sketch the locus for |z − 1 − i| = 2. This can be
written as |z − (1 + i)| = 2, which is a circle centre (1, 1) and radius 2. Once the
boundary line (the circle) is drawn, the appropriate region must be shaded.
Since the inequality is ‘less than or equal to’, the circle must be solid and everything
inside the circle is shaded.
π, we first draw
For arg (z − 1 − i) ≤ __
4
Im (z)
π ⇒ arg (z − (1 + i)) = __
π
arg (z − 1 − i) = __
4
4
The locus of z is a half-line, starting
π
arg (z – 1 – i) =
4
at (1, 1) (but excluding (1, 1)) and making
2
Region
π radians with the positive
an angle of __
4
real axis. After drawing this line, look at
(1, 1)
Re (z)
the inequality and shade the appropriate
0
region. Remember, if the equality is
included the boundary line must be solid
and if the boundary line is excluded
|z – 1 – i| = 2
draw it as a broken line.
E X A M P L E 47
Shade the region in the Argand diagram representing the set of complex numbers z
satisfying the condition |z − 4 − 5i| ≤ 4. Find the greatest and least value of |z|.
SOLUTION
First we draw |z − 4 − 5i| = 4 ⇒
|z − (4 + 5i)| = 4, which is a circle
centre (4, 5) and radius 4. The region
to be shaded is |z − (4 + 5i)| ≤ 4, i.e.
inside the circle and the boundary is left
solid as this is included in the inequality.
Since all the complex numbers z
satisfying the inequality are either
on or inside the circle, the complex
number with the largest modulus is
the number starting at the origin,
passing through the centre and ending
on the circumference of the circle OB.
The complex number with the smallest
modulus is OA.
Im (z)
9
B
8
7
4
6
5
(4, 5)
4
3
2
A
1
Re (z)
0
_______
1
2
Length of OC = √42 + 52 which is the length of 4 + 5i
___
= √41
___
Greatest value of |z| = √41 + 4 = 10.4 (1 d.p.)
___
Least value of |z| = √41 − 4 = 2.4 (1 d.p.)
32
C
3
4
5
6
7
8
9
MODULE 1tCHAPTER 1
E X A M P L E 48
Sketch on an Argand diagram the set of points representing all complex numbers z
satisfying the inequality |z − 3 − 4i| ≤ 2. Find the least value of arg (z).
SOLUTION
First we sketch |z − 3 − 4i| = 2, which is
a circle centre (3, 4) and radius 2. We next
shade the inside of the circle, leaving the
boundary line as solid.
Im (z)
B
The complex number with the smallest
argument is OA (the tangent to the circle).
We need to find β.
2
(3, 4)
C
2
5
Since OC is the complex
_______number 3 + 4i, the
length of OC = √32 + 42 = 5.
α
O
A
4
β
3
Re (z)
Using the diagram,
2
sin α = __
5
α = sin−1 __2 = 0.412 radians
5
4
__
sin (α + β) =
5
α + β = sin−1 __4 = 0.927
5
β = sin−1 __4 − sin−1 __2 = 0.927 − 0.412 = 0.515 radians (3 d.p.)
5
5
The least value of arg (z) = 0.515 radians.
( )
( )
( )
( )
Intersecting loci
E X A M P L E 49
On a single Argand diagram, sketch the following loci.
π
(a) arg (z − 1) = __
2
π
(b) arg (z) = __
3
Hence or otherwise find the exact value of z satisfying both equations.
SOLUTION
π, the locus of z is
(a) For arg (z − 1) = __
2
a half-line starting at (1, 0) [excluding
π radians
(1, 0)] and making an angle of __
2
with the positive real axis.
π, the locus of z is a half(b) For arg (z) = __
3
line starting at (0, 0) [excluding (0, 0)]
π radians with
and making an angle of __
3
the positive real axis.
Let the point of intersection be a + bi.
From the diagram
b ⇒ b = tan __
π = __
π = √__
a = 1, tan ( __
3
)
(
3
1
3)
__
∴ The point of intersection is z = 1 + i √3 .
4
Im (z)
arg (z – 1) =
arg (z) =
3
2
1
–1
0
–1
2
3
(a, b)
π
b
Re (z)
3
–2
π
π
a 1
2
3
4
–2
33
M O DUL E 1
E X A M P L E 50
On a single Argand diagram sketch the loci given by
π
(b) arg (z) = __
4
(a) |z − 2 − 2i| = 1
Hence, find the exact values of all complex numbers z satisfying both (a) and (b).
SOLUTION
(a) |z − 2 − 2i| = 1 ⇒ |z − (2 + 2i)| = 1
4
The locus of z is a circle with centre (2, 2) and
radius 1.
π, the locus of z is a half-line
(b) For arg (z) = __
4
starting at (0, 0) [excluding (0, 0)] and makπ radians with the positive
ing an angle of __
4
real axis.
_______
From the
diagram there
are two points of
intersection, A, B.
Length of OC = √
22
+
22
__
3
–2
__
= √8 = 2√2
–2
__
π
a = (2√2 − 1) cos ( __
4)
__
__
__
√2
√2
___
___
=2−
a = (2√2 − 1)
2
2
( )
Since the triangle is isosceles a = b.
Therefore the first point of intersection is
(
)
To find the second point of intersection, c + id:
__
Length of OB = 2√2 + 1
π = _______
__c
cos ( __
4 ) 2√2 + 1
__
π
c = (2 √2 + 1) cos ( __
4)
__
__
(2)
__
√2
√2
c = (2 √2 + 1) ___
= 2 + ___
2
Since the triangle is isosceles c = d.
Therefore the second point of intersection is
__
(
__
√2
√2
+ i 2 + ___
2 + ___
2
2
)
The two points of intersection are
__
(
__
)
__
(
__
)
√2
√2
√2
√2
+ i 2 − ___
+ i 2 + ___
and 2 + ___
.
2 − ___
2
34
2
2
0
–1
__
__
–1
2
B
(c, d)
C
(2, 2)
(a, b)
1
Length of OA = 2√2 − 1
π = _______
__a
cos ( __
4 ) 2√2 − 1
__
1
2
To find the first point of intersection, a + ib
√2
√2
+ i 2 − ___
2 − ___
2
2
Im (z)
π
4
a1
A
b
2
Re (z)
3
4
MODULE 1tCHAPTER 1
Cartesian form of loci
__
E X A M P L E 51
Find in Cartesian form the equation of the locus of z where |z − 1 − i| = √2 .
Describe the locus of z.
SOLUTION
Since z is a variable complex number,
let z = x + iy.
__
Substituting into |z − 1 − i| = √2 , we have
Recall
__
|x + iy − 1 − i| = √2
__
_______
| x + iy | = √ x2 + y2
__
⇒| (x − i) + i(y − 1) |
⇒ |(x − 1) + i(y − 1)| = √2
________________
√(x − 1)2 + (y − 1)2 = √2
________________
= √(x − 1)2 + (y − 1)2
Squaring both sides: (x − 1)2 + (y − 1)2 = 2
The Cartesian equation is (x − 1)2 + (y − 1)2 = 2.
__
It represents a circle with centre (1, 1) and radius √2 .
E X A M P L E 52
Find in Cartesian form the locus of z where |z − 1 + i| = |z + 2 + 3i|.
SOLUTION
Since z is a variable complex number we replace z by z = x + iy in
|z − 1 + i| = |z + 2 + 3i| to get
|x + iy − 1 + i| = |x + iy + 2 + 3i|
|(x − 1) + i(y + 1)| = |(x + 2) + i(3 + y)|
________________
________________
√(x − 1)2 + (y + 1)2 = √(x + 2)2 + (3 + y)2
Squaring both sides: (x − 1)2 + (y + 1)2 = (x + 2)2 + (3 + y)2
Expanding both sides
x2 − 2x + 1 + y2 + 2y + 1 = x2 + 4x + 4 + 9 + 6y + y2
0 = 6x + 4y + 11
The Cartesian equation is 6x + 4y + 11 = 0
E X A M P L E 53
The locus of a variable complex number z is |z − 2 + i| = 2|z − 1 − i|.
Find the locus of z in Cartesian form.
SOLUTION
Let z = x + iy
Substituting into |z − 2 + i| = 2|z − 1 − i| gives
|x + iy − 2 + i| = 2|x + iy − 1 − i|
Rearranging
|x − 2 + i(y + 1)| = 2|(x − 1) + i(y − 1)|
________________
________________
√(x − 2)2 + (y + 1)2 = 2√(x − 1)2 + (y − 1)2
Squaring both sides
(x − 2)2 + (y + 1)2 = 4[(x − 1)2 + (y − 1)2]
∴ x2 − 4x + 4 + y2 + 2y + 1 = 4x2 − 8x + 4 + 4y2 − 8y + 4
0 = 3x2 + 3y2 − 10y − 4x + 3
The Cartesian equation is 3x2 + 3y2 − 10y − 4x + 3 = 0
35
M O DUL E 1
EXERCISE 1D
1
2
3
4
5
Sketch the locus of z in each of the following.
(a) |z| = 3
(b) |z − i| = 4
(c) |z + 4| = 2
(d) |z − 1 + 2i| = 5
(e) |z + 1 + 3i| = 6
(f) |z + 2 − 4i| = 7
Sketch the locus of z in each of the following.
(a) |z − 1 − i| = |z − 1 + 2i|
(b) |z − 3 + i| = |z + 1 + 2i|
(c) |z − 3i| = |z|
(d) |z + 2| = |z − 2|
|z + 1 + 4i|
(f) |z −1 − 7i| = |z + 1 + i|
(e) __________ = 1
|z −1 − 2i|
Sketch the locus of z in each of the following.
π
π
(b) arg (z) = − __
(a) arg (z) = __
6
2
π
π
(d) arg (z − i) = ___
(c) arg (z − 1) = __
4
12
2π
3π
(e) arg (z − 3 + 2i) = ___
(f) arg (z − 3 − 4i) = −___
3
4
Sketch the locus of z in each of the following.
(a) z = 1 + 2i + λ(1 − 3i)
λ∊ℝ
(b) z = 1 − 2i + λ(3 + 2i)
λ∊ℝ
(c) z = i + λ(4 + i)
λ∊ℝ
(d) z = 3 − 2i + λ(5 + 2i)
λ∊ℝ
(e) z = 1 − 4i + λ(−1 − 3i)
λ∊ℝ
(f) z = 2 + λ(4 + 2i)
λ∊ℝ
Sketch on a separate Argand diagram the region that represents each of the
following.
(a) |z − 2| ≤ 3
(b) |z − 3| < | z − i|
(c) |z − 3| ≤ 2
6
(d) |z − 2i| ⩽ | z + 3 − i|
2π
π
(f) arg ( z − 1 + 3i) ≤ ___
(e) arg (z − i) ≥ __
4
3
Sketch and describe the locus of z in each of the following.
(a) |z − 1 + 2i| = 2
7
8
(b) |z + 3 + 2i| = |z − 1 − i|
3π
π
(c) arg (z − 1 + i) = __
(d) arg (z − 2 − 3i) = ___
4
3
Describe and sketch on a separate diagram the following loci of z.
2π
(a) |z + 2 + 3i| = 5
(b) arg (z − 2 − 2i) = − ___
3
(c) |z − 3 − i| = |z + 4 + 2i|
(d) z = (1 + i) + λ(−3 + 5i)
λ∊ℝ
On a single diagram, sketch the loci given by
(a) |z − 2| = 3
(b) |z − 2 − 2i| = |z|
Find the point of intersection of the two loci.
9
36
Find the Cartesian equation of the locus of z where |z − 3 + i| = |z + 1 + 2i|.
MODULE 1tCHAPTER 1
10 Find in Cartesian form the equation of the locus of z where |z − 2 + 3i | = 4.
Describe the locus of z.
11 Find in Cartesian form the equation of the locus of z where
2π . Describe the locus of z.
arg (z − 3 − 4i) = − ___
3
12 Sketch on separate diagrams the locus of z such that
π
(c) |z − 2 + 5i| = 3
(a) |z + 2| = |z − 1|
(b) arg (z − i)= __
4
13 By drawing the locus of |z − (3 + 3i)| = |z| and |z − 3| = 4 on an Argand
diagram, find the exact values of all the complex numbers satisfying both loci.
14 Sketch on an Argand diagram the locus of z where |z − 4 − 2i| = 1.
Hence find the complex number z with the smallest argument. Find also the
complex number z with the largest possible argument.
π on an Argand
15 Draw the line |z − 2i| = |z − 4| and the half-line arg (z − 1) = __
4
diagram. Hence find the complex number which satisfies both equations.
16 Indicate on an Argand diagram the set of points satisfying the condition
1. Find the greatest and least value of arg (z).
|z − 1 − i| = __
2
SUMMARY
Complex numbers
z = x + iy where i = √–1
z = x + iy
Loci in the Argand diagram
Re (z) = x, Im(z) = y
r =z= √x2 + y2
z – c= r, c ∈ℂ, r ∈ℝ
Circle centre c, radius r
Conjugate of z = z, or z*
z = x – iy
y
arg (z) = θ = tan–1 x
θ is measured from the
positive real axis,
θ is in radians and –π < θ ≤ π
z z = x2 + y2
if z1 = a + bi, z2 = c + di
then
z1 + z2 = (a + c) + i(b + d)
z1 – z2 = (a – c) + i(b – d)
z1z2 = (a + bi) (c + di)
= (ac – bd) + (ad + bc)i
()
z = r(cos θ + i sin θ )
z = reiθ
z – a= z – b, a, b ∈ℝ
Perpendicular bisector
of the line joining a to b
arg (z – a) = θ, a ∈ℝ
Half-line starting at a
(excluding a) making an
angle of θ radians with the
positive real axis
De Moivre’s theorem
z = a + λ b, a, b ∈ℂ, λ ∈ℝ
For a real number n
Straight line passing through
(cos θ + i sin θ )n = cos nθ + i sin nθ
a and parallel to b
z1 a + bi a + bi c – di
=
=
×
z2 c + di c + di c – di
z1 = z2
a = b, c = d
37
M O DUL E 1
Checklist
Can you do these?
■ Express complex numbers in the form a + ib, a, b, ∊ ℝ.
■ Calculate the square root of a complex number.
■ Carry out the algebra of complex numbers (add, subtract, multiply, divide).
■ Calculate the complex roots of a polynomial.
■ Find the modulus of a complex number.
■ Find the argument of a complex number.
■ Understand the properties of modulus and argument.
■ Interpret the modulus and argument of complex numbers.
■ Represent complex numbers on an Argand diagram (including sums,
differences, products and quotients).
■ Identify and sketch the locus on an Argand diagram for |z − c| = r, c ∊ ℂ and r ∊ ℝ.
■ Identify and sketch the locus on an Argand diagram for |z − a| = |z − b|, a, b ∊ ℂ.
■ Identify and sketch the locus on an Argand diagram for arg (z − a) = θ, a ∊ ℂ
and θ in radians.
■ Identify and sketch the locus on an Argand diagram for z = a + λb, a, b ∊ ℂ
and λ ∊ ℝ.
■ Convert a locus to Cartesian form.
■ Use de Moivre’s theorem for n ∊ ℤ+.
■ Establish that eiθ = cos θ + i sin θ.
Review exercise 1
1
2
3
−1 + 2i.
(a) Simplify _______
3+i
(b) Find the modulus and argument of −5 + 12i.
5 − 12i, find the modulus and argument of z.
Given that z = _______
3 + 4i
Find the square root of 16 − 30i in the form x + iy where x, y ∊ R.
4
Show that 1 + 4i is a root of the equation 2z3 − z2 + 32z + 17 = 0.
Hence find all the roots of the equation.
5
2
(a) Find the roots
__ of the equation z − 2z + 6 = 0, giving your answers in the
form a ± i √b , where a and b are integers.
(b) Show these roots on an Argand diagram.
(c) For each root, find (i) the modulus (ii) the argument.
38
MODULE 1tCHAPTER 1
6
__
A complex number z satisfies the inequality |z − (−2 + 2 √3 i)| ≤ 2.
Sketch the locus of z on an Argand diagram. Find
(a) the least possible value of |z|,
7
(b) the greatest possible value of arg (z).
Express each of the following complex numbers in exponential form.
__
(b) sin α − i cos α
(a) 1 − √3 i
(c) 1 + cos 2θ + i sin 2θ
8
Show that 3 + 2i is a root of the equation 3z3 − 16z2 + 27z + 26 = 0.
Hence find all the roots of the equation.
9
Describe and sketch the following loci on the same Argand diagram.
π
(a) |z| = |z − 4|
(b) arg (z − i) = __
4
Hence find the complex number which satisfies both equations.
__
10 Use de Moivre’s theorem to express (√3 − i)6 in the form a + bi.
3 + 4i .
11 The complex number w is given by w = ______
1 − 2i
(a) Express w in the form a + bi, where a and b are real.
(b) Sketch an Argand diagram showing the point representing w.
Show on the same diagram the locus of the point z such that |z − w| = 1.
(c) Find the greatest value of arg z for points on this locus.
__
12 Show that z = 1 + √2 i is a root of the equation 2z3 + z2 − 4z + 15 = 0.
Hence find all the roots of the equation.
13 (a) Given that z1 = 1 − i, find z18.
(b) Given also that z1z2 = 5 + 12i, find z2 in the form c + di.
θ
__
θ .
(c) Show that eiθ = 2e i 2 cos __
2
14 (a) Sketch on an Argand diagram the set of points representing all complex
numbers z satisfying both the inequalities |z − i| ≤ 1 and |z − i| ≤ |z|.
( )
(b) Find the square root of −5 + 12i. Hence solve z2 + 4z = 9 − 12i = 0.
eiα where 0 < α < 2π.
15 The complex number z is given by z = ______
1 − eiα
1
α
1
__
__
__
Show that z = − + cot ( ) i.
2
2
2
(
)
16 Find the modulus and argument of −7 + 8i.
Hence express (−7 + 8i)8 in the form x + yi.
1
17 (a) If w = 4 − 3i, express w + __
w in the form a + bi.
(b) Find the square roots of 4i.
(c) Sketch on an Argand diagram the set of points satisfying |z| < |z − 1| and
π < arg z < __
π.
− __
4
4
18 Describe the locus given by |z − 1| = |z + i| and |z − (3 − 3i)| = 2 and sketch
both loci on an Argand diagram.
Find, in an exact form, the points of intersections of the loci.
39
M O DUL E 1
19 (a) Evaluate (1 − i)15.
2π
___
5i
e giving your answer correct to three significant figures.
(b) Evaluate ___
3π
___
e4i
cot θ − i in exponential form.
20 Express ________
cot θ + i
21 Show that the Cartesian equation of the locus of z where
|z − 1 − i| = 2|z − 2 + 3i|
(
) (
)
7 2 + y + ___
13 2 = ___
68 .
is x − __
3
3
9
Clearly describe the locus of z.
nα
___
α e 2 i.
22 Prove that for all n ∊ ℤ and α ∊ ℝ, (1 + cos α + i sin α)n = 2n cosn ( __
)
2
23 Use de Moivre’s theorem to show that cos 5θ = 16
− 20
+ 5 cos θ.
__
1 + √5 .
π = ______
By considering the equation cos 5θ = 0, show that cos ( __
5)
4
π
___
2
Hence deduce the exact value of cos ( ).
10
cos5 θ
40
cos3 θ
MODULE 1tCHAPTER 2
CHAPTER 2
Differentiation
At the end of this chapter you should be able to:
■ find the derivative of e f (x) where f (x) is a function of x
■ find the derivative of ln f (x) where f (x) is a function of x
■ use the chain rule to obtain gradients of tangent and normals
■ carry out implicit differentiation
■ differentiate combinations of polynomial, trigonometric, exponential and
logarithmic functions
■ differentiate inverse trigonometric functions
■ find the second derivative f ″(x) of a function
■ carry out parametric differentiation
■ find the first partial derivative of u = f(x, y) and w = f(x, y, z)
■ find the second partial derivative of u = f(x, y) and w = f(x, y, z).
KEYWORDS/TERMS
EFSJWBUJWFtUSJHPOPNFUSJDGVODUJPOtQPMZOPNJBMt
DIBJOSVMFtDPNQPTJUFGVODUJPOtFYQPOFOUJBM
GVODUJPOtMPHBSJUINJDGVODUJPOtQSPEVDUSVMFt
RVPUJFOUSVMFtUBOHFOUtOPSNBMtJNQMJDJU
EJČFSFOUJBUJPOtTFDPOEEFSJWBUJWFtQBSBNFUSJD
EJČFSFOUJBMtJOWFSTFUSJHPOPNFUSJDGVODUJPOt
QBSUJBMEFSJWBUJWFtĕSTUQBSUJBMEFSJWBUJWFtTFDPOE
PSEFSQBSUJBMEFSJWBUJWF
41
M O DUL E 1
Standard differentials
Differentiation was introduced in Unit 1. We found derivatives of functions using
the definition (from first principles) and we spent time differentiating products
and quotients of trigonometric functions and polynomials. The table of standard
differentials is given below with the extension for Unit 2 included. Familiarise
yourself with this table. In this table a and b are constants.
Function
Derivative
xn
nx n−1
(ax + b)n
na(ax + b)n−1
ex
ex
eax+b
sin x
aeax+b
1
__
x
a
______
ax + b
cos x
cos x
− sin x
tan x
sec2 x
sec x
sec x tan x
cosec x
− cosec x cot x
cot x
− cosec2 x
arcsin x
1
_______
______
ln x
ln (ax + b)
√1 − x2
−1
_______
______
√1 − x2
1
______
1 + x2
arccos x
arctan x
Product rule y = uv
u
Quotient rule y = __
v
dy
dv + v ___
du
___
= u ___
dx
dx
dx
dv
du − u ___
v ___
dy ___________
___
= dx 2 dx
dx
v
Differentiation of ln x
Let f (x) = ln x
f (x + h) − f (x)
By definition f ′(x) = limh→0 _____________
h
f (x) = ln(x + h)
ln(x + h) − ln x
f ′(x) = lim ______________
h
h→0
x+h
ln _____
= lim _________
x
h→0
h
h
1 ln 1 + __
= lim __
x
h
h→0
[
42
(
(
)
]
)
MODULE 1tCHAPTER 2
h ⇒ h = xt. As h → 0, t → 0, therefore
Let t = __
x
Remember
1
__
lim (1 + t) t = e
t→0
[
1 ln(1 + t)
f ′(x) = lim __
xt
t→0
1 ln(1 + t)__1t
= lim __
x
t→0
]
1
1 ln e = __
= __
x
x
1.
Hence when f (x) = ln x, f ′(x) = __
x
Differentiation of ex
Let y = ex. Taking logs to base e, we get
ln y = x
Differentiating with respect to x gives
dy
dy
1 ___
x
__
___
y dx = 1 therefore dx = y = e
dy
Hence when y = ex, ___ = ex .
dx
EXAMPLE 1
Differentiate with respect to x:
SOLUTION
(a) y = e4x
(a) y = e4x, (b) y = 3e−2x, (c) y = 6e3x+5
dy
___
= 4e4x
dx
(b) y = 3e−2x
dy
___
= −6e−2x
dx
(c) y = 6e3x+5
dy
___
= 6 × 3e3x+5 = 18e3x+5
dx
Chain rule (function of a function rule)
Remember
d (e x) = e x
___
dx
d (eax+b) =
and ___
dx
aeax+b where
a and b are
constants.
The chain rule is used to differentiate a composite function or a function of a
function. When using the chain rule we need to keep in mind that one function is
inside the other and we let u be the function inside.
Let y = f g (x) and u = g (x)
dy dy du
Then ___ = ___ × ___
dx du dx
Let us start with differentiating exponential functions.
43
M O DUL E 1
2+3
EXAMPLE 2
Differentiate y = e4x
SOLUTION
This is a function of a function with 4x 2 + 3 being the function inside the
exponential function.
Use the chain
rule.
From the table
on page 42
d (e x ) = e x
___
dx
with respect to x.
We let u = 4x2 + 3
dy
du = 8x
Then y = eu and ___ = eu, ___
du
dx
Using the chain rule:
2+3
y = e4x
← inside
↑
outside
dy ___
dy ___
___
=
× du
dx du dx
= 8x eu
Substituting u = 4x2 + 3 gives
dy
2
___
= 8x e4x +3
dx
EXAMPLE 3
Find the gradient function of y = esin x + 2 cos x
SOLUTION
This function is a composite function with
sin x + 2 cos x being the function inside the
exponential function.
Let u = sin x + 2cos x
dy
du = cos x − 2 sin x
∴ y = eu, ___ = eu and ___
du
dx
dy ___
dy ___
___
=
× du = (cos x − 2 sin x)eu
dx du dx
inside
↓
y = esin x + 2 cos x
↑
outside
dy
Using ___ = f ′g(x) × g′(x)
dx
dy
___
= (cos x − 2 sin x)e sin x + 2 cos x
dx
↑
↑
f ′g(x)
g′(x)
Substituting u = sin x + 2 cos x gives
dy
___
= (cos x − 2 sin x)esin x + 2 cos x
dx
EXAMPLE 4
SOLUTION
44
3 + 2x
Find the gradient of the curve y = 5e3x
when x = 0.
dy
We need to find ___ when x = 0.
dx
3
Since y = e3x +2x, let u = 3x 3 + 2x, and y = 5eu
du = 9x2 + 2
___
dx
dy
___
= 5eu
du
dy ___
dy ___
___
=
× du = 5 (9x2 + 2)eu
dx du dx
3
= 5 (9x2 + 2)e3x +2x
dy
3
When x = 0, ___ = 5(9(0)2 + 2) e3(0) +2(0) = 10e0 = 10
dx
g(x)
↓
3 + 2x
y = 5e3x
↑
f
dy
___
= f ′g(x) × g′(x)
dx
3
= 5(9x2 + 2)e3x + 2x
↑
↑
g′(x)
f ′g(x)
MODULE 1tCHAPTER 2
(a) Find the gradient function of (i) y = 4e3x2+7x+2 (ii) y = etan x
Try these 2.1
π.
(b) Find the gradient of the tangent to the curve y = e3 cos x − sin x at x = __
2
(c) Given that y = 5esec x, find the rate of change of y with respect to x.
d [ln x] = __
1 and
Moving on to logarithmic functions, recall that ___
x
dx
d [ln (ax + b)] = ______
a , where a and b are constants.
___
dx
ax + b
EXAMPLE 5
Find the derivative of y = ln (6x + 3).
SOLUTION
y = ln (6x + 3) is a function of a function with 6x + 3 inside the logarithmic function.
We let u = 6x + 3, y = ln u
g
↓
y = ln (6x + 3)
↑
f
g′
↓
dy ______
___
= 1 ×6
dx 6x + 3
↑
f ′g
6
= ______
6x + 3
dy __
du = 6 and ___
___
=1
dx
du u
dy dy du
Using ___ = ___ × ___
dx du dx
6
1
__
= __
u×6=u
Substituting u = 6x +3,
dy ______
___
= 6
dx 6x + 3
EXAMPLE 6
Find the derivative of y = ln (cos x − sin x) with respect to x.
SOLUTION
y = ln (cos x − sin x) is a function of a function with cos x − sin x inside the
logarithmic function.
We let u = cos x − sin x and y = ln u
y = ln (cos x − sin x)
dy __
du = −sin x − cos x and ___
1
___
=u
dy ___________
1
___
dx
du
× (−sin x − cos x)
=
dx cos x − sin x
dy
dy
du
___ = ___ × ___
dx du dx
−sin x − cos x
= ____________
dy __
cos x − sin x
1
___
= u × (−sin x − cos x)
dx
−sin x − cos x
= ____________
(substituting u = cos x − sin x)
cos x − sin x
EXAMPLE 7
Differentiate y = ln (ex + 6) with respect to x.
SOLUTION
Let u = ex + 6
∴ y = ln u
Hence
dy __
du = ex and ___
___
=1
dx
du u
dy ___
dy ___
___
=
× du
dx du dx
ex
1
1
______
______
x
x
= __
u × e = ex + 6 × e = ex + 6
(since u = ex + 6)
45
M O DUL E 1
(
)
EXAMPLE 8
x + 1 with respect to x.
Differentiate y = ln _____
x+2
SOLUTION
Using rules of logs will make it easier to differentiate the function.
(
)
x + 1 = ln (x + 1) − ln (x + 2)
y = ln _____
x+2
( ln ( __ab )= ln a − ln b )
dy
1
1 − _____
Hence ___ = _____
dx x + 1 x + 2
_______
EXAMPLE 9
Differentiate y = ln √2x2 + 3 .
SOLUTION
y = ln √2x2 + 3 = ln (2x 2 + 3) 2
_______
1
__
Using the rules of logs (ln x n = n ln x)
1 ln (2x2 + 3)
function inside
y = __
2
Differentiating using the chain rule
[
]
dy __
2x
1
___
(4x) = _______
= 1 _______
dx 2 2x2 + 3
2x2 + 3
Try these 2.2
f
g
↓
↓
1 ln (2x2 + 3)
y = __
2
dy __
1
___
× 4x
= 1 × _______
dx 2 2x2 + 3
↑
↑
f ′g
g′
Differentiate the following functions with respect to x.
(a) y = ln (4x2 + 3x + 2)
(b) y = ln (tan 2x)
1
(c) y = ln (2x + 1)__2
Differentiating exponential functions of the form
y = ax
Let y = ax. Taking logs to base e,
ln y = ln ax
Use the chain
rule to differentiate ln y since
this is a composite function.
inside
↓
d [ln(y)]
___
dx
↑
outside
dy
d (y) = ___
___
dx
dx
dy
d
1 ___
___[ln y] = __
y dx
dx
46
∴ ln y = x ln a
Differentiate both sides with respect to x: since we are differentiating with respect to
dy
1 ___
x, the derivative of ln y is __
y dx .
dy
1 ___
__
y dx = ln a
dy
∴ ___ = y ln a
dx
dy
Since y = ax, replacing this in ___, we get
dx
dy
___
x
= a ln a
dx
dy
Hence if y = ax then ___ = ax ln a.
dx
MODULE 1tCHAPTER 2
E X A M P L E 10
dy
Given that y = 2x, find ___ when x = 2.
dx
SOLUTION
Let y = 2x
If the derivatives
are known they
can be used as
standard results,
e.g.
d ( 3x ) = 3x ln 3.
___
dx
Taking logs to base e,
ln y = ln 2x
∴ ln y = x ln 2
Differentiating both sides with respect to x:
dy
1 ___
__
y dx = ln 2
dy
∴ ___ = y ln 2
dx
dy
Since y = 2x, replacing this in ___, we get
dx
dy
___
= 2x ln 2
dx
dy
Hence if y = 2x then ___ = 2x ln 2.
dx
Replacing x = 2,
dy
___
= 22 ln 2 = 4 ln 2
dx
Differentiating logarithms of the form y = loga x
Converting y = loga x to exponential form
x = ay
Taking logs to base e
ln x = ln a y
ln x = y ln a
Differentiating with respect to x
dy
1 = ___
__
x dx ln a
dy _____
___
= 1
dx x ln a
dy
1 .
Hence if y = loga x then ___ = _____
dx x ln a
E X A M P L E 11
dy
Given that y = log10 x, find ___.
dx
SOLUTION
Converting y = log10 x to exponential form
x = 10 y
d log x = _____
1
___
( a )
Taking logs to base e
ln x = ln 10 y
ln x = y ln 10
Remember
dx
x ln a
d log x = ______
1
___
dx ( 10 ) x ln 10
(by rules of logs)
47
M O DUL E 1
Differentiating with respect to x
dy
1 = ___
__
x dx ln 10
dy ______
___
= 1
dx x ln 10
E X A M P L E 12
SOLUTION
dy
Given that y = log3 x, find ___.
dx
dy
1 . Using this known standard
If y = loga x then ___ = _____
dx x ln a
result, many questions can be answered quickly in the
following manner.
dy
1 .
If y = log3 x then ___ = _____
dx x ln 3
(
)
E X A M P L E 13
dy _____________
2x + 1 , show that ___
−5
Given that y = ln ______
=
.
x−2
dx (2x + 1)(x − 2)
SOLUTION
Using rules of logs
(
)
2x + 1 = ln (2x + 1) − ln (x − 2)
y = ln ______
x−2
Differentiating with respect to x, we have
dy ______
1
___
= 2 − _____
dx 2x + 1 x − 2
dy _________________
2(x − 2) − (2x + 1) _____________
−5
___
=
=
(2x + 1)(x − 2)
(2x + 1)(x − 2)
dx
E X A M P L E 14
Find the gradient of the curve y = xx at x = 1.
SOLUTION
y = xx
Did you try this?
d ( x x ) = (x − 1) x x–1
___
dx
Why is it incorrect?
Taking logs to base e and using rules of logs
ln y = ln xx = x ln x
Differentiating with respect to x, we have
dy
1
1 ___
__
__
y dx = x ( x ) + ln x = 1 + ln x
dy
∴ ___ = xx ( 1 + ln x )
dx
dy
Substituting x = 1, ___ = 11 ( 1 + ln 1 ) = 1
dx
48
This is useful when
answering multiple
choice questions.
Remember
( )
a = ln a − ln b
ln __
b
Using rules of logs
and differentiating is
much faster than
differentiating
directly.
MODULE 1tCHAPTER 2
_____
E X A M P L E 15
dy ______
1 − x , show that ___
Given that y = ln _____
= 1 .
1+x
dx x2 − 1
√
_____
SOLUTION
Remember
When finding
the derivatives
of log functions
applying the
rules of logs first
may make the
differential easier
to cope with.
1−x
y = ln _____
1+x
1 − x __12
y = ln _____
1+x
√
(
)
Using rules of logs
1 {ln (1 − x) − ln (1 + x)}
y = __
2
Differentiating with respect to x, we have
dy __
−1 − _____
1
___
= 1 _____
dx 2 1 − x 1 + x
{
{
}
−1(1 + x) − (1 − x)
1 __________________
= __
2
(1 − x)(1 + x)
{
1 ______
−2
= __
2 1 − x2
}
}
−1 = ______
1
= ______
1 − x2 x2 − 1
Try these 2.3
Differentiate the following with respect to x.
(a) y = 5x
(b) y = 12x
(c) y = log6 5x
(d) y = log10 (2x)
EXERCISE 2A
1
Find the derivative of the following functions.
(a) y = ln (4x − 5)
(b) y = ln (x2 + 2x + 4)
(c) y = ln (3x2 + 2)
2x + 1
(d) y = ln ______
x−2
(e) y = ln(4x + 7)6
(f)
(
5x − 3
(g) y = ln _______
(3x+5)4
)
(
)
4x + 1
y = ln ( ______
7x − 2 )
______
(h) y = ln √3 − 4x
(i) y = ln cos3 x
2
Find the rate of change of y with respect to x for the following functions.
(a) y = 4e x
(c) y = e x
(b) y = 7e5x + 2
2
3+3x − 2
______
(d) y = e√4x − 1
(e) y = ecos x − sin 2x
(f) y = etan 4x
(g) y = e6 cos 6x
(h) y = 5ecos 4x + 3
______
(i) y = 4e−√ x2 + cos x
49
M O DUL E 1
3
Find the gradient function for each of the following.
(a) y = 7x
(b) y = 6x
(c) y = (4x + 1)x
(d) y = (1 − 2x)x+1
(e) y = log2 (2x + 1)
(f) y = log3 x 2
_____
(g) y = log10 √x + 2
(h) y = 5 log4 (2 − 3x)
4
Given that y = 4x, find the rate of change of y with respect to x.
5
dθ when t = 2.
6 . Find the value of ___
Two variables θ and t are related by θ = __
dt
3t
6
Find the gradient of the curve y = e2x−x when x = 1.
7
8
2
dμ
dμ
Show that ___ = 4 tan t when μ = ln (3 cos4 t) and find the value of ___ when
dt
dt
π
__
t= .
4
Find the gradient of the curve f (x) = x2x + 1 at x = 1.
(
)
dy
4x + 2 find ___
Given that y = ln ______
.
3x − 1
dx
__
dy
sin2 x .
π when y = ln ________
10 Show that ___ = 2 √2 at x = __
4
1
+
cos x
dx
9
(
)
Differentiation of combinations of functions
In this section we are going to find the derivatives of functions by combining the
product rule, quotient rule and chain rule with the table of standard derivatives.
The product rule and quotient rule are given below.
dy
dv + v ___
du .
Product rule: If y = uv where u and v are functions of x then ___ = u ___
dx
dx
dx
du − u ___
dv
___
dy v___________
dx
dx .
u
__
___
=
Quotient rule: If y = v where u and v are functions of x then
dx
v2
E X A M P L E 16
dy
Find ___ when y = ex ln x.
dx
SOLUTION
Since y = ex ln x is a product, we use the product rule to differentiate the function.
Use the product
rule.
50
Let u = ex, v = ln x
dv = __
1
___
dx x
dy
dv + v ___
du , we get
Substituting into ___ = u ___
dx
dx
dx
dy
1 + (ln x)ex
___ = ex __
x
dx
1 + ln x
= ex ( __
)
x
du = ex,
___
dx
MODULE 1tCHAPTER 2
E X A M P L E 17
dy
x+1 .
Find ___ when y = _______
dx
2x2 + 1
SOLUTION
x+1
y = _______
2x2 + 1
Use the quotient
rule.
du = 1
Let u = x + 1, ___
dx
dv = 4x
v = 2x2 + 1, ___
dx
dv
du − u ___
v ___
dy __________
dx
dx
___
, we get
Substituting into
=
dx
v2
dy ______________________
(2x2 + 1)(1) − (x + 1)(4x)
___
=
dx
(2x2 + 1)2
2x + 1 − 4x − 4x
= ________________
(2x2 + 1)2
2
2
(expanding and simplifying the numerator)
−2x − 4x + 1 = ___________
1 − 4x − 2x
= _____________
(2x2 + 1)2
(2x2 + 1)2
2
2
Differentiation of combinations involving trigonometric functions
E X A M P L E 18
Differentiate y = sin (4x + 6) with respect to x.
SOLUTION
y = sin (4x + 6) is a function of a function.
Use the chain
rule.
Let u = 4x + 6, then y = sin u
dy
du = 4 and ___
___
= cos u
dx
du
dy dy du
, we get
Substituting into ___ = ___ × ___
dx du dx
dy
___
= 4 cos u = 4 cos (4x + 6)
dx
E X A M P L E 19
Differentiate y = sin4 x with respect to x.
SOLUTION
Let u = sin x, then y = u4
dy
du = cos x and ___
___
= 4u3
dx
du
dy ___
dy ___
___
=
× du
dx du dx
Use the chain
rule.
= 4u3 cos x
= 4 sin3 x cos x
E X A M P L E 20
Find the gradient function of y = tan6 (4x + 5).
SOLUTION
Rewrite the function as
y = [tan (4x + 5)]6
Use the chain
rule.
51
M O DUL E 1
Let u = tan (4x + 5), y = u6
dy
du = 4 sec2(4x + 5) and ___
___
= 6u5
dx
du
dy ___
dy ___
___
=
× du
dx du dx
= 6u5 × 4 sec2 (4x + 5)
= 24 u5 sec2 (4x + 5)
= 24 (tan5 (4x + 5)) sec2 (4x + 5)
E X A M P L E 21
Differentiate y = cos 3ϕ tan 3ϕ with respect to ϕ.
SOLUTION
y = cos 3ϕ tan 3ϕ is a product of two functions of ϕ, so we can use the product rule.
Let u = cos 3ϕ, v = tan 3ϕ
du = −3 sin 3ϕ, ___
dv = 3 sec2 3ϕ
___
dϕ
dϕ
dy
dv + v ___
du
Substituting into the product rule ___ = u ___
dϕ
dϕ
dϕ
dy
___
= 3 sec2 3ϕ cos 3ϕ − 3 sin 3ϕ tan 3ϕ
dϕ
sin 3ϕ
= 3 (tan2 3ϕ + 1) cos 3ϕ − sin 3ϕ ______
cos 3ϕ
(
(
sin2 3ϕ
sin2 3ϕ
= 3 ______
cos 3ϕ + cos 3ϕ − ______
2
cos 3ϕ
cos 3ϕ
)
)
= 3 cos 3ϕ
E X A M P L E 22
ex
Find the gradient function of the curve y = __________
.
cos (2x) + 6
SOLUTION
e
y = __________
is a quotient of two functions of x.
cos (2x) + 6
dv
du − u ___
v ___
dx
dx
Using the quotient rule y = ___________
v2
x
where u = ex
du = ex
___
dx
v = cos (2x) + 6
dv = −2 sin (2x)
___
dx
dy _____________________________
(cos (2x) + 6) ex − ex (−2 sin (2x))
___
=
dx
(cos (2x) + 6)2
ex(cos (2x) + 2 sin (2x) + 6)
= _______________________
(cos (2x) + 6)2
52
OR
y = cos 3ϕ tan 3ϕ
sin 3ϕ
= cos 3ϕ ______
cos 3ϕ
y = sin 3ϕ
dy
___
= 3cos 3ϕ
dϕ
MODULE 1tCHAPTER 2
E X A M P L E 23
Find the gradient of the curve θ = 6 cos ( 3t ) e4t
SOLUTION
θ = 6 cos (3t) e4t
2+5
2+5
at t = 0.
represents a product of two functions of t.
Using the product rule
du = −18 sin (3t)
let u = 6 cos ( 3t ), then ___
dt
2
and v = e4t +5
Using the chain rule
w = 4t2 + 5,
dw = 8t,
___
dt
v = ew
dv = ew
___
dw
dv = ___
dv × ___
dw
___
dt dw
dt
2
= 8tew = 8te4t +5
dθ = 6 cos (3t) (8te4t
∴ ___
dt
= 6e4t
2+5
2+5
) + e4t
2+5
(−18 sin (3t))
[8t cos (3t) − 3 sin (3t)]
When t = 0,
dθ = 6e4(02)+5 [8(0) cos (3(0)) − 3 sin (3 (0))] = 0
___
dt
Try these 2.4
(a) Find the derivative of
(i) y = e2x cos x
(ii) y = 4 ln x sin (3x + 2)
(b) Find the gradient of the following functions at the given point.
π
(i) θ = tan (t + 3) sec t, when t = __
4
4t−1
3
(ii) θ = e
(t + 2), when t = 0
EXERCISE 2B
1
Find the derivatives of the following functions.
(a) (2x + 3) ln x
(b) (4x − 1) sin x2
(c) 3 ln x cos x
(d) e4x+1 cos 4x
(e) e cos x ln 4x
(f) ln3 x tan x2
_________
(g) √ln (3x + 5) sin x
(h) x ln sin (4 − 2x)
(i) 2x sec 2x
(j) 3−x ln x2
53
M O DUL E 1
2
Differentiate the following functions.
cos x
(a) y = ________
sin x + 1
ln (3x − 5)
(b) y = _________
x+2
2x + 4
(c) y = _________
ln (2x + 4)
e
(d) y = ________
2 + sin x
2sin 5x
(e) y = ______
ex + 2
x + sec 2x
(f) y = _________
e3x − 4
(g) y = sin3 x + tan2 x
cos3 (2x)
(h) y = _______
sin2 x
x
3
dθ when t = π.
π sin 2 (t − π), find the value of ___
Given that θ = cos ( t + __
2)
dt
4
Find the coordinates of the point on the curve y = 4 xex at which the gradient is 0.
5
dy
Given that y = 4x, show that ___ = 32 ln 2 when x = 2.
dx
6
dy
Find the values of x for which ___ = 0, where y = x 2x.
dx
7
8
9
__
dθ = 2 √2 e2t sin 2t + __
Given that θ = e2t sin 2t, show that ___
( π4 ).
dt
dθ when t = __
π.
Hence find the value of ___
8
dt
dy
Show that y = 4e−2 when ___ = 0 where y = x2 ex.
dx
dy
5 when ___
Given that y = ln (sin x − cos x), show that tan x = __
= 4.
3
dx
dx = ___
2__ .
10 A function x is given by x = ln (sec t + tan t). Show that x = __21 ln 3 when ___
dt
√3
Tangents and normals
Gradients of tangents and normals
Consider the function y = f (x) with a point (x, y) lying on the graph of the
function. The tangent line to the function at (x, y) is a line that touches the curve
at one point. Both the graph of y = f (x) and the tangent line pass through this point
and the gradient of the tangent line and the gradient of the function have the same
value at this point. The gradient of the tangent at x = a is the
y
dy
value of ___ when x = a. The normal is perpendicular to the
dx
normal
dy
tangent. If the gradient of the tangent is ___ then the gradient
P(x, y)
dx
1
___
dy since the product of the gradients of
of the normal is − ___
tangent x
dx
perpendicular lines is −1.
54
MODULE 1tCHAPTER 2
E X A M P L E 24
SOLUTION
Find the gradient of the tangent to the curve y = (2x + 1)x−2 at the point x = 0.
dy
The gradient of the tangent is ___ at this point.
dx
y = (2x + 1)x−2
Taking logs to base e, we get
ln y = ln {(2x + 1) x−2}
ln y = (x − 2) ln (2x + 1)
Differentiating with respect to x
dy _______
2(x − 2)
1 ___
__
y dx = 2x + 1 + ln (2x + 1)
Note
Use the product rule, where
u = x − 2, v = ln (2x + 1).
du = 1, ___
dv = ______
2
___
dx
dx 2x + 1
When x = 0, y = (2(0) + 1)0−2 = 1
Substituting into the derivative
dy ________
2(0 − 2)
___
=
+ ln (2(0) + 1)
dx 2(0) + 1
−4 + ln 1 = −4
= ___
1
E X A M P L E 25
Find the gradient of the tangent to the curve y = (x + 2) ln x at the point x = 1.
SOLUTION
dy
The gradient of the tangent is ___ at x = 1.
dx
Using the product rule with u = x + 2 and v = ln x
dv = __
1
and ___
dx x
dy
1 + (1) ln x
We have ___ = (x + 2) __
x
dx
Substituting x = 1
dy
1 + (1) ln 1 = 3
___
= (1 + 2) __
1
dx
du = 1
___
dx
∴ The gradient of the tangent is 3.
E X A M P L E 26
SOLUTION
(
)
x + 2 at the point x = 2.
Find the gradient of the normal to the curve y = ln _____
x−1
dy
We find the gradient of the tangent first, i.e. ___ when x = 2.
dx
x
+
2
_____
y = ln
x −1
(
)
Using rules of logs
y = ln (x + 2) − ln (x − 1)
Differentiating with respect to x
dy _____
1
___
= 1 − _____
dx x + 2 x − 1
55
M O DUL E 1
dy
3
1 − _____
1 = __
1 − 1 = − __
When x = 2, ___ = _____
4
4
2
+
2
2
−
1
dx
1
4
Gradient of the normal = − ________________ = __
gradient of tangent 3
Equations of tangents and normals
Let m be the gradient of the tangent at the point (x1, y1). Using the equation of a
straight line, the equation of the tangent is y − y1 = m (x − x1).
Since the tangent and normal are perpendicular to each other the gradient of the
1
1
__
normal is − __
m and the equation of the normal is y − y1 = − m (x − x1).
E X A M P L E 27
π.
Find the equation of the tangent to the curve y = x cos x at the point x = __
2
SOLUTION
y = xcos x
Find the gradient of the tangent
ln y = ln xcos x = (cos x) ln x
Differentiating with respect to x
dy
1 ___
1
__
__
y dx = cos x ( x ) − (sin x) ln x
π, y = __
π cos __π2 = 1, since cos __
When x = __
(
( π2 ) = 0 and ( __π2 )0 = 1.
2
2)
( )
dy
π ____
π ln __
π
1 ___
1 − sin __
__
= cos ( __
2 ) π/2
2 ( 2)
1 dx
dy
π
∴ ___ = − ln __
2
dx
Hence the equation of the tangent is
π x − __
π
y − 1 = − ln __
2(
2)
π x − __
π
y = 1 − ln __
2(
2)
π + 1 + __
π ln __
π
y = −x ln __
2
2
2
E X A M P L E 28
SOLUTION
Find the equation of the tangent and the equation of the normal to the curve
π.
y = x2 cos x when x = __
2
π by differentiating
We find the gradient of the curve at x = __
2
y = x2cos x
Using the product rule
u = x2,
56
v = cos x
du = 2x, ___
dv = −sin x
___
dx
dx
dy
___
= −x2sin x + 2x cos x
dx
MODULE 1tCHAPTER 2
( )
dy
π2 sin __
π + 2 __
___
= − ___
( π2 ) cos __π2
2
2
dx
π2
= − ___
4
2
π, y = __
When x = __
( π2 ) cos __π2 = 0
2
π, 0 , m = − ___
π2
∴ At ( __
4
2 )
π,
When x = __
2
The equation of the tangent is
π2 x − __
π
y − 0 = −___
4 (
2)
π2 x + ___
π3
y = − ___
4
8
π , the gradient of the normal is ___
4 .
Since the gradient of the tangent is − ___
4
π2
The equation of the normal is
2
π
4 x − __
y − 0 = ___
2)
π2 (
2
4 x − __
y = ___
π
π2
E X A M P L E 29
Find the equation of the tangent to the curve y = ex cos x at the point x = 0.
SOLUTION
y = ex cos x
dy
dv + v ___
du
___
= u ___
dx
dx
dx
We use the product rule to find the gradient function.
du = ex
___
u = ex,
dx
dv = −sin x
___
v = cos x,
dx
dy
___
= −ex sin x + ex cos x
dx
dy
When x = 0, ___ = −e 0 sin 0 + e0 cos 0
dx
=1
We need the value of y: when x = 0, y = e0 cos 0 = 1
dy
∴ At (0, 1), ___ = 1
dx
The equation of the tangent is
y − 1 = 1(x − 0) = x
y=x+1
Try these 2.5
x − 1 at the point x = 2.
(a) Find the equation of the tangent to the curve y = ______
x+1
(b) Find the equation of the normal to the curve y = ex sin x at the point x = 0.
(c) Find the equation of the tangent to the curve and the equation of the normal to
the curve y = x2 ln (x + 1) when x = 1. Give your answer in exact form.
57
M O DUL E 1
EXERCISE 2C
In questions 1–4, find the gradient of the tangent to the curve at the given point.
x2 + 2 , 2, _____
6
π, 0
1 y = x cos x, ( __
2 y = ______
2 )
x ln x
2 ln 2
π, 0
3 y = (4x + 2) e4x−1, __41 , 3
4 y = ln sin 2x, ( __
4 )
(
)
( )
In questions 5 and 6, find the gradient of the normal to the curve at the given
value of x.
e x+3
, x=1
5 y = x3 e x 3 + 3x, x = 1
6 y = _________
ln (2x + 1)
7 Given that the gradient of the normal to a curve is − __41 at the point (2, 3),
find the equation of the tangent to the curve at (2, 3).
8 Find the equation of the tangent to the curve y = x2x+1 at x = __21 .
9 Find the equation of the tangent to the curve y = x2ex at the point (1, e).
Leave your answer in terms of e.
10 Given that y = (x2 + 1) cos 2x, show that the equation of the tangent to the
curve at (0, 1) is y = 1.
11 Show that the equation of the normal to the curve y = sin x e cos x at (0, 0) is
ey + x = 0.
dy
dx
π.
equation of the tangent to the curve at x = __
4
−1
13 Find the equation of the normal to the curve y = etan x at the point x = 1.
dy
14 The equation of a curve is given by y = ln (cos 2x). Find ___ and the equation of
dx
π.
the normal to the curve at x = __
6
x + 2 at the points
15 Find the equations of the normals to the curve y = _________
x2 + x − 2
where the curve cuts the x-axis.
12 Given that y = 8 cos 2x sin 2x, show that ___ = 16 cos 4x. Hence find the
16 The equation of a curve is y = x 3 + 6x 2 + 11x + 6. Find
(a) the gradient at the point (1, 1),
(b) the x-coordinate of the point at which the tangent to the curve is parallel to
the tangent at (1, 1).
Implicit differentiation
The functions we have worked with so far have all been given by equations of the
form y = f(x). A function of this form is an explicit function. For example, the
______
2x + 3 , y = e x+2 are all explicit functions.
2
functions y = 4x + 3x − 2, y = ______
x+4
There are functions in which y cannot be written explicitly in terms of x. In these
functions y is said to be in implicit form. For example, in equations such as
x2 + 3xy − 4y3 = 4x and x4 + ex+y − x2y = 6y, y is defined implicitly as a function
dy
of x. The technique for finding ___ for implicit functions is called implicit differentation.
dx
√
58
MODULE 1tCHAPTER 2
E X A M P L E 30
SOLUTION
Differentiate y2 with respect to x.
dy
d [y2] × ___
d [y2] = ___
___
dy
dx
dx
dy
= 2y ___
dx
dy
We make use of the chain rule, keeping in mind that the differential of y is ___ .
dx
E X A M P L E 31
Differentiate x2y2 with respect to x.
SOLUTION
[x2y2] is a product of two functions of x, so we need to use the product rule.
Let u = x2, v = y2
dy
du = 2x, ___
dv = 2y ___
___
dx
dx
dx
dy
d [x2 y2] = x2 2y ___
+ y2[2x]
∴ ___
dx
dx
dy
= 2x 2 y ___ + 2xy 2
dx
[
]
y = uv
dy
dv + v ___
du
⇒ ___ = u ___
dx
dx
dx
E X A M P L E 32
dy
Find ___ in terms of x and y for the equation x2 + 3y + 2xy2 = 4.
dx
SOLUTION
x2 + 3y + 2xy2 = 4
Differentiating term by term, we have
d [x2] = 2x
___
dx
dy
d [3y] = 3 ___
___
dx
dx
(
)
dy
d [2xy2] = 2x 2y ___
___
+ y2 (2), using the product rule with u = 2x and v = y2
dx
dx
dy
= 4xy ___ + 2y2
dx
d [4] = 0
___
dx
∴ Differential of x2 + 3y + 2xy2 = 4 is
dy
dy
2x + 3 ___ + 4xy ___ + 2y2 = 0
dx
dx
dy
Making ___ the subject of the formula
dx
dy
dy
3 ___ + 4xy ___ = −2y2 − 2x
dx
dx
dy
___
[3 + 4xy] = −2y2 − 2x
dx
dy _________
−2y2 − 2x
___
=
3 + 4xy
dx
59
M O DUL E 1
(
dy
1, 1 .
Given that 3xy + 4x2 y3 = 5x, find the value of ___ at __
2
dx
SOLUTION
3xy + 4x2y3 = 5x
Differentiating each term,
dy
d [3xy] = 3x ___
___
+ 3y
(product rule)
dx
dx
dy
d [4x2y3] = 4x2 3y2 ___
___
+ y3(8x)
(product rule)
dx
dx
dy
= 12x2y2 ___ + 8xy3
dx
d
___ [5x] = 5
dx
(
)
∴ The differential of 3xy + 4x2y3 = 5x is
dy
dy
3x ___ + 3y + 12x2y2 ___ + 8xy3 = 5
dx
dx
dy
Making ___ the subject of the formula
dx
dy
dy
3x ___ + 12x2y2 ___ = 5 − 8xy3 − 3y
dx
dx
dy
(3x + 12x2y2) ___ = 5 − 8xy3 − 3y
dx
dy ____________
5 − 8xy3 − 3y
___
=
dx
3x + 12x2y2
1 , y = 1, we get
Substituting x = __
2
1 (1)3 − 3(1)
5 − 8 __
dy _________________
2
___
=
1 + 12 __
1 2 (1)2
dx
3 __
2
2
5−4−3
= _________
3+3
__
2
4
= − __
9
( )
( )
( )
E X A M P L E 34
Find the gradient of ex+y + 3x2 − 2y = 1 at (0, 0).
SOLUTION
ex+y + 3x2 − 2y = 1
Differentiate each term.
d [ex+y] (This term is a function of a function.)
___
dx
Let u = x + y
dy
du = 1 + ___
___
dx
dx
d [eu] = eu
___
du
60
)
E X A M P L E 33
MODULE 1tCHAPTER 2
( )
( 1 + ___dxdy )
dy
d [ex+y] = eu × 1 + ___
∴ ___
dx
dx
= ex+y
(substituting u = x + y)
d [3x2] = 6x
Now ___
dx
dy
d [−2y] = −2 ___
___
dx
dx
d [1] = 0
___
dx
The differential of ex+y + 3x2 − 2y = 1 is
dy
dy
ex+y 1 + ___ + 6x − 2 ___ = 0
dx
dx
(
)
dy
dy
ex+y + ex+y ___ + 6x − 2 ___ = 0
dx
dx
dy
dy
ex+y ___ − 2 ___ = −6x − ex+y
dx
dx
dy x+y
___
[e − 2] = −6x − e x+y
dx
dy __________
− ex+y
___
= −6x
x+y
dx
e −2
When x = 0, y = 0,
Try these 2.6
dy __________
−6(0) − e0 ___
___
= −1 = 1
=
0
−1
dx
e −2
dy
dx
(i) 3x2y2 + 4x = 6x
(a) Find ___ for
(ii) 6x2y + 2x2y2 = 4xy
(iii) 7x3 + 4y = 3y2
(b) Find the gradient of the following at the given point.
(i) 4e3x+y − 2x2 = 4 at (0, 0)
π
(ii) cos (xy) − 3x 4 + 3 = 0 at ( 1, __
2)
E X A M P L E 35
Find the equation of the tangent and the equation of the normal to the curve
xy2 − 3x2y − 4x = 0 at the point (1, 4).
SOLUTION
Differentiating xy2 − 3x2y − 4x = 0, we get
dy
d [xy2] = y2 + x 2y ___
___
dx
dx
dy
d [3x2y] = 3 x2 ___
___
+ 2xy
dx
dx
d [4x] = 4
___
dx
dy
dy
∴ y2 + x 2y ___ − 3 x2 ___ + 2xy − 4 = 0
dx
dx
(
(
(
) (
)
)
)
61
M O DUL E 1
Substituting x = 1, y = 4
(
) (
)
dy
dy
42 + 2(4) ___ − 3 ___ + 2(4) − 4 = 0
dx
dx
dy
dy
16 + 8 ___ − 3 ___ − 24 − 4 = 0
dx
dx
dy
5 ___ = 12
dx
dy ___
___
= 12
5
dx
12 is
The equation of the tangent at x = 1, y = 4 with gradient ___
5
12 (x − 1)
y − 4 = ___
5
12 + 4
12
___
y = x − ___
5
5
8
12
___
__
x+
y=
5
5
5
12 , the gradient of the normal is − ___
Since the gradient of the tangent is ___
12 .
5
The equation of the normal is
5
y − 4 = − ___
12 (x − 1)
5
5
___
y = − ___
12 x + 12 + 4
5
53
___
y = − ___
12 x + 12
Differentiation of inverse trigonometric functions
Differentiation of y = sin−1x
Let y = sin−1 x, where −1 ⩽ x ⩽ 1.
f f −1(x) = x
sin sin−1(x) = x
Applying the sine function to both sides
sin y = sin sin−1 x
sin y = x
(1)
dy
d [y] = ___
Differentiating two sides with respect to x and recalling ___
dx
dx
dy
cos y ___ = 1
dx
dy
1
∴ ___ = ____
dx cos y
To obtain cos y in terms of x, use the identity cos2 y + sin2 y = 1
cos 2 y = 1 − sin2 y
________
cos y = √ 1 − sin2 y
From (1) above, sin y = x
______
∴ cos y = √ 1 − x2
dy
1
______
∴ ___ = _______
dx √1 − x2
d [sin−1 x] = ________
1
______
Hence ___
dx
√1 − x2
62
MODULE 1tCHAPTER 2
Differentiation of y = tan−1x
Let y = tan−1 x
Applying the tan function to both sides
tan y = x
d [tan x] = sec2x
___
dx
Differentiating with respect to (w.r.t.) x
dy
sec2 y ___ = 1
dx
dy _____
___
= 1
dx sec2 y
Recall that sec2 y = 1 + tan2 y
∴ sec2 y = 1 + x2, since x = tan y.
dy ______
___
= 1
dx 1 + x2
d [tan−1 (x)] = ______
1
___
dx
1 + x2
Try these 2.7
dy
Find ___ when
dx
(a) y = cos−1 x
(c) y = cot−1 x
(b) y = cosec−1 x
(d) y = sec−1 x
E X A M P L E 36
dy
Find ___ when y = sin−1 (2x).
dx
SOLUTION
y = sin−1 (2x) is a function of a function.
Let u = 2x, y = sin−1 u
dy ________
du = 2, ___
1
___
= ______
dx
du √ 1 − u2
dy ___
dy ___
2
___
______
=
× du = ________
dx du dx
1
√ − u2
2
2
________
_______
= _________
= __________
2
1
−
(2x)
1
−
4x2
√
√
E X A M P L E 37
dy
Given that y = tan−1 (x2 + 1) find ___.
dx
SOLUTION
y = tan−1 (x2 + 1) is a function of a function.
OR
Take sine of both sides.
sin y = 2x
Differentiate w.r.t. x:
dy
cos y ___ = 2
dx
dy __________
2
2
___
_______
= ________
= ________
dx √1 − sin2y √1 − 4x2
Let u = x2 + 1, y = tan−1 (u)
dy ______
du = 2x, ___
___
= 1
dx
du 1 + u2
dy ___
dy ___
2x
2x = ___________
___
=
× du = ______
dx du dx 1+ u2 1 + (x2 + 1)2
63
M O DUL E 1
E X A M P L E 38
Find the derivative of θ = t2 sin−1 (t).
SOLUTION
Applying the product rule
u = t2, v = sin−1 (t)
du = 2t, ___
dv = _______
1
___
______
dt
dt
√1 − t2
2
dθ = _______
t
___
______ + 2t sin−1 (t)
dt
√1 − t2
(using the product rule)
EXERCISE 2D
dy
In questions 1–8 , differentiate each function with respect to x and hence find ___ in
dx
terms of x and y.
1
x4 + xy3 − y2 = 2
2
y = xe2y
3
cos (xy) + 4x2 = 7x
4
x2 + xy + y3 = 0
5
x(x2 + y2) = y3
1 + __
1 = __
1
__
2
2
9
x
y
6
xe y = x2 + 2
8
sin x tan y = 4
7
In questions 9 –1 4 , find the gradient of the tangent to the curve at the given point.
9
x4 − 2x2y + 3y = 2 at (1, 1)
10 xy − 2x + y2 = 4 at (0, 2)
11 x2y3 + 4xy = 7x at ( __83 , 2 )
12 ex+y − 3xy − 2 = y at (ln 2, 0)
13 3x2y2 − 2y3 = −4 at (1, 2)
14 (xy − y3)3 = 5y2 + 22 at (4, 1)
15 Find the equation of the tangent to the curve x2 + 4y2 − 2xy = 7 at the point
(1, −1).
dy
dx
16 Given that ex + y + cos x = 4y + 2, show that ___ = __31 when x = 0 and y = 0.
17 Find the equation of the tangent to the curve xy2 + y3 = 6x + 3y at (1, 2).
18 Find the equation of the tangent to the curve xy + x2 = y2 that is parallel to
(a) the x-axis
(b) the y-axis.
19 A computer is programmed to draw the graph of the function
(x2 + y2)3 = 64x2y2. Find the gradient of the tangent to the curve at (2, 0.56).
20 Find by implicit differentiation the four points on the curve
Remember
arcsin ≡ sin−1
64
(x2 + y2)2 = x2 − y2 where the tangent line is parallel to the x-axis and the
two points where the tangent line is parallel to the y-axis.
21 Differentiate the following functions with respect to x.
arccos ≡ cos−1
(a) y = arcsin (x2 + 2x +1)
(b) y = arccos (x2 + 1)
arctan ≡ tan−1
(c) y = arctan (3x2 + 5x +2)
(d) y = x arcsin (4x)
MODULE 1tCHAPTER 2
Second derivatives
d 2y
Recall that ___2 or f ″(x) is the second derivative of y w.r.t. x or the second
dx
derivative of the function x. To find the second derivative, we differentiate the
first derivative.
E X A M P L E 39
If f(x) = 4xex, find f ″(x).
SOLUTION
f(x) = 4xex
u = 4x, v = ex
du = 4, ___
dv = ex
___
dx
dx
Using the product rule
f′(x) = 4xex + 4ex
Differentiating again w.r.t. x
f ″(x) = 4xex+4ex + 4ex
= 4xex + 8ex
= 4ex (x + 2)
E X A M P L E 40
d2y
dy
Given that y = 4e−2x + e−3x show that ___2 + 5 ___ + 6y = 0.
dx
dx
SOLUTION
y = 4e−2x + e−3x
dy
___
= −8e−2x − 3e−3x
dx
d 2y
___
= 16e−2x + 9e−3x
dx2
d 2y
dy
___
+ 5___ + 6y
2
dx
dx
= 16e−2x + 9e−3x + 5[−8e−2x − 3e−3x] + 6[4e−2x + e−3x]
= 16e−2x + 9e−3x − 40e−2x − 15e−3x + 24e−2x + 6e−3x
= 16e−2x − 40e−2x + 24e−2x + 9e−3x − 15e−3x + 6e−3x
= 40e−2x − 40e−2x + 15e−3x − 15e−3x
=0
d 2y
dy
Hence ___2 + 5 ___ + 6y = 0
dx
dx
E X A M P L E 41
d2y
If y = ex cos 2x, find ___2 .
dx
SOLUTION
y = ex cos 2x
u = ex, v = cos 2x
du = ex, ___
dv = −2 sin 2x
___
dx
dx
65
M O DUL E 1
dy
___
= −2ex sin 2x + ex cos 2x
dx
= ex(−2 sin 2x + cos 2x)
Differentiating again w.r.t. x
d2y
___
= ex(−4 cos 2x − 2 sin 2x) + ex(−2 sin 2x + cos 2x)
dx2
= ex(−4 cos 2x − 2 sin 2x − 2 sin 2x + cos 2x)
= ex(−3 cos 2x − 4 sin 2x)
E X A M P L E 42
d2y
dy
Given that y = ex(cos 2x + sin 2x) show that ___2 − 2 ___ + 5y = 0.
dx
dx
SOLUTION
y = ex(cos 2x + sin 2x)
Using the product rule
dy
___
= ex(−2 sin 2x + 2 cos 2x) + ex(cos 2x + sin 2x)
dx
= ex(−2 sin 2x + sin 2x + 2 cos 2x + cos 2x)
= ex(−sin 2x + 3 cos 2x)
Differentiating again
d 2y
___
= ex(−2 cos 2x − 6 sin 2x) + ex(−sin 2x + 3 cos 2x)
dx2
= ex(−2 cos 2x + 3 cos 2x − 6 sin 2x − sin 2x)
= ex(cos 2x − 7 sin 2x)
dy
d2y
∴ ___2 − 2___ + 5y
dx
dx
x
= e (cos 2x − 7 sin 2x) − 2ex(−sin 2x + 3 cos 2x) + 5ex(cos 2x + sin 2x)
= ex cos 2x − 7ex sin 2x + 2ex sin 2x − 6ex cos 2x + 5ex cos 2x + 5ex sin 2x
= 6ex cos 2x − 6ex cos 2x + 7ex sin 2x − 7ex sin 2x
=0
d2y
dy
Hence ___2 − 2 ___ + 5y = 0
dx
dx
EXERCISE 2E
1
d2y
x2 . Find ___
An equation is given by y = _____
.
x+1
dx2
2
dy
d2y
Show that y = (2x + 5)e3x satisfies the equation ___2 − 6 ___ + 9y = 0.
dx
dx
3
d2y dy
Given that y = Ae−2x + Bex, show that ___2 + ___ − 2y = 0.
dx
dx
4
2(y − 1)
d2y _______
x , show that ___
.
Given that y = _____
=
2
x+1
dx
(x + 1)2
5
66
The relationship between the two variables x and y is given by
d2y
dy
y = ex[Acos 2x + B sin 2x]. Show that ___2 − 2 ___ + 5y = 0.
dx
dx
MODULE 1tCHAPTER 2
6
7
dy
2x , show that ___
Given that y = ______
= 2(1 − y)2. Hence show that
2x
+
1
dx
2
dy
___
= −8(1 − y)3.
dx2
dy
d2y
x+1
Find the values of ___ and ___2 at x = 0, where y = __________
.
dx
dx
x2 + 2x + 5
Parametric differentiation
First derivative of parametric equations
Parametric equations were introduced in Unit 1, Module 2. If the coordinates of
a point P(x, y) are given as x = f(t), y = g(t) where t is a third variable called the
parameter of the equation, the equations x = f(t), y = g(t) are called the parametric
equations and the parametric differential is
dy ___
dy ___
___
=
÷ dx
dx dt
dt
E X A M P L E 43
Given that the parametric equations of a curve are
x = 4t 2 + 5, y = 6t2 + t3
dy
find ___ in terms of t.
dx
SOLUTION
dy ___
dy ___
___
× dt
=
dx dt dx
E X A M P L E 44
SOLUTION
dy
dx and ___
.
First find ___
dt
dt
x = 4t2 + 5
dx = 8t
___
dt
y = 6t2 + t3
dy
___
= 12t + 3t2
dt
dy ___
dy ___
___
=
÷ dx
dx dt
dt
12t + ___
3t2
12t + 3t2 = ___
= ________
8t
8t
8t
3t
3 + __
= __
2 8
dy 3 __
+ 3t
Hence ___ = __
dx 2 8
dy
Find the value of ___ when t = 1 for the equation defined as
dx
3
x = 4t − 3, y = 7t2 + 5t +1.
Since the curve is defined parametrically
dy ___
dy ___
___
=
÷ dx
dx dt
dt
67
M O DUL E 1
Since x = 4t3− 3
dx = 12t2
___
dt
Since y = 7t2 + 5t + 1
dy
___
= 14t + 5
dt
dy ___
dy dx
___
=
÷ ___
dx dt
dt
dy _______
___
= 14t +2 5
dx
12t
dy 14(1) + 5 ___
= 19
When t = 1 ___ = ________
12
dx
12(1)2
E X A M P L E 45
dy
Given that x = tan−1(t) and y = t3, find ___ in terms of t.
dx
SOLUTION
dy
dx = ______
1 , ___
___
= 3t2
2
dt 1 + t
dt
Remember
d tan−1x ______
1
___
dx
1 + x2
dy dy dx
Since ___ = ___ ÷ ___
, we have
dx dt
dt
2
dy ______
___
= 3t
1
dx _____
1 + t2
= 3t2(1 + t2)
E X A M P L E 46
Find the equation of the tangent to the curve x = t + 2, y = t2 at the point t = 1.
SOLUTION
To find the equation of the tangent at t = 1, we need to find the values of x and y
when t = 1, and the gradient of the tangent when t = 1.
When t = 1, substituting into x = t + 2, y = t2 gives
x = 1 + 2, y = 12
∴ x = 3, y = 1
To find the
dy
gradient, find ___.
dx
dy ___
dy dx
___
= ÷ ___
dx dt dt
Since x = t + 2, y = t2
dy
dx = 1, ___
___
= 2t
dt
dt
dy dy dx
Since ___ = ___ ÷ ___
, we have
dx dt
dt
dy __
___
= 2t = 2t
1
dx
When t = 1
dy
___
= 2(1) = 2
dx
The equation of the tangent at x = 3, y = 1 and gradient 2 is
y − 1 = 2(x − 3)
y = 2x − 6 + 1
y = 2x − 5
68
MODULE 1tCHAPTER 2
E X A M P L E 47
SOLUTION
Find the equation of the normal to the curve x = 2 sin t, y = cos 2t at the point with
π.
parameter __
2
π means the point when t = __
π.
The point with parameter __
2
2
π, x = 2 sin __
π, y = cos 2 __
When t = __
( π2 ) = cos π
2
2
i.e. x = 2, y = −1.
π,
We need to find the gradient of the tangent at t = __
2
dy
dx = 2 cos t, ___ = −2 sin 2t
___
dt
dt
(
)
dy ___
dy ___
___
=
÷ dx
dx dt
dt
dy
2 sin 2t
∴ ___ = −______
2 cos t
dx
dy
4 sin t cos t
___
= −_________
2 cos t
dx
(sin 2t = 2 sin t cos t)
dy
___
= −2 sin t
dx
dy
π, ___
π = −2
When t = __
= −2 sin __
2 dx
2
Since the
gradient of the
tangent is −2,
the gradient of
1.
the normal is __
2
E X A M P L E 48
1 is
The equation of the normal at x = 2, y = −1 and gradient __
2
1
__
y + 1 = (x − 2)
2
1x − 1 − 1
y = __
2
1x − 2
y = __
2
Find the coordinates of the stationary point on the curve
x = 25 − 15t + 9t2,
SOLUTION
y = 6t − t2.
dy
Stationary points exist where ___ = 0.
dx
dy dy dx
Since the equation is in parametric form ___ = ___ ÷ ___
dx dt
dt
dy
dx = −15 + 18t and ___ = 6 − 2t
___
dt
dt
dy
6 − 2t
Hence ___ = _________
dx −15 + 18t
dy
6 − 2t = 0
___
= 0 ⇒ _________
−15 + 18t
dx
∴ 6 − 2t = 0
t=3
69
M O DUL E 1
When t = 3, x = 25 − 15(3) + 9(3)2,
y = 6(3) − (3)2
x = 25 − 45 + 81 = 61, y = 18 − 9 = 9
The coordinates of the stationary point are (61, 9).
Second derivative of parametric equations
For parametric equations the second derivative can be found using
dy
d ___
__
2y
d
dt dx
___ = ______
(by the chain rule)
dx
dx2
___
dt
[ ]
E X A M P L E 49
SOLUTION
d2y
Given that x = t2 + 5t − 4, y = t3 + 3t + 1, find ___2 .
dx
[ ]
dy
d ___
__
2y
d
dy
dt dx we first find ___
Since ___2 = _______
where
dx
dx
dx
___
dt
dy
dy
dx
___ = ___ ÷ ___
dx dt
dt
dy
dx 2t + 5 and ___
Now ___
=
= 3t2 + 3
dt
dt
dy ______
3t2 + 3
___
= 2t + 5
dx
[ ]
[
]
dy
(2t + 5)(6t) − (3t2 + 3)(2)
d _______
3t2 + 3
d ___
______________________
= __
(using the quotient rule)
Now __
=
dt dx
dt 2t + 5
(2t + 5)2
12t2 + 30t − 6t2 − 6 ____________
6t2 − 6 + 30t
= _________________
=
2
(2t + 5)
(2t + 5)2
6t2 − 6 + 30t
____________
(2t + 5)2
6t2 − 6 + 30t
Hence ___2 = ____________ = ____________
2t
+
5
dx
(2t + 5)3
d 2y
E X A M P L E 50
SOLUTION
d2y
Given that x = arcsin t, y = t2 + 2t + 1, show that ___2 = −2(2t2 + t − 1).
dx
[ ]
dy
d ___
__
2y
d
dy
dt dx we first find ___
Since ___2 = ______
where
dx
dx
dx
___
dt
dy
dy
dx
___ = ___ ÷ ___
dx dt
dt
dy
dx _______
1
___
= ______ , ___ = 2t + 2
dt
√1 − t2 dt
______
dy _______
2t + 2
___
(2t + 2) √ 1 − t2
=
=
1
dx _______
______
√1 − t2
______
dy
d ___
d
__
= __ [ (2t + 2) √1 − t2 ]
dt dx
dt
[ ]
70
MODULE 1tCHAPTER 2
______
______
t(2t + 2)
1 (−2t)(1 − t2)− __12 = 2 √1 − t2 − ________
______
= 2 √1 − t2 + (2t + 2) __
2
√1 − t2
( )
2(1 − t2) − 2t2 −2t ____________
−4t2 ______
− 2t + 2
______
= ________________
=
2
√1 − t2
√1 − t
[ ]
−4t2 ______
− 2t + 2
____________
dy
d ___
___
√1 − t2
dt dx
___ _______
= −4t2 − 2t + 2 = −2(2t2 + t − 1)
=
= ____________
2
1
dx
_______
dx
___
______
dt
√1 − t2
d 2y
EXERCISE 2F
dy
d 2y
In questions 1–8, find ___ and ____2 in terms of t.
dx
dx
1
__
1 x = t + 1, y = 2t − 1
2 x = 6t2 + 1, y = 2t2 + 3
3
x = t3, y = t2
4
x = t + 2, y = 2t2 − t − 1
5
x = et, y = 7e2t
6
x = 5 cos t, y = 4 sin t
7
x = cos3 t, y = 4 sin3 t
8
x = t sin t, y = t cos t
In questions 9–14, find the value of the gradient of the curve at the given point.
9
x = et, y = e−t, t = 0
11 x = 3t, y = 1 − t, t = 4
6
13 x = t(2t + 1)2, y = ________
______ , t = 1
√ 4t + 3
10 x = 2t2 + 1, y = 3t2 + 2, t = 1
5t , y = 0, t = 2
12 x = ______
2t + 1
14 x = 3t, y = log2 (2t + 1), t = 1
15 A curve is given by the parametric equation x = t2 + 2t, y = t3 − 3t + 1.
dy
Find ___ in terms of t. Hence find
dx
(a) the equation of the tangent and normal for t = 0
(b) the value of t for the turning points.
1 , y = 1 − 2t, find the points on the curve where the
16 Given that x = _____
1+t
gradient is 1. Hence write down the equations of the tangent at these points.
17 The parametric equations of a curve are x = t(t2 + 1)3, y = t2 + 1.
Find and simplify the equation of the tangent to the curve at the point with
parameter t = 3.
18 Find the equation of the tangent to the curve x = sec θ, y = tan θ at the point
with parameter α.
19 Given that the parametric equations of a curve are x = 3(θ − sin θ),
dy
dy
π.
1 θ. Hence find the value of ___
y = 3(1 − cos θ), show that ___ = cot __
when x = __
2
3
dx
dx
__
__
dy
___
20 If x = √2 (1 − cos θ) and y = √2 sin θ, show that = cot θ and find the
dx
π.
equation of the tangent at θ = __
4
71
M O DUL E 1
Partial derivatives
In Unit 1 when finding the derivative as the instantaneous rate of change of one
variable with respect to another, only one independent variable and one dependent
variable were present. By definition
f (x + h) − f (x)
f´(x) = limh→0 _____________
h
We can extend this result to a function of two or more variables. We find the
derivative of the function with respect to one variable while holding the other
variable constant. This is called a partial derivative.
First order partial derivatives
D EFIN IT IO N
The first partial derivatives of the function z = f(x, y) are the two functions defined by
f(x + h, y) − f(x, y)
fx(x, y) = lim _________________
h
f(x,
y
+
h)
− f(x, y)
fy(x, y) = lim _________________
h→0
h
h→0
wherever these limits exist.
fx(x, y) is the first partial derivative with respect to x and fy(x, y) is the first partial
derivative with respect to y.
Notation for partial derivatives
If z = f(x, y) then the first partial derivative with respect to x is represented by
∂f
∂z = ___
∂ [f(x, y)]
___
= fx(x, y) = ___
∂x ∂x
∂x
and the first partial derivative with respect to y is represented by
∂f
∂z = ___
∂ [f(x, y)]
___
= fy(x, y) = ___
∂y ∂y
∂y
∂f
To find ___ differentiate the function with respect to x while keeping y constant, and to
∂x
∂f
find ___ differentiate the function with respect to y while keeping x constant.
∂y
E X A M P L E 51
SOLUTION
72
∂f
∂f
Find the partial derivatives ___ and ___ of the function f(x, y) = x3 + 4x2y3 − y4.
∂x
∂y
∂f
We can differentiate term by term and then add to find each derivative. To find ___ we
∂x
differentiate with respect to x while keeping y constant.
∂ (x3) = 3x2
___
∂x
∂ (4x2y3) = 8xy3
∂ (x2) = 2x and our
___
(4y3 is treated as a constant so we find ___
∂x
∂x
derivative becomes 4y3 × 2x = 8xy3)
∂ (y4) = 0
___
(since y4 is treated as a constant when differentiating with respect to x)
∂x
∂f
Hence ___ = 3x2 + 8xy3
∂x
∂f
To find ___, we differentiate with respect to y while keeping x constant
∂y
MODULE 1tCHAPTER 2
∂ (x3) = 0
___
∂y
∂ (4x2y3) = 12x2y2
___
∂x
(since x is treated as a constant when we are differentiating
with respect to y).
∂ (y3) = 3y2 and
(4x2 is treated as a constant so we find ___
∂y
our derivative becomes 4x2 × 3y2 = 12x2y2).
∂ (y4) = 4y3
___
∂y
∂f
Hence ___ = 12x2y2 − 4y3
∂y
E X A M P L E 52
∂z and (b) ___
∂z if z = (x2 + y) cos (2x2y).
Find (a) ___
∂y
∂x
SOLUTION
(a) Since we have the product of two functions of x, we use the product rule:
Let u = x2 + y, v = cos (2x2y)
∂u
___
∂v
___
∂x = 2x, ∂x = −TJO x y) × xy)
∂z = −xy x + y) TJO xy) +x DPT xy)
∴ ___
∂x
(b) Using the product rule with u = x2 + y, v = cos (2x2y)
∂u
___
∂v
___
2
2
∂y = 1, ∂y = −2x sin (2x y)
∂z = −2(x2 + y)x2 sin (2x2y) + cos (2x2y).
Hence ___
∂y
Second order partial derivatives
∂f
∂f
Since ___ and ___ are functions of x and y, we can derive partial derivatives of each of
∂x
∂y
these. We can thus find partial derivatives of higher orders. The four possible second
order partial derivatives of f(x, y) are
∂2f
∂f
∂ ___
fxx = (fx)x = ___2 = ___
∂
x
∂
x
∂x
∂f
∂2f
∂ ___
fxy = (fx)y = ___
= _____
∂y ∂x
∂y∂x
( )
( )
( )
∂f
∂f
∂ ___
= (f ) = ___
= ___
(
)
∂y ∂y
∂y
∂f
∂2f
∂ ___
= _____
fyx = (fy)x = ___
∂x ∂y
∂x∂y
2
fyy
y y
2
The function fxy is the second order partial derivative of f with respect to x first and
then with respect to y, and fyx is the second order partial derivative of f with respect
to y first and then with respect to x. Partial derivatives fxy and fyx are equal if and only
if both the function and the partial derivatives are continuous.
E X A M P L E 53
SOLUTION
∂ z = _____
∂z.
Give that z = ln (2xy + y2) show that _____
∂x∂y ∂y∂x
2
2
z = ln (2xy + y2)
2y
∂z = ________
___
∂x 2xy + y2
73
M O DUL E 1
(
2y
∂2z = ___
∂ ________
_____
∂x∂y ∂y 2xy + y2
)
(2xy + y2) (2) − (2y) (2x + 2y)
= __________________________
(2xy + y2)2
(using the quotient rule)
4xy + 2y2 − 4xy − 4y2
2y2
__________
=
−
= ___________________
(2xy + y2)2
(2xy + y2)2
2x + 2y
∂z = ________
___
∂y
2xy + y2
2x + 2y
∂2z = ___
∂ ________
____
∂x∂y
(
∂x 2xy + y2
)
(2xy + y2)(2) − (2x + 2y)(2y)
= _________________________
(2xy + y2)2
2y2
4xy + 2y2 − 4xy − 4y2
__________
=
−
= ___________________
(2xy + y2)2
(2xy + y2)2
∂ z = ____
∂z.
Hence ____
∂x∂y ∂y∂x
2
E X A M P L E 54
SOLUTION
2
∂2f
∂2f
The function f (x, y) = e−3x cos y. Find ___2 and ___2.
∂x
∂y
Differentiating with respect to x, keeping y constant
∂f
___
= −3e−3x cos y
∂x
Differentiating again with respect to x, while y is kept constant
∂2f
___
= 9e−3x cos y
∂x2
Differentiating with respect to y, keeping x constant
∂f
___
= −e−3x sin y
∂y
Differentiating again with respect to y, while x is kept constant
∂2f
___
= −e−3x cos y
∂y2
Applications of partial derivatives
74
E X A M P L E 55
Two resistors R1 and R2 are placed in parallel. They have a combined resistance R,
1 + __
1 . Find ____
∂R and ____
∂R .
given by R = __
∂R1
∂R2
R1 R2
SOLUTION
Keeping R2 constant and differentiating with respect to R1, we get
1
∂R = − ___
___
∂R1
R21
Keeping R1 constant and differentiating with respect to R2, we get
1
∂R = − __
___
∂R2
R22
MODULE 1tCHAPTER 2
E X A M P L E 56
An electric circuit has parallel resistances R1 and R2. The current i through R1 can be
found from
IR2
i = _______
R1 + R2
∂i .
where I is the total current. Find ___
∂R2
SOLUTION
Using the quotient rule with R1 constant, let
u = IR2, v = R1 + R2
∂u = I, ___
∂v = 1
___
∂R2
∂R2
(R1 + R2)I − (IR2)(1) _____________
IR + IR2 − IR2 _________
IR1
∂i = __________________
___
= 1
=
∂R2
E X A M P L E 57
(R1 + R2)2
(R1 + R2)2
(R1 + R2)2
The temperature θ of a thin piece of metal at any point (x, y) is given by
60
θ = __________
x2 + y2 + 1
where θ is measured in degrees Celsius and x, y are measured in metres. Find the rate
of change of the temperature with respect to the distance in the x-direction. Find also
this rate of change at the point (2, 3).
SOLUTION
∂θ , so we keep y constant and differentiate.
We need to find ___
∂x
60
= 60(x2 + y2 + 1)−1,
Since θ = __________
x2 + y2 + 1
using the chain rule, we get
120x
∂θ = −60(2x)(x2 + y2 + 1)−2 = − ____________
___
∂x
(x2 + y2 + 1)2
When x = 2, y = 3,
120(2)
∂θ = − _______________
60
___
= −___
∂x
49
((2)2 + (3)2 + 1)2
E X A M P L E 58
The volume of a right circular cone of height h cm and base radius r cm is given by
1 πr2h.
V = __
3
∂2V when r = 2 cm.
∂2V when h = 6 cm
(b) ____
Find (a) ____
2
∂r
∂h2
SOLUTION
(a) V = __31 πr2h
Keeping h constant and differentiating twice with respect to r, we get
∂V __
___
2
∂r = 3 πrh
∂2V = __
2 πh
____
∂r2 3
When h = 6 cm,
∂2V = __
2 π(6) = 4π
____
∂r2 3
75
M O DUL E 1
(b) Keeping r constant and differentiating twice with respect to h, we get
∂V __
___
= 1 πr2
∂h 3
∂2V = 0
____
∂h2
∂2V = 0.
Hence, when r = 2 cm, ____
∂h2
Laplace’s equation
∂2u + ___
∂2u = 0 where u = f(x, y) is called Laplace’s
The partial differential equation ___
2
∂x
∂y2
equation. The solutions derived from this equation are called harmonic functions
and they play an important role in electrical and heat conduction.
E X A M P L E 59
SOLUTION
∂ z + ___
∂ z = 0.
Show that the equation z = x2 − y2 satisfies the equation ___
∂x2 ∂y2
2
2
z = x2 − y2
∂z = 2x
___
(since we treat y as a constant)
∂x
∂2z = 2
___
∂x2
Also
∂z = −2y
___
(treating x as a constant)
∂y
∂2z = −2
___
∂y2
Now
∂2z + ___
∂2z = 2 − 2 = 0
___
2
∂x
∂y2
∂2z + ___
∂2z = 0
Hence z = x2 − y2 satisfies the equation ___
2
∂x
∂y2
Cobb–Douglas function
Cobb and Douglas modelled the total production P of an economic system as a
function of labour L and investment K. The function used to model production is
P(L, K) = bLα K1– α
where P is the total production, L is the amount of labour and K is the amount of
∂P is the rate at which production changes
capital invested. The partial derivative ___
∂L
with respect to labour and is called the marginal productivity of labour. The partial
∂P is the rate of production with respect to capital and is called the
derivative ___
∂K
marginal productivity of capital.
E X A M P L E 60
Suppose that, at a certain factory, output is given by the Cobb–Douglas production
function.
1
__
2
__
Q = 70K 3 L 3 units, where K is the capital investment measured in units of TT $1000
and L is the size of the labour fource measured in worker-hours.
76
MODULE 1tCHAPTER 2
(a) What is the output if the capital investment is TT $1 728 000 and
729 worker-hours of labour are used?
(b) Does the output double if both the capital investment and the size of the labour
force in (a) doubles?
∂Q
when the capital expenditure is
(c) Find the marginal productivity of capital ___
∂K
TT $1 728 000 and the level of labour is 729 worker-hours.
SOLUTION
(a) When K = 1728, L = 729
1
__
2
__
Q = 70(1728)3(729)3 = 70(12)(81) = 68 040 units
(b) When K = 2(1728), L = 2(729)
1
__
2
__
( 1)
__
1
__
( 2) (
__
2
__
)
Q = 70(2(1728))3 (2(729))3 = 70 23 (1728)3 23 7293 = 2(70)(12)(81)
= 2(68 040) units
Hence the output will double when the capital investment and the worker-hours
are doubled.
1
__
2
__
(c) Differentiating Q = 70K3 L3 with respect to K and keeping L constant, we get
∂Q
1 K 3 L3
___
= 70 __
∂K
(3)
–2
___
2
__
When K = 1728, L = 729
__
∂Q
105
1 (1728)___
___
3 (729) 3 = ___
= 70 __
∂K
(3)
–2
2
8
Function of three variables
We can extend our results to three or more variables and find the corresponding
partial derivatives.
E X A M P L E 61
Given that t = 4x2 + 3y2 + 2z2 + 6x2yz, find
∂2t
∂2t
∂2t
(a) ___
(b) ___
(c) ___
2
2
∂x
∂y
∂z2
SOLUTION
(a) t = 4x2 + 3y2 + 2z2 + 6x2yz
%JČFSFOUJBUJOHXJUISFTQFDUUPx LFFQJOHyBOEzDPOTUBOU XFHFU
∂t = 8x +xyz
___
∂x
∂ t = 8 + 12yz
___
2
∂x2
(b) t = 4x2 + 3y2 + 2z2 + 6x2yz
%JČFSFOUJBUJOHXJUISFTQFDUUPy LFFQJOHxBOEzDPOTUBOU XFHFU
∂t = 6y + 6xz
___
∂y
∂t = 6
___
∂y
77
M O DUL E 1
(c) t =x + 3y +z + 6xyz
%JČFSFOUJBUJOHXJUISFTQFDUUPz LFFQJOHxBOEyDPOTUBOU XFHFU
∂t =z + 6xy
__
∂z
∂t =
___
∂z
E X A M P L E 62
Find the value of:
∂t
(a) ___
∂x
∂2t
(b) ___
2
∂z
∂2t
(c) ____
∂z∂y
at the point (1, −1, 2) where t = 6xy + y3 + z3 − x3y.
SOLUTION
∂t = 6y − 3x2y
(a) ___
∂x
8IFOx = 1, y = −1
∂t = −1) − −1)
___
∂x
∂t = −6 + 3 = −3
___
∂x
∂t = 3z
(b) __
∂z
∂t = 6z
___
∂z
8IFOz = 1,
∂ t = = 6
___
∂z
[ ]
∂t = __
∂t ___
∂t
(c) ____
∂z∂y ∂z ∂y
∂ 6x + 3y2 − x3
= __
)
∂z (
= 6 − 3x2
When x = 1,
∂2t = 6 − 3(1)2
____
∂z∂y
=6−3=3
E X A M P L E 63
SOLUTION
78
∂w, ___
∂w and ___
∂w .
Let w = 2x2 + yz. Find ___
∂z ∂y
∂x
w = 2x2 + yz
∂w = y, keeping x and y constant.
___
∂z
∂w = z, keeping x and z constant.
___
∂y
∂w = 4x, ___
∂ (yz) = 0, keeping y and z constant.
___
∂x
∂x
MODULE 1tCHAPTER 2
E X A M P L E 64
SOLUTION
Let w = 6x2y2z2 + 3xz3 + 4x2y2. Find
∂w
∂w
(a) ___
(b) ___
∂x
∂y
∂w
(c) ___
∂z
(a) Keeping y and z constant, and differentiating with respect to x, we get
∂w = 12xy2z2 + 3z3 + 8xy2
___
∂x
(b) Keeping x and z constant, we get
∂w = 12x2yz2 + 8x2y
___
∂y
(c) Keeping x and y constant, we get
∂w = 12x2y2z + 9xz2
___
∂z
EXERCISE 2G
∂w and ___
∂w.
Given that w = x2y − 2x + y3, find the first partial derivative ___
∂x
∂y
∂
z
∂
z
___
___
2
3
2
2
4
2 Find ∂x and ∂y if z = (x y + 3y )(x + y ) .
3 Find the first partial derivative of z = cos(xy2) + exy+y2.
∂2z + ___
∂2z = 0.
4 Given that z = ex cos y, show that ___
2
∂x
∂y2
2
2
∂ z + ___
∂ z = 0.
5 If z = x2 − y2, show that ___
∂x2 ∂y2
∂2z , ____
∂2z , ____
∂2z and ___
∂2z for each of the functions z.
In questions 6–10, find ___
2
∂
y
∂
x
∂
x
∂
y
∂x
∂y2
2
2
2
3
6 z = 3x − 4x y + y .
1
7
z = sin 4x cos 6y.
8
z = sin−1(xy).
y2
x − __
z = __
.
y3 x4
10 z = tan−1(xy).
9
∂z .
11 If z = ex +xy+t, find ∂___z2, ∂___z2, ∂___z2 and ______
∂x∂y∂t
2
2
2
∂x ∂y
12 If w =
e–z
2
∂t
2
∂ w + ____
∂ w = ___
∂w.
(sin x + cos y), show that ____
∂z
∂x2
∂y2
2
2
79
M O DUL E 1
SUMMARY
Differentiation
d [xn] = nxn−1
dx
d [(ax + b)n] = na(ax + b)n−1
dx
Chain rule
d [In x] = 1
x
dx
Parametric differentiation
d [y] = dy
dx
dx
x = f(t), y = g(t)
dy
= f’(g(x)) × g’(x)
dx
d [ex ] = ex
dx
d [eax + b ] = aeax + b
dx
Implicit differentiation
y = fg(x)
Product rule
y = uv
dy
= u dv = v du
dx
dx
dx
dy
dy dx
=
÷
dx
dt dt
d [y2] = 2y dy
dx
dx
d [In(ax + b)] = a
dx
ax + b
d [sin x] = cos x
dx
Quotient rule
y=u
v
d [sin(ax + b)] = a cos (ax + b)
dx
d [cos x] = −sin x
dx
d [cos(ax + b)] = −a sin (ax + b)
dx
dy
=
dx
v u − u dv
dx
dx
v2
d [xy3] = x 3y2 dy + y3(1)
dx
dx
(
)
dy
= 3xy2
dx
[ ]
d2y
d dy
=
÷ dx
dt
dx2 dt dx
+ y3
Partial derivatives
u = f(x, y)
d [tan x] = sec2x
dx
d [sec x] = sec x tan x
dx
d [cosec x] = −cosec x cot x
dx
First partial derivative
∂u
∂u
and
∂x
∂y
d [cot x] = −cosec2x
dx
Second partial derivatives
1
d [sin−1(x)] =
dx
√1 − x2
d [cos−1x] = −1
dx
√1 − x2
1
d [tan−1(x)] =
dx
√1 + x2
d [ax ] = ax In x
dx
d [log x] = 1
a
x ln a
dx
80
∂ 2u , ∂ 2u , ∂ 2 u , ∂ 2u
∂ x2 ∂ 2y ∂ x ∂ y ∂ y ∂ x
MODULE 1tCHAPTER 2
Checklist
Can you do these?
■ Differentiate e f (x).
■ Differentiate ln f(x).
■ Use the chain rule for gradient.
■ Differentiate trigonometric functions.
■ Differentiate polynomials.
■ Differentiate combinations of polynomial, trigonometric, exponential and
logarithmic functions.
■ Differentiate implicit functions.
■ Differentiate functions represented parametrically.
■ Differentiate inverse trigonometric functions.
■ Find the equation of the tangent to a curve defined parametrically.
■ Find the equation of the tangent or normal to a curve defined implicitly.
■ Find the second derivative of a function.
■ Find the first partial derivative of a function.
■ Find the second partial derivative of a function.
Review exercise 2
1
2
3
______
d 2y
Given that x = sin−1 t, y = √1 − t2 find and simplify ___2 .
dx
dy
___
3
Find the value of
at t = 0 when x = (4t − 1) , y = 2t.
dx
Differentiate the following functions with respect to x.
4x + 3
(a) ln ______
(b) y = 3x ln x
2x − 5
__ is 3.
Show that the gradient of the tangent to the curve y = ln sin3x at x = π
4
2
A curve is represented parametrically by x = 4t − 2, y = 4t + 3.
(
4
5
)
(a) Calculate the length of the chord that joins the points with parameters
t = 0 and t = 1.
(b) Find the equation of the tangent and normal to the curve at t = 1.
6
Find the gradient of the normal to the curve at the given point.
1
(a) y = sin−1(4x), x = __
4
1x , x = 4
(b) y = tan−1 __
2
−1
(c) y = x tan (2x + 1), x = 0
( )
81
M O DUL E 1
7
8
9
dy
d2y
Given that y = x tan−1 x, show that (1 + x2)___2 + 2(x − 1) ___ = 2.
dx
dx
Given that y = ex(A cos 3x + B sin 3x), where A and B are constants, show that
d 2y
dy
___
− 2 ___ + 10y = 0.
dx
dx2
A curve is given parametrically by x = t 3 + 2t, y = t 4 + 2t.
dy
d2y
(a) Find ___ and ___2 in terms of t.
dx
dx
(b) Show that there is no tangent to the curve that is vertical.
dy
dx
10 The equation of a curve is given by xy3 + 3xy − x2 = 4x. Find ___ as a function
of x and y.
11 Find the equation of the tangent to the curve x2 + y2 − 4xy = 6 at the point (1, −1).
12 Find all the points on the graph of x2 + y2 = 4x + 4y at which the tangent is
parallel to the x-axis.
13 The parametric equations of a curve are x = a cos3 θ, y = a sin3 θ.
dy
Show that ___ = −tan θ.
dx
π is √__
Show also that the equation of the normal at θ = __
3 x − y = a.
6
______
3x + 8 at the
14 Find the equation of the normal to the curve y = ln 3 ______
x+2
√
point x = 0.
15 The parametric equations of a curve are x = 4(2 cos θ − cos 2θ),
y = 4(2 sin θ − sin 2θ).
dy
3θ .
(a) Prove that ___ = tan ___
2
dx
2
dy
(b) Find and simplify ___2 .
dx
π.
(c) Find the equation of the normal to the curve at θ = __
3
16 The equation of a curve is given by x2y2 + 4y = x.
dy 1 − 2xy2
.
(a) Show that ___ = ________
dx 4 + 2x2y
(b) Hence find the gradient of the curve at (0, 0).
(c) Find the equation of the normal to the curve at (0, 0).
dy
dx
17 Given that x = θ − sin θ, y = 1 − cos θ, find ___ , simplifying your answer as far
d2y
1
θ
4 __
as possible. Hence show that ___2 = −__
4 cosec 2 .
dx
( )
18 Given that the variables x and y are related by y = x + exy find the value of
2
dy
___
when y = 0.
dx2
d2y
dx
2θsin θ .
19 Given that x = θ + sin θ, y = θ2 + 2cos θ, show that ___2 = __________
3
82
(1 + cos θ)
MODULE 1tCHAPTER 2
dy
dx
20 Given that 2x+y = x3 + 3y, find ___.
21 Find the second partial derivaties of f(x, y) = 6x3 + 12x2y2 − 3y3.
22 Show that the function f (x, y) = ex sin y satisfies the equation
fxx + fyy = 0
23 Find the first partial derivatives of the functions:
(a) w = xy2z3 + 3yz
(b) w = ln(x + 3y + 4z)
24 The total resistance R produced by three conductors with resistance R1, R2, R3
connected in a parallel electrical circuit is given by the formula
1 + __
1 + __
1
1 = __
__
R R1 R2 R3
∂2R .
∂R and ____
Find ___
∂R1
∂R21
25 Let f(x, y) = x2 sin (yz). Find fxx, fyy, fzz and fxyz.
83
M O DUL E 1
CHAPTER 3
Partial Fractions
At the end of this chapter you should be able to:
■ separate a fraction with unrepeated linear factors in the denominator
into its parts
■ separate a fraction with repeated linear factors in the denominator into
its parts
■ separate a fraction with unrepeated quadratic factors in the denominator into
its parts
■ separate a fraction with repeated quadratic factors in the denominator into
its parts
■ separate an improper fraction into its parts.
KEYWORDS/TERMS
SBUJPOBMGSBDUJPOtQSPQFSGSBDUJPOtJNQSPQFS
GSBDUJPOtQBSUJBMGSBDUJPOtMJOFBSGBDUPSt
RVBESBUJDGBDUPStSFQFBUFEGBDUPS
84
MODULE 1tCHAPTER 3
DE FIN ITI ON
A rational fraction
is one in which the
numerator and
denominator are
polynomials.
A proper fraction
is one in which
the degree of the
numerator is less
than the degree
of the denominator. An improper
fraction is one in
which the degree
of the numerator
is greater than
or equal to the
degree of the
denominator.
3
2 and _____
_____
x+1
x+2
are the parts of
5x + 7
____________
(x + 1)(x + 2)
Rational fractions
The sum or difference of a number of rational fractions can be combined to
form one fraction in the following way:
2(x + 2) + 3(x + 1)
3 ≡ _________________
2 + _____
_____
x+1 x+2
(x + 1)(x + 2)
(finding the LCM, (x + 1)(x + 2))
2x + 4 + 3x + 3
= ______________
(x + 1)(x + 2)
5x + 7
= ____________
(x + 1)(x + 2)
Rational fractions can be either proper fractions or improper fractions.
Rational fractions with factors in the denominator can be separated into parts.
The process of separating rational fractions into their parts is called partial
fractions. All improper fractions must be converted to mixed fractions before
separating into parts. For proper fractions, the denominator must be factorised
into a product of linear factors and quadratic factors and the partial fractions
will depend on the form of these factors. The denominator can contain
(a) unrepeated linear factors, (b) repeated linear factors, (c) unrepeated
quadratic factors, or (d) repeated quadratic factors.
Proper fractions: Unrepeated linear factors
Note
Since there are
three factors
in the denominator there must be
three corresponding fractions.
The denominator of the proper fraction consists of unrepeated linear factors of the
form ax + b. For every factor of the form ax + b there is a corresponding fraction of
A , where A is a constant to be found.
the form ______
ax + b
P(x)
C
A + ______
B + ______
For example, _____________________ ≡ ______
(ax + b)(cx + d)(ex + f ) ax + b cx + d ex + f
There are several ways of finding the constants A, B and C. The following examples
demonstrate the different methods that can be used.
EXAMPLE 1
5x + 7
Separate ____________
into partial fractions.
(x + 1)(x + 2)
SOLUTION
Since the denominator consists of two distinct linear factors, we have
Remember
Every distinct
linear factor in
the denominator
must have a
fraction associated with it.
5x + 7
A + _____
B
____________
≡ _____
(x + 1)(x + 2) x + 1 x + 2
Multiply both sides of the identity by the denominator of the left-hand side to
obtain
5x + 7
A × (x + 1)(x + 2) + _____
B × (x + 1)(x + 2)
____________
× (x + 1)(x + 2) ≡ _____
x+1
x+2
(x + 1)(x + 2)
∴ 5x + 7 ≡ A(x + 2) + B(x + 1)
[1]
85
M O DUL E 1
Note
The first method
of finding A and
B is to substitute
values for x on
both sides of the
identity. The values of x are found
from the zeros of
the denominator
i.e. x + 2 = 0 and
x + 1 = 0 give
x = −2 and
x = −1.
EXAMPLE 2
SOLUTION
Substitute x = −2 into [1] to eliminate A
5(−2) + 7 = A(−2 + 2) + B(−2 + 1)
−3 = −B
B=3
Substitute x = −1 into [1] to eliminate B
5(−1) + 7 = A(−1 + 2) + B(−1 + 1)
A=2
5x + 7
3
2 + ______
So ____________
≡ ______
x+1
x+2
(x + 1)(x + 2)
2x + 1
Separate _____________
into partial fractions.
(x − 3)(3x − 1)
Since the denominator of this proper fraction consists of two distinct linear factors,
we have
2x + 1
A + ______
B
_____________
≡ _____
(x − 3)(3x − 1) x − 3 3x − 1
Note
This method for
finding A and
B is comparing
coefficients. The
coefficients of
x and the constants on both
sides of the
identity must be
equal.
Multiplying both sides of the identity by (x − 3)(3x − 1) gives
2x + 1 ≡ A(3x − 1) + B(x − 3)
= 3Ax − A + Bx − 3B
∴ 2x + 1 = 3Ax + Bx − A − 3B
Equating coefficients of x, we have
3A + B = 2
Equating constants, we have
−A − 3B = 1
Multiplying the second equation by 3 and adding to the first gives
Two equations
are formed. To
find A and B we
solve the equations simultaneously.
−8B = 5
−5
B = ___
8
Substituting into the second equation gives
15 = 1
−A + ___
8
7
A = __
8
So
−5
_7
__
8
2x + 1
_____
______
_____________
≡
+ 8
(x − 3)(3x − 1)
x−3
3x − 1
7
5
≡ _______
− ________
8(x − 3) 8(3x − 1)
86
MODULE 1tCHAPTER 3
EXAMPLE 3
6x − 7x − 1 into partial fractions.
Separate _____________
(x2 − 1)(x − 2)
SOLUTION
The first term in the denominator can be factorised into x2 − 1 = (x − 1)(x + 1).
2
6x − 7x − 1
6x − 7x − 1 ≡ __________________
∴ _____________
(x2 − 1)(x − 2) (x − 1)(x + 1)(x − 2)
2
Substituting
values of x
for which the
denominator is
zero will normally
be the fastest way
of finding the
constants.
2
The denominator of this fraction consists of three distinct linear factors.
C
6x2 − 7x − 1
A + _____
B + _____
≡ _____
∴ __________________
(x − 1)(x + 1)(x − 2) x − 1 x + 1 x − 2
Multiplying throughout by (x − 1)(x + 1)(x − 2) gives
6x2 − 7x − 1 ≡ A (x + 1)(x − 2) + B (x − 1)(x − 2) + C (x + 1)(x − 1)
To find A, B and C, we can substitute x = −1, x = 1 and x = 2 into the equation.
When x = −1,
6(−1)2 − 7(−1) − 1 = B (−1 − 1)(−1 − 2)
6 + 7 − 1 = 6B
12 = 6B
B=2
When x = 1,
6(1)2 − 7(1) − 1 = A(1 + 1)(1 − 2)
6 − 7 − 1 = −2A
−2 = −2A
A=1
When x = 2,
6(2)2 − 7(2) − 1 = C (2 + 1)(2 − 1)
24 − 14 − 1 = 3C
9 = 3C
C=3
3
1 + _____
2 + _____
6x2 − 7x − 1
∴ __________________
≡ _____
(x − 1)(x + 1)(x − 2) x − 1 x + 1 x − 2
EXAMPLE 4
6x + 10x + 2 into partial fractions.
Separate ______________
(2x2 + 5x + 2)x
SOLUTION
The denominator can be factorised into 2x2 + 5x + 2 ≡ (2x + 1)(x + 2).
2
6x + 10x + 2
6x + 10x + 2 ≡ ______________
∴ ______________
(2x2 + 5x + 2)x x(2x + 1)(x + 2)
2
2
Since we have three distinct factors in the denominator:
C
6x2 + 10x + 2 ≡ __
B + _____
A + ______
______________
2x + 1 x + 2
x(2x + 1)(x + 2) x
87
M O DUL E 1
Multiplying throughout by x(2x + 1)(x + 2), we get
6x2 + 10x + 2 ≡ A(2x + 1)(x + 2) + Bx (x + 2) + Cx (2x + 1)
1
To find A, B and C we can substitute x = 0, x = − __
2 , x = −2 into the equation.
When x = 0,
6(0)2 + 10(0) + 2 = A(2(0) + 1)((0) + 2)
2 = 2A
∴A=1
1
__
When x = − 2 ,
( )
( )(
( )
12
1
1
1
__
__
__
6 −__
2 + 10 −2 + 2 = B −2 − 2 + 2
3B
6 − 5 + 2 = − ___
__
4
4
3B
6
___
− __
4=− 4
When x = −2,
)
B=2
6 (−2)2 + 10 (−2) + 2 = C (−2)(2(−2) + 1)
6 = 6C
C=1
+ 10x + 2 ≡ __
1 + ______
2 + _____
1
∴ ______________
(2x2 + 5x + 2)x x 2x + 1 x + 2
6x2
Try these 3.1
Separate the following into partial fractions.
x+2
(a) ____________
(x − 3)(x + 1)
4
(b) _____________
(2x + 1)(x + 2)
x
(c) __________
x2 + 5x + 6
x+2
(d) ______________
x(3x2 + 4x + 1)
Ask yourself
Is the fraction proper or improper, can the denominator
be factorised, are the factors in the denominator distinct,
what are the parts, what values of x can be used to find
the constants?
Proper fractions: Repeated linear factors
The denominator of the proper fraction contains repeated linear factors of the form
(cx + d)n. For each factor repeated n times there is a corresponding sum of n
fractions of the form
λ3
λn
λ1
λ2
______
+ ________
+ . . . + ________
+ ________
2
3
cx + d (cx + d)
(cx + d)n
(cx + d)
where λ1, λ2, λ3, . . . , λn are constants to be found.
P(x)
C
A + ______
B + ________
For example, _______________2 ≡ ______
ax + b cx + d (cx + d)2
(ax + b)(cx + d)
Note the sum of two fractions associated with the repeated linear factor.
88
MODULE 1tCHAPTER 3
EXAMPLE 5
SOLUTION
2x + 1
Separate _____________
into partial fractions.
(x + 2)(x + 1)2
Since the denominator consists of a distinct linear factor and a repeated linear factor,
we have
C
2x + 1
A + _____
B + _______
_____________
≡ _____
(x + 2)(x + 1)2 x + 2 x + 1 (x + 1)2
(one fraction for the distinct linear factor and the sum of two fractions for the
repeated linear factor).
Multiply both sides of the identity by the denominator of the left-hand side.
Note
The values of x
that are chosen
to find A, B and C
are −1, −2 and
0. The first two
values are found
from the zeros of
the denominator
(x + 1 = 0 and
x + 2 = 0) and
the third value
can be any arbitrary value of x.
2x + 1 = A(x + 1)2 + B(x + 2)(x + 1) + C(x + 2)
[1]
Substitute x = −1 into [1]
2(−1) + 1 = A(−1 + 1)2 + B(−1 + 2)(−1 + 1) + C(−1 + 2)
−2 + 1 = A(0)2 + B(1)(0) + C(1)
−1 = C
Substitute x = −2 into [1]
2(−2) + 1 = A(−2 + 1)2 + B(−2 + 2)(−2 + 1) + C(−2 + 2)
−4 + 1 = A(−1)2 + B(0)(−1) + C(0)
−3 = A
To find B we can substitute any other value for x.
Substitute x = 0, A = −3, C = −1 into [1]
2(0) + 1 = −3(0 + 1)2 + B(0 + 2)(0 + 1) + (−1)(0 + 2)
1 = −3 + 2B − 2
6 = 2B
3=B
Hence
2x + 1
−3 + _____
3 − _____
3 + _______
3 − _______
−1 = _____
1
_____________
≡ _____
(x + 2)(x + 1)2 x + 2 x + 1 (x + 1)2 x + 1 x + 2 (x + 1)2
EXAMPLE 6
SOLUTION
3x − 1 into partial fractions.
Separate ________
x2(x + 1)
2
Since the denominator of this fraction consists of one repeated factor and one
distinct linear factor, we have
C
3x2 − 1 ≡ __
B + _____
A + __
________
x2(x + 1) x x2 x +1
Multiplying throughout by x2(x + 1), we have
3x2 − 1 ≡ Ax(x + 1) + B(x + 1) + Cx2
When x = 0,
−1 = B(0 + 1)
B = −1
89
M O DUL E 1
When x = −1,
3(−1)2 − 1 = C(−1)2
C=2
Equating coefficients of x2, we have
3=A+C
Substituting C = 2, we have
3=A+2
A=1
3x2 − 1 ≡ __
1 + _____
2
1 − __
∴ ________
x2(x + 1) x x2 x +1
EXAMPLE 7
SOLUTION
A × (x + 1)3 x
_____
x+1
= A(x + 1)2 x
B × (x + 1)3 x
_______
(x + 1)2
= B(x + 1) x
x2 + x − 1 into partial fractions.
Separate _________
(x + 1)3 x
Since the denominator of this fraction consists of one repeated linear factor and one
distinct linear factor, we have
C
x2 + x − 1 ≡ _____
B
A + _______
D.
_________
+ _______
+ __
(x + 1)3x
x + 1 (x + 1)2 (x + 1)3 x
Multiplying throughout by (x + 1)3x, we have
x2 + x − 1 ≡ Ax(x + 1)2 + Bx(x + 1) + Cx + D(x + 1)3
= Ax(x2 + 2x + 1) + Bx2 + Bx + Cx + D(x + 1)(x2 + 2x + 1)
C × (x + 1)3 x
_______
(x + 1)3
= Cx
= Ax3 + 2Ax2 + Ax + Bx2 + Bx + Cx + Dx3 + 2Dx2 + Dx + Dx2
+ 2Dx + D
= x3(A + D) + x2(2A + B + 3D) + x(A + B + C + 3D) + D
D × (x + 1)3 x
__
x
= D(x + 1)3
When x = 0,
−1 = D
When x = −1,
(−1)2 + (−1) − 1 = C (−1)
C=1
Equating coefficients of x3, we have
0=A+D
Substituting D = −1, we have
0=A−1
A=1
Equating coefficients of x2, we have
1 = 2A + B + 3D
90
MODULE 1tCHAPTER 3
Substituting D = −1, A = 1, we have
1=2+B−3
B=2
x2 + x − 1 ≡ _____
1 + _______
1
1
2
+ _______
− __
∴ _________
x + 1 (x + 1)2 (x + 1)3 x
(x + 1)3 x
Try these 3.2
Separate the following into partial fractions.
6x2 − x − 2
(a) __________
2
Ask yourself
6x2 + 5x + 3
(b) ___________
2
Is the fraction proper, is the denominator factorised,
are the factors linear, are the factors repeated, what
are the fractions associated with the factors of the
denominator, what is the fastest method of finding
the constants?
x (x − 1)
x(2x + 1)
3x2 − 3x − 2
(c) __________________
2
(x + 1)(x − 2x + 1)
Proper fractions: Unrepeated quadratic factors
The denominator of the proper fraction contains unrepeated quadratic factors of
the form ax2 + bx + c. For each factor of this form in the denominator there is a
Ax + B .
corresponding fraction of the form ___________
ax2 + bx + c
P(x)
Bx + C
A + ___________
≡ ______
For example ___________________
ax + b cx2 + dx + e
(ax + b)(cx2 + dx + e)
where A, B and C are constants to be found.
EXAMPLE 8
4
Separate ______________
into partial fractions.
(x + 1)(2x2 + 1)
SOLUTION
First check that the quadratic factor does not factorise. This fraction consists of one
distinct linear factor and one quadratic factor in the denominator, so we have
Bx + C
A + _______
4
______________
≡ _____
(x + 1)(2x2 + 1) x + 1 2x2 + 1
Multiply both sides of the identity by (x + 1)(2x2 + 1)
4 ≡ A(2x2 + 1) + (Bx + C)(x + 1)
[1]
Substitute x = −1 into [1]
4 = A(2(−1)2 + 1) + (B(−1) + C)((−1) + 1)
4 = 3A
4
A = __
3
Substitute x = 0 into [1]
4 = A(2(0)2 +1) + (B(0) + C)(0 + 1)
4=A+C
4+C
4 = __
3
8
4 = __
C = 4 − __
3 3
91
M O DUL E 1
Note
Since we have
found A and C,
to find B we can
equate coefficients or substitute any other
value for x.
Equating coefficients of x2, we
get 0 = 2A + B.
Substitute x = 1 and also the values for A and C into [1]
(
)
8 (1 + 1)
4 (2(1)2 + 1) + B(1) + __
4 = __
3
3
16
4
__
___
4 = (3) + 2B +
3
3
16
___
2B = − 3
8
B = − __
3
Hence
−8 x + __
8
4
__
___
8(1 − x)
3
3
3
4
4
_____
________
______________
+
+ _________
≡
≡ _______
2 + 1)
2
(x + 1)(2x2 + 1) x + 1
3(2x
3(x
+
1)
2x + 1
EXAMPLE 9
2x2 + 1 into partial fractions.
Separate ________
x(x2 + 1)
SOLUTION
The denominator consists of one linear factor and one quadratic factor, so we have
Bx + C
2x2 + 1 ≡ __
A + _______
________
x(x2 + 1) x
x2 + 1
Multiplying throughout by x(x2 + 1), we have
2x2 + 1 ≡ A(x2 + 1) + (Bx + C)x
= Ax2 + A + Bx2 + Cx
= (A + B)x2 + Cx + A
Substituting x = 0, we have
1=A
Equating coefficients of x2, we have
2=A+B
∴2=1+B
(Substituting A = 1)
B=1
Equating coefficients of x, we have
C=0
2x2 + 1 ≡ __
x
1 + ______
∴ ________
x(x2 + 1) x x2 + 1
Try these 3.3
Separate the following into partial fractions.
3
(a) ________
2
x(x + 1)
x2 + 4x − 1
(b) __________________
2
(x − 2)(x + 2x + 3)
−2x − 1
(c) _________________
2
(x + 1)(x + x + 1)
Ask yourself
Is the fraction proper, is the denominator factorised, are the factors linear or quadratic or
both, are the factors repeated, what are the fractions associated with the factors of the
denominator, what is the fastest method of finding the constants?
92
MODULE 1tCHAPTER 3
Proper fractions: Repeated quadratic factors
The denominator of the proper fraction consists of repeated quadratic factors of
the form (ax2 + bx + c)2. For each factor of this form in the denominator there
Cx + D
Ax + B + _____________
corresponds the sum of two fractions of the form ___________
ax2 + bx + c (ax2 + bx + c)2
where A, B, C and D are constants to be found.
E X A M P L E 10
SOLUTION
2x + x + 8 into partial fractions.
Separate ______________
(x2 + 4)2(x + 1)
2
We have a repeated quadratic factor and a linear factor in the denominator, therefore
the partial fractions are:
Cx + D + _____
Ax + B + ________
2x2 + x + 8 ≡ _______
E
______________
2
(x + 4)2(x + 1)
x2 + 4
(x2 + 4)2 x + 1
Multiplying throughout by (x2 + 4)2(x + 1) gives
2x2 + x + 8 ≡ (Ax + B)(x2 + 4)(x + 1) + (Cx + D)(x + 1) + E(x2 + 4)2
Substituting x = −1 gives
9
9 = 25E ⇒ E = ___
25
Equating coefficients of x4 gives
0=A+E
9
∴ A = −E = −___
25
9
Equating coefficients of x3 gives 0 = A + B ⇒ B = −A = ___
25
9 + D + 16 ___
9 ⇒ D = __
4
When x = 0, 8 = 4B + D + 16E ⇒ 8 = 4 ___
5
25
25
2
Equating coefficients of x gives 2 = 4A + B + C + 8E
9 + C + ___
36 + ___
72 ⇒ C = __
1
⇒ 2 = − ___
5
25 25
25
Hence
( )
( )
9
9
__
__
_1 x + _4
−__
25 x + 25
5
5
25
2x2 + x + 8 ≡ _________
________
_____
______________
+
+
(x2 + 4)2(x + 1)
x2 + 4
(x2 + 4)2 x + 1
9
This simplifies to
9(1 − x)
9
x + 4 + ________
2x2 + x + 8 ≡ _________
______________
+ _________
2
(x + 4)2(x + 1) 25(x2 + 4) 5(x2 + 4)2 25(x + 1)
Try these 3.4
Separate the following into partial fractions.
x4 + 1
(a) _________
2
2
x(x + 1)
1 − x + 2x2 − x3
(b) ______________
2
2
x(x + 1)
3x4 + 5x3 + 7x2 + 2x + 1
(c) _____________________
2
2
x(x + x + 1)
93
M O DUL E 1
Improper fractions
If the degree of the numerator is equal to or higher than that of the denominator, the
expression is an improper fraction. Change the fraction to a mixed fraction and then
separate into partial fractions.
E X A M P L E 11
x3 − 1
Separate _____________
into partial fractions.
(x + 1)(x2 + 1)
SOLUTION
Since the fraction is an improper fraction, by long division we get
1
x3 + x2 + x + 1 )‾‾‾‾‾‾‾
x3 + 0x2 + 0x − 1
x3 + x2 + x + 1
− x2 − x − 2
x +x+2
x −1
≡ 1 − _____________
∴ _____________
(x + 1)(x2 +1)
(x + 1)(x2 + 1)
3
2
x + x + 2 into partial fractions gives:
Separating _____________
(x + 1)(x2 + 1)
2
Bx + C
x2 + x + 2 ≡ _____
A + _______
_____________
(x + 1)(x2 + 1) x + 1
x2 + 1
x2 + x + 2 ≡ A(x2 + 1) + (Bx + C)(x + 1)
Substitute x = −1 into [1]
1 − 1 + 2 = A(2) + (−B + C)(0)
2A = 2
A=1
Substitute x = 0 into [1]
2=A+C
Since A = 1
2=1+C
C=1
Substitute x = 1, A = 1 and C = 1 into [1]
1 + 1 + 2 = (1)(1 + 1) + (B(1) + 1)(1 + 1)
4 = 2 + 2B + 2
B=0
x2 + x + 2 ≡ _____
1
1 + ______
Hence _____________
(x + 1)(x2 + 1) x + 1 x2 + 1
x +x+2
x −1
≡ 1 − _____________
Since _____________
(x + 1)(x2 + 1)
(x + 1)(x2 + 1)
3
2
x3 − 1
1
1 − ______
_____________
≡ 1 − _____
x + 1 x2 + 1
(x + 1)(x2 +1)
94
[1]
MODULE 1tCHAPTER 3
E X A M P L E 12
SOLUTION
x3 − x2 − 1 into partial fractions.
Separate ____________
(x + 1)(x − 2)
x − x − 1 is an improper fraction we need to divide first.
Since ____________
(x + 1)(x − 2)
x
‾‾‾‾‾‾
‾‾‾
x2 − x − 2 )‾
−1
x3 − x2 + 0x
3
2
x − x − 2x
2x − 1
3
2
x − x − 1 ≡ x + ____________
2x − 1
∴ ____________
(x + 1)(x − 2)
(x + 1)(x − 2)
3
2
2x − 1
Separating ____________
into partial fractions, we have
(x + 1)(x − 2)
2x − 1
B
A + _____
____________
≡ _____
(x + 1)(x − 2) x + 1 x − 2
⇒ 2x − 1 ≡ A(x − 2) + B(x + 1)
When x = 2,
2(2) − 1 = B(2 + 1)
3 = 3B
B=1
When x = −1,
2(−1) − 1 = A(−1 − 2)
−3 = −3A
A=1
2x − 1
1 + _____
1
∴ ____________
≡ _____
(x + 1)(x − 2) x + 1 x − 2
Hence
x3 − x2 − 1 ≡ x + _____
1 + _____
1
____________
x+1 x−2
(x + 1)(x − 2)
Try these 3.5
Separate the following into partial fractions.
x2
(a) ____________
Ask yourself
(x + 1)(x + 2)
x3
(b) _______
2
(x + 1)
x3 + 6x2 + 8x + 2
(c) _______________
x(x + 1)(x + 2)
Is the fraction proper, is the denominator factorised, are the factors linear or quadratic or both, are
the factors repeated, what are the fractions associated with the factors of the denominator, what is
the fastest method of finding the constants?
Extension
Investigation: Can you work out what happens if there
is a repeated quadratic fraction?
95
M O DUL E 1
EXERCISE 3A
In questions 1–25, separate the expression into partial fractions.
1
2
____________
(x + 1)(x + 2)
2
3x
____________
(x − 1)(x + 2)
3
4x
______________
(3x + 2)(2x + 1)
4
2x + 3
____________
(x − 1)(x − 2)
5
x
___________
2x2 − 5x + 2
6
4
___________
x2 + 7x + 12
7
x+1
________
x2(x + 2)
8
x
_____________
(x + 1)2(x + 3)
9
3x + 5
_____________
x(x2 − 4x + 4)
5
10 ______________
2 2
(x + 1) (x − 1)
7
11 ______________
2
5x + 2
12 _________
2
2x + 1
13 ________
2
3x − 2
14 _________
2
2
x
15 ______________
2
4x
16 _________________
2
2
17 ______________
2
7x − 5
18 ___________________
2
12x2 − 3x − 10
19 ______________
x2 + 8x
20 ____________
12x2 + 10x + 15
21 ______________
2
−11x − 15
22 ___________________
2
5x2 + 8x + 5
23 __________________
2
1
24 ______
3
x (3x − 1)
(x − 4) (2x + 1)
(x + 1)x
(x + 4)x
(x + 2)(4x + 3)
(x + x + 1)(x + 1)
(4x + 2)(x + 2)
(3x + 1)(2x + 5x + 4)
(2x + 1)(3x − 4)
(x + 2)(x − 1)
(x + 2) (1 − x)
(2x + 1)(2x − 5x − 3)
(2x + 1)(x + x + 1)
x −8
2x2 − 3x + 5
25 _____________
2
(x + 1)(x + 4)
C .
4x + 4x + 1 in the form A + __
B + _____
26 Write the fraction ___________
x x+1
2
x(x + 1)
x + 6x + 7x + 2 into partial fractions.
27 Separate _______________
3
2
(x + 2)(x + 4)
96
MODULE 1tCHAPTER 3
Cx + D where
x + x + 2x + 1 can be written in the form A + __
B + _______
28 Show that ______________
2
2
3
2
x(x + 1)
x
x +1
A, B, C and D are constants.
29 (a) Separate the following into partial fractions.
x2 + 3x + 3
(i) __________
(x + 1)2
x+1
(ii) _______
x4 − 16
x − 3x − 4x − 9x − 6 in partial fractions.
(b) Write _____________________
(x − 4)(x2 + x + 1)
4
3
2
30 Given that A(x2 + x + 1) + (Bx + C)(x + 2) = 2x2 + 4x + 3, find A, B and C.
2x + 4x + 3
into partial fractions.
Hence separate _________________
(x + 2)(x2 + x + 1)
2
SUMMARY
Partial fractions
Improper
fractions
Proper
fractions
Unrepeated
linear factors
in the
denominator
Repeated
linear factors
in the
denominator
Unrepeated
quadratic
factors in the
denominator
Repeated
quadratic
factors in the
denominator
λ1
λ2
λ3
λn
A
Ax + B
Ax + B +
Cx + D
+ ... +
+
+
(cx + d)n ax2 + bx + c ax2 + bx + c (ax2 + bx + c)2
ax + b cx + d (cx + d)2 (cx + d)3
Divide and separate
into the quotient
plus the remainder
Go to proper
fractions
97
M O DUL E 1
Checklist
Can you do these?
■ Express an improper fraction as a mixed fraction.
■ Separate a proper fraction with distinct linear factors in the denominator into
partial fractions.
■ Separate a proper fraction with repeated linear factors in the denominator into
partial fractions.
■ Separate a proper fraction with (unrepeated or repeated) quadratic factors in
the denominator into partial fractions.
■ Separate an improper fraction into partial fractions.
98
MODULE 1tCHAPTER 4
CHAPTER 4
Integration
At the end of this chapter you should be able to:
■ carry out integration by recognition
■ carry out integration by substitution
■ carry out integration by parts
■ carry out integration by using partial fractions
■ integrate trigonometric functions.
KEYWORDS/TERMS
JOUFHSBUJPOtEJČFSFOUJBUJPOtDPOTUBOUPGJOUFHSBUJPOt
SFDPHOJUJPOtQBSUJBMGSBDUJPOTtTVCTUJUVUJPOt
JOUFHSBUJPOCZQBSUTtMJNJUTPGJOUFHSBUJPO
99
M O DUL E 1
Integration is the reverse of differentiation. Recall that whenever we integrate a
function without limits we need to add a constant of integration.
There are four methods of integration that we use for the CAPE syllabus: integration
by recognition, integration using partial fractions, integration by substitution and
integration by parts.
Recognition makes use of a set of standard integrals that you must be familiar with. As
soon as you look at the function that has to be integrated you should recognise its form
and be able to write down the result. For integration using partial fractions the function
being integrated is a rational function that can be split into parts. Once the split is done,
the resulting fractions will be simpler to integrate than the original function. Substitution is generally used when integrating composite functions. When a substitution is
used, the original function is replaced and the resulting function will be of a form that
can be integrated quite easily. Integration by parts makes use of the product rule for
differentiation and is used to integrate some products of functions. All methods of
integration break down the function being integrated to a simpler form.
Integration by recognition
Integrating by recognition makes use of the standard list of integrals in this table.
Function
xn
_____ + c,
n+1
(ax + b)n
(ax + b)n+1
1 __________
__
+ c,
a
n+1
[
n ≠ −1
]
n ≠ −1
1
__
x
ln | x | + c
1
______
ax + b
1 |
__
|
a ln ax + b + c
ex
ex + c
e ax+b
1 ax+b + c
__
ae
sin x
− cos x + c
sin (ax + b)
1
−__
a cos (ax + b) + c
cos x
sin x + c
cos (ax + b)
1
__
a sin (ax + b) + c
tan x
100
Integral
x n+1
−ln | cos x| + c
or
ln | sec x| + c
tan (ax + b)
1
__
a ln sec (ax + b) + c
sec x
ln | sec x + tan x| + c
sec2 x
tan x + c
MODULE 1tCHAPTER 4
Function
Integral
cosec x
x +c
ln tan __
2
cot x
ln | sin x | + c
f ′ (x)
____
f (x)
ln | f (x) | + c
| ( )|
[ f (x) ]
________
+ c,
n+1
f ′ (x)[ f (x) ]n
n+1
n ≠ −1
f ′ (x) e f (x)
e f (x) + c
1
________
______
√a2 − x2
1
_______
a2 + x2
x
sin−1 ( __
a) + c
1
__
x
−1 __
a tan ( a ) + c
∫
EXAMPLE 1
1
Find _______
d x.
(3x + 1)
SOLUTION
1 is of the form ______
1 . Therefore the integral is a standard
The function ______
3x + 1
ax + b
integral.
∫
1
1 d x = __
Using ______
a ln | ax + b | + c where a = 3, b = 1,
ax + b
1 ln | 3x + 1 | + c
1 d x = __
we get ______
3
3x +1
∫
∫
EXAMPLE 2
1
Find ________
d x.
(3x + 1)3
SOLUTION
This integral is of the form (ax + b)n where a = 3, b = 1, n = −3.
1
________
d x = (3x + 1)−3 d x
(3x + 1)3
(3x + 1)−2
1 _________
= __
+c
3
−2
1
1 (3x + 1)−2 + c = −_________
+c
= −__
6
6(3x + 1)2
∫
∫
∫
∫
EXAMPLE 3
Find e4x+5 d x.
SOLUTION
1 ax+b + c.
From the table e ax+b d x = __
ae
∫
Substituting a = 4, b = 5, we get
1 e 4x+5 + c
e 4x+5 d x = __
4
∫
∫
EXAMPLE 4
Find e3−2x d x.
SOLUTION
1 ax+b + c
This is of the form e ax+b d x = __
ae
∫
where a = −2, b = 3.
1 e3−2x + c
e3−2x d x = −__
2
∫
101
M O DUL E 1
EXAMPLE 5
dy
Given that y = 3x, find ___ and hence find 3x d x.
dx
SOLUTION
y = 3x
Using the chain
rule:
d ln y
___
[ ]
dx
dy
d ln y × ___
= ___
[ ] dx
dy
dy
1 ___
= __
y dx
∫
Taking ln on both sides, we have
ln y = ln 3x
= x ln 3
Differentiating both sides with respect to x, we have
dy
1 ___
__
y d x = ln 3
dy
Therefore ___ = y ln 3
dx
Substituting y = 3x, we have
dy
___
= 3x ln 3
dx
To find the integral we can use integration as the reverse of differentiation.
Since
d [3x] = 3x ln 3
___
dx
integrating both sides with respect to x, we get
∫
3x = 3x ln 3 d x
∫
= (ln 3) 3x d x
Hence
3 +c
∫3x dx = ___
ln 3
x
∫
EXAMPLE 6
Find ax d x, where a is a constant.
SOLUTION
Let y = ax
Taking ln on both sides, we have
ln y = ln ax = x ln a
Differentiating both sides with respect to x, we have
dy
1 ___
__
y d x = ln a
dy
Therefore ___ = y ln a
dx
Substituting y = ax, we have
dy
___
= ax ln a
dx
dy
d ln y = __
___
[ ] 1___
y dx
dx
d [ x ln a ] = ln a
___
dx
To find the integral we can use integration as the reverse of differentiation.
Since
d [ax] = ax ln a
___
dx
102
MODULE 1tCHAPTER 4
integrating both sides with respect to x, we get
∫
ax = ax ln a d x
∫
= (ln a) ax d x
Hence
a +c
∫ax dx = ____
ln a
x
Now let us use this result to integrate for particular values of a.
EXAMPLE 7
SOLUTION
Show that
3 .
∫0 4x dx = ___
ln 4
1
∫
x
a + c = ax d x
Using ____
ln a
where a = 4, we have
1
4x 1
4x d x = ___
ln 4 0
0
1
4
40
= ___ − ___
ln 4 ln 4
1
4 − ___
= ___
ln 4 ln 4
3
= ___
ln 4
This result is worth remembering.
[ ]
∫
EXAMPLE 8
π d x.
Find tan ( 3x + __
2)
SOLUTION
1
Using the standard form for tan (ax + b) d x = __
a ln sec (ax + b) + c
∫
∫
∫tan ( 3x + __π2 ) dx = __31 ln sec ( 3x + __π2 ) + c
EXAMPLE 9
SOLUTION
∫
1
Find ____________
π d x.
cosec ( x + __
2)
1
Recall that ______
cosec x = sin x
π
1
__
∴ ____________
π = sin ( x + 2 )
cosec ( x + __
2)
1
Using sin (ax + b) d x = −__
a cos (ax + b) + c
∫
π
π
1
__
__
π d x = ∫sin ( x + 2 ) d x = −cos ( x + 2 ) + c
∫ ____________
cosec ( x + __
2)
Try these 4.1
Find the following integrals.
(a) ∫e5x−2 d x
(b) ∫e2−7x d x
(c)
(d)
(e)
∫ cos ( 3x − __π2 ) dx
∫27x dx
(f)
∫ tan ( 5x + __π2 ) dx
1
∫0 5x dx
103
M O DUL E 1
When the numerator is the differential of the denominator
f ′ (x)
Now let us look at the form _____ d x = ln | f (x) | + c.
f (x)
Note that this is a general form where the numerator is the differential of the
denominator; f (x) can be any function of x.
∫
E X A M P L E 10
SOLUTION
∫
2x d x.
Integrate ______
1 + x2
d [1 + x2] = 2x
___
dx
Let f (x) = 1 + x2
f ′(x) = 2x
f ′ (x)
This integral is of the form ____ d x = ln | f (x) | + c.
f (x)
2x
∴ ______2 d x = ln | 1 + x2 | + c
1+x
∫
∫
E X A M P L E 11
SOLUTION
x+2
d x.
Integrate __________
x2 + 4x + 1
∫
d [x2 + 4x + 1] = 2x + 4.
___
dx
f ′ (x)
This is of the form _____ d x.
f (x)
We can rewrite the integral as
∫
f′
Note
1 × 2 = 1. Our
__
2
function has not
changed.
↓
x+2
2(x + 2)
d x = __ ∫ __________
dx
∫ __________
2 x2 + 4x + 1
x2 + 4x + 1
1
↑
f
x+2
1 ln [x2 + 4x + 1] + c
∴ __________
= __
2
x + 4x + 1 2
∫
∫
E X A M P L E 12
Show that tan x d x = ln |sec x| + c
SOLUTION
sin x
Writing tan x = ____
cos x
cos x d x
∫ tan x dx = ∫ ____
−sin x d x,
= −∫ ______
cos x
sin x
d [cos x] = −sin x
___
dx
= −ln |cos x| + c
= ln |(cos x)−1| + c, using rules of logarithms
= ln |sec x| + c
E X A M P L E 13
SOLUTION
∫
ex d x.
Find ______
1 + ex
f ′(x)
d [1 + ex] = ex the integral is of the form ____
Since ___
d x = ln | f (x) | + c.
dx
f (x)
ex d x = ln |1 + ex| + c
∴ ______
1 + ex
∫
∫
104
MODULE 1tCHAPTER 4
1
__
∫ ln x
x d x.
___
E X A M P L E 14
Find
SOLUTION
1
d [ln x] = __
___
x
dx
f ′(x)
d x = ln | f (x) | + c
∫ ____
f (x)
∴
Try these 4.2
1
__
∫ ln x
x d x = ln | ln x | + c
___
Find the following integrals.
(a)
x dx
∫ ______
4
x +5
(b)
x dx
∫ ______
2
x −1
(c)
cos x d x
∫ ____
sin x
(d)
dx
∫ ___________
3x2 + 2x + 1
3
3x + 1
The form ∫ f ′(x)[ f (x)]n dx, n ≠ −1
The form f ′(x) [ f (x) ]n is also a useful general form of an integral. The differential
of the function inside the brackets must be multiplied by the function raised to the
power of n.
∫
[ f (x)]n+1
f ′(x) [ f (x)]n d x = ________ + c,
n+1
n ≠ −1.
Let us see how this works.
∫
1
__
E X A M P L E 15
Integrate 2x (1 + x2) 2 d x.
SOLUTION
Let f (x) = 1 + x2
f ′(x) = 2x,
1
n = __
2
(to fit the form f ′(x) [ f (x)]n)
[ f (x)]n+1
Using f ′(x) [ f (x)]n d x = ________ + c, n ≠ −1
n+1
∫
∫ 2x (1 +
1
__
x2) 2
3
__
(1 + x2) 2
d x = ________ + c
3
__
2
2 (1 + x2) __32 + c
= __
3
∫
E X A M P L E 16
Integrate cos x sin4 x d x.
SOLUTION
Let f (x) = sin x
f ′(x) = cos x, n = 4
∫
∫
∴ f ′(x) [f (x)]n d x = cos x (sin x)4 d x
5
sin x + c
= _____
5
105
M O DUL E 1
∫
E X A M P L E 17
Find tan4 x sec2 x d x.
SOLUTION
Let f (x) = tan x
f ′(x) = sec2 x,
n=4
∫ f ′(x) [ f (x)]n dx = ∫tan4 x sec2 x dx
∫
= sec2 x (tan x)4 d x
tan5 x + c
= _____
5
∫
E X A M P L E 18
1 (ln x) d x.
Find __
x
SOLUTION
Let f (x) = ln x
1, n = 1
f ′(x) = __
x
∫ f ′(x) [ f (x)]n dx = ∫ __1x (ln x)1 dx
(ln x)2
= ______ + c
2
Try these 4.3
Find the following integrals.
x
______
(a) _______
dx
√1 + x2
∫
(b) ∫sin x cos7 x d x
2x + 1
dx
∫ _____________
2
(2x + 2x + 3)3
(c)
The form ∫ f ′(x) e f (x)dx
In the general form f ′(x) e f (x), again the differential of the index is multiplied
by e f (x).
∫ f ′(x) e f (x) dx = e f (x) + c
∫
2
E X A M P L E 19
Find x e x d x.
SOLUTION
Let f (x) = x2
f ′(x) = 2x
∫
∫
∴ f ′(x) e f (x) d x = 2xe x d x = e x + c
2
∫
2
2
Since we are interested in xe x d x we can write the function as
∫xe x dx = __21 ∫2xe x dx = __21 e x + c
2
106
2
2
MODULE 1tCHAPTER 4
∫
E X A M P L E 20
Find sec2x e tan x d x.
SOLUTION
Let f (x) = tan x
f ′(x) = sec2 x
∫
∫
∴ f ′(x) e f (x) d x = sec2 x e tan x d x
= e tan x + c
∫
E X A M P L E 21
Find cos x e sin x d x.
SOLUTION
Let f (x) = sin x
f ′(x) = cos x
∫
∫
∴ f ′(x) e f (x) d x = cos x esin x d x
= e sin x + c
Try these 4.4
These functions
can be integrated
with careful
recognition.
Find the following integrals.
(a)
______ e sin
∫ _______
√1 − x2
(c)
∫x2e x +1 dx
1
−1x
1 e tan−1 x d x
(b) ∫ ______
2
dx
1+x
(d) ∫sin x ecos x d x
3
EXERCISE 4A
Remember
These can all
be integrated
by recognition.
You should be
able to map
each function to
one in the table
given at the
beginning of the
chapter.
Write down the integrals of the following functions.
1
∫e7x dx
2
∫e4x+2 dx
3
∫e5−2x dx
4
dx
∫ ______
4x + 5
5
dx
∫ ______
7x − 2
6
2 dx
∫ ______
4 − 3x
7
∫ tan ( 2x + __π4 )
8
∫ sec2 ( __π2 − 3x ) dx
π dx
∫ ___________
sec ( 2x − __
)
10
1
dx
∫ ___________
cosec (x + 2)
11
1
dx
∫ ___________
cos2 (3x + 1)
12
∫6x2e x dx
13
∫
14
1 e x dx
∫ ___
√x
9
3
1
4
sin x ecos x d x
1
3
__
__
√
107
M O DUL E 1
15
∫xe −x dx
16
∫(e3x − ex)2 dx
17
x dx
∫______
x2 + 9
18
cos x d x
∫ _________
2 sin x + 1
19
4 sec x d x
∫__________
2 tan x − 5
20
2x d x
∫ ______
5 − x3
21
e
dx
∫ ______
e3x + 1
22
arcsin
x dx
______
∫ _______
√1 − x2
23
∫ √tan2 3x + 1 dx
24
sin ( __
x ) dx
∫ __
x2
25
∫sin x cos4 x dx
26
∫esin 4x cos 4x dx
27
∫e t t2 dt
28
x dx
∫0 ______
x2 + 9
2
2
3x
__________
3
2
1
1
1
Integration by substitution
The method of substitution is used to simplify the functions into a standard form
and then integrate these functions. When using a substitution, each function in the
integral must be replaced by a new variable; if there are limits within the integral it is
advisable to change the limits as well.
∫
_____
E X A M P L E 22
Using the substitution u = 1 + x find x √1 + x d x.
SOLUTION
Step 1
We first find a replacement for d x by differentiating u.
Since u = 1 + x
du = 1
___
dx
du = dx
Ask yourself
Why is u = 1 + x being used as
the substitution?
_____
Step 2 Change √1 + x
Since u = 1 + x
_____
__
√1 + x = √u
Step 3 Change x
Again u = 1 + x ⇒ x = u − 1.
_____
__
Substituting du = d x, √1 + x = √u, x = u − 1, we have
_____
__
∫ x √1 + x dx = ∫ (u − 1) √u du
∫
1
__
= (u − 1)u 2 du
108
MODULE 1tCHAPTER 4
Multiplying the brackets, we have
∫(u − 1)u 2 du = ∫ u 2 − u 2 du
3
__
1
__
5
__
2
1
__
3
__
2
u +c
u − ___
= ___
5
3
__
__
2
2
2 u __23 + c
2 u __25 − __
= __
5
3
Substituting u = 1 + x, we have
_____
∫ x √x + 1 dx = __52 (1 + x) 2 − __32 (1 + x) 2 + c
5
__
3
__
E X A M P L E 23
x + 1 d x.
______
Using the substitution u = 2x + 1, find ________
√2x + 1
SOLUTION
Since we are using a substitution we need to convert all our x’s to u’s.
∫
x + 1 dx
∫ ________
√2x + 1
Ask yourself
______
Let u = 2x + 1
Notice
that
______
√ 2x + 1 is a
composite
function. We
use the function
inside as the
substitution.
Why is 2x + 1 being used as the
substitution?
Starting with d x
du = 2
___
dx
du = 2 dx
1 du = dx
__
2
1 d u.
We will replace d x by __
2
______
__
Next: √2x + 1 = √u since u = 2x + 1
______
__
We will replace √2x + 1 by √u
Since u = 2x + 1, making x the subject of the formula
u − 1 = 2x
1 u − __
1
x = __
2
2
1 + 1 = __
1 u + __
1
1 u − __
∴ x + 1 = __
2
2
2
2
1 u + __
1
We will replace x + 1 by __
2
2
We have:
1 u + __
1
x + 1 = __
2
2
1 du
dx = __
2
______
__
√ 2x + 1 = √u
Now that we have changed all our functions, let us substitute:
∫
x + 1 dx =
_______
______
√2x + 1
∫
1
__
1
__
u+
2 __ 2 __
_______
( 1 ) du
√u
2
At this stage we must recognise that this is a standard function that can be integrated
by multiplying out the bracket.
u + 1 du
1 _____
= __
1
__
4
u2
1 (u + 1) u −__21 d u
= __
4
∫
∫
109
M O DUL E 1
1
= __
4
∫( u__21 + u − 12 ) du
__
[
3
__
1
__
]
[
]
u 2 + ___
u 2 + c = __
1 __
2 u __23 + 2u __21 + c
1 ___
= __
4 __
4 3
3
1
__
2
2
We now replace
u to get back
our integral as a
function of x.
Substituting u = 2x + 1
x + 1 d x = __
2 (2x + 1) 2 + 2(2x + 1) 2 + c
1 __
]
∫ _______
2 [3
√2x + 1
3
__
______
3
__
1
__
1
__
1 (2x + 1) 2 + (2x + 1) 2 + c
= __
3
Let us evaluate a definite integral using substitution.
1
__
x
d x using the substitution u = 1 − x2.
∫02 _______
√1 − x2
E X A M P L E 24
Evaluate
SOLUTION
x
______
dx
∫02 _______
√1 − x2
______
_1
Step 1
Replace d x by a function of u
Since u = 1 − x2
du = −2x
___
dx
1
−__
2 du = x d x
Remember
With substitutions
you should:
t Change dx
to du.
t Change all
functions of x
to functions
of u.
t Change the
limits when
the integral
is a definite
integral.
t If there are no
limits, remember to convert
back to the
original
function.
110
1
The integral contains x d x, which we can replace by −__
2 du.
Step 2
_____
Convert √1 − x2 to a function of u.
______
__
Now √1 − x2 = √u since u = 1 − x2
Step 3
Convert the limits to limits of u
To change the limits we use u = 1 − x2
When x = 0, u = 1 − 02 = 1
( )
3
1 2 = 1 − __
1 = __
1 , u = 1 − __
When x = __
4 4
2
2
1 du = x d x
∴ −__
2
______
__
√1 − x2 = √u
x=0⇒u=1
3
1 ⇒ u = __
x = __
4
2
MODULE 1tCHAPTER 4
Substituting, we get
−1 d u
∫02 √1 − x2 dx = ∫14 ____
2 √u
1
__
x
_______
______
3
__
__
=
1 du
∫__43 ____
2 √u
=
∫__43 ( __12 u−2 )du
1
__
1
(switching the limits and changing the sign)
1
__
[ ( )]
1
__
u2
= ____
1
2 __
2
1
3
__
4
[ ()]
3
= 1 − __
4
1
__
2
__
√3
= 1 −___
2
Try these 4.5
(a) Find ∫0 xe x2 d x using the substitution u = x2.
1
4x + 1 d x using the substitution u = x + 2.
(b) Evaluate ∫0 _______
2
1
(x + 2)
EXERCISE 4B
∫
1
x
Find the integral ________
d x using the substitution u = 4x + 2.
(4x + 2)3
2
x
_______
Use the substitution u = 6x2 + 8 to find ________
d x.
√6x2 + 8
3
Using the substitution u = 2x − 1, prove that
4
Given that y = x3 show that
integral.
5
Evaluate
6
Using the substitution u = x2 + 9, prove that
7
3 + 12x + 2
8x___________
Find _____________
d x using the substitution u = x4 + 3x2 + x.
√x4 + 3x2 + x
∫
2
8.
d x = ___
∫1 ________
25
(2x − 1)3
3
x
1
3x d x = ______
d y. Hence evaluate the
6
∫0 ______
∫
1
+
y2
1+x
0
2
8
x + 1 d x using u = 3x − 2.
∫1 ________
√ 3x − 2
2
______
∫0
4
______
98 .
x √x2 + 9 d x = ___
3
∫
∫
x d x using u = x2.
Find the integral ______
1 + x4
In questions 9–21, find the integrals using a suitable substitution.
8
9
________
∫sin x √cos x + 1 dx
10 ∫(2x + 1)(4x − 1)5 d x
111
M O DUL E 1
When solving
questions 9–21,
look back at
questions 1– 8
and identity why
that particular
substitution was
used. This will
help you choose
substitutions for
questions 9–21.
11
ln tan x d x
∫ ________
x
12 ∫x3e x 4+5 d x
13
x dx
∫ ______
1 + x4
14 ∫9xe 4−3x2 d x
15
∫
16
∫
17
sec 4x
dx
∫ ____________
(1 − 3 tan 4x)5
18
∫
19
∫ x cos ( x2 + __π2 ) dx
20
x + 1 dx
∫ _________
x2 + x − 1
21
cos 3x d x
∫ _________
4 + sin 3x
−1
3
x (x2 + 4)8 d x
2
2
__
x 3 dx
______
1
__
1 + x3
−1
3
_______
______
e sin (x) d x
2
√1 − x
Integration by parts
Our third method of integration is derived directly from the product rule for differentiation. Integration by parts is used to integrate some products of functions of x.
d [uv] = u ___
du
dv + v ___
Recall that ___
dx
dx
dx
Integrating both sides with respect to x, we have
∫
∫
dv d x + v ___
du d x
uv = u ___
dx
dx
∫
∫
dv d x = uv − v ___
du d x
∴ u ___
dx
dx
This is the formula for integrating by parts. Let us see how to use it.
E X A M P L E 25
SOLUTION
Remember
∫x e
x2
dx is done
by recognition
∫
but xex dx is
integrated by
parts.
112
∫
Find xex d x.
∫
∫
dv d x = uv − v ___
du d x.
Using integration by parts u ___
dx
dx
Compare the left-hand side of the formula with your integral.
dv d x ≡ xex d x
∫u ___
∫
dx
dv .
We must assign one of the functions to u and the other to ___
dx
dv = ex
∴ Let u = x, ___
dx
Look at the formula on the right-hand side
∫
du d x
uv − v ___
dx
du.
We need u, v, ___
dx
du we differentiate u = x.
To find ___
dx
dv = ex.
To find v we integrate ___
dx
dv = ex
We have u = x ___
dx
du = 1 v = ex
___
dx
Be careful with
your choice of
functions for
u and v. If we
switch the
functions,
dv = x,
u = ex, ___
dx
du = ex,
___
dx
1 x2, our
v = __
2
integral
becomes
∫
1 x2ex − __
1x2 exdx
__
2
2
which is more
complex than the
one we started
with.
MODULE 1tCHAPTER 4
Substituting into
dv d x = uv − v ___
dx
∫u ___
∫ du
dx
dx
∫xex dx = xex − ∫ex dx
∫
∴ xex d x = xex − ex + c
Let us try this again with fewer explanations.
∫
E X A M P L E 26
Find x sin x dx.
SOLUTION
dv d x = uv − v ___
d x.
∫u ___
∫ du
dx
dx
∫
∫
dv d x ≡ x sin x d x
Comparing u ___
dx
dv = sin x
Let u = x, ___
dx
du = 1,
∴ ___
v = −cos x
dx
dv d x = uv − v ___
du d x,
Substituting into u ___
dx
dx
∫
∫
∫x sin x dx = −x cos x − ∫−cos x dx
= −x cos x + sin x + c
Functions such as ln x, arccos x and arctan x can be integrated by parts.
∫
E X A M P L E 27
Use integration by parts to find ln x d x.
SOLUTION
Since we need two functions for integrating by parts we write ln x = 1 × ln x.
∫ln x dx = ∫1 ln x dx.
dv d x ≡ 1 ln x d x, in this case we must let u = ln x
Comparing with ∫u ___
∫
dx
(since we know the differential of ln x and we are going to integrate ln x).
dv = 1
Let u = ln x, ___
dx
du = __
1,
___
v=x
dx x
dv d x = uv − v ___
du dx
Substituting into u ___
dx
dx
∫
∫
∫ln x dx = x ln x − ∫ __1x × x dx
∫
= x ln x − 1 d x
= x ln x − x + c
113
M O DUL E 1
∫
E X A M P L E 28
Find tan−1 x d x.
SOLUTION
We use tan−1 x = 1 × tan−1 x.
dv d x ≡ 1 tan−1 (x) d x
∫u ___
∫
dx
Remember
x dx
∫ ______
1 + x2
1 ______
2x dx
= __
2 ∫ 1 + x2
1 ln (1 + x2) + c
= __
2
f' dx = ln f
since ∫ __
f
dv = 1
u = tan−1 x, ___
dx
du = ______
1 ,
___
v=x
dx 1 + x2
dv d x = uv − v ___
du d x
Substituting into u ___
dx
dx
∫
∫
x dx
∫tan−1 x dx = x tan−1 x − ∫ ______
1 + x2
1 ln (1 + x2) + c
= x tan−1 x − __
2
∫0 x2ex dx.
1
E X A M P L E 29
Evaluate
SOLUTION
d x = ∫ x2ex d x
∫u ___
dx
0
1
dv
Let u = x2,
dv = ex
___
dx
du = 2x, v = ex
___
dx
∫
∫
dv d x = uv − v ___
du d x
Substituting into u ___
dx
dx
∫0 x2ex dx = [x2ex]01 − ∫0 2x ex dx
1
= [ 12e1 − 02e0 ] − 2 ∫ x ex d x
0
1
= e − 2 ∫ x ex d x
0
1
Remember
You must decide
on the functions
dv.
for u and ___
dx
Since we are
dv we
integrating ___
dx
must know the
integral of this
function.
1
We need to integrate
dv = ex
Let u = x, ___
dx
du = 1, v = ex
___
dx
∫0 x ex dx by parts again.
1
∫0 x ex dx = [x ex]10 − ∫0 ex dx
1
1
= [x ex − ex]10 = (1e1 − e1) − (0e0 − e0)
=1
∴
∫0 x2ex dx = e − 2 (1) = e − 2
1
When we have a definite integral, we need to evaluate the integral at each stage or we
can integrate until the end and then substitute the limits.
114
MODULE 1tCHAPTER 4
∫
E X A M P L E 30
Find ex cos x d x.
SOLUTION
Using integration by parts
dv = cos x
u = ex, ___
dx
du = ex, v = sin x
___
dx
∫
∫
∴ ex cos x dx = ex sin x − ex sin x dx
∫
Now we find ex sin x dx using integration by parts.
dv = sin x
Let u = ex, ___
dx
du = ex, v = −cos x
___
dx
∫
∫
∴ ex sin x dx = −ex cos x + ex cos x dx
∫
∫
Substituting ex sin x dx = −ex cos x + ex cos x dx into
∫ex cos x dx = ex sin x − ∫ex sin x dx
gives
∫ex cos x dx = ex sin x − [ −ex cos x + ∫ex cos x dx ]
∫ex cos x dx = ex sin x + ex cos x − ∫ex cos x dx
∫ex cos x dx + ∫ex cos x dx = ex sin x + ex cos x
∫
2 ex cos x dx = ex sin x + ex cos x
∫
1 [ ex sin x + ex cos x ] + c
Hence ex cos x dx = __
2
dv be the
Notice we need to integrate by parts twice at each stage. We let u = ex and ___
dx
trigonometric function.
Try these 4.6
Find
(a) ∫x cos x d x
π
__
(b)
∫02 x2 sin x dx
(c) ∫sin−1 (x) d x
EXERCISE 4C
In questions 1–8, use integration by parts to find the integrals.
1
∫1 x ln x dx
2
∫x2 cos x dx
3
∫ x 2 ln x dx
4
∫02 x sin 2x dx
5
∫x e2 x dx
6
∫x2 ln x dx
2
1
__
π
__
115
M O DUL E 1
7
∫x3 ln x dx
9
Use integration by parts to find
∫x3 arctan x dx
8
_____
∫0 x √1 + x dx.
1
10 Show that ∫x2e−3x d x = −__31 e−3x [ x2 + __23 x + __92 ] + c.
11 Show that ∫ x2 ln x d x = __98 ln 2 − __97.
2
0
_______
12 Show that ∫arccos (2x) d x = x arccos (2x) − __21 √1 − 4x2 + c.
13 Use integration by parts to find ∫x2 ln (5x) d x.
14 Show that ∫ (ln x)2 d x = e + 2.
e
π
__
2 ex
0
15 Find ∫
1
sin x d x.
π − __
1 ln 2.
16 Show that ∫ arctan x d x = __
4 2
1
0
17 Use integration by parts to show that
______
18
4 (x3 + 1) 2 + c.
∫x5 √1 + x3 dx = __92 x3 (x3 + 1) 2 − ___
45
π
__
3
Show that ∫ 2 3x cos 2x d x = −__
2.
0
3
__
5
__
__
16 ln 4 − ___
28 .
19 Show that ∫ √θ ln θ d θ = ___
3
9
4
1
20 Use integration by parts to show that ∫ ln x2 d x = ln 16 − 2.
2
1
Integration using partial fractions
P (x)
To integrate rational functions _____ you need to separate into partial fractions
Q (x)
and then integrate. At this stage it is advisable to review your knowledge of partial
fractions. (Improper fractions must be divided out first and written as a mixed
fraction and then separated into partial fractions.)
E X A M P L E 31
SOLUTION
∫
x
Determine ____________
d x.
(x + 1)(x + 2)
x
Separating ____________
into partial fractions
(x + 1)(x + 2)
x
A + _____
B
____________
≡ _____
(x + 1)(x + 2) x + 1 x + 2
Multiplying by (x + 1)(x + 2) gives
x ≡ A(x + 2) + B (x + 1)
When x = −1, −1 = A (−1 + 2)
−1 = A
When x = −2, −2 = B (−2 + 1)
B=2
116
MODULE 1tCHAPTER 4
x
−1 + _____
2
∴ ____________
≡ _____
x+1 x+2
(x + 1)(x + 2)
We can also write
−1 and _____
2 are two standard integrals.
_____
x+1
x+2
x
dx
∫____________
(x + 1) (x + 2)
x
−1 + _____
2 dx
d x = ∫ _____
∫ ____________
x+1 x+2
(x + 1)(x + 2)
= −ln (x + 1) + 2 ln(x + 2) + ln c
= ln (x + 1)−1 + ln(x + 2)2 + ln c
= −ln |x + 1| + 2 ln |x + 2| + c
= ln c(x + 2)2 (x + 1)−1
c(x + 2)2
= ln ________
x+1
E X A M P L E 32
x+1
Determine _____________
d x.
(x + 2)(x2 − 4)
SOLUTION
x+1
We separate _____________
into partial fractions as follows.
(x + 2)(x2 − 4)
∫
This is a proper fraction, so we need to factorise the denominator
x2 − 4 = (x − 2)(x + 2)
x+1
x+1
∴ _____________
≡ _____________
(x + 2)(x2 − 4) (x + 2)2 (x − 2)
We now have a repeated linear factor in the denominator and a distinct linear factor.
C
x+1
A + _______
B
______________
≡ _____
+ ______
(x + 2)2 (x − 2) x + 2 (x + 2)2 x − 2
Multiply by (x + 2)2 (x − 2) ⇒ x + 1 ≡ A (x + 2)(x − 2) + B (x − 2) + C (x + 2)2
When x = 2, 2 + 1 = C (2 + 2)2
3
so C = ___
16
When x = −2, −2 + 1 = B (−2 − 2)
3 = 16C
−1 = −4B
1
so B = __
4
Comparing coefficients of x2:
0=A+C
There are no
terms in x 2 on
the LHS. On the
RHS
A × x × x = Ax 2
and C(x + 2)2
gives Cx 2
∴0=A+C
3
0 = A + ___
16
3
A = −___
16
3
3
1
___
__
−___
16
16
4
x
+
1
_______
_____
_____
_____________
+
≡
+
∴
(x + 2)2 (x − 2) x + 2 (x + 2)2 x − 2
3
3
1
___
__
−___
16
16
x+1
4
_______
_____
_____
_____________
+
dx =
+
dx
(x + 2)2 (x − 2)
x + 2 (x + 2)2 x − 2
∫
∫
∫
3
3
1
__
___
−2
= −___
16 ln |x + 2| + 4 (x + 2) d x + 16 ln |x − 2| + c
3
3
1
__
___
−1
= −___
16 ln |x + 2| − 4 (x + 2) + 16 ln |x − 2| + c
1
3
3
________
___
= −___
16 ln |x + 2| − 4 (x + 2) + 16 ln |x − 2| + c
117
M O DUL E 1
E X A M P L E 33
x
Separate _____________
into partial fractions and hence show that
(x + 1)(x2 + 1)
1 _____________
x
1
π
__
d x = −__
2
4 ln 2 + 8 .
0 (x + 1)(x + 1)
∫
SOLUTION
Bx + C
A + _______
x
_____________
≡ _____
2
x
+
1
x2 + 1
(x + 1)(x + 1)
⇒ x ≡ A (x2 + 1) + (Bx + C)(x + 1)
When x = − 1, − 1= A (2)
1
A = −__
2
When x = 0, 0 = A + C
1+C
0 = −__
2
1
C = __
2
Equating the coefficients of x2
0=A+B
1+B
0 = −__
2
1
B = __
2
Note
_1 x + _1
−_12
x
2
_____________
_____
_______
∴
≡
+ 22
2
x
+
1
(x + 1)(x + 1)
x +1
∫0
1
1 dx =
∫ ______
1 + x2
tan−1 (x) + c
(Standard form)
x
______
dx
1 + x2
1 ______
2x dx
= __
2 1 + x2
1 ln (1 + x2) + c
= __
2
f' (x)
____
dx
f (x)
∫
∫
∫
= ln f (x) + c
E X A M P L E 34
SOLUTION
x
_____________
dx =
(x + 1)(x2 + 1)
∫0
1
_1 x + _1
−_12
2
2
_____
dx
+ _______
x+1
x2 + 1
_1 x + _1
_1 x
_1
2
2
2
______
______
______
into 2
+ 22
Separating 2
x +1
x +1 x +1
1
= −__
2
x d x + __
1 ______
1 ______
1 dx
1 d x + __
∫0 _____
2 ∫0 x2 + 1
2 ∫0 x2 + 1
x+1
1
1
[
1
1
1
__
__
2
−1
= −__
2 ln (x + 1) + 4 ln (x + 1) + 2 tan (x)
[
] [
]
1
0
1
1
1
1
1
__
__
__
__
−1
−1
= −__
2 ln 2 + 4 ln 2 + 2 tan (1) − − 2 ln 1 + 4 ln 1 + tan (0)
π
1 ln 2 + __
= −__
8
4
Evaluate
]
x
dx
∫3 __________
x2 − 3x + 2
4
2
x2
__________
is an improper fraction
2
x − 3x + 2
By long division
x2
3x − 2
__________
≡ 1 + __________
2
2
x − 3x + 2
x − 3x + 2
3x − 2 ≡ ____________
3x − 2
__________
x2 − 3x + 2 (x − 2)(x − 1)
118
1
1
x2 − 3x+2 ) x2
−x2 − 3x + 2
3x − 2
2
x
__________
∴ 2
x − 3x + 2
3x − 2
= 1 + __________
x2 − 3x + 2
MODULE 1tCHAPTER 4
Separating into partial fractions
3x − 2
B
A + _____
____________
≡ _____
(x − 2)(x − 1) x − 2 x − 1
3x − 2 ≡ A(x − 1) + B (x − 2)
When x = 1, then 1 = −B ⇒ B = −1
When x = 2, then 6 − 2 = A ⇒ A = 4
3x − 2 ≡ _____
4 − _____
1
__________
x2 − 3x + 2 x − 2 x − 1
x2
4 − _____
1
≡ 1 + _____
Hence __________
2
x−2 x−1
x − 3x + 2
∴
x
4 − _____
1 dx
d x = ∫ ( 1 + _____
∫3 __________
x −2 x − 1)
x2 − 3x + 2
3
4
2
4
= [x + 4 ln (x − 2) − ln (x − 1)]43
= (4 + 4 ln 2 − ln 3) − (3 + 4 ln 1 − ln 2)
= 4 − 3 + 4 ln 2 + ln 2 − ln 3
= 1 + 5 ln 2 − ln 3
( )
32
= 1 + ln ___
3
E X A M P L E 35
Bx + C .
3x + 1 ≡ __
A + _______
Find the values of A, B and C for which ________
x (x2 + 1) x
x2 + 1
2
3x2 + 1 dx = ln (cx (x2 + 1)).
Hence show that ________
x (x2 + 1)
∫
SOLUTION
Separating into partial fractions
Bx + C
3x2 + 1 ≡ __
A + _______
_________
x (x2 + 1) x
x2 + 1
⇒ 3x2 + 1 ≡ A (x2 + 1) + (Bx + C)(x)
Substituting x = 0 ⇒ 1 = A
Equating coefficients of x2 ⇒ 3 = A + B
3=1+B
B=2
Equating coefficients of x ⇒ 0 = C
3x2 + 1 ≡ __
2x
1 + ______
∴ _________
x (x2 + 1) x x2 + 1
3x + 1 d x = __
2x d x
∫ _________
∫ ( 1x + ______
x (x2 + 1)
x2 + 1 )
2
= ln x + ln (x2 + 1) + ln c
= ln (cx (x2 + 1))
119
M O DUL E 1
Try these 4.7
Find
(a)
Remember
x
dx
∫ ___________
x2 + 7x + 12
2x dx
∫______
1− x
2
3x2 + x + 1 d x
______________
(x + 1)(2x2 + 1)
(b)
∫
(c)
x
dx
∫ __________
x2 + 3x + 2
can be done by
recognition or
partial fractions.
Recognition is
much faster.
3
EXERCISE 4D
Find the integrals in questions 1–23.
x + 2 dx
∫ _____
x
1
5x + 7 d x
∫ ______
2x − 1
x3 + x + 2 d x
∫ _________
x+1
3
5
2
4
6
2x + 3 d x
∫ ______
x−2
x dx
∫ _____
x−2
1
dx
∫ ____________
(x + 2)(x − 3)
2
7
4
dx
∫ ____________
(x − 3)(x − 7)
8
3x
dx
∫ _____________
(2x + 3)(x + 1)
9
x
dx
∫ __________
x2 + 5x + 6
10
x+2
dx
∫ ___________
3x2 − 8x + 4
11
5x − 2 d x
∫ __________
6x2 + x − 2
12
4x
dx
∫ _____________
(x2 − 4)(x − 3)
13
4x + 2
dx
∫ __________________
(x − 1)(x + 2)(x + 3)
14
8x + 2x − 24 d x
∫ ______________
(x2 + 4x)(x − 2)
15
3x + 8x − 8
dx
∫ ___________________
(x + 2)(2x2 − 3x − 2)
16
43 − 22x − 3x
dx
∫ ___________________
(2x2 − 7x + 3)(x + 2)
17
1
dx
∫ ________
x (x2 + 1)
18
1 dx
∫ ______
x3 − 8
19
5x − 4x + 4
dx
∫ __________________
(x + 2)(x2 − 2x + 2)
20
4x + 4x + 1 d x
∫ ___________
x (x + 1)
21
x + 3x − 2 d x
∫ __________
x2 − 1
22
x + x + 2x + 1 d x
∫ ______________
x (x2 + 1)
23
x + 3x + 3 d x
∫ __________
x2 + 2x + 1
2
2
4
2
2
2
3
2
2
Ask yourself
Is the differential of the denominator contained in the numerator?
2x
______
dx = −ln | 1 − x2 | + c
1 − x2
d [ 1 − x2 ] = −2x
i.e. ___
dx
∫
Instead of partial fractions you can use recognition to integrate this function.
120
MODULE 1tCHAPTER 4
Bx + C , find A, B and C.
3x + 3x + 2 ≡ ______
A + _______
24 Given that ______________
2
2
2x + 1
2
(2x + 1)(x + 1)
x +1
3x2 + 3x + 2 d x = __
π
1 ln 6 + __
∫0 ______________
4
2
(2x + 1)(x2 + 1)
1
Hence show that
4x + 5x + 6 into partial fractions. Hence find ______________
25 Separate _____________
∫ 4x + 5x2+ 6 dx.
2
2
2
(x + 2)(x + 9)
(x + 2)(x + 9)
1 ln (x2 + 2) + c.
x + x + x + 2 d x = arctan x + __
26 Show that ∫______________
4
2
2
3
2
x + 3x + 2
3x2 − 5
27 Evaluate ∫0 _________________
d x.
2
2
(x + x + 1)(x − 2)
Integration of trigonometric functions
Integration of trigonometric functions makes use of the trigonometric identities to
convert more complicated functions to standard integrals. Trigonometric substitutions
∫
______
allow us to find integrals such as √ 9 − x2 dx. Before you make a trigonometric substitution you will need to be familiar with all the identities and integration of a wider
variety of products and powers of trigonometric functions. At this stage you should
review the trigonometric identities. Here is a list of integrals of trigonometric functions.
Function
Integral/procedure
sin x
− cos x + c
cos x
sin x + c
tan x
ln |sec x| or
− ln |cos x| + c
sec x
ln |sec x + tan x| + c
cosec x
x +c
ln tan __
2
cot x
ln |sin x| + c
sin x cosn x, n ≠ −1
1
n+1
−_____
n + 1 cos x + c
cos x sinn x, n ≠ −1
1 sinn+1x + c
_____
n+1
1 tann+1x + c
_____
n+1
sec2 x tann x, n ≠ −1
| ( )|
2sin Px cos Qx
convert to sin(Px + Qx) + sin(Px − Qx) and integrate
2cos Px sin Qx
convert to sin(Px + Qx) − sin(Px − Qx) and integrate
2cos Px cos Qx
convert to cos(Px + Qx) + cos(Px − Qx) and integrate
−2sin Px sin Qx
convert to cos(Px + Qx) − cos(Px − Qx) and integrate
121
M O DUL E 1
E X A M P L E 36
π d x.
Determine sin ( 4x − __
2)
SOLUTION
∫sin ( 4x − __π2 ) dx = −__41 cos ( 4x − __π2 ) + c
E X A M P L E 37
π d x.
Determine sec2 ( x − __
4)
SOLUTION
∫sec2 ( x − __π4 ) dx = tan ( x − __π4 ) + c
E X A M P L E 38
π d x.
Find cos 3x + __
6
SOLUTION
∫cos ( 3x + __π6 ) dx = __31 sin ( 3x + __π6 ) + c
E X A M P L E 39
Find sin x cos2 x d x.
SOLUTION
sin x cos2 x is of the form f ′(x)[f (x)]n
∫
∫
∫ (
)
∫
Let f (x) = cos x
f ′(x) = −sin x
∫ f ′(x) [ f (x)]n dx = −∫−sin x cos2 x dx
−cos3 x + c
= _______
3
∫
E X A M P L E 40
Find sin x cos6 x d x.
SOLUTION
Let f (x) = cos x
f ′(x) = −sin x, n = 6
∫ f ′(x) [ f (x)]n dx = −∫−sin x cos6 x dx
cos7 x
= −_____
7 +c
∫
E X A M P L E 41
Find sin x cosn x d x, n ≠ −1.
SOLUTION
Let f (x) = cos x
f ′(x) = −sin x
∫ f ′(x) [ f (x)]n dx = −∫−sin x cosn x dx
x
cos
∫sin x cosn x dx = −_______
n + 1 + c,
n+1
122
n ≠ −1
MODULE 1tCHAPTER 4
∫
E X A M P L E 42
Find cos x sinn x d x, n ≠ −1.
SOLUTION
Let f (x) = sin x
f ′(x) = cos x
x + c,
sin
∫ f ′(x) [ f (x)]n dx = ∫cos x sinn x dx = _______
n+1
n+1
E X A M P L E 43
SOLUTION
n ≠ −1
∫
Determine tann x sec2 x d x, n ≠ −1.
d tan x = sec2 x
Since ___
dx
tann+1 x + c,
tann x sec2 x d x = _______
n+1
∫
n ≠ −1
Integrating sin2 x and cos2 x
∫
E X A M P L E 44
Find sin2 x d x.
SOLUTION
1 − cos 2x
Since sin2 x = _________
2
cos 2x = 1 − 2 sin2x
∫ sin2 x dx = ∫ ( __21 − __12 cos 2x ) dx
1 x − __
1 sin 2x + c
= __
4
2
∫
E X A M P L E 45
Find cos2 x d x.
SOLUTION
1 + __
1 cos 2x
Since cos2 x = __
2 2
∫ cos2 x dx = ∫( __21 + __21 cos 2x ) dx
1 x + __
1 sin 2x + c
= __
4
2
Integrating sin3x and cos3x
∫
E X A M P L E 46
Find sin3 x d x.
SOLUTION
Writing sin3 x = sin x sin2 x
∫
∫
we have sin3 x d x = sin x sin2 x d x
Substituting
sin2 x
= 1 − cos2 x
123
M O DUL E 1
∫ sin3 x dx = ∫ sin x (1 − cos2 x) dx
= ∫ (sin x − sin x cos2 x) d x
3
cos x + c
= −cos x + _____
3
∫ sin x cos2 x dx
f'
n
↓
↓
cos3 x
= − − sin x (cos x)2 dx = −_____
3
↑
∫
f
∫
E X A M P L E 47
Find cos3 x d x.
SOLUTION
The procedure is the same as that for integrating sin3 x.
Write cos3 x = cos x cos2 x
Substitute cos2 x = 1 − sin2 x
∴ cos3 x = cos x (1 − sin2 x)
∫ cos3 x dx = ∫ cos x (1 − sin2 x) dx
sin3 x + c
= ∫ (cos x − cos x sin2 x) d x = sin x − _____
3
We can integrate sin5 x, cos5 x, sin7 x, cos7 x, etc. using the same procedure as for
sin3 x and cos3 x.
E X A M P L E 48
SOLUTION
∫
Find sin4 x d x.
1 − __
1 cos 2x and write
We use sin2 x = __
2 2
1 − __
1 cos 2x 2
sin4 x = (sin2 x)2 = __
2 2
(
∫
∴ sin4 x d x =
)
∫ ( __41 − __21 cos 2x + __41 cos 2 2x ) dx
1 + __
1 cos 2(2x)
Using the double angle formula cos2 2x = __
2 2
1 cos 4x
1 + __
= __
2 2
∫
∫[ __41 − __12 cos 2x + __14 ( __21 + __21 cos 4x ) ] dx
1 − __
1 cos 2x + __
1 + __
1 cos 4x d x
= ∫( __
)
4 2
8 8
3 − __
1 cos 2x + __
1 cos 4x d x
= ∫ ( __
)
8 2
8
∴ sin4 x d x =
3 x − __
1 sin 2x + ___
1 sin 4x + c
= __
4
32
8
We can integrate cos4 x, sin6 x, cos6 x etc. in a similar manner.
Try these 4.8
Find
(a)
124
∫ sin5 x dx
(b)
∫ cos5 x dx
(c)
∫ cos4 x dx
MODULE 1tCHAPTER 4
Integrating powers of tan x
∫
E X A M P L E 49
Find tan2 x d x.
SOLUTION
Write tan2 x = sec2 x − 1
∫ tan2 x dx = ∫ (sec2 x − 1) dx
= tan x − x + c
∫
E X A M P L E 50
Determine tan3 x d x.
SOLUTION
We write tan3 x = tan x tan2 x
= tan x (sec2 x − 1)
Recall
= tan x sec2 x − tan x
∫
n
∫
∴ tan3 x d x = tan x sec2 x − tan x d x
2
tan2 x + c
∴ tan x sec2 x dx = _____
2
∫
tan2 x − ln (sec x) + c
= _____
2
Try these 4.9
tann+1 x
+ c, n ≠ −1
∫ tan x sec x dx = _______
n+1
Find
(a)
∫ tan4 x dx
(b) ∫ tan5 x d x
Integrating products of sines and cosines
Recall
2 sin P cos Q = sin (P + Q) + sin (P − Q)
2 cos P sin Q = sin (P + Q) − sin (P − Q)
2 cos P cos Q = cos (P + Q) + cos (P − Q)
−2 sin P sin Q = cos (P + Q) − cos (P − Q)
Let us use these to integrate the following.
∫
E X A M P L E 51
Find cos 4x sin 2x d x.
SOLUTION
Using 2 cos P sin Q = sin (P + Q) − sin (P − Q)
2 cos 4x sin 2x = sin (4x + 2x) − sin (4x − 2x)
∫
2 cos 4x sin 2x = sin 6x − sin 2x
1 sin 6x − __
1 sin 2x
∴ cos 4x sin 2x = __
2
2
1 sin 6x − __
1 sin 2x d x
cos 4x sin 2x d x = __
2
2
1 cos 6x + __
1 cos 2x + c
= − ___
4
12
∫(
)
125
M O DUL E 1
∫
E X A M P L E 52
Find cos 5x cos 3x d x
SOLUTION
Using 2 cos P cos Q = cos (P + Q) + cos (P − Q)
2 cos 5x cos 3x = cos (5x + 3x) + cos (5x − 3x)
∫
2 cos 5x cos 3x = cos 8x + cos 2x
1 cos 8x + __
1 cos 2x
cos 5x cos 3x = __
2
2
1
1
__
cos 5x cos 3x d x = __
2 cos 8x + 2 cos 2x d x
1 sin 8x + __
1 sin 2x + c
= ___
4
16
∫(
π
__
2
0
∫
)
E X A M P L E 53
Find
SOLUTION
Using 2 sin P sin Q = cos (P − Q) − cos (P + Q)
4 sin 6x sin 2x d x.
2 sin 6x sin 2x = cos (6x − 2x) − cos (6x + 2x)
2 sin 6x sin 2x = cos 4x − cos 8x
4 sin 6x sin 2x = 2 cos 4x − 2 cos 8x
π
__
2
0
∫
4 sin 6x sin 2x d x =
π
__
∫02 (2 cos 4x − 2 cos 8x) dx
[
1 sin 4x − __
1 sin 8x
= __
4
2
π
__
2
0
]
[
1 sin 4 __
= __
( π2 ) − __41 sin 8 ( __π2 ) − __21 sin 4 (0) − __41 sin 8 (0)
2
]
=0
Try these 4.10
Find
(a)
∫ cos 6x sin 3x dx
(b)
∫ cos 8x cos 2x dx
(c)
∫ sin 10x sin x dx
Finding integrals using the standard forms
x
x
1
1
1
__
_______
−1 __
_______
dx = sin−1 ( __
a ) + c and ∫ a2 + x2 dx = a tan ( a ) + c
∫ ________
2
2
√a − x
∫
E X A M P L E 54
x
1
_______
d x = sin−1 ( __
Using the substitution x = a sinθ, show that ________
a ) + c.
√a2 − x2
SOLUTION
Let x = a sinθ
d x = a cosθ dθ
_______
___________
√a2 − x2 = √a2 − a2 sin2 θ
____________
= √ a2 (1 − sin2 θ )
126
MODULE 1tCHAPTER 4
_______
= √ a2 cos2 θ
= a cos θ
d x = ∫ ______ a cos θ dθ
∫ ________
2 − x2
a cos θ
a
√
1
_______
1
∫
= 1 dθ = θ + c
Converting to a function of x, since x = a sin θ, we get
x
sin θ = __
a
x
θ = sin−1 ( __
a)
x +c
1
_______
d x = sin−1 ( __
∴ ________
)
a
2
2
√a − x
∫
∫
E X A M P L E 55
x + c, using the substitution x = a tan θ.
1 d x = __
1 tan−1 __
Show that _______
(
)
2
2
a
a
a +x
SOLUTION
Let x = a tan θ
d x = a sec2 θ dθ
a2 + x2 = a2 + (a tan θ )2
= a2 + a2 tan2 θ
= a2 (1 + tan2 θ )
= a2 sec2θ
(substituting 1 + tan2 θ = sec2θ)
θ dθ
1 d x = _______
∫ _______
∫ aa2sec
a2 + x2
sec2 θ
2
1
1
__
= ∫ __
a dθ = a θ + c
Since x = a tan θ
x
tan θ = __
a
x
θ = tan−1 ( __
a)
∫
x
1 d x = __
1
−1 __
∴ _______
a tan ( a ) + c
a2 + x2
∫
E X A M P L E 56
1
________
Find _________
d x.
√4 − 25x2
SOLUTION
1
1
________
_______
in the form ________
we get
Writing _________
2
2
√4 − 25x
√a − x2
1
1
_________
________
___________
= ____________
4
___
√4 − 25x2
25
− x2
25
√ (
)
127
M O DUL E 1
1
________
= _____________
___
2
√ 25 __ 2 − x2
5
1
__
5
_________
________
=
2 2 − x2
__
5
√( )
√( )
1 _________
1
1
d x = __
∫__________
∫
5
2
2
2
__
4
−
25
x
√
√( 5 ) − x2
________
________
()
x +c
1 sin−1 __
= __
_2
5
5
( )
5x + c
1 sin−1 ___
= __
5
2
∫
E X A M P L E 57
1
d x.
Determine __________
x2 +2x + 2
SOLUTION
Complete the square of x2 + 2x + 2:
x2 + 2x + 2 = (x + 1)2 + 1
(
)
x+1 +c
1
1
d x = ___________
dx = tan−1 _____
∴ __________
1
1 + (x + 1)2
x2 + 2x + 2
∫
∫
= tan−1 (x + 1) + c
∫
E X A M P L E 58
1 d x.
Find ______
4 + x2
SOLUTION
1 d x = ______
∫ ______
∫ 22 +1 x2 dx
4 + x2
∫
x
1 d x = __
1
−1 __
Since _______
a tan ( a ) + c, substituting a = 2 gives
a2 + x2
1 d x = __
1 tan−1 __
______
( 2x ) + c
2
22 + x2
∫
∫
E X A M P L E 59
1
Find _______
d x.
4 + 9x2
SOLUTION
1
1 , we get
Writing the function _______
in the form _______
4 + 9x2
a2 + x2
1
1
1 ________
d x = __
dx
∫ _________
∫
2
9
4
2
__
__
2
9 +x
+ x2
(9
∫
)
(3)
x
1 d x = __
1
2
__
−1 __
Since ______
a tan ( a ) + c, substituting a = 3 gives
a2 +x2
∫
[
( )]
x +c
1 ________
1
1 __
1 tan−1 __
__
d x = __
_2
9 ( _2 )2 + x2
9 _2
3
3
3
3 tan−1
1 × __
= __
9 2
3x
1 tan−1 ___
= __
2
6
3x + c
( ___
2)
( )+c
128
MODULE 1tCHAPTER 4
∫
E X A M P L E 60
1
_______
d x.
Find ________
√4 − 9x2
SOLUTION
4 − x2
√4 − 9x2 = 9 __
9
_________
_______
√(
__
= √9
)
_______
√( __9 − x )
4
2
________
=3
∫
√( __3 ) − x
2
2
2
∫
1
1
_______
________
d x = ___________
∴ ________
dx
2
2
2
__
√4 − 9x
3
− x2
3
√( )
()
x +c
1 sin−1 __
= __
3
2
__
3
( )
3x + c
1 sin−1 ___
= __
3
2
d x.
∫−1 _____________
√−x2 + 2x + 8
1
____________
1
E X A M P L E 61
Evaluate
SOLUTION
Complete the square:
−x2 + 2x + 8 = −(x2 − 2x) + 8
= −((x − 1)2 − 1) + 8
= 9 − (x − 1)2
∴
1
____________
1
____________
___________
dx =
∫−1 _____________
∫−1 √9 − (x − 1)2 dx
√−x2 + 2x + 8
1
1
[
(
x−1
= sin−1 _____
3
)]
1
−1
( )
2
= sin−1 (0) − sin−1 −__
3
= 0.729 73 = 0.730 (3 d.p.)
∫
E X A M P L E 62
1
d x.
Determine __________
x2 + 4x + 6
SOLUTION
Complete the square:
x2 + 4x + 6 = (x2 + 4x + (2)2) + 6 − 22
= (x + 2)2 + 6 − 4
= 2 + (x + 2)2
∫
∫
1
1
d x = ___________
dx
∴ __________
x2 + 4x +6
2 + (x + 2)2
129
M O DUL E 1
∫
1
__
= ______________
dx
( √ 2 )2 + (x + 2)2
(
)
x +__ 2 + c
1__ tan−1 _____
= ___
√2
√2
∫
E X A M P L E 63
1
Find ________
dx using the substitution x = 2 tan θ.
3
__
( x2 + 4 )2
SOLUTION
x = 2 tan θ
dx = 2 sec2 θ dθ
3
__
3
__
( x2 + 4 )2 = ( 4 tan2θ + 4 )2
3
__
= [ 4( 1 + tan2θ ) ]2
3
__
= ( 4 sec2 θ )2
= 8 sec3θ
Now
2 sec θ dθ
1
dx = ∫ _______
∫________
3
8
sec3 θ
( 2
)2
2
x +4
__
∫
1 ____
1 dθ
= __
4 sec θ
∫
1 cos θ dθ
= __
4
1 sin θ + c
= __
4
x
Since tan θ = __
2
x
______
sin θ = _______
√4 + x2
∴
E X A M P L E 64
SOLUTION
x
1 _______
1
+c
= __
∫________
3
4
√4 + x2
( 2
)2
x +4
______
__
________
√9 − 4x2 dx using the substitution x = __3 sin θ.
Find _________
2
x2
∫
3 sin θ
x = __
2
3 cos θ dθ
dx = __
2
_____________
_______
√9 − 4x2 = 9 − 4 __23 sin θ 2
_____________
9 sin2 θ
= 9 − 4 __
4
√
√
(
( )
__________
= √9 − 9 sin2 θ
___________
= √9(1 − sin2 θ)
_______
= √9 cos2 θ
= 3 cos θ
130
)
√4 + x2
θ
2
x
MODULE 1tCHAPTER 4
_______
3 cos θ dθ
√9 − 4x2 d x = ________
3 cos θ × __
∴ ________
3 sin θ 2 2
__
x2
2
2
cos2 θ
_____
= __
3 × 3 sin2 θ dθ
∫
∫
(
)
∫
∫
∫
= 2 cot2 θ dθ = 2 (cosec2 θ − 1)dθ
= −2 cot θ − 2θ + c
3 sin θ
Since x = __
2
2x
sin θ = __
3
θ = sin−1 __2 x
3
_______
_______
2
9 − 4x2
√
9
−
4x
2x + c
√
________
∴ ________
dx
=
−
− 2 sin−1 __
x
2
3 )
x
3
θ
( )
Try these 4.11
√9 − 4x2
(
∫
2x
cot θ =
√9 − 4x2
2x
Find
(a)
1 dx
∫ ______
9 + x2
(b)
1
dx
∫ ________
4 + 25x2
(c)
4
dx
∫ ____________
9x2 + 6x + 16
(d)
1
dx
∫ __________
2
√−x − 2x
________
Look at these fractions:
1 dx,
x dx, _____
∫_____
∫_____
∫x 1+ 1 dx
x –1
x –1
2
2
2
Can you identify which method of integration is best for each one?
EXERCISE 4E
∫
1
Find tan3x sec2x dx.
2
1 sin 10x + c.
1 sin 4x − ___
Show that sin 7x sin 3x dx = __
8
20
3
1
dx
Find (a) ________
25x2 + 4
4
1 sin 4x − ___
1 sin3 2x + c.
1 x − ___
Show that cos2 x sin4 x dx = ___
84
16
64
∫
∫
∫
∫
1
(b) ________
dx
16x2 + 9
1
(c) _______
dx
2x2 + 6
∫
π
__
8.
∫04 tan2 x sec4 x dx = ___
15
5
Show that
6
1
dx.
By completing the square of x2 + 6x + 13, find ___________
x2 + 6x + 13
7
Using the substitution x = 2 sin θ, find √4 − x2 dx.
8
Find (a) cos 8x cos 6x dx
∫
∫
∫
∫
______
(b) sin 7x cos 3x dx
∫
(c) cos 6x sin 2x dx
131
M O DUL E 1
9 Evaluate ∫
1 ____________
1
2 + 4x + 10
4x
0
dx.
1
1
1
___________
___________
__________
dx (b) ∫____________
10 Find (a) ∫____________
dx
dx (c) ∫____________
2
2
2
√3 + 2x − x
√5 − 4x − x
√7 − 6x − x
cos θ dθ.
1
______
11 By using the substitution x = 2 tan θ, show that ∫_________
dx = ∫______
2
2
2
∫
1
______
dx.
Hence find _________
x2 √ 4 + x2
x √4 + x
4 sin θ
________
12 Find ∫√1 − cos x dx.
1
13 Evaluate ∫
2
x dx.
______
+ x2
sin x
________
dx.
14 Find _________
√ 1 + cos x
01
∫
1
1 cosec2 θ dθ.
______
15 By using the substitution x = 3 sin θ, show that ∫_________
dx = ∫__
2
2
9
x √9 − x
∫
1
______
Hence find _________
dx.
2
x √ 9 − x2
132
MODULE 1tCHAPTER 4
SUMMARY
Integration
Recognition
Substitution
Partial fractions
By parts
Use the rules of partial
fractions to split up the
function and then integrate.
∫u dx dx = uv –∫v dx dx
∫x dx = xn+1 + c, n ≠ –1
n+1
n
∫(a + b) dx = a1 (a n+ +b) 1
n+1
n
+ c, n ≠ –1
∫ x dx = ln |x| + c
1
∫ ax + b dx = a ln |ax + b| + c
1
1
∫e dx = e + c
∫e dx = a1 e
x
x
ax + b
∫fg(x)dx
Using u = g(x) convert
all functions of x to functions
of u. If there are any limits,
change the limits. Using the
substitution, integrate your
function of u.
ax + b + c
dv
du
Be careful in your choice
of function for u and dv .
dx
ln x, arcsin x, arccos x
arctan x are integrated
by parts.
∫sin x dx = –cos x + c
∫sin(ax + b) dx = – a1 cos(ax + b) + c
∫cos x = sin x + c
∫cos(ax + b) dx = a1 sin(ax + b) + c
∫tan x dx = ln|sec x | + c
∫cot x dx = ln|sin x | + c
∫sec x dx = ln|sec x + tan x | + c
∫cosec x dx = ln|tan (2)| + c
x
∫sec x dx = tan x + c
2
∫cosec x dx = –cot x + c
2
∫ √a 1– x
2
∫a
2
1
2 + x2
x
dx = sin–1 a + c
()
1 tan–1 x + c
dx = a
a
()
∫ f (x) dx = ln|f(x)| + c
f’(x)
n+1
∫f’(x)[f (x)] dx = [f (x)]
n +1
n
+ c, n ≠ –1
∫f’(x)e
f (x)dx = ef (x) + c
133
M O DUL E 1
Checklist
Can you do these?
■ Integrate standard functions using the table of integrals.
■ Integrate exponential functions.
■ Integrate logarithmic functions.
f ′(x)
, f ′(x)[ f (x)]n, f ′(x) e f (x).
■ Find integrals of the form ____
f (x)
■ Integrate by substitution.
■ Integrate by parts.
■ Integrate using partial fractions.
■ Integrate trigonometric functions.
■ Integrate inverse trigonometric functions.
1
1
_______
and ________
.
■ Integrate functions of the form _______
2
2
a2 + x2
√a − x
Review e x e r c i s e 4
1
_____
∫
(a) Find x √ 1 + x d x.
(b) Evaluate
(i)
∫
2 cos2 4x d x
Find x cosec2 (x) d x.
3
Evaluate
4
2
(ii)
(ln x)
∫1 ______
x dx
(b)
1
dx
∫0 ____________
(1 + x)(2 − x)
2
∫
2
(a)
∫1
2
2x + 1 d x
______
x−2
1
Find
(a)
1
∫ ________
3 d x using the substitution x = 3sec θ
__
2
(x − 9) 2
(b)
∫ x (1 + 3x2) 2 dx
1
__
______
5
Use the substitution x =
6
Show that
7
Find the following integrals
(a)
134
π
__
4
0
2 sin2
θ to evaluate
∫4 x2 − 5x + 6 dx = ln ( 9 ).
5
2
__________
∫ cos2 4x dx
(b)
4 − x d x.
∫0 √______
2x
1
2
16
___
∫ ln (x + 4) dx
(c)
∫ x e3x dx
MODULE 1tCHAPTER 4
∫1 x3 ln x dx.
e
8
Evaluate
9
Find the exact value of
∫0 4x dx.
4
C .
Ax + B + _____
5
= _______
10 Find the values of A, B and C for which _____________
2
2
x+1
(x + 1)(x + 4)
Hence find the exact value of
x +4
5
d x.
∫0 _____________
(x + 1)(x2 + 4)
2
C
1
A + _______
B
_____
11 Express ______________
in the form _____
2
2 + x − 2 where A, B and
x+1
(x + 1) (x − 2)
(x + 1)
∫
1
d x.
C are constants. Hence find ______________
2
(x + 1) (x − 2)
__
1
12 Show, by means of the substitution x = 3 tan θ, that ∫ ________
2
2 dx
√3
0
1
= __
3
π
__
6
0
∫
cos2 θ d θ. Hence find the exact value of
∫0
(x + 9)
__
√3
1
________
d x.
(x2 + 9)2
1
_____________
d x.
13 Find ∫_______________
2
√−x − 6x + 16
14 By using the substitution x = 4cos2 θ + 7sin2 θ, evaluate
7
1
______________
____________
∫4 √(x − 4)(7 − x) dx.
15 Show that ∫
1 π
___
12
0
__
(4 − π) √2
x sin 3x d x = __________.
72
C
15 − 13x + 4x = _____
A + _______
B
_____
16 Given that ______________
2
2 + 4 − x , find the constants
1−x
2
(1 − x) (4 − x)
(1 − x)
A, B and C.
3
15 − 13x + 4x2 d x, giving the exact value in terms of
Hence evaluate _____________
2
2 (1 − x) (4 − x)
logarithms.
∫
17 Find the exact value of ∫6 π sec2 2x d x.
1
__
0
Hence find the exact value of
∫06 π tan2 2x dx.
1
__
1 1 − x2
18 Use the substitution x = tan θ to evaluate ∫ ________
d x.
2 2
0
(1 + x )
+x−7
19 Separate __________________
into partial fractions.
2
x2
(x + 2x + 2)(x − 1)
x2 + x − 7
Hence find __________________
d x.
2
(x + 2x + 2)(x − 1)
∫
π
__
20 Evaluate ∫ 2 ex cos x d x.
0
135
M O DUL E 1
CHAPTER 5
Reduction Formulae
At the end of this chapter you should be able to:
■ derive reduction formulae for sinn x, cosn x and tann x
■ derive reduction formulae for other functions using integration by parts
■ use reduction formulae to find specific integrals.
KEYWORDS/TERMS
SFEVDUJPOGPSNVMBtJOUFHSBUJPOCZQBSUT
136
MODULE 1tCHAPTER 5
A reduction formula is used to facilitate integrals that cannot easily be found.
Integration by parts can be used to find one integral in terms of a simpler integral
of the same form. Using the method of integration by parts, it is sometimes possible
to express such an integral in terms of a similar integral where n is replaced by
(n − 1), or sometimes (n − 2). The relationship between the two integrals is called a
‘reduction formula’. A reduction formula is normally derived by changing the form
of the integral to a product which can be used to integrate by parts.
The reduction formula will take the repeats out of the integrals. If we have to find
∫x8ex dx we will have to integrate by parts eight times to get the result. Instead of
integrating so many times we can find a formula for ∫xnex d x and use it to find
the integral. A reduction formula is used in the same way that we use an iterative
formula (i.e., to determine a specific, current numerical value by using previous
values obtained from the formula).
∫
Reduction formula for sinnx dx
∫
EXAMPLE 1
Let In = sinn x d x. Find a reduction formula for In.
SOLUTION
Write sinn x = sinn−1 x sin x
∫
We have In = sinn−1 x sin x d x
dv = sin x
Using integration by parts, let u = sinn−1 x, ___
dx
To find the differential of sinn−1 x we use the chain rule (bring down the power,
reduce the power by 1 and multiply by the differential of sin x)
du = (n − 1) sinn−2 x cos x, v = −cos x.
___
dx
Substituting into
dv d x = uv − v ___
du d x
u ___
dx
dx
we get
∫
∫
∫
In = −cos x sinn−1 x − (−cos x)(n − 1) sinn−2 x cos x d x
∫
= −cos x sinn−1 x + (n − 1) sinn−2 x cos 2 x d x
Substituting cos 2 x = 1 − sin2 x
∫
= −cos x sinn−1 x + (n − 1) ∫(sinn−2 x − sinn x) d x (expanding brackets)
= −cos x sinn−1 x + (n − 1) ∫sinn−2 d x − (n − 1) ∫sinn x d x
Since In = ∫sinn x d x, replacing n by n − 2 gives
In−2 = ∫sinn−2 x d x
In = −cos x sinn−1 x + (n − 1) sinn−2 x (1 − sin2 x) d x
∴ In = −cos x sinn−1 x + (n − 1)In−2 − (n − 1)In
137
M O DUL E 1
Making In the subject of the formula
In + (n − 1)In = −cos x sinn−1 x + (n − 1)In−2
∴ n In = −cos x sinn−1 x + (n − 1)In−2,
A reduction formula for In =
∫
sinn x
(1 + (n − 1) = n)
d x is therefore
1 n−1
n−1 I ,
_____
In = −__
n sin x cos x +
n−2
n
(
n≥2
)
n≥2
∫
Reduction formula for cosnx dx
∫
EXAMPLE 2
Obtain a reduction formula for In = cosn x d x and hence determine I4.
SOLUTION
Let In = cosn x d x
∫
We write cosn x = cosn−1 x cos x
dv = cos x
Using integration by parts, let u = cosn−1 x, ___
dx
To differentiate u we use the chain rule.
du = (n − 1) cosn−2 x (−sin x), v = sin x
___
dx
In = sin x cosn−1 x − sin x [(n − 1) cosn−2 x (−sin x)] d x
∫
Use the integration by
parts formula:
dv dx = uv − ∫v ___
du dx
∫u ___
dx
dx
∫
= sin x cosn−1 x + (n − 1) cosn−2 x sin2 x d x
Substituting
sin2 x
=1−
cos 2 x,
we have
∫
= sin x cosn−1 x + (n − 1) ∫(cosn−2 x − cosn x) d x
= sin x cosn−1 x + (n − 1) ∫cosn−2 x d x − (n − 1) ∫cosn x d x
Since In = ∫cos n x d x, replacing n by n − 2 we get
In−2 = ∫cos n−2 x d x
In = sin x cosn−1 x + (n − 1) cosn−2 x (1 − cos2 x) d x
∴ In = sin x cosn−1 x + (n − 1)In−2 − (n − 1)In
⇒ In + (n − 1) In = sin x cosn−1 x + (n − 1)In−2
Since In + (n − 1) In = nIn
nIn = sin x cosn−1 x + (n − 1)In−2,
n≥2
∫
The reduction formula for In = cosn x d x is
nIn = sin x
cosn−1 x
+ (n − 1)In−2,
n≥2
To find I4, substitute n = 4 in the reduction formula:
4I4 = sin x cos3 x + 3 I2
3I
1 sin x cos 3 x + __
I4 = __
4
4 2
Substituting n = 2 in the reduction formula:
138
2I2 = sin x cos x + I0
1 sin x cos x + __
1I
I2 = __
2
2 0
MODULE 1tCHAPTER 5
We cannot use the formula to find I0.
∫
∫
Since In = cosn x d x, when n = 0, I0 = cos0 x d x
∫
I0 = 1 d x = x + c
1 sin x cos x + __
1 (x + c)
∴ I2 = __
2
2
3 __
1 sin x cos3 x + __
1 sin x cos x + __
1 x + __
1c
I4 = __
4
4 2
2
2
3 sin x cos x + __
3x + A
1 sin x cos3 x + __
I4 = __
4
8
8
[
]
∫
Reduction formula for tannx dx
∫
EXAMPLE 3
Obtain a reduction formula for In = tann x d x.
SOLUTION
Write tann x = tann−2 x tan2 x
∫
In= tann−2 x tan2 x d x
Substitute tan2 x = sec2 x − 1
∫
= ∫(tann−2 x sec2 x − tann−2 x) d x
= ∫tann−2 x sec2 x d x − ∫tann−2 x d x
In = tann−2 x (sec2 x − 1) d x
Recall from Chapter 2 that the differential of tan x is sec2 x.
f(x) = tan x
f ′(x) = sec2 x
From Chapter 4,
1
n+1 + c,
∫ f ′(x)[ f(x)]n dx = _____
n + 1 [f(x)]
tann−1 x + c
So ∫tann−2 x sec2 x dx = _______
n−1
n
Since In = ∫tan x d x
In−2 = ∫tann−2 x d x
n−1
tan x − I ,
∴ In = _______
n−2
n−1
n≠1
n≥2
Other reduction formulae
EXAMPLE 4
Given that In =
SOLUTION
In =
∫0 x ne x dx, show that In = e − nIn−1, for n ≥ 1. Hence, find I3.
1
∫0 xnex dx
1
Integrating by parts, let
dv = ex
u = xn, ___
dx
139
M O DUL E 1
du = nx n−1,
___
v = ex
dx
dv d x = uv − v ___
du d x, we have
Substituting into u ___
dx
dx
∫
∫
∫0 nxn−1 ex dx
1
= (1ne1 − 0ne0) − ∫ nxn−1 ex d x
0
1
n−1
e x dx
∴ In = e − n ∫ x
0
1
Since In = ∫ xne x d x, replacing n by n − 1, we have
0
1
In−1 = ∫ xn−1 ex d x
0
1
In = [xnex]0 −
1
∴ In = e − nIn−1,
n≥1
Substituting n = 1, 2, 3, we get
n = 3 ⇒ I3 = e − 3I2
n = 2 ⇒ I2 = e − 2I1
n = 1 ⇒ I1 = e − I0
To find I0 we use
∫0 x ne x dx
In =
1
When n = 0, I0 =
1
∫0 x 0e x dx = ∫0 e xdx
1
= [e x]0
= e1 − e0 = e − 1
1
Substituting I0 = e − 1
I1 = e − [e − 1] = 1
I2 = e − 2(1) = e − 2
I3 = e − 3[e − 2] = e − 3e + 6
= 6 − 2e
EXAMPLE 5
Given that In =
determine I4.
SOLUTION
In =
π
__
2
∫0 sinn x dx find a reduction formula for In and use this formula to
π
__
2
∫0 sinn x dx
Writing sinn x = sinn−1 x sin x
In =
π
__
2
∫0 sinn−1 x sin x dx
Integrating by parts, let
dv = sin x
u = sinn−1 x, ___
dx
du = (n − 1) sinn−2 x cos x, v = −cos x
___
dx
π
__
π
__
2
In = [ −sinn−1 x cos x ] 02 + (n − 1) sinn−2 x cos2 x d x
∫0
cos2x = 1 − sin2x
π cos __
π − (−sinn−1 0 cos 0) + (n − 1)
= −sinn−1 __
2
2
(
= (n − 1)
140
)
π
__
2
∫0
sinn−2 x d x − (n − 1)
π
__
2
∫0 sinn x dx
π
__
2
∫0 sinn−2 x (1 − sin2 x) dx
MODULE 1tCHAPTER 5
In = (n − 1)In−2 − (n − 1) In
In + (n − 1)In = (n − 1)In−2
∴ nIn = (n − 1)In−2
n−1 I ,
In = _____
n−2
n
(
)
n≥2
3I
4 − 1 I = __
To find I4 we substitute n = 4 ⇒ I4 = _____
4−2
4
4 2
1
__
Substitute n = 2 ⇒ I2 = I0
2
3
3I
1
__
__
So
I4 =
I = __
4 2 0
8 0
(
)
( )
To find I0 we go back to In =
I0 =
π
__
2
∫0
sin0 x d x =
π
__
2
∫0 sinn x dx and replace n by 0:
π
__
2
∫0 1 dx = [x]0 = __π2
π
__
2
3 __
π = ___
3π
∴ I4 = __
8 2
16
( )
EXAMPLE 6
Determine a reduction formula for In =
evaluate
SOLU TION
In =
∫0 (1 + x3)4 dx.
1
∫0(1 + x3)n dx, where n is an integer. Hence
1
∫0 (1 + x3)n dx
1
dv = 1
___
dx
du = n(3x2)(1 + x3)n−1, v = x
___
dx
Let u = (1 + x3)n
∫0 n(3x2)(1 + x3)n−1x dx
1
= 1(1 + 13)n − 0 − n ∫ 3x3(1 + x3)n−1 d x
0
1
In = [ x(1 + x3)n ] 0 −
Remember
x3
Separate ______3
1+x
1
into 1 − ______
1 + x3
or
1
x3 + 1)x3
x3 + 1
−1
3
x
∴ ______
1 + x3
1
= 1 − ______
1 + x3
1
(integration by parts)
3x3(1 + x3)n
We have 3x3 (1 + x3)n−1 = ___________
1 + x3
1
3
3x (1 + x3)n d x
In = 2n − n ______
3
01+x
∫
x3 = _________
x3 + 1 − 1 = ______
x3 + 1 − ______
1 = 1 − ______
1
Now ______
3
3
1+x
x +1
x3 + 1 x3 + 1
x3 + 1
∴ In = 2n − 3n
1
(1 + x3)n d x
∫0 ( 1 − ______
1 + x3 )
1
= 2n − 3n
∫0 [ (1 + x3)n − (1 + x3)n−1 ] dx
= 2n − 3n
∫0 (1 + x3)n dx + 3n ∫0 (1 + x3)n−1 dx
1
1
1
= 2n − 3n In + 3nIn−1
3nIn + In = 2n + 3nIn−1
(3n + 1)In = 2n + 3nIn−1,
Since
n ≥ 1, which is a reduction formula for the integral
∫0(1 + x3)4 dx = I4 we can use the reduction formula to find this integral.
1
141
M O DUL E 1
Using (3n + 1) In = 2n + 3nIn−1
n = 4 ⇒ (12 + 1)I4 = 24 + 3(4) I3
13I4 = 16 + 12I3
n = 3 ⇒ 10I3 = 23 + 3(3)I2
10I3 = 8 + 9I2
n = 2 ⇒ 7I2 = 22 + 3(2)I1
7I2 = 4 + 6I1
n = 1 ⇒ 4I1 = 2 + 3I0
∫0(1 + x3)n dx
1
1
1
I0 = ∫ (1 + x3)0 d x = ∫ 1 d x = [x]0 = 1
0
0
1
Since In =
∴ 4I1 = 2 + 3 (1)
4I1 = 5
5
I1 = __
4
5 = ___
23
7I2 = 4 + 6 __
4
2
23
I2 = ___
14
23
10I3 = 8 + 9 ___
14
( )
( )
1 8 + 9 ___
I = ___
( 1423 ) ]
10 [
3
[
207
1 ____
112 + ____
= ___
10 14
14
]
319
= ____
140
319
13I4 = 16 + 12 ____
140
957
1 16 + ____
I4 = ___
13
35
( )
[
]
1517
= _____
455
EXERCISE 5A
In questions 1–5, find a reduction formula for each of the integrals.
142
∫
2
In = xn sin x d x
In =
∫0(4 − x)n dx
4
In =
5
In =
∫04 cosn x dx
6
Obtain a reduction formula for In =
1
In = xne2x d x
3
2
∫
∫0 xne2x dx
1
π
__
π
__
4
0
∫
secn x d x and hence determine I4.
MODULE 1tCHAPTER 5
7
(a) Use the substitution u = 1 +
(b) Given that In =
∫0
__
√3
(n + 2) In = 8(3)
x2 to
show that
∫0
__
√3
______
7.
x √1 + x2 d x = __
3
______
xn √1 + x2 d x, prove that
n−1
____
2
− (n − 1) In−2,
n ≥ 2.
(c) Find I5.
8
Given that In =
n + 1 I − __
1 e−1.
∫0 xne−x dx, show that In + 2 = ( _____
2 ) n 2
1
2
Hence find I5 in terms of e.
9
Show that In = nIn−1 −1, n ≥ 1 where In =
∫0ex (1 − x)n dx.
1
Use the reduction formula to show that I4 = 24e − 65.
2
1
10 Given that In = ∫ (1 + x2)n dx, n ≥ 1, show that In = ______
I
+ 1 − ______
1 + 2n n−1
1 + 2n
1
Find I4.
n
0
(
)
11 Show that 2nIn+1 = (2n − 1)In + 2−n, where In = ∫ (1 + x2)−n d x.
1
0
Use the reduction formula to find I4.
π
__
1__ n + (n − 1)I . Hence, find
12 Given that In = ∫ 4 sinn x d x, show that nIn = ___
n−2
( √2 )
0
the volume formed when the region bounded by the curve y = sin2 x, the lines
π and the x-axis is rotated through 2π radians about the x-axis.
x = 0, x = __
4
x (ln x)n − __
nI .
13 Let In = ∫x (ln x)n d x. Show that In = __
2
2 n−1
2
[
]
3x2 __
2 (ln x)3 − (ln x)2 + ln x − __
1 + c.
Hence, show that I3 = ___
2
4 3
14 Given that In = ∫ x 2 e
1
n
__
−1x
___
2
0
1
__
d x, show that In = −2e − 2 − nIn−2. Evaluate I4.
e
15 Derive a reduction formula for ∫ (ln x)n d x and use this formula to find I4 in
terms of e.
1
__
π
√ 2 _______
4n − π __
π n−1 − n(n − 1)I where I = __4 xn sin x d x.
16 Show that In = ___
∫0
n−2
n
(4)
4
2
(
)
Find the exact value of I2.
17 Derive a reduction formula for In = ∫ xne2x d x. Use your reduction formula to
1
0
find the volume formed when the region bounded by the curve y = x2ex, the
x-axis, the lines x = 0 and x = 1 is rotated through 2π radians about the x-axis.
Leave your answer in terms of e.
143
M O DUL E 1
SUMMARY
Reduction formulae
∫
= ∫sin
∫
In = tann x dx
In = sinn x dx
n – 1x sin x dx
∫
= –cos x
sinn – 1 x +
= –cos x
sinn – 1 x + (n – 1)
= –cos x sinn – 1
∫
= tann – 2 x tan2 x dx
(n – 1)sinn – 2 x cos2 x dx
∫
x + (n – 1)∫sin
∫
= tann – 2 x (sec2 x – 1)dx
sinn – 2 x(1 – sin2 x) dx
n – 2 x dx – (n – 1)
∫
sinn x dx
= –cos x sinn – 1 x + (n – 1)In – 2 – (n – 1) In
∫
=
In + (n – 1)In = –cos x sinn – 1 x + (n – 1)In – 2
nIn = –cos x sinn – 1 x + (n – 1)In – 2
In =
1 cos x sinn – 1 x + n – 1 I
n n–2
n
( )
∫
Reduction formula for cosn x dx is
derived in a similar manner.
Checklist
Can you do these?
■ Derive a reduction formula for sinn x.
■ Derive a reduction formula for cosn x.
■ Derive a reduction formula for tann x.
■ Derive a reduction formula by integrating by parts.
■ Derive a reduction formula for a definite integral.
■ Use a reduction formula to find integrals (e.g. I4, I3).
144
∫
= tann – 2 x sec2 x dx – tann – 2 x dx
tann – 1 x
– In – 2
n –1
MODULE 1tCHAPTER 6
CHAPTER 6
Trapezoidal Rule (Trapezium Rule)
At the end of this chapter you should be able to:
■ use the trapezium rule to estimate integrals
■ identify the width of an interval given the number of intervals
■ identify the values of x to be used in the trapezium rule
■ identify when the trapezium rule underestimates the area under a curve
■ identify when the trapezium rule overestimates the area under a curve.
KEYWORDS/TERMS
BSFBVOEFSBDVSWFtFRVBMTVCJOUFSWBMTt
USBQF[JVNSVMFtPWFSFTUJNBUFtVOEFSFTUJNBUF
145
M O DUL E 1
The area under a curve
We can approximate the area under a curve by dividing the region into n equal
subintervals and forming trapezia of equal width within each subinterval. The area
under the curve can then be estimated by adding the areas of the n trapezia.
y
d d d d
x0 = a x1 x2 x3 x4
Area under the curve y = f (x) from a to b =
d
x
xn – 1 xn = b
∫a f (x) d x.
b
Dividing the interval [a, b] into n equal
subintervals, we have
b−a
d = _____
n
where d is the width of one interval.
Notes
(i) x0 = a, xn = b.
(ii) The region
must be split
into n equal
subintervals.
(iii) If there are n
intervals we
will have
(n + 1)
x-coordinates.
(iv) The trapezium rule
may either
overestimate
or underestimate the
area under
the curve.
(v) We get a
better
approximation
if we take
more trapezia.
Remember
1 (sum of parallel sides)
Area of trapezium = __
2
× perpendicular height
Finding the area of each trapezium, we have
d [ f (x ) + f (x )]
1 [ f (x ) + f (x )]d = __
1st: Area = __
0
1
0
1
2
2
d [ f (x ) + f (x )]
1 [ f (x ) + f (x )]d = __
2nd: Area = __
1
2
1
2
2
2
d [ f (x ) + f (x )]
3rd: Area = __
2
3
2
b
f (x) d x ≈ sum of the areas of the trapezia
parallel sides are
f (x1) and f (x2)
∫a
∫a f (x) dx ≈ __21 d [ f (x0) + f (x1)] + __21 d [ f (x1) + f (x2)] + __21 d [ f (x2) + f (x3)] + . . .
b
1 d [ f (x ) + f (x )]
+ __
n−1
n
2
1 d [ f (x ) + f (x ) + f (x ) + f (x ) + f (x ) + . . . + f (x )
= __
0
1
1
2
2
n−1
2
+ f (xn−1) + f (xn)]
d [ f (x ) + 2f (x ) + 2f (x ) + . . . + 2f (x ) + f (x )]
= __
0
1
2
n−1
n
2
d [ f (x ) + f (x ) + 2 [ f (x ) + f (x ) + . . . + f (x )]]
= __
0
n
1
2
n−1
2
This gives the trapezium rule (sometimes called the trapezoidal rule):
∫a f (x) dx ≈ __d2 [ f (x0) + f (xn) + 2 [ f (x1) + f (x2) + . . . + f (xn−1)]]
( __21 interval width ) [1st y-value + last y-value + twice (sum of all y-values in between)]
b
Let us use the rule to estimate some integrals.
146
parallel sides are
f (x0) and f (x1)
MODULE 1tCHAPTER 6
EXAMPLE 1
SOLUTION
1 and the
Estimate the area of the region bounded by the curve y = ______
1 + x2
lines x = 0, x = 1 using the trapezium rule with five trapezia.
1
Let f (x) = ______
1 + x2
Dividing the interval into five subintervals we get
1 − 0 = __
1
d = _____
5
5
1 until we reach 1.
∴ Starting at x = 0 we can get the other values of x by adding __
5
1
5
0
2
5
3
5
4
5
1
1 , we have
Drawing up a table of x and f (x) = ______
1 + x2
x
f (x)
0
1 =1
______
1
__
5
( )
2
__
5
3
__
5
4
__
5
1
1 + 02
1
________
= 0.96154
1 2
1 + __
5
1
________
= 0.86207
2 2
1 + __
5
1
________
= 0.73529
3 2
1 + __
5
( )
( )
1
________
= 0.609 76
4 2
1 + __
5
1 = 0. 5
______
1 + 12
( )
Using the trapezium rule:
b
1 d [ f (x ) + f (x ) + 2[f (x ) + f (x ) + . . . + f (x )]]
f (x) d x ≈ __
0
n
1
2
n−1
2
a
∫
1 d x = __
1 __
1 [(1 + 0.5) + 2(0.961 54 + 0.862 07 + 0.735 29 + 0.609 76)]
∫0 ______
2 (5)
1 + x2
1
= 0.784 (3 d.p.)
EXAMPLE 2
SOLUTION
Find an estimate of the integral
subintervals.
1 d x, using the trapezium rule with six equal
∫0 _____
x+1
1
1 . Since we have six intervals the width of one interval is
Let f (x) = _____
x+1
1 − 0 = __
b − a = _____
1
d = _____
6
6
n
1 , starting from 0 until we reach 1.
The x-values can be found by adding __
6
3 , __
5 , 1.
1 , __
2 , __
4 , __
∴ x = 0, __
6 6 6 6 6
147
M O DUL E 1
Drawing up a table for x and f (x), we have
1
f (x) = _____
x+1
1 =1
_____
0+1
1 = 0.857 14
_____
1
1 + __
6
1 = 0.75
_____
2
1 + __
6
1 = 0.666 67
_____
3
1 + __
6
1 = 0.6
_____
4
1 + __
6
1 = 0.545 45
_____
5
1 + __
6
1 = 0.5
_____
6
1 + __
6
x
0
1
__
6
2
__
6
3
__
6
4
__
6
5
__
6
1
Using the trapezium rule:
∫a f (x) dx ≈ __21 d [ f (x0) + f (xn) + 2 [f (x1) + f (x2) + . . . + f (xn−1)]]
b
we get
1 d x ≈ __
1 __
1 [1 + 0.5 + 2(0.857 14 + 0.75 + 0.666 67 + 0.6 + 0.545 45)]
∫0 _____
2 (6)
x+1
1
8.338 52
= _______
12
= 0.695 (3 d.p.)
EXAMPLE 3
Given that I =
5
∫0 2−x dx, find an estimate for I using the trapezium rule with
four intervals. By sketching the graph of y = 2−x show that the trapezium rule
overestimates the value of I.
SOLUTION
Since we have four intervals, the width of each interval is
5 − 0 = 1.25
d = _____
4
The values of x are found by adding 1.25 starting from 0, therefore
x = 0, 1.25, 2.5, 3.75, 5.
x
148
f (x) = 2−x
0
2−0 = 1
1.25
2−1.25 = 0.420 45
2.5
2−2.5 = 0.176 78
3.75
2−3.75 = 0.074 33
5
2−5 = 0.031 25
MODULE 1tCHAPTER 6
5
∫0 2−x dx ≈ __12 (1.25)[1 + 0.031 25 + 2(0.420 45 + 0.176 78 + 0.074 33)]
= 1.483 98
= 1.484 (3 d.p.)
y
1
y = 2–x
x
0
1.25
2.5
3.75
5
The shaded region above the curve is included in the trapezium rule, so we have an
overestimate.
EXAMPLE 4
Using the substitution u = x2 + 1 find the value of
∫0 x √x2 + 1 dx.
2
______
Find an estimate of the integral using the trapezium rule with eight equal
subintervals. Compare the two values.
∫0 x √x2 + 1 dx
2
SOLUTION
______
Since u = x2 + 1
du = 2x d x
1 du = x d x
∴ __
2
______
__
√x2 + 1 = √u
Changing limits:
when x = 0, u = 02 + 1 = 1
when x = 2, u = 22 + 1 = 5
∴
∫0
2
______
x √ x2 + 1 d x =
5
∫
1 √__
__
u du =
2
1
∫
[
]
[
]
1 u __21 du = __
1 × __
2 u __23 5 = __
1 5 __23 − 1 = 3.393 45
__
2
2
3
3
1
1
5
Using the trapezium rule with eight intervals:
2 − 0 = __
1
d = _____
4
8
The x-values are x = 0, 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, 2.
______
Let f (x) = x √ x2 + 1
149
M O DUL E 1
Table of values:
______
f (x) = x √ x2 + 1
x
______
0
0 √ 02 + 1 = 0
0.25
0.25 √ 0.252 + 1 = 0.257 69
0.5
0.5 √0.52 + 1 = 0.559 02
0.75
0.75 √ 0.752 + 1 = 0.937 50
1
1 √ 12 + 1 = 1.414 21
1.25
1.25 √ 1.252 + 1 = 2.000 98
1.5
1.5 √1.52 + 1 = 2.704 16
1.75
1.75 √ 1.752 + 1 = 3.527 24
2
2 √ 22 + 1 = 4.472 14
________
_______
________
______
________
_______
________
______
______
∫0 x √x2 + 1 dx ≈ __12 ( __41 ) [0 + 4.472 14 + 2 (0.257 69 + 0.559 02 + 0.937 50 + 1.414 21
2
+ 2.000 98 + 2.704 16 + 3.527 24)]
= 3.409 22
When rounded to one decimal place we get the same estimate of 3.4 as we found
using the integral.
The percentage error in using the trapezium rule to estimate the integral is
3.409 22 − 3.393 45 × 100 = 0.465% (3 d.p.)
________________
3.393 45
EXAMPLE 5
Estimate
SOLUTION
_______
∫0 √16 − x2 dx using the trapezium rule with x = 1, x = 2, x = 3 and x = 4.
4
_______
x
f (x) = √ 16 − x2
0
√16 − 02 = 4
1
√16 − 12 = √15
2
√16 − 22 = √12
3
√16 − 32 = √7
4
√16 − 42 = 0
_______
_______
___
_______
___
_______
__
_______
_______
∫0 √16 − x2 dx ≈ __21 (1)[4 + 0 + 2 (√15 + √12 + √7 )] = 11.982 84 = 11.983 (3 d.p.)
4
___
___
__
EXERCISE 6A
In questions 1–8, use the trapezium rule with the given number of subintervals to
approximate the given integral.
4
1
2
1 d x, n = 8
______
e x d x, n = 4
1
2
3
0
01+x
∫
150
∫
MODULE 1tCHAPTER 6
3
ln x d x, n = 6
∫1 _____
1+x
4
5
∫0
6
∫0 x2 sin x dx, n = 4
7
∫1 ln (x3 + 2) dx, n = 3
8
∫2 xex dx, n = 3
9
Find an approximate value for
2
1
__
e −√x d x, n = 8
2
10 Show that ∫
1
0
1
d x, n = 6
∫0 __________
4
x + x2 + 1
6
π
5
______
∫0 √x2 + 1 dx using six subintervals.
1
π. Use the trapezium rule with six intervals to find an
1 d x = __
_____
4
1+x2
approximation of
1 d x. Hence estimate π to 3 d.p.
∫0 ______
1 + x2
1
2
11 Find an approximate value of ∫ __1x d x using the trapezium rule with
1
(a) four intervals
(b) eight intervals.
Find the percentage error in the approximations by evaluating
∫1 __1x dx.
2
π
__
12 Use the trapezium rule with two intervals to estimate ∫ 3 sin2 x d x.
Find an exact value of
0
π
__
∫03 sin2 x dx.
13 The region bounded by the curve y = e cos x, the x-axis and the lines x = 0 and
3π ,
π, __
π, ___
x = π, is denoted by R. Use the trapezium rule with ordinates at x = 0, __
4 2 4
π to estimate the area of R, giving three decimal places in your answer.
π
__
1
________
d x.
14 Use the trapezium rule with five intervals to estimate ∫ 2 _________
0
√ 1 + cos x
15 Use the trapezium rule, with ordinates at x = −1, x = −__12 , x = 0, x = __12 and
x = 1, to estimate the value of
in your answer.
_________
∫−1 √ln (3 + x) dx giving three significant figures
1
16 Determine the approximate area between the curve y = x3 + x2 − 4x − 4, the
ordinates x = −2 and x = −1 and the x-axis by applying the trapezium rule
with four intervals. Compare the result obtained by this method with the true
area obtained by integration.
151
M O DUL E 1
SUMMARY
Trapezium rule
Used to estimate definite integrals.
b
∫ a f(x) dx ≈ d2 [f(x0) + f(xn) + 2[f(x1) + f(x2) + ... + f(xn–1)]]
–a
d=bn
Formula must use the upper and
lower limit of the integral.
n = number of intervals
The interval [a, b] must be divided
into n equal subintervals.
The trapezium rule can either
overestimate or underestimate
the integral.
Overestimate
Underestimate
Checklist
Can you do these?
■ Use the trapezium rule to estimate integrals.
■ Identify the width of an interval given the number of intervals.
■ Identify the values of x to be used in the trapezium rule.
■ Identify when the trapezium rule underestimates the area under a curve.
■ Identify when the trapezium rule overestimates the area under a curve.
152
MODULE 1tCHAPTER 6
Module 1 Tests
Module 1 Test 1
1
(a) Differentiate with respect to x
πx
(i) 6e x+1 sin ( __
4 )
[3]
______
(ii) sin−1 √1 − 2x
[4]
dθ when t = 2
(b) Given that θ = 4−t, find ___
dt
8x
+
10
___________
(c) (i) Express 2
in partial fractions.
2x + 5x − 3
2
15
8x + 10
___
(ii) Hence, show that ___________
2 + 5x − 3 d x = 2 ln 4
2x
1
( )
∫
2
−π ≤ tan−1 x ≤ __
π , prove
(d) Given that y = tan−1 x, where ____
2
2
dy ______
1
___
that
=
d x 1 + x2
x
______
d x.
(a) (i) Using the substitution u = 1 − x2, find _______
√1 − x2
1
(ii) Hence, find sin−1 x d x.
(b) If In =
π
__
2 n
t
0
∫
∫
∫0
[4]
[4]
[4]
[6]
[4]
[4]
π − n(n − 1)I for n ≥ 2.
cos t dt, prove that In = ( __
n−2
2)
n
Hence find I4.
[9]
(c) The trapezium rule, with two intervals of equal width, is to be used to find
an approximate value for ∫20 e−2x d x. Explain, with the aid of a sketch,
why the approximation will be greater than the exact value of the integral.
Calculate the approximate value and the exact value, giving each answer
correct to 3 d.p.
[8]
3
(a) The parametric equations of a curve are defined in terms of θ by
x = 4 + 2 cos θ, y = 2 cos 2θ
dy
(i) Show that ___ = 4 cos θ.
dx
[4]
π
[4]
(ii) Find the equation of the tangent to the curve at θ = __
2
(b) Given that variables x and y are related by the equation y2 + sin (xy) = 2,
dy
π , y = 1.
show that ___ = 0 when x = __
[7]
2
dx
(c) On a single Argand diagram sketch the loci given by
(i) |z − 2 − 3i| = 2
[3]
π
[3]
(ii) arg (z − 2 − 3i) = __
4
Hence, find the exact value of the complex number z that satisfies both
(i) and (ii).
[4]
153
M O DUL E 1
Module 1 Test 2
1
dy
(a) Find ___ if
dx
(i) y = tan3 3x + 4 cos2 x
[3]
______
1 + x2
(ii) y = ln ______
2+x
√
[4]
__________
(iii) y = √ sin (x2 + 4)
[4]
d2y
(b) Given that y = sin−1 (2 − x), find ___2 .
dx
2+ x + 4
3x
_____________
in partial fractions.
(c) (i) Express 2
(x + 1)(x + 1)
(ii) Hence, find
2
(a) If In =
3x + x + 4 d x.
∫0 _____________
(x2 + 1)(x + 1)
1
2
[4]
[5]
[5]
∫0 xne3x dx, show that In = __31 e3 − __n3 In−1 for n ≥ 1.
1
∫0 x4e3x dx.
1
Hence find
[10]
(b) The parametric equations of a curve are defined by x = ln (2t + 1),
y = t2 − 1. Find the equation of the normal to the curve
when t = 1.
[8]
dy
2
(c) Given that xy + 2x2y2 = 3x, show that ___ = −__
5 when x = 1 and y = 1.
dx
Hence, find the equation of the tangent at (1, 1) giving your answer in
the form ax + by = c where a, b and c are integers.
[7]
3
(a) Find the exact value of
e
∫1 x3 ln x dx
[5]
(b) Use de Moivre’s theorem to show that
4 tan θ − 4 tan θ
tan 4θ = _________________
1 − 6 tan2 θ + tan4 θ
3
Hence or otherwise solve the equation
x4 − 6x2 + 1 = 0
nπ .
giving your answer in the form tan ___
8
(c) (i) Find complex numbers v = x + iy such that x, y ∈ ℝ and
v 2 = 3 + 4i.
( )
[10]
[5]
(ii) Hence, or otherwise, solve for z the equation
z2 − (4 + 3i)z + 1 + 5i = 0
154
[5]
2
Sequences, Series and
Approximations
155
M O DUL E 2
CHAPTER 7
Sequences
At the end of this chapter you should be able to:
■ define a sequence
■ identify the different types of sequences
■ decide whether a sequence converges or diverges
■ find the value to which a sequence converges
■ find the terms of a sequence
■ identify the general term of a sequence
■ define a sequence as a recursive relation
■ find the terms of a sequence given a recurrence relation.
KEYWORDS/TERMS
TFRVFODFtDPOWFSHFOUtEJWFSHFOUtPTDJMMBUJOHt
QFSJPEJDtBMUFSOBUJOHtHFOFSBMUFSNtSFDVSSFODF
SFMBUJPOtMJNJUtMJNJUMBXTGPSTFRVFODFT
156
MODULE 2tCHAPTER 7
A sequence is a function whose domain is the set of all positive integers, while a
series is a list of numbers added together. A sequence can be written as
(i) a list of terms (ii) a formula (or function) or (iii) a recurrence relation
Notation: If un is a sequence, it is denoted by {un} or (un). In the sequence {un}, the
terms are u1, u2, u3, u4, . . .
We can graph a sequence since a sequence represents a function. A finite sequence
contains only a finite number of terms. An infinite sequence is unending.
Types of sequence
The different types of sequences can be described as: convergent, divergent,
oscillating, periodic or alternating.
Convergent sequences
A sequence {un} is convergent if the values approach a fixed point as n increases, that
is, lim un = l
n→∞
un
n
Divergent sequences
Any sequence which does not converge is called a divergent sequence. A sequence
which diverges can (i) diverge to positive infinity, (ii) diverge to negative infinity.
un
n
Oscillating sequences
Oscillating sequences are neither convergent nor divergent. For example, the
sequence −1, 2, −3, 4, −5, . . . is an oscillating sequence, as is the sequence 1, 0,
3, 0, 5, 0, 7, . . . Oscillating sequences can be divided into two types:
(i) an oscillating finite sequence which is bounded but not convergent
157
M O DUL E 2
(ii) an oscillating infinite sequence which is neither bounded nor diverges to +∞ or −∞.
Bounded and oscillating
un
un
Oscillating sequence
which is neither
bounded nor
diverges to +∞ or –∞.
n
n
Periodic sequences
A periodic sequence is a sequence which repeats its terms regularly. The smallest
interval with which the sequence repeats itself is the period of the sequence.
The sequence 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, . . . is a periodic sequence with period 3.
Alternating sequences
An alternating sequence is a sequence with terms that alternate between positive and
negative values. Alternating sequences can be convergent, divergent or oscillating.
For example, the sequence −1, 0, 3, −0.3, 0, 0.3, −0.03, . . . is alternating and
convergent; the sequence −1, 1, −1, 1, −1, . . . is oscillating.
The terms of a sequence
1 , __
1 , __
1 , __
1 , … can be written as u = __
1
The sequence 1, __
n
n . This represents a rule for the
2 3 4 5
nth term of a sequence (or the general term of the sequence); the list of terms can be
replaced by the formula for the nth term.
EXAMPLE 1
Given the general term of the sequence un = 4n + 2, write down the first four terms
of the sequence.
SOLUTION
un = 4n + 2
Substituting n = 1,
u1 = 4(1) + 2
=4+2
=6
n = 2,
u2 = 4(2) + 2
=8+2
= 10
n = 3, u3 = 4(3) + 2
= 12 + 2
158
= 14
MODULE 2tCHAPTER 7
n = 4, u4 = 4(4) + 2
= 16 + 2
= 18
The first four terms of the sequence are 6, 10, 14 and 18.
EXAMPLE 2
4 .
Identify the first five terms of the sequence un = ______
3n + 2
SOLUTION
4
un = ______
3n + 2
4
4 = __
4
= _____
n = 1, u1 = ________
3(1) + 2 3 + 2 5
n = 2,
4
1
4 = __
4 = __
u2 = ________
= _____
3(2) + 2 6 + 2 8 2
4
4 = ___
4
n = 3, u3 = ________
= _____
3(3) + 2 9 + 2 11
n = 4,
4
4 = ___
4 = __
2
u4 = ________
= ______
3(4) + 2 12 + 2 14 7
n = 5,
4
4 = ___
4
u5 = ________
= ______
3(5) + 2 15 + 2 17
1 , ___
4 , __
4.
4 , __
2 , ___
∴ The first five terms of the sequence are __
5 2 11 7 17
EXAMPLE 3
1
Write down the first six terms of the sequence un = (−1)n+1 __
n .
SOLUTION
1 = (−1)2 (1) = 1
n = 1, u1= (−1)1+1 __
1
( )
n = 2, u2= (−1)2+1
n = 3, u3= (−1)3+1
n = 4, u4= (−1)4+1
n = 5,
u5= (−1)5+1
n = 6,
u6 = (−1)6+1
( )
( __21 ) = (−1) ( __21 ) = −__21
( __31 ) = (−1) ( __31 ) = __31
( __41 ) = (−1) ( __41 ) = −__41
( __51 ) = (−1) ( __51 ) = __15
( __61 ) = (−1) ( __61 ) = −__61
3
4
5
6
7
1 __
1 __
1 __
1 __
1
∴ The first six terms of the sequence are 1, −__
2, 3, −4, 5, −6.
Try these 7.1
Write down the first six terms of the following sequences.
(−1)n
n
(a) un= 4n + 3
(b) un= _____
(c)
un = ________
n+2
3(n + 1)
159
M O DUL E 2
Finding the general term of a sequence
by identifying a pattern
We should be able to write down the general term of a sequence when given a list of
its terms. To do this, we look for a connection between the subscript that identifies
the term and the term itself.
EXAMPLE 4
1 , __
1 , ___
1 , __
1,…
Find the general term of the sequence 1, __
3 9 27 81
SOLUTION
The first term is u1 = 1.
Note
You can always
check your result
by listing the
terms derived
from your
formula.
We can write 1 as 30.
1 = __
1=1
∴ u1 = __
30 1
1 = __
1
The second term is u2 = __
31 3
1 = __
1
The third term is u3 = __
32 9
Notice that all the terms can be written as 1 divided by 3 to the power of some integer.
Notice also that the power of 3 is always one less than the subscript n.
1
Hence, the general term, un = ____
3n−1
EXAMPLE 5
Find the nth term of the sequence 2, 4, 8, 16, 32, 64, 128, …
SOLUTION
All the terms of this sequence can be written as powers of 2.
∴ u1 = 21
The subscript n and the index are the same. Let us see if this continues.
u2 = 4 = 22
u3 = 8 = 23
∴ un = 2n
EXAMPLE 6
Find the nth term of the sequence 4, 12, 20, 28, 36, …
SOLUTION
This sequence increases by a constant value of 8.
Recall that the general term of a sequence that increases by a constant value is of the form
un = an + b
where a = 8 in this case.
un = 8n + b
We can find b by using u1
u1 = 8 (1) + b = 8 + b
but u1 = 4
∴8+b=4
b = 4 − 8 = −4
∴ un = 8n − 4
160
MODULE 2tCHAPTER 7
EXAMPLE 7
1 , _____
1 , _____
1 , _____
1 ,…
Find the nth term of the sequence _____
1×2 2×3 3×4 4×5
SOLUTION
The denominator consists of a product with
1st term: 1, 2, 3, 4, … general term is n
2nd term: 2, 3, 4, 5, … general term is n + 1
1
∴ un = ________
n(n + 1)
Try these 7.2
Find the nth term of these sequences.
(a) 6, 9, 12, 15, 18, 21, …
1 , _____
1 , _____
1 , _____
1 ,…
(b) ____
3(7) 5(11) 7(15) 9(19)
(c) 3, 9, 27, 81, 243, …
A sequence defined as a recurrence relation
A third way of defining a sequence is to specify a value for one of the terms and
identify the nth term by an equation (a recurrence relation) involving one (or more)
of the preceding terms. A sequence that is defined using previous terms is called a
recursive sequence.
EXAMPLE 8
Given that u1, u2, u3, . . . are the terms of a sequence with u1 = 2 and un+1 = 4un − 1,
find the first four terms of the sequence.
SOLU TION
Since un+1 = 4un − 1
Substituting n = 1 ⇒ u1+1 = 4u1 − 1
u2 = 4u1 − 1
But u1 = 2 ⇒ u2 = 4(2) − 1 = 8 − 1 = 7
Substituting n = 2, we get
u2+1 = 4u2 − 1
u3 = 4u2 − 1
u3 = 4(7) − 1 = 28 − 1 = 27
Substituting n = 3, we get
u3+1 = 4u3 − 1
u4 = 4u3 − 1
u4 = 4(27) − 1 = 108 − 1 = 107
∴ The first four terms of this sequence are 2, 7, 27, 107.
161
M O DUL E 2
EXAMPLE 9
Write down the first five terms of the sequence un = nun−1 where u1 = 1.
SOLUTION
When n = 2, u2 = 2(u2−1)
u2 = 2u1 = 2 (1) = 2
n = 3:
u3 = 3u2 = 3 (2) = 6
n = 4:
u4 = 4u3 = 4 (6) = 24
n = 5:
u5 = 5u4 = 5 (24) = 120
Hence, the first five terms are 1, 2, 6, 24, 120.
E X A M P L E 10
SOLUTION
4 . Given that u = 3,
A sequence is generated by the recurrence relation un+1 = ______
1
un + 2
find u2, u3, u4 and u5.
4
un+1 = ______
un + 2
4 = __
4
n = 1: u2 = ______
u1 + 2 5
20 = ___
10
4 = ___
4 = _____
n = 2: u3 = ______
7
u2 + 2 __
4 + 2 14
5
28 = __
7
4 = ___
4 = ______
n = 3: u4 = ______
u3 + 2 ___
10 + 2 24 6
7
24
4
_____
______
n = 4: u5 =
= 4 = ___
7 + 2 19
u4 + 2 __
6
10
7 , u = ___
4
24 .
__
___
Hence, u2 = , u3 = , u4 = __
5
7
6 5 19
Try these 7.3
Find the first five terms of these sequences.
2n
(a) un+1 = __
un , u1 = 1
(b) un = 3un−1 − 2, u1 = 4
n+2 , u =1
(c) un+1 = _______
1
3u + 1
n
Convergence of a sequence
Note
If lim un does
n→∞
not exist or is
infinite we say
the sequence
diverges.
162
The sequence u1, u2, u3, … converges to a real number l, or has limit l provided
that un can be made as close to l as possible. This can be done by choosing n to be
sufficiently large. We write
lim un = l
n→∞
If the sequence does not converge, then the sequence diverges.
MODULE 2tCHAPTER 7
The limit laws for sequences are similar to the limit laws for functions studied in Unit 1:
If the limits lim un = l1 and lim vn = l2 exist, then
n→∞
n→∞
(a) n→∞
lim (un + vn) = lim un + lim vn = l1 + l2
n→∞
n→∞
(b) n→∞
lim un vn = l1 l2
u
l
n
__1
(c) n→∞
lim ___
vn = l
2
(d) n→∞
lim kun = k lim un, where k is a constant.
n→∞
2
E X A M P L E 11
8n
Let un = ____________
. Find lim un.
n→∞
6n2 + 3n + 2
SOLU TION
8n
un = ____________
6n2 + 3n + 2
2
8n2
lim un = lim ____________
2
n→∞
n→∞ 6n + 3n + 2
Dividing the numerator and denominator by n2, we get
8
lim u = lim __________
n→∞ n
n→∞
3
2
__
6 + n + __
n2
lim 8
n→∞
= _______________________
1 + 2 lim __
1
lim 6 + 3 lim __
n→∞
n→∞ n
n→∞ n2
8
= _____________
6 + 3(0) + 2(0)
8 = __
4
= __
6 3
4.
Hence, un converges to __
3
______
E X A M P L E 12
4n − 1 . Is u convergent?
Let un = ______
n
n+1
SOLU TION
4n − 1
lim u = lim ______
n→∞ n
n→∞ n + 1
√
(
)
4n − 1
= ( lim ______
n+1 )
1
__
2
1
__
2
n→∞
1
4 − __
n
= lim _____
n→∞
1
1 + __
n
(
4−0
= _____
1+0
(
)
1
__
2
)
1
__
2
__
= √4 = 2
Hence, un is convergent and converges to 2.
163
M O DUL E 2
We can apply l’Hôpital’s rule to find the limit of a sequence in the same way we apply
l’Hôpital’s rule to find the limit of a function.
Rule
L’Hôpital’s rule was covered in Unit 1:
f (x) 0
Suppose that we have lim ____ = __
or any indeterminate form, then
x→a g (x)
0
f (x)
f ′(x)
lim ____ = lim ____.
x→a g (x)
x→a g′(x)
If we have an indeterminate form, we can find the limit by differentiating the
numerator and the denominator and then take the limit. If the new function is still
indeterminate, we can differentiate again and evaluate the limit.
E X A M P L E 13
6n converges and find its limit if it does
Determine whether the sequence un = ______
2n − 1
converge.
SOLUTION
To decide if the sequence converges, we find lim un.
n→∞
6n
lim u = lim ______
n→∞ n
n→∞ 2n − 1
__ , we can apply l’Hôpital’s rule.
Since the function is of the form ∞
∞
6n = lim __
6=3
lim ______
2n − 1 n→∞ 2
n→∞
∴ lim un = 3
n→∞
Hence, un converges and its limit is 3.
Try these 7.4
Determine whether the following sequences converge and find the limit of those that
do converge.
n
n2 − 3n + 6
n3
(a) un = 1 + __98
(b) un = ___________
(c) un = _______
2
2
2n + 5
6n + 1
()
EXERCISE 7A
In questions 1–9, write down the first five terms of each sequence.
2
3
un = 2n + 1
n−1
un = _____
n
5
un = (−1)n (2n − 1)
6
7
n
un = __
en
9
un = (−1)n+1 n3
1
2
4
8
un = n2 + 3
5 n
un = __
4
2n
un = __
n
3n
un = ______
2n + 1
( )
In questions 10–19, write down the nth term of the given sequence.
164
1,…
10 __12 , __14 , __16 , __18 , ___
10
1,…
1 , ___
11 __21 , __14 , __18 , ___
16 32
12 __12 , __13 , __14 , __15 , __16 , …
13 1, −1, 1, −1, 1, −1, …
MODULE 2tCHAPTER 7
14 1, −3, 5, −7, 9, −11, 13, …
1 , _____
1 , _______
1 , _______
1 , _______
1 ,…
15 _____
4 × 5 7 × 9 10 × 13 13 × 17 16 × 21
8 , ____
16 , …
4 , ____
16 __25 , ___
25 125 625
17 5 × 8, 7 × 11, 9 × 14, 11 × 17, 13 × 20, …
18 1, 2, 6, 24, 120, 720, …
19 8, −16, 32, −64, 128, −256, …
In questions 20–29, the sequences are defined recursively. Find the first four terms of
each sequence.
20 un = 2 + 4un−1, u1 = 3
21 un = 4n + 2un−1, u1 = 1
22 un = (un−1)(un−2), u1 = 1, u2 = 2
23 un+1 = un + d, u1 = a
24 un = −3un−1, u1 = __13
25 un = (n − 1) − un−1, u1 = 5
__
________
26 un = √3 + √2 + un−1, u1 = 1
2u
n−1
27 un = _____
, u1 = 2
n+2
3u
n−1
28 un = ______
, u1 = 1
2n + 1
5u
n
n−1
29 un = _____
2 , u1 = 4
In questions 30–39, decide whether the sequence converges or diverges.
If the sequence converges, find the limit of the sequence.
4n + 3
30 un = ______
2n − 1
3n − 5
31 un = ______
6n + 7
n + 4n + 1
32 un = __________
2
6n + 5
33 un = _______
n+2
− 3n + 4
34 un = ___________
2
35 un = __32
2
2n + 3
n3
n −n+6
n
36 un = 1 + __97
______
6n − 1
38 un = ______
5n + 2
()
√
2
()
n
(ln n)2
37 un = ______
n
1−n
39 un = _______
2
3
2 + 3n
40 A sequence {un} of real numbers satisfies un+1 un = 4(−1)n, u1 = 2.
Show that un+2 = −un. Find the first five terms of the sequence.
3 , n ≥ 1.
41 A sequence is generated by the recurrence relation un+1 = ______
un − 5
5 ± √__
Given that u2 = 4u1, show that the possible values of u1 are __
7.
2
165
M O DUL E 2
SUMMARY
Sequences
A finite sequence contains only a
finite number of terms
A sequence can be written as a list
An infinite sequence is unending
In a convergent sequence the
values get closer and closer
to a fixed value
lim un = l
n→∞
Any sequence which does
not converge to a fixed value
is called divergent
A periodic sequence repeats at
regular intervals
An oscillating sequence is
(i) bounded and not convergent,
or (ii) unbounded and does not
diverge to +∞ or –∞
An alternating sequence is of
the form un = (–1)n an where
an is a non-negative real number
Checklist
Can you do these?
■ Define a sequence.
■ Identify the different types of sequences.
■ Decide whether a sequence converges or diverges.
■ Find the value to which a sequence converges.
■ Find the terms of a sequence.
■ Identify the general term of a sequence.
■ Define a sequence as a recursive relation.
■ Find the terms of a sequence given a recurrence relation.
166
as a function of n, i.e.
un = f(n)
or as a recursive relation
un + 1 = f(un)
MODULE 2tCHAPTER 8
CHAPTER 8
Series
At the end of this chapter you should be able to:
■ identify the nth partial sum of a series
■ write a series in sigma notation
■ identify a convergent series
n
n
n
■ find the sum of a series using the standard results for ∑1 r, ∑1 r 2, ∑1 r 3
■ use the summation laws
■ use the method of differences to find the sum of a series
■ test for convergence of a series.
KEYWORDS/TERMS
TVNtTFSJFTtTJHNBOPUBUJPOt
TVNNBUJPOMBXTtDPOWFSHFODFt
EJWFSHFODFUFTUtJOUFHSBMUFTU
167
M O DUL E 2
The sum of the terms of a sequence is called a series. An infinite series is the sum of
the terms of an infinite sequence. Therefore, for the sequence u1, u2, u3, . . . , the
corresponding infinite series is u1 + u2 + u3 + . . . .
For the infinite series u1 + u2 + u3 + . . . , Sn represent the sum of the first n terms.
∴ S1 = u1 (the first term)
S2 = u1 + u2 (the sum of the first two terms)
S3 = u1 + u2 + u3 (the sum of the first three terms)
Sn = u1 + u2 + u3 + . . . + un (the sum of the first n terms)
S1, S2, S3, . . . , Sn form a sequence and each term of this sequence is called a partial sum.
S1 is called the first partial sum, S2 is the second partial sum, . . . , Sn is the
nth partial sum.
Writing a series in sigma notation (∑)
The sigma notation was introduced in Unit 1 and it may be a good idea to revise
this. Let us look at ‘∑’ again.
Recall that ∑ represents ‘the sum of ’ and is used to write a series. The series
∞
u1 + u2 + u3 + . . . can be written as ∑r = 1 ur. The general term of the series is found
and put inside the summation sign. The series starts when r = 1 and is infinite.
The first value of r is the lower limit in the summation sign and the largest value of r
is the upper limit.
EXAMPLE 1
Write the series 2 + 4 + 8 + 16 + 32 + . . . using sigma notation.
SOLUTION
Let ur be the rth term of the series.
u1 = 2 = 21
u2 = 4 = 22
u3 = 8 = 23
ur = 2r
Since the series starts at u1 = 21, the lower limit in the sigma notation is r = 1.
Since the series is infinite, the upper limit is ∞.
∴ 2 + 4 + 8 + 16 + 32 + . . . =
∑2
r
r=1
EXAMPLE 2
Write the series 3 × 7 + 5 × 12 + 7 × 17 + 9 × 22 + . . . in sigma notation.
SOLUTION
Let us look for a pattern using the first term in the various products: 3, 5, 7, 9, . . .
Each term goes up by 2. Therefore, the general form is 2r + b. Since the first term is 3,
i.e. 3 = 2 × 1 + b, b = 1.
∴ 2r + b = 2r + 1.
168
∞
MODULE 2tCHAPTER 8
Now 7, 12, 17, 22, . . . goes up by 5. Therefore, 5r + c represents the terms in this
sequence. Since the first term is 7, i.e. 7 = 5 × 1 + c, c = 2.
Hence, the general term is 5r + 2.
∴ (2r + 1)(5r + 2) represents the general term of this sequence.
Hence, 3 × 7 + 5 × 12 + 7 × 17 + 9 × 22 + . . . =
∞
∑(2r + 1)(5r + 2)
r =1
EXAMPLE 3
5 + 1 + ___
17 + . . .
11 + ___
14 + ___
Write in sigma notation: 1 + __
4
16 32 64
SOLUTION
5 + __
8 + ___
7 + ___
17 + . . .
2 + __
11 + ___
First we rewrite as __
2 4 8 16 16 64
The numerator is of the form 3r + a where a = −1 since
3 + a = 2 ⇒ a = −1 and the denominator is 2r.
(
3r − 1
∴ ur = ______
2r
)
7 + ___
17 + . . . =
5 + 1 + ___
11 + ___
Hence, 1 + __
4
16 16
64
Try these 8.1
∞
3r − 1
∑( ______
2 )
r =1
r
Write the following in sigma notation.
1 + ___
1 + ____
1 +...
(a) 1 + __31 + __91 + ___
27 81 243
1 + _____
1 + _____
1 + _____
1 +...
(b) _____
2×3 3×4 4×5 5×6
(c) 9 + 13 + 17 + 21 + 25 + . . .
Sum of a series
The following standard results can be used to find the sum of many series.
n
n(n + 1)
∑ r = ________
2
r =1
n
∑r
r =1
n
2
n (n + 1)(2n + 1)
= _______________
6
∑
n2(n + 1)2
r 3 = _________
4
r =1
Recall the summation laws:
n
(i)
∑c = n × c where c is a constant.
r=1
15
Example:
∑2 = 2 × 15 = 30
r =1
169
M O DUL E 2
∑ cur = c ∑ ur where c is a constant.
Example: ∑ 4r = 4 ∑ r
∑(ur + vr) = ∑ur + ∑vr
(ii)
(iii)
n
∑
Example:
r=1
n
(r2 − r) =
∑
r=1
n
r2 −
∑r
r=1
Note: There is no rule for the
product or quotient.
∑(ur vr) ≠ ∑ur ∑vr
u
∑ur
∑( __vrr ) ≠ _____
∑vr
25
EXAMPLE 4
Find
∑r
r=1
n
SOLUTION
Using
n (n + 1)
and substituting n = 25, we get
∑ r = ________
2
r=1
25
25 (25 + 1) ______
25(26)
=
= 25 × 13 = 325
∑ r = __________
2
2
r=1
30
∑(r − 2)
EXAMPLE 5
Find
SOLUTION
Separating into two series, we have
r=1
30
∑
r=1
30
(r − 2) =
30
∑ ∑2
r−
r=1
r=1
n
Recall that
n (n + 1)
∑r = ________
2
r=1
30
30 (30 + 1) ______
30(31)
=
= 31 × 15 = 465
∑ r = __________
2
2
Also recall that ∑c = n × c
∴ ∑ 2 = 30 × 2 = 60
∴
r=1
n
r=1
30
r=1
30
∴
30
∑ ∑2 = 465 − 60 = 405
r=1
r−
r=1
30
Hence,
∑(r − 2) = 405
r=1
40
Find
SOLUTION
First we need to expand r(r + 2) = r 2 + 2r
∴
170
∑r (r + 2)
EXAMPLE 6
r=1
40
40
r=1
r=1
∑r(r + 2) = ∑(r + 2r)
2
MODULE 2tCHAPTER 8
Separating, we have
40
∑
r=1
40
r(r + 2) =
n
∑r
Using
2
r=1
∑
r=1
∑r
r=1
n (n + 1)(2n + 1)
= _______________
6
40
∑r
n = 40 ⇒
40
r2 + 2
r=1
2
40(41)(81)
= _________ = 22 140
6
n
Also using
n (n + 1)
∑r = ________
2
r=1
40
n = 40 ⇒
40 (41)
= 820
∑r = ______
2
r=1
40
∑ r(r + 2) = 22 140 + 2(820) = 22 140 + 1640 = 23 780
∴
r=1
All our summations started at r = 1 and all the standard results start at r = 1.
What if the lower limit changes? Let us see what happens.
25
∑r
EXAMPLE 7
Find
SOLUTION
Our lower limit is at r = 11.
r = 11
We can rearrange the summation to start at r = 1 as follows:
25
25
10
r = 11
r=1
r=1
∑r = ∑r − ∑r
i.e. if we sum from r = 1 to r = 25 and subtract the sum from r = 1 to r = 10, then
we are left with the sum from r = 11 to r = 25. We can now apply our standard
results to the RHS.
25
25
10
r = 11
r=1
r=1
∑r = ∑r − ∑r
25 (25 + 1) 10 (10 + 1)
= __________ − __________
2
2
25 (26) 10 (11)
= ______ − ______
2
2
= 25 × 13 − 55 = 325 − 55 = 270
20
∑r (r − 1)
EXAMPLE 8
Evaluate
SOLUTION
We expand r(r − 1) = r 2 − r
r=8
20
∴
∑
r=8
r (r − 1) =
20
∑ (r − r)
2
r=8
171
M O DUL E 2
Since our lower limit is 8, we need to rearrange as follows:
20
∑
r=8
20
(r 2
− r) =
∑
=
∑
r=1
20
r=1
7
(r2
− r) −
∑(r
20
r2 −
2
r=1
− r)
7
7
∑ ∑ ∑r
r=1
r−
r=1
r2 +
r=1
20(21)(41) 20(21) 7(8)(15) 7(8)
= _________ − ______ − _______ + ____
2
2
6
6
= 2870 − 210 − 140 + 28 = 2548
Try these 8.2
Find the sum of
15
∑
(a)
r=1
12
(b)
r2
∑
r=1
25
22
3r (r + 1)
(c)
∑
r(r + 3)
(d)
r = 11
∑r
3
r=8
Sum of a series in terms of n
n
∑(r + 4)
EXAMPLE 9
Find the sum of
SOLUTION
Separating, we have
n
∑
r=1
r=1
n
(r + 4) =
n
n(n + 1)
+ 4n
∑ ∑4 = ________
2
r=1
r+
r=1
n (n + 9)
n [(n + 1) + 8] = __
= __
2
2
(factorising)
n
∑ r (r − 1)
E X A M P L E 10
Find the sum of
SOLUTION
Expanding the bracket, we have
n
r=1
n
∑r(r − 1) = ∑
r=1
r=1
n
(r2
− r) =
n
∑ − ∑r
r=1
r2
r=1
Substituting the standard results,
n(n + 1)(2n + 1) n(n + 1)
= _______________ − ________
2
6
Factorising, we have
n
n(n + 1) (2n + 1 − 3)
∑r(r − 1) = ___________________
6
r=1
n(n + 1)(2n − 2)
= _______________
6
2n(n + 1)(n − 1)
= _______________
6
n(n + 1)(n − 1)
= ______________
3
172
(separating)
n
n(n + 1)(2n + 1)
∑r = ______________
6
n(n + 1)
∑r = _______
2
2
r=1
n
r=1
MODULE 2tCHAPTER 8
n
E X A M P L E 11
Find and simplify
n
SOLUTION
∑
r=1
∑r (r − 1)
2
r=1
n
r 2(r − 1) =
n
∑
r=1
(r 3 − r 2) =
n
∑ ∑r
r=1
r3 −
2
r=1
n2 (n + 1)2 n (n + 1)(2n + 1)
= _________ − _______________
4
6
3n2 (n + 1)2 − 2n(n + 1) (2n + 1)
= ____________________________
12
n (n + 1) [3n(n + 1) − 2(2n + 1)]
= ____________________________
12
n (n + 1) (3n2 + 3n − 4n − 2)
=__________________________
12
n (n + 1) [3n2 − n − 2]
= ___________________
12
n (n + 1)(3n + 2)(n − 1)
= _____________________
12
2n
E X A M P L E 12
SOLUTION
Find and simplify
∑(3r + 2)
r=1
2n
2n
2n
r=1
r=1
r=1
∑(3r + 2) = 3∑r + ∑2
Since the upper
limit is 2n,
remember to
replace n by 2n
in the standard
result.
3(2n)(2n + 1)
= ____________ + 2 (2n)
2
= 3n(2n + 1) + 4n
= n[3(2n + 1) + 4)]
= n[6n + 3 + 4]
= n[6n + 7]
2n
E X A M P L E 13
Show that
∑ r(r −1) = __n3 (7n − 1)
2
r=n+1
2n
SOLUTION
∑
2n
r(r − 1) =
r=n+1
∑ (r − r)
2
r=n+1
The lower limit is (n + 1); we need to rearrange the summation to start at 1.
2n
∴
∑
2n
(r 2
r=n+1
− r) =
∑
=
∑
r=1
2n
r=1
n
(r2
− r) −
∑(r − r)
2
r=1
2n
r2 −
n
∑ ∑
r=1
r−
r=1
r2 +
n
∑r
r=1
(expanding)
173
M O DUL E 2
Substituting the standard results,
2n
2n(2n +1)(2(2n) + 1) __________
2n(2n + 1) _______________
n(n + 1)
n(n + 1)(2n + 1) ________
−
+
−
∑ r(r + 1) = __________________
2
2
6
6
r=n+1
n(n + 1)(2n + 1) n(n + 1)
n(2n + 1)(4n + 1)
= ________________ − n(2n + 1) − _______________ + ________
3
2
6
Factorising, we have
2n
∑ r(r + 1) = __n6 [2(2n + 1)(4n + 1) − 6(2n + 1) − (n + 1)(2n + 1) + 3(n + 1)]
r=n+1
n [2(8n2 + 6n + 1) − 12n − 6 − (2n2 + 3n + 1) + 3n + 3]
= __
6
n [(16n2 − 2n2) + (12n − 12n − 3n + 3n) + 2 − 6 − 1 + 3]
= __
6
n
__
= (14n2 − 2)
6
2n
= ___ (7n2 − 1)
6
n (7n2 −1)
= __
3
2n
∑ (3r + 4) = __n2 (9n + 11)
E X A M P L E 14
Show that
SOLUTION
We need to rearrange the summation so that the lower limit will start at 1.
r=n+1
2n
2n
n
r=n+1
r=1
r=1
∑ (3r + 4) = ∑(3r + 4) − ∑(3r + 4)
2n
2n
∑ ∑
=3
r=1
r+
r=1
n
n
∑ ∑4
4−3
r=1
r−
r=1
3n(n + 1)
3(2n)(2n + 1)
= ____________ + 4(2n) − _________ − 4(n)
2
2
3n(n + 1)
= 3n(2n + 1) + 8n − _________ − 4n
2
n
= __ [6(2n + 1) + 16 − 3(n + 1) − 8]
2
n (12n + 6 + 16 − 3n − 3 − 8)
= __
2
n (9n + 11)
= __
2
Try these 8.3
Find the sum of
n
(a)
∑(4r − 2)
r=1
n
(b)
2n
(c)
174
∑ r (4r − 1)
r=n+1
∑r (2r − 1)
r=1
2n
(d)
∑ r (r
r=n+1
2
+ 4)
MODULE 2tCHAPTER 8
EXERCISE 8A
In questions 1–10, write the series in sigma notation.
1 +...
1 1 + __21 + __41 + __18 + ___
16
2
1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + . . .
3
3 − 5 + 7 − 9 + 11 − 13 + 15 − 17 + . . .
4
12 + 20 + 28 + 36 + 44 + 52 + 60 + 68 + . . .
5
6
1 + __
1 + __
1 + ___
1 + ___
1 +...
__
2 5 8 11 14
1 + __
1 + ___
1 + ___
1 +...
__
2 5 10 17
7
1 + __
1 + __
2 + ___
4 +...
__
2 3 9 27
8
3 + ___
3 + ___
3 +...
3 + __
4 16 64
9
96 + ____
384 + . . .
24 + ___
6 + ___
7
49 343
18 − ____
54 + . . .
10 2 − __56 + ___
25 125
In questions 11–20, identify the indicated term of the series.
n
11
13
15
17
19
∑
∑r , the 12th term
∑6r (r − 1), the 25th term
r , the (k − 2)th term
∑_____
r+2
3 , the (k + 1)th term
∑ _______
4r + 5
r=1
17
2r, the 10th term
3
r=1
50
r=1
n
r=1
n
r
2
r=1
n
12
14
16
18
20
∑r (r + 1), the 8th term
∑r (3r − 1), the 20th term
∑3 , the (k + 1)th term
3r + 1 , the (k + 1)th term
∑______
r+4
2 (r − 1)
, the (k + 1)th term
∑________
r+5
r=1
n
r=1
n
r−1
r=1
n
r=1
n
r
r=1
In questions 21–29, find the summation of the series.
36
60
21
∑r
23
∑3r
25
27
29
r=1
20
3
r=1
10
∑7r(r + 1)
∑(r + 2r − 3)
r=1
15
2
r=1
13
∑r (2r
r=1
2
22
∑r
24
∑(6r + 4)
26
28
2
r=1
22
r=1
25
∑6r(r − 2)
∑(2r − 1)
r=1
10
2
r=1
− 1)
175
M O DUL E 2
In questions 30–34, find the sum of the series.
28
40
∑r
31
32
∑r
33
34
∑r(r + 1)
30
r = 10
15
2
r=7
17
∑(r − 1)
r=8
30
∑(2r + 5)
r = 12
r=5
In questions 35–45, find and simplify the sum of the series.
n
n
35
∑(r + 6)
36
∑(r + 5r)
37
∑r(2r − 1)
38
∑4r(r + 2)
39
∑(r + 1)(r − 3)
40
∑r(r − 1)(r − 2)
41
∑(r + 2)(r − 4)
42
∑ 3r (r + 1)
44
43
45
r=1
n
r=1
n
r=1
r=n
r=1
2n
r=n+1
2n
2
r=1
n
r=1
n
r=1
2n
∑ (r − 2)
r=n+1
2n
∑ (r
r=n+1
3
− r)
∑ 2r(r − 1)
r=n+1
2n
46 Prove that
∑ (4r − 3) = n(15n + 14n + 3n − 3)
47 Show that
∑(4r + 1) = __31 n(4n + 6n + 5)
3
3
2
r=n+1
n
2
r=1
n
∑
2
20
∑
n (3n − 1). Hence find
48 Show that
(3r − 2) = __
(3r − 2).
2
r=1
r = 10
n
49 Show that
∑
40
∑
n(n + 1)(4n − 1)
r(2r − 1) = _______________. Hence find
r(2r − 1).
6
r=1
r = 15
Method of differences
The method of differences is used to find the sum of a series whose general term can
be written as the difference of terms. If we can write ur = f (r + 1) − f (r), then the
sum of the series can be found as follows.
n
n
∑u = ∑ {f (r + 1) − f (r)}
r=1
r
r=1
= f (2) − f (1) + f (3) − f (2) + f (4) − f (3) + . . . + f ((n − 2) + 1)
− f (n − 2) + f ((n − 1) + 1) − f (n − 1) + f (n + 1) − f (n)
= f (n + 1) − f (1)
176
MODULE 2tCHAPTER 8
Notice that pairs of terms sum to zero except f(n + 1) and f (1).
n
∴
∑u = f (n + 1) − f (1)
r
r=1
E X A M P L E 15
1 − _____
1
1 . Hence find
= __
Show that _______
r(r + 1) r r + 1
SOLUTION
Separate into partial fractions
n
1 .
∑ _______
r(r + 1)
r=1
B
1
A + _____
_______
≡ __
r
r+1
r(r + 1)
⇒ 1 ≡ A(r + 1) + B(r)
[r = 0] ⇒ 1 = A
[r = −1] ⇒ 1 = −B
B = −1
1
1
1 − _____
∴ _______
= __
r(r + 1) r r + 1
n
Hence,
∑
r=1
1
_______
=
r(r + 1)
n
1
∑( __1r − _____
r + 1)
r=1
Expanding the series and summing, we have:
) (
) ( )
1 − __
1 − _____
1
1 + __
+ ( _____
n − 1 n) (n n + 1)
(
(
1 + __
1 − __
1 + __
1 − __
1 + … + _____
1 − _____
1
= 1 − __
2
2 3
3 4
n − 2 n −1
)
1
= 1 − _____
n+1
n
1
1
_______
= 1 − _____
∴
n+1
r = 1 r(r + 1)
∑
E X A M P L E 16
Show that (r + 1)2 − r2 = 2r + 1. Hence find
n
(a)
SOLUTION
20
∑(2r + 1)
(b)
r=1
∑(2r + 1)
r=1
(a) (r + 1)2 − r2 = r2 + 2r + 1 − r2 = 2r + 1
n
So
Substitute values for r
into (r + 1)2 − r2
When r = n, we have
(n + 1)2 − n2
When r = n − 1, we
have (n − 1 + 1)2 −
(n − 1)2 = n2 − (n − 1)2
∑[(r +
r=1
n
1)2
−
r2]
=
∑(2r + 1)
r=1
Using the method of differences
n
∑[(r + 1) − r ] = (2 − 1 ) + (3 − 2 ) + (4
2
r=1
2
2
2
2
2
2
− 32 ) + . . . +
[(n − 1)2 − (n − 2)2 ] + [n2 − (n − 1)2 ] + [(n + 1)2 − n2 ]
= (n + 1)2 − 12
= (n + 1)2 − 1
177
M O DUL E 2
n
∑(2r + 1) = (n + 1) − 1
∴
2
r=1
= n2 + 2n + 1 − 1
= n2 + 2n
n
(b) When n = 20,
∑(2r + 1) = 20 + 2(20)
2
r=1
= 400 + 40 = 440
E X A M P L E 17
1
Separate ____________
into partial fractions. Hence show that
(r − 1)(r + 1)
n
(3n + 2)(n − 1)
2
= ______________
∑____________
(r − 1)(r + 1)
2n (n + 1)
r=2
SOLUTION
A + _____
B
1
____________
≡ _____
(r − 1)(r + 1) r − 1 r + 1
⇒ 1 = A(r + 1) + B(r − 1)
[r = 1] ⇒ 1 = 2A
1=A
__
2
[r = −1] ⇒ 1 = −2B
1
−__
2=B
1
1
1
−_______
∴ ____________
= _______
(r − 1)(r + 1) 2(r − 1) 2(r + 1)
n
∑
r=2
1
____________
=
(r − 1)(r + 1)
n
1
1
− _______
∑[ _______
2(r − 1) 2(r + 1) ]
r=2
n
1
= __
2
1 − _____
1
∑( _____
r − 1 r + 1)
r=2
[
]
3 − __
1 __
1 − _____
1
= __
2 [2 n n + 1]
1 1 + __
1 − __
1 − _____
1
= __
2
2 n n+1
[
]
3n + 3n − 2n − 2 − 2n
1 _____________________
= __
]
2[
2n(n + 1)
3n − n − 2
1 ___________
= __
2 [ 2n(n + 1) ]
+ 2)(n − 1)
1 ______________
= __
[2 (3n2n(n
+ 1) ]
3n(n + 1) − 2(n + 1) − 2n
1 _______________________
= __
2
2n(n + 1)
2
2
n
⇒2
(3n + 2)(n − 1)
1
= ______________
∑ ____________
(r − 1)(r + 1)
2n(n + 1)
r=2
n
Hence,
178
(3n + 2)(n − 1)
2
= ______________
∑ ____________
(r − 1)(r + 1)
2n(n + 1)
r=2
Substitute values for r
1 − _____
1 and add:
into ____
r−1 r+1
1
r=2:
1 − __
3
1
1
__
r=3:
+ − __
2 4
1 − __
1
r=4:
+ __
3 5
//
/
1 − _____
1
r = n − 2 : + _____
n−3 n−1
1 − __
1
r = n − 1 : + _____
n−2 n
1
1 − _____
r=n:
+ _____
n−1 n+1
MODULE 2tCHAPTER 8
E X A M P L E 18
Given that 3r(r + 1) = r(r + 1)(r + 2) − (r − 1)(r)(r + 1), show that
n
∑r (r + 1) = __13 n(n + 1)(n + 2)
r=1
SOLUTION
Since 3r(r + 1) = r(r + 1)(r + 2) − (r − 1)(r)(r + 1)
n
∑
r=1
n
3r(r + 1) =
∑[r (r + 1)(r + 2) − (r − 1)(r)(r + 1)]
r=1
Expanding the right-hand side, we have
= [(1)(2)(3) − (0)(1)(2)] + [(2)(3)(4) − (1)(2)(3) ] + [(3)(4)(5) − (2)(3)(4) ] + . . .
+[(n − 1)n(n + 1) − (n − 2)(n − 1)n] + [n(n + 1)(n + 2) − (n − 1)n(n + 1)]
= n(n + 1)(n + 2)
n
∴3
∑r (r + 1) = n(n + 1)(n + 2)
∑r (r + 1) = __31 n(n + 1)(n + 2)
r=1
n
r=1
E X A M P L E 19
1
1
__
__
2
2
1
________
_____________
______________
Show that
−
.
=
n(n + 1)(n + 2) n(n + 1) (n + 1)(n + 2)
∞
N
∑
∑
1
1
______________
______________
and deduce
.
Find
n(n
+
1)(n
+
2)
n(n
+
1)(n
+ 2)
n=1
n=1
SOLUTION
1
__
1
__
n(n + 1)
(n + 1)(n + 2)
1
__
1
__
(n + 2) − (n)
2
2
2
2
________
− _____________
= ______________
n(n + 1)(n + 2)
1n
1 n + 1 − __
__
2
2
______________
=
n(n + 1)(n + 2)
1
= ______________
n(n + 1)(n + 2)
∑[
1
1
__
__
N
2
2
1
________
_____________
______________
−
∴
=
(n + 1)(n + 2)
n = 1 n(n + 1)(n + 2)
n = 1 n(n + 1)
N
∑
N
∑[
]
1
1
1
________
− _____________
= __
2 n = 1 n(n + 1) (n + 1)(n + 2)
]
[
1 ____
1 − ____
1 + ____
1 − ____
1 +...
= __
2 1(2) 2(3) 2(3) 3(4)
1
1
1
1
− ________
+ ________
− _____________
+ ________
(N − 1)N N(N + 1) N(N + 1) (N + 1)(N +2)
[
1 __
1 − _____________
1
= __
2 2 (N + 1)(N + 2)
]
]
1 − ______________
1
= __
4 2(N + 1)(N + 2)
179
M O DUL E 2
As N → ∞, (N + 1)(N + 2) → ∞
1 − ______________
1
1
1
Therefore ______________
→ 0 and __
→ __
4 2(N + 1)(N + 2)
4
2(N + 1)(N + 2)
∞
Hence,
1
1.
= __
∑ ______________
n(n + 1)(n + 2) 4
n=1
n
E X A M P L E 20
Find
∑[
r=1
]
1 − _______
1
__
. Hence find
r 2 (r + 1)2
∞
1
∑ [ __r1 − _______
(r + 1) ]
.
2
r=1
2
n
SOLUTION
1
1 − __
1 + __
1 − __
1 + . . . + _______
= __
∑ [ __r1 − _______
( (n −1 1) − __n1 )
(r + 1) ] ( 1
2 ) (2
3 )
2
r=1
2
2
2
2
(
1 − _______
1
+ __
n2 (n + 1)2
2
2
2
)
1
= 1 − _______
(n + 1)2
1
As n → ∞, (n + 1)2 → ∞ so _______
→0
(n + 1)2
∴
Try these 8.4
∞
1
=1
∑ [ __r1 − _______
(r + 1) ]
r=1
(a)
2
2
1
Express __________
in partial fractions. Hence, find
r 2 + 5r + 6
n
1
.
∑ __________
r + 5r + 6
r=1
2
∞
Deduce
1
.
∑ __________
r + 5r + 6
r=1
2
(b) Given that 2r ≡ r(r + 1) − (r − 1)r, find
n
∑r.
r=1
1
(c) Express ______________
in partial fractions. Find
(2r + 1)(2r + 3)
∞
1
∑______________
(2r + 1)(2r + 3)
r=1
Convergence of a series
-FUSn = u1 + u2 + u3 + . . . + un*GMJN Sn = S UIFTFSJFTJTDPOWFSHFOUBOEUIF
n→∞
TFSJFTDPOWFSHFTUPS*GMJN Sn = ∞PSMJN Sn = −∞PSMJN SnEPFTOPUFYJTU
n→∞
n→∞
n→∞
UIFOUIFTFSJFTJTEJWFSHFOU
n
E X A M P L E 21
SOLUTION
Given that
r=1
n(n + 1)
Let Sn = ________
2
n(n + 1)
MJN Sn =MJN ________
=∞
n→∞
n→∞
2
4JODFMJN Sn = ∞ UIFTFSJFTEJWFSHFT
n→∞
180
n(n + 1)
, is the series convergent?
∑r = ________
2
MODULE 2tCHAPTER 8
n
3 − __
1 __
1
1 + _____
1 , is this series convergent?
= __
∑( ______
r − 1) 4 2 (n n + 1)
E X A M P L E 22
Given that
SOLUTION
3 − __
1
1 __
1 + _____
Let Sn = __
4 2 n n+1
3 − __
1 __
1 + _____
1
MJN S = MJN __
n →∞ n
n→∞ 4
2 n n+1
1−_
1 MJN _____
1
3−_
1 MJN __
= MJN _
n→∞ 4
2 n→∞ n 2 n→∞ n + 1
3
1 = 0 and MJN _____
1
=_
since MJN __
=0
n→∞ n
n→∞ n + 1
4
3.
3, the series converges to the value __
Since MJN Sn = __
n→∞
4
4
2
r=2
(
[
)
)]
(
(
( )
(
) )
Tests for convergence of a series
There are many tests we can use to decide whether a series converges or diverges. In this
section we will discuss the divergence test, the integral test and D’Alembert’s ratio test.
Theorem
n
If
∑u converges then lim
n=1
n
n→∞
un = 0.
Note: The converse is not true, i.e. if lim un = 0 the series does not necessarily
n→∞
converge.
Integral test
Suppose that f(x) is a positive decreasing function for x ⩾ k and that f(n) = un.
∞
If
∞
n
∫k f(x)dx is convergent then ∑un is also convergent and if ∫k f(x)dx is divergent
∞
then
∑u is also divergent.
n=1
n=k
n
Note that this test does not give the value that the series converges to if it converges;
it simply tells us whether the series converges or diverges.
∞
∑( __n1 ) diverges.
E X A M P L E 23
Show that
SOLUTION
1,
Let f(x) = __
x
n=1
∞
∞
∫1 f(x)dx = ∫1 __x1 dx = ∞
1 + (1) __
1 + (1) __
1 + . . . + (1) __
1
...
Area of the rectangles = (1)(1) + (1) __
n +
2
2
3
1 + __
1 + __
1+...
1 + __
= 1 + __
2 3 4 5
∞
1
__
=
n
()
()
()
( )
∑( )
n=1
181
M O DUL E 2
Since the rectangles overestimate the area under the curve,
∞
∞
y
∑( ) ∫ x
⇒
1
__
n >
n=1
∞
1
__
1 dx = ∞
y = 1x
∞
( __n1 ) > ∞ ⇒ ∑
( __n1 ) = ∞
∑
n=1
n=1
∞
Hence the series
∑ ( __n1 ) diverges.
n=1
This series is known as the
harmonic series.
∞
∑ ne
E X A M P L E 24
Determine if
SOLUTION
Let f(x) = xe−x
−n2
n=0
x
0
1
2
3
4
5
6
7
8
9
10
11
converges.
2
∞
∞
∫0 f(x)dx = ∫0 xe−x dx
1 e−x ∞
= [ − __
]0
2
2
2
1
= __
2
Since
∞
∫0
∞
f(x)dx converges,
Divergence test
If lim un ≠ 0 then
n→∞
∑ ne
−n2
n=0
will also converge.
∑u will diverge.
n
Note
The converse is
not true.
n
n
convergent?
∑( _____
n + 1)
E X A M P L E 25
Is
SOLUTION
n
1
lim _____
= lim 1 − _____
n→∞
n+1
n+1
=1
n
_____
Since lim
≠ 0, the series diverges.
n→∞ n + 1
n=1
n→∞
(
(
)
(
)
)
D'Alembert’s ratio test
∞
D’Alembert’s ratio test states that for a series of positive terms of the form
u
| |
u
(ii) if lim | _____
u | > 1, then the series diverges;
u
(iii) if lim | _____
u | = 1, then we need to test further.
n+1
_____
(i) if nlim
un < 1, then the series converges;
→∞
n+1
n→∞
n
n+1
n→∞
182
n
∑u
n=0
n
MODULE 2tCHAPTER 8
∞
2
converges.
∑ ____
3
n−1
n
E X A M P L E 26
Determine whether the series
SOLUTION
2n−1 ,
2n
Let un = ____
un+1 = ____
n
3
3n+1
Now
2n
____
u
3n
3n+1 = ____
n+1
2n × ____
2n−(n−1) = __
2
____
_____
= _______
n−1
n−1
n+1
un = ____
3
2
2
3
3n+1−n
3n
un+1
> 1 ⇒ the series converges.
∴ lim ____
n → ∞ un
|
n=1
|
EXERCISE 8B
n
1
2
Find
1
.
∑________
(4r − 1)
2
r=1
1 − 2r .
1 . Show that f (r) − f (r − 1) = ________
Let f(r) = __
r2n
r2(r − 1)2
1 − 2r .
_________
Hence find
2
2
r = 1 r (r − 1)
∑
3
Given that f (r + 1) = (r + 2)2, show that f(r + 1) − f(r) = 2r + 3.
n
4
n
∑ r.
3r + 3r + 1 find
3r + 3r + 1 .
1 − _______
1
___________
Given that __
≡ ___________
∑
r
(r + 1)
r (r + 1)
r (r + 1)
Using this result find
∑
r=1
2r + 3. Deduce
2
3
3
3
3
r=1
n
r=1
2
3
3
Show that the series converges and find the sum to infinity.
5
Show that (r + 4)(r + 5)(r + 6) − (r + 3)(r + 4)(r + 5) = 3(r + 4)(r + 5).
n
∑(r + 4)(r + 5). Decide whether this series converges or diverges.
1
1
Express ___________
in partial fractions. Hence find ∑ ___________
.
4r + 8r + 3
4r + 8r + 3
Hence find
6
r=1
2
n
r=1
Show that Sn converges and find the sum to infinity.
7
1
Express ____________
in partial fractions. Hence find
(r − 1)(r − 2)
n
2
n
1
.
∑____________
(r − 1)(r − 2)
r=3
3
.
∑ ______________
(3r − 1)(3r + 2)
8
Find
9
−4
A + ______
B find A and B.
Given that ______________
≡ ______
(4r + 1)(4r − 3) 4r + 1 4r − 3
r=1
n
−4
. Deduce the sum to infinity of the series.
∑______________
(4r + 1)(4r − 3)
Hence find
10 Prove that
r=1
n
n .
2
= _____
∑____________
(r + 1)(r + 2) n + 2
r=1
N
11 Find
∑ [e
nx
n=1
− e(n−1)x].
183
M O DUL E 2
SUMMARY
Series
Sum of series
n
∑c = nc
Method of differences
Convergent series
r =1
n
n
r =1
r =1
Let ∑vr = ∑{f(r + 1) – f (r)}
∑cur = c ∑ur
∑(ur + vr ) = ∑ur + ∑vr
n
∑r =
r =1
n(n + 1)
2
Expand the series
lim S
n→∞ n
=l
Divergent series: lim Sn = ∞, lim Sn = –∞ or
n→∞
lim S
n→∞ n
n→∞
cannot be determined
Cancel terms
D’Alembert’s ratio test
Find the required sum
n
∑r2 =
r =1
n(n + 1) (2n + 1)
6
∞
∑un converges
n=0
when lim
n→∞
n
∑r3 =
r =1
n2(n + 1)2
4
If lim
n→∞
2n
2n
n
r=1
r=1
∑ur = ∑ur – ∑ur
r = n+1
If lim
n→∞
|uu |<1
n +1
n
| u u | > 1, the series diverges
n +1
n
| u u | = 1, test further.
n +1
n
Integral test
Let f(x) be a positive decreasing function, x ≥ k,
∞
and f(n) = un. If ∫k f(x) dx is convergent
∞
∞
u
is
also
convergent
and if ∫k f(x) dx is
then ∑ n
n=k
∞
divergent then ∑ un is also divergent.
n=k
Divergence test
If lim un ≠ 0 then ∑un
n→∞
will diverge. The converse is not true.
Checklist
Can you do these?
■ Identify the nth partial sum of a series.
■ Write a series in sigma notation.
■ Identify a convergent series.
n
n
n
■ Find the sum of a series using the standard results for ∑1 r, ∑1 r 2, ∑1 r 3.
■ Use the summation laws.
■ Use the method of differences to find the sum of a series.
■ Identify whether a series converges or diverges.
184
MODULE 2tCHAPTER 9
CHAPTER 9
Principle of Mathematical Induction
(PMI): Sequences and Series
At the end of this chapter you should be able to:
■ prove statements true for sequences using mathematical induction.
■ prove statements true for a series using mathematical induction
KEYWORDS/TERMS
TFRVFODFTtTFSJFTtNBUIFNBUJDBM
JOEVDUJPO
185
M O DUL E 2
The principle of mathematical induction (PMI) was introduced in Unit 1.
The four steps for PMI are
Step 1
Prove the statement is true for n = 1.
Step 2
Assume the statement is true for n = k.
Step 3
Prove the statement is true for n = k + 1.
Step 4
Deduce that, using PMI, the statement is true for all integers.
PMI and sequences
EXAMPLE 1
A sequence u1, u2, u3, . . . of integers is defined by u1 = 1 and un+1 = 2un + 3.
Prove by induction that, for all n ≥ 1, un= 2n+1 − 3.
SOLUTION
RTP: un = 2n+1 − 3
Note
RTP stands for
‘it is Required To
Prove that’.
The problem gives u1= 1 and un+1 = 2un + 3. We will use this information in our
proof.
When n = 1, u1 = 21+1 − 3
= 22 − 3
=4−3
=1
∴ When n = 1, un = 2n+1 − 3 is true.
Assume true for n = k, i.e. uk = 2k+1 − 3
RTP:
true for n = k + 1, i.e. uk+1 = 2(k+1)+1 − 3
Proof: Since un+1 = 2un + 3,
when n = k, uk+1 = 2uk + 3
By our assumption uk = 2k+1 − 3
Substituting into uk+1, we have
uk+1 = 2[2k+1 − 3] + 3
= 2 × 2k+1 − 3 × 2 + 3
186
MODULE 2tCHAPTER 9
= 2(k+1)+1 − 6 + 3
= 2(k+1)+1 − 3
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, un = 2n+1 − 3 is true for all integers.
EXAMPLE 2
un
A sequence u1, u2, u3, . . . is defined by u1 = 1 and un+1= ______
.
un+ 2
1 .
Show by induction that, for all n ≥ 1, un = ______
2n − 1
SOLUTION
1 for all n ≥ 1.
RTP: un = ______
2n − 1
un
We have u1 = 1 and un+1 = ______
un + 2
1 =1
When n = 1, u1 = ______
21 − 1
1 is true
∴ when n = 1, un= ______
2n − 1
1
Assume true for n = k, i.e. uk = ______
2k − 1
1
true for n = k + 1, i.e. uk+1 = ________
2k + 1 − 1
un
Proof: Since un+1 = ______
un + 2
uk
uk+1 = ______
uk + 2
RTP:
1 , we have
Substituting uk = ______
k
2 −1
uk+1
1
______
1
______
1 +2
______
2k − 1
1 + 2 (2k − 1)
____________
2k − 1
2k − 1 = ____________
2k − 1
= _________
1
______
2k − 1
= ___________
1 + 2k+1 − 2
___________
2k − 1
1 × ________
2 −1
= ______
2k − 1 2k+1 − 1
k
1
= ________
2k + 1 − 1
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, the statement is true for all integers.
187
M O DUL E 2
EXAMPLE 3
Prove that 2n > 1 + n for all n ≥ 2.
SOLUTION
When n = 2, 22 = 4
⇒ 2n > 1 + n for n = 2
(since 4 > 3)
Assume true for n = k, i.e. 2k > 1 + k
RTP:
Proof:
true for n = k + 1, i.e. 2k+1 > (k + 1) + 1
2k+1 = 2k(2)
Since 2k > 1 + k
2 × 2k > 2(1 + k)
2k+1 > 2(k + 1)
2k+1 > 2k + 2
i.e. 2k+1 > (k + 2) + k
k > 0 ⇒ (k + 2) + k > k + 2
∴ 2k+1 > k + 2
2k+1 > (k + 1) + 1
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, the statement is true for all n ≥ 2.
EXAMPLE 4
1 , where a < __
1.
A sequence of positive numbers is defined by an+1 = an2 + __
1
4
2
1 for all positive integers n.
Prove by induction that an+1 < __
2
Hence show that an+1 > an.
SOLUTION
1 , ∀ n ≥ 1.
RTP: an+1 < __
2
1
Proof: Since an+1 = an2 + __
4
1
when n = 1, a2 = a 21 + __
4
1 , a2 < __
1
Since a1 < __
2 1 4
1 < __
1 + __
1
⇒ a21 + __
4 4 4
1 __
1
⇒ a21 +__
4<2
188
Remember
The symbol ∀
means ‘for all’.
MODULE 2tCHAPTER 9
1
Since a2 = a21 + __
4
1
a2 < __
2
1 for n = 1
Hence an+1 < __
2
1
Assume that the statement is true for n = k, that is, ak+1 < __
2
1
RTP: the statement is true for n = k + 1, that is, ak+2 < __
2
1
Proof: Since ak+1 < __
2
1
(squaring both sides)
a2k+1 < __
4
1 < __
1 + __
1
a2k+1 + __
4 4 4
1 , we have a < __
1
Since ak+2 = a2k+1 + __
k+2
4
2
1
Hence, when n = k + 1, an+1 < __
2
1 , ∀ n ≥ 1.
Therefore by PMI, an+1 < __
2
Next, we need to show that an+1 > an
1
Since an+1 = a2n + __
4
1 −a
an+1 − an = an2 + __
n
4
1 2
= an− __
(completing the square)
2
1 2 > 0, because when we square a number it is positive and also
Now an − __
2
1
an < __
2
(
(
)
)
∴ an+1 − an > 0
⇒ an+1 > an
Try these 9.1
(a) Prove by the principle of mathematical induction that
n × (n − 1) × . . . × 3 × 2 × 1 > 2n for any integer n ≥ 4.
(b) A sequence u1, u2, …, un is defined as follows
u1 = 1, u2 = 2 and un+2 = un+1 + un for n ≥ 1.
Prove by induction that u2n+1 − un un+2 = (−1)n−1 for all positive integers n.
189
M O DUL E 2
PMI and series
n
∑
n(n2 − 1)
r(r − 1) = ________.
3
r=1
EXAMPLE 5
Prove by induction that
SOLUTION
When n = 1, LHS = 1(1 − 1) = 0
1(12 − 1) 1(0)
RHS = ________ = ____ = 0
3
3
∴ LHS = RHS
n
n(n − 1)
when n = 1.
∑r(r − 1) = ________
3
k(k − 1)
Assume true for n = k, i.e. ∑r (r − 1) = ________
3
Hence,
2
r=1
k
2
r=1
RTP:
k+1
∑
r=1
true for n = k + 1, i.e.
(k + 1)( (k + 1)2 − 1 )
r (r − 1) = __________________
3
k
Proof: The LHS can be split into
∑r (r − 1) + (k + 1)th term
r=1
The (k + 1)th term can be found by substituting r = k + 1 into r(r − 1).
∴ (k + 1)th term = (k + 1)(k + 1 − 1)
= k(k + 1)
k+1
k
∑r (r − 1) = ∑r (r − 1) + k(k + 1)
r=1
r=1
k
Substituting
k (k − 1)
, we have
∑r (r − 1) = ________
3
2
r=1
k+1
∑
k (k2 − 1)
r (r − 1) = ________ + k(k + 1)
3
r=1
k(k + 1)(k − 1)
= _____________ + k(k + 1)
3
k(k + 1)
= ________ ( k − 1 + 3 )
3
k(k + 1)(k + 2)
= ______________
3
(k + 1)(k2 + 2k)
= ______________
3
(k + 1) ( (k + 1)2 −1 )
= __________________
3
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, the statement is true for all integers.
190
MODULE 2tCHAPTER 9
n
n .
1
= _____
∑_______
r (r + 1) n + 1
EXAMPLE 6
Prove by induction that
SOLUTION
1
1
When n = 1, LHS = ________
= __
1(1 + 1) 2
r=1
1
1 = __
RHS = _____
1+1 2
n
Hence,
n for n = 1.
1
= _____
∑ _______
r(r + 1) n + 1
r=1
k
Assume true for n = k, i.e.
k
1
= _____
∑_______
r(r + 1) k + 1
r=1
k+1
RTP:
true for n = k + 1, i.e.
k+1
Proof:
∑
r=1
k+1
1
= __________
∑ _______
r(r + 1) (k + 1) + 1
r=1
1
_______
can be split into
r(r + 1)
k
1
+ (k + 1)th term.
∑ _______
r(r + 1)
r=1
The (k + 1)th term can be found by replacing r = k + 1 into
1
1
_______
= ____________
r(r + 1) (k + 1)(k + 2)
k+1
∴
k
1
1
+ ____________
∑ r (r + 1) ∑ _______
r(r + 1) (k + 1)(k + 2)
r=1
1
_______
=
r=1
k
1
+ ____________
= _______
(k + 1) (k + 1)(k + 2)
k(k + 2) + 1
= ____________
(k + 1)(k + 2)
k + 2k + 1
= ____________
(k + 1)(k + 2)
2
(k + 1)2
= ____________
(k + 1)(k + 2)
k+1
= _____
k+2
k+1
= __________
(k + 1) + 1
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, the statement is true for all integers.
191
M O DUL E 2
n
2n + 1 for n ≥ 2.
3 − ________
2
= __
∑____________
(r − 1)(r + 1) 2 n(n + 1)
EXAMPLE 7
Prove by induction that
SOLUTION
We start by proving the statement is true for n = 2.
r=2
2
2
= __
When n = 2, LHS = _____________
(2 − 1)(2 + 1) 3
2(2) + 1 __
5 = __
3 − ________
2
= 3 − __
RHS = __
2 2(2 + 1) 2 6 3
∴ LHS = RHS
n
Hence,
2n + 1
3 − ________
2
= __
∑ ____________
(r − 1)(r + 1) 2 n(n + 1)
r=2
for n = 2.
k
Assume true for n = k, i.e.
RTP:
2k + 1
3 − ________
2
= __
∑ ____________
(r − 1)(r + 1) 2 k(k + 1)
r=2
true for n = k + 1, i.e.
k+1
2(k + 1) + 1
3 − ________________
2
= __
∑ ____________
2
(r − 1)(r + 1)
(k + 1)(k + 1 + 1)
r=2
k+1
Proof:
∑
2
____________
=
r = 2 (r − 1)(r + 1)
k
2
2
+ ________
∑ ____________
(r − 1)(r + 1) k(k + 2)
r=2
2k + 1 + _______
3 − _______
2
= __
2 k(k + 1) k(k + 2)
[
2k + 1 − _____
3 − __
1 ______
2
= __
2 k k+1
k+2
]
[
(2k + 1)(k + 2) − 2(k + 1)
3 − __
1 _______________________
= __
2 k
(k + 1)(k + 2)
[
]
[
]
3 − __
2k2 + 3k
1 ____________
= __
2 k (k + 1)(k + 2)
k(2k + 3)
3 − __
1 ____________
= __
2 k (k + 1)(k + 2)
]
2(k + 1) + 1
3 − ________________
= __
2 (k + 1)(k + 1 + 1)
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, the statement is true for all n ≥ 2.
192
MODULE 2tCHAPTER 9
n
∑3
1 (3n − 1).
= __
2
EXAMPLE 8
Prove by induction that
SOLUTION
When n = 1, LHS = 31−1 = 30 = 1
r−1
r=1
1 (3 − 1) = __
1 (2) = 1
1 (31 − 1) = __
RHS = __
2
2
2
∴ LHS = RHS
n
Hence,
∑3
r−1
r=1
1 (3n − 1) when n = 1.
= __
2
k
Assume true for n = k, i.e.
∑3
1 (3k − 1)
= __
2
r−1
r=1
k +1
true for n = k + 1, i.e.
RTP:
k+1
r=1
r−1
r=1
1 (3k+1 − 1)
= __
2
k
∑
Proof:
∑3
3r−1 =
∑3
r−1
r=1
+ 3(k + 1) −1
1 ( 3k − 1 ) + 3k
= __
2
1 [ 3k − 1 + 2(3k) ]
= __
2
1 [ 3(3k) − 1 ]
= __
2
1 [ 3k+1 − 1 ]
= __
2
Hence, when true for n = k, the statement is also true for n = k + 1.
Hence, by PMI, the statement is true for all integers.
EXERCISE 9A
Prove the statement in questions 1–10 by PMI.
n
1
2
3
4
n(n + 3)
∑(r + 1) = ________
2
∑(r + 1)(r + 2) = __n3 (n + 6n + 11)
∑(2r − 1) = n
∑2 = 2(2 − 1)
r=1
n
2
r=1
n
2
r=1
n
r
n
5
n
r=1
∑ ( __31 ) = __21(1 − 3
r=1
r
−n)
193
M O DUL E 2
n
6
∑(4r + 5) = n(2n + 7)
r=1
n
7
n(n + 1)(4n − 1)
∑r(2r − 1) = _______________
6
r=1
n
8
∑(3r + 1)(r + 1) = __n2 (2n + 7n + 7)
2
r=1
n
9
∑r(r
2
r=1
n(n + 1)(n2 + n + 2)
+ 1) = __________________
4
n
10
∑(2r + 1)(4r − 1) = __n3 (8n + 15n + 4)
2
r=1
11 A sequence whose first term is a1 = 1 is defined by an+1 = 5an + 2.
(a) Find the first five terms of the sequence.
3 (5n−1) − __
1.
(b) Prove by induction that an= __
2
2
12 A sequence is defined by a0 = 2 and an+1 = 1 − 2an for n ≥ 0.
5 (−2)n for all non-negative integers n.
1 + __
Prove by induction that an = __
3 3
13 A sequence whose first term is such that a1= 1 is defined by
6 . Prove by induction that a < 4 for all n ≥ 1.
an+1 = 5 − _____
n
an + 2
14 The sequence of real numbers u1, u2, . . . is such that u1 = 1 and
(
)
1 3
3n−1
un+1 = un + __
u . Prove by induction that, for n ≥ 1, un ≥ 2 .
n
15 The sequence of real numbers u1, u2, . . . is such that u1 = 1 and
(
)
1 2
2n−1
un+1 = un + __
u . Prove by induction that, for n ≥ 1, un ≥ 2 .
n
16 Given a sequence a1, a2, . . . , an is such that a1 = 1 and an = an−1 + 3, n ≥ 2,
prove by induction that an= 3n − 2, for all positive integers n.
17 Prove by the principle of mathematical induction than 2n > n for all positive
integers n.
18 A sequence of real numbers a0, a1, a2, . . . , an is defined by a0= 0 and
an+1 = 2n − an for n = 1, 2, 3, . . . Prove by induction that 2an = 2n − 1 +
(−1)n for n ≥ 1.
194
MODULE 2tCHAPTER 9
SUMMARY
PMI
Prove the statement true for n = 1
Assume the statement true for n = k
Prove the statement true for n = k + 1
Hence by PMI the statement is true
Checklist
Can you do these?
■ Prove statements true for sequences using mathematical induction.
■ Prove statements true for series using mathematical induction.
195
M O DUL E 2
CHAPTER 10
Binomial Theorem
At the end of this chapter you should be able to:
■ use the factorial notation
■ use nCr
■ find the binomial expansion of (a + b)n for a positive integer n
■ find the term independent of x in an expansion
■ use the expansion for any real number n
■ find the region for which the expansion is valid
■ use binomial expansion and partial fractions.
KEYWORDS/TERMS
CJOPNJBMtGBDUPSJBMtDPNCJOBUJPOtJOEFQFOEFOUt
QBSUJBMGSBDUJPOT
196
MODULE 2tCHAPTER 10
DE FIN ITI ON
A binomial
expression is an
expression with
two terms,
e.g. a + b.
Pascal’s triangle
Look at the coefficients when we raise a binomial expression to a positive integer
power:
(a + b)0 = 1
1
(a + b)1 = 1a + 1b
Pascal’s triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1
(a + b)2 = 1a2 + 2ab + 1b2
1
1
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
2
1
3
1
3
1
The triangle representing the coefficients of the terms is known as Pascal’s
triangle. The first row of Pascal’s triangle represent (a + b)0 = 1, the second row
is the coefficients of a + b, the third row is the coefficients of (a + b)2, and so on.
We can use the triangle to obtain the coefficients of the terms of any expansion.
The row below is found by adding the terms to the left and right of the row above.
Example: Row 2 is
(0)
1
1
(0)
Row 3 becomes
0+1
1+1
1+0
i.e.
1
2
1
Row 4 becomes
0+1
1+2
2+1
1+0
i.e.
1
3
3
1
As the powers of a + b increase, so does the size of the triangle. After a while the
triangle becomes very large and not so useful. The alternative to Pascal’s triangle is
the binomial expansion.
Factorial notation
If we arrange two letters, A and B, in order we get
AB
BA
The number of arrangements is 2 × 1 = 2.
If we arrange three letters, A, B and C, in order we get
ABC
ACB
BAC
BCA
CAB
CBA
Number of arrangements = 3 × 2 × 1 = 6.
We represent 2 × 1 = 2!
(2! is read as ‘2 factorial’)
3 × 2 × 1 = 3!
4 × 3 × 2 × 1 = 4!
and
n × . . . × 4 × 3 × 2 × 1 = n!
197
M O DUL E 2
Therefore n!, the product of the first n natural numbers, represents the number of
ways of arranging n distinct objects in order.
Also 0! = 1 by definition.
EXAMPLE 1
8!
Find __
6!
SOLUTION
8!
__
6! =
8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 56
____________________________
6×5×4×3×2×1
8 × 7 × 6! = 56
8! = __________
or __
6!
6!
Now (n − 1)! = (n − 1) × . . . × 4 × 3 × 2 × 1
n! = n × (n − 1) × . . . × 4 × 3 × 2 × 1
∴ n! = n(n − 1)!
EXAMPLE 2
Simplify: (n + 1)! − (n − 1)!
SOLUTION
(n + 1)! = (n + 1) × n × (n − 1) × . . . × 3 × 2 × 1
= (n + 1)(n)(n − 1)!
∴ (n + 1)! − (n − 1)! = (n + 1)n(n − 1)! − (n − 1)!
= (n − 1)![(n + 1)(n) − 1] = (n − 1)![n2 + n − 1]
EXAMPLE 3
2n + _____
n−1
Express as a single fraction _______
n!
(n + 1)!
SOLUTION
2n + (n + 1)(n − 1)
n − 1 = __________________
2n + _____
_______
(n + 1)!
(n + 1)!
n!
2n + n − 1
= ___________
2
(n + 1)!
+ 2n − 1
= ___________
(n + 1)!
n2
EXAMPLE 4
n + 2 − _______
n
Express as a single fraction _______
(n + 3)! (n + 2)!
SOLUTION
n
n + 2 − _______
_______
(n + 3)!
n + 2 − n(n + 3)
= ________________
(n + 2)!
(n + 3)!
n + 2 − − 3n
= _______________
(n + 3)!
n2
−n − 2n + 2
= _____________
(n + 3)!
2
Try these 10.1
198
Simplify
n!
(a) _______
(n + 3)!
(b) n! + (n + 1)! + (n − 2)!
Note
Since (n + 1)!
= (n + 1)n!, the
LCM is (n + 1)!
Note
(n + 3)! = (n + 3)
× (n + 2) × . . . ×
3× 2× 1
= (n + 3) (n + 2)!
MODULE 2tCHAPTER 10
Combinations
Instead of finding the number of ways of arranging n distinct objects in order, we can
look at the number of ways of choosing r objects from n distinct objects without any
particular order. This is called combination. We write nCr and read this in a number
of ways: ‘n C r’, ‘combine r objects from n distinct objects’ or ‘choose r from n distinct
objects’.
EXAMPLE 5
Find the number of ways of choosing two letters out of A and B.
SOLUTION
There is only one way: choose AB.
EXAMPLE 6
Find the number of ways of choosing two out of four different letters.
SOLUTION
Suppose the letters are ABCD. We can choose
AB AC AD BC BD CD
i.e. there are six ways of choosing two out of four different letters.
We can write this as 4C2 = 6
The number of ways of choosing two out of four distinct objects = 4C2 = 6.
EXAMPLE 7
Find the number of ways of choosing four out five different objects.
SOLUTION
Suppose the objects are ABCDE.
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
ABCD ABDE BCDE ACDE ABCE
5 ways
So 5C4 = 5.
General formula for nCr
4!
4!
Notice that 4C2 = 6 = _________
= ______
(4 − 2)!2! 2! × 2!
5!
Also 5C4 = 5 = _________
(5 − 4)!4!
Let us look at nC0
n!
n! = 1
nC = _________
= __
0
(n − 0)!0! n!
n × (n − 1)!
n!
nC = _________
= ___________ = n
1
(n − 1)!1!
(n − 1)!1!
nC
2
n(n − 1) × (n − 2)! n(n − 1)
n!
= _________
= _________________ = ________
2
(n − 2)!2!
(n − 2)!2!
In general, the number of ways of choosing r out of n distinct objects is
n!
nC = ________
r
(n − r)!r!
199
M O DUL E 2
EXAMPLE 8
Prove that nCr = nCn−r
SOLUTION
nC
r
Note
n!
= ________
(n − r)!r!
n!
= ___________________
(n − r)!(n − (n − r))!
= nCn−r
Since r! = (n − (n − r))!
n!
_________________
nC
n−r = (n − (n − r))!(n − r)!
EXAMPLE 9
Simplify nC3 + nC4
SOLUTION
nC
3
n!
= _________
(n − 3)!3!
nC
4
n!
= _________
(n − 4)!4!
nC
3
n!
n!
+ nC4 = _________
+ _________
(n − 3)!3! (n − 4)!4!
4(n!) + (n − 3)n! n![4 + n − 3]
= _______________ = ____________
(n − 3)!4!
(n − 3)!4!
(n + 1)(n!) (n + 1)(n)(n − 1)(n − 2)(n − 3)!
= __________ = ____________________________
(n − 3)!4!
(n − 3)!4!
(n + 1)n (n − 1)(n − 2)
= _____________________
24
Since
4! = 4 × 3! and
(n−3)! = (n−3)
× (n−4)! the
LCM is (n−3)! 4!
E X A M P L E 10
(n − 1)(n2 − 2n + 6)
Prove that nC2 + n−1C3 = __________________
6
SOLUTION
nC
2
n!
= _________
(n − 2)!2!
(n − 1)!
(n − 1)!
n−1C = _____________
= _________
3
(n − 1 − 3)!3! (n − 4)!3!
nC
2
(n − 1)!
_______
=
(n −2)
(n
− 1)(n − 2)!
____________
=n−1
(n − 2)!
(n − 2)! = (n − 2)(n − 3)(n − 4)!
3! = 3 × 2! LCM = (n − 2)!3!
(n − 1)!
(n − 4)!3!
n!
+ _________
+ n−1C3 = _________
(n − 2)!2!
3(n!) + (n − 2)(n − 3)(n − 1)!
= __________________________
(n − 2)!3!
3n(n − 1)! + (n2 − 5n + 6)(n − 1)!
= ______________________________
(n − 2)!3!
(n − 1)!(3n + n2 − 5n + 6)
= _______________________
(n − 2)!3!
(n − 1)(n2 − 2n + 6)
= __________________
6
Binomial theorem for any positive integer n
For any positive integer n:
(a + x)n = an + nC1 an−1 x + nC2 an−2x2 + nC3 an−3x3 + ∙ ∙ ∙ + xn
for all values of x.
200
MODULE 2tCHAPTER 10
PROOF
Combinatorial proof
(a + x)n = (a + x)(a + x)(a + x) × . . . × (a + x)
n times
The first term on the RHS is found by multiplying a n times, which gives an. The term
in x is found by multiplying (n − 1) a’s and one x. Since there are n brackets the x
must come from one of these brackets. The number of ways of choosing 1 out of n
distinct brackets is nC1. Therefore the second term is nC1 an−1x
Similarly the third term is found by multiplying (n − 2) a’s and two xs from two
brackets: nC2 an−2x2, and so on
∴ (a + x)n = an + nC an−1x + nC an−2x2 + . . . + xn
1
2
Proof by induction
(a + x)n = an + nC1 an−1 x + . . . + xn
RTP:
When n = 1, (a + x)1 = a + x
∴ (a + x)n = an + nC1 an−1x + . . . + xn for n = 1
Assume true for n = k, i.e.
(a + x)k = ak + kC1 ak−1 x + . . . + xk
true for n = k + 1, i.e.
RTP:
(a +
x)k+1
= ak+1 + k+1C1 akx + k+1C2 ak−1x2 + . . . + xk+1
Proof:
 a + x)k+1 = (a + x)k (a + x)
= ( ak + kC1 ak−1 x + . . . + xk ) (a + x)
= ak+1 + ak x + kC1 ak x + kC1 ak−1 x2 + kC2 ak−1 x2 + . . . + xk+1
= ak+1 + ( 1 + kC1 ) ak x + ( kC1 + kC2 ) ak−1 x2 + . . . + ( kCr−1 + kCr ) ak−r+1 x r
+ . . . + xk+1
We need to show that kCr−1 + kCr = k+1Cr
Proof:
kC
r −1
k!
k!
+ ________
+ kCr = _________________
(k − r + 1)!(r − 1)! (k − r)!r!
r(k!) + (k − r + 1)k! k![r + k − r + 1]
= _________________ = _______________
(k − r + 1)!r!
r!(k − r + 1)!
(k + 1)!
k![k + 1]
= ____________ = ____________ = k+1Cr
r!(k − r + 1)! r!(k + 1 − r)!
So (a + x)k+1 = ak+1 + ( k+1C1 ) akx + k+1C2ak−1x2 + . . . + k+1Cr ak−r+1xr + . . . + xk+1
Alternative proof of k+1Cr = kCr + kCr −1
First we mark one of (k + 1) items X. We then choose our r items either with or without X.
The number of ways of choosing the r items without X is kCr (i.e. we choose the r items
out of the remaining k items).
The number of ways of choosing the r items with X is kCr × 1C1 = kCr −1 (i.e. we choose
the r − 1 items out of the remaining k items and we choose the object X).
Therefore the number of ways of choosing the r items = the number of ways of
choosing the r items without X plus the number of ways of choosing the r items with
X = k+1Cr = kCr + kCr −1.
201
M O DUL E 2
Now that we have looked at the proof of the binomial expansion, let us look at some
of the properties of the series.
(a + x)n = an + nC1an−1x + nC2 an−2 x2 + . . . + xn
=
n
nC an−r xr, for all values of x.
r
∑
r=0
(i) The series is finite.
(ii) The series consists of (n + 1) terms.
(iii) The term ur+1 = nCr an−rxr.
(iv) Coefficient of xr = nCran−r.
E X A M P L E 11
Write down the first three terms in the expansion in ascending powers of x of:
(a) (2 + x)6
SOLUTION
(b) (1 + 2x)8
(a) Comparing (a + x)n with (2 + x)6, we see that
a = 2, x ≡ x, n = 6
Substituting in
(a + x)n = an + nC1an−1 x + nC2an−2 x2 + . . . + xn
we get
(2 + x)6 = 26 + 6C1(2)5(x) + 6C2(2)4(x2) + . . .
(up to the third term)
= 64 + 192x + 240x2 + . . .
(b) Comparing (1 + 2x)8 with (a + x)n, we see that
a = 1, n = 8 and x is replaced by 2x
Substituting in (a + x)n = an + nC1 an−1 + . . . + xn
In the third
term you must
square 2x.
we get
(1 + 2x)8 = 18 + 8C117(2x) + 8C216(2x)2 + . . .
(up to the third term)
= 1 + 16x + 112x2 + . . .
E X A M P L E 12
Obtain the expansion of (2 + 3x)4 in ascending powers of x.
SOLUTION
Since n is a positive integer,
use (a + x)n = an + nC1an−1 x + nC2 an−2 x2 + . . . + xn
where a = 2, n = 4 and x is replaced by 3x.
We have
(2 + 3x)4 = 24 + 4C123(3x) + 4C2(2)2 (3x)2 + 4C3(2)(3x)3 + (3x)4
= 16 + 96x + 216x2 + 216x3 + 81x4
202
MODULE 2tCHAPTER 10
8FDBOUBLFPVUTQFDJĕDUFSNTGSPNUIFFYQBOTJPOJGOFFEFEćFGPSNUIBUNBLFT
UIFUBTLFBTJFSJT a + x)n = ∑nr=0 nCr an−r xr
E X A M P L E 13
Find the third term of the expansion (2 + 3x)4.
SOLUTION
(2 + 3x)4 =
R em e m b e r
The series starts at
r = 0. So the first
term occurs when
r = 0, the second
term when r = 1
and so on.
n
4C (2)4−r(3x)r
r
∑
r=0
The third term occurs when r = 2.
∴ The third term is 4C2(2)4−2(3x)2 = 216x2
E X A M P L E 14
Find the tenth term of the expansion (1 − 2x)18.
SOLUTION
Write (1 − 2x)18 in summation form
(1 − 2x)18 =
18
18C 118−r(−2x)r
r
∑
r=0
The tenth term occurs when r = 9
i.e. the tenth term is 18C9(1)9(−2x)9 = −24 893 440x9
The term independent of x in an expansion
The term independent of x is the constant term in the expansion.
(
)
8
E X A M P L E 15
1 .
Find the term independent of x in the expansion of x + ___
2x
SOLUTION
1 =
Let us write this as x + ___
2x
(
8
1
) ∑ C (x) ( ___
2x )
8
8
r=0
r
8−r
r
The power of x in the term independent of x is 0.
1 r = x8−r ____
1
Now x8−r ___
2x
2r xr
= x8−r × x−r × 2−r
( )
(
)
= x 8−2r2−r
The power of x must be zero.
∴ 8 − 2r = 0
2r = 8
r=4
The term independent of x occurs when r = 4
1 4 = 8C × __
14
i.e. 8C4(x)4 ___
4
2x
2
35
70 = ___
= ___
8
16
( )
( )
203
M O DUL E 2
(
)
6
E X A M P L E 16
1 .
Find the term independent of x in the expansion of 2x2 − __
x2
SOLUTION
(
6
) ∑
6
1 =
2x2 − __
x2
r=0
6C (2x2)6−r
r
( )
1
−__
x2
r
Let us look at
1 r
(2x2)6−r − __2 = 26−r x2(6−r)(−1)r(x−2)r
x
= 26−r x12−2r(−1)r x−2r
( )
= x12−4r(2)6−r(−1)r
In the term independent of x the power of x is zero.
∴ 12 − 4r = 0
4r = 12
r=3
∴ The term independent of x is
1 3 = 6C 23(−1)3 = −160
6C (2x2)6−3 −__
3
3
x2
( )
EXERCISE 10A
1
Expand
(a) (1 + x)4
(b) (1 − x)5
(c) (1 + 2x)5
(d)
( 1 − __23x )
4
(e) (3 + x)6
2
Obtain the expansion of (3 − 2x)5 in ascending powers of x.
3
Write down the first four terms in the expansion of (1 − 2x)9, simplifying the
coefficients.
4
Write down the first three terms of the expansion of
1 x 10
(a) (4 − x)10
(b) (1 − x)15
(c) 2 + __
3
6
Find the coefficient of x in each of the following.
(
5
(a) (1 − 3x)10
6
(b) (2 + 3x)12
(c) (1 − 2x)9
(d) (3 + x)15
In the expansion of (1 − 3x)8, find
(a) the number of terms
7
)
(b) the term in x5
(c) the fifth term.
Find
(a) the fifth term in the expansion of (1 + 4x)7
(b) the sixth term in the expansion of (2 − x)11
2x
(c) the seventh term in the expansion of 3 − __
3
(
204
)
12
8
Find the coefficient of x3 in the expansion of (1 − x)(1 − 2x)6.
9
Given that the expansion of (2 − x)(3 + x)5 is a + bx + cx2 up to the third
term. Find the values of a, b and c.
MODULE 2tCHAPTER 10
10 The coefficient of the third term in the expansion of (2 − x)5 is equal to half the
coefficient of the fourth term in the expansion of (1 + ax)6. Find a.
In questions 11–14, find the term independent of x.
11
( __2x − x )8
13
( __x2 + 4x )
(x
)
1 − 3x
( ___
)
2x
3 − 5x3 10
12 __
2
2 15
14
3
4 6
2
15 Find the first three terms of the expansion, in ascending powers of x, of
(a) (1 + 2x)5
(b) (1 − 3x)5
Hence obtain the coefficient of x2 in the expansion of (1 − x − 6x2)5.
16 Write down and simplify the first three terms of the expansion of (2 + x)5 in
ascending powers of x. Given that the coefficients of x and x2 in the expansion
of (1 + px + qx2) (2 + x)5 are both 16, calculate the value of p and of q.
17 Find the first three terms of the expansion, in ascending powers of x, of
(1 − 4x)7. Hence obtain the coefficient of x2 in the expansion of
(1 + 2x − 3x2) (1 − 4x)7.
18 Write down and simplify the first three terms of the expansion, in ascending
powers of x, of
1x 6
(b) (2 + x)6
(a) 1 − __
4
Hence, or otherwise, obtain the coefficient of x2 in the expansion
3x − __
1 x2 6.
of 2 + __
4
4
19 Find, in its simplest form, the coefficient of y4 in the expansion of
1y 6
4 10
(a) 2 − __
(b) y − __
y
3
(
)
(
)
(
)
(
)
20 Given that the expansion of (1 − 2x)2 (1 + px)8 in ascending powers of x is
1 + 20x + qx2 + . . . , calculate the value of p and the value of q.
Extension of the binomial expansion
'PSBOZSFBMOVNCFSn
n n−
n n − n − 2)
+ x)n =+ nx + ________ x2 + ______________ x3 + 2!
3!
QSPWJEFEUIBU−< x <
]x]< /PUFUIBUUIFFYQBOTJPOJTJOĕOJUF
"OFYQBOTJPOPGUIFGPSN a + x)nNVTUCFSFXSJUUFOCFGPSFVTJOHUIFFYQBOTJPOBCPWF
x n
xn
__
n
(a + x)n = [ a ( 1 + __
a )] = a (1 + a)
x
This expansion is valid for −1 < __
a < 1, i.e. for −a < x < a.
205
M O DUL E 2
E X A M P L E 17
SOLUTION
Remember
When substituting, put (−x) in
brackets and
(−x)2 = x2.
1
__
Find the first three terms of the expansion (1 − x) 2. Write down the values of x
for which the expansion is valid.
n(n − 1)
1,
Using (1 + x)n = 1 + nx + ________ x2 + . . . where x ≡ −x, n = __
2
2!
(1 −
1
__
x) 2
( )( )
1
1 −__
__
2
2
1
__
________
(−x)2 + . . .
= 1 + (−x) +
2
2!
1 x2 + . . .
1 x − __
= 1 − __
2
8
Expansion is valid for −1 < −x < 1, i.e. for −1 < x < 1.
E X A M P L E 18
1 , stating the values of x for
Obtain the first four terms in the expansion of ______
1 − 2x
which the expansion is valid.
SOLUTION
1 = (1 − 2x)−1
______
1 − 2x
Since we are
replacing x
by −2x in our
expansion, the
region for which
the expansion is
valid is
−1 <−2x < 1
n n −
n n − n − 2)
6TJOH + x)n =+ nx + ________ x2 + ______________ x3 + 2!
3!
XIFSFn = −BOExJOUIFCJOPNJBMFYQSFTTJPOJTSFQMBDFECZ−2x,
we have
− −2)
− − −3)
− 2x)− =+ − −2x) + _________
−2x)2 + _____________
−2x)3 +
2!
3!
= 1 + 2x + 4x2 + 8x3 + . . .
1 < x < __
1.
where the expansion is valid for −1 < −2x < 1, i.e. for −__
2
2
1
__
E X A M P L E 19
Find the first four terms in the expansion of (4 + x)2, stating the values of x for
which the expansion is valid.
SOLUTION
Since (4 + x)2 is not in the form (1 + x)n, we write it in that form and then expand.
1
__
x __12 = 4__12 1 + __
x __21 = 2 1 + __
1 x __12
(4 + x)2 = 4( 1 + __
( 4)
4)
4
1
__
[
(
]
)
n(n − 1)
n(n − 1)(n − 2)
Using (1 + x)n = 1 + nx + ________x2 + ______________x3 + . . .
2!
3!
1 and x JOUIFCJOPNJBMFYQSFTTJPOJTSFQMBDFECZ __
1x
where n = __
4
2
1 __
1−1
1 __
1 − 1 __
1−2
__
__
1
__
2 2
2 2
2
1
1
1
1
1x
__________
________________
__
__
__
__
__
2
2
x +
x +
2 1+ x =2 1+
2!
3!
4
4
4
2 4
(
)
[
[
( )( )
( )(
( )( )
)
( )(
( )
( )( ) ( )
)(
3
1
1
1 −__
1 −__
__
__
__
2
2 ___
2 − 2 ___
2
2
x
1
1 x3
____________
________
__
+
=2 1+ x+
8
2
16
6
( 64 )
1 x ___
1 x2 + ___
1 x3
= 2 + __
512
4 − 64
1 x < 1, i.e. for −4 < x < 4.
This expansion is valid for −1 < − __
4
206
)
]
( )
3
]
MODULE 2tCHAPTER 10
E X A M P L E 20
Obtain the1 first three terms in the expansion, in ascending powers of x, of
__
(16 − 3x)4 and state the values of x for which the expansion is valid.
SOLUTION
We need to write (16 − 3x)4 in the form (1 + x)n and then expand.
1
__
[ (
1
__
3x
(16 − 3x)4 = 16 1 − ___
16
)]
1
__
4
(
)
(
1
__
3 x __14 = 2 1 − ___
3 x
= 164 1 − ___
16
16
)
1
__
4
n(n − 1)
Using (1 + x)n = 1 + nx + ________x2 + . . .
2!
3
1 and x JOUIFCJOPNJBMFYQSFTTJPOJTSFQMBDFECZ −___
where n = __
16 x
4
3
1 −__
__
1
__
4
3
3 2
4
3
1
___
___
__
_______
4
−___
x
we have 1 − x = 1 + −16 x +
16
4
16
2!
(
(
)
( )( )
)
(
)
3 x − _____
27 x2
= 1 − ___
8192
64
1
__
3x
∴ (16 − 3x)4 = 2 1 − ___
16
(
)
1
__
4
3 x − _____
27 x2
= 2 − ___
32
4096
3
16
___
16
___
This expansion is valid for −1 < −___
16 x < 1, i.e. for − 3 < x < 3 .
1
__
E X A M P L E 21
Obtain the first three terms in the expansion of (32 − 5x)5, stating the values of x for
which the expansion is valid.
SOLUTION
Write (32 − 5x)5 in the form (1 + x)n.
1
__
1
__
[ (
)]
(
)
5x
(32 − 5x)5 = 32 1 − ___
32
1
__
5x
= 325 1 − ___
32
1
__
5
1
__
5
(
5x
= 2 1 − ___
32
)
1
__
5
n(n − 1)
Using (1 + x)n = 1 + nx + ________ x2 + . . .
2!
5
1
with n = __, and x JOUIFCJOPNJBMFYQSFTTJPOSFQMBDFECZ −___
32 x
5
4
1 −__
__
1
__
5
5 2
5
5
5
1
___
___
__
_______
5
− x +
−___
x +...
2 1− x =2 1+
32
5
32
32
2!
(
)
[
( )(
[
( )( )
)
1 x − ___
1 x2
= 2 1 − ___
512
32
(
)
]
]
1 x − ___
1 x2
= 2 − ___
256
16
5
32
32
___
___
This expansion is valid for −1 < −___
32 x < 1, i.e. for − 5 < x < 5 .
207
M O DUL E 2
Approximations and the binomial expansion
E X A M P L E 22
SOLUTION
3
__
3
__
Using the first three terms of the expansion of (1 + x)2, estimate (1.02)2 to four
places of decimals.
3
__
n(n − 1)
Expanding (1 + x)2 using (1 + x)n = 1 + nx + ________x2 + . . .
2!
3
__
with n = , we get
2
3 __
1
__
3
__
2
2
3
____
__
(1 + x)2 = 1 + x +
x2 + . . .
2
2!
()
Any term containing x3 and
above will not affect our
result to 4 d.p.
The term in x3 gives
1
3 __
1 −__
__
2 3
1 3
2 2
__________
x = −___
16 x
3!
When x = 0.02, we have
1 (0.02)3 = − 0.000 0005
−___
16
3 x + __
3 x2 + . . .
= 1 + __
2
8
Substituting x = 0.02, we get
3
__
3 (0.02) + __
3 (0.02)2
(1 + 0.02)2 ≈ 1 + __
2
8
= 1 + 0.03 + 0.000 15
( )( )( )
= 1.0302 (4 d.p.)
3
__
∴ (1.02)2 ≈ 1.0302
5
_______
5 ___
E X A M P L E 23
Use the expansion of √ (1 + x) to estimate √ 33 to five places of decimals.
SOLUTION
√(1 + x) = (1 + x)5
5
_______
1
__
n(n − 1)
1 , we have
Using (1 + x)n = 1 + nx + ________x2 + . . . with n = __
5
2!
9
4
4
1 −__
1 −__
__
__
__
1
__
5 2
5 −5 3 . . .
5
5
1
__
______
__________
5
x +
x +
(1 + x) = 1 + x +
5
2!
3!
( )(
( )
6 x3 + . . .
1 x − ___
2 x2 + ____
= 1 + __
5
25
125
)
(up to the fourth term)
Since 33 = 32 + 1
5 ___
5 ______
√ 33 = √ 32 + 1
1
__
= (32 + 1)5
)]
[ (
1
= 32 ( 1 + ___
32 )
1
= 2( 1 + ___
32 )
1
__
1
= 32 1 + ___
32
5
1
__
1
__
5
5
1
__
5
1
__
6 x3 with x = ___
1 x − ___
2 x2 + ____
1 we have
Using (1 + x)5 = 1 + __
5
25
125
32
6 ___
1 __51 = 1 + __
1 ___
1 − ___
2 ___
1 2 + ____
1 3
1 + ___
5
32
32
25 32
125 32
=
1.006
173
34
5 ___
So √ 33 = 2 × 1.006 173 34 = 2.012 35 (to 5 d.p.)
(
208
)
( )
( )
( )
MODULE 2tCHAPTER 10
_____
E X A M P L E 24
1 + x in ascending powers of x up to and including the term in x2
Expand _____
1−x
181 .
1 , show that √__
where |x| < 1. By assuming x = __
5 ≈ ____
9
81
√
Note
1 + x = (1 + x)__12(1 − x)−__21
_____
1−x
Terms above
Expanding, we have
x2 have been
1
1 −__
ignored when
__
1
__
2 2 ...
2
1
1
1
______
__
__
__
2
expanding
.
.
.
2
(1 + x) = 1 + x +
=1+ x− x +
x +
2
2
8
2!
brackets.
3
1
__
−__
1
−__
1 (−x) + _________
2 − 2 (−x)2 + . . . = 1 + __
3 x2 + . . .
1 x + __
(1 − x) 2 = 1 + −__
2
2
8
2!
_____
SOLUTION
√
( )
( )( )
( )
)(
(
)
1
1
__
__
3x2
1x2 1 + __
1x − __
1 x + __
(1 + x) 2 (1 − x)− 2 ≈ 1 + __
2
8
2
8
3 x2 + __
1 x + __
1x2 − __
1x + __
1 x2
≈ 1 + __
4
2
8
2
8
1 x2
≈ 1 + x + __
2
_____
1 + x = 1 + x + __
1 x2 up to the term in x2.
Therefore _____
1− x
2
1 on both sides we have
Substituting x = __
9
√
_____
√
1
__
1+
9
_____
1
1 − __
9
1 __
1 + __
1
≈ 1 + __
9 2 9
()
2
___
√
10
___
9
___
8
__
1
1 + ___
≈ 1 + __
9 162
9
___
162 + 18 + 1
162
√___8 ≈ ____________
10
__
181
√__4 ≈ ____
162
5
__
181
√ 5__
___
≈ ____
162
√4
__
181
162
__
181
162
181
81
√ 5 ≈ ____ × √ 4 = ___ × 2 = ___
Partial fractions and the binomial expansion
We can expand rational functions by first separating into partial fractions and then
using the binomial expansion.
209
M O DUL E 2
E X A M P L E 25
x
up to and including the term in x2.
Find the expansion of ____________
(x + 1)(x − 2)
SOLUTION
x
into partial fractions, we have
Separating ____________
(x + 1)(x − 2)
x
B
A + _____
____________
≡ _____
x+
x + x − 2)
x−2
x ≡ A x − 2) + B x +
2
x = 2 ⇒ 2 = 3B, B = __
3
x = −⇒ −= −3A, A = __
3
2
__
__
3
3
x
_____
_____
____________
=
+
x + x − 2) x + x − 2
__
2
__
x +
x−2
3 BOE_____
3
-FUVTFYQBOE_____
__
3 = __
+ x)−
_____
x +
2
__
3
− −2) 2 + − x) + _________
x) +
= __
3
2!
x + __
x2 + . . .
− __
= __
3 3
3
[
]
3 = __
2 −2 + x)−
_____
x−2
3
x −
2 −2 − __
= __
3
2
x −
2 −2)− − __
= __
3
2
− −2) __
+ − − __
= −__
( −2x )2 + ( 2x ) + _________
3
2!
x2 + x + __
+ __
= −__
4
2
3
− __
x2 + x − ___
= − __
3 6
( (
))
(
[
)
]
"EEJOHUIFUXPFYQBOTJPOT XFIBWF
__
2
__
3 + _____
3 = __
x2 x + __
x2 − __
x − ___
− __
− __
_____
3
3
3 6
3
3 2
___
≈ − __
6 x + x
2
__
≈ − __
2x + 4x
The region for which the expansion is valid is the overlap (intersection) of the regions.
1 (1 + x)−1
__
is valid for −1 < x < 1
3
2 (−2 + x)−1 is valid for −2 < x < 2
__
3
Overlapping region is −1 < x < 1.
x +
210
x−2
3
MODULE 2tCHAPTER 10
E X A M P L E 26
SOLUTION
B x + C , find A, B and C. Hence find
2 + x + 2x = _____
A + _______
Given that f(x) = _____________
(2 + x)(1 + x2) 2 + x
1 + x2
UIFFYQBOTJPOPGf x VQUPBOEJODMVEJOHUIFUFSNJOx34UBUFUIFWBMVFTPGxGPS
XIJDIUIFFYQBOTJPOJTWBMJE
2
Bx + C
A + _______
2 + x + 2x ≡ _____
_____________
2
2+x
(2 + x)(1 + x2)
1 + x2
⇒ 2 + x + 2x2 ≡ A x2 + + Bx + C + x)
x = −2 ⇒ 2 − 2 + −2)2 = A +
8 = 5A
8
A = __
5
x = 0 ⇒ 2 = A + 2C
8 + 2C
2 = __
5
8 = __
− __
2
2C = ___
5
5 5
C = __
5
$PNQBSJOHDPFďDJFOUTPGx2HJWFT 2 = A + B
8 = __
2
B = 2 − __
5 5
8
2x + __
__
__
5
5
2 + x + 2x2 ≡ _____
5 + ______
_____________
2+x
+ x + x2)
+ x2
8
__
8 + x)−
&YQBOE_____
5 = __
2+x
5
[
8 2 + __
x
= __
5 (
2)
]−
−
8 [2−] + __
= __
( 2x )
5
Note
This expansion
is valid for
x < 1, or
−1 < __
2
−2 < x < 2.
[
− −2) __
− − −3) __
2
3
4 + − __
= __
( 2x ) + _________
( 2x ) + ______________
( 2x ) + . . .
5
2!
3!
[
4 − __
x2 − __
x + __
x3
= __
5
4
2
8
This expansion is
valid for 0 < x2 < 1
i.e. −1 < x < 1.
]
4 − __
2x + __
x2 − ___
x3
= __
5 5
5
2 x + __
__
5
5 = __
+ x2)−
2 x + __
_______
&YQBOE
5
5
+ x2
(
)
− −2)
+ − x ) + _________
2x + __
x ) )+...
= ( __
5
5) (
2!
− x + x )
2 x + __
≈ ( __
5
5)
2
Note
]
2
2 2
4
2x3 + __
x2
− __
2 x − __
≈ __
5
5
5 5
x2 − __
+ __
2 x3
2 x − __
≈ __
5 5
5
5
211
M O DUL E 2
8
__
Adding
2
__
2+x
1 + x2
Overlap
–2
–1
0
1
1
__
x+
5 + ______
5
5 ≈ __
4 − __
2 x + __
1 x2 − ___
1 x3 + __
1 + __
2 x − __
1 x2 − __
2 x3, up to x3
_____
2
5
5
5
5 x3
≈ 1 − ___
10
1 x3
≈ 1 − __
2
5
10
5
5
5
We can find the region for which the function f (x) is valid by taking the overlap of
the regions −2 < x < 2 and −1 < x < 1, which is −1 < x < 1.
Hence f (x) is valid for −1 < x < 1.
Exercise 10B
In questions 1–8, write down the first four terms of the expansion, stating the values
of x for which the expansion is valid.
1
1
__
__
1
1 (1 + x)3
2 (1 + x)−2
3 _______
4 (4 − x)2
3
(1 + x)
3 ______
1
1
1
______
_____
5 2x + 3
6 √1 − 3x
7 2−
8 ________
x
(2x − 1)2
9 Expand (1 − x)−1 + (1 − 2x)−2 as a series of ascending powers of x as far as the
term in x3.
10 Find the first four terms in the expansion of
(1 + x)2
2+x
(b) _____
(c) _______
2−x
2+x
2
11 Obtain the expansion of _______
up to the term in x8.
1 − 3x2
1_____
−x
(a) ______
√1 + x
____
12 Use a binomial expansion to find the value of √1.02 to 4 decimal places.
1 to 4 decimal places.
13 Use the binomial theorem to find the value of ______
(0.97)4
14 Find the first three terms in the series expansion, in ascending powers of x,
of (1 + 2x)−2.
(
)
1 − x 2 ≈1 − 6x + 21x2.
Hence, or otherwise, show that when x is small ______
1 + 2x
x2 + 2x + 3
in partial fractions. Hence, or otherwise,
15 Express f(x) = __________________
(1 + x)(2 + x)(1 − x)
find the expansion of f(x) up to and including the term in x2.
3
__
16 Given that the first four terms of the expansion of (1 − x)2 are
3 x2 + bx3, find the value of a and the value of b.
1 + ax + __
8
Bx + C where A, B and C
6x + 4
A + _______
17 Express _______________
in the form ______
1 − 2x 1 + 3x2
(1 − 2x)(1 + 3x2)
are constants.
6x + 4
Hence, or otherwise, find the expansion of _______________
in ascending
(1 − 2x)(1 + 3x2)
powers of x, up to and including the term in x3.
1
__
18 Find the expansion of (1 + x)5 up to and including the term in x3. Hence
5 ___
estimate √31 correct to 4 decimal places.
212
MODULE 2tCHAPTER 10
SUMMARY
Binomial
expansion
Notation
n! = n × ... × 3 × 2 × 1
0! = 1
n!
nC =
r
(n – r)!r!
For any positive
integer n,
(a + x)n = an + nC1an – 1x
+ nC2an – 2x2 + ... + x n
= ∑nr = 0 n Cr an – rx r
(r + 1)th term = n Cr an – rx r
For any real number n,
(1 + x)n = 1 + nx
+ n(n – 1) x2 + ...
2!
Expansion is valid for
–1 < x < 1
( )
For (a + x)n write as an 1 + xa
n
and then expand.
The expansion is valid
for –a < x < a.
Checklist
Can you do these?
■ Use n! to find values and simplify terms.
■ Use nCr to simplify terms.
■ Use the binomial expansion to expand (a + b)n for positive integer values of n.
■ Use ∑nr=0 nCran−r to find terms of expansions.
■ Find the term independent of x in an expansion.
■ Find an expansion for any real number n.
■ Write down the region for which an expansion is valid.
■ Use partial fractions to find an expansion.
Review exercise 10
1
Given that the coefficient of x2 in the expansion of (a + x)5 + (1 − 2x)4 is 664,
calculate the value of a.
2
In the expansion of (p + 2x)7, where p is a positive constant, the coefficients of
x2 and x3 are equal. Find the value of p.
1 x n, in ascending powers of x as far as the term
The expansion of (4 − 3x) 1 − __
4
in x2 is 4 − 13x + px2. Find the value of p and of n.
3
4
(
)
Given that the coefficients of x3 and x4 in the expansion of (3 − 2x)30 are p and
p
q respectively, evaluate __
q.
213
M O DUL E 2
5
6
7
8
9
Given that the coefficient of x in the expansion of (5 + px)(3 − x)6 is zero, find
the value of p.
4x − 1
A + ______
B , find A and B. Hence obtain the
If _____________
≡ _____
(1 − x)(1 + 2x) 1 − x 1 + 2x
4x −1
expansion of _____________
in ascending powers of x as far as the term in x3.
(1 − x)(1 + 2x)
For what values of x is the expansion valid?
15 − 7x − x2 into partial fractions. Deduce the expansion of the
Separate _____________
(x + 2)(3 − x)2
expression as far as the term in x2. State the values of x for which the expansion
is valid.
1 5.
Find the coefficient of x in the expansion of ( x − __
x)
1
__
Obtain the first four ____
terms in the expansion of (1 − 2x)2. Use your expansion to
obtain the value of √0.98 correct to 4 d.p.
10 Expand (1 + x)6. Hence, obtain the value of (1.02)6 correct to 5 d.p.
11 Obtain the first four terms of the expansion of (1 − x)−2. Hence obtain the
1
to five decimal places.
value of _______
(0.998)2
2
__
12 Find the coefficient of x4 in the expansion of (1 + 3x)− 3.
1
__
13 Obtain
the first four terms in the expansion of (4 + x)2. Hence obtain the value
_____
of √4.004 to four places of decimals.
14 Find the term independent of x in the expansion of ( x2 + __2x ) .
4 9.
15 Find the term independent of x in the expansion of 3x − __
2
12
(
(
)
8
x
)
2__ .
− ___
√x
17 The first three terms in the expansion of (2 + px)5 are given as
2 x + qx2. Find the value of p and of q.
32 + 26__
3 _____
1 + x in ascending powers of x up to and including the term in x2.
18 Expand _____
1−x
___
663 .
Use your expansion to show that √11 ≈ ____
200
______
1 + pt
19 The first three terms in the expansions of √1 + 2t and ______
in ascending
1 + qt
powers of t are the same. Find the values of p and of q, assuming that t is
sufficiently small for both expansions to be valid.
16 Find the coefficient of x2 in the expansion of
__
√x
√
3 ______
20 Expand √1 − 3x in ascending powers of x as far as the term in x3 and state for
what values of x this expansion is valid.
214
MODULE 2tCHAPTER 11
CHAPTER 11
Arithmetic and Geometric Progressions
At the end of this chapter you should be able to:
■ identify a sequence as an arithmetic sequence
■ use the formula for the nth term of an arithmetic sequence
■ know and use the formula for the last term of an arithmetic sequence
■ find the first term, the common difference and the nth term of an arithmetic
progression
■ know and use the formula for the sum of the first n terms of an arithmetic
progression
■ prove that a sequence is an arithmetic sequence
■ identify a sequence as a geometric sequence
■ use the formula for the nth term of a geometric sequence
■ know and use the formula for the last term of a geometric sequence
■ find the first term, the common ratio and the nth term of a geometric
progression
■ know and use the formula for the sum of the first n terms of a geometric
progression
■ prove that a sequence is a geometric progression
■ find the sum to infinity of a geometric progression
■ understand and use the condition for a geometric progression to converge.
KEYWORDS/TERMS
BSJUINFUJDQSPHSFTTJPOtDPNNPOEJČFSFODFtĕSTU
UFSNtnUIUFSNtTFSJFTtHFPNFUSJDQSPHSFTTJPOt
DPNNPOSBUJPtTVNUPJOĕOJUZtDPOWFSHFOU
215
M O DUL E 2
Arithmetic progressions
An arithmetic progression (AP) is a sequence in which each term is obtained from
the preceding one by adding a constant.
For example 1, 2, 3, 4, 5, . . . ; 2, 5, 8, 11, 14, . . . ; 8, 12, 16, 20, 24, . . . are arithmetic
progressions.
The difference between consecutive terms is called the common difference.
For a general arithmetic progression, let the first term be a, the common difference d,
and the number of terms n. The terms of the progression are
a, a + d, a + 2d, a + 3d, . . . , a + (n − 1)d
The nth term of the progression is
Tn = a + (n − 1)d
The last term of the progression is
l = a + (n − 1)d
EXAMPLE 1
Find the common difference of the following arithmetic progressions.
(a) 2, 7, 12, 17, 22, . . .
(b) 8, 15, 22, 29, 36, . . .
(c) x, 2x, 3x, 4x, . . .
SOLUTION
(a) 2, 7, 12, 17, 22, . . .
d=7−2=5
(b) 8, 15, 22, 29, 36, . . .
d = 29 − 22 = 7
(c) x, 2x, 3x, 4x, . . .
Since the sequence is an AP the
difference for any consecutive
pair will be the same.
d = 4x − 3x = x
EXAMPLE 2
Write down the nth term and the eighth term of the arithmetic sequence
8, 15, 22, 29, 36, . . .
SOLUTION
The first term a = 8
Common difference d = 7
Tn = a + (n − 1)d
= 8 + 7(n − 1)
= 8 + 7n − 7
= 7n + 1
216
MODULE 2tCHAPTER 11
The eighth term is T8 = 7(8) + 1 = 56 + 1 = 57
EXAMPLE 3
Write down the 20th term of the arithmetic sequence
4, 9, 14, 19, 24, . . .
SOLUTION
We can find the nth term as follows
a = 4, d = 9 − 4 = 5
Tn = a + (n − 1)d
= 4 + 5(n − 1)
= 4 + 5n − 5
= 5n − 1
When n = 20, T20 = 5(20) − 1 = 100 − 1 = 99
Hence the 20th term is 99.
EXAMPLE 4
The fourth term of an AP is 20 and the 8th term is 60. Find the 15th term.
SOLUTION
T4 = 20
T8 = 60
Writing as equations, since Tn = a + (n − 1)d,
when n = 4,
T4 = a + 3d
when n = 8, T8 = a + 7d
∴ a + 3d = 20
[1]
a + 7d = 60
[2]
[2] − [1] ⇒ 4d = 40
Solve simultaneous equations [1]
and [2] to find a
and d.
d = 10
Substituting d = 10 in [1],
a + 3(10) = 20
a = 20 − 30 = −10
∴ Tn = −10 + (n − 1)10
= −10 + 10n − 10 = 10n − 20
T15 = 10(15) − 20 = 130
∴ The 15th term is 130.
217
M O DUL E 2
EXAMPLE 5
A contractor in Trinidad brings in 200 workers from Asia at the beginning of January
2010. The contractor continues to recruit workers for the next two years. He brings in
10 workers at the beginning of each month. How many workers has this contractor
brought in during the two-year period?
SOLUTION
n = 24, a = 200, d = 10
T24 = 200 + (23)(10) = 200 + 230 = 430
The contractor has brought in 430 workers.
Sum of the first n terms of an AP
Rememb er
Sn represents the
sum of the first n
terms.
For an AP the sum of the first n terms, denoted by Sn, is
n [2a + (n − 1)d]
Sn = __
2
We can prove this result as follows.
The terms of the sequence are
a, a + d, a + 2d, . . . , a + (n − 2)d, a + (n − 1)d
The sum of the first n terms, known as an arithmetic series, is
Sn = a + (a + d) + (a + 2d) + . . . + (a + (n − 2)d) + (a + (n − 1)d)
[1]
Reversing the order of the series
Sn = (a + (n − 1)d) + (a + (n − 2)d) + . . . + (a + d) + a
Adding the n terms of [1] and [2] we have
Sn + Sn = [a + a + (n − 1)d] + [(a + d) + a + (n − 2)d] + . . .
+ [a + (n − 1)d + a]
∴ 2Sn = 2a + (n − 1)d + 2a + (n − 1)d + . . . + 2a + (n − 1)d
n times
Sn =
n [2a + (n − 1)d]
__
2
or
n [a + l].
Sn = __
2
218
2Sn = n[2a + (n − 1)d]
n [2a + (n − 1)d]
∴ Sn = __
2
We can also write this result as
n [a + a + (n − 1)d] or
Sn = __
2
n [a + l] where l is the last term.
Sn = __
2
EXAMPLE 6
Find the sum of the first n natural numbers.
SOLUTION
The first n natural numbers are 1, 2, 3, . . . , n
This is an AP with a = 1, d = 1 and number of terms = n
n [2a + (n − 1)d]
Using Sn = __
2
n [2(1) + (n − 1)(1)]
Sn = __
2
[2]
MODULE 2tCHAPTER 11
n [2 + n − 1]
= __
2
n
__
= (n + 1)
2
n
__
Hence Sn = (n + 1).
2
EXAMPLE 7
An arithmetic progression consisting of 30 terms has a sum of 6000 and the tenth
term is 90. Find the first term and the common difference.
SOLUTION
n [2a + (n − 1)d]
Using Sn = __
2
30[2a + 29d]
S30 = ___
2
6000 = 15[2a + 29d]
6000 = 2a + 29d
____
15
400 = 2a + 29d
[1]
T10 = 90
a + 9d = 90
[2] × 2 gives
[2]
2a + 18d = 180
[1] − [3] gives
[3]
11d = 220
d = 20
Substituting d = 20 into [2]
a + 9(20) = 90
a = −90
Hence a = −90, d = 20.
EXAMPLE 8
Given that the sum to 30 terms of an AP is 1425 and the tenth term is 31, find a and d.
SOLUTION
S30 = 1425
Using
n [2a + (n − 1)d]
Sn = __
2
30 [2a + 29d]
S30 = ___
2
∴ 1425 = 15[2a + 29d]
95 = 2a + 29d
[1]
T10 = 31
Tn = a + (n − 1)d
T10 = a + 9d
31 = a + 9d
62 = 2a + 18d
[2]
219
M O DUL E 2
Subtracting [2] from [1]
33 = 11d
d=3
Substituting d = 3 into equation [1]
95 = 2a + 29(3)
95 = 2a + 87
2a = 8
a=4
∴ a = 4, d = 3
EXAMPLE 9
The eighth term of an AP is 35 and the sum of the first 20 terms is 900. Find the tenth
term of the series.
SOLUTION
T8 = 35
Since Tn = a + (n − 1)d
T8 = a + 7d
∴ a + 7d = 35
[1]
S20 = 900
n [2a + (n − 1)d]
Since Sn = __
2
20 [2a + 19d]
S20 = ___
2
∴ 10(2a + 19d) = 900
2a + 19d = 90
[2]
[1] × 2 gives 2a + 14d = 70
[3]
[2] − [3] gives 5d = 20
d=4
a = 35 − 7(4)
a=7
T10 = a + 9d = 7 + 9(4) = 43
Proving that a sequence is an AP
To prove that a sequence is an arithmetic sequence we find Tn − Tn − 1.
If Tn − Tn − 1 is a constant then the sequence is an arithmetic progression.
220
MODULE 2tCHAPTER 11
E X A M P L E 10
Prove that 2, 5, 8, 11, . . . is an arithmetic progression.
SOLUTION
T1 = 2
T2 = 5
T3 = 8
Tn = 3n − 1
∴ Tn − 1 = 3(n − 1) − 1 = 3n − 4
Tn − Tn − 1 = (3n − 1) − (3n − 4)
=3
Since Tn − Tn − 1 is a constant the sequence is an AP with common difference 3.
E X A M P L E 11
Given that Sn = n[5 + 2n], find Tn and Tn − 1. Prove that the sequence is an AP.
SOLUTION
Sn = 5n + 2n2
Sn − 1 = 5(n − 1) + 2(n − 1)2
Tn = Sn − Sn − 1 = 5n + 2n2 − [5(n − 1) + 2(n − 1)2]
= 5n + 2n2 − [5n − 5 + 2n2 − 4n + 2]
= 4n + 3
Tn − 1 = 4(n − 1) + 3 = 4n − 1
Tn − Tn − 1 = (4n + 3) − (4n − 1) = 4
Since Tn − Tn − 1 is a constant the sequence is an AP with d = 4.
EXERCISE 11A
1
Find the common difference of the following arithmetic sequences.
(a) 4, 7, 10, 13, 16, 19, . . .
(c) 8, 6, 4, 2, 0, −2, . . .
(e) −5, −8, −11, −14, . . .
(g) x, 4x, 7x, 10x, . . .
(i) −0.5, −0.75, −1, −1.25, . . .
2
(b) 5, 5.5, 6, 6.5, 7, 7.5, . . .
13 , . . .
7, __
5 , ___
1, __
1 , __
(d) __
8 2 8 4 8
(f) 10, 20, 30, 40, 50, . . .
7 , __
5, ___
13 , . . .
1, ___
(h) __
3 12 6 12
(j) 8, 8.25, 8.50, 8.75, . . .
For the following arithmetic progressions, find the indicated terms.
(a) 5, 7, 9, 11, . . .
T10 and T15
(b) 8, 12, 16, 20, 24, . . .
(c) 2, 5, 8, 11, 14, . . .
T8 and T12
T7 and T9
(d) −20, −17, −14, −11, . . .
T11 and T13
(e) −8, −8.5, −9, −9.5, . . . T8 and T10
5 __
3
−3, −1, − __
(f) ___
4 , − 2, . . . T6 and T9
4
221
M O DUL E 2
2, 1, . . . T and T
1, __
(g) __
10
15
3 3
(h) 13, 9, 5, . . . T8 and T13
(i) 400, 394, 388, . . .
(j) 5, 11, 17, . . .
3
4
T5 and T16
T12 and T13
Find an expression for the nth term in each of the following arithmetic progressions.
(a) 2, 3, 4, 5, 6, . . .
(b) 4, 8, 12, 16, . . .
(c) 14, 21, 28, 35, . . .
17, ___
12 , ___
1, ___
22 , . . .
(e) __
5 35 35 35
(g) −15, −12, −9, . . .
6, ___
82 , ___
110, . . .
(i) __
7 63 63
(d) 998, 992, 986, 980, . . .
(h) 452, 404, 356, . . .
9 , ___
5 , ___
13 , . . .
(j) ___
12 12 12
Find the number of terms in the following arithmetic progressions.
(a) 1, 5, 9, 13, 17, . . . , 117
(c) −4, −6, −8, . . . , −202
(e) 400, 394, 388, . . . , 106
49
3 , . . . , ___
1, __
2 , __
(g) __
9 9 9
9
(i) 98, 105, 112, . . . , 231
5
(f) 4x, 10x, 16x, . . .
(b) 6, 9, 12, 15, . . . , 183
5, . . . , ___
65
1, 1, __
(d) __
3 3
3
(f) 4x, 10x, 16x, . . . , 148x
(h) 64, 72, 80, 88, . . . , 680
(j) 15, 30, 45, . . . , 225
Find the sum of the following arithmetic progressions.
(The last five sequences are from question 4.)
(a) 5, 11, 17, 23, . . . , 599
(b) 7, 10, 13, . . . , 79
(c) 540, 536, 532, . . . , 324
(d) 8, 8.25, 8.5, . . . , 15.5
(e) 60, 53, 46, . . . , −3
(f) 4x, 10x, 16x, . . . , 148x
49
3 , . . . , ___
2 , __
1, __
(g) __
9 9 9
9
(h) 64, 72, 80, 88, . . . , 680
(i) 98, 105, 112, . . . , 231
(j) 15, 30, 45, . . . , 225
6
Find the sum of all the odd numbers between 100 and 250.
7
Find the sum of all the even numbers between 81 and 151.
8
An arithmetic progression has a fifth term of 28 and the sum of the first five
terms is 1000. Find the first term and the common difference of the series.
9
The tenth term of an arithmetic progression is 75 and the sum of the first fifteen
terms is 1035. Find
(a) the first term of the arithmetic progression
(b) the common difference
(c) the fifteenth term of the progression.
222
MODULE 2tCHAPTER 11
10 The twelfth term of an arithmetic progression is − __34 and the sum of the first
forty terms is −115. Find
(a) the first term of the AP
(b) the tenth term of the AP.
11 The twentieth term of an AP is 97 and the sum of the first sixty terms is 8970.
Find
(a) the common difference
(b) the fifth term of the AP.
12 An arithmetic progression has an eighth term of −32 and the sum of the
first twenty terms is −890. Find the first term and the common difference
of the AP.
13 Given that the fifth term of an AP is 2 and the sum of the first seventeen terms
is 170. Find the first term and the common difference.
14 The third term of an AP is four times the sixth term and the sum of the first ten
terms is 150. Find
(a) the first term and the common difference
(b) the least number of terms of the AP which must be taken for the sum to be
negative.
15 Find the nth term of the series
log 2 + log x + log 4 + log x2 + log 8 + log x3 + . . .
Show that the series is arithmetic.
Find the sum of the first n terms of the series.
16 Given that the sum of the first n terms of a series is Sn = 2n2 + 3n,
(a) find the nth term of the series
(b) prove that the sequence is an AP
(c) find the first term and the common difference of the sequence.
17 Given that the sum of the first n terms of a sequence is Sn = 3n2 + 6n, show that
Tn = 6n + 3. Hence prove that the sequence is an AP. Find the first term and
the common difference of the sequence.
18 Given that the sum of the first n terms of a sequence is Sn = − __32n2 + __12n,
(a) find the nth term of the sequence
(b) prove that the sequence is an AP
(c) find the first term and the common difference of the sequence.
223
M O DUL E 2
19 Given that the sum of the first n terms of a sequence is Sn = − __72n2 − __12n,
(a) find the nth term of the sequence
(b) prove that the sequence is an AP
(c) find the first term and the common difference of the sequence.
20 Given that x2, 5x, 7x − 4 are three consecutive terms of an arithmetic progression
and that x is positive, find the value of x.
Geometric progressions
A geometric progression (GP) is a sequence in which the ratio of each term to
the term before is a constant. The constant is known as the common ratio of the
series.
For a general GP, let a be the first term, r be the common ratio and n be the number
of terms of the progression.
The terms of the progression are a, ar, ar2, ar3, ar4, . . . , arn−1.
The nth term is Tn = arn−1.
The last term is l = arn−1.
E X A M P L E 12
Find the number of terms in the geometric progression 2, 6, 18, . . . , 1458.
SOLUTION
a = 2,
6 = 3,
r = __
2
arn−1 = 1458
l = 1458
2(3)n−1 = 1458
3n−1 = 729
3n−1 = 36
n−1=6
n=7
E X A M P L E 13
1 , __
1 , __
1, . . .
Find the 10th term of the geometric progression 1, __
2 4 8
SOLUTION
The first term is a = 1
1
__
2 = __
1
The common ratio is r = __
1 2
Using the nth term as Tn = arn−1 with n = 10
1
T10 = (1) __
2
10−1
( )
224
1 9 = ____
1
= __
2
512
( )
MODULE 2tCHAPTER 11
E X A M P L E 14
Given that the first term of a GP is 4 and the fifth term is 64 and that r > 0, find the
common ratio of the progression.
SOLU TION
Since a = 4
and Tn = arn−1, using n = 5 and T5 = 64 gives
64 = 4(r)4
64
4
___
4
∴ r = √ 16 = 2
∴ r 4 = ___ = 16
E X A M P L E 15
5 and the fourth term is ___
5 , find the first term
Given that the second term of a GP is __
4
64
and the eighth term of the GP if r > 0.
SOLU TION
Since Tn = arn−1,
we have T2 = ar1
5
∴ ar = __
4
T4 = ar3
5
64
Dividing [2] by [1]
∴ ar3 = ___
[1]
[2]
5
___
64
ar3 = ___
___
ar
5
__
4
___
1 so r = ___
1 = __
1
r 2 = ___
16
16 4
1 into [1]
Substituting r = __
4
5 so a = 5
1 = __
a __
4
4
The first term, a = 5
√
( )
5
1 7 = ______
T8 = ar7 = 5 __
4
16 384
( )
E X A M P L E 16
Bristol Rovers Sports Club of Beaucarro Road receives a donation from a business
organisation on a yearly basis. The donation started in the year 2001 and the sum
of money received was $5000. Every year after that Bristol Rovers receives 90% of
the donation in the preceeding year. Calculate the year in which the value of the
donation first falls below $1000.
SOLUTION
The first term is the amount donated in 2001,
a = $5000
r = 0.90
225
M O DUL E 2
Tn = arn−1 = 5000(0.9n − 1)
Since Tn < 1000
5000(0.9n−1) < 1000
1000
0.9n−1 < _____
5000
1
0.9n−1 < __
5
Taking logs to base 10
1
lg 0.9n−1 < lg __
5
( )
1
∴ (n − 1)lg 0.9 < lg ( __
5)
Note
lg 0.9 < 0
so the inequality
sign reverses.
1
lg __
5
_____
n − 1>
lg 0.9
1
__
lg
5
n > 1 + _____
lg 0.9
n > 16.3
The year in which the donation falls below $1000 is 2017.
EXAMPLE 17
On 1 January 2004 Rajeev opened a savings account with $1000. Interest, at 6% of
the amount in the account at the time, is added each year on 1 January, starting in
2005. Given that Rajeev does not withdraw any of the funds in the account, find the
year in which there will be more than $5200 in the account after the interest has
been added.
SOLUTION
1 Jan 2004: Amount in the account = $1000
1 Jan 2005: Amount in the account = $1000 + 0.06(1000) = 1000(1 + 0.06)
= (1.06)(1000)
1 Jan 2006: Amount in the account = $(1.06)(1000) + 0.06 [(1.06)(1000)]
= [1.06(1000)][1 + 0.06]
= 1.06(1000)(1.06) = 1.062(1000)
1 Jan 2007: Amount in the account = 1.063(1000)
∴ Tn = 1000(1.06)n−1
We need to find n for which
Tn > 5200
1000(1.06)n−1 > 5200
5200
(1.06)n−1 > _____
1000
(1.06)n−1 > 5.2
226
MODULE 2tCHAPTER 11
Taking logs to base 10
lg (1.06)n−1 > lg 5.2
(n − 1)lg 1.06 > lg 5.2
lg 5.2
n − 1 > ______ so n > 29.3
lg 1.06
Hence the first year in which there will be more than $5200 is 2034.
Sum of the first n terms of a GP (Sn)
Let a, ar, ar2, . . . , arn−2, arn−1 be the first n terms of a GP.
The sum of the first n terms, known as a geometric series, is
Sn = a + ar + ar2 + . . . + arn−2 + arn−1
× r ⇒ rSn = ar + ar2 + ar3 + . . . + arn−1 + arn
[1]
[2]
[1] − [2] ⇒ Sn − rSn = a − arn
Sn(1− r) = a(1 − rn)
a(1 − rn)
Sn = ________
1−r
∴ Sum of the first n terms of a GP is
a(1 − rn)
Sn = ________
where r ≠ 1.
1−r
r≠1
E X A M P L E 18
Find the sum of the first ten terms of the geometric progression 1, 2, 4, 8, . . .
SOLU TION
a = 1, r = 2, n = 10
a(1 − rn)
Using Sn = ________
1−r
1(1 − (2)10)
S10 = __________
1−2
= −(1 − 210)
= 210 − 1 = 1023
E X A M P L E 19
SOLU TION
1 , __
1, . . . .
Find the sum of the first ten terms of the GP 1, __
2 4
a(1 − rn)
Using Sn = ________
1−r
1
__
1 and n = 10.
__
where a = 1, r = 2 = __
1 2
1 10
1 1 − __
2
1023 = 1.998
1 10 = _____
S10 = __________ = 2 1 − __
2
512
1
__
1−
2
( ())
( ())
227
M O DUL E 2
E X A M P L E 20
SOLUTION
3 and the common ratio is __
1 . Find the sum of the first
The third term of a GP is ___
4
16
eight terms of the progression.
3 and r = __
1
Tn = arn−1 where n = 3, T3 = ___
4
16
T3 = ar2
3
1 2 = ___
∴ a __
(4)
16
3
a
___ = ___
16 16
∴
a=3
a(1 − r 8)
S8 = ________
1−r
1 8
3 1− __
4
S8 = _________ = 4.00 (2 d.p.)
__
1−1
4
( ( ))
E X A M P L E 21
Find the 10th term and the sum of the first n terms of the geometric series
1 + ___
1 +...
1 + __
1 + __
3 9 27
SOLUTION
1
a = 1, r = __
3
n−1
1
1
= __
Tn = a __
3
3
()
()
n−1
1 10−1 = __
1 9 = ______
1
∴ T10 = __
(3)
(3)
19 683
a(1 − rn)
Using Sn = ________
1−r
1 n
1n
__
1 1− __
1−
3
3
Sn = _________ = _______
1
2
__
__
1−
3
3
( ( ))
3 1 − __
1
= __
2
3
()
n
( ( ))
E X A M P L E 22
1.
The 2nd term of a geometric series is −6 and the 5th term is 20 __
4
Find the sum of the first eight terms of the series.
SOLUTION
The nth term of a GP is Tn = arn−1
When n = 2, T2 = ar
ar = −6
When n = 5,
1
ar4 = 20__
4
228
[1]
T5 = ar5−1 = ar4
[2]
MODULE 2tCHAPTER 11
1
20__
ar4 = ____
4
[2] ÷ [1] gives ___
ar
−6
81 = ___
27
r 3 = _______
4 × −6 −8
____
3 27
−3
= ___
r = ___
−8
2
Substituting in [1]
√
−3 = −6
a ( ___
2 )
−2 = 4
a = −6 × ___
3
a(1 − rn)
Sn = ________
1−r
−3 8
4 1 − ___
2
S8 = ___________
−3
___
1−
2
( ( ))
( )
2 1 − ___
= 4 × __
( −32 ) ) = −39.406
5(
8
Hence S8 = 39.406 (3 d.p.).
Sum to infinity
Consider the geometric series a + ar + ar2 + . . .
a (1 − rn)
The sum of the first n terms is Sn = _____
1−r
a remains unchanged and (1 − rn) changes according to the value of r.
As n → ∞, _____
1−r
If −1 < r < 1, rn → 0 as n → ∞
a
∴ Sn → _____ as n → ∞, i.e. the series converges.
1−r
a if −1 < r < 1.
Hence the sum to infinity, S∞ = _____
1−r
For any other value of r, the series is not convergent and the sum to infinity does
not exist.
E X A M P L E 23
SOLUTION
1 + __
1 + __
1 + ___
1 +...
Find the sum to infinity of the geometric series 1 + __
2 4 8 16
1
a = 1, r = __
2
a
∴ S∞ = _____
1−r
1 =2
1 = __
= _____
1
1
__
__
1−
2 2
229
M O DUL E 2
E X A M P L E 24
Express 0.363 636 36 . . . as a fraction.
SOLUTION
0.363 636 36 = 0.36 + 0.0036 + 0.000 036 + . . .
= 36(0.01 + 0.0001 + 0.000 001 + . . .)
0.01 + 0.0001 + 0.000 001 + . . . is a geometric series with a = 0.01
0.0001 = 0.01
r = ______
0.01
36 12 = ___
0.01
4
∴ 0.363 636 36 = 36 ________ = ___ = ___
1 − 0.01
99 33 11
(
E X A M P L E 25
SOLUTION
)
2 and a sum to infinity of 99.
A geometric series has a common ratio of __
3
Find the first term of this series.
2
r = __
3
a
S∞ = _____
1−r
a = 3a
S∞ = _____
2
1 − __
3
∴ 99 = 3a
a = 33
Proving that a sequence is a GP
Tn
To prove that a sequence is a GP we need to prove that ____
is a constant.
Tn−1
E X A M P L E 26
SOLUTION
1 , ___
1 , ___
1 , . . . is a geometric progression and find the
Prove that the sequence __
4 16 64
common ratio.
1 , T = __
1 , T = __
1
T1 = __
4 2 42 3 43
1
Tn = __
4n
1
Tn − 1 = ____
n−1
4
1
__
n
Tn
4
1 × 4n − 1 = 4n−1−n = 4−1 = __
1
____
____
=
= __
4
1
4n
Tn−1 ____
4n−1
Tn
1.
Since ____
is a constant, this sequence is a GP with r = __
4
Tn−1
E X A M P L E 27
3 1 − __
1
Given that Sn = __
2
3
SOLUTION
3 1 − __
1
Sn = __
2
3
( ( ) ), find T and prove that this sequence is a GP.
Sn − 1
230
n
n
( ( ))
3 1 − __
= __
( 31 ) )
2(
n−1
n
MODULE 2tCHAPTER 11
Tn = Sn − Sn−1
Note
n
3 − __
3 __
3 + __
3 __
1 − __
1
= __
2 2 3
2 2 3
n−1
()
()
1
= ( __
( __23 )[ 1 − __13 ]
3)
1
1
⇒ T = ( __
= ( __
3)
3)
1
T = ( __
3)
T
( __31 ) = __1
1
____ = ______
= ( __
(3)
T
1
3)
__
(3)
Tn = Sn − Sn−1
n−1
n−1
n−1
n
n−2
n−1
n−1
n−1−(n−2)
n
1
n−2
n−1
Tn
1.
is constant, the sequence is a GP with common ratio __
Since _____
3
Tn−1
Convergence of a geometric series
Recall that a geometric series converges if and only if −1 < r < 1.
E X A M P L E 28
Determine whether the geometric series
∞
1
3( __
∑
3)
r =1
r−1
1 + 3 __
1 2+ . . .
= 3 + 3 __
3
3
() ()
converges or diverges. If it converges, find its sum.
SOLU TION
1 ÷ 3 = __
1
Common ratio = 3 __
3
3
()
Since −1 < r < 1, the series converges.
3 = __
3 = __
9
a = _____
S∞ = _____
1 − r 1 − __
1 __
2 2
3 3
E X A M P L E 29
2x .
The common ratio, r, of a geometric series is given by r = _____
x−1
Find all the values of x of which the series converges.
SOLUTION
The series converges if −1 < r < 1.
2x < 1
i.e. −1 < _____
x−1
2x < 1
For _____
x−1
2x −1 < 0
_____
x−1
2x − (x − 1)
___________
<0
x−1
x+1<0
_____
x−1
231
M O DUL E 2
Using a sign table:
x+1
x−1
x+1
______
x < −1
−ve
−ve
+ve
−1 < x <1
+ve
−ve
−ve
x>1
+ve
+ve
+ve
3x − 1
x−1
3x − 1
______
x− 1
1
x < __
3
−ve
−ve
+ve
1<x<1
__
3
x>1
+ve
−ve
−ve
+ve
+ve
+ve
x−1
∴ {x: −1 < x < 1}
2x > −1
For _____
x−1
2x + 1 > 0
_____
x−1
2x + x − 1 > 0
__________
x−1
3x − 1 > 0
______
x−1
Sign table:
{
}
1 ∪ {x: x > 1}
∴ x: x < __
3
The region for which both inequalities hold is the overlap of the two regions.
1 .
Therefore the series converges in the region x: −1 < x < __
3
{
E X A M P L E 30
}
1 + __
1 + __
1 + ___
1 + . . .,
Given the series __
5 54 57 510
(a) show that the series is geometric
(b) find the sum of the series to n terms
(c) find the sum to infinity of the series if the series converges.
SOLUTION
1
(a) Tn = _____
3n−2
5
Tn−1
1
1
= ________
= _____
53(n−1)−2 53n−5
1
_____
3n−2
T
5
53n−5
n
____ = _____ = _____
1
Tn−1 _____
53n−2
53n−5
= 53n−5−(3n−2)
1
= 5−3 = __
53
232
Power of 5 in the denominator:
1, 4, 7, 10, . . . goes up by 3, so
the general term for the power
of 5 is 3n − 2.
1 .
Hence Tn = _____
53n−2
MODULE 2tCHAPTER 11
Tn
1.
_____
is a constant, so the series is geometric with r = __
Tn − 1
53
1
1 1 − ___
__
3n
5
a(1 − rn) _________
5
________
(b) Sn = 1 − r =
1
1 − __
53
1 × ___
125
1
___
= __
5 124 1 − 53n
(
(
25 1 − ___
1
= ____
124
53n
(
)
)
)
1 which is in the region −1 < r < 1, the series
(c) Since the common ratio is r = ____
125
converges.
a where a = __
1 , r = ____
1 , we get
Using S∞ = _____
5
1−r
125
1
__
5
25
_______
= ____
S∞ =
124
1
____
1−
125
E X A M P L E 31
3 (3n − 1),
Given that the sum of the first n term of a series is Sn = __
2
(a) find the nth term of the series
(b) show that the series is geometric
(c) is the series convergent or divergent?
SOLU TION
(a) Sn = __23 (3n − 1)
Replacing n by n − 1,
3 (3n−1 − 1)
Sn − 1 = __
2
3 (3n) − __
3 − __
3 (3n−1) + __
3 = __
3 (3n) − __
3 (3n−1)
Tn = Sn − Sn−1 = __
2
2 2
2 2
2
3 (3n−1)(3 − 1)
= __
2
= 3(3n−1) = 3n
(b) Tn−1 = 3n−1
T
Tn−1
n
3 = 3n−(n−1) = 3
n
____
= ____
3n−1
∴ The series is geometric with r = 3
(c) The series diverges since r > 1.
EXERCISE 11B
1
Find the common ratio for the following geometric progressions.
1 , __
1 , __
1 , ___
1,...
(a) 2, 4, 8, 16, . . .
(b) __
2 4 8 16
233
M O DUL E 2
(c) 3, 9, 27, 81, 243, . . .
5, . . .
5 , __
(e) 15, 5, __
3 9
(g) 5, 20, 80, 320, . . .
1 , ___
1 , 1, 10, . . .
(i) ____
100 10
2
1 , ___
1 , ___
1 , __
1,...
(d) __
3 9 27 81
3 __
3
(f) −6, 3, −__
2, 6, . . .
1 , ___
1 , ___
1 , ____
1 ,...
(h) __
3 12 48 192
1,...
1 , ___
(j) 6, 1, __
6 36
For the following geometric progressions, find the indicated terms.
5 , __
5 , . . . T and T
(a) 15, 5, __
10
15
3 9
1 , ___
1 , ___
1 , ____
1 , . . . T and T
(b) __
8
12
3 12 48 192
(c) 2, 4, 8, 16, . . . T7 and T9
(d) 3, 9, 27, 81, . . . T11 and T13
1 , ___
1 , ___
1 , ___
1 , . . . T and T
(e) __
8
10
8 16 32 64
1 , ___
1 , . . . T and T
(f) 6, 1, __
6
9
6 36
(g) x, x2, x3, x4, . . . T10 and T15
1 , 1, 5, 25, . . . T and T
(h) __
8
13
5
(i) 16, 4, 1, 0.25, . . . T5 and T16
(j) 2.25, 0.75, 0.25, . . . T12 and T13
3
4
5
Find an expression for the nth term in each of the following geometric progressions.
8 , ___
16 , ___
32 , . . .
(a) 2, 4, 8, 16, 32, . . .
(b) __
9 9 9
1 , ___
1,...
1 , __
(d) 10, 5, 2.5, . . .
(c) 1, __
3 9 27
1 , __
1 , 2, . . .
(e) __
(f) 6, 3, 1.5, . . .
8 2
5, . . .
3 , ___
3,...
5 , __
(h) 12, 3, __
(g) 15, 5, __
4 16
3 9
5 , ___
10 , ___
40 , . . .
1 , __
2 , __
4, . . .
(i) __
(j) __
5 5 5
6 3 3
Find the number of terms in the following geometric progressions.
2 , __
2 , . . . , _____
2
(a) 2, __
3 9
6561
5
(b) 10, 5, 2.5, . . . , ____
512
3 , . . . , ______
3
3 , ___
(c) 3, __
5 25
78 125
For the following GPs, find the sum of the indicated number of terms.
(a) 1, 3, 9, 27, . . . (10 terms)
(c) −12, 24, −48, . . . (7 terms)
1 , ___
1 , . . . (6 terms)
(e) 6, 1, __
6 36
3 , . . . (9 terms)
3 , ___
(g) 3, __
5 25
234
(b) 8, 4, 2, 1, 0.5, . . . (15 terms)
1 , __
2 , __
4 , . . . (12 terms)
(d) __
5 5 5
(f) 10, 5, 2.5, . . . (8 terms)
10 , ___
40 , . . . (10 terms)
5 , ___
(h) __
6 3 3
MODULE 2tCHAPTER 11
6
7
Write down the first five terms of the geometric progression which has a first
1.
term of 5 and a common ratio of __
4
Find the 7th and 10th terms of the geometric progression with first term 2 and
common ratio 6.
8
How many terms in the GP 4, 3.6, 3.24, . . . are needed so that the sum exceeds 35?
9
A geometric progression has a first term of a and a common ratio r. Given that
5 , calculate
the second term of the progression is 20 and the fifth term is ___
16
(a) r
(b) a
(c) the sum to infinity.
A second geometric progression is formed by squaring each term of the first
geometric progression. Find the sum to infinity of this progression.
6075
10 The first and the fifth terms of a geometric progression are 600 and _____
32
respectively. Find
(a) the values of the second and the third terms
(b) the sum to infinity of the progression.
11 The sum of an infinite geometric progression is 1000 and the common ratio is
0.2. Calculate
(a) the first term
(b) the fifteenth term
(c) the least number of terms of the progression whose sum exceeds 990.
12 The fourth term of a geometric progression is −96 and the seventh term is 768.
Calculate
(a) the common ratio
(b) the first term
(c) the sum of the first ten terms.
13 Given that y + 6, y, y − 3 are three consecutive terms of a GP, calculate the value of
(a) y
(b) the common ratio.
4x .
14 The common ratio, r, of a geometric series is given by r = ______
2
Find all the values of x for which the series converges.
3+x
5x .
15 The common ratio, r, of a geometric series is given by r = ______
2
Find all the values of x for which the series converges.
1 + __
1 +...
16 Given the series __21 + __
23 25
(a) show that the series is geometric
4+x
(b) find the sum of the first n terms
(c) deduce the sum to infinity if it exists.
235
M O DUL E 2
17 Given that the sum of the first n terms of a series, s, is Sn = 4 1 − __41
n
( ( ))
(a) find the nth term of s
(b) show that s is geometric
(c) find the sum to infinity of s.
18 Given that the sum of the first n terms of a series, s, is Sn = 4 1 − __31
n
( ( ))
(a) find the nth term of s
(b) show that s is geometric
(c) find the sum to infinity of s.
19 On 3 June 1964, a person opened a savings account with $1500. Interest, at 7%
of the amount in the account at the time, was added each year on 3 June, starting in 1965. Given that no withdrawals were made, find the year in which there
was more than $7000 in the account after the interest had been added.
20 Michael enjoys texting his friends on a daily basis. His parents assign him 480
texts in total for a period of time. On the first day he uses 35 texts. On each
subsequent day he texts 95% of the number of times he texted the previous day,
until all 480 texts have been used up. He begins to text on 6 December. Find
(a) the number of texts he sent on 16 December
(b) the date on which he uses up all 480 texts.
SUMMARY
Arithmetic progression (AP) and geometric progression (GP)
AP
GP
An AP is a sequence
in which each term after
the first is found by adding
a constant (common difference)
to the previous term.
A GP is a sequence
in which each term after
the first is found by multiplying
the previous term by a
constant (common ratio).
The terms of an AP are
a, a + d, a + 2d, a + 3d, ..., a + (n – 1)d.
a: first term, r: common ratio
n: no. of terms, the terms are
a, ar, ar2, ..., ar n – 1
a: first term, d common difference
Tn = a + (n – 1)d
l = last term, l = a + (n – 1)d
Sn = n [2a + (n – 1)d]
2
Sn = n [a + l ]
2
236
Tn = ar n –1
n
Sn = a(1 – r ), r ≠ 1,
1–r
A GP is convergent iff
–1<r<1
S∞ = a
1–r
MODULE 2tCHAPTER 11
Checklist
Can you do these?
■ Identify a sequence as an arithmetic sequence.
■ Use the formula for the nth term of an arithmetic sequence.
■ Use the formula for the last term of an arithmetic sequence.
■ Find the first term, the common difference and the nth term of an arithmetic
progression.
■ Use the formula for the sum of the first n terms of an arithmetic progression.
■ Prove that a sequence is an arithmetic sequence.
■ Identify a sequence as a geometric sequence.
■ Use the formula for the nth term of a geometric sequence.
■ Use the formula for the last term of a geometric sequence.
■ Find the first term, the common ratio and the nth term of a geometric progression.
■ Use the formula for the sum of the first n terms of a geometric progression.
■ Prove that a sequence is a geometric progression.
■ Find the sum to infinity of a geometric progression.
■ Understand and use the condition for a geometric progression to converge.
Review e x e r c i s e 1 1
1
An oil company in Trinidad hires a contractor to bore a well. The contractor
charges $1000 for the first 5 m, $1250 for the next 5 m, $1500 for the next 5 m,
and so on.
(a) What is the cost of drilling 200 m deep?
(b) What is the depth of a well which cost $10 000 to bore?
2
2 x m.
A ball dropped from a height of x m rebounds to a height of __
3
If the ball is dropped from a height of 4 m, find
(a) the height of rebound after the 6th bounce
(b) the total distance covered before the ball comes to rest.
3
Kimberly sets a pendulum swinging, the first oscillation is 45° and each
2 of the one before it. What is the total angle described
succeeding oscillation is __
3
by the pendulum before it stops?
237
M O DUL E 2
4
If $1000 is invested each year at 5% interest compounded annually, what would
be the total amount of the investment after 10 years?
5
A ball is dropped from a height of 10 m. Each time it strikes the ground, it
bounces up three-quarters of the previous height.
(a) What height will the ball bounce up to after it strikes the ground for the
third time?
(b) How high will it bounce after it strikes the ground for the nth time?
(c) What is the total distance travelled by the ball before it stops
bouncing?
6
Determine whether the infinite geometric series 128 − 64 + 32 − 16 + . . .
converges or diverges. If it converges, find its sum.
7
A 1997 Nissan Sentra sold for $130 000 in Trinidad and Tobago.
If the vehicle loses 15% of its value each year, how much will it be worth
after 10 years?
8
The value of a certain type of vehicle when it is new is $200 000 TT and it
depreciates by 10% each year. Find the value of the vehicle when it is five
years old.
9
At the beginning of each year Sally invests $8000 in a finance institution that
pays 5% per annum compound interest. Calculate the amount of money she will
have at the end of the 10th year.
10 A piece of string of length 8 m is cut into pieces in such a way that the length of
each piece follows an arithmetic sequence. If the length of the shortest piece
of string is 10 cm and the longest piece of string is 15 cm, find how many pieces
of string can be cut.
n−1
2
11 The sum of the first n terms of a series is given by Sn = 6 − ____
n−1 .
3
(a) Find the nth term of the series.
(b) Show that the series is geometric.
(c) If the series converges, find the sum to infinity.
12 The sum of the first n terms of a series is given by Sn = __21 (3n2 + n).
(a) Find the nth term of the series.
(b) Show that the series is arithmetic.
(c) Find the first term and the common difference of the progression.
13 The sum of the first four terms of a geometric series is five times the sum of
the first term and the third term. Find the common ratio of the series given
that r > 1.
238
MODULE 2tCHAPTER 11
14 The sum of the first n terms of a series is given by Sn = 14n − 2n2.
(a) Find the nth term of the series.
(b) Show that the series is arithmetic.
(c) Find the first term and the common difference of the series.
15 The sum of the first n terms of a series is given by Sn = __25 (n2 − 3n).
(a) Find the nth terms of the series.
(b) Show that the series is arithmetic.
(c) Find the first term and the common difference of the series.
n
16 The sum of the first n terms of a series is given by Sn = 90 1 − __31 .
( ( ))
(a) Find the nth term of the series.
(b) Show that the series is arithmetic.
(c) Find the first term and the common ratio of the series.
17 The sum of the first n terms of a series is given by Sn = 24(2n − 1).
(a) Find the nth term of the series.
(b) Show that the series is geometric.
(c) Find the first term and the common ratio of the series.
1
125 1 − −__
18 The sum of the first n terms of a series is given by Sn = ___
5
2
(a) Find the nth term of the series.
(
n
( ) ).
(b) Show that the series is geometric.
(c) Find the common ratio of the series.
19 If x, y, 10 is an AP and y, x, 10 is a GP where x ≠ y, determine the value of x and
the value of y.
20 The sum of the first 10 terms of an AP is 615 and the tenth term is 48.
Find the first term and the common difference of the progression.
21 The sum of the first fifteen terms of an AP is 1560 and the seventh term is 102.
Find the first term and the common difference of the progression.
22 Find the 10th term of an arithmetic progression given that the sum of the
166 and the fifth term is ___
40 .
first 12 terms is ____
9
3
239
M O DUL E 2
CHAPTER 12
Numerical Techniques
At the end of this chapter you should be able to:
■ test for the existence of a root
■ use the intermediate value theorem
■ use interval bisection to find the root of an equation
■ derive and use linear interpolation to find a root of an equation
■ explain how the Newton–Raphson method works
■ find successive approximations for any root
■ use the Newton–Raphson method to find successive approximations to
f (x) = 0 where f (x) is differentiable
■ understand when the Newton–Raphson method fails to converge to a root of
the equation.
KEYWORDS/TERMS
JOUFSNFEJBUFWBMVFUIFPSFNtSPPUTtMJOFBS
JOUFSQPMBUJPOtJUFSBUJWFt/FXUPOo3BQITPONFUIPEt
BQQSPYJNBUJPOtEJČFSFOUJBCMF
240
MODULE 2tCHAPTER 12
Note
The IMVT can be
applied to a continuous function
or a function that
is continuous
over a specified
interval.
The intermediate value theorem (IMVT)
Given a function f(x) that is continuous over the closed interval [a, b], let d be any
number between f(a) and f (b), that is, d lies in the interval [ f(a), f (b)]. Then there
must be at least one value c in the interval [a, b] such that f(c) = d. This result is
known as the intermediate value theorem (IMVT).
y
f(b)
d = f(c)
f(a)
x
a
c
b
EXAMPLE 1
Use the IMVT to verify that f(x) = (x − 2)2 − 7 has a root between 4 and 5.
SOLUTION
f (4) = (4 − 2)2 − 7 = 4 − 7 = −3
f (5) = (5 − 2)2 − 7 = 9 − 7 = 2
Since 0 lies between f (4) and f (5), and f(x) is continuous, by the IMVT there
must be some α such that f(α) = 0 and 4 < α < 5.
∴ There is a root between 4 and 5.
EXAMPLE 2
Show that there is a root of the equation 4x3 − 6x2 + 3x − 2 = 0 between 1 and 2.
SOLUTION
f(x) = 4x3 − 6x2 + 3x − 2
f (1) = 4(1)3 − 6(1)2 + 3(1) − 2
= 4 − 6 + 3 − 2 = −1
f (2) = 4(2)3 − 6(2)2 + 3(2) − 2
= 32 − 24 + 6 − 2 = 12
Thus f (1) < 0 < f (2)
Since f(x) is continuous, by the IMVT there must be some α such that f(α) = 0.
∴ The equation has at least one root α in the interval 1 < α < 2.
Finding the roots of an equation
Remember
The roots of an
equation y = f(x)
are the values of x
for which f(x) = 0.
In mathematics and in practical problems it is often necessary to find the roots of
an equation. Although we have many methods for solving equations, some equations cannot be solved using the methods we discussed in previous chapters and we
cannot find exact values for their roots. Numerical solutions to different degrees of
accuracy may be required, depending on the situation involved. For example, there
241
M O DUL E 2
are no algebraic methods for solving equations such as 2sin x − x = 0, 4 ln x = x
or 10x = ex. Four ways of solving equations of this form to a required degree of
accuracy are: a graphical method, interval bisection, linear interpolation and
the Newton–Raphson method. The graphical method and interval bisection are
discussed briefly while linear interpolation and the Newton–Raphson method are
explained in detail.
Graphical solution of equations
The roots of the equation f (x) = 0 are the values of x where the curve y = f(x) cuts
the x-axis. We can find solutions to this equation by drawing the graph of y = f (x) on
graph paper and read off where the graph intersects the x-axis. The use of a graphical
calculator will give the solution to a required degree of accuracy. When using this
method all the terms of the equation must be on one side of the equality.
EXAMPLE 3
Solve graphically to two decimal places the equation f(x) = 0 where f(x) = 10x − e x.
SOLUTION
An equation f (x) − g (x) = 0 can be solved graphically by drawing the graphs of
y = f (x) and y = g (x) and finding the x-coordinates of the points of intersections of
the two graphs.
1 ex and
To solve the equation 10x − ex = 0, we can write the equation as x = ___
10
1 ex. The solution to the equation is the point of
draw the graphs of y = x and y = ___
10
intersection of the two graphs.
4
y
y = 1 ex
10
3
2
1
x
–2
–1
1
2
3
4
–1
y = x –2
The solutions are x = 0.11 and x = 3.58 (2 d.p.)
Interval bisection
One method of finding the root of an equation is interval bisection. Suppose an equation f (x) = 0 has a root in the interval a < x < b, we can locate the root as accurately
as desired by bisecting the interval as follows: find the midpoint of the interval
a + b, next find f(x ). If f(x ) f (a) < 0 then, by the IMVT, the root lies in the
x m = _____
m
m
2
interval a < x < xm; if this product is positive then the root will be in the interval
xm < x < b. Select the interval which contains the root and repeat the process.
The strategy is repeated until the root is found to the required degree of accuracy.
If the interval contains more than one root, this process is very difficult.
242
MODULE 2tCHAPTER 12
EXAMPLE 4
Show that the equation f (x) = x2 − 4x + 1 = 0 has a root in the interval 3 < x < 4.
Hence find the root to one decimal place using interval bisection.
SOLUTION
f (3) = 32 − 4(3) + 1 = 9 − 12 + 1 = −2
f (4) = 42 − 4(4) + 1 = 1
Since f(x) is a
continous function, by the IMVT,
when f(a)f(b) < 0
there exists a root
in the interval
[a, b].
Since f (x) is continuous, by the IMVT, there exists a root in the interval 3 < x < 4.
3 + 4 = 3.5
Midpoint of the interval = _____
2
2
f (3.5) = 3.5 − 4(3.5) + 1 = − 0.75
Since f (3.5) f(4) < 0, the root lies in the interval 3.5 < x < 4.
3.5 + 4 = 3.75
Midpoint of the interval = _______
2
2
f (3.75) = (3.75) − 4(3.75) + 1 = 0.0625
Since f (3.5)f (3.75) < 0 the root lies in the interval 3.5 < x < 3.75.
3.5 + 3.75 = 3.625
Midpoint of the interval = _________
2
2
f (3.625) = (3.625) − 4(3.625) + 1 = −0.359 375
Since f (3.625) f (3.75) < 0, the root lies in the interval 3.625 < x < 3.75
3.625 + 3.75 = 3.6875
Midpoint of the interval = ___________
2
2
f (3.6875) = (3.6875) − 4(3.6875) + 1 = −0.152 343 75
Since f (3.625)f (3.6875) > 0, the root lies in the interval 3.6875 < x < 3.75
Continuing this process, we get 3.719 < x < 3.734.
When the upper bound is rounded to one decimal place the value is 3.7 and when
the lower bound is rounded to one decimal place the value is 3.7.
Hence the root is 3.7 to 1 d.p.
Linear interpolation
Linear interpolationJTBUFDIOJRVFVTFEUP
FTUJNBUFVOLOPXOWBMVFTUIBUMJFCFUXFFOUXP
LOPXOWBMVFT-JOFBSJOUFSQPMBUJPOBTTVNFT
UIBUUIFSBUFPGDIBOHFCFUXFFOLOPXOWBMVFT
JTDPOTUBOUBOEDBOCFDBMDVMBUFEVTJOHUIF
HSBEJFOUPGUIFMJOFKPJOJOHUIFUXPLOPXO
QPJOUT8FEFSJWFUIFMJOFBSJOUFSQPMBUJPO
GPSNVMBBTGPMMPXT
-FUf(x CFBDPOUJOVPVTGVODUJPOBOEMFU
f(a) < 0 and f(b) >#ZUIF*.75 UIFSF
FYJTUTαTVDIUIBUf(α) =MJFTCFUXFFOf(a)
and f(b UIBUJT f(x) =IBTBSPPUJOUIF
interval [a b>
(b, f(b)) Q
a
x1
x
R
b
y = f(x)
(a, f(a))
P
*OUIFEJBHSBNUIFMJOFPQDSPTTFTUIFxBYJTBU3
243
M O DUL E 2
4JODFHSBEJFOUPG13= HSBEJFOUPG32
| f(a)|
| f(b)|
______
= ______
x1 − a b − x1
3FBSSBOHFUIJTSFTVMUUPNBLFx1UIFTVCKFDUPGUIFGPSNVMB
(b − x1)| f(a)| = (x1 − a)| f(b)|
b| f(a)| − x1 | f(a)| = x1 | f(b)| − a| f(b)|
b| f(a)| + a| f(b)| = x1 | f(b)| + x1 | f(a)|
= x1 [| f(b)| + | f(a)|]
b| f(a)| + a| f(b)|
∴ x1 = ______________
| f(b)| + | f(a)|
8FOPXIBWFx1 BĕSTUBQQSPYJNBUJPOUPUIFSPPU α JOUFSNTPGUIFLOPXOQPJOUT
8FDBOUIFOĕOEf(x1 BOEEFUFSNJOFJGUIFSPPUMJFTJOUIFJOUFSWBM a x1) or (x1 b -JOFBSBQQSPYJNBUJPODBOUIFOCFVTFEBHBJOUPĕOEBTNBMMFSJOUFSWBMXIJDI
DPOUBJOTUIFSPPUPGUIFFRVBUJPOα8FDBODPOUJOVFUIJTQSPDFTTVOUJMXFĕOEUIF
root αUPUIFEFHSFFPGBDDVSBDZUIBUJTSFRVJSFE
EXAMPLE 5
Show that the equation x3 − x2 = 2 − 10x has a root between x = 0 and x = 1.
Use linear interpolation to find this root correct to two decimal places.
SOLUTION
x3 − x2 = 2 − 10x
⇒ x3 − x2 + 10x − 2 = 0
-FUf(x) = x3 − x2 + 10x − XIFSFf(x JTBDPOUJOPVTGVODUJPO
/PXf(0) = 0 − 0 + 10(0) − 2 = −2
f(1) = 12 − 13 + 10(1) − 2 = 1 − 1 + 10 − 2 = 8
Note
We continue
using the linear
interpolation
result until
we get two
consecutive
values that are
the same when
rounded to
the degree of
accuracy that is
required.
244
4JODFf(0)f(1) < CZUIF*.75UIFSFFYJTUTx = αXIFSFf(α) = JFUIFSFJTBSPPU
CFUXFFOx = 0 and x = /PXf(x) = x3 − x2 + 10x − 2
f(0) = − f(1) = 8
6TJOHUIFMJOFBSJOUFSQPMBUJPOGPSNVMB XFIBWF
a| f(b)| + b| f(a)|
x1 = ______________
| f(a)| + | f(b)|
a = b = | f(a)| = | f(0)| = − ]f(b)| = | f(1)| = 8
0(8) + (1)(2) ___
2 = __
1
= 10
∴ x1 = ___________
5 = 8+2
MODULE 2tCHAPTER 12
1 3
1 2
−
4
1 = __
1
__
____
__
/PXf __
5 − 5 + 10 5 − 2 = 125
5
1 f(1) <CZUIF*.75UIFSFFYJTUTx = α in the interval __
1 TVDIUIBU
4JODFf __
5
5
f(α) = JFBSPPUMJFTCFUXFFO__
1 BOE
5
1
6TJOHUIFMJOFBSJOUFSQPMBUJPOGPSNVMBBHBJO XJUIa = __
5 b = XFHFUBTFDPOE
BQQSPYJNBUJPO
( ) ( ) ( )
( )
We continue using the linear interpolation result
until we get consecutive values
that are the same
when rounded to
the degree of accuracy required.
( )
[ ]
1
4
__
____
5 (8) + (1) 125
_______________
= x2 =
4 +8
____
125
( )
( )
4JODFCPUIx1 = BOEx2 = UPUXPEFDJNBMQMBDFT UIFSPPUJTUPEQ
EXAMPLE 6
4IPXUIBUUIFFRVBUJPOTJOx − x = IBTBUMFBTUPOFSPPUCFUXFFOBOE
)FODFĕOEUIFSPPUUPUXPEFDJNBMQMBDFT VTJOHMJOFBSJOUFSQPMBUJPO
SOLUTION
-FUf(x) = TJOx − x
Remember we
are working in
radians.
f = TJO − = f = TJO − = − 4JODFf(x JTDPOUJOVPVTBOEf f < CZUIF*.75 JOUIFJOUFSWBM< >
UIFSFFYJTUTx = α TVDIUIBUf(α) = )FODFUIFSFJTBUMFBTUPOFSPPUJOUIFJOUFSWBM< >
/PXf(x) = TJOx − x
For x = f = For x = f = −
6TJOHUIFMJOFBSJOUFSQPMBUJPOGPSNVMB XFIBWF
a| f(b)| + b| f(a)|
x1 = ______________
| f(a)| + | f(b)|
+ = _____________________________
+ = /PXf = TJO − = 4JODFf f < UIFSPPUMJFTbFUXFFOBOE
+ ∴ x2 = ___________________________________
+ = 4JODFx1 = BOEx2 = UPUXPEFDJNBMQMBDFT UIFSPPUJTUPEQ
245
M O DUL E 2
EXAMPLE 7
Show that the equation ex = 3x has a root between 0 and 1. Hence find the root of
this equation to two decimal places, using linear interpolation.
SOLUTION
-FUf(x) = ex − 3x
f(0) = e0 − 3(0) = 1
f(1) = e1 − 3(1) = −
4JODFf(0) f(1) < CZUIF*.75 UIFSFFYJTUTx = α such that f(α) = 0 in the interval
[0, 1]. )FODFUIFSFJTBSPPUJOUIFJOUFSWBM< >
ćFMJOFBSJOUFSQPMBUJPOGPSNVMBTUBUFT
b| f(a)| + a| f(b)|
x1 = ______________
| f(b)| + | f(a)|
-FUa = f = e − = -FUb = f(1) = e1 − 3(1) = − 4VCTUJUVUJOHUPĕOEx1 XFIBWF
+ x1 = _________________________
= + f = e − = − 4JODFf f < CZUIF*.75 UIFSPPUMJFTCFUXFFOBOE
6TJOH
b| f(a)| + a| f(b)|
x2 = ______________
| f(b)| + | f(a)|
8FHFU
+ x2 = _______________________________ = + 8FOFYUĕOEx3
f = e − = −
4JODF
f f < CZUIF*.75 UIFSPPUMJFTCFUXFFOBOE
6TJOHUIFMJOFBSJOUFSQPMBUJPOGPSNVMBBHBJO XFIBWF
+ x3 = _________________________________ = + /PXx1 = x2 = x3 = UPEQ
)FODFUIFSPPUPGUIFFRVBUJPOJTUPUXPEFDJNBMQMBDFT
246
MODULE 2tCHAPTER 12
Newton–Raphson method for finding the roots
of an equation
An iterative formula is a formula used
to produce a sequence of values, so that
each successive value is dependent on the
preceding value. The Newton–Raphson
method for approximating the roots of an
equation makes use of an iterative formula
based on an initial value. The formula
works in the following way.
Tangent
(x 1, f(x1))
Tangent
(x 2, f(x2))
Let x = α be a root of the equation f (x) = 0.
Let x1 be a first approximation to α. The
tangent to the curve at x1 cuts the x-axis at
x2 which is closer to α than x1. The second
approximation x2 can be found in the following manner.
When x = x1, y = f (x1)
α x3
x2
x1
∴ The co-ordinates at x1 are (x1, f (x1))
y = f (x)
dy
___
= f ′(x)
dx
dy
At x = x1 the gradient of the tangent is ___ = f ′(x1).
dx
The equation of the tangent at x = x1 is
f (x) − f (x1) = f ′(x1)[x − x1]
f (x) − f (x1) = x f ′(x1) − x1 f ′(x1)
This line cuts the x-axis when x = x2, f (x) = 0
∴ −f (x1) = x2 f ′(x1) − x1 f ′(x1)
x2 f′(x1) = x1 f ′(x1) − f (x1)
f (x1)
⇒ x2 = x1 − _____
f ′(x1)
We now have a second approximation to the root of the equation, x2 , where
f (x1)
x2 = x1 − _____
f ′(x1)
We can repeat this process at x2 and arrive at
f (x2)
x3 = x2 − _____
f′(x2)
and so on until
f (xn)
xn+1 = xn − _____
f ′(xn)
This result is known as the Newton–Raphson (N–R) formula.
247
M O DUL E 2
When using the N–R method:
(i)
A first approximation must be known (this is either given or chosen).
(ii) The accuracy of the result needed will tell us when to stop using the
approximation (if the root is needed to 2 d.p., the iteration must be done until
two consecutive approximations give the same value when rounded to 2 d.p.).
(iii) Accuracy is important, so we work to at least two more decimal places or
significant figures than is required.
(iv) When working with trigonometric functions, remember to put your calculator
in radian mode.
EXAMPLE 8
Given that f(x) = 2 sin x − x show that f (x) = 0 has a root in the interval 1.5 to 2.
2 sin xn − 2xn cos xn
Using the N–R formula show that xn +1 = ________________
1 − 2 cos xn
Find the root to four decimal places.
SOLUTION
Remember to
work in radians.
Since f (x) = 2 sin x − x
f(1.5) = 2 sin (1.5) − 1.5 = 0.494 99
f(2) = 2 sin 2 − 2 = −0.181 41
Since f (1.5) is positive and f (2) is negative, by the IMVT, there is a value α such
that 2 sin x − x = 0 where 1 < α < 2.
f (x) = 2 sin x − x
f ′(x) = 2 cos x − 1
By N–R
f (xn)
xn + 1 = xn − _____
f ′(xn)
f (xn) = 2 sin xn − xn
f′(xn) = 2 cos xn − 1
2 sin xn − xn
xn + 1 = xn − __________
2 cos xn − 1
2xn cos xn − xn − 2 sin xn + xn
= _________________________
2 cos xn − 1
2xn cos xn − 2 sin xn
= ________________
2 cos xn − 1
2 sin xn − 2xn cos xn
= ________________
1 − 2 cos xn
Choosing a value between 1.5 and 2, let the first approximation be 1.65
x1 = 1.65
248
MODULE 2tCHAPTER 12
2 sin x1 − 2x1cos x1
x2 = ________________
1 − 2 cos x1
2 sin 1.65 − 2(1.65) cos 1.65
= _______________________
1 − 2 cos (1.65)
= 1.946 769 (6 d.p.)
2 sin 1.946 769 − 2(1.946 769) cos 1.946 769
x3 = ___________________________________ = 1.896 913
1 − 2 cos 1.946 769
2 sin 1.896 913 − 2(1.896 913) cos 1.896 913
x4 = ___________________________________ = 1.895 495
1 − 2 cos 1.896 913
2 sin 1.895 495 − 2(1.895 495) cos 1.895 495
x5 = ___________________________________ = 1.895 494
1 − 2 cos 1.895 495
To four decimal places, x4 = 1.8955, x5 = 1.8955
Two consecutive values are the same, therefore the root is 1.8955 to four decimal places.
EXAMPLE 9
Given the equation x3 = 1 − x, show that the equation has a root in the interval
2xn3 + 1
[0, 1]. Use the N–R formula to show that xn+1 = _______
3x2n + 1
Hence find the root to two decimal places.
SOLUTION
Since x3 = 1 − x
x3 + x − 1 = 0
Let f (x) = x3 + x − 1
f (0) = 03 + 0 − 1 = −1
f (1) = 13 + 1 − 1 = 2 − 1 = 1
Since f (0) = −1 and f (1) = 1, there exists x = α such that
f (α) = 0 (IMVT)
∴ The equation has a root in the interval [0, 1].
f (x) = x3 + x − 1
f ′(x) = 3x2 + 1
By N–R:
f (xn)
xn+1 = xn − _____
f ′(xn)
x3 + x − 1
3x n + 1
n
n
= xn − __________
2
xn(3xn2 + 1) − (x3n + xn − 1)
= _______________________
3x2n + 1
3xn3 + xn − x3n − xn + 1
= ___________________
3x2n + 1
2xn3 + 1
= _______
3x 2n + 1
249
M O DUL E 2
Let us choose a starting value of x1 = 0.5
2x13 + 1 __________
2(0.5)3 + 1 ____
x2 = _______
=
= 1.25 = 0.7143
3x21 + 1 3(0.5)2 + 1 1.75
2(0.7143)3 + 1
x3 = ____________
= 0.6832
3(0.7143)2 + 1
2(0.6832)3 + 1
x4 = ____________
= 0.6823
3(0.6832)2 + 1
Hence the root is 0.68 to two decimal places.
E X A M P L E 10
f (x) = x3 + 3x2 + 5x + 9
(a) Show that there is at least one root of the equation f(x) = 0 in the interval
[−3, −2].
2x3 + 3x 2 − 9
3x n + 6xn + 5
n
n
(b) Using the N–R formula, show that xn+1= ____________
.
2
(c) Hence find the root of the equation in the interval [−3, −2] to two decimal
places.
SOLUTION
(a) f(x) = x3 + 3x2 + 5x + 9
f(−2) = (−2)3 + 3(−2)2 + 5(−2) + 9 = −8 + 12 − 10 + 9 = 3
f(−3) = (−3)3 + 3(−3)2 + 5(−3) + 9 = −27 + 27 − 15 + 9 = −6
By the IMVT there exists a root in the interval [−3, −2].
(b) f (x) = x3 + 3x2 + 5x + 9
f ′(x) = 3x2 + 6x + 5
f(xn)
x3n + 3x2n + 5xn + 9
= xn − ________________
xn+1 = xn − _____
f ′(xn)
3x2n + 6xn + 5
xn(3x2n + 6xn + 5) − (xn3 + 3x2n + 5xn + 9)
= ___________________________________
3x2n + 6xn + 5
3xn3 + 6x2n + 5xn − xn3 − 3x2n − 5xn − 9 ____________
2xn3 + 3x2n − 9
________________________________
= 2
=
3xn + 6xn + 5
3x2n + 6xn + 5
(c) Let the first approximation x1 = −2.5
2(−2.5)3 + 3(−2.5)2 − 9
= −2.4571
x2 = _____________________
3(−2.5)2 + 6(−2.5) + 5
2(−2.4571)3 + 3(−2.4571)2 − 9
= −2.4562
x3 = ___________________________
3(−2.4571)2 + 6(−2.4571) + 5
∴ α = −2.46 (2 d.p.)
250
MODULE 2tCHAPTER 12
E X A M P L E 11
Show graphically that the equation ex = 25x − 10 has exactly two roots.
Find an interval of unit width which contains the smaller root of the equation.
Use the Newton–Raphson method to find this root of the equation to two
decimal places.
y
SOLUTION
y = 25x – 10
y = ex
x
Since the graphs intersect at two points there are two solutions to the equation.
Let f(x) = ex + 10 − 25x
f(0) = e0 + 10 − 25(0) = 11
f(1) = e + 10 − 25 = −12.2817
By the IMVT there is a root in the interval (0, 1).
f(x) = ex + 10 − 25x
f′(x) = ex − 25
f (xn)
exn + 10 − 25xn
= xn − _____________
xn+1 = xn − _____
exn − 25
f ′(xn)
xn exn − 25xn − exn − 10 + 25xn
= ___________________________
exn − 25
xn
xn
xn e − e − 10
= ______________
exn − 25
Using 0.5 as the first approximation, we have
x1 = 0.5
0.5e − e −10 = 0.4635
x2 = _______________
e0.5 − 25
0.4635e0.4635 − e0.4635 − 10
x3 = ______________________
= 0.4635
e 0.4635 − 25
So the root is 0.46 to 2 d.p.
0.5
0.5
The Newton–Raphson formula does not always converge to the root of the equation.
The N–R method may not converge if
(i) the starting value is too far from the root of the equation;
(ii) the starting value is close to a stationary point;
251
M O DUL E 2
(iii) the same values keep recurring – for example if you start with x1 = 1, obtain
x2 = 1.6, then x3 = 1 and x4 = 1.6 and so on, the sequence enters an infinite
cycle;
(iv) if the function is not continuously differentiable in the neighbourhood of the root.
EXERCISE 12A
In questions 1–6, use the IMVT to show that there is a root of the function in the
given interval.
1 x3 + 3x2 + 5x + 5 = 0 [−2, −1] 2 __12 e2 = x + 1 [1, 2]
3 2 ln x = 3 − x [1, 2]
4 x + 3 sin x = 2 [0.45, 0.55]
5
x4 + x = 4
6
[1, 2]
2x3 + 1 = 6x2 − 3x
[2, 3]
In questions 7–12, use the IMVT to find an interval [n, n + 1], where n ∊ ℤ which
contains at least one root of the equation.
__
7
cos x = √ x
8
9
x4 − x = 1
10 x3 + 2x = 4
11 x4 = 20
x2 + 2 = 5x
12 tan2 x = 1 − x2
13 Apply the Newton–Raphson method to the equation x2 = 1000 to show that
_____
1000 . Use this result to find √1000 correct to six decimal
1 x + ____
xn+1 = __
xn
2 n
places.
(
)
14 Show that 2x3 − 6x2 = −3x − 1 has a root in the interval [2, 3]. Find an
approximation to the root to six decimal places by using the Newton–Raphson
method.
15 (a) Show that the equation 2 ln x + x = 2 has a root in the interval [1, 2].
4xn − 2xnln xn
.
(b) Use the Newton–Raphson method to show that xn+1 = ____________
2 + xn
(c) Use the result in part (b) to find the root of the equation in part (a) to
four decimal places.
16 (a) Find an interval [n, n + 1] where n ∊ ℤ which contains a root of the
π.
equation t + sin t = __
3
(b) Using the Newton–Raphson method, show that
π
tn cos tn − sin tn + __
3.
_________________
tn+1 =
1 + cos tn
(c) Use the iterative formula to find t2 and t3 to four decimal places.
17 (a) By sketching the graphs of y = 5 − θ and y = 5e−θ, show that there are two
roots to the equation 5 − θ = 5e−θ.
(b) Use the IMVT to show that the positive root of the equation is between 4 and 5.
(c) Use the Newton–Raphson method to find the positive root to two decimal
places.
252
MODULE 2tCHAPTER 12
18 (a) Show that the equation x2 = 4 cos x has a positive root in the interval [1, 2].
(b) Use the Newton–Raphson method to show that
x2n + 4xnsin xn + 4 cos xn
xn+1 = _____________________
2xn + 4 sin xn
(c) Hence find the positive root of the equation to two decimal places.
19 (a) By sketching suitable graphs, show that the equation 2x − 1 = 3 cos x has
exactly one real root.
(b) Use the IMVT to show that this root is between 1 and 2.
(c) Use the Newton–Raphson method to find the root to three decimal places.
20 Show that there is a root of the equation 7x3 + 34x2 = 4 − 23x between 0 and 1.
Hence find the root to two decimal places using linear interpolation.
21 Show that the equation x − 2 = x5 has a root between −1 and −2. Use linear
interpolation to obtain two approximations to the root of this equation.
22 Find an interval which contains the root of the equation x3 = 3x − 1. Hence
find this root using linear interpolation. Give your answer to three decimal places.
23 The equation f(x) = x³ − 3x² − 4x + 11 = 0 has a root in the interval [3, 4] find
this root to three decimal places, using the linear interpolation formula.
24 Show that the equation f(x) = 3x + sin x − ex = 0 has a root in the interval
[0, 1]. Find this root to two decimal places using linear interpolation.
SUMMARY
Roots of an equation
IMVT
f(x) is continuous over [a, b], let
d ∈[f(a), f(b)] then there must
be a c such that f(c) = d.
N–R formula
xn + 1 = xn –
f(xn)
f ’(xn)
Used to show the existence of
a root in an interval.
α x3
x2
x1
Convergence of N–R
Linear interpolation
N–R may not converge if
t the starting value is
too far from the root
t the starting value is
close to a stationary
point
t the sequence enters
an infinite cycle
tthe function is
not continuously
differentiable.
If α is in [a, b] then an
approximation to α is
a|f(b)| + b|f(a)|
x1 =
|f(a)| + |f(b)|
A first approximation
must be known.
The accuracy of the
result will indicate
when to end the
iterative process.
253
M O DUL E 2
Checklist
Can you do these?
■ Test for the existence of a root.
■ Use the intermediate value theorem.
■ Use interval bisection to find the root of an equation.
■ Use linear interpolation to find the root of an equation.
■ Explain how the Newton–Raphson method works.
■ Find successive approximations for any root.
■ Use the Newton–Raphson method to find successive approximations to f (x) = 0
where f (x) is differentiable.
■ Understand when the Newton–Raphson method fails to converge to a root of
the equation.
254
MODULE 2tCHAPTER 13
CHAPTER 13
Power Series
At the end of this chapter you should be able to:
■ find the Taylor series for a function f(x)
■ identify the properties of the Maclaurin series
■ find the Maclaurin expansion for certain functions
■ know the values of x for which the expansion is valid.
KEYWORDS/TERMS
QPMZOPNJBMtWBMJEtQPXFSTFSJFTt5BZMPSTFSJFTt
5BZMPSTFYQBOTJPOtEFHSFFt.BDMBVSJOFYQBOTJPO
255
M O DUL E 2
Power series and functions
1 BTBpolynomial in x BT
6TJOHUIFCJOPNJBMFYQBOTJPOXFDBOXSJUFf(x) = ______
1x
1 − __
GPMMPXT
2
1 x –1
1 = 1 − __
______
2
1
__
1− x
2
3
(−1)(−2)
(−1)(−2)(−3) __
1 x + _________
1 2 _____________
= 1 + (−1) − __
− __
− 21 x + …
2
2x +
2!
3!
1 x2 + __
1 x3 + … GPS−2 < x < 2.
1 x + __
= 1 + __
4
2
8
1 3 …
1
1 2 __
__
)FODFf(x) = 1 + __
2 x + 4 x + 8 x + GPS−2 < x < 2.
(
)
(
)
(
)
(
)
ćFDPOEJUJPO−2 < x < 2 UFMMTVTUIBUUIFSJHIUIBOETJEFJTFRVBMUPUIFMFęIBOE
TJEFJGBOEPOMZJGx JTXJUIJOUIJTSBOHF*Gx JTPVUTJEFUIJTSBOHFUIFOUIFMFęIBOE
TJEFBOEUIFSJHIUIBOETJEFXJMMOPUCFFRVBM8FTBZUIBUUIFTFSJFTDPOWFSHFTGPS
−2 < x < BOEUIBUUIFFYQBOTJPOJTvalidGPSUIJTSFHJPOPOMZ
.BOZGVODUJPOTDBOCFXSJUUFOJOUIFGPSNa0 + a1x + a2x2 + a3x3 + … UIBUJT BT
BOJOĕOJUFQPMZOPNJBMJOx.ćFTFQPMZOPNJBMTBSFDBMMFEpower series XJUIDFOUSF
"QPXFSTFSJFTXJUIDFOUSFc JTPGUIFGPSN
a0 + a1(x − c) + a2(x − c)2 + a3(x − c)3 + …
'PSNBOZGVODUJPOTUIBUDBOCFSFQSFTFOUFEBTBQPXFSTFSJFTUIFSFQSFTFOUBUJPO
JTWBMJEGPSBSBOHFPGWBMVFTPGx POMZćFCJOPNJBMFYQBOTJPOBMMPXTVTUPXSJUF
GVODUJPOTPGx BTBQPXFSTFSJFT5BZMPSTFYQBOTJPOBOE.BDMBVSJOTFYQBOTJPOBMMPX
VTUPXSJUFBXJEFSTFUPGGVODUJPOTPGx BTQPXFSTFSJFT
Taylor expansion
A Taylor series BMMPXTGVODUJPOTUPCFXSJUUFOBTBQPXFSTFSJFTPSBTBQPMZOPNJBM
1PXFSTFSJFTHJWFBDMPTFBQQSPYJNBUJPOUPUIFPSJHJOBMGVODUJPOBOEBSFFBTZUPEJG
2
GFSFOUJBUFBOEJOUFHSBUF"OJOUFHSBMPGUIFGPSN∫e−x dx DBOCFGBDJMJUBUFECZUIFVTF
PGB5BZMPSTFSJFTFYQBOTJPO*OPSEFSUPĕOETVDIBTFSJFT UIFGPMMPXJOHDPOEJUJPOT
IBWFUPCFJOQMBDF
1. The function f(x) has to be infinitely differentiable (that is, we can find the first
derivative, second derivative, third derivative, and so on).
2. The function f(x) has to be defined in a region near the value x = a.
-FUf(x CFBGVODUJPOUIBUNBZCFFYQBOEFEBTBQPMZOPNJBMPGUIFGPSN
Pn(x) = c0 + c1(x − a) + c2(x − a)2 + … + cn(x − a)n
XIFSFc0, c1, c2, …, cn BSFBMMDPOTUBOUT
ćFWBMVFPGPn(x BUaBOEJUTnEFSJWBUJWFTBSFFRVJWBMFOUUPUIFWBMVFPGf(x BUx = a
BOEJUTnEFSJWBUJWFTBUx = a JF
Pn(a) = f(a),
(n)
P′n(a) = f ′(a), P″n(a) = f ″(a), …, P(n)
n (a) = f (a)
4JODF
Pn(x) = c0 + c1(x − a) + c2(x − a)2 + … + cn(x − a)n
256
[1]
MODULE 2tCHAPTER 13
XIFOx = a XFIBWFPn(a) = c0 = f(a)
%JČFSFOUJBUJOH<> XFIBWF
P′n(x) = c1 + 2c2(x − a) + 3c3(x − a)2 + … + ncn(x − a)n−1
P″n(x) = 2c2 + 2 × 3c3(x − a) + … + (n − 1)ncn(x − a)n−1
P‴n (x) = 3!c3 + … + (n − 2)(n − 1)(n)cn(x − a)n−3
8IFOx = a XFIBWFP′n(a) = c1 = f ′(a)
1 P″ (a) = __
1 f ″(a)
P″n(a) = 2c2 ⇒ c2 = __
2! n
2!
1 P‴ (a) = __
1 f ‴(a)
P‴n (a) = 3!c3 ⇒ c3 = __
3! n
3!
f ″(a)
f ‴(a)
f (n)(a)
f(x) = f(a) + (x − a)f ′(a) + (x − a)2 _____ + (x − a)3 _____ + … + (x − a)n _____ + …
3!
n!
2!
∞
(k)(a)
f_____
=
(x − a)k
k!
k=0
∞
f (k)(a)
_____
"Tn → ∞, f(x) =
(x − a)kJTDBMMFEUIFTaylor expansion or Taylor
k!
k=0
series PGUIFGVODUJPOćJTFYQBOTJPODPOWFSHFTSBQJEMZGPSWBMVFTPGxUIBUBSFDMPTF
UPaćFdegreePSPSEFSPGUIFQPMZOPNJBMJTUIFIJHIFTUQPXFSPG x − a JOUIF
polynomial.
∑
∑
EXAMPLE 1
'JOEUIF5BZMPSTFSJFTFYQBOTJPOPGf(x) = exXJUIa = 2.
SOLUTION
'JSTUXFĕOEUIFEFSJWBUJWFTBOEFWBMVBUFFBDIBUx =
f(x) = ex,
f(2) = e2
f (1)(x) = ex,
f (1)(2) = e2
f (2)(x) = ex,
f (2)(2) = e2
f (3)(x) = ex,
f (3)(2) = e2
f (4)(x) = ex,
f (4)(2) = e2
6TJOHUIF5BZMPSTFSJFTFYQBOTJPO XFIBWF
f (2)(a)
f (3)(a)
f(x) = f(a) + (x − a) f (1)(a) + (x − a)2 ______ + (x − a)3 ______ + …
2!
3!
BOETVCTUJUVUJOHXJUIa = XFIBWF
(x − 2)4 2 …
(x − 2)3 2 _______
e2 + _______
ex = e2 + (x − 2)e2 + (x − 2)2 __
e+
e +
2!
3!
4!
(x − 2)2 (x − 2)3 (x − 2)4
= e2 1 + (x − 2) + _______ + _______ + _______ + …
2!
3!
4!
∞
r
(x − 2)
_______
= e2
r!
r=0
[
]
∑
)FODFUIF5BZMPSTFSJFTFYQBOTJPOPGf(x) = exXJUIa =JT
∞
(x − 2)r
_______
ex = e2
r!
r=0
∑
257
M O DUL E 2
EXAMPLE 2
SOLUTION
$BMDVMBUFUIFBQQSPYJNBUFWBMVFPGTJO¡UPFJHIUEFDJNBMQMBDFTVTJOHB5BZMPS
TFSJFTPGEFHSFF
π. We also need to convert ¡ to
We need to find the Taylor series of sin x about x = __
4
radians. Since we are looking for an expansion of degree 3, we find the first three
π.
derivatives and their values at
x = __
__
4
2
√
π
__
___
-FUf(x) = sin x, f ( ) =
4
2
__
π
√
π
__
__
___
f ′(x) = cos x, f ′( 4 ) = cos = 2
4
2
__
2
√
π
π
___
__
f ″(x) = −sin x, f ″( 4 ) = −TJO ( __ ) = − 2
4
__
√2
π
π
___
__
__
f ‴(x) = −cos x, f ‴( 4 ) = −cos ( ) = − 2
4
ćF5BZMPSTFSJFTFYQBOTJPOHJWFT
(x − a)2
(x − a)3
_______
f(x) = f(a) + (x − a) f ′(a) + _______
f
″(a)
+
f ‴(a) + …
2!
3!
π 2 __
π 3 __
__
__
x − __
x − __
(
(
√2
√2
4 ) ___
4 ) ___
√2
π
√2
___
__
________
________
___
…
−
∴ TJOx = 2 + ( x − 4 )
−
2
2 +
2
3!
2
π2
π3
__
x − __
x − __
(
)
(
4
4)
2
√
π
__
________ − ________
= ___
+…
2 1 + (x − 4 ) −
2
π + ____
π
/PX¡ = __
4
180 SBEJBOT TJODF¡ = ¡ + ¡
π
π + ____
)FODFTJO¡ = TJO __
180
4
π + ____
π − __
π2
π + ____
π − __
π3
__
__
__
(
)
(
2
√___
4
4
4
4) +…
180
180
π
π
π
__
____
__
______________
______________
= 2 1+ ( +
−( ) −
−
)
4 180
4
2
π 2
π 3
__
____
____
( 180 ) ______
( 100 ) …
√2
π
____
______
= ___
2 1 + 180 − 2! − +
[
π radians = ¡
π
⇒ ¡ = ____
180 rad
45π __
π
____
¡ = 180
= 4 rad
__
[
[
]
(
)
[
]
]
√2 [
]
= ___
2 1 + 0.017 453 29 − 0.000 152 308 7 −
= EQ
)FODFTJO¡= EQ
258
EXAMPLE 3
1 with centre a = 3.
Find the third order Taylor expansion for f(x) = _____
x+2
SOLUTION
We find the first three derivatives and evaluate these at x = 3.
1 , f(3) = _____
1 = __
1
f(x) = _____
x+2
3+2 5
1 = (x + 2)−1 VTJOHUIFDIBJOSVMF XFIBWF
4JODFf(x) = _____
x+2
−1 = ___
−1
−1 , f ′(3) = _______
f ′(x) = _______
25
(x + 2)2
(3 + 2)2
2 ,
f ″(x) = _______
(x + 2)3
2
2
f ″(3) = _______
= ____
(3 + 2)3 125
− ,
f ‴(x) = _______
(x + 2)4
− = ___
− = ____
−
f ‴(3) = _______
(3 + 2)4
54
]
MODULE 2tCHAPTER 13
6TJOHUIF5BZMPSTFSJFTFYQBOTJPO
(x − a)2
(x − a)3
f(x) = f(a) + (x − a) f ′(a) + _______ f ″(a) + _______ f ‴(a) + …
2!
3!
XJUIa = XFIBWF
(x − 3)2 ____
(x − 3)3 ____
− + …
−1 + _______
2 + _______
1 + (x − 3) ___
1 = __
_____
x+2 5
25
125
2!
3!
1
1 (x − 3) + ____
1
1 − ___
____
2
3
…
= __
125 (x − 3) − (x − 3) +
5 25
1 XJUIDFOUSFJT
)FODFUIFUIJSEPSEFS5BZMPSTFSJFTGPSf(x) = _____
x+2
1 = __
1 (x − 3) + ____
1 (x − 3)3 + …
1 − ___
1 (x − 3)2 − ____
_____
125
x + 2 5 25
(
)
(
)
( )
__
EXAMPLE 4
'JOEUIFĕSTUGPVSOPO[FSPUFSNTPGUIF5BZMPSFYQBOTJPOPGf(x)
= √x BCPVU
___
x =)FODFFWBMVBUF√9.3 .
SOLUTION
f(x) = √x ,
__
__
f(a) = √9 = 3
1 −__21
f ′(x) = __
2x ,
1
1 −__12 _____
1
__
f ′(a) = __
2 (9) = 2 × 3 = 1 −__23
f ″(x) = − __
4x ,
3 −__25
f ‴(x) = __
8x ,
Notes
(i) The higher
the degree
of the Taylor
polynomial
the better it
approximates
to the given
function.
(ii) The Taylor
expansion
centred at 0
is called the
Maclaurin
expansion.
1 −__23 ____
−1
f ″(a) = − __
4 (9) = 108
3 −__25 ____
1
f ‴(a) = __
8 (9) = 6TJOH5BZMPSFYQBOTJPO
(x − a)3
(x − a)2
_______
f
″(a)
+
f ‴(a) + …
f(x) = f(a) + (x − a) f ′(a) + _______
2!
3!
XIFOa = XFIBWF
__
√x
(x − 9)2 ____
(x − 9)3 ____
1 + _______
−1 + _______
1 +…
= 3 + (x − 9) __
2!
3!
108
1
1
1
____
_____
2
3
…
= 3 + __
(x − 9) − (x − 9) + 3888 (x − 9) +
( )
( )
( )
4VCTUJUVUJOHx =JOUPUIFFYQBOTJPOHJWFT
___
1
1
1
__
____
_____
√9.3 = 3 + (9.3 − 9) − (9.3 − 9)2 + 3888 (9.3 − 9)3
= 3 + 0.05 − + = EQ
___
Hence √9.3 = 3.049 59 (5 d.p.)
The Maclaurin expansion
ćF.BDMBVSJOTFSJFTJTVTFEUPBQQSPYJNBUFWBMVFTPGBGVODUJPOf (x ćF
BQQSPYJNBUJPOJTWFSZBDDVSBUF BTMPOHBTUIFx-WBMVFTBSFDMPTFUPćF.BDMBVSJO
TFSJFTHFOFSBUFTQPMZOPNJBMTćFTFSJFTJTJOĕOJUFBOEXIFOBQQSPYJNBUJOHf (x)
ZPVXJMMVTFBEFĕOJUFOVNCFSPGUFSNTUIFNPSFUFSNTZPVVTFUIFCFUUFSZPVS
BQQSPYJNBUJPOXJMMCF
259
M O DUL E 2
Notes
(i) f (x) must be
differentiable.
(ii) f (x) must
exist at x = 0.
(iii) The derivatives of f (x)
must exist at
x = 0.
(iv) The series
is valid only
within certain
values of x.
(v) f (n) (0) is the
nth derivative
evaluated at
x = 0.
-FUf (x CFBEJČFSFOUJBCMFGVODUJPOXIJDIDBOCFXSJUUFOBTBQPXFSTFSJFTPGx.
ćFO‫ڀ‬UIF.BDMBVSJOTFSJFTGPSf(x JT
x3 f ‴(0) + . . . + __
xn f (n)(0) + . . .
x2 f ″(0) + __
f(x) = f (0) + x f ′(0) + __
2!
3!
n!
=
∞
∑
(n)
n
f (0)x
_______
n=0
n!
ćF.BDMBVSJOFYQBOTJPOJTPCUBJOFEBTGPMMPXT
-FUf (x) = a + a1x + a2x2 + a3x3 + QPXFSTFSJFT
f (0) = a + a1(0) + a2(0)2 + . . . = a
f ′(x) = a1 + 2a2x + 3a3x2 + . . .
f ′(0) = a1
f ″(x) = 2a2 +a3x + . . .
f ″(0)
⇒ a2 = _____
2!
f ″(0) = 2a2
f ‴(x) =a3 + . . .
f ‴(0) f ‴(0)
⇒ a3 = _____ = _____
3!
f ‴(0)
f ″(0)
4VCTUJUVUJOHa = f(0), a1 = f ′(0), a2 = _____, a3 = _____
3 JOUPUIFPSJHJOBMQPXFS
2
TFSJFTXFIBWF
f ‴(0) =a3
2
3
x f ‴(0) + . . .
x f ″(0) + __
f (x) = f (0) + x f ′(0) + __
2!
3!
EXAMPLE 5
Find the Maclaurin series for f(x) = e x.
SOLUTION
Since f (x) = ex
f (0) = e0 = 1
f ′(x) = e x
f ′(0) = e0 = 1
f ″(x) = e x
f ″(0) = e0 = 1
f ‴(x) = e x
f ‴(0) = e0 = 1
6TJOHUIF.BDMBVSJOFYQBOTJPO
2
3
x f ‴(0) + . . .
x f ″(0) + __
f (x) = f (0) + xf ′(0) + __
2!
3!
BOETVCTUJUVUJOHf (0) = 1, f ′(0) = 1, f ″(0) = 1, f ‴(0) = XFIBWF
2
3
x + __
x +...
f (x) = 1 + x + __
2! 3!
260
MODULE 2tCHAPTER 13
x3 + __
x4 + . . .
x2 + __
)FODFex = 1 + x + __
2! 3! 4!
ćJTFYQBOTJPOJTWBMJEGPSBMMWBMVFTPGx JF∀x ∈ ℝ.
EXAMPLE 6
Find the Maclaurin series for f (x) = sin x up to and including the term in x5.
SOLUTION
f (x) = sin x
x = 0 ⇒ f (0) = sin 0 = 0
f (1) (x) =DPTx
x = 0 ⇒ f (1)(0) = cos 0 = 1
f (2)(x) = −TJOx
x = 0 ⇒ f (2)(0) = −sin 0 = 0
f (3)(x) = −DPTx
x = 0 ⇒ f (3)(0) = −cos 0 = −1
f (4)(x) =TJOx
x = 0 ⇒ f (4)(0) = sin 0 = 0
f (5)(0)=DPTx
x = 0 ⇒ f (5)(0) = cos 0 = 1
x2 f (2)(0) + __
x3 f (3)(0) + . . .
4VCTUJUVUJOHJOUPf (x) = f (0) + x f (1)(0) + __
2!
3!
x3 (−1) + __
x4 (0) + __
x5 (1) + . . .
x2 (0) + __
XFIBWFTJOx = 0 + x(1) + __
2!
3!
4!
5!
3
5
x + __
x − . . . ∀x ∈ ℝ.
= x − __
3! 5!
EXAMPLE 7
Find the Maclaurin series for f (x) = cos x up to and including the term in x 6 and use
this series to find an approximate value of cos (0.1).
SOLUTION
By the Maclaurin expansion
2
3
x f ‴(0) + . . .
x f ″(0) + __
f (x) = f (0) + x f ′(0) + __
2!
3!
4JODFXFBSFFYQBOEJOHVQUPx XFOFFEUPĕOEVQUPUIFTJYUIEFSJWBUJWF
f (x) = DPTx
x = 0 ⇒ f (0) = cos 0 = 1
f (1)(x) = −TJOx
x = 0 ⇒ f (1)(0) = −sin 0 = 0
f (2)(x) = −DPTx
x = 0 ⇒ f (2)(0) = −cos 0 = −1
261
M O DUL E 2
f (3)(x) =TJOx
x = 0 ⇒ f (3)(0) = sin 0 = 0
f (4)(x) =DPTx
x = 0 ⇒ f (4)(0) = cos 0 = 1
f (5)(x) = −TJOx
x = 0 ⇒ f (5)(0) = −sin 0 = 0
f (x) = −DPTx
x = 0 ⇒ f (0) = −cos 0 = −1
∴ f (0) = 1
f (1)(0) = 0
f (2)(0) = −1
f (3)(0) = 0
f (4)(0) = 1
f (5)(0) = 0
f (0) = −1
4VCTUJUVUJOHJOUPUIF.BDMBVSJOFYQBOTJPO XFHFU
x2 (−1) + __
x3 (0) + __
x4 (1) + __
x5 (0) + __
x (−1) + . . .
f (x) = 1 + x(0) + __
2!
3!
4!
5!
2
4
x
x
x
∴ DPTx = 1 − __ + __ − __ + . . ., ∀x ∈ ℝ
2! 4! 4VCTUJUVUJOHx = XFHFU
(0.1)2 (0.1)4 (0.1)
DPT = 1 − _____ + _____ − _____ + . . .
2!
4!
= 1 − 0.005 + 0.000 004 177 − 0.000 000 001 39 = 0.995 004 2
262
EXAMPLE 8
1
Find the Maclaurin series for f (x) = ______
up to and including the term in x4.
(x + 2)2
Verify the expansion by using the binomial expansion.
SOLUTION
x2 f (2)(0) + __
x3 f (3)(0) + __
x4 f (4) (0) + . . .
f (x) = f (0) + x f (1)(0) + __
2!
3!
4!
f (x) = (x + 2)−2
1
f (0) = (0 + 2)−2 = 2−2 = __
4
(1)
−3
f (x) = −2(x + 2)
(using the chain rule)
1
−2
f (1)(0) = −2(0 + 2)−3 = ___ = − __
4
8
(2)
−4
f (x) = 6(x + 2)
3
6 = __
f (2) (0) = 6 (0 + 2)−4 = ___
16 8
f (3) (x) = −24 (x + 2)−5
3
____ = − __
f (3)(0) = −24 (0 + 2)−5 = −24
4
32
MODULE 2tCHAPTER 13
f (4)(x) = 120(x + 2)−6
15
120 = ___
f (4)(0) = 120(0 + 2)−6 = ____
8
2
x __
3 + __
x3 ___
−3 + __
15
x4 ___
1 − __
1 x + __
1
= __
∴ _______
2
4 4
4! 8
2! 8
3! 4
(x + 2)
5 x4 − . . .
3x2 − __
1 x + ___
1 − __
1 x3 + ___
= __
64
4 4
8
16
( )
( )
( )
ćFCJOPNJBMFYQBOTJPOJT
1 x −2
(2 + x)−2 = 2 1 + __
2
1 x −2
= 2−2 1 + __
2
(−2)(−3) __
(−2)(−3)(−4) __
1 1 + (−2) __
1 x + _________
1 x 2 + _____________
1x
= __
4
2
2
2
2!
3!
((
[
))
)
(
( )
( )
( )
(−2)(−3)(−4)(−5) 1 4 . . .
+ _________________ __
x +
2
4!
5 x4 + . . .
3 x2 − __
4 x3 + ___
1 1 − x + __
= __
4
4
8
16
5 x4 + . . .
3 x2 − __
1 x + ___
1 x3 + ___
1 − __
= __
4 4
8
16
64
(
)
( )
3
]
)FODFUIFFYQBOTJPOTBSFUIFTBNF
EXAMPLE 9
Find the first two non-zero terms in the power series expansion of ln(1 + sin 2x).
SOLUTION
We use
x2 f (2)(0) + __
x4 f (4)(0) + . . .
x3 f (3)(0) + __
f (x) = f (0) + x f (1)(0) + __
3!
2!
4!
XIFSFf (x) = ln (1 + sin 2x)
8IFOx = 0, f (0) = ln (1 + sin 2(0)) = ln 1 = 0
%JČFSFOUJBUJOHf (x XSUx
2 cos 2x
f (1)(x) = ________
(chain rule)
1 + sin 2x
2 cos 0 = 2
f (1)(0) = _______
1 + sin 0
(1 + sin 2x)(−4 sin 2x) − 2 cos 2x(2 cos 2x)
(quotient rule)
f (2)(x) = ___________________________________
(1 + sin 2x)2
−4 sin 2x − 4
−4 sin 2x − 4 sin2 2x − 4 cos2 2x = ____________
= __________________________
(1 + sin 2x)2
(1 + sin 2x)2
−4 sin 0 − 4 = −4
f (2)(0) = ___________
(1 + sin 0)2
4VCTUJUVUJOH f (0) = 0, f (1)(0) = 2, f (2)(0) = −4JOUP
Can you use the
expansions of
ln(1 + x) and
sin x to find the
expansion of
ln(1 + sin 2x)?
2
x f (2)(0) + . . .
f (x) = f (0) + xf (1)(0) + __
2!
XFHFU
2
x (−4)
f (x) = 0 + x(2) + __
2!
Hence ln (1 + sin 2x) = 2x − 2x2
263
M O DUL E 2
E X A M P L E 10
Find the first five terms in the power series expansion of ln (1 + 2x), stating the range
of values of x for which the expansion is valid.
)FODF XSJUFEPXOUIFĕSTUĕWFUFSNTPGUIFFYQBOTJPOln (1 − 2x).
x3 + ___
x5 + . . .
1 + 2x = 4x + ___
4IPXUIBUMO ______
5
1 − 2x
3
(
SOLUTION
)
Let f (x) = ln (1 + 2x)
2
f (1)(x) = ______
= 2(1 + 2x)−1
1 + 2x
f (2)(x) = −4(1 + 2x)−2
f (3)(x) = 16(1 + 2x)−3
f (4)(x) = −96(1 + 2x)−4
f (5)(x) = 768(1 + 2x)−5
f (0) = ln (1 + 2(0)) = 0
2
=2
f (1)(0) = _______
1 + 2(0)
f (2)(0) = −4(1 + 2(0))−2 = −4
f (3)(0) = 16(1 + 2(0))−3 = 16
f (4)(0) = −96(1 + 2(0))−4 = −96
f (5)(0) = 768(1 + 2(0))−5 = 768
4VCTUJUVUJOHJOUP
x2 f (2)(0) + __
x3 f (3)(0) + __
x4 f (4)(0) + . . .
f (x) = f (0) + x f (1)(0) + __
2!
3!
4!
XFHFU
x3 (16) + __
x4 (−96) + __
x5 (768) + . . .
x2 (−4) + __
f (x) = 0 + x(2) + __
2!
3!
4!
5!
8 x3 − 4x4 + ___
32 x5 + . . .
= 2x − 2x2 + __
5
3
8 x3 − 4x4 + ___
32x5
)FODFMO(1 + 2x) = 2x − 2x2 + __
5
3
1 < x ≤ __
1.
ćJTFYQBOTJPOJTWBMJEGPS−__
2
2
3FQMBDJOHxCZ−xJOUIFFYQBOTJPOPGMO + 2x XFHFU
8 x3 − 4x4 − ___
32 x5
ln (1 − 2x) = −2x − 2x2 − __
5
3
1 + 2x = ln (1 + 2x) − ln (1 − 2x)
/PXMO _______
1 − 2x
(
)
(
)
8 x3 − 4x4 + ___
32 x5 − −2x − 2x2 − __
8 x3 − 4x4 − ___
32 x5 + . . .
= 2x − 2x2 + __
5
5
3
3
32 x5 + . . .
8 x3 + 2 ___
= 4x + 2 __
5
3
x5 + . . .
x3 + ___
= 4x + ___
5
3
( )
264
(
)
MODULE 2tCHAPTER 13
Maclaurin expansions of some common functions
2
3
x +...
x + __
ex = 1 + x + __
2! 3!
3
, ∀x ∈ ℝ
5
x + __
x − . . . , ∀x ∈ ℝ
sin x = x − __
3! 5!
x2 + __
x4 − __
x6 + . . . , ∀x ∈ ℝ
cos x = 1 − __
2! 4! 6!
n(n − 1)
(1 + x)n = 1 + nx + ________ x2 + . . . , −1 < x < 1
2!
2
3
4
x − __
x +...,
x + __
ln (1 + x) = x − __
4
2
3
x5 − __
x7 + . . . ,
x3 + __
tan−1(x) = x − __
5
7
3
−1 < x ≤ 1
−1 ≤ x ≤ 1
EXERCISE 13A
In questions 1–7, find the Maclaurin series for the function up to the term in x4.
_____
1
f(x) = ln (1 + x)
2 f(x) = √4 + x
3
f(x) = e4x
4 f(x) = (1 + x)n
5
f(x) = tan x
6 f(x) = sin−1 x
7
f(x) = tan−1 x
Given that y = excos x show that
dy
d2y
dy
(a) ___ = y − ex sin x
(b) ___2 = 2 ___ − y
dx
dx
dx
Hence find the Maclaurin expansion up to and including the term in x3.
dy
9 Given that y = esin x show that ___ = y cos x. Find the Maclaurin series of y up to
dx
and including the term in x4.
π up to and including the
10 Show that the Maclaurin series for y = cos ( x + __
2)
3
5
x
x
__
__
5
term in x is y = −x + − .
3! 5!
11 (a) Find the Maclaurin series of ln (a + x), where a is a constant, up to and
including the term in x5.
8
(
)
(b) Obtain an expansion for ln (a − x) up to the term in x5.
x ___
x3
x5
a+x
__
___
...
(c) Show that ln ( _____
a − x ) = 2 a + 3a3 + 5a5 +
12 Find the first four non-zero terms of the Maclaurin series of f (x) = sin2 x.
[
]
13 Obtain the Maclaurin expansion of ex sin−1 (x), up to and including the term in
x4. Hence show that, when x is small, ex sin−1 x ≈ x + x2.
14 Write down the first four terms of the Maclaurin expansion of
1 .
(a) e2x (b) _____
1+x
2x
e
up to the term in x4.
Hence find the Maclaurin expansion of _____
1+x
265
M O DUL E 2
1
__
15 (a) Find the Maclaurin expansion of y = (1 + x) 4 as____
far as the term is x4.
4
Use the expansion to find an approximation of √1.03 correct to five
decimal places.
(b) Write down the Maclaurin expansion of ln (1 + x) up to the term in x4.
Use this expansion to identify the number of decimal places to which
ln(1.02) can be obtained correctly.
16 Given that f(x) = ln x, find the Taylor series for f(x) centred at a = 2.
17 Given that f(x) = cos x, find the Taylor series of degree 3 for f(x) centred at
__. Hence find cos 31¡ to five decimal places.
a=π
6
18 Find the first three non-zero terms in the Taylor series of f(x) = ex cos x, about
x = 0. Hence estimate
π
__
∫03 excos x dx correct to three decimal places.
1__ about
19 Find the first four non-zero terms of the Taylor’s expansion of f(x) = ___
√x
1 correct to four decimal places.
___
x = 4. Hence evaluate ____
√4.5
SUMMARY
Taylor and Maclaurin series
Taylor series
Maclaurin series
(i) f(x) must be infinitely
differentiable
(ii) f(x) must be defined
in a region near x = a.
f(x) = f(0) + xf ’(0) +
+
=Σ
(x n)
n!
x2
f ”(0) + ...
2!
f (n)(0) + ...
∞
n=0
f (n)(0)x n
n!
Taylor’s expansion about x = a is
2
f(x) = f(a) + (x – a) f ’(a) + (x – a) f ”(a) +
2!
(x – a)3
f ’’’(a) + ...
3!
∞ (x – a)n
=Σ
f (n) (a)
n = 0 n!
(i) f(x) must be differentiable
(ii) f(x) and its derivatives
must exist at x = 0
2
3
e x = 1 + x + x + x + ..., ∀x ∈ℝ
2!
3!
3
5
sin x = x – x + x – ..., ∀x ∈ℝ
3!
5!
2
4
6
(iii) The expansion is valid
cos x = 1 – x + x – x + ..., ∀x ∈ℝ
only within intervals
2!
4!
6!
for the differentiable
n(n – 1) 2
(1 + x)n = 1 + nx +
x + ..., –1 < x < 1
function.
2!
2
3
4
In(1 + x) = x – x + x – x +..., –1 < x ≤1
2
3
4
3
5
7
tan–1(x) = x – x + x – x +..., –1 ≤ x ≤1
3
5
7
Checklist
Can you do these?
■ Find the Taylor series for a function f(x).
■ Identify the properties of the Maclaurin expansion.
■ Write down the Maclaurin expansion for f (x).
■ Use the expansion for f (x) to find the Maclaurin expansion for ex, sin x, etc.
■ Identify the values of x for which the expansion is valid.
266
MODULE 2tCHAPTER 13
Module 2 Tests
Module 2 Test 1
1
(a) The coefficients of y and y2 in the expansion (1 + ay)(1 + by)8 are 0 and
−36 respectively. Find the values of a and b, given that a is negative and
b is positive.
[7]
(b) A sequence {un} is defined by the recurrence relation
1
un+1 = un + 2n + __
2 , u1 = 2, n ∈ ℕ.
(i) State the first four terms of the sequence.
2
[3]
(ii) Prove by mathematical induction that
2n2 − n + 3
un = ___________
[7]
2
(c) Obtain the first three non-zero terms in the power series expansion
in x of ln(1 + sin
x).
[8]
_____
1
+
x
_____
(a) (i) Expand
in ascending powers of x up to and including
1−x
the term in x2. State the set of values of x for which the expansion is
valid.
[7]
___
663
(ii) Hence show that √11 ≈ ____.
[3]
200
(b) Given that f(r) = r(r + 1), show that f(r + 1) − f(r) = 2(r + 1). Hence
√
n
find
∑2 r + 1 .
(
)
[6]
r=1
(c) Prove by mathematical induction that
n
n
1
________
= ________.
2
(2n + 1)
r=1 (4r − 1)
∑
3
[9]
The function f is given by f(x) = 4x4 − 16x + 1.
(a) When x > 1, show that f is strictly increasing.
[4]
(b) Show that f(x) = 0 has a root in each of the intervals [0, 1] and [1, 2].
[4]
(c) Show that f(x) = 0 has no other roots in the interval [1, 2].
[3]
(d) If xn is an approximation to the roots f(x) = 0 in the interval [1, 2],
show that the Newton–Raphson method gives
12x4n − 1
_________
xn+1 =
16x3n − 16
(e) Find the root in the interval [1, 2] to two decimal places.
[9]
[5]
267
M O DUL E 2
Module 2 Test 2
1
(a) (i)
1
2
1
Show that _______
= _____ − _____
( r2 − 1 ) r − 1 r + 1
[1]
n
2
∑______
r −1
2
[6]
(iii) Find also the sum to infinity, if it exists.
[2]
(ii) Hence, find
r=2
(b) Write down the first four terms of the sequence defined by a1 = 3 and
an = an−1 + 7 for n ≥ 1. Hence write down a conjecture for an in terms
of n. Use induction to prove that your conjecture is true.
[9]
(c) Prove by the principle of mathematical induction that 2n + 1 < 2n for
all positive integers n ≥ 3.
[7]
2
3 ______
(a) Find the expansion of √1 − 3x in ascending powers of x as far as the
term in x3 and state the values of x for which the expansion is valid.
3 __
By using a suitable value for x, estimate √3 to five decimal places.
[9]
(b) Show that 2x3 + x2 = 2 has a root in the interval [0.5, 1]. Use the linear
interpolation method to find this root, correct to two decimal places.
[9]
dy
(c) Given that y(0) = 2 and that ___ = 4y2 + 7, find the Maclaurin series of
dx
y up to and including the term in x3.
[7]
3
(a) (i)
Write down the first three non-zero terms of the Maclaurin
expansion of sin x and of cos x.
[3]
(ii) Use the binomial expansion and the expansion of cos x in part (i)
to find the expansion of (cos x)−1 ignoring terms in x5 and higher. [7]
(iii) Hence, find the expansion of tan x up to and including the term
in x5.
[5]
(iv) Using the expansion of tan x find the value of tan 0.001 to five
decimal places.
[3]
(b) Find the first three terms of Taylor’s expansion of f(x) = cos x about
π
x = __
3 . Hence estimate cos 61° to five decimal places.
268
[7]
3
Counting, Matrices
and Differential
Equations
269
M O DUL E 3
CHAPTER 14
Permutations and Combinations
At the end of this chapter you should be able to:
■ use the counting principles
■ understand the term permutation
■ find the number of ways of arranging n different objects
■ find the number of ways of arranging r out of n different objects
■ find the number of arrangements of n objects not all different
■ find the number of ways of arranging objects with restrictions
■ find the number of ways of choosing r out of n different objects
■ find the number of ways of choosing r out of n objects that are not all distinct
■ understand the difference between a permutation and a combination.
KEYWORDS/TERMS
DPVOUJOHQSJODJQMFTtNVUVBMMZFYDMVTJWFt
FYIBVTUJWFtQFSNVUBUJPOtBSSBOHFNFOUt
EJTUJODUtSFQFBUFEtDPNCJOBUJPO
270
MODULE 3tCHAPTER 14
The counting principles
ćFSFBSFUXPCBTJDcounting principlesUIFNVMUJQMJDBUJPOSVMFBOEUIF
BEEJUJPO‫ڀ‬SVMF
Notes
(i) This rule
makes use
of nonoverlapping
events.
(ii) The rule can
be extended
to three or
more events.
Multiplication rule
*GFWFOUADBOPDDVSJOmQPTTJCMFXBZTBOEXIFOUIJTIBTCFFOEPOFFWFOUBDBO
occur in n QPTTJCMFXBZT UIFSFBSFm × nQPTTJCMFXBZTGPSCPUIFWFOUTABOEB
UPPDDVS
-FUVTMPPLBUIPXUIFNVMUJQMJDBUJPOSVMFJTVTFE
4VQQPTFXFQMBDFGPVSEJČFSFOUDPMPVSFENBSCMFTJOBCBH8FXBOUUPĕOEUIF
OVNCFSPGXBZTPGUBLJOHPVUUXPPGUIFNBSCMFTJGXFBSFUBLJOHPVUPOFBUBUJNF
XJUIPVUSFQMBDFNFOU0OUIFĕSTUESBXXFDBOUBLFPVUBOZPGUIFGPVSDPMPVSTBęFS
UBLJOHPVUPOFDPMPVSXFIBWFUISFFSFNBJOJOHDPMPVSTBOEXFDBOUBLFPVUBOZPG
UIFTFGPSUIFTFDPOEESBX
ćFOVNCFSPGXBZTPGUBLJOHPVUUXPNBSCMFT= 4 × 3 =
EXAMPLE 1
A box contains six red marbles, four blue marbles and four green marbles. In how many
ways can you select two red marbles in two selections (select without replacement)?
SOLUTION
Since there are six red marbles, we can select any of the six on the first draw.
For the second selection we can choose any of the five remaining red marbles.
Since we need both red marbles the number of selections = 6 × 5 = 30.
Notes
(i) For this rule,
events A
and B must
be mutually
exclusive.
(ii) The rule can
be extended
to three or
more mutually exclusive
events.
5XPFWFOUTABOEBBSFmutually exclusiveJGBOEPOMZJGUIFZDBOOPUPDDVSBUUIF
TBNFUJNF'PSNVUVBMMZFYDMVTJWFFWFOUTA ∩ B = ∅
"TFUPGFWFOUTJTexhaustiveJGBUMFBTUPOFPGUIFFWFOUTNVTUPDDVS8IFO
U PTTJOH‫ڀ‬BDPJO UIFFWFOUTIFBE ) PSUBJM 5 BSFFYIBVTUJWFTJODFPOFPSUIF
PUIFS‫ڀ‬NVTUPDDVS
Addition rule
ćFBEEJUJPOSVMFJTVTFEXIFOXFBSFMPPLJOHBUPOFFWFOUorBOPUIFS
*GFWFOUADBOPDDVSJOmQPTTJCMFXBZTBOEFWFOUBDBOPDDVSJOnQPTTJCMFXBZT UIFSFBSFm + nXBZTGPSFJUIFSFWFOUAPSFWFOUBUPPDDVS
EXAMPLE 2
A box contains six red marbles, four blue marbles and four green marbles. In how
many ways can you select one red or one blue marble if only one selection is made?
SOLUTION
There are six red marbles and four blue marbles, therefore there are 6 + 4 = 10 red
or blue marbles in all.
ćFOVNCFSPGXBZTPGTFMFDUJOHBSFEPSCMVFNBSCMF=
271
M O DUL E 3
Permutations
A permutationJTBOarrangementPGPCKFDUTJOBTQFDJĕDPSEFS
EXAMPLE 3
Find the number of permutations of the three letters A, B and C.
SOLUTION
Method 1
8FOFFEUPBSSBOHFUISFFEJČFSFOUMFUUFST
8FDBOQMBDFBOZPGUIFUISFFMFUUFSTJOUIFĕSTUQPTJUJPO
After doing this we can place any of the remaining two letters in the second position.
The remaining letter is placed in the third position.
The number of permutations = 3 × 2 × 1 = 3! = 6.
Method 2
8FDBOMJTUBMMUIFEJČFSFOUBSSBOHFNFOUTBTGPMMPXT
ABC
ACB
BAC
BCA
CAB
CBA
There are 6 different arrangements.
Method 3: ćFCPYNFUIPE
4JODFXFBSFBSSBOHJOHUISFFMFUUFSTXFDBOESBXUISFFCPYFTBOEQMBDFUIFMFUUFSTJO
UIFCPYFTPOFBUBUJNFBOZPGUIFUISFFMFUUFSTHPJOUIFĕSTUCPY BOZPGUIFSFNBJOJOHUXPJOUIFTFDPOECPYBOEUIFMBTUJOUIFUIJSECPY
3
2
1
ćFOVNCFSPGBSSBOHFNFOUT= 3 × 2 × 1 =
Permutations of n distinct objects
ćFOVNCFSPGQFSNVUBUJPOTPGnEJČFSFOU distinct PCKFDUTJTn!
Note
All objects must
be different and
all objects must
be arranged.
ćFOPUBUJPOVTFEGPSUIFQFSNVUBUJPOPGnEJTUJODUPCKFDUTJTnPn = n
ćJTJTSFBEBTAn P n AQFSNVUFn out of nEJČFSFOUPCKFDUTPSAUIFOVNCFSPG
EJČFSFOUBSSBOHFNFOUTPGn out of nEJTUJODUPCKFDUT
8FDBOEFSJWFUIJTGPSNVMBCZMPPLJOHBUUIFCPYNFUIPE
4JODFXFBSFBSSBOHJOHnPCKFDUTXFESBXnCPYFTBOEĕMMUIFNPOFBUBUJNF8FDBO
QMBDFBOZPGUIFnPCKFDUTJOUIFĕSTUCPY BOZPGUIF n − SFNBJOJOHPCKFDUTJOUIF
TFDPOECPYBOETPPOVOUJMXFĕMMUIFMBTUCPYXJUIUIFMBTUPCKFDU
n
n–1
n–2
...
...
...
...
3
2
1
ćFOVNCFSPGBSSBOHFNFOUT= n × n − × n − × × 3 × 2 × 1
= n!
272
MODULE 3tCHAPTER 14
EXAMPLE 4
Find the number of permutations of the letters of the word IPOD.
SOLUTION
Method 1
4JODF*10%DPOTJTUTPGGPVSEJČFSFOUMFUUFSTUIFOVNCFSPGBSSBOHFNFOUTJT=
Method 2
8FDBOVTFUIFCPYNFUIPEBOEBSSBOHFUIFMFUUFSTVTJOHGPVSCPYFT*OUIFĕSTU
CPYXFDBOQMBDFBOZPGUIFGPVSMFUUFST BOZPGUIFUISFFSFNBJOJOHJOUIFTFDPOE
CPY BOZPGUIFUXPSFNBJOJOHJOUIFUIJSECPYBOEUIFMBTUMFUUFSHPFTJOUIFMBTU
CPYBTTIPXO
4
3
2
1
The total number of arrangements = 4 × 3 × 2 × 1 = 4! = 24.
EXAMPLE 5
Find the number of different arrangements of the digits of the number 123 479.
SOLUTION
Method 1
Ask yourself
Is order important?
Are all the
objects being
arranged?
ćFOVNCFS DPOTJTUTPGTJYEJČFSFOUEJHJUT
ćFOVNCFSPGQFSNVUBUJPOTPGnEJČFSFOUPCKFDUT= n!
8IFO n = XFIBWF
ćFOVNCFSPGEJČFSFOUQFSNVUBUJPOTPGUIFTJYEJHJUT= 6! =
Method 2 ćFCPYNFUIPE
"OZPGUIFTJYEJHJUTXJMMHPJOUIFĕSTUCPY BOZPGUIFSFNBJOJOHĕWFJOUIFTFDPOE
CPY BOETPPOVOUJMUIFMBTUEJHJUSFNBJOTUPQMBDFJOUIFMBTUCPY
6
5
4
3
2
1
/VNCFSPGBSSBOHFNFOUT= 6 × 5 × 4 × 3 × 2 × 1 =
EXAMPLE 6
Find the number of permutations of the letters of the name TRISHAN.
SOLUTION
Since there are seven different letters, the number of arrangements = 7! = 5040.
EXAMPLE 7
Four boys and five girls are to form a line at the school cafeteria. In how many ways
can this line be formed?
SOLUTION
The total number of people to form the line = 4 + 5 = 9.
4JODFXFBSFEFBMJOHXJUIQFPQMFUIFZBSFBMMEJČFSFOUBOEUIFOVNCFSPG
BSSBOHFNFOUT= 9! =
273
M O DUL E 3
Try these 14.1
(a) Find the total number of 6-digit numbers that can be formed using all the digits
of the number 423 579.
(b) Calculate the total number of permutations of the letters of the word DOUBLES.
(c) Calculate the total number of ways of arranging 4 boys and 2 girls in a row.
Permutation of r out of n distinct objects
n! ćFOVNCFSPGQFSNVUBUJPOTPGr out of nEJTUJODUPCKFDUT= nPr = _______
n−r EXAMPLE 8
Find the number of different arrangements of four letters of the word NUMBER.
SOLU TION
Method 1
Note
Your calculator has
nP . You can use
n
this to evaluate 6P4.
/6.#&3IBTTJYEJČFSFOUMFUUFSTBOEXFOFFEUPBSSBOHFGPVSPGUIFTFMFUUFST
Therefore r =BOEn =
6! = __
6! = 6 × 5 × 4 × 3 =
ćFOVNCFSPGQFSNVUBUJPOT= 6P4 = _______
− 2!
Method 2
8FDBOVTFUIFCPYNFUIPEUPBSSBOHFUIFGPVSMFUUFSTBTGPMMPXTTJODFXFBSF
BSSBOHJOHGPVSMFUUFSTXFESBXGPVSCPYFT*OUIFĕSTUCPYXFDBOQMBDFBOZPGUIFTJY
MFUUFST"ęFSVTJOHPOFMFUUFSJOUIFĕSTUCPYXFIBWFĕWFSFNBJOJOHMFUUFSTUIBUDBO
HPJOUIFTFDPOECPY"OETPPOVOUJMBMMGPVSCPYFTBSFĕMMFE
6
5
4
3
ćFUPUBMOVNCFSPGBSSBOHFNFOUT= 6 × 5 × 4 × 3 =
EXAMPLE 9
Carl has seven different paintings of the Caroni swamp. He decides to hang three of
these paintings in a row on the wall. In how many ways can Carl arrange three of his
paintings?
SOLUTION
Method 1
$BSMIBTTFWFOEJČFSFOUQBJOUJOHTBOEIFOFFETUPBSSBOHFUISFFPGUIFTFQBJOUJOHT
Therefore r =BOEn =
7! = __
7! = 7 × 6 × 5 =
ćFOVNCFSPGQFSNVUBUJPOT= 7P3 = _______
− 4!
Method 2
8FDBOVTFUIFCPYNFUIPEUPBSSBOHFUIFUISFFQBJOUJOHTBTGPMMPXTTJODFXFBSF
BSSBOHJOHUISFFQBJOUJOHTXFESBXUISFFCPYFT*OUIFĕSTUCPYXFDBOQMBDFBOZPG
UIFTFWFOQBJOUJOHT"ęFSVTJOHPOFQBJOUJOHJOUIFĕSTUCPYXFIBWFTJYSFNBJOJOH
QBJOUJOHTUIBUDBOHPJOUIFTFDPOECPY
274
MODULE 3tCHAPTER 14
"OEUIFMBTUCPYDBOCFĕMMFEXJUIBOZPGUIFĕWFSFNBJOJOHQBJOUJOHT
7
6
5
ćFUPUBMOVNCFSPGBSSBOHFNFOUT= 7 × 6 × 5 =
E X A M P L E 10
How many five-digit numbers can be formed from the digits 1, 2, 3, 5, 7 and 8?
SOLUTION
We have six different digits and we need to arrange five. Using nPr where n = 6 and
r = 5, the number of permutations = 6P5 = 720.
ćFCPYNFUIPEDBOBMTPCFVTFE
E X A M P L E 11
A mathematics competition has 10 competitors and four different prizes are assigned.
In how many ways can the prizes be assigned to the competitors if each person
can only win one prize?
SOLUTION
Since there are four prizes to be assigned to any of 10 competitors, the number of
assignments = 10P4 = 5040.
E X A M P L E 12
Dipchand has eight different coloured markers on his desk and one of his students,
Mitera, decides to arrange two of these markers in a row. How many possible
arrangements are there?
SOLUTION
The markers are all different and we are arranging 2 out of 8.
The number of arrangements = 8P2 = 56.
Try these 14.2
Ask yourself
Is the order
important?
Are all objects
being arranged?
Do you have
repeated objects?
(a) Ryan has seven Harry Potter books and has space for six books on his shelf.
In how many ways can Ryan arrange six of these books?
(b) Michael has eight friends and four chairs. In how many ways can Michael
arrange four of his friends on the chairs if the chairs are all in a row?
(c) Rajeev has 20 different DVDs and decides to arrange 12 of them in a line.
In how many ways can Rajeev arrange his 12 DVDs?
Permutations with repeated objects
8FIBWFCFFOEFBMJOHXJUIPCKFDUTUIBUBSFBMMEJTUJODU/PXMFUVTMPPLBUXIBU
IBQQFOTXIFOXFIBWFrepeatedPCKFDUTBNPOHUIFTFUXFBSFBSSBOHJOH
n!
ćFOVNCFSPGQFSNVUBUJPOTPGnPCKFDUTOPUBMMEJTUJODUJT_________________
r1! × r2! ×× rk!
XIFSFr1 + r2 + + rk = nBOEr1JTUIFOVNCFSPGPCKFDUTPGPOFLJOE r2JTUIF
OVNCFSPGPCKFDUTPGBEJČFSFOULJOEBOETPPO
275
M O DUL E 3
E X A M P L E 13
Find the number of arrangements of the letters of the word DIFFERENT.
SOLUTION
The word DIFFERENT has repeated letters therefore we cannot use n! as the number
of arrangements.
%*''&3&/5IBTOJOFMFUUFSTPGXIJDIUIFSFBSFUXP'TBOEUXP&T BMMPUIFSMFUUFST
BSFEJTUJODU
n!
XIFSFn = r1 = r2 = XFIBWF
Using ________________
r1! × r2! ×× rk!
9! =
OVNCFSPGQFSNVUBUJPOT= ______
2! × 2!
E X A M P L E 14
Alex has four blue marbles and six red marbles, all indistinguishable except for the
colour. In how many ways can he arrange all 10 marbles in a row?
SOLUTION
Total no. of marbles = 10
/PPGSFENBSCMFT= 6
/PPGCMVFNBSCMFT= 4
∴ n = r1 = r2 = 4
4JODFUIFSFBSFPCKFDUTPGUIFTBNFLJOE
10! =
OVNCFSPGBSSBOHFNFOUT= ____
6! 4!
E X A M P L E 15
The local fruit POMMERAC is popular for making chow and curry. In how many
ways can we arrange the letters of the word POMMERAC?
SOLUTION
POMMERAC consists of eight letters including two Ms.
n!
Using ________________
XIFSFn = r1 = XFIBWF
r1! × r2! ×× rk!
8! =
OVNCFSPGBSSBOHFNFOUT= __
2!
E X A M P L E 16
Calculate the number of permutations of the letters of the word DIFFERENTIATION.
SOLUTION
DIFFERENTIATION consists of 15 letters, not all distinct.
8FIBWF*T 'T &T /T 5T
n!
Using ________________
XIFSFn = r1 = r2 = r3 = r4 = r5 =
r1! × r2! ×× rk!
15!
=
OVNCFSPGBSSBOHFNFOUT= _________
3! 2! 2! 2! 2!
276
E X A M P L E 17
Calculate the number of different 8-letter arrangements that can be made with the
letters of the word TRINIDAD.
SOLUTION
Since we have two Ds and two Is,
8! =
OVNCFSPGBSSBOHFNFOUT= ____
2! 2!
MODULE 3tCHAPTER 14
Try these 14.3
(a) Find the number of arrangements of the digits of the number 2 221 344.
(b) Calculate the number of arrangements of the letters of the word ADDITIONAL.
(c) In how many ways can we arrange the letters of the word MATHEMATICS?
Permutations with restrictions
*OTPNFQFSNVUBUJPOTXFIBWFSFTUSJDUJPOTPOUIFBSSBOHFNFOUT5PĕOEUIFOVNCFS
PGBSSBOHFNFOUTXFUBLFBDDPVOUPGUIFSFTUSJDUJPOTĕSTUBOEUIFOBTTJHOUIFPUIFS
PCKFDUTUPCFBSSBOHFEJOBOZQBSUJDVMBSPSEFS
E X A M P L E 18
Find the total number of arrangements of the letters of the word SINGULAR with all
three vowels next to each other.
SOLUTION
SINGULAR consists of three vowels and five consonants. We treat the three vowels as
one object, and the five consonants as five different objects.
Ask yourself
Does order matter?
How many letters
are being used?
Are there repeated
letters?
E X A M P L E 19
8FOPXBSSBOHFUIFTJYPCKFDUTJOXBZT
ćFUISFFWPXFMTDBOBMTPCFBSSBOHFEJOXBZT
4JODFXFNVTUBSSBOHFUIFWPXFMTBOEUIFTJYPCKFDUTUPHFUIFS
UPUBMOPPGBSSBOHFNFOUT= 3! × 6! =
Five boys and four girls are to stand in a line. Find how many ways can this be done if
(a) there is no restriction in how they stand next to each other,
(b) all the girls are to stand next to each other,
(c) all the boys are to stand next to each other,
(d) no two boys stand next to each other.
SOLUTION
Ask yourself
What are the
restrictions in the
problem?
How do we
deal with the
restrictions?
Are there any
repeated objects?
(a) There are nine people in total to arrange.
Since there is no restriction, we can arrange nine people in 9! = 362 880 ways.
(b) We can separate the nine people into five boys and four girls.
Our restriction is that all four girls must be together.
Let us put the four girls together and treat them as one.
Treating the five boys as five separate units, we now have six units to arrange
in 6! ways.
6! ways
G1 G2 G3 G4 B1 B2 B3 B4 B5
4! ways
For every arrangement of the six units, we can arrange the girls in 4! ways.
Total no. of arrangements with all four girls next to each other = 6! × 4! = 17 280.
277
M O DUL E 3
(c) The restriction in this case is that all five boys must be next to each other.
Treating the five boys as one unit and the four girls as four separate units, we
have five units to arrange in 5! ways.
5! ways
B1 B2 B3 B4 B5 G1 G2 G3 G4
5! ways
For every arrangement of the five units the boys can be arranged in 5! ways.
∴ Total no. of arrangements = 5! × 5! = 14 400
(d) Since no two boys are to stand next to each other we must start our arrangement with a boy. Any of the five boys can go in the first position followed by any
of the four girls in the second position and so on until all positions are filled,
alternating with boy followed by girl.
B
G
B
G
B
G
B
G
B
5
4
4
3
3
2
2
1
1
5PUBMOPPGBSSBOHFNFOUT= 5! × 4! = 2880
E X A M P L E 20
How many arrangement of the letters of the word PLAYERS start with a vowel and
end with a vowel?
SOLUTION
PLAYERS consists of two vowels and five consonants.
8FOFFEUPUBLFBDDPVOUPGUIFSFTUSJDUJPOĕSTU UIBUJT XFNVTUTUBSUXJUIBWPXFM
BOEFOEXJUIBWPXFM
6TJOHUIFCPYNFUIPE XFESBXTFWFOCPYFTTJODFXFBSFBSSBOHJOHTFWFOMFUUFST"OZ
PGUIFUXPWPXFMTDBOCFQMBDFEJOUIFĕSTUCPYBęFSVTJOHPOFWPXFMUIFSFNBJOJOH
WPXFMHPFTJOUIFMBTUCPY8FDBOQMBDFBOZPGUIFĕWFDPOTPOBOUTJOUIFTFDPOE
CPYGPMMPXFECZBOZPGUIFSFNBJOJOHGPVSDPOTPOBOUTJOUIFUIJSECPYVOUJMBMMUIF
DPOTPOBOUTBSFVTFEVQ
consonants
2
vowel
5
4
3
2
1
1
vowel
5PUBMOPPGBSSBOHFNFOUT= 2 × 5 × 4 × 3 × 2 × 1 × 1 = 240
278
E X A M P L E 21
How many odd numbers can be formed from the digits of the number 245 681?
SOLUTION
We have a restriction: the number must be odd. The number will be odd if it ends in
a 1 or 5. We can place any of the two digits 1 or 5 in the last position.
MODULE 3tCHAPTER 14
5
4
3
2
1
5 remaining digits
2
either the 1 or the 5
ćFSFBSFĕWFSFNBJOJOHEJHJUTBOEUIFTFDBOCFBSSBOHFEJOUIFĕSTUĕWFCPYFTJOXBZT
5PUBMOPPGBSSBOHFNFOUT= 5! × 2 = 240
E X A M P L E 22
What is the total number of arrangements of the letters of the word MODULAR
(a) if there are no restrictions on the arrangements
(b) if the vowels are next to each other
(c) if the first letter is L and the last letter is a vowel?
SOLUTION
(a) The word MODULAR consists of seven different letters.
The no. of arrangements of the seven letters = 7! = 5040.
(b) Separating the letters into vowels and consonants: there are three vowels and
four consonants.
We can place the three vowels together to form 1 unit.
The four consonants are 4 units.
OUA
MDLR
Total no. of arrangements of 5 units = 5!
5! ways
O U A
M D L R
OUA → 3! ways
The three vowels can be arranged in 3! ways.
Total no. of arrangements = 3! × 5! = 720
7 letters
3 vowels
4 consonants
Treat as 1 unit
4 units
Arranged in 3! ways
5 units
Arranged in 5! ways
7 letters
(c)
L
3 vowels
3 consonants
Place one letter in the first box (the letter L).
Place any of the three vowels in the last box.
279
M O DUL E 3
We now have five letters for the other positions.
1
5
4
3
2
1
3
Total no. of arrangements = 1 × 5 × 4 × 3 × 2 × 1 × 3 = 360.
E X A M P L E 23
Three boys, Alvin, Carl and Shiva, and three girls, Aliyah, Alyssa and Letitia, go to
movietowne in Port of Spain. The six friends sit in one row, adjacent to each other.
(a) In how many ways can they seat themselves in six seats if there is no restriction
on the seating arrangements?
(b) In how many of these arrangements will Aliyah sit next to Alyssa?
(c) Alvin does not want to sit next to Carl. In how many ways can the friends
arrange themselves so that Alvin and Carl are not next to each other?
SOLUTION
(a) Since there is no restriction on seating arrangements six people can be arranged
in 6! ways.
The six friends can arrange themselves in 720 ways.
(b) We are now restricted in our arrangements in that Alyssa and Aliyah must sit
next to each other. We treat Alyssa and Aliyah as one and the other four friends
as four different units. We now have five units to arrange in 5! ways.
Alyssa and Aliyah can arrange themselves in 2! ways.
2! ways
Alyssa Aliyah
Four others
5! ways
The total number of arrangements with Alyssa and Aliyah next to each other =
5! × 2! = 240.
(c) The number of arrangements with Alvin and Carl next to each other is the same
as that with Alyssa and Aliyah next to each other = 240.
The total number of arrangements with no restriction = 720.
FOVNCFSPGBSSBOHFNFOUTXJUI"MWJOBOE$BSMTFQBSBUFE=UIFUPUBMOVNCFS
ć
PGBSSBOHFNFOUT−UIFOVNCFSPGBSSBOHFNFOUTXJUI$BSMBOE"MWJOOFYUUP
FBDIPUIFS= 720 − 240 =
OR 8FDBOBSSBOHFUIFGPVSPUIFSGSJFOETĕSTUJOXBZTBOEUIFOQMBDF$BSM
BOE"MWJOJOUIFBWBJMBCMFQPTJUJPOT
Alyssa
280
Aliyah
Shiva
Letitia
FSFBSFĕWFBWBJMBCMFQPTJUJPOTGPS$BSMBOE"MWJO XFDBOBTTJHOUIFTF
ć
in 5P2 =XBZT
ćFUPUBMOVNCFSPGBSSBOHFNFOUTXJUI"MWJOBOE$BSMTFQBSBUF= 20 × 4! =
MODULE 3tCHAPTER 14
Permutations with restrictions and repetition
E X A M P L E 24
Find the number of different permutations of the letters of the word REPETITION.
)PXNBOZPGUIFTFQFSNVUBUJPOTTUBSUBOEFOEXJUIUIFMFUUFS5
SOLUTION
REPETITION has 10 letters including 2 Es, 2 Ts and 2 Is.
10! = 453 600
4JODFXFIBWFSFQFBUFEMFUUFSTUIFOPPGQFSNVUBUJPOT= ______
2! 2! 2!
-FUVTBTTVNFUIBUUIFMFUUFSTBSFEJTUJODU
ćFBSSBOHFNFOUXJMMCFBOZPGUIFUXP5TJOUIFĕSTUQPTJUJPOBOEUIFSFNBJOJOH
5JOUIFMBTUQPTJUJPO
8FIBWFFJHIUSFNBJOJOHMFUUFSTUPBSSBOHFJOUIFNJEEMFBT
2
8
7
6
5
4
3
2
1
1
T
T
-FUVTEFBMXJUIUIFSFQFUJUJPOTJODFUIFSFBSF5T &TBOE*TXFEJWJEFCZ
2! × 2! × 2!
2! 8! =
∴ /PPGQFSNVUBUJPOT= ______
2! 2! 2!
E X A M P L E 25
The digits of the number 123 411 are to be arranged so that the resulting number is
odd. How many different odd numbers can be formed?
SOLUTION
The resulting number will be odd if it ends in a 1 or 3.
Remember
When dealing
with repeated objects, arrange the
objects as if they
were different
and then divide
by the factorial
of the number of
repeated objects.
E X A M P L E 26
*GUIFEJHJUTXFSFBMMEJČFSFOUXFDPVMEBSSBOHFUIFNJOUIFGPMMPXJOHNBOOFSćF
MBTUQPTJUJPODBOCFPDDVQJFECZBOZPGUIFGPVSPEEEJHJUT
8FDBOQMBDFBOZPGUIFSFNBJOJOHĕWFEJHJUTJOUIFĕSTUQPTJUJPO GPVSJOUIFTFDPOE
QPTJUJPO UISFFJOUIFUIJSEQPTJUJPO FUD
5
4
3
2
1
4
4JODFXFIBWFUISFFTXFEJWJEFCZ
5! × 4 = 80
∴ /PPGQFSNVUBUJPOT= ______
3!
Find the total number of arrangements using all of the letters in the word
MATHEMATICS.
(a) How many of these arrangements start with M and end with M?
(b) How many of these arrangements have all the vowels together?
SOLUTION
MATHEMATICS has 11 letters including 2 Ms, 2 As, and 2 Ts.
11! =
ćFUPUBMOVNCFSPGBSSBOHFNFOUT= ______
2! 2! 2!
281
M O DUL E 3
(a) Treating the letters as if they are all different, we can place any of the two Ms in
the first position and the remaining M in the last position. The other nine letters
can be placed in the middle in 9! ways.
2 × 9! = 90 720.
The number of permutations with M at the beginning and end = ______
2! 2! 2!
(b) Treating the four vowels as one unit, we have eight units to arrange in
8! = 10 080 ways.
____
2! 2!
4! = 12 ways (note that there are two As).
We can arrange the vowels in __
2!
The total number of arrangements = 10 080 × 12 = 120 960
E X A M P L E 27
How many numbers can be formed using the digits 1, 3, 3, 4, 5, 7, 8?
(a) How many of these numbers are odd?
(b) How many of these numbers are even?
SOLUTION
Not all the digits are distinct; we have two 3s and seven digits in total.
7! = 2520
5PUBMOPPGBSSBOHFNFOUT= __
2!
(a) Let us treat all digits as if they are distinct.
For an odd number we must end with 1, 3, 3, 5 or 7, i.e. we can place any of the
five digits in the last box.
Remember
Find the number
of arrangements
assuming the digits were different,
then divide by 2!
8FDBOBSSBOHFUIFPUIFSTJYEJHJUTJOXBZT
6
5
4
3
2
1
5
4JODFXFIBWFUXPTXFOPXEJWJEFCZ
6! × 5 = 1800
∴ /PPGPEEOVNCFST= ______
2!
(b) The number will be even if it ends in 4 or 8.
ćFSFBSFUXPEJHJUTUPQMBDFJOUIFMBTUCPY
6! × 2 = 720
∴ /PPGFWFOOVNCFST= ______
2!
03
/PPGFWFOOVNCFST=UPUBMOPPGOVNCFST−OPPGPEEOVNCFST
= 2520 − 1800 = 720
E X A M P L E 28
Calculate the number of permutations of the letters A, A, B, C, D, E when
(a) there are no restrictions
(b) the arrangement starts and ends with the letter A
(c) the two letters A are together.
SOLUTION
(a) There are six letters, with two As.
6! = 360.
No. of arrangements = __
2!
282
MODULE 3tCHAPTER 14
(b) We can place any of the two As in the first position and the remaining A in the
last position. The other letters can be arranged in the middle in 4! ways.
48 = 24
4! × 2 = ___
The number of arrangements = ______
2
2
(c) Placing the two As together and treating them as one unit, we can arrange the
five units in 5! ways.
The number of arrangements = 5! = 120.
Try these 14.4
(a) (i) Find the number of arrangements of the letters in the word HISTORICAL.
(ii) How many of these arrangements will end with an I?
(b) (i) How many different 6-digit numbers can be formed from the digits 4, 4, 7, 8, 6, 6?
(ii) How many of these numbers are even?
EXERCISE 14A
1
Find the number of permutations of the letters in the word MANGO.
2
Brian places the digits 3, 1, 4, 7, 8, 9 in any order. How many different numbers
can Brian form with these 6 digits?
3
Anslem decides to write out all the different arrangements of the letters of the
word EXAMPLE. How many different arrangements will there be?
4
Rajeev, Ryan, Aaron and Ian run a race for four prizes. In how many ways can
the four prizes be distributed if each person can only win one prize?
5
How many numbers can be formed from the digits 4, 2, 8, 9 if repetition is not
allowed?
6
How many different four-letter arrangements can be formed using the letters of
the word EDUCATION?
7
In how many ways can the 1st prize, 2nd prize and 3rd prize be awarded to a
group of 25 people? Each person can be awarded only one prize.
8
On a flight from Port of Spain to Mia on Caribbean Airlines, a row consists of
six seats. Five friends are assigned to this row. In how many ways can the friends
be seated if they are allowed to sit without restrictions?
9
Find the number of 5-digit numbers that can be formed from the digits
2, 3, 4, 5, 7, 8, 9.
10 How many different arrangements are there of the letters of the word
TRIGONOMETRY?
11 How many permutations are there of the letters of the word POSSIBILITIES?
12 What is the number of arrangements of the letters of the word JUNEPLUM?
(a) In how many of these arrangements are the two Us together?
(b) In how many of these arrangements are the two Us not together?
13 In how many ways can the letters of the words CELEBRATION be arranged?
How many of these arrangements will begin with EE?
283
M O DUL E 3
14 In how many ways can the letters of the word PROBABILITY be arranged?
In how many of these arrangements will the two Is be together?
15 The digits of the number 24 329 875 are rearranged so that the resulting number
is even. Find the number of ways in which this can be done.
16 Find the number of arrangements of the letters of the word POMMECYTHERE
in which
(a) there are no restrictions
(b) the two letters M are next to each other
(c) the two letters M are not next to each other.
17 In how many ways can four boys and three girls form a line?
In how many ways can they line up if a boy is to be at each end?
18 Find how many even numbers between 2000 and 6000 can be written down
using the digits 1, 3, 4, 6 if
(a) no two digits can occur more than once in any number
(b) repetition of digits is allowed.
19 Calculate the number of arrangements of the letters of the word RANDOM if
(a) all the consonants are together
(b) each arrangement begins with a consonant and ends with a vowel.
20 Eight students, Stuart, Alex, Partap, Kimberly, Mitera, Safiyyah, Kiara and Kiev,
are chosen to attend a conference on leadership. Calculate the total number of
different permutations when
(a) the students are seated in a row without restrictions
(b) Partap, Kimberly and Mitera are seated next to each other
(c) Safiyyah refuses to sit next to Alex.
21 The national bird of Trinidad and Tobago is the HUMMING bird. In how many
ways can the letters of the word HUMMING be arranged if
(a) there are no restrictions
(b) the two letter Ms are together
(c) the letter Ms are not together.
22 Find the number of different arrangements of the seven letters of the word
JAMAICA in which
(a) all three letter As are consecutive
(b) the first letter is J and the last letter is C.
23 Calculate the number of different permutations, taking all the letters at the same
time, of the letters of the word CONCLUSION if
(a) all the consonants are together
(b) the arrangement begins with a consonant and ends with a vowel
284
MODULE 3tCHAPTER 14
(c) all the vowels are together
(d) the two Cs are together.
24 Calculate the number of different 5-digit numbers that can be formed from the
digits 1, 2, 3, 4, 1 if
(a) the numbers must be odd
(b) the numbers must be even
(c) the numbers must end with a 3.
Combinations
A combinationJTBTFMFDUJPOPGPCKFDUTJOXIJDIUIFPSEFSPGTFMFDUJPOEPFTOPUNBUUFS
ćFOVNCFSPGEJČFSFOUDPNCJOBUJPOTPGrEJTUJODUPCKFDUTPVUPGnEJTUJODUPCKFDUTJT
n!
nC = _________
ćJTSFTVMUPOMZIPMETGPSPCKFDUTUIBUBSFEJČFSFOU
r
n − r r!
E X A M P L E 29
In how many ways can we choose two letters from A, B and C?
SOLUTION
Method 1
4JODFUIFSFBSFUISFFEJČFSFOUMFUUFSTUPDIPPTFGSPNBOEUIFPSEFSPGTFMFDUJPO
EPFTOPUNBUUFS XFDBOVTFnCrXIFSFn = r =
∴ /PPGDIPJDFT= 3C2 =
Method 2
8FDBONBLFBMJTU "# "$ #$
∴ ćFSFBSFUISFFDIPJDFT
E X A M P L E 30
Calculate the number of ways of choosing 4 boys out of 6 boys to form a team for a
mathematics competition.
SOLUTION
Since the order of selection does not matter we use combinations.
Using nCr XJUIn = r =
/PPGDIPJDFT= 6C4 =
E X A M P L E 31
Calculate the number of ways of choosing 5 letters from the letters of the word
TRIANGLES.
SOLUTION
Since order does not matter, this is a combination. There are 9 different letters and we
need to choose 5.
No. of choices = 9C5 = 126.
285
M O DUL E 3
E X A M P L E 32
A school committee is to consist of 4 boys and 3 girls. In how many ways can the
school choose this committee out of 12 students consisting of 7 boys and 5 girls?
SOLUTION
We have 12 students consisting of 7 boys and 5 girls.
Separate the
group of 12 into
boys and girls,
then choose from
the separate
groups. Multiply
since we need all
7 people.
E X A M P L E 33
12 students
7 boys
5 girls
Choose 4 boys
Choose 3 girls
7C
4
5C
3
No. of choices = 7C4 × 5C3 (multiplication rule)
= 350
A team of 6 members is to be selected from 8 boys and 6 girls. Find the number of
ways in which this can be done if
(a) there are no restrictions
(b) the team has exactly 4 girls
(c) the team has at least 1 girl.
SOLUTION
The order does not matter so we use combinations.
(a) Since we have no restrictions we can put all the boys and girls together and
choose from the total number of people.
Total no. of people = 8 + 6 = 14.
No. of ways of choosing 6 = 14C6 = 3003.
(b) We need to have exactly 4 girls and therefore 2 boys to form the team.
We can separate and choose as follows.
14 people
8 boys
6 girls
Choose 2
Choose 4
8C
2
OR our choices
are: 5B 1G, 4B 2G,
3B 3G, 2B 4G,
1B 5G, 0B 6G.
No. of choices
with at least 1
= 8C5 6C1 + 8C4 6C2
+ 8C3 6C3
+ 8C2 6C4
+ 8C1 6C5
+ 6C6 = 2975
286
6C
4
No. of choices with exactly 4 girls = 8C2 × 6C4 = 420
(c) At least 1 girl means 1 girl or more than 1 girl.
We can find the number of committees with 0 girls and subtract from the
number of committees with no restrictions.
14 people
8 boys
Choose 6
8C
6
6 girls
Choose 0
6C
0
MODULE 3tCHAPTER 14
No. of committees with no girls = 8C6 × 6C0 = 8C6 = 28
No. of committees with at least 1 girl = 3003 − 28 = 2975
E X A M P L E 34
Four letters are chosen at random from the word PROBLEMATIC. Find the number
of choices which contain four consonants.
SOLUTION
PROBLEMATIC has seven different consonants and we need to choose four.
7! = 35.
No. of choices = 7C4 = ____
4! 3!
E X A M P L E 35
Bristol Rovers Sports Club consists of twelve members including the President,
Treasurer and Secretary. The Club is asked to send a team of six members to
represent them at a football tournament. Calculate the number of different ways in
which the team can be formed if it must contain
(a) the President, Secretary and Treasurer
(b) the President or the Treasurer, but not both.
SOLUTION
(a) Since the team must contain the President, Secretary and Treasurer we must
choose these three and there is only one way of doing this.
We need three more players from the remaining nine; these three can be chosen
in 9C3 ways.
The total number of choices = 1 × 9C3 = 84.
Remember
Separate into the
different groupings and then
choose what you
need from each
group.
(b) We can separate the 12 members into two groups: the President and Treasurer
and ten others.
We need to choose one of the President or Treasurer and this can be done
in 2C1 ways.
We need to choose the remaining five members from the ten others.
This can be done in 10C5 ways.
The total number of choices = 2C1 × 10C5 =504
Combinations with repetition
When there are repeated objects to choose from we cannot use nCr directly since this
works when all objects are different. Let us work out how to deal with repetition by
considering some simple examples.
E X A M P L E 36
Find the number of ways of choosing two letters from the word ALL.
SOLUTION
Method 1
The word ALL contains two Ls. We can choose two letters by choosing one or two of
the letters L.
The number of ways of choosing two letters with one L = 1 (i.e. choose the A and one L).
287
M O DUL E 3
Note
If you used 3C2 ,
you assumed
that all the letters
are different and
your result will
be 3, which is
incorrect.
The number of ways of choosing two letters using both Ls = 1.
The total number of ways of choosing two letters = the number of ways of choosing
one L + the number of ways of choosing two Ls = 1 + 1 = 2.
Method 2
By listing them we can see that our choices are AL or LL.
The number of ways of choosing two letters from ALL = 2.
E X A M P L E 37
Find the number of different choices of three of the letters of the word BOOKS.
SOLUTION
We separate our five letters into two Os and three others (which are distinct).
Total number of ways of choosing three letters = number of ways of choosing three
letters with none of the Os + number of ways of choosing three letters with one letter
O + number of ways of choosing three letters with two letter Os.
Number of ways of choosing three letters with none of the Os:
Since we need three letters from three distinct letters (B, K, S), the number of choices
with no Os = 3C3 = 1
Number of ways of choosing three letters with one letter O:
Note
If you tried
5C = 10, this is
3
incorrect since
the letters are not
all different.
We need two letters from B, K, S and we need one O.
Number of choices with one letter O = 3C2 × 1C1 = 3
Number of ways of choosing three letters with two letter Os:
We need the two Os and one letter from B, K, S.
The number of choices with the two letter Os = 3C1 = 3
The total number of choices = 1 + 3 + 3 = 7.
Try these 14.5
Find the number of ways of choosing three letters from the word DIFFERENT.
EXERCISE 14B
288
1
In how many ways can we choose four letters from the word SIMPLE?
2
Given the number 94 685 321, in how many ways can we choose five digits from
this number?
3
Ryan has seven Harry Potter books and he wants to lend a friend any two of the
books. How many choices does Ryan have?
4
In how many ways can a relay team of four athletes be chosen from a group of
20 athletes?
5
6
In how many ways can two cars be chosen from 10 different cars?
Six pupils are to be chosen to form a team for a competition. If 12 pupils are
under consideration for selection, find the number of ways in which the team
can be chosen.
MODULE 3tCHAPTER 14
7
During a World Cup cricket tournament, 11 players are to be chosen to play the
first match for the West Indies. There are 25 available players.
How many choices does the coach have?
8
A menu consists of five dishes: fried rice, macaroni pie, curry chicken, stew peas
and stew chicken. Anjali decides to choose three dishes for her dinner.
In how many ways can Anjali choose her dinner?
9
Ten friends attend a dinner at their graduation. Upon arrival there are three free
tables. One table has two seats, one has three seats and the largest table can seat
five. In how many ways can the 10 friends be seated, assuming that the order of
seating at a table is immaterial?
10 A group of five adults and four children is to be formed from eight adults and
six children. How many possible groups are there?
11 A team of five people is to be selected from six women and seven men.
Find the number of possible teams if
(a) there are no restrictions
(b) there is at least one woman in the team
(c) one of the men refuses to be in a team with one of the women.
Find the number of teams which include this particular woman or man,
but not both.
12 A committee of six is to be selected from seven men and eight women.
Calculate the number of different possible ways of forming the committee if
(a) there is no restriction
(b) there are equal numbers of men and women on the committee
(c) there is at least one man on the committee.
13 GULABJAMOON is a popular Indian sweet among Trinidadians.
In how many ways can Radha choose four letters from the letters of
GULABJAMOON if
(a) she must choose the two Os
(b) she must choose one A and one O?
14 Naparima Boys and Naparima Girls High School decide to enter a debating
competition together. Naparima Boys submits eight names to the recommending committee and Naparima Girls submits 10 names. The debating team
consists of five members.
How many choices will the committee have if
(a) there is no restriction on choice
(b) the team contains exactly two boys
(c) the team consists of more boys than girls
(d) the team consists of at least one boy?
289
M O DUL E 3
15 A typical menu for Divali at a Hindu home in Trinidad consists of the following.
Appetisers
Main dishes
Desserts
Saheena
Roti
Sweet rice
Katchouri
Rice
Gulabjamoon
Pholouri
Curry Channa
Khurma
Curry Mango
Peera
Pumpkin
Prasad
Chataigne
Bodi
Sadhara invited her friends home and they were allowed to choose their
combinations for their meal.
(a) Linda decided to choose Roti and three dishes from the main course,
two appetisers and four desserts. How many choices did Linda have?
(b) Mylene decided to choose Rice and one appetiser, two main dishes and
two desserts. How many choices did Mylene have?
(c) Arlene decided on Roti or Rice but not both. She also wanted two appetisers,
three further main dishes and two desserts. How many choices did Arlene
have?
16 A mathematics class consists of four girls and seven boys. In how many ways can a
team of four represent their class for the Mathematics Olympiad if
(a) there is no restriction on choice
(b) there must be at least one boy in the team
(c) there must be an equal number of boys and girls in the team?
17 An integration worksheet consists of 20 problems.
Five of the problems are integration by parts, four problems are integration by
substitution, six problems are integration by partial fractions and five problems are
integration by recognition. Each student has to do four problems for homework.
(a) Find the number of choices each student has if they must do one problem
of each type.
(b) How many choices does each student have if they must do exactly two
integration by parts questions?
(c) How many choices does each student have if they do all four questions of
one type?
290
MODULE 3tCHAPTER 14
SUMMARY
Permutations and combinations
Permutations
Combinations
Order matters
Order does not matter
No. of permutations of n distinct objects = n!
No. of permutations of r out of n distinct objects = nPr
No. of ways of choosing r
out of n distincts objects = nCr
Permutations with repetition
No. of permutations of n objects
n!
not all distinct =
r1 !r2 !×...×rk !
Counting principles
Event A OR B
Addition rule
Events A AND B
Multiplication rule
Checklist
Can you do these?
■ Use the counting principles.
■ Find the number of ways of arranging n different objects.
■ Find the number of ways of arranging r out of n different objects.
■ Find the number of arrangements of n objects not all different.
■ Find the number of ways of arranging objects with restrictions.
■ Find the number of ways of choosing r out of n different objects.
■ Find the number of ways of choosing r out of n objects that are not all distinct.
■ Distinguish between a permutation and a combination.
Review e x e r c i s e 1 4
1
Find the number of different arrangements of the letters of the word
PARAMETRIC.
2
Sandra is going to arrange ten paintings in a row on a wall. In how many ways
can this be done?
291
M O DUL E 3
3
(a) Calculate the number of arrangements of the letters of the word EQUATION.
(b) How many of the arrangements in part (a) begin and end with a vowel?
4
From a sixth form of 25 boys and 28 girls, two boys and two girls are to be
chosen to represent their school. How many possible selections are there?
5
A group of 16 people is to make a journey to Maracas beach in 4 cars, with
4 people in each car. Each car is driven by its owner. Find the number of ways in
which the remaining 12 people may be allocated to the cars.
6
In how many ways can a committee of four boys and four girls be seated in a row if
(a) they can sit in any position
(b) no one is seated next to a person of the same sex?
7
Four men and two women sit in a line on stools in front of a table.
(a) In how many ways can they arrange themselves so that the two women are
together?
(b) In how many ways can they sit if the two women are not together?
8
The digits of the number 41 138 216 are rearranged so that the resulting number
is odd. Find the number of ways in which this can be done.
9
Calculate the number of permutations of the 9 letters of the word BACKSPACE
when
(a) there are no restrictions
(b) A is the first letter and C is the second letter
(c) the two As are next to each other.
10 Nine women and six men are standing in a queue.
(a) How many arrangements are possible if any individual can stand in any
position?
(b) In how many of the arrangements will all six men be standing next to one
another?
(c) In how many of the arrangements will no two men be standing next to one
another?
11 In how many ways can four different letters be chosen from the word TANGENT?
12 Four boys, Kevin, Shastri, Rajeev, Dinesh, and three girls, Siddi, Sandra and
Sharmila, arrange to go to the cinema. The seven people sit in a row next to each
other. In how many ways can this be done if
(a) there is no restriction in the seating arrangements
(b) all three girls sit next to each other
(c) Siddi refuses to sit next to Kevin.
13 D and N flower shop sells 10 different varieties of roses for Valentine’s day. A
customer wishes to buy 6 roses, all of different varieties.
(a) Calculate the number of ways she can make her selection.
Of the varieties, 3 are purple, 5 are red and 2 are blue. Calculate the number of
ways in which her selection of 6 roses could contain
292
MODULE 3tCHAPTER 14
(b) no purple roses
(c) at least one rose of each colour.
14 (a) Find the number of different arrangements of the letters of the word TOBAGO.
Find the number of these arrangements which
(b) begin with T
(c) have the letter A at one end and the letter G at the other end.
Four of the letters of the word TOBAGO are selected at random.
Find the number of different combinations if
(d) there is no restriction on the letters selected
(e) the two Os must be selected.
15 Calculate the number of different six-digit numbers which can be formed using
the digits 0, 1, 2, 3, 4, 5 without repetition and assuming a number cannot begin
with 0. How many of these six-digit numbers are even?
16 Ten friends watch a movie at Movietowne in Chaguanas. There are eight boys
and two girls. In how many ways can the friends sit in a row consisting of
10 seats if the girls must sit at the ends?
17 A committee of three is to be formed in the village council at Beaucarro Road. If
the council members consist of 10 men and their wives, in how many ways can
the committee be formed if no husband serves on the committee with his wife?
18 Find the number of arrangements of the letters of the word REVERSE.
In how many of these arrangements are the V and S separate?
19 Find the total number of arrangements of the letters of the word ISOSCELES.
In how many of these arrangements are the two Es separate?
20 Find the total number of arrangements of the letters of the word CHAGUANAS.
How many of these arrangements begin with an A and end with the letter S?
21 In how many ways can four letters be chosen from the word ADVANCE?
How many of these will contain the two As?
22 How many numbers of five or six digits can be formed from the digits 1, 2, 3,
3, 3, 4?
23 Find the number of different arrangements of the letters of the word FURTHER.
Find also the number of ways of choosing four letters from the word FURTHER.
24 (a) Find the number of ways of arranging the letters of the word
STATISTIC.
(b) In how many of these arrangements will the three Ts be together?
25 Find the number of arrangements of the letters of the word AIRFLOWS.
Find also the number of arrangements with none of the vowels next to
each other.
293
M O DUL E 3
CHAPTER 15
Probability
At the end of this chapter you should be able to:
■ identify a sample space
■ identify the number of possible outcomes in a sample space
■ define probability
■ calculate probability
■ understand and use the fact that 0 P(A)
1
■ demonstrate that the total probability of an event space is 1
■ use the property that the total probability for all possible events is 1
■ use the property that P(A′) = 1 − P(A)
■ use the property that P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
■ use the property that P(A ∪ B) = P(A) + P(B) for mutually exclusive events
■ use the property P(A ∩ B) = P(A) × P(B), where A and B are independent
events
■ use Venn diagrams and tree diagrams to solve probability questions
■ find conditional probabilities.
KEYWORDS/TERMS
TUBUJTUJDBMFYQFSJNFOUtPVUDPNFtTBNQMFQPJOUt
TBNQMFTQBDFtFWFOUtNVUVBMMZFYDMVTJWFt
FRVBMMZMJLFMZtFYIBVTUJWFtQSPCBCJMJUJZt
DPNQMFNFOUtVOJPOtJOUFSTFDUJPOtJOEFQFOEFOU
294
MODULE 3tCHAPTER 15
Sample space and sample points
A statistical experimentJTBOFYQFSJNFOUJOXIJDIUIFoutcomeDBOOPUCFQSFEJDUFE
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JOUFSFTUFEJOUIFOVNCFSMBOEJOHGBDFVQ XJMMCF\ ^*GUIFEJFJT
GBJSUIFOFBDIPVUDPNFJTFRVBMMZMJLFMZ IBTUIFTBNFDIBODFPGPDDVSSJOH (ii) 8IFOUPTTJOHBDPJOUIFQPTTJCMFPVUDPNFTBSF\) 5 ^
(iii) 8IFOUPTTJOHUXPDPJOT UIFQPTTJCMFPVUDPNFTBSF\)) 55 )5 5) ^
(iv) *GBCBHDPOUBJOTUXPSFENBSCMFTBOEUXPHSFFONBSCMFTBOEUXPNBSCMFTBSF
UBLFO‫ڀ‬PVUGSPNUIFCBH UIFQPTTJCMFPVUDPNFTGPSUIJTFYQFSJNFOUBSF
\3( (3 33 ((^
&BDIQPTTJCMFPVUDPNFJOBTUBUJTUJDBMFYQFSJNFOUJTDBMMFEBsample point
The sample space S JTBMJTUPGBMMQPTTJCMFPVUDPNFTJOBTUBUJTUJDBMFYQFSJNFOU
ćFOVNCFSPGTBNQMFQPJOUTPSUIFOVNCFSPGQPTTJCMFPVUDPNFTJOUIFTBNQMF
TQBDFJTEFOPUFECZn S -FUVTJEFOUJGZTBNQMFTQBDFTBOEUIFOVNCFSPGPVUDPNFTJOBTBNQMFTQBDF
EXAMPLE 1
8IBUJTUIFTBNQMFTQBDFXIFOUISFFDPJOTBSFUPTTFEBOEXIBUJTUIFOVNCFSPG
PVUDPNFTPGUIJTFYQFSJNFOU
SOLUTION
8IFOMJTUJOHFWFOUT USZUPXPSLTZTUFNBUJDBMMZ
*GZPVTUBSUXJUIUISFFIFBETUIFODIBOHFUIFMBTU)UP5BOELFFQJOHBMMIFBETBUUIF
GSPOUNPWFBMPOHXJUIZPVSMJTU
H
H
H
T
T
T
H
T
H
H
T
H
T
H
T
T
H
T
H
H
H
T
T
T
'JSTUPVUDPNF)))
4XJUDIUIFMBTU)UP5
8FOPXIBWF))5 TXJUDIUIFTFDPOE)UP5XIJDIHJWFT)5) OPXNPWFPOUPUXP
UBJMTBUUIFFOEBOEUIFOBUBJMBUUIFCFHJOOJOHVOUJMBMMPVUDPNFTBSFMJTUFE TFFMJTU S =\))) ))5 )5) )55 5)) 5)5 55) 555 ^
n S = 8
EXAMPLE 2
"CBHDPOUBJOTGPVSNBSLFST DPMPVSFESFE PSBOHF HSFFOBOECMBDL0OFNBSLFSJT
ESBXOGSPNUIFCBH&BDINBSLFSIBTUIFTBNFDIBODFPGCFJOHTFMFDUFE8IBUJTUIF
TBNQMFTQBDFBOEIPXNBOZQPTTJCJMJUJFTBSFUIFSFJOUIFTBNQMFTQBDF
SOLUTION
ćFSFBSFGPVSQPTTJCJMJUJFTBOEUIFTBNQMFTQBDFJT
S =\R O G B^
n S =
295
M O DUL E 3
EXAMPLE 3
/JSBWIBTBCBHXJUIGPVSNBSCMFTPGEJČFSFOUDPMPVSTSFE HSFFO CMVFBOEZFMMPX
)F‫ڀ‬EFDJEFTUPUBLFUXPPGUIFNBSCMFTPVUPGUIFCBHBTGPMMPXT
(a) )FUBLFTPOFNBSCMFPVU OPUFTUIFDPMPVSPGUIJTNBSCMFBOEUIFOreplacesUIF
NBSCMF UBLFTPVUUIFTFDPOENBSCMFBOEOPUFTUIFDPMPVS
(b) /JSBWEFDJEFTUPUBLFPVUUIFĕSTUNBSCMF OPUFUIFDPMPVS BOEUIFOUBLFPVUUIF
TFDPOENBSCMFwithoutSFQMBDJOHUIFĕSTUBOEOPUFJUTDPMPVS
'PS(a) BOE(b)JEFOUJGZUIFTBNQMFTQBDFBOEUIFOVNCFSPGTBNQMFQPJOUTJOUIF
TBNQMFTQBDF
SOLUTION
(a) /JSBWJTTBNQMJOHXJUISFQMBDFNFOUBOEIJTPVUDPNFTBSF
S =\3# 3( 3: 33 #3 #( #: ## (# (: (( (3 :# :( :3 ::^
n S =
(b) *OUIJTDBTF/JSBWJTTBNQMJOHXJUIPVUSFQMBDFNFOUBOEIFDBOOPUHFUUXP
NBSCMFTPGUIFTBNFDPMPVSćFTBNQMFTQBDFJTOPX
S =\3# 3( 3: #3 #( #: (# (: (3 :# :( :3^
n S =
Try these 15.1
*EFOUJGZUIFTBNQMFTQBDFBOEUIFOVNCFSPGQPTTJCJMJUJFTJOUIFTBNQMFTQBDFGPSUIF
GPMMPXJOHFYQFSJNFOUT
(a) 5XPEJDFBSFSPMMFEBOEUIFTVNPGUIFTDPSFTPOUIFGBDFTMBOEJOHVQJTOPUFE
(b) "DPJOBOEBEJFBSFUPTTFEUPHFUIFSBOEXFBSFJOUFSFTUFEJOXIBUTIPXTPOUIF
GBDFPGUIFDPJOBOEUIFGBDFPGUIFEJFMBOEJOHVQ
(c) 'PVSSFENBSCMFTBOEGPVSHSFFONBSCMFTBSFQMBDFEJOBCBHćFNBSCMFTBSF
ESBXOPOFCZPOFBOE3ZBOJTJOUFSFTUFEJOUIFESBXJOHPGBSFENBSCMFCFGPSF
UIFĕSTUHSFFOJTESBXO
Events: mutually exclusive; equally likely
An eventJTBTVCTFUPGUIFTBNQMFTQBDF
'PSFYBNQMF\)) ^JTBOFWFOUXIFOUPTTJOHUXPDPJOT
5XPFWFOUTABOEB are mutually exclusiveJGUIFZDBOOPUPDDVSUPHFUIFS
ćFPVUDPNFTPGBOFYQFSJNFOUBSFequally likelyUPPDDVSXIFOUIFZIBWFUIFTBNF
DIBODFPGPDDVSSJOH'PSFYBNQMF
(a) XIFOZPVUPTTBGBJSDPJO ZPVBSFFRVBMMZMJLFMZUPHFUBIFBEPSBUBJM
(b) XIFOZPVSPMMBGBJSEJF ZPVBSFFRVBMMZMJLFMZUPSPMMB PS
&WFOUTBSFexhaustiveJGUIFZDPNCJOFUPHJWFUIFFOUJSFQSPCBCJMJUZTQBDF
Probability
*GZPVUPTTBEJF XIBUBSFUIFQPTTJCMFPVUDPNFT 8IBUJTZPVSDIBODFPGHFUUJOH
B *GZPVSBOTXFSJTUIBUZPVIBWFBJODIBODFPGHFUUJOHB ZPVBSFFTUJNBUJOH
296
MODULE 3tCHAPTER 15
UIFDIBODFPSprobabilityPGBDFSUBJOFWFOUIBQQFOJOHćFFWFOUJOUIJTDBTFJTUIBU
PGHFUUJOHB.BOZHBNFTBSFQMBZFEXJUIUPTTJOHBEJFBOETUBSUJOHXJUIB
*GZPVUISPXBEJFUJNFT IPXNBOZTEPZPVFYQFDUUPHFU )PXNBOZTEPZPV
BDUVBMMZHFU JGZPVUPTTUIJTEJF
EXAMPLE 4
,IBEJOFUPTTFEBEJFUJNFTGPSIFS"QQMJFE.BUIFNBUJDT6OJUJOUFSOBM
B TTFTTNFOUBOETIFPCUBJOFETJYFT8IBUJTUIFFTUJNBUFPGUIFQSPCBCJMJUZPGIFS
HFUUJOHB
(a) CBTFEPOIFSSFTVMU (b) CBTFEPOIFSFYQFDUBUJPO
SOLUTION
(a) #BTFEPOIFSSFTVMU
ćFQSPCBCJMJUZPGHFUUJOHBTJYCBTFEPOIFSSFTVMUDBOCFGPVOECZUBLJOHUIF
OVNCFSPGTJYFTTIFPCUBJOFEEJWJEFECZUIFUPUBMOVNCFSPGUPTTFTNBEF
= ___
,IBEJOFTQSPCBCJMJUZCBTFEPOIFSSFTVMUJT1 HFUUJOHB = ___
(b) #BTFEPOIFSFYQFDUBUJPO
"TTVNJOHIFSEJFJTGBJS 1 HFUUJOHB = __
ćFDIBODFPSprobabilityPGBOFWFOUIBQQFOJOHJTUIFOVNCFSPGXBZTJO
XIJDIUIFFWFOUDBOIBQQFOEJWJEFECZUIFUPUBMOVNCFSPGFWFOUT
8FDBOXSJUFUIJTBT
OVNCFSPGUJNFTAPDDVST
1 FWFOUAPDDVSSJOH = _____________________
UPUBMOVNCFSPGPVUDPNFT
E X A M P L E 5
'JOEUIFQSPCBCJMJUZPGHFUUJOH
(a) FYBDUMZUISFFIFBET (b) FYBDUMZUXPIFBET
XIFOUISFFDPJOTBSFUPTTFEPODF
SOLUTION
The sample space is
S =\))) ))5 )5) )55 5)) 5)5 55) 555 ^
n S = 8
(a) ćFUPUBMOVNCFSPGPVUDPNFT=
ćFOVNCFSPGUJNFTUISFFIFBETPDDVST=
6TJOHUIFEFĕOJUJPO
OVNCFSPGUJNFTAPDDVST
1 FWFOUAPDDVSSJOH = _____________________
UPUBMOVNCFSPGPVUDPNFT
XFHFU1 ))) = __
8
(b) 'SPNUIFTBNQMFTQBDFUIFSFBSFthreePVUDPNFTGPSFYBDUMZUXPIFBET
1 FYBDUMZUXPIFBET = __
8
297
M O DUL E 3
EXAMPLE 6
ćFSFBSFHSFFONBSLFSTBOESFENBSLFSTJOBCBH"MMJTPOQJDLTPOFNBSLFS
GSPNUIFCBHXJUIPVUMPPLJOHBUUIFN8IBUJTUIFQSPCBCJMJUZPG"MMJTPOHFUUJOHBSFE
NBSLFS
SOLU TION
/VNCFSPGNBSLFSTJOUIFCBH=+ 8 =
/VNCFSPGSFENBSLFST= 8
OVNCFSPGUJNFTAPDDVST
1 FWFOUAPDDVSSJOH = _____________________
UPUBMOVNCFSPGPVUDPNFT
8 = __
1 "MMJTPOESBXJOHBSFENBSLFS = ___
Rules of probability
Remember
A′ is the complement of A, i.e. A′ consists of all the outcomes not in A.
A ∪ B is the union of A and B, i.e. all the outcomes in A or B or both.
A ∩ B is the intersection of A and B, i.e. all the outcomes in both A and B.
Rule 1
'PSBOZFWFOUA ≤1 A ≤ BQSPCBCJMJUZDBOOPUCFOFHBUJWFPSHSFBUFSUIBO Rule 2
1 S = UIFQSPCBCJMJUZPGUIFTBNQMFTQBDF S PDDVSSJOHJT
Rule 3
U
A
A’
Since A′BOEANBLFVQUIFTBNQMFTQBDF XFIBWF1 A +1 A′ =1 S
TP1 A +1 A′ =
∴ 1 A′ =−1 A Rule 4
4JODFUIFFNQUZTFU ∅ EPFTOPUDPOUBJOBOZFMFNFOUT
1 ∅ =
Rule 5
"7FOOEJBHSBNGPSUXPTFUTABOEB
TVHHFTUTUIBU UPĕOEUIFTJ[FPGA ∪ B XFBEEUIFTJ[FPGABOEUIFTJ[FPGB CVUTJODFUIJTJOWPMWFTBEEJOHA ∩ B
UXJDF XFTVCUSBDUA ∩ B 1 A ∪ B =1 A +1 B −1 A ∩ B
298
A
B
A∩B
U
MODULE 3tCHAPTER 15
*GABOEBBSFNVUVBMMZFYDMVTJWFUIFO
1 A ∩ B =BOE1 A ∪ B =1 A +1 B A
B
U
8FDBOFYUFOEUIJTSFTVMUUPUISFFFWFOUT
1 A∪B∪C =1 A +1 B +1 C −1 A ∩ B − 1 A ∩ C −1 B ∩ C +1 A ∩ B ∩ C
Rule 6
A
B
U
1 A =1 A ∩ B′ +1 A ∩ B
A ∩ B’
Rule 7
'PSFYIBVTUJWFFWFOUT 1 A +1 A ++1 An =
Rule 8
%F.PSHBOTMBXT
A
B
U
A ∪ B ′ = A′ ∩ B′
A
B
U
A ∩ B ′ = A′ ∪ B′
DE FIN ITI ON
Two events A and
B are independent
if and only if
P(A ∩ B) = P(A) ×
P(B). If A and B
are independent
events then A' and
B' are independent
events, A and B'
are independent
events and A' and
B are independent
events.
Conditional probability
1 A∩B
JTEFĕOFEUPCFUIFDPOEJUJPOBMQSPCBCJMJUZUIBU
*G1 B > UIFO1 A|B = ________
1B
FWFOUAPDDVSTHJWFOUIBUFWFOUBIBTPDDVSSFE8IFOXFDPOEJUJPOPOB XFBSF
BTTVNJOHUIBUFWFOUBIBTPDDVSSFETPUIBUBCFDPNFTUIFOFXQSPCBCJMJUZTQBDF
4PNFSFTVMUTGPSDPOEJUJPOBMQSPCBCJMJUZ
(i) 1 A ∩ B =1 B|A ×1 A =1 A|B ×1 B GPSBOZFWFOUTABOEB
(ii) 1 A′|B =− 1 A|B
(iii)1 A ∪ B|C =1 A|C +1 B|C −1 A ∩ B|C GPSFWFOUTA BBOEC
299
M O DUL E 3
EXAMPLE 7
1 B = __
BOE1 A ∪ B = __
ćFFWFOUTABOEBBSFTVDIUIBU1 A = __
'JOE(a)1 A ∩ B (b)1 B|A SOLUTION
(a) 6TJOH1 A ∪ B =1 A +1 B −1 A ∩ B XFIBWF
EXAMPLE 8
= __
+ __
−1 A ∩ B
__
+ __
− __
= __
1 A ∩ B = __
1 B∩A
(b) 1 B|A = ________
1A
4JODF1 A ∩ B =1 B ∩ A XFIBWF1 B ∩ A = __
__
__
∴ 1 B|A = = __
__
5XPJOEFQFOEFOUFWFOUTABOEBBSFTVDIUIBU1 A =BOE1 B =
(a) $BMDVMBUF1 A ∩ B (b) 'JOE1 A ∪ B SOLUTION
(a) Since ABOEBBSFJOEFQFOEFOU 1 A ∩ B =1 A ×1 B
∴ 1 A ∩ B =×=
(b) 1 A ∪ B =1 A +1 B −1 A ∩ B =+−=
EXAMPLE 9
5XPFWFOUTABOEBBSFLOPXOUPCFNVUVBMMZFYDMVTJWF1 A =BOE1 B =
'JOE(a) 1 ABOEB (b) 1 APSB
SOLUTION
(a) Since ABOEBBSFNVUVBMMZFYDMVTJWF 1 A ∩ B =
(b) 1 A ∪ B =1 A +1 B −1 A ∩ B =+−=
E X A M P L E 10
4VQQPTFUIBUABOEBBSFFWFOUTBOEXFLOPXUIBU1 A = 1 B = 1 A ∩ B =$BMDVMBUF(a)1 A′ (b)1 A ∪ B (c)1 A|B
SOLUTION
(a) 1 A′ =−1 A =−=
(b) 1 A ∪ B =1 A +1 B −1 A ∩ B =+ −=
1 A∩B
1B
= __
(c) 1 A|B = ________ = ___
E X A M P L E 11
SOLUTION
300
ABOEBBSFFWFOUTTVDIUIBU1 A ∪ B = 1 A ∩ B =BOE1 A|B =
'JOE1 A BOE1 B′ 1 A∩B
1 A|B = ________
1B
= ____
1B
=
= __
1 B = ___
1 A ∪ B =1 A +1 B −1 A ∩ B
MODULE 3tCHAPTER 15
=1 A + −
1 A = +−=
1 B′ = − 1 B = −=
E X A M P L E 12
5XPGBJSEJDFBSFUISPXO
XJTUIFFWFOU ćFTDPSFPOUIFĕSTUEJFJTEJWJTJCMFCZ
YJTUIFFWFOU ćFTVNPGUIFTDPSFTJT
4IPXJOHBMMXPSLJOHDMFBSMZ EFUFSNJOFXIFUIFSXBOEYBSFJOEFQFOEFOU
'JOE1 X ∪ Y SOLUTION
Since X = {2, 4, 6}
and Y = {(2, 6),
(6, 2), (3, 5), (5, 3),
(4, 4)},
X ∩ Y = {(2, 6),
(6, 2), (4, 4)}
= __
PVUDPNFT BOEBSFEJWJTJCMFCZ PVUPGTJYQPTTJCJMJUJFT
1 X = __
ćFSFBSF×=QPTTJCJMJUJFTXIFOUPTTJOHUIFUXPEJDF
ćFQBJSTPGTDPSFTUIBUHJWFUIFTVNPGBSF 1 Y = ___
= ___
TJODFUIFSFBSFQPTTJCJMJUJFTPVUPG
1 X ∩ Y = ___
× ___
= ___
1 X × 1 Y = __
4JODF1 X ∩ Y ≠1 X × 1 Y XBOEYBSFOPUJOEFQFOEFOU
1 X ∪ Y =1 X + 1 Y − 1 X ∩ Y
− ___
= __
+ ___
= ___
= __
E X A M P L E 13
*OBTVSWFZPGTJYUIGPSNTUVEFOUT UIFGPMMPXJOHEBUBXFSFPCUBJOFE
UPPL1IZTJDT UPPL.BUIFNBUJDT UPPL$IFNJTUSZ UPPL1IZTJDTBOE
.BUIFNBUJDT UPPL.BUIFNBUJDTBOE$IFNJTUSZ UPPL1IZTJDTBOE$IFNJTUSZ UPPLBMMUISFFTVCKFDUT'JOEUIFQSPCBCJMJUZPGBTUVEFOUUBLJOH
(a) OPOFPGUIFTVCKFDUT
(b) .BUIFNBUJDT CVUOPU1IZTJDTPS$IFNJTUSZ
(c) 1IZTJDTBOE.BUIFNBUJDTCVUOPU$IFNJTUSZ
SOLUTION
"7FOOEJBHSBNJTOFFEFEGPSTPMWJOHUIJTQSPCMFN
U
To complete
the diagram,
start with the
information that
P∩M∩C=6
then work out
the other numbers from the
information in
the question.
P
M
28
16
8
6
10
14
26
12
C
(a) 'SPNUIF7FOOEJBHSBNUIFOVNCFSPGTUVEFOUTUBLJOHOPOFPGUIFTVCKFDUT=
= ___
1 UBLJOHOPOFPGUIFTVCKFDUT = ___
301
M O DUL E 3
(b) /VNCFSPGTUVEFOUTUBLJOH.BUIFNBUJDT CVUOPU1IZTJDTPS$IFNJTUSZ= 8
8 = ___
1 UBLJOH.BUIFNBUJDT CVUOPU1IZTJDTPG$IFNJTUSZ = ____
(c) /VNCFSPGTUVEFOUTUBLJOH1IZTJDTBOE.BUIFNBUJDTCVUOPU$IFNJTUSZ=
= ___
1 UBLJOH1IZTJDTBOE.BUIFNBUJDTCVUOPU$IFNJTUSZ = ____
Tree diagrams
"USFFEJBHSBNDBOCFVTFEUPĕOEUIFQPTTJCMFPVUDPNFTJOBTBNQMFTQBDFBOEUP
ĕOEQSPCBCJMJUJFTPGDPNCJOFEFWFOUTJOBOFYQFSJNFOUćFLFZGFBUVSFTPGBUSFF
EJBHSBNBSFBTGPMMPXT
(i) 1SPCBCJMJUJFTGPSFWFOUTBSFXSJUUFOPOUIFCSBODIFT UIFTVNPGUIFQSPCBCJMJUJFT
POUIFCSBODIFTGSPNBQPJOUJT (ii) ćFQSPCBCJMJUZPGBOPVUDPNFJOUIFTBNQMFTQBDFJTUIFQSPEVDUPGBMM
QPTTJCJMJUJFTBMPOHUIFQBUIUIBUSFQSFTFOUTUIFPVUDPNFPOUIFEJBHSBN
(iii)5PĕOEUIFQSPCBCJMJUZPGPOFPVUDPNFPSBOPUIFS XFBEEUIFJSQSPCBCJMJUJFT
)FSFJTBUSFFEJBHSBNGPSUPTTJOHBGBJSDPJOUISFFUJNFT
Ist toss
2nd toss
3rd toss
H
(12 )
HHH
1
2
()
H
T
H
(12 )
H
T
(12 )
T
T
(12 )
H
(12 )
HT H
(12 )
T HT
1
2
()
T
()
T
(12 )
(12 )
THH
TT H
TT H
"GBNJMZIBTUXPDIJMESFO%SBXBUSFFEJBHSBNTIPXJOHUIFQPTTJCMFPVUDPNFT
BTTVNJOHUIBUUIFQSPCBCJMJUZPGBCPZJTBOEUIFQSPCBCJMJUZPGBHJSMJT
Ist child
2nd child
B(0.5)
BB
B(0.5)
G(0.5)
B(0.5)
BG
GB
G(0.5)
G(0.5)
302
H HT
HTT
1
T
2
SOLUTION
(12 )
(12 ) H (1 )
2
H
E X A M P L E 14
Outcome
GG
MODULE 3tCHAPTER 15
E X A M P L E 15
"SMFOFIBTUXPJEFOUJDBMCBHT FBDIIBWJOHUIFTBNFDIBODFPGCFJOHDIPTFO#BH
DPOUBJOTSFECBMMTBOEHSFFOCBMMT BOECBHDPOUBJOTSFECBMMTBOEHSFFOCBMMT
"SMFOFTFMFDUTBCBHBUSBOEPNBOEUBLFTPVUBTJOHMFCBMM%SBXBUSFFEJBHSBNUP
S FQSFTFOUUIFQPTTJCMFPVUDPNFTJOUIJTFYQFSJNFOU'JOEUIFQSPCBCJMJUZUIBUUIFCBMM
DIPTFOJTSFE
SOLUTION
Bag
Ball
R
Bag 1
(12 )
G
R
1
Bag 2
2
()
G
(24 )
(24 )
(35 )
(25 )
R
G
R
G
1 SFECBMMDIPTFO =1 ESBXJOHSFECBMMGSPNFJUIFSCBH
=1 ESBXJOHSFEGSPNCBH +1 ESBXJOHSFEGSPNCBH
+ __
+ __
× __
= ___
= ___
× __
= __
E X A M P L E 16
"NBOVGBDUVSFSNBLFTDPNQBDUEJTDT"RVBMJUZDPOUSPMPďDFSJTIJSFEUPDIFDLUIF
RVBMJUZPGUIFEJTDTQSPEVDFEćFPďDFSUFTUFEBSBOEPNTBNQMFPGUIFEJTDTGSPNB
MBSHFCBUDIBOEDBMDVMBUFEUIFQSPCBCJMJUZPGBEJTDCFJOHEFGFDUJWFBT
%FFQBLCVZTUISFFPGUIFEJTDTNBEFCZUIFNBOVGBDUVSFS
(a) %SBXBUSFFEJBHSBNUPSFQSFTFOUUIFJOGPSNBUJPO
(b) $BMDVMBUFUIFQSPCBCJMJUZUIBUBMMUISFFEJTDTBSFEFGFDUJWF
(c) $BMDVMBUFUIFQSPCBCJMJUZUIBUBUMFBTUUXPPGUIFEJTDTBSFGBVMUZ
SOLUTION
(a)
Disc 1
Disc 2
Disc 3
Defective (0.015)
DDD
Defective (0.015)
Defective (0.015)
Not defective (0.985)
DDN
Defective (0.015)
DND
Not defective (0.985)
DNN
Defective (0.015)
NDD
Not defective (0.985)
Defective (0.015)
Not defective (0.985)
Not defective (0.985)
NDN
Defective (0.015)
NND
Not defective (0.985)
NNN
Not defective (0.985)
303
M O DUL E 3
(b) 1 BMMUISFFEJTDTBSFEFGFDUJWF =××= (c) 1 BUMFBTUUXPEJTDTBSFEFGFDUJWF =1 UXPBSFEFGFDUJWF +1 UISFFBSFEFGFDUJWF
=××+××+××
+××
= E X A M P L E 17
"CBHDPOUBJOTUISFFHSFFONBSCMFTBOETJYCMVFNBSCMFT5XPNBSCMFTBSFESBXOBU
SBOEPNBOEUBLFOPVUPGUIFCBHPOFBUBUJNF*GUIFĕSTUNBSCMFJTHSFFO UIFNBSCMF
JTSFUVSOFEUPUIFCBH*GUIFĕSTUNBSCMFJTCMVF UIFNBSCMFJTSFNPWFEGSPNUIFCBH
'JOEUIFQSPCBCJMJUZUIBUUIFUXPNBSCMFTBSF
(a) UIFTBNFDPMPVS (b) EJČFSFOUDPMPVST
SOLUTION
%SBXBUSFFEJBHSBNUPTIPXUIFPVUDPNFT
1st marble
2nd marble
B
B
(58 )
6
9
()
3
G 9
()
6
B 9
3
G 8
()
()
3
G 9
()
BB
BG
GB
GG
(a) 1 TBNFDPMPVS =1 CPUIHSFFO +1 CPUICMVF
× __
+ __
× __
= ___
= __
8 (b) 1 EJČFSFOUDPMPVST =1 #( +1 (#
+ __
× __
= ___
× __
= __
8 EXERCISE 15A
1
ćFGPVSLJOHTBSFSFNPWFEGSPNBEFDLPGDBSET"DPJOJTUPTTFEBOEPOFPGUIF
LJOHTJTDIPTFO%SBXBUSFFEJBHSBNUPJMMVTUSBUFUIFTBNQMFTQBDF
8IBUJTUIFQSPCBCJMJUZPGHFUUJOHIFBETPOUIFDPJOBOEUIFLJOHPGEJBNPOET
2
"DPJOJTCJBTFETPUIBUJUIBTBDIBODFPGMBOEJOHPOIFBET*GJUJTUISPXO
UISFFUJNFT ĕOEUIFQSPCBCJMJUZPGHFUUJOH
3
304
B UISFFIFBET C UXPUBJMTBOEBIFBE D BUMFBTUPOFUBJM
"CBHDPOUBJOTUISFFCMVFCBMMTBOEĕWFXIJUFCBMMT"MFYQJDLTBCBMMBUSBOEPN
GSPNUIFCBHBOESFQMBDFTJUCBDLJOUIFCBH)FNJYFTUIFCBMMTJOUIFCBHBOE
UIFOQJDLTBOPUIFSCBMMBUSBOEPNGSPNUIFCBH
B $POTUSVDUBQSPCBCJMJUZUSFFPGUIFQSPCMFN
MODULE 3tCHAPTER 15
C $BMDVMBUFUIFQSPCBCJMJUZUIBU"MFYQJDLT
J UXPCMVFCBMMT
JJ BCMVFCBMMJOIJTTFDPOEESBX
JJJ BXIJUFCBMMPOIJTTFDPOEESBXHJWFOUIBUIFESBXTBXIJUFCBMMPOIJT
ĕSTUESBX
4
#BH"DPOUBJOTCBMMTPGXIJDIBSFSFEBOEBSFHSFFO#BH#DPOUBJOTCBMMT
PGXIJDIBSFSFEBOEBSFHSFFO"CBMMJTESBXOBUSBOEPNGSPNFBDICBH
B %SBXBQSPCBCJMJUZUSFFEJBHSBNUPTIPXBMMUIFPVUDPNFTPGUIFFYQFSJNFOU
C 'JOEUIFQSPCBCJMJUZUIBU
J CPUICBMMTBSFSFE
JJJ POFCBMMJTHSFFOBOEPOFCBMMJTSFE
JW BUMFBTUPOFCBMMJTSFE
5
JJ CPUICBMMTBSFHSFFO
"CPYDPOUBJOTGPVSSFEBOEUXPCMVFNBSCMFT"NBSCMFJTESBXOBUSBOEPN
BOEUIFOSFQMBDFE"TFDPOENBSCMFJTUIFOESBXOBUSBOEPN‫ڀ‬
B %SBXBQSPCBCJMJUZUSFFEJBHSBNUPTIPXBMMUIFQPTTJCMFPVUDPNFT
C $BMDVMBUFUIFQSPCBCJMJUZPGHFUUJOH
6
7
8
9
J BUMFBTUPOFCMVF‫ڀ‬NBSCMF
JJ POFSFENBSCMFBOEPOFCMVF‫ڀ‬NBSCMF
JJJ UXPNBSCMFTPGUIFTBNFDPMPVS
JW B CMVFNBSCMFPOUIFTFDPOEESBXHJWFOUIBUBSFENBSCMFJTESBXOPO
UIFĕSTUESBX
B * GFWFOUTXBOEYBSFTVDIUIBUUIFZBSFJOEFQFOEFOUBOE1 X =BOE
1 Y = ĕOE J 1 X ∩ Y JJ 1 X ∪ Y
C "SFFWFOUTXBOEYNVUVBMMZFYDMVTJWF
1 B = __
BOE1 A = __
ĕOE B 1 B|A C 1 A ∩ B
*G1 A|B = __
8
1 A|B = __
1 B = __
&WFOUTABOEBBSFTVDIUIBU1 A = __
8
'JOE B 1 A ∩ B C 1 B|A
&WFOUTABOEBBSFTVDIUIBU1 A =BOE1 B =*GABOEB are
JOEFQFOEFOUFWFOUT ĕOE B 1 A ∩ B C 1 A ∩ B′ D 1 A′ ∩ B′
10 5XPJOEFQFOEFOUFWFOUTXBOEYBSFTVDIUIBU1 X =BOE1 Y =
&WBMVBUF
B 1 X ∩ Y C 1 X|Y D 1 X ∪ Y
11 5XPJOEFQFOEFOUFWFOUTFBOEGBSFTVDIUIBU1 F = __ 1 F ∩ G′ = __ 'JOE B 1 F ∩ G C 1 G D 1 G|F
1 F|G = __
12 QFPQMF NBMFBOEGFNBMF XFSFBTLFEJGUIFZXFSFJOGBWPVSPGPSBHBJOTUUIF
4VNNJUPGUIF"NFSJDBTCFJOHIFMEJO5SJOJEBE0GUIFNBMFT XFSFJO
GBWPVS XIFSFBTGFNBMFTXFSFBHBJOTU
305
M O DUL E 3
*GBQFSTPOJTTFMFDUFEBUSBOEPNGSPNUIFTFQFPQMF ĕOEUIFGPMMPXJOH
QSPCBCJMJUJFT
B ćFZBSFJOGBWPVSPGUIF4VNNJU
C ćFZBSFJOGBWPVSPGUIF4VNNJUHJWFOUIBUUIFQFSTPOJTGFNBMF
D ćFZBSFNBMFBOEBHBJOTUUIF4VNNJU
E ćFZBSFJOGBWPVSPGUIF4VNNJUPSUIFZBSFGFNBMF
F "SFUIFFWFOUTNBMFBOEJOGBWPVSJOEFQFOEFOU
13 "DBSDPNQBOZEJEBTVSWFZPOUIFEJČFSFOUNPEFMTPGDBSTTPMEEVSJOHB
POFNPOUIQFSJPEćFUBCMFCFMPXTIPXTUIFNPEFMTTPMEUPDVTUPNFST
Car model
Male
Female
Toyota
65
55
Nissan
40
36
Honda
33
31
"DVTUPNFSJTTFMFDUFEBUSBOEPN'JOEUIFQSPCBCJMJUZUIBUUIJTDVTUPNFSJT
B BNBMF
C BNBMFHJWFOUIFDVTUPNFSCPVHIUB/JTTBO
D BGFNBMFXIPQVSDIBTFEB)POEB
E BNBMFXIPQVSDIBTFEB5PZPUBPSBGFNBMFXIPQVSDIBTFEB)POEB
14 ćFUBCMFCFMPXTIPXTUIFNPOUIMZTBMBSJFTPGFNQMPZFFTBUBVOJWFSTJUZJO
5SJOJEBEBOE5PCBHP
Less than $8000
$8000 to $10 000
$10 000 or more
Male
40
45
50
Female
35
30
60
"OFNQMPZFFJTTFMFDUFEBUSBOEPNGSPNUIJTVOJWFSTJUZ'JOEUIFQSPCBCJMJUZ
UIBUUIJTFNQMPZFF
B FBSOTBTBMBSZMFTTUIBO QFSNPOUI
C JTBNBMFBOEFBSOTBTBMBSZPGBUMFBTU QFSNPOUI
D JTBGFNBMFPSFBSOTMFTTUIBOQFSNPOUI
15 "SBOEPNTBNQMFPGBEVMUTXBTUBLFO BOEUIFZXFSFBTLFEXIFUIFSUIFZ
QSFGFSXBUDIJOHOFXTPSNPWJFTPOUFMFWJTJPO0GUIFN BSFNBMFBOE
QSFGFSXBUDIJOHNPWJFTPGUIFGFNBMFTQSFGFSXBUDIJOHOFXT*GBOBEVMUJT
TFMFDUFEBUSBOEPNGSPNUIJTHSPVQ ĕOEUIFQSPCBCJMJUZUIBUUIJTBEVMU
B JTBNBMFPSQSFGFSTOFXT
C JTBNBMFPSBGFNBMFXIPQSFGFSTNPWJFT
16 "SBOEPNTBNQMFPGTUVEFOUTXBTUBLFO BOEUIFZXFSFBTLFEXIFUIFSUIFZ
QSFGFSXBUDIJOHBDUJPONPWJFTPSDPNFEZNPWJFTPOUFMFWJTJPOćFGPMMPXJOH
UBCMFHJWFTUIFJOGPSNBUJPODPMMFDUFEGSPNUIFTBNQMF
306
MODULE 3tCHAPTER 15
Prefers watching action
Prefers watching comedy
Male
96
24
Female
45
85
B * GPOFBEVMUJTTFMFDUFEBUSBOEPNGSPNUIJTHSPVQ ĕOEUIFQSPCBCJMJUZUIBU
UIJTBEVMU
J QSFGFSTXBUDIJOHDPNFEZ
JJ JTBGFNBMFBOEQSFGFSTXBUDIJOHDPNFEZ
JJJ QSFGFSTXBUDIJOHBDUJPOPSJTBNBMF
C "SFUIFFWFOUTGFNBMFBOEQSFGFSTXBUDIJOHBDUJPOJOEFQFOEFOU
"SFUIFZNVUVBMMZFYDMVTJWF &YQMBJOZPVSBOTXFSDMFBSMZ
Probability and permutations
E X A M P L E 18
SOLUTION
ćFMFUUFSTPGUIFXPSE'*/"-BSFBSSBOHFEBUSBOEPN8IBUJTUIFQSPCBCJMJUZUIBU
UIFUXPWPXFMTBSFUPHFUIFS
OVNCFSPGUJNFTAPDDVST
#ZEFĕOJUJPO1 FWFOUAPDDVSSJOH = _____________________
UPUBMOVNCFSPGPVUDPNFT
4JODFPSEFSJTJNQPSUBOUXFDBOĕOEUIFOVNFSBUPSBOEEFOPNJOBUPSVTJOH
QFSNVUBUJPOT
ćFSFBSFĕWFEJČFSFOUMFUUFSTJO'*/"-BOEXFDBOBSSBOHFEJČFSFOUMFUUFSTJO
‫ڀ‬XBZT
ćFUPUBMOVNCFSPGPVUDPNFT==
AJTUIFFWFOUAUIFUXPWPXFMTBSFUPHFUIFS
8FDBOĕOEUIFOVNCFSPGUJNFTAPDDVSTBTGPMMPXT-FUVTQMBDFUIFUXPWPXFMT"*
UPHFUIFSBOEUSFBUUIFNBTVOJUćFMFUUFST' / -BSFTFQBSBUFVOJUTBOEXFDBO
BSSBOHFGPVSVOJUTJOXBZT
ćF"*VOJUDBOCFBSSBOHFEJOXBZT
6TJOHUIFDPVOUJOHQSJODJQMF XFIBWF
ćFOVNCFSPGUJNFTAPDDVST=×=
= __
1 UXPWPXFMTUPHFUIFS = ____
E X A M P L E 19
ćFMFUUFSTPGUIFXPSE1"35*$6-"3BSFBSSBOHFEBUSBOEPN
8IBUJTUIFQSPCBCJMJUZUIBU
(a) U IFUXPMFUUFS"TBSFUPHFUIFS (b) BMMUIFWPXFMTBSFUPHFUIFS
SOLUTION
4JODFXFBSFPSEFSJOHUIFPCKFDUTXFVTFQFSNVUBUJPOTUPDBMDVMBUFUIFOVNFSBUPS
BOEUIFEFOPNJOBUPS
1"35*$6-"3JTBMFUUFSXPSEBOEDPOUBJOTUXP"TBOEUXP3T TPXFIBWF
SFQFBUFEPCKFDUT
307
M O DUL E 3
ćFOVNCFSPGQPTTJCMFBSSBOHFNFOUTPGUIFMFUUFSTPGUIFXPSE1"35*$6-"3JT
=
____
(a) 5
SFBUJOHUIF"TBTVOJU XFIBWFVOJUTUPBSSBOHFXJUI3T
=
ćFOVNCFSPGBSSBOHFNFOUTXJUIUIF"TUPHFUIFS= __
OVNCFSPGUJNFTAPDDVST
_____________________
1 FWFOUAPDDVSSJOH =
UPUBMOVNCFSPGPVUDPNFT
=
1 "TUPHFUIFS = _______
(b) 8FIBWFGPVSWPXFMT" " *BOE6
4JODFUIFSFBSF"TXFDBOBSSBOHFUIFWPXFMTJO__
XBZT=
1MBDFUIFWPXFMTUPHFUIFSBOEUSFBUUIFNBTVOJU
=XBZT
8FOPXIBWFVOJUTUPBSSBOHFJO__
ćFOVNCFSPGBSSBOHFNFOUTXJUIUIFWPXFMTUPHFUIFS=×=
OVNCFSPGUJNFTAPDDVST
1 FWFOUAPDDVSSJOH = _____________________
UPUBMOVNCFSPGPVUDPNFT
=
1 BMMWPXFMTBSFUPHFUIFS = _______
Note
We divide by 2!
since we have
2 Rs.
E X A M P L E 20
'PVSEJHJUOVNCFSTBSFUPCFGPSNFEGSPNUIFEJHJUT XJUIPVU
SFQFUJUJPO 'JOEUIFQSPCBCJMJUZUIBU
(a) UIFSFTVMUJOHOVNCFSJTPEE
(b) UIFSFTVMUJOHOVNCFSTUBSUTXJUIBBOEFOETXJUIB
SOLUTION
4JODFPSEFSJTJNQPSUBOUXFDBOĕOEUIFUPUBMOVNCFSPGPVUDPNFTBOEUIFOVNCFS
PGGBWPVSBCMFPVUDPNFTCZVTJOHQFSNVUBUJPOT
4JODFBMMUIFEJHJUTBSFEJČFSFOUXFDBOPSEFSGPVSPGUIFTFJOP =XBZT
(a) 8FDBOVTFUIFCPYNFUIPEUPĕOEUIFOVNCFSPGPEEOVNCFSTBTGPMMPXT
SBXJOHGPVSCPYFTXFDBOQMBDFBOZPGUIFGPVSEJHJUT JOUIFMBTU
%
QPTJUJPO TJODFUIJTXJMMNBLFUIFOVNCFSPEE
OZPGUIFSFNBJOJOHĕWFEJHJUTDBOHPJOUIFĕSTUCPY BOETPPOGPSUIF
"
SFNBJOJOHCPYFT
5
4
3
4
ćFOVNCFSPGPEEOVNCFST=×××=
OPPGPEEOVNCFST
1 UIFSFTVMUJOHOVNCFSJTPEE = _____________________________
UPUBMOVNCFSPGGPVSEJHJUOVNCFST
= __
= ____
(b) 8FDBOVTFUIFCPYNFUIPEUPĕOEUIFOVNCFSPGOVNCFSTTUBSUJOHXJUIB
‫ڀ‬BOEFOEJOHXJUIB
308
8FNVTUQMBDFUIFEJHJUJOUIFĕSTUCPYBOEUIFJOUIFMBTUCPY
MODULE 3tCHAPTER 15
FDBOQMBDFBOZPGUIFSFNBJOJOHEJHJUTJOUIFTFDPOECPYBOEBOZPGUIF
8
SFNBJOJOHEJHJUTJOUIFUIJSECPY
1
4
3
1
ćFOVNCFSPGPVUDPNFTXJUIBĕSTUBOEBMBTU=×× × = OVNCFSPGPVUDPNFTXJUIBGJSTUBOEBMBTU
1 BGJSTUBOEBMBTU = _____________________________________
UPUBMOVNCFSPGGPVSEJHJUOVNCFST
___
= ___
= E X A M P L E 21
4JYGSJFOET UXPCPZTBOEGPVSHJSMT BMMTJUJOBSPX8IBUJTUIFQSPCBCJMJUZPG
(a) UIFUXPCPZTTJUUJOHOFYUUPFBDIPUIFS
(b) UIFUXPCPZTOPUTJUUJOHOFYUUPFBDIPUIFS
SOLUTION
(a) /VNCFSPGXBZTPGBSSBOHJOHQFPQMF==
/VNCFSPGBSSBOHFNFOUTXJUIUXPCPZTUPHFUIFS=× = 1 UXPCPZTOFYUUPFBDIPUIFS
OPPGBSSBOHFNFOUTXJUIUIFUXPCPZTOFYUUPFBDIPUIFS __
= ______________________________________________ = ___
= UPUBMOVNCFSPGBSSBOHFNFOUT
(b) 1 UXPCPZTOPUOFYUUPFBDIPUIFS =−1 UXPCPZTOFYUUPFBDIPUIFS
= __
=− __
E X A M P L E 22
4BOESBUBLFTIFSTFWFODIJMESFO DPOTJTUJOHPGGPVSCPZTBOEUISFFHJSMT GPSBQIPUPBU
UIF$BSPOJ4XBNQ4IFBTLTUIFNUPGPSNBMJOF'JOEUIFQSPCBCJMJUZUIBU
(a) BMMGPVSCPZTTUBOEOFYUUPFBDIPUIFS
(b) BMMUISFFHJSMTTUBOEOFYUUPFBDIPUIFS
(c) OPUXPCPZTTUBOEOFYUUPFBDIPUIFS
SOLUTION
ćFSFBSFTFWFODIJMESFOBOEXFDBOBSSBOHFUIFNJO=XBZT
(a) 5SFBUJOHUIFGPVSCPZTBTPOFVOJUBOEUIFUISFFHJSMTBTTFQBSBUFVOJUT XFDBO
BSSBOHFUIFGPVSVOJUTJOXBZT
8FDBOBSSBOHFUIFGPVSCPZTJOXBZT
/VNCFSPGPVUDPNFTXJUIBMMGPVSCPZTOFYUUPFBDIPUIFS=× =
1 BMMGPVSCPZTOFYUUPFBDIPUIFS OVNCFSPGPVUDPNFTXJUIBMMGPVSCPZTOFYUUPFBDIPUIFS
= ______________________________________________
UPUBMOVNCFSPGBSSBOHFNFOUT
= ____
= 309
M O DUL E 3
(b) 5SFBUJOHUIFUISFFHJSMTBTPOFVOJUBOEUIFGPVSCPZTBTGPVSVOJUT XFDBO
BSSBOHFĕWFVOJUTJOXBZT
/VNCFSPGBSSBOHFNFOUTXJUIUIFUISFFHJSMTOFYUUPFBDIPUIFS=×=
1 BMMUISFFHJSMTOFYUUPFBDIPUIFS
OVNCFSPGBSSBOHFNFOUTXJUIUISFFHJSMTOFYUUPFBDIPUIFS
__
= ________________________________________________ = ____
= UPUBMOVNCFSPGBSSBOHFNFOUT
(c) 8JUIOPUXPCPZTTUBOEJOHOFYUUPFBDIPUIFSXFDBOBSSBOHFUIFDIJMESFOBT
GPMMPXT
#(#(#(#
ćFOVNCFSPGBSSBOHFNFOUT=× × × × × × = 1 OPUXPCPZTOFYUUPFBDIPUIFS OPPGBSSBOHFNFOUTXJUIOPUXPCPZTOFYUUPFBDIPUIFS
= ______________________________________________ = ____
= UPUBMOVNCFSPGBSSBOHFNFOUT
Try these 15.2
(a) ' JOEUIFQSPCBCJMJUZPGUIFSFTVMUJOHOVNCFSCFJOHPEEXIFOUISFFPGUIFEJHJUT
PGUIFOVNCFSBSFBSSBOHFEJOPSEFS
(b) ćFMFUUFSTPGUIFXPSE$"-"-00BSFBSSBOHFEJOPSEFS
'JOEUIFQSPCBCJMJUZUIBU
J UIFBSSBOHFNFOUTUBSUTXJUIBO0BOEFOETXJUIBO0
JJ UIFUXP-TBSFUPHFUIFS
EXERCISE 15B
1
B *OIPXNBOZXBZTDBOUIFMFUUFSTPGUIFXPSE4$)0-"45*$CFBSSBOHFE C 8IBUJTUIFQSPCBCJMJUZUIBUBOBSSBOHFNFOUCFHJOTXJUIUIFMFUUFS0
2
B * OIPXNBOZXBZTDBOUIFMFUUFSTPGUIFXPSE$"/%-&CFBSSBOHFEJOB
MJOF C8
IBUJTUIFQSPCBCJMJUZUIBUBOBSSBOHFNFOUCFHJOTXJUIB$BOEFOETXJUI
BO&
3
ćFMFUUFSTPGUIFXPSE46*5$"4&BSFBSSBOHFEBUSBOEPN
8IBUJTUIFQSPCBCJMJUZUIBU
B
C UIFUXP4TBSFOPUUPHFUIFS
4
310
"4$)0-"45*$BQUJUVEFUFTUXBTBENJOJTUFSFEUPBHSPVQPGTUVEFOUT
UIFUXP4TBSFUPHFUIFS
*GUIFMFUUFSTJOUIFXPSE5&-&1)0/&BSFBSSBOHFEBUSBOEPN ĕOEUIF
QSPCBCJMJUZUIBUUIFBSSBOHFNFOUTUBSUTXJUIBO&BOEFOETXJUIBO&
MODULE 3tCHAPTER 15
5
B )PXNBOZEJČFSFOUBSSBOHFNFOUTBSFUIFSFPGUIFMFUUFSTPGUIFXPSE%*(*$&-
C 8IBUJTUIFQSPCBCJMJUZUIBUUIFBSSBOHFNFOUTUBSUTXJUIUIFMFUUFS*BOE
FOETXJUIUIFMFUUFS*
6
B ćFEJHJUTPGUIFOVNCFSDBOCFBSSBOHFEJOIPXNBOZXBZT C 8IBUJTUIFQSPCBCJMJUZUIBUUIFBSSBOHFNFOUJTBOPEEOVNCFS
7
"HSPVQPGĕWFNFOBOETJYXPNFOBSFUPCFBSSBOHFEJOBSPX
'JOEUIFQSPCBCJMJUZUIBU
B BMMUIFNFOTUBOEOFYUUPFBDIPUIFS
C BMMUIFXPNFOTUBOEOFYUUPFBDIPUIFS
D OPUXPXPNFOTUBOEOFYUUPFBDIPUIFS
8
"BSPOBSSBOHFTUIFMFUUFSTPGUIFXPSE4$"/%"-064'JOEUIFQSPCBCJMJUZUIBU
B BMMUIFWPXFMTBSFUPHFUIFS
C UIFBSSBOHFNFOUTUBSUTXJUIBO"BOEFOETXJUIBO"
D UIFUXP"TBOEUIFUXP4TBSFUPHFUIFS
E UIFBSSBOHFNFOUFOETXJUIUIFMFUUFS"PSUIFMFUUFS%
9
$BMDVMBUFUIFOVNCFSPGEJČFSFOUXBZTJOXIJDIUIFMFUUFSTPGUIFXPSE
45"5*45*$4DBOCFBSSBOHFE8IBUJTUIFQSPCBCJMJUZUIBU
B BOBSSBOHFNFOUCFHJOTBOEFOETXJUIUIFMFUUFS4
C UIFUISFF5TBSFUPHFUIFS
D UIFBSSBOHFNFOUCFHJOTXJUIUIFMFUUFS*
10 'JOEUIFQSPCBCJMJUZPGBOPEEOVNCFSHSFBUFSUIBOCFJOHGPSNFEGSPNUIF
EJHJUT BOEJG
B SFQFUJUJPOJTBMMPXFE C SFQFUJUJPOJTOPUBMMPXFE
Probability and combinations
E X A M P L E 23
'JOEUIFQSPCBCJMJUZPGDIPPTJOHUXPDPOTPOBOUTGSPNUIFMFUUFSTPGUIFXPSE
/6.#&3
SOLUTION
ćFXPSEOVNCFSDPOTJTUTPGTJYEJČFSFOUMFUUFST
ćFUPUBMOVNCFSPGDIPJDFTGPSBOZUXPMFUUFST= C
4JODFUIFXPSE/6.#&3DPOTJTUTPGGPVSDPOTPOBOUT UIFOVNCFSPGXBZTPG
D IPPTJOHUXPPVUPGUIFGPVSDPOTPOBOUT= C
OPPGXBZTPGDIPPTJOHUXPDPOTPOBOUT
1 DIPPTJOHUXPDPOTPOBOUT = _______________________________________
UPUBMOVNCFSPGXBZTPGDIPPTJOHBOZUXPMFUUFST
C
___
__
= ___
C = = 311
M O DUL E 3
E X A M P L E 24
'PVSTUVEFOUTBSFUPCFDIPTFOUPSFQSFTFOUBTDIPPMJODFOUSBM5SJOJEBEJOB
BUIFNBUJDTDPNQFUJUJPOćFTUVEFOUTBSFUPCFDIPTFOGSPNTJYCPZTBOEGPVSHJSMT
.
'JOEUIFQSPCBCJMJUZUIBU
(a) BMMGPVSTUVEFOUTDIPTFOBSFHJSMT
(b) BMMGPVSTUVEFOUTDIPTFOBSFCPZT
(c) UIFTUVEFOUTDIPTFODPOTJTUPGFYBDUMZUXPCPZTBOEFYBDUMZUXPHJSMT
(d) UIFSFBSFNPSFHJSMTUIBOCPZTSFQSFTFOUJOHUIFTDIPPM
SOLUTION
10 students
6 boys
4 girls
/VNCFSPGXBZTPGDIPPTJOHPVUPGTUVEFOUTXJUIPVUSFTUSJDUJPOT= C
(a) /VNCFSPGXBZTPGDIPPTJOHPVUPGHJSMT= C
OPPGXBZTPGDIPPTJOHHJSMT
1 DIPPTJOHHJSMT = _______________________________________
OVNCFSPGXBZTPGDIPPTJOHPVUPGTUVEFOUT
C
____
= ____
C = (b) /VNCFSPGXBZTPGDIPPTJOHPVUPGCPZT= C
OPPGXBZTPGDIPPTJOHCPZT
1 DIPPTJOHCPZT = _______________________________________
OVNCFSPGXBZTPGDIPPTJOHPVUPGTUVEFOUT
C
____
___
= ____
C = = (c)
10 students
6 boys
4 girls
6C
2
4C
2
/VNCFSPGXBZTPGDIPPTJOHPVUPGCPZT= C
/VNCFSPGXBZTPGDIPPTJOHPVUPGHJSMT= C
1 DIPPTJOHFYBDUMZCPZTBOEFYBDUMZHJSMT OPPGXBZTPGDIPPTJOHPVUPGCPZT×OPPGXBZTPGDIPPTJOHPVUPGHJSMT
= _____________________________________________________________
OPPGXBZTPGDIPPTJOHPVUPGTUVEFOUTXJUIPVUSFTUSJDUJPOT
C × C
× = ___
= ______
= ________
C
(d) 8FOFFEUIFOVNCFSPGDIPJDFTXJUINPSFHJSMTUIBOCPZT
312
4JODFXFIBWFUPDIPPTFGPVSQFPQMFUIFSFNVTUCF
HJSMTBOECPZT 03 HJSMTBOECPZ
MODULE 3tCHAPTER 15
/PPGDIPJDFTXJUIHJSMT= C
/PPGDIPJDFTXJUIHJSMT= C × C
ćFOVNCFSPGDIPJDFTXJUINPSFHJSMTUIBOCPZT= C+ C × C
1 DIPPTJOHNPSFHJSMTUIBOCPZT
OPPGDIPJDFTXJUINPSFHJSMTUIBOCPZT
= ___________________________________________________
OPPGXBZTPGDIPPTJOHPVUPGTUVEFOUTXJUIPVUSFTUSJDUJPOT
C + C × C
= ___
+ = ____
= ______
= ______________
C
E X A M P L E 25
"TPDDFSUFBNDPOTJTUTPGTJYUFFOQMBZFSTPGXIJDIUXPBSFHPBMLFFQFST"UFBNPG
FMFWFONVTUCFDIPTFOUPQMBZBHBNFBHBJOTU5SJOJEBEBOE5PCBHP
'JOEUIFQSPCBCJMJUZPGDIPPTJOHUIFUFBNJG
(a) UIFSFNVTUCFPOFHPBMLFFQFS
(b) UIFGPVSNPTUFYQFSJFODFEQMBZFSTNVTUCFPOUIFUFBN
SOLUTION
8FOFFEUPDIPPTFQMBZFSTGPSUIFUFBN
16 players
(a) 4JODFUIFSFBSFQMBZFSTGSPNXIJDIUP
DIPPTF UIFOVNCFSPGXBZTPGDIPPTJOH
QMBZFSTGSPN= C.
2 goalkeepers
14 others
2C
1
14C
10
8FOFFEPOFHPBMLFFQFSGSPNUIFUXPBOE
GSPNUIFSFNBJOJOH
/VNCFSPGXBZTPGDIPPTJOHPOFHPBMLFFQFS= C × C
OPPGXBZTPGDIPPTJOHPOFHPBMLFFQFS
1 DIPPTJOHPOFHPBMLFFQFS = __________________________________
OPPGXBZTPGDIPPTJOHQMBZFSTGSPN
C × C
= ___
= __________
C
(b) 8FDBOTFQBSBUFUIFQMBZFSTBTTIPXO
/VNCFSPGXBZTPGDIPPTJOHUIFGPVSNPTU
FYQFSJFODFE= C× C
1 DIPPTJOHUIFGPVSNPTUFYQFSJFODFE
QMBZFST
16 players
4 most experienced
12 others
4C
4
12C
7
OPPGXBZTPGDIPPTJOHUIFGPVSNPTUFYQFSJFODFEQMBZFST
= _______________________________________________
OPPGXBZTPGDIPPTJOHQMBZFSTGSPN
C × C
= ___
= _________
C
313
M O DUL E 3
E X A M P L E 26
3BKFFWQMBDFTGPVSSFECBMMT TJYCMVFCBMMTBOEUISFFPSBOHFCBMMTJOBCBH)FBTLTIJT
CSPUIFS5SJTIBOUPDIPPTFUISFFCBMMTGSPNUIFCBHBUSBOEPNBOEXJUIPVUSFQMBDFNFOU
8IBUJTUIFQSPCBCJMJUZUIBU5SJTIBODIPPTFT
(a) BMMSFECBMMT (b) FYBDUMZUXPPSBOHFCBMMT (c) BMMCBMMTPGUIFTBNFDPMPVS
SOLUTION
5PUBMOVNCFSPGCBMMT=
13 balls
4 red
6 blue
3 orange
8FOFFEUPDIPPTFUISFFCBMMT
/VNCFSPGXBZTPGDIPPTJOHCBMMTXJUIPVUSFTUSJDUJPOT= C =
(a) 5SJTIBOOFFETUPDIPPTFSFECBMMTGSPNSFECBMMT
/VNCFSPGXBZTPGDIPPTJOHSFECBMMTGSPN‫ڀ‬SFECBMMT= C
OPPGXBZTPGDIPPTJOHSFECBMMTGSPNSFECBMMT
1 DIPPTJOHSFECBMMT = _________________________________________
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MODULE 3tCHAPTER 15
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M O DUL E 3
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MODULE 3tCHAPTER 15
SUMMARY
Probability
Sample points: each possible outcome
in a statistical experiment
Sample space(s): list of all possible
outcomes in a statistical experiment
Probability of an event A occurring
P(A) = no. of times A occurs
total no. of outcomes
0 ≤ P(A) ≤ 1
P(S) = 1
n(S) = number of elements in the
sample space
P(A’) = 1 – P(A)
Two events are mutually exclusive
if they cannot occur together.
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
The events of an experiment are equally likely
to occur when they have the same chance
of occurrence.
Events are exhaustive if they combine
to give the entire probability space.
De Morgan’s laws
P(A ∪ B)’ = P(A’ ∩ B’)
P(A ∩ B)’ = P(A’ ∪ B’)
P(A ∩ B) = P(A) × P(B) ⇔ A and B
are independent
A|B is A conditional on B
P(A|B) = P(A ∩ B) , P(B) > 0
P(B)
Checklist
Can you do these?
■ Identify a sample space.
■ Identify the number of possible outcomes in a sample space.
■ Define probability.
■ Calculate the probability of different events occurring.
■ Use the fact that 0
P(A)
1.
■ Demonstrate that the total probability of an event space is 1.
■ Use the property that the total probability for all possible events is 1.
■ Use the property that P(A′) = 1 − P(A).
■ Use the property that P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
■ Use the property that P(A ∪ B) = P(A) + P(B) for mutually exclusive events.
317
M O DUL E 3
■ Use the property P(A ∩ B) = P(A) × P(B), where A and B are independent
events.
■ Use Venn diagrams and tree diagrams to solve probability questions.
■ Find conditional probabilities.
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MODULE 3tCHAPTER 15
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MODULE 3tCHAPTER 16
CHAPTER 16
Matrices
At the end of this chapter you should be able to:
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
identify the order of a matrix
identify equal matrices
multiply a matrix by a scalar
identify matrices conformable to addition, subtraction or multiplication
add, subtract and multiply matrices
use the properties of matrix addition
use the properties of matrix multiplication
identify the identity matrix for an n × n matrix
find the transpose of a matrix
know and use the properties of the transpose
identify symmetric and skew-symmetric matrices
find the determinant of a matrix (2 × 2 and 3 × 3)
recall and use the properties of determinants
determine singular matrices
determine non-singular matrices
solve simultaneous equations using determinants
find the inverse of a matrix (2 × 2 and 3 × 3)
identify the matrix of cofactors
identify a system of linear equations
solve a system of linear equations using the inverse of a matrix
row reduce a matrix
find the inverse of a matrix using row reduction
write down an augmented matrix
solve simultaneous equations using row reduction
decide whether a system of equations has (i) one solution, (ii) an infinite set of
solutions or (iii) no solutions
■ solve application problems.
KEYWORDS/TERMS
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321
M O DUL E 3
Matrices: elements and order
A matrix is a rectangular array of numbers enclosed by brackets. The numbers or
functions in a matrix are called the elements of the matrix and the size or order
of the matrix is defined as m × n where m is the number of rows in the matrix
and n is the number of columns in the matrix. We refer to m × n as the order of
the matrix.
m × n is read as
‘m by n’.
An m × n matrix can be written as
(
a11
a12 . . .
a1n
am1
am2 . . .
amn
t
t
t
t
t
t
)
a11 represents the element in the first row and first column,
a12 represents the element in the first row and second column,
and so on. In general, aij represents the element in the ith row and jth column.
( 21
)
3
.
−1
EXAMPLE 1
Write down the order of the matrix
SOLUTION
This matrix has two rows and two columns,
Read 2 × 2 as ‘2
by 2’.
∴ the order of the matrix is 2 × 2.
2 × 2 = 4 and
there are 4
elements in the
matrix.
(
)
EXAMPLE 2
2
1
What is the order of the matrix −1 −1 ?
3
2
SOLUTION
This matrix has three rows and two columns,
A ‘3 by 2’ matrix
has 6 elements.
Note
∴ the order of the matrix is 3 × 2.
(
0
4
1
)
6
5 ?
1
EXAMPLE 3
1
What is the order of the matrix 3
0
SOLUTION
This matrix has three rows and three columns,
∴ the order of the matrix is 3 × 3.
Square matrices
A matrix of order n × n is called a square matrix; the number of rows is equal to the
number of columns.
( 42 53 ) is a square matrix of order 2 × 2.
322
MODULE 3tCHAPTER 16
(
6
5
3
)
0
2 is a square matrix of order 3 × 3.
0
1
2
4
Equal matrices
Two matrices A and B are said to be equal if they have the same order and each of the
corresponding elements are equal.
Zero matrix
A matrix with all elements 0 is called the zero matrix.
(
0
The zero matrix of order 3 × 3 is written as 0
0
( 24
) (
3
2
=
4
x+2
)
0
0
0
0
0.
0
)
3
.
3
EXAMPLE 4
Find the value of x for which
SOLUTION
Since the two matrices are equal the corresponding elements are equal
x+2=3⇒x=1
Note
Matrices can
only be added or
subtracted if they
are of the same
order.
Addition and subtraction of matrices
If A and B are two m × n matrices then C = A + B where each element of C is the
sum of the corresponding elements of A and B. Two matrices of the same order are
said to be conformable for addition or subtraction while two matrices of different
orders cannot be added or subtracted.
( 21
) (
)
3
4 3
+
.
2 5
−1
EXAMPLE 5
Find
SOLUTION
To add two matrices we add the corresponding elements.
( 21
EXAMPLE 6
) (
) (
) (
3
6 6
3+3
4 3
2+4
+
=
=
2 5
3 4
−1
1 + 2 −1 + 5
Evaluate the following.
(a)
( −11 32 ) + ( 42 12 )
(
1 1 3
) (
(b)
1 3
1
0
3
(c) 4 2 1 − 2 3 −1
SOLUTION
)
5 1
3
1
)
( −12
1
3
) (
3
5
+
4
2
1
0
1
3
)
(a) Add corresponding elements.
( −11 23 ) + ( 42 12 ) = ( −11 ++ 42
) (
5
2+1
=
1
3+2
3
5
)
(b) Add corresponding elements.
( −12
) (
5
1 3
+
3 4
2
1
0
) (
) (
2+5 1+1 3+1
7
1
=
=
3
1
−1 + 2 3 + 0 4 + 3
2
3
4
7
)
323
M O DUL E 3
(c) Subtract corresponding elements.
(
1
4
5
) (
3
1 3
1 − 2 3
3
0 1
1
2
1
) (
1
1−1
−1 = 4 − 2
3
5−0
) (
3−1
1−3
0
2 − 3 1 − (−1) = 2
5
1−1
3−3
−2 2
−1 2
0 0
Multiplication of a matrix by a scalar
A scalar is a quantity that has magnitude only. Examples of scalar quantities are
speed, time and temperature. A scalar quantity has no directional component.
To find αA where α ∈ ℝ, we multiply each element of A by α.
EXAMPLE 7
Find 2
SOLUTION
2
( 34
( 43
1
2
1
2
)
−1
.
−5
) (
2×3
−1
=
−5
2×4
) (
6
2 × 1 2 × −1
=
8
2 × 2 2 × −5
2
4
−2
−10
)
Properties of matrix addition
DEFINI TI ON
−A is called the
negative of matrix
A and is found by
multiplying each
element of A by
−1. When we add
a matrix to its
negative we get
the zero matrix.
EXAMPLE 8
Let A, B and C be matrices conformable for addition.
(i)
Matrix addition is commutative
A+B=B+A
(ii) Matrix addition is associative
A + (B + C) = (A + B) + C
(iii) Scalar multiplication is distributive over matrix addition
k(A + B) = kA + kB, k ∈ ℝ.
(
2
Given that A = 3
1
) ( )
3
4
−1 , B = 4
3
0
( )
2
1
2 and C = 1
2
2
4
6 , find
3
(a) A + B (b) A + 2B − C (c) 2A − B + 2C
SOLUTION
) ( )
(
) ( )
( ) ( ) ( )
( ) ( ) ( )
(
) (
(
2
(a) A + B = 3
1
3
4
−1 + 4
3
0
2
2
2
2+3
5
4+2
= 3 + 4 −1 + 2 = 7
4
0+2
1+3
6
1
2
2
4
3
2
1
1
0
3
2
2 3
4
(b) A + 2B − C = 3 −1 + 2 4 2 − 1 6
2
= 3
1
6
4
−1 + 8
0
6
2+6−1
= 3+8−1
1+6−2
324
1
4
4 − 1
4
2
4
6
3
7
4+4−4
−1 + 4 − 6 = 10
5
0+4−3
4
−3
1
)
)
MODULE 3tCHAPTER 16
(
2
4
) ( ) ( )
) ( )
3
2
1
4
(c) 2A − B + 2C = 2 3 −1 − 4 2 + 2 1 6
3
2
2 3
8−2+8
3
−2 − 2 + 12 = 4
3
0−2+6
0
1
4−3+2
= 6−4+2
2−3+4
(
Try these 16.1
Given that A =
( 24
(a) A − B − C
1
3
(
)
0
1
2
1
,B=
2
−1
(b) 2A + 2B − 3C
)
14
8
4
(
4
−4
2
and C =
5
3 −2
)
1
, find
1
(c) 3A − B + 4C.
Matrix multiplication
The product AB, in that order, of the m × n matrix A and the n × l matrix B is the
m × l matrix C.
Notes
(i) AB exists if and only if the number of columns of A is equal to the number of rows
of B, that is A and B are conformable for matrix multiplication.
(ii) When finding the product AB the rows of A are multiplied by the columns of B.
(iii) Matrix multiplication is not commutative, i.e. AB ≠ BA.
EXAMPLE 9
Find the product AB where A =
SOLUTION
AB =
( 13 52 ) and B = ( −12 12 ).
( 31 25 ) ( −12 12 )
To find the product we multiply rows by columns in the following way:
The first element in the product AB is found by multiplying the first row of A by the
first column of B
(1
2)
( −12 ) = 1 × 2 + 2 × (−1) = 2 − 2 = 0
(0 )
The element in the first row second column of AB is found by multiplying the first
row of A by the second column of B
(1
2)
( 12 ) = 1 × 1 + 2 × 2 = 5
(0
5
)
The element in the second row first column of AB is found by multiplying the second
row of A by the first column of B
(3
5)
( −12 ) = 3 × 2 + 5 × (−1) = 1
( 01
5
)
325
M O DUL E 3
The element in the second row second column of AB is found by multiplying the
second row of A by the second column of B
(3
5)
( 21 ) = 3 × 1 + 5 × 2 = 13
(
∴ AB = 01
E X A M P L E 10
5
13
)
Can we multiply the following matrices? If we can, find the product.
( )
SOLUTION
()
−2
3
−2
(a) (4 3 1)
1
(b) 3 (4 −2 0)
(c)
2
(
1
4
−4
)(
2 0
6 1
8 1
−2
3
4
)
(a) Matrices are conformable for multiplication if the number of columns of the
first is equal to the number of rows of the second.
The first matrix is of order 1 × 3
The second matrix is of order 3 × 1
The number of columns of the first matrix equals the number of rows of the
second matrix.
A 1 × 3 matrix
multiplied by a
3 × 1 matrix gives
a 1 × 1 matrix.
∴ The matrices are conformable for multiplication.
(4
3
1)
( )
−2
3 = (4 × −2 + 3 × 3 + 1 × −2) = (−1)
−2
We get a 1 × 1 matrix as the product.
(b) The first matrix is of order 3 × 1
A 3 × 1 matrix
multiplied by a
1 × 3 matrix gives
a 3 × 3 matrix.
The second matrix is of order 1 × 3
∴ The matrices are conformable for multiplication.
()
1
3 (4 −2
2
(
) (
1×0
4 −2 0
3 × 0 = 12 −6 0
8 −4 0
2×0
1 × 4 1 × −2
0) = 3 × 4 3 × −2
2 × 4 2 × −2
)
We get a 3 × 3 matrix as the product.
(c) The first matrix is of order 3 × 2
The second matrix is of order 3 × 2
Since the number of columns of the first matrix (3) is not equal to the number
of rows of the second matrix (2), the matrices are not conformable for
multiplication.
E X A M P L E 11
326
(
1
Let A = 2
1
)
(
2 3
1 1
0 1 and B = −1 2
3 3
−1 1
)
2
−1 . Find AB.
2
MODULE 3tCHAPTER 16
SOLUTION
The element in the first row first column of AB is found by multiplying the first row
of A by the first column of B
(1
2
( )
( )
8
1
3) −1 = 1 × 1 + 2 × (−1) + 3 × 3 = 8
3
The element in the first row second column of AB is found by multiplying the first
row of A by the second column of B
(1
2
(
()
1
3) 2 = 1 × 1 + 2 × 2 + 3 × 3 = 14
3
)
8 14
The element in the first row third column of AB is found by multiplying the first row
of A by the third column of B
(1
2
(
( )
2
3) −1 = 1 × 2 + 2 × (−1) + 3 × 2 = 6
2
)
8 14 6
At this stage we can find any element in AB by looking at the position of the element
we need. If we want the element in the third row third column of AB we can multiply
the third row of A by the third column of B
(1
−1
(
5
)
(
8 14 6
5
5
)
(
8 14 6
5 5
5
)
(
8 14 6
5 5 6
5
)
(
8
5
5
14 6
5 6
5
)
(
8
5
5
14
5
2
6
6
5
)
( )
2
1) −1 = 1 × 2 + (−1) × (−1) + 1 × 2 = 5
2
8 14 6
We can fill out the rest as follows.
2nd row 1st column:
(2
0
( )
1
1) −1 = 2 × 1 + 0 × (−1) + 1 × 3 = 5
3
2nd row 2nd column:
(2
0
()
1
1) 2 = 2 × 1 + 0 × 2 + 1 × 3 = 5
3
2nd row 3rd column:
2
(2 0 1) −1 = 2 × 2 + 0 × (−1) + 1 × 2 = 6
2
( )
3rd row 1st column:
(1
−1
( )
1
1) −1 = 1 × 1 + (−1) × (−1) + 1 × 3 = 5
3
3rd row 2nd column:
(1
−1
()
(
1
1) 2 = 1 × 1 + (−1) × 2 + 1 × 3 = 2
3
8 14
Hence AB = 5 5
5 2
6
6
5
)
327
M O DUL E 3
When multiplying matrices the element aij of AB can be found by multiplying the ith
row of A by the jth column of B.
E X A M P L E 12
Find the element in the second row second column of the product
( 23
SOLUTION
)(
)
1 2 2
1 4
3 4 4.
−1 2 −1 2 2
Since we need the element in the second row second column, we multiply the second
row of the first matrix by the second column of the second matrix.
(3
()
2
2) 4 = 3 × 2 + (− 1) × 4 + 2 × 2 = 6
2
−1
E X A M P L E 13
Francis Fashions sells men’s shirts for TT$70, ties for TT$40, and suits for TT$1200.
Last month the store had sales consisting of 200 shirts, 100 ties and 25 suits.
What was the total revenue due to these sales?
SOLUTION
Total = (70
( )
200
1200) 100 = (70)(200) + (40)(100) + (1200)(25) = 48 000
25
The total revenue is TT$ 48 000.
40
Properties of matrix multiplication
If A is conformable to B for matrix multiplication, B is not necessarily conformable to
A for multiplication. Generally AB ≠ BA. Let A, B and C be conformable matrices for
addition and multiplication, then
(i) A(B + C) = AB + BC
Try these 16.2
(
(ii) (A + B)C = AC + BC
) (
1
1
2 1
Let A = −1 2
3 ,B= 0
3 1 −1
1
2
0
1
)
−1
2
4 and C =
1
3
(
0
−1
(iii) A(BC) = (AB)C
)
3
.
4
Find the following products if they exist.
(a) AB
(d) CA
(b) BA
(e) BC
(c) AC
(f) CB
Identity matrix
The leading
diagonal is the
diagonal from
the top left-hand
corner to the
bottom righthand corner of
the matrix.
328
Let I be the n × n identity matrix then AI = IA = A for any n × n matrix A.
The identity matrix is the matrix whose leading diagonal consists of 1s and all other
elements in the matrix are 0s.
1 0
.
The 2 × 2 identity matrix is
0 1
1 0 0
The 3 × 3 identity matrix is 0 1 0 .
0 0 1
(
(
)
)
MODULE 3tCHAPTER 16
( 35 24 ) ( 01 01 ).
E X A M P L E 14
Find
SOLUTION
( 53 24 ) ( 01 10 ) = ( 35 ×× 11 ++ 24 ×× 00
) (
3
3×0+2×1
=
5
5×0+4×1
2
4
)
( 53 24 ) ( 01 01 ) = ( 35 24 ) since AI = A
Multiplication of square matrices
E X A M P L E 15
Let A =
( −12 11 ).
Find A2 + A + 2I, where I is the 2 × 2 identity matrix.
SOLUTION
∴ A2
E X A M P L E 16
SOLUTION
( −12 11 ) ( −12 11 ) = ( −33 30 )
+ A + 2I = ( 3 3 ) + ( 2 1 ) + 2 ( 1 0 )
0 1
−3 0
−1 1
= ( 3 + 2 + 2 3 + 1 + 0) = ( 7 4)
−4 3
−3 −1 + 0 0 + 1 + 2
A2 = AA =
(
)
1 1 −1
Given that A = 0 1
1 , find A2 and A3.
1
0 0
Hence find 2A3 + A2 + A − I, where I is the 3 × 3 identity matrix.
(
1
A2 = AA = 0
0
1
1
0
)(
)(
−1 1 1
1 0 1
1 0 0
(
1 1 −1 1
A3 = AA2 = 0 1
1 0
1 0
0 0
(
(
(
E X A M P L E 17
The matrix D is
(a) Find (i) D2
2
1
0
2
1
0
1
−1
2 = 0
1
0
0
1
3 + 0
1
0
3
1
0
1
2A3 + A2 + A − I = 2 0
0
) (
)
) ( )
) (
) (
) (
) (
)
1
−1
1 = 0
1
0
2
= 0
0
6
2
0
0
1
6 + 0
0
2
3
= 0
0
9
3
0
−2
9
3
0
3
1
1
−1
2 + 0
1
0
2
1
0
2
1
0
3
1
0
−1
2
1
1
−1
2 + 0
1
0
1
1
0
1
1
0
) (
) (
1
−1
1 − 0
1
0
1
−1
1 − 0
1
0
0 0
1 0
0 1
0
1
0
0
0
1
)
)
( 02 02 ).
(ii) D3
(b) Deduce a matrix for Dn.
SOLUTION
(0 2)(0 2) (0 4)
8 0
2 0 4 0
(ii) D = (
=
0 2)(0 4) (0 8)
(a) (i) D2 = DD = 2 0 2 0 = 4 0
3
329
M O DUL E 3
(0 2)
4 0
2
D =(
=
0 4) (0
(b) Since D = 2 0 and
We can find An if
and only if A is a
square matrix.
2
2
0
22
)
we can deduce that Dn =
D3 =
( 20
n
0
2n
( 80 08 ) = ( 20
3
0
23
)
)
Transpose of a matrix
The transpose of a matrix A is found by interchanging the rows and columns of A
and is denoted by AT or A′.
( 32 98 ).
E X A M P L E 18
Find AT where A =
SOLUTION
Exchanging the rows and columns
3 2
AT =
9 8
(
E X A M P L E 19
SOLUTION
)
(
2
Write down AT where A = 3
7
2 3 7
AT = 4 2 5
5 6 1
(
)
4
2
5
)
5
6 .
1
()
2
The first row (2 4 5) becomes 4
5
()
()
3
The second row (3 2 6) becomes 2
6
7
The third row (7 5 1) becomes 5
1
Properties of the transpose of a matrix
(i) (AT)T = A
(ii) (kA)T = kAT, k ∈ℝ
(iii) (AB)T = B T AT
Symmetric and skew-symmetric matrices
A square matrix A is called symmetric if and only if AT = A.
The matrix
( 12 21 ) is symmetric since the transpose is also ( 12 21 ) .
A square matrix A is called skew-symmetric if and only if AT = −A.
330
MODULE 3tCHAPTER 16
Determinant of a square matrix
Take the product
of the leading
diagonal and
subtract the
product of the
other diagonal.
Determinant of a 2 × 2 matrix
(
:(
a11
as a
21
)
a12
a22 . The determinant of A is denoted by |A| or det(A) and is defined
a11
Let A = a
21
):
a12
a22 = a11 a22 − a12 a21.
( 25 47 ).
E X A M P L E 20
Find the determinant of A =
SOLUTION
|A| = (2 × 7) − (4 × 5) = 14 − 20 = −6
E X A M P L E 21
Find the determinant of A =
SOLUTION
|A| = (3 × 4) − (2 × (−5)) = 12 + 10 = 22
E X A M P L E 22
Given that
SOLU TION
:( ):
( −53 24 ).
(: a3 a2 ) : = 3, find the value(s) of a.
a 2 = a × a − 2 × 3 = a2 − 6
3 a
2
a −6=3
a2 = 9
a = 3 or
Try these 16.3
a = −3
Find the determinant of each of the following matrices.
(
(3 2)
(b) B = 63 4
(a) A = 4 2
2
)
(6 5)
(c) C = b 5
Determinant of a 3 × 3 matrix
(
a11
If A = a21
a31
)
a12
a22
a32
a13
a23 then the determinant of A is
a33
:
a23
a21 a23
a21 a22
−
a
+
a
12 a
13 a
a33
31 a33
31 a32
a22
det(A) = a11 a
32
: :
: :
:
5PHFUUIFĕSTUUFSNXFUBLFa11 DPWFSVQUIFSPXBOEDPMVNODPOUBJOJOHa11BOE
NVMUJQMZa11CZUIFEFUFSNJOBOUPGUIFSFNBJOJOH× 2 matrix.
'PSUIFTFDPOEUFSN XFDPWFSVQUIFSPXBOEDPMVNODPOUBJOJOHa12:
(
a11 a12 a13
a21 a22 a23
a31 a32 a33
)
331
M O DUL E 3
BOENVMUJQMZa12CZUIFEFUFSNJOBOUPGUIFSFNBJOJOH×NBUSJY HJWJOH
a
a
a12 a21 a23
31
33
:
:
BOETJNJMBSMZGPSUIFMBTUUFSNa13.
(
)
: : : :
E X A M P L E 23
1
Find the determinant of A = 0
2
SOLUTION
|A| = 1
: :
0
1 4
− (−2)
1 2
2
−2
1
1
3
4 .
2
0 1
4
+3
2
2 1
= 1(1 × 2 − 4 × 1) + 2(0 × 2 − 4 × 2) + 3(0 × 1 − 1 × 2)
= −2 − 16 − 6 = −24
(
E X A M P L E 24
3
Find the determinant of A = 4
2
SOLUTION
|A| = 3
: : : : : :
1 0
4 0
4
−1
+1
2 7
2 7
2
)
1
0.
7
1
1
2
1
2
= 3(1 × 7 − 0 × 2) − 1(4 × 7 − 0 × 2) + 1(4 × 2 − 1 × 2)
= 21 − 28 + 6 = −1
(
E X A M P L E 25
1
Find the determinant of B = 2
3
SOLUTION
|B| = 1
:
:
:
−1
1
1
)
2
−1 .
−4
: : :
−1
2 −1
2
− (−1)
+2
−4
3 −4
3
1
1
1
1
= 1(1 × (−4) − (−1) × 1) + (2 × (−4) − (−1) × 3) + 2(2 × 1 − 1 × 3)
= −3 − 5 − 2 = −10
Try these 16.4
Find the determinant of the following matrices.
(
1
0
1
4
7
6
(a) A = 2 −3 5
)
(
4
1
−2
2
3
3
(b) B = 2 1 −2
)
(
0
1
3
3
2
1
(c) C = 5 4 3
)
Properties of determinants
(i) If every element of a row or column of a matrix A is 0 then |A| = 0.
(ii) If A is a square matrix then |AT| = |A|.
(iii) If two rows or columns of A are identical then |A| = 0.
(iv) If two rows or columns of a matrix are interchanged, the sign of the value of the
determinant is changed.
(v) Let k ∈ ℝ; then kA = knA, where A is an n × n matrix.
(vi) For any two matrices A and B conformable to matrix multiplication, AB = A B.
332
MODULE 3tCHAPTER 16
Try these 16.5
(
a
b
c
g
h
i
)
(a) Let A = d e f . Show that |AT| = |A|.
(
a
b
c
d
e
f
)
(b) Show that if A = d e f then A = 0.
EXERCISE 16A
In questions 1–6, evaluate the sums and differences.
1
( 12 43 ) + ( −14 33 )
3
(4
5
(
2
5) + (3
2
(
(
4
8)
) (
1
−2 −1
5 + −2
3
−1
8 10
0
1
3
2
−2 −4
( 12
2
1
4
7
)
6
) (
3
0
4
2 1
− 1
−1 4 1
1
)
) ( )
) (
0
3
1
4 −3 − 1
4
9
2
2
5
3
5 6 1
6 2
4
3 2 2 − −4 −2 1
3 2
−6
1 1 1
)
Questions 7–14 refer to the following matrices.
A=
(
) (
0
2 1
1
1 3
1 ,B= 4
5
−1 2 −1
1
3
1
(
)
0
1
2 and C = 3
3
2
1
4
1
−2
1
1
)
Find
7 2A
11 4A + 2B − 3C
8 3A + 2B
9 A+B+C
12 B − A − 2C 13 4C − __12 A
10
14
A − 2C
1C
1 B − __
__
4
2
In questions 15–20, find the products of the matrices.
(
4
3
1
1
15 (−1 2 1) 2 −1
17
( 61 12 ) ( −11 24 )
19
(
4
3
−1
2 2
1 1
2 2
)(
)
−1 0
1 2
3 1
(4
) (1
16 2 −11 3 0 1
2
1
)
18 (4 6) ( 10 −12 )
1
1
1
)
20
(
1 2
3 1
−1 1
)(
−1 2
4 1
−1 1
1 3
1 0
−2 2
)
In questions 21–25, find the products AB and BA if possible.
( )
1
21 A = (1 2 4), B = −1
(
(
−2
22 A = 3
2
23 A =
1
2
−1
3
1 4
4 0 , B = (1
1 0
)
) (
−1
0
3 0
1
1 2 , B = 1 −1
1 0
1
0
2)
4
1
0
)
333
M O DUL E 3
(
−2 1
24 A = ( −21 40 11 ), B = 1 0
−1 5
(
)
)
1
25 A = 2 1 2 , B = −3
3 1 3
1 1
2 4
−4 −1 0
In questions 26–30, find the determinant of the matrix.
26 2 1
27 −43 −22
28
3 1
(
29
)
( )
( x0 12x )
(
30
)
( a2
1
4a
( 54
0
−1
)
)
In questions 31–35, find the determinant of the matrix.
31
34
(
(
)
1 0 1
2 1 2
1 3 1
1
4
x
x
3 −2
x −2
1
32
)
35
(
(
−3
1
1
4 1
−2 1
0 0
2
0
2
1
−4
4
6
2
1
)
)
(
33
a
−2
3
0
2
−1
In questions 36–40, write down the transpose of the matrix.
1 1
36 4 5
37 1 −2 1
38 0 1
3 2
3 −1 4
2 1
3 4
−1
2
2 −7
39 3 21 1
40
6 −5
4
4 1
1
3
1
9
(
(
)
(
)
(
)
(
1
41 Given that A = 2
3
(a) A = AT
)
(
)
(
−1 2
1
0 1 and B = 1
4
1 1
1
2
2
0
−1
2
2
3
1
)
)
)
3
−1 , verify that
2
(b) AB = A B
(
1
42 (a) Given that A = 1
1
−1
2
1
)
0
1 find (i) A2 (ii) A3.
1
(b) Hence find A3 + 2A2 − A − 2I, where I is the 3 × 3 identity matrix.
(
1 −1
0
1
0
43 Let A = 2
)
(
2
1
1 and B = −1
2
3
)
4
2
1
1
1 . Find A2 + 2AB + B2.
1
Singular and non-singular matrices
An n × n matrix A is singular if and only if A = 0.
E X A M P L E 26
Find the value of x for which the matrix A =
SOLU TION
For a singular matrix A = 0
Now A = (1)(2x) − (4)(2)
= 2x − 8
334
( 12
)
4
is singular.
2x
MODULE 3tCHAPTER 16
A matrix is singular if and only if its
determinant is 0.
Since A = 0, we have
2x − 8 = 0
x=4
A matrix A is non-singular if and only if A ≠ 0.
E X A M P L E 27
Find the set of values of a for which the matrix A =
SOLUTION
Since A is non-singular, we have |A| ≠ 0.
( a3 23 ) is non-singular.
|A| = (a)(3) − (2)(3)
= 3a − 6
Since |A| ≠ 0, we have 3a − 6 ≠ 0
3a ≠ 6
a≠2
E X A M P L E 28
SOLUTION
(
1 2
Determine whether the matrix −1 2
2 1
)
1
3 is singular.
2
If the matrix is singular then its determinant is 0. Let us find the determinant of the
matrix.
1 2 1
2 3
−1 3
−1 2
−2
+1
−1 2 3 = 1
1
2
2
2
2 1
2 1 2
:
Since
:
:
: : :
: :
:
= 1(4 − 3) − 2(−2 − 6) + 1(−1 − 4) = 1 + 16 − 5 = 12
1 2
−1 2
2 1
:
1
3 ≠ 0 the matrix is not singular.
2
Solving equations using determinants
(Cramer’s rule)
Solve the simultaneous equations
a11x + a12 y = b1
[1]
a21x + a22 y = b2
[2]
Equation [1] multiplied by a21 gives
a21a11x + a21a12 y = a21b1
[3]
Equation [2] multiplied by a11 gives
a11a21x + a11a22 y = a11b2
[4]
335
M O DUL E 3
Equation [4] − equation [3] gives
a11a22 y − a21a12 y = a11b2 − a21b1
∴ y(a11a22 − a21a12 ) = a11b2 − a21b1
: :
a11 b
1
a12 b
a
b
−
a
b
2
11 2
21 1
________
y = _____________
a11a22 − a21a12 = a11 a12
a21 a22
: :
Similarly, we get
: :
b1 a12
b2 a22
x = ________
a a
: :
The coefficient matrix is the matrix formed from the coefficients
of x and y in the equations. For the equations
12
a11x + a12y = b1
a21 a22
a21x + a22y = b2
11
a
a
the coefficient matrix is a11 a12
21
21
(
)
Notice that for both the x-value and the y-value, the denominator is the determinant
of the coefficient matrix.
In the numerator for the x-value, the first column of the matrix consists of the values
on the right-hand side of the coefficient equations and the second column the coefficients of y.
For the y-value, the first column of the numerator consists of the coefficients of x and
the second column contains the values on the right-hand side.
This result is known as Cramer’s rule.
E X A M P L E 29
Solve these simultaneous equations using Cramer’s rule.
2x + y = 3
3x − 2y = 1
SOLUTION
:
:
:
:
3
1
1
−2
−7 = 1
−6 − 1 = ___
x = ________ = _______
−4 − 3 −7
2
1
3 −2
2 3
3 1 = _______
2 − 9 = ___
−7 = 1
y = ________
−4 − 3 −7
2
1
3 −2
Hence x = 1, y = 1.
:
: :
:
The coefficient matrix is 2
3
1.
−2
1
∴ The denominator of x and y is 2
3 −2
(
|
For the numerator of x replace the
first column of the coefficient matrix
with 3 .
1
()
For the numerator of y replace the
second column of the coefficient
matrix with 3 .
1
()
336
)
|
MODULE 3tCHAPTER 16
E X A M P L E 30
Use Cramer’s rule to solve the simultaneous equations
4x + 5y = 10
3x − 4y = −8
:
SOLUTION
( 34
10
5
:
−40 − (−40) ____
−8 −4 = ____________
x = _________
= 0 =0
:
:
−31
−16 − 15
5
4
3 −4
4 10
−32 − (30) ____
3 −8 = __________
= −62 = 2
y = ________
−31
−16 − 15
5
4
3 −4
5
−4 is
:
:
)
the coefficient
matrix.
:
:
For the numerator of x replace
the first column of the coefficient
matrix with 10 .
−8
( )
For the numerator of y replace the
second column of the coefficient
matrix with 10 .
−8
( )
Hence x = 0, y = 2.
Try these 16.6
Solve the following pairs of simultaneous equations using Cramer’s rule.
(a) x + 3y = 5
(b) 2x − 4y = 2
4x + y = 9
3x − 7y = 4
Using Cramer’s rule to solve three equations in three unknowns
The coefficient
matrix is
a11 a12 a13
a21 a22 a23
a31 a32 a33
(
For the set of equations
a11x + a12 y + a13z = b1
)
a21x + a22 y + a23z = b2
a31x + a32 y + a33z = b3
using Cramer’s rule, we have
: : : : : :
b1 a12 a13
a11 b1 a13
b2 a22 a23
a21 b2 a23
b3 a32 a33
a31 b3 a33
_____________
x = _____________
a11 a12 a13 , y = a11 a12 a13 ,
a21 a22 a23
a21 a22 a23
a31 a32 a33
a31 a32 a33
:
: :
a11 a12 b1
a21 a22 b2
a31 a32 b3
z = _____________
a11 a12 a13
a21 a22 a23
a31 a32 a33
: :
:
Note the positions of b1, b2, b3 in the numerators of x, y and z. In the value of x,
b1, b2, b3 replaces the coefficient of x and similarly for y and z. The denominator is the
determinant of the coefficient matrix.
E X A M P L E 31
Use Cramer’s rule to solve the following simultaneous equations.
x + 2y + 3z = 1
2x − y + z = 2
x + 2y + z = 1
337
M O DUL E 3
: :
By Cramer’s rule
SOLUTION
:
:
: : : :
: : : :
:
:
2 3
1
1
2 1
2 −1
2 −1 1
−2
+3
1 −1
1(−3) − (2)(1) + 3(5)
2
2
1
1
1
1
1
1
2
x = _________ = ____________________________
= ___________________
1
2 3
1
2 1
2 −1
1(−3) − 2(1) + 3(5)
−1
1
−2
+3
2 −1 1
2 1
2
1 1
1
2 1
1
10 = 1
= ___
10
1 1 3
2 2 1
1 1 1 = ___
0 =0
y = _________
10
1 2 3
2 −1 1
1 2 1
: :
: :
: :
: :
Note
Since two
columns are
the same in the
numerator, the
determinant is 0.
1
2 1
2 −1 2
1
2 1 = ___
0 =0
z = __________
10
2 3
1
2 −1 1
2 1
1
∴ x = 1, y = 0, z = 0.
E X A M P L E 32
Solve these simultaneous equations.
2x + y + 3z = 1
4x − 3y + z = 7
x + 2y + z = 5
SOLUTION
Using Cramer’s rule, we have
:
:
:
:
:
:
: : : : :
: : : : :
1 3
1
−3 1
7 1
7 −3
7 −3 1
−
+3
1
1(−5) − (2) + 3(29) 80
2
5
5 1
5
2 1 = __________________________
2 1
x = __________
=4
= _________________ = ___
1
4
1
4
1 3
2
2(−5) − (3) + 3(11) 20
−3
−3
2
−
+
3
4 −3 1
2 1
2
1 1
1
1
1 1
: :
: : : : : :
: :: : : :
2 1 3
7 1
4 1
4 7
4 7 1
−
+3
2
2(2) − (3) + 3(13)
5 1
1 5 1 = _________________________
1 1
1 5 = _________________
40 = 2
y = __________
= ___
20
−3 1
1 3
4 1
4 −3
2
2(−5)
−
(3)
+
3(11)
2
−
+3
4 −3 1
2 1
1 1
1 2
1
2 1
:
:
:
:
:
:
:
: : : : :
: : : : :
2 1 1
−3 7
4 7
4 −3
4 −3 7
−
+
2
2(−29) − (13) + (11) ____
5
5
5
1
2
2
1
1
2 = __________________
__________
___________________________
z=
=
= −60 = −3
20
−3 1
1 3
4 1
4 −3
2
2(−5)
−
(3)
+
3(11)
2
−
+3
4 −3 1
2 1
1 1
1
2
1
2 1
Try these 16.7
:
Hence x = 4, y = 2, z = −3.
Use Cramer’s rule to solve the following simultaneous equations.
(a) 3x + 4y − 2z = 9
338
(b) 4x − 5y + 2z = 6
5x + y − z = 6
x+y+z=2
2x + y − 3z = 0
7x + 2y − 2z = 5
MODULE 3tCHAPTER 16
Inverse of a matrix
Inverse of a 2 × 2 matrix
(
)
a
Let A = c
b
, then the inverse of A, denoted by A−1, is
d
(
−b
a
1
d
A−1 = ___
|A| −c
)
provided that : A : ≠ 0. We interchange the elements of the leading diagonal and
d −b .
change the signs of the other two elements to get −c
a
(
( 23 54 ).
E X A M P L E 33
Find the inverse of A =
SOLUTION
|A| = 3 × 5 − 4 × 2 = 15 − 8 = 7
1
d −b
Since A−1 = ___
a
|A| −c
(
(
Given that A =
SOLU TION
1
d
A−1 = ___
|A| −c
( 12
(
First find the determinant of A.
)
1 5 −4
A−1 = __
7 −2
3
E X A M P L E 34
)
Exchange the terms in the leading
diagonal and change the sign of the
other two terms.
)
)
1 , find the inverse of A.
−2
−b
a
)
|A| = (1)(−2) − (1)(2)
= −2 − 2 = −4
∴
a11 a12 a13
a21 a22 a23
a31 a32 a33
)
Cofactor of a11 is
a
a
+ a22 a23
32
33
|
|
For a12:
(
a11 a12 a13
a21 a22 a23
a31 a32 a33
)
Cofactor of a12 is
|
a
(
−2
___
−4
−2 −1
1
___
=
=
1
−4 −2
−2
___
−4
(
)
−1
___
)( )
1
1
__
__
−4
2
4
=
1 −__
1
__
1
___
4
2
−4
Cofactors of a 3 × 3 matrix
For a11:
(
A−1
a
− a21 a23
31
33
|
The cofactor of an element of a matrix is found as follows:
(i) Ignore the row or column containing the element and find the determinant of
the remaining matrix.
(ii) Give the cofactor a positive or negative sign, depending on the position of the
element in the matrix. For a 3 × 3 matrix the pattern of signs is:
:
+ − +
− + −
+ − +
In the matrix
(
:
a11 a12 a13
A = a21 a22 a23
a31 a32 a33
)
339
M O DUL E 3
the signed cofactor of a11 is found by ignoring the row and column containing a11
a22 a23
, the signed
and finding the determinant of the remaining matrix, i.e. + a
32 a33
a21 a23
a21 a22
and
the
signed
cofactor
of
a
is
+
cofactor of a12 is − a
13
a33
a31 a32 .
31
:
:
:
: : :
: : :
:
For the second row, the signed cofactors of a21, a22, a23 are
:
:
: :
: :
a13
a11
,
+
a33
a31
a12
− a
32
:
:
a13
a11 a12
,
−
a33
a31 a32 respectively.
For the third row, the signed cofactors a31, a32, a33 are:
a13
a11
,
−
a23
a21
a12
+ a
22
E X A M P L E 35
Find the matrix of cofactors of
(
SOLUTION
a13
a11 a12
,
+
a23
a21 a22 respectively.
1 2 1
−1 3 2
1 1 4
A11 =
)
: :
: :
: :
: :
: :
: :
: :
: :
: :
3 2
= 12 − 2 = 10
1 4
−1
A12 = −
1
A13 =
2
1
1
= −(8 − 1) = −7
4
1 1
=4−1=3
1 4
A23 = −
A31 =
1
1
2
= −(1 − 2) = 1
1
2 1
=4−3=1
3 2
A32 = −
1
= −(2 + 1) = −3
2
1
−1
1 2 =3+2=5
−1 3
6
10
Matrix of cofactors = −7
3
1 −3
A33 =
E X A M P L E 36
SOLUTION
340
(
Find the matrix of cofactors of
(
−1 2
1 0
3 1
A11 =
|
2
= −(−4 − 2) = 6
4
−1 3
= −1 − 3 = −4
1 1
A21 = −
A22 =
Ignore the first row and first column
and find 3 2 .
1 4
:
2
−1
4
)
:
0 −1
= 0 − (−1) = 1
4
1
−4
1
5
)
|
MODULE 3tCHAPTER 16
:
A12 = −
A13 =
: :
: :
: :
1 0
=1−0=1
3 1
2
1
2
= −(8 − 2) = −6
4
−1
3
2
= −4 − 6 = −10
4
A21 = −
A22 =
:
A23 = − −13
A31 =
:
2 = −(−1 − 6) = 7
1
: :
2
2 = −2 − 0 = −2
0 −1
:
A32 = − −11
A33 =
:
1 −1
= −(4 + 3) = −7
3
4
:
−1
1
:
2 = −(1 − 2) = 1
−1
:
2
= 0 − 2 = −2
0
(
1
Matrix of cofactors = −6
−2
−7
1
7
−10
1 −2
)
Inverse of a 3 × 3 matrix
The adjoint is the
transpose of
the matrix of
cofactors.
(
a11
a
Let A = 21
a31
a12
a22
a32
)
a13
a23 , then
a33
1 adj(A)
A−1 = ___
|A|
where adj(A) = adjoint of A = (matrix of cofactors)T.
(
E X A M P L E 37
1 0
Find the inverse of A = 2 2
1 2
SOLUTION
|A| = 1
:
: :
2
1
2
− (0)
2 −1
1
:
)
−1
1.
−1
: :
1
2
+ (−1)
1
−1
2
2
= 1(−2 − 2) − 0 − 1(4 − 2)
= −4 − 2 = −6
Cofactors:
A11 =
:
A12 = −
A13 =
:
2
1
= −2 − 2 = −4
2 −1
:
2
1
: :
:
1
= −(−2 − 1) = 3
−1
2 2
=4−2=2
1 2
341
M O DUL E 3
A21 = −
A22 =
:
Note
You can check if
the inverse is correct by showing
that AA−1 = I.
E X A M P L E 38
SOLUTION
:
−1
= −(0 + 2) = −2
−1
0
2
:
: :
: :
: :
: :
−1
= −1 + 1 = 0
−1
1
1
A23 = −
A31 =
:
0
= −(2 − 0) = −2
2
1
1
−1 = 0 + 2 = 2
1
0
2
−1
A32 = − 1
= −(1 + 2) = −3
2
1
1 0
=2−0=2
A33 =
2 2
3
−4
2
Matrix of cofactors = −2
0 −2
2 −3
2
The transpose of the matrix of cofactors is found by interchanging the rows and
columns.
−2
2
4 2 −2
1 −4
1 −3
= __
A−1 = ___
3
0
−3
3
6 −2 02 −2
−6 2 −2
2
(
)
) (
(
(
)
: : : : : :
3 1
Find the inverse of A = 4 2
1 0
|A| = 3
1
2.
1
2 2
4 2
4 2
− (1)
+ (1)
0 1
1 1
1 0
= 3(2 − 0) − 1(4 − 2) + 1(0 − 2)
=6−2−2=2
Cofactors:
2 2
A11 =
=2−0=2
0 1
4 2
= −(4 − 2) = −2
A12 = −
1 1
A13 =
: :
: :
: :
: :
: :
: :
: :
: :
: :
4 2
= 0 − 2 = −2
1 0
A21 = −
A22 =
3
1
A23 = −
A31 =
342
3
4
1
= −(1 − 0) = −1
1
1
=3−1=2
1
3
1
1
2
A32 = −
A33 =
1
0
1
= −(0 − 1) = 1
0
1
=2−2=0
2
3
4
1
= −(6 − 4) = −2
2
1
=6−4=2
2
)
MODULE 3tCHAPTER 16
Note
An n × n matrix
has an inverse if
and only if the
matrix is nonsingular.
(
2 −2 −2
Matrix of cofactors = −1
2
1
0 −2
2
)
The transpose of the matrix of cofactors is found by interchanging the rows and
columns.
1
0
1 − __
0
2
2
−1
1
__
−1
A = −2
2 −2 = −1
1 −1
2 −2
1
1
2
__
−1
1
2
(
(
)
)
Properties of inverses
(i) (A−1)−1 = A
(ii) (AB)−1 = B−1A−1
(iii) A−1A = AA−1 = I
Systems of linear equations
The set of equations
a11x + a12 y + a13z = b1
a21x + a22 y + a23z = b2
a31x + a32 y + a33z = b3
can be written as
(
a11
a21
a31
a12
a22
a32
(
a11
where a21
a31
)( )
()
b1
a13 x
a23 y = b2
a33 z
b3
)
a12 a13
a22 a23 is called the coefficient matrix. We can represent the equation as:
a32 a33
(
a11
AX = B where A = a21
a31
a12
a22
a32
)
()
b1
a13
x
a23 , X = y and B = b2 .
z
a33
b3
()
Since AX = B, premultiplying both sides by A−1 we get
A−1AX = A−1B
Since A−1A = I, we have IX = A−1B
∴ X = A−1B
To solve the set of equations
a11x + a12 y + a13z = b1
a21x + a22 y + a23z = b2
a31x + a32 y + a33z = b3
343
M O DUL E 3
(i) Write the equations in the form AX = B.
(ii) Find the inverse of A.
(iii) Premultiply B by A−1 to get X.
E X A M P L E 39
Solve the simultaneous equations
4x + y + 3z = 4
2x − y + 2z = 1
3x + 2y + z = 3
SOLUTION
We can write the equation in the form AX = B, where
x
4
1 3
4
A = 2 −1 2 , X = y and B = 1
z
2 1
3
3
(
) ()
()
X = A−1B
First find A−1:
−1 2
2 2
2 −1
|A|= 4
−1
+3
3 1
3
2 1
2
= 4(−1 − 4) − 1(2 − 6) + 3(4 + 3)
: : : :
:
= −20 + 4 + 21 = 5
The cofactors of A are
A11 =
: :
: :
: :
: :
: :
: :
: :
: :
: :
−1 2
= −1 − 4 = −5
2 1
A12 = −
A13 =
1
2
3
= −(1 − 6) = 5
1
3
= 4 − 9 = −5
1
4
3
A23 = −
A31 =
2
= −(2 − 6) = 4
1
2 −1
=4+3=7
3
2
A21 =−
A22 =
2
3
4
3
1
= −(8 − 3) = −5
2
1 3 =2+3=5
−1 2
A32 = −
4
2
3
= −(8 − 6) = −2
2
4
2
1 = −4 − 2 = −6
−1
−5
7
4
Matrix of cofactors = 5 −5 −5
5 −2 −6
A33 =
adj(A) =
344
(
5
−5
4 −5
7 −5
(
5
−2
−6
)
)
:
MODULE 3tCHAPTER 16
1 adj(A)
A−1 = ___
|A|
5
5
1 −5
∴ A−1 = __
−5
4
−2
5 7 −5 −6
(
)
Since X = A−1B, we have
() (
x
1 −5
y = __
5 47
z
5
5
−5 −2
−5 −6
(
() ()
−20 + 5 + 15
16 − 5 − 6
28 − 5 − 18
1 05 = 10
= __
5 5
1
1
= __
5
)( )
)
4
1
3
∴ x = 0, y = 1, z = 1.
E X A M P L E 40
(a) Write these equations in the form AX = B.
x+y+z=3
2x + y + 3z = 6
7x − 4y + z = 4
(b) Find A−1.
(c) Hence solve the simultaneous equations.
SOLUTION
(
1
(a) 2
7
)( ) ( )
(
) ()
3
1 1 x
1 3 y = 6 is of the form AX = B where
−4 1 z
4
()
x
3
1 1
1 3 , X = y and B = 6 .
z
−4 1
4
1
A= 2
7
1 adj(A)
(b) A−1 = ___
|A|
:
|A| = 1
1
−4
: : : :
3 −1 2
1
7
3
+ 2
7
1
:
1 = 13 + 19 − 15 = 17
−4
The cofactors of A are
A11 =
:
:
: :
: :
: :
: :
1 3
= 13
−4 1
A12 = −
A13 =
2
7
A21 = −
A22 =
1
7
2
7
3
= 19
1
1
= −15
−4
1 1
= −5
−4 1
1
= −6
1
345
M O DUL E 3
:
A23 = −
1
7
: :
: :
: :
1
A31 =
1
:
1
= 11
−4
1
=2
3
1 1
= −1
A32 = −
2 3
1 1
= −1
A33 =
2 1
(
13 19
Matrix of cofactors = −5 −6
2 −1
(
)
13 −5
2
19 −6 −1
−15 11 −1
1
___
−1
A =
adj(A)
|A|
13 −5
2
1
∴ A−1 = ___
19
−6
−1
17 −15 11 −1
adj (A) =
(
−15
11
−1
)
)
(c) Since X = A−1B, we have
() (
)( )
(
)
( ) ()
x
13
1
y = ___
19
17 −15
z
−5
−6
11
2 3
−1 6
−1 4
39 − 30 + 8
1
= ___
57 − 36 − 4
17 −45
+ 66 − 4
1
1 17
= 1
= ___
17
17 17
1
∴ x = 1, y = 1, z = 1.
Try these 16.8
(a) Solve the following simultaneous equations.
x + 2y + 4z = 14
3x − y − z = 3
x + 5y + 2z = 9
(b) Find the values of x, y and z satisfying the following equations.
3x − 2y + z = −2
4x + y + 7z = 14
x + y + 2z = 6.
Row reduction to echelon form
Two matrices A and B are row equivalent if and only if B can be obtained from A
using the following operations, called row reduction.
346
MODULE 3tCHAPTER 16
● Ri ↔ Rj that is, we can interchange rows.
● Ri → cRi that is, we can multiply a row by a scalar quantity.
● Ri → aRi + bRj that is, Ri is a linear combination of Ri and Rj .
A matrix is in row echelon form if the number of zeros before the first non-zero
element in each row increases from row to row, the leading diagonal consists of 1s
and all elements below the leading diagonal consists of zeros.
(
E X A M P L E 41
SOLU TION
1
0
0
a
1
0
)
b
c is a matrix in row echelon form.
1
(
)
−1
4 to echelon form.
2
2 1
Row reduce the matrix A = 1 1
3 1
We can leave row 1 alone and perform the following two operations to make the first
elements of rows 2 and 3 zero.
R2 → 2R2 − R1
2(1
2(3
R3 → 2R3 − 3R1
1
1
4) − (2 1 −1) = (0 1 9)
2) −3(2 1 −1) = (0 −1 7)
The row matrix becomes
(
2
0
0
1
1
−1
−1
9
7
)
We next leave rows 1 and 2 alone and make the second element in row 3 zero.
We can make this element zero by the following operation:
R3 → R3 + R2
−1
(0
7) + (0
1
9) = (0
0
16)
A becomes
(
2
0
0
1
1
0
−1
9
16
)
We can now make the elements in the leading diagonal 1 by performing the
following operations:
1R
R1 → __
2 1
1R
R3 → ___
16 3
1
1
__
1 __
2 −2
∴ After row reduction A becomes 0 1
9
0 0
1
(
E X A M P L E 42
SOLU TION
(
3
Reduce the matrix 2
4
2
−1
2
)
)
4
2 to echelon form.
2
We can leave row 1 alone and perform the following two operations to make the first
elements of rows 2 and 3 zero.
R2 → 3R2 − 2R1
R3 → 3R3 − 4R1
347
M O DUL E 3
The row matrix becomes
(
3
0
0
4
2
−7 −2
−2 −10
)
We next leave rows 1 and 2 alone and make the second element in row 3 zero.
We can make this element zero by the following operation:
R3 → 7R3 − 2R2
A becomes
(
3
0
0
2
4
−7 −2
0 −66
)
We can now make the elements in the leading diagonal 1 by performing the
following operations:
1R
R1 → __
3 1
1R
R2 → −__
7 2
1 R
R3 → −___
66 3
( )
1
∴ After row reduction, A becomes 0
0
2
__
4
__
3
1
0
3
2
__
7
1
A step-by-step elimination will keep you on track for reducing to echelon form.
With some practice you may be able to do the reduction using fewer steps.
Remember to leave row 1 alone, make the first element in row 2 and the first element
in row 3 zero. Next make the second element in row 3 zero and finally make the
elements in the leading diagonal 1.
Try these 16.9
Reduce the following matrices to echelon form.
0
2
1 5
4 1
(a) 1 −2 3
(b) 2 3 −1
(c)
2
1 2
2 5
2
(
(
)
)
(
−1
1
4
2
3
1
1
2
2
)
Finding the inverse of a matrix by row reduction
An augmented matrix is a matrix formed by joining the columns of two matrices.
1
2 −1 3
Given the matrices A = 4
1 2 and B = 2 the augmented matrix is denoted
1
0
1 1
3
1
2 −1
by (A | B) = 4
1 2 2.
0
1 1 1
The augmented matrix is useful when solving systems of linear equations and finding
the inverse of a matrix.
(
(
)
)
()
To find the inverse of the matrix A by row reduction we first form the augmented
matrix (A | I ) and row reduce until the augmented matrix becomes (I | A−1).
348
MODULE 3tCHAPTER 16
(
2
3
1
)
E X A M P L E 43
1
Find the inverse of A = −1
2
SOLU TION
Forming the augmented matrix, we have
(
:
1 2 1 1
−1 3 2 0
2 1 4 0
0
0
1
0
1
0
1
2 using row reduction.
4
)
/PXXFSPXSFEVDFUPPCUBJO I|A−1).
R2 → R2 + R1
Use row operations to obtain 0 in the first
column of rows 2 and 3.
R3 → R3 − 2R1
(
1
0
0
2
5
−3
:
0 0
1 0
0 1
1 1
3 1
2 −2
1R
R2 → __
5 2
1R
R3 → −__
3 3
1 2 1
3
0 1 __
5
−2
___
0 1 3
:
(
)
Use row operations to obtain 1s in the
second column of rows 2 and 3.
1
1
__
5
2
__
3
0
0
0
−1
___
3
1
__
5
0
R1 → R1 − 2R2
:
)
R3 → R3 − R2
(
(
−1
___
5
3
__
0 1
5
−19
0 0 ____
15
15
R3 → −___
19 R3
−1
1 0 ___
5
__
0 1 53
1
0
0
0
1
:
3
__
5
1
__
5
7
___
15
3
__
5
1
__
5
−7
___
19
−2
___
0
5
1
__
5
−1
___
5
0
−1
___
3
−2
___
5
1
__
5
3
___
19
0
0
5
___
19
1R
R1 → R1 + __
5 3
3
R2 → R2 − __
5 R3
(
:
10
0 0 ___
19
8
0 1 0 ___
19
−7
0 0 1 ___
19
1
)
)
Use row operations to obtain 0 in column 2
of rows 1 and 3.
Use a row operation to obtain 1 in column
3 of row 3.
Use row operations to obtain 0 in column 3
of rows 1 and 2.
−7
___
19
2
___
19
3
___
19
1
___
19
−3
___
19
5
___
19
)
Notice that the identity matrix is now on
the left.
Hence the inverse of the matrix is
(
1
1 108 −7
A−1 = ___
−3
2
19 −7
5
3
)
349
M O DUL E 3
E X A M P L E 44
2
Use row reduction to find the inverse of the matrix A = 1
3
SOLUTION
Forming the augmented matrix, that is, (A|I), we have
(
(
2
1
3
1
0
1
:
0 0
1 0
0 1
2 1
−1 0
4 0
1
0
1
2
−1 .
4
)
)
Now we row reduce to obtain (I|A−1).
Interchanging rows 1 and 2:
(
1
2
3
0
1
1
:
−1 0
2 1
4 0
1 0
0 0
0 1
)
R2 → R2 − 2R1
Use row operations to obtain 0 in column 1
of rows 2 and 3.
R3 → R3 − 3R1
(
1
0
0
0
1
1
:
:
−1 0
4 1
7 0
1 0
−2 0
−3 1
)
R3 → R3 − R2
(
1
0
0
0
1
0
−1 0
4 1
3 −1
1R
R3 → __
3 3
1 0 −1 0
4 1
0 1
−1
1 ___
0 0
3
R1 → R1 + R3
:
(
1 0
−2 0
−1 1
)
0
0
1
__
3
1
−2
−1
___
3
Use a row operation to obtain 0 in column
2 of row 3.
Use a row operation to obtain 1 in column
3 of row 3.
)
Use row operations to obtain 0 in column 3
of rows 1 and 2.
R2 → R2 − 4R3
(
1
0
0
1
0
0
:
−1
0 ___
3
7
0 __
3
−1
1 ___
3
2
__
3
−2
___
3
−1
___
3
1
__
3
−4
___
3
1
__
3
)
The left side of the matrix is the identity 3 × 3 matrix, therefore the right side is the
inverse of A.
(
1 −1
Hence A−1 = __
7
3 −1
E X A M P L E 45
350
2
−2
−1
1
−4
1
(
3
−2
2
2
Find the inverse of A = 1
3
)
)
2
2.
1
MODULE 3tCHAPTER 16
SOLU TION
The augmented matrix is
(
2
1
3
:
3 2 1
−2 2 0
2 1 0
1R
R1 → __
2 1
0 0
1 0
0 1
)
Use row operations to obtain 1 in column 1
of row 1 and zeros in column 1 of rows 2
and 3.
R2 → 2R2 − R1
R3 → 2R3 − 3R1
(
1
0
0
:
1R
R2 → −__
7 2
1R
R3 → −__
5 3
(
1
)
3
1
__
__
1
2
2 0 0
−7
2 −1 2 0
−5 −4 −3 0 2
3
__
0
2
1
0
1
1
−2
___
7
4
__
5
:
:
:
1
__
2
1
__
7
3
__
5
Use row operations to obtain 1s in column
2 of rows 2 and 3.
0
−2
___
7
0
0
0
−2
___
5
)
3R
R1 → R1 − __
2 2
R3 → R3 − R2
(
1
0
0
1
0
0
10
___
2
__
7
7
−2 __
1
___
7
7
38 ___
16
___
35 35
35R
R3 → ___
38 3
(
1
0
0
1
0
0
10
___
2
__
7
7
−2 __
1
___
7
7
8
1 ___
19
Use row operations to obtain 0 in column 2
of rows 1 and 3.
3
__
7
−2
___
7
2
__
7
0
0
−2
___
5
Use a row operation to obtain 1 in column 3
of row 3.
3
__
7
−2
___
7
5
___
19
0
0
−7
___
19
10R
R1 → R1 − ___
7 3
2
__
R2 → R2 + R3
7
(
1
0
0
1
0
0
:
−6
0 ___
19
5
___
0 19
8
1 ___
19
)
)
Use row operations to obtain 0 in column 3
of rows 1 and 2.
1
___
19
−4
___
19
5
___
19
10
___
19
−2
___
19
−7
___
19
)
351
M O DUL E 3
The left side of the matrix is the identity 3 × 3 matrix, therefore the right side is the
inverse of A.
1
Hence A−1 = ___
19
(
−6
5
8
1 10
−4 −2
5 −7
)
Try these 16.10 Find the inverse of each of the following matrices by row reduction.
(
1
3
5
3
2
5
(a) 2 −1 1
)
(b)
(
−2 1 5
1 0 1
1 7 1
)
Solving simultaneous equations using row reduction
Given AX = B where A represents the matrix of coefficients, we can solve these
equations by first forming the augmented matrix (A|B) and then reducing this matrix
to echelon form. Let us see how this works.
E X A M P L E 46
Solve the simultaneous equations
3x + 2y − z = 11
x + 4y − 5z = 7
2x + 3y + z = 9
SOLUTION
Writing the equations in matrix form we have
(
3
1
2
2
4
3
)( ) ( )
−1 x
11
−5 y = 7
1 z
9
Form the augmented matrix:
(
3
1
2
2
4
3
:)
−1 11
−5 7
1 9
We now reduce this matrix to echelon form.
Leaving row 1 alone, we can make the first elements in row 2 and row 3 zero by the
following operations.
R2 → 3R2 − R1
R3 → 3R3 − 2R1
(
3
0
0
2
10
5
:)
−1 11
−14 10
5 5
Leaving rows 1 and 2 alone, we make the second element in row 3 zero using
R3 → 2R3 − R2
Our matrix becomes
(
352
3
0
0
2
10
0
:
−1 11
−14 10
24 0
)
MODULE 3tCHAPTER 16
We can now make the elements in the leading diagonal all 1s as follows.
1R
R1 → __
3 1
1R
R2 → ___
10 2
1R
R3 → ___
24 3
The matrix now reduces to
(
1
0
0
:)
2 ___
11
−1 ___
__
3
3
3
−14
____
1 10
1
1
0
0
(
2
1 __
3
We can now rewrite the equations as 0 1
0 0
Our equations reduce to
2 y − __
1 z = ___
11
x + __
3
3
3
14
___
y− z=1
10
z=0
−1
___
)( ) ( )
11
___
x
3
3
−14 y = 1
____
10
z
0
1
Substituting z = 0 into the second equation we get
14 (0) = 1
y − ___
10
y=1
Substituting z = 0 and y = 1 in the first equation
2 (1) − __
1(0) = ___
11
x + __
3
3
3
11
2 = ___
x + __
3
3
9=3
x = __
3
Hence x = 3, y = 1 and z = 0.
E X A M P L E 47
Write this set of equations in matrix form.
x + 4y − 5z = 13
2x + 3y + z = −1
2x + y − z = 1
Form the augmented matrix and solve the equations by reducing the augmented
matrix to echelon form.
SOLUTION
The equations in matrix form is
(
1
2
2
4
3
1
)( ) ( )
−5 x
13
1 y = −1
−1 z
1
353
M O DUL E 3
The augmented matrix is
(
1
2
2
4
3
1
:
−5 13
1 −1
−1 1
)
We now reduce this matrix to echelon form.
Leaving row 1 alone, we make the first elements in rows 2 and 3 zero by the
following row operations
R2 → R2 − 2R1
R3 → R3 − 2R1
(
1
0
0
4
−5
−7
:
−5 13
11 −27
9 −25
)
Leaving rows 1 and 2 alone, we make the second element in row 3 zero:
R3 → 5R3 − 7R2
(
1
0
0
:
4 −5 13
−5
11 −27
0 −32 64
)
We now make the elements in the leading diagonal 1
−1 R
R2 → ___
5 2
−1
R3 → ___ R3
32
The matrix reduces to
(
1
0
0
4
1
0
−5
:)
13
27
−11 ___
____
5
1
5
−2
(
1
We can now rewrite the equations as 0
0
∴ x + 4y − 5z = 13
4
1
0
−5
−11
____
5
1
27
11z = ___
y − ___
5
5
z = −2
Substituting z = −2 into the second equation
27
11 (−2) = ___
y − ___
5
5
y=1
Substituting z = −2, y = 1 into the first equation
x + 4(1) − 5(−2) = 13
x + 14 = 13
x = −1
Hence x = −1, y = 1 and z = −2.
354
)( ) ( )
x
y =
z
13
27
___
5
−2
MODULE 3tCHAPTER 16
Try these 16.11 Solve the sets of simultaneous equations by row reducing the augmented matrix.
(a) 2x + y + 6z = 2
(b) 3x + y + z = 3
x − 4y + z = −6
4x − 2y + 3z = 4
4x + 2y − z = 17
2x + 4y − z = 2
Systems of linear equations with two unknowns
A system of equations is a set of equations in two or more variables. A solution of
a system of equations is a set of values of the variables satisfying each equation. To
solve a set of equations is to find all solutions of the equations. The term consistent
is used for a system of equations that has at least one solution and inconsistent for
a system of equations that has no solutions. A system of equations is linear if all the
equations are linear.
In a system of equations with two unknowns each equation represents the equation
of a line. We can find solutions to these equations by elimination or substitution.
Two lines can do one of the following:
(i) intersect at one point
(ii) be parallel to each other
(iii) coincide.
Intersecting lines
For two lines that intersect at one point the equations are said to be consistent and
independent of each other.
E X A M P L E 48
Solve this set of equations using a matrix method and represent this on a diagram.
x−y=1
x + 2y = 4
SOLUTION
Writing the equations in matrix form, we have
)( ) ( )
( 11
−1 x = 1
4
2 y
( y ) = ( 11
−1
2
(
)
x
) ( 14 )
−1
det 1 −1 = (1)(2) − (−1)(1) = 3
1
2
( 11
−1
2
)
−1
(
1 2 1
= __
3 −1 1
)
x
( y ) = __13 ( −12 11 ) ( 41 ) = __31 ( 36 )= ( 21 )
∴ x = 2, y = 1.
355
M O DUL E 3
These two equations represent two straight lines which intersect at (2, 1).
4
y
x–y=1
3
2
1
(2, 1)
–4
–3
–2
0
–1
1
2
x
3
4
–1
x + 2y = 4
–2
–3
–4
E X A M P L E 49
Solve the simultaneous equations 3x − y = 4 and 2x + y = 1 and represent them on
the xy plane.
SOLUTION
Write the equations in matrix form:
( 32
)( ) ( )
−1 x = 4 .
1 y
1
Form the augmented matrix:
( 32
−1 4
11
)
4
Row reduce to echelon form:
(
:
3 −1 4
R2 → 3R2 − 2R1 gives
0 5 −5
(
1 R gives 3 −1 4
R2 → __
5 2
0 1 −1
We now have
(
)( ) ( )
3 −1 x
4
=
y
0 1
−1
)
)
2
1
x
–4
–3
–2
–1
0
–1
1
2
(1, –1)
3
4
5
–2
–3
y = −1
–4
Hence 3x + 1 = 4, x = 1
3x – y = 4
3
∴ 3x − y = 4
and
y
2x + y = 1
–5
∴ x = 1, y = −1
Parallel lines
If two lines are parallel, the system of equations has no solutions because the lines
never intersect; the equations are said to be inconsistent.
356
MODULE 3tCHAPTER 16
E X A M P L E 50
Decide whether this system of equations is consistent.
x + 2y = 3
2x + 4y = 4
SOLUTION
Write the equations in matrix form:
3
1 2 x
=
4
2 4 y
Form the augmented matrix:
( )( ) ( )
( 12
:
(
:
y
x + 2y = 3 3
)
2 3
4 4
Row reduce to echelon form:
3
1 2
R2 → R2 − 2R1 gives
0 0 −2
Rewriting, we have
3
1 2 x
=
0 0 y
−2
∴ x + 2y = 3
(
4
2
2x + 4y = 4
1
)
x
–4
–3
–2
)( ) ( )
–1
0
1
2
3
4
5
–1
–2
–3
and 0 = −2, which is impossible.
The equations have no solutions and hence the equations are inconsistent.
In this case the lines are parallel and do not intersect.
In the reduced form of the augmented matrix one row of the coefficient matrix has
all zeros and hence the equations are inconsistent.
Lines that coincide
When two lines coincide, the system of equations is consistent and there is an infinite
set of equations.
E X A M P L E 51
Solve the simultaneous equations
3x + 2y = 4
6x + 4y = 8
SOLUTION
Writing in matrix form, we have
3 2 x
4
=
8
6 4 y
Forming the augmented matrix, we get
3 2 4
6 4 8
Row reducing,
3 2 4
R2 → R2 − 2R1 gives
0 0 0
The set of equations is
(
)( ) ( )
(
:)
(
:)
( 30 20 ) ( xy ) = ( 04 )
357
M O DUL E 3
giving one equation 3x + 2y = 4, which means the two lines are coincident.
Let y = λ where λ ∈ ℝ.
We get 3x + 2λ = 4
4 − 2λ
x = ______
3
Hence there is an infinite set of solutions and the solutions are
y = λ where λ ∈ ℝ
4 − 2λ
x = ______
3
In this particular case the augmented matrix consists of a whole row of zeros which
indicates that we have an infinite set of solutions to the equations.
When solving equations row reducing the augmented matrix can help us identify if
the set of equations is consistent or inconsistent, and if they are consistent whether
there is an infinite set of solutions or one solution.
(i) If the augmented matrix has no complete row of zeros then the two lines
intersect at a point and there is one solution to the simultaneous equations.
(ii) If the augmented matrix consists of a row of zeros in the coefficient matrix only,
then the equations are inconsistent and there is no solution to the equations. In
this case the lines are parallel and do not intersect.
(iii) If the augmented matrix consists of a whole row of zeros then the equations
have an infinite set of solutions and the lines coincide.
Systems of linear equations with three unknowns
A system of three linear equations containing three unknowns has either
(i) a unique solution
(ii) no solutions, or
(iii) an infinite set of solutions.
For the set of equations
a11x + a12 y + a13z = b1
a21x + a22 y + a23z = b2
a31x + a32 y + a33z = b3
(
) ()
()
a11 a12 a13
b1
x
we have AX = B where A = a21 a22 a23 , X = y and b = b2 .
a31 a32 a33
z
b3
Unique solution
If the determinant of the coefficient matrix is not equal to zero then the equations
have a unique solution. If we row reduce the augmented matrix to echelon form and
the matrix has no complete row of zeros then the equations are consistent and there
is a unique solution.
358
MODULE 3tCHAPTER 16
E X A M P L E 52
Show that the following set of equations has a unique solution.
2x + y + 3z = 10
x−y+z=1
4x + y + z = 8
SOLUTION
Method 1
Writing the equation in matrix form, we have
2
1 3 x
10
1 −1 1 y = 1
8
4
1 1 z
)( ) ( )
(
The coefficient matrix is
2
1 3
1 −1 1
4
1 1
(
)
The determinant of the coefficient matrix is
2
1 3
1
1 1
1
−1
+3
1 −1 1 = 2 −11
1
1
4
4
4
1 1
|
|
|
|
|
|
|
−1
= 2(−2) − 1 (−3) + 3(5) = 14
1
|
Since the determinant of the coefficient matrix is not equal to zero, the equations
have a unique solution.
Method 2
We row reduce the augmented matrix to echelon form:
3 10
2
1
1 1
1 −1
4
1
1 8
R2 → 2R2 − R1
:
(
)
R3 → R3 − 2R1
3 10
2
1
0 −3 −1 −8
0 −1 −5 −12
R3 → 3R3 − R2
:
(
(
:
)
)
3 10
1
−3
−1 −8
−1 −14 −28
−1 R
R3 → ___
14 3
3 10
2
1
0 −3 −1 −8
1 2
0
0
Since the augmented matrix has no complete row of zeros, the equations have a
unique solution.
(
E X A M P L E 53
2
0
0
:
)
Decide whether the following equations have a unique solution.
3x + 2y + z = 6
x − 3y + 2z = 0
2x + y + z = 4
359
M O DUL E 3
SOLUTION
The coefficient matrix is
3
2 1
1 −3 2
2
1 1
The determinant of the coefficient matrix is
3
2 1
−3 2
1 2
1 −3
−2
+1
= 3(−5) − 2 (−3) − 1(7) = −16
1 −3 2 = 3
1
1
1
1
2
2
2
1 1
Since the determinant of the coefficient matrix is not equal to zero, the equations
have a unique solution.
(
)
:
:
|
|
| |
|
|
We can also row reduce the augmented matrix to echelon form to make a decision on
unique solutions.
E X A M P L E 54
Show that these equations have a unique solution and find this solution.
x + y + 2z = 1
x−y+z=1
2x + y + 3z = 4
SOLUTION
We can write the equations as
(
)( ) ( )
1 1 2 x
1
1 −1 1 y = 1
4
2 1 3 z
The augmented matrix is
(
1
1
2
1 2
−1 1
1 3
:)
1
1
4
Row reducing to echelon form:
R2 → R2 − R1
R3 → R3 − 2R1
gives
(
1
0
0
:)
1
0
2
1
2
−2 −1
−1 −1
R3 → 2R3 − R2
gives
(
1
0
0
:)
1
2 1
−2 −1 0
0 −1 4
Therefore
(
360
)( ) ( )
1
1
2 x
1
0 −2 −1 y = 0
4
0
0 −1 z
MODULE 3tCHAPTER 16
We have
−z = 4, z = −4
−2y − z = 0,
2y = −(−4), y = 2
x + y + 2z = 1, x + 2 − 8 = 1, x = 7
Hence x = 7, y = 2, z = −4
No solutions
When the augmented matrix (A|B) is reduced to echelon form and the coefficient
matrix (only) contains a row of zeros then the set of equations has no solutions.
E X A M P L E 55
Show that this set of equations has no solutions.
x + 2y + 3z = 3
4x + y − z = 5
3x − 2y + 5z = 6
SOLUTION
Writing the equations in matrix form, we have
(
1
4
3
)( ) ( )
3 x
3
2
1 −1 y = 5
5 z
−2
6
Forming the augmented matrix, we get
(
1
4
3
:)
3 3
2
1 −1 5
5 6
−2
Row reducing to echelon form:
R2 → R2 − 4R1
R3 → R3 − 3R1
(
1
2
0 −7
0 −8
3 3
−13 −7
−4 −3
)
R3 → 7R3 − 8R2
(
1
0
0
2
0
0
:
3
3
−13 −7
76 35
)
R3 → 13R3 + 76R2
(
1
0
0
2
0
0
:
3
3
−13 −7
0 −77
)
Since the last row of the coefficient matrix consists of a row of zeros, there are no
solutions.
361
M O DUL E 3
If we rewrite the equations we get
(
1
0
0
2
0
0
)( ) ( )
3 x
3
−13 y = −7
−77
0 z
∴ x + 2y + 3z = 3
−13z = −7
0 = −77
The last equation is impossible and hence there are no solutions to the equations.
E X A M P L E 56
Show that this set of equations has no solutions.
2x + 3y − z = 3
5x − y + 3z = 8
6x + 9y − 3z = 4
SOLUTION
Writing the equations in matrix form, we have
(
2
5
6
3
−1
9
)( ) ( )
3
−1 x
3 y = 8
4
−3 z
Forming the augmented matrix, we get
(
2
5
6
3 −1 3
3 8
−1
9 −3 4
)
Row reducing to echelon form,
R2 → 2R2 − 5R1
R3 → R3 − 3R1
(
2
0
0
:
3 −1 3
−17 11 1
0 −5
0
)
Since the last row of the coefficient matrix consists of a row of zeros, there are no
solutions.
Infinite set of solutions
When the augmented matrix (A|B) is reduced to echelon form and the augmented
matrix contains at least one row of zeros then the set of equations has an infinite set
of solutions.
E X A M P L E 57
Show that this set of equations has an infinite set of solutions.
x+y+z=4
2x + y − 3z = 2
4x + 3y − z = 10
362
MODULE 3tCHAPTER 16
SOLU TION
Writing the equations in matrix form, we have
(
1
2
4
1
1
3
)( ) ( )
1 x
4
−3 y = 2
10
−1 z
The augmented matrix is
(
1
2
4
1
1
3
:
1 4
−3 2
−1 10
)
Row reducing to echelon form, we get
R2 → R2 − 2R1
R3 → R3 − 4R1
(
1
0
0
R3 → R3 − R2
(
1
0
0
:
:
1
1 4
−1 −5 −6
−1 −5 −6
1
1 4
−1 −5 −6
0
0 0
)
)
Since the augmented matrix contains a row of zeros, the equations have an infinite
set of solutions.
E X A M P L E 58
Show that this set of equations has an infinite set of solutions.
3x + 2y + z = 4
2x + y − 3z = 2
7x + 4y − 5z = 8
SOLU TION
Writing the equations in matrix form, we have
(
)( ) ( )
3 2
1 x
4
2 1 −3 y = 2
7 4 −5 z
8
The augmented matrix is
(
3
2
7
2
1
4
:)
1 4
−3 2
−5 8
Row reducing to echelon form, we get
R2 → 3R2 − 2R1
R3 → 3R3 − 7R1
(
3
0
0
:
2
1
4
−1 −11 −2
−2 −22 −4
)
363
M O DUL E 3
R3 → R3 − 2R2
(
1
0
0
:
1
4
1
−1 −11 −2
0
0 0
)
Since the augmented matrix contains a row of zeros, the equations have an infinite
set of solutions.
E X A M P L E 59
Show that the set of equations
x+z=6
2x + y − 3z = 2
8x + 3y − 7z = 18.
has an infinite set of solutions. Hence find the solutions.
SOLUTION
Writing the equations in matrix form, we have
(
1
2
8
0
1
3
)( ) ( )
6
1 x
−3 y = 2
−7 z
18
The augmented matrix is
(
1
2
8
0
1
3
:
1 6
−3 2
−7 18
)
Row reducing to echelon form, we get
R2 → R2 − 2R1
R3 → R3 − 8R1
(
1
0
0
0
1
3
:
6
1
−5 −10
−15 −30
)
R3 → R3 − 3R2
(
1
0
0
0
1
0
:
6
1
−5 −10
0
0
)
Since the augmented matrix contains a row of zeros, the equations have an infinite
set of solutions.
Rewriting the equations, we have
(
1
0
0
0
1
0
)( ) ( )
6
1 x
−5 y = −10
0
0 z
x+z=6
y − 5z = −10
364
MODULE 3tCHAPTER 16
Since x, y and z can be any real number, let z = λ where λ ∈ ℝ.
Substituting into the equations, we have
y − 5λ = −10
y = 5λ − 10
x+λ=6
x = −λ + 6
Hence the solutions are
x = −λ + 6
y = 5λ − 10
z = λ where λ ∈ ℝ.
The three original equations represent three planes in three dimensions. The solutions
x = −λ + 6, y = 5λ − 10, z = λ
represent the parametric equations of a line in three dimensions. The planes intersect
in the line.
When the determinant of the coefficient matrix is zero, the set of equations has
either no solutions or an infinite set of solutions. To decide which one is true, reduce
the augmented matrix to echelon form and if the coefficient matrix alone has a row
of zeros then there are no solutions and if there is at least one row of zeros in the
augmented matrix then there is an infinite set of solutions.
Solution of linear equations in three unknowns:
geometrical interpretation
Recall from Unit 1 that an equation of the form ax + by + cz = D represents the
a
equation of a plane in three dimensions with the vector b
c
being a vector perpendicular (or normal) to the plane.
()
The system of equations
a11x + a12y + a13z = b1
Π1
a21x + a22y + a23z = b2
Π2
a31x + a32y + a33z = b3
Π3
represents three planes. These planes can
(i) intersect at one point
(ii) intersect at an infinite set of points, or
(iii) not intersect.
365
M O DUL E 3
The following table gives a geometrical interpretation of each.
Type of system
Consistent and
independent
Number of
solutions
1
Behaviour of planes
All three planes intersect at one point
Point of intersection
Π1
Π2
Π3
Consistent and
dependent
Infinite
(i) All three planes intersect in a line. In this case one of the
equations is a linear combination of the other two.
Π3
Line of intersection
of the three planes
Π2
Π1
(ii) All three planes are identical.
Π1, Π2, Π3
Inconsistent
None
(i) All three planes are parallel and distinct.
Π1
Π2
Π3
(ii) Any two of the planes are parallel and distinct.
Π1
Π1 is parallel to Π3
Π2
Π3
(iii) One plane is parallel to the line of intersection of the other two.
Π3 and Π2
intersect in a line
Π2
Π1
Π3
366
MODULE 3tCHAPTER 16
Applications of matrices
E X A M P L E 60
The supply function for a commodity is given by q s(x) = ax2 + bx + c, where a, b
and c are constants. When x = 1, the quantity supplied is 5; when x = 2, the quantity
supplied is 12; when x = 3, the quantity supplied is 23. Use a matrix method to find
the values of a, b and c.
SOLUTION
qs(1) = a(1)2 + b(1) + c = 5
qs(2) = a(2)2 + b(2) + c = 12
qs(3) = a(3)2 + b(3) + c = 23
We get three equations to solve simultaneously:
a+b+c=5
4a + 2b + c = 12
9a + 3b + c = 23
Writing the equations in matrix form
5
1 1 1 a
4 2 1 b = 12
23
9 3 1 c
)( ) ( )
() ( ) ( )
: :: :: :: :
(
)
( ) (
)
() (
)( ) ( )
(
a
1 1 1
b = 4 2 1
c
9 3 1
−1
5
12
23
1 1 1
2 1
4 1
4 2
−
+
= −1 + 5 − 6 = −2
4 2 1 =
3
1
9
1
9 3
9 3 1
5 −6
−1
Matrix of cofactors = 2 −8
6
−1
3 −2
1 1 1
4 2 1
9 3 1
−1
1 −1
= −__
2 5
−6
a
1 −1
__
b = −2 5
c
−6
2 −1
−8
3
6 −2
2 −1 5
2
−8
3 12 = 1
2
6 −2 23
Hence a = 2, b = 1, c = 2
The equation is q s(x) = 2x 2 + x + 2
E X A M P L E 61
A 16% solution, a 22% solution and a 36% solution of an acid are to be mixed to
get 300 ml of a 24% solution. If the volume of acid from the 16% solution equals
half the volume of acid from the other two solutions, write down three equations
satisfying the conditions given and solve the equations to find how much of each
is needed.
SOLUTION
Let x be the volume of 16% solution, y be the volume of 22% solution and z be the
volume of 36% solution needed.
367
M O DUL E 3
Now x + y = z = 300 since the total volume is 300 ml.
24 × 300 = 72
0.16x + 0.22y + 0.36z = ___
100
1 (0.22y + 0.36z) = 0
and 0.16x − __
2
Therefore the equations are
x + y + z = 300
0.16x + 0.22y + 0.36z = 72
0.16x − 0.11y − 0.18z = 0
Writing the equations in matrix form, we have
(
1
0.16
0.16
1
0.22
−0.11
)( ) ( )
300
1 x
0.36 y = 72
0
−0.18 z
Forming the augmented matrix, we get
(
1
0.16
0.16
1
0.22
−0.11
:
1 300
0.36 72
0
−0.18
R2 → R2 − 0.16R1
R3 → R3 − 0.16R1
(
1
0
0
1
0.06
−0.27
R3 → 0.06R3 + 0.27R2
(
(
:
1 300
0.20 24
−0.34 −48
:
)
)
)
300
1
1
0.06 0.20 24
0 0.0336 36
x
300
1
1
1
y = 24
0.20
0.06
0
3.6
0
0.0336 z
0
1
0
0
)( ) ( )
0.0336z = 3.6 ⇒ z = 107.14
0.06y + 0.20z = 24
0.06y + 0.20(107.14) = 24
y = 42.86
x + y + z = 300
x + 42.86 + 107.14 = 300
x = 150
Hence 150 ml of the 16% solution, 42.86 ml of the 22% solution and 107.14 ml of the
36% solution are needed.
E X A M P L E 62
368
A popular carnival band sells three types of costumes. The costumes are made at the
Mas-camp in Port-of-Spain. The owner of the band makes cheap costumes, mediumpriced costumes and expensive costumes. The making of the costumes involves
MODULE 3tCHAPTER 16
fabric, labour, buttons and machine time. The following table shows the units of
input required per costume for each type of costume.
Cheap
Medium-priced
Expensive
Fabric
5
6
8
Labour
20
25
30
Buttons
15
20
22
7
9
12
Machine time
The owner makes the three types of costumes and uses 270 units of fabric, 1050 units
of labour and 790 buttons. How many of each type of costume does the owner make?
What is the corresponding machine time used?
SOLUTION
Let x be the number of cheap costumes made, y the number of medium-priced
costumes made, z the number of expensive costumes made.
Since 270 units of fabrics are used we have
5x + 6y + 8z = 270
For labour, we have
20x + 25y + 30z = 1050
For buttons,
15x + 20y + 22z = 790
Writing the equations in matrix form, we have
(
6
25
20
5
20
15
)( ) ( )
8 x
270
30 y = 1050
22 z
790
Forming the augmented matrix and reducing gives:
(
:
6 8 270
25 30 1050
20 22 790
5
20
15
)
R2 → R2 − 4R1
R3 → R3 − 3R1
(
5
0
0
6
1
2
:
)
:
)
8 270
−2 −30
−2 −20
R3 → R3 − 2R2
(
5
0
0
6
1
0
8 270
−2 −30
2 40
369
M O DUL E 3
We now have
(
6
1
0
5
0
0
)( ) ( )
270
8 x
−2 y = −30
2 z
40
The equations are
2z = 40 ⇒ z = 20
y − 2z = −30 ⇒ y − 40 = −30 ⇒ y = 10
5x + 6y + 8z = 270
5x + 60 + 160 = 270
x = 10
Hence the owner made 10 cheap costumes, 10 medium-priced costumes and
20 expensive costumes.
Machine time used = 7 × 10 + 9 × 10 + 12 × 20 = 400 units.
EXERCISE 16B
In questions 1–5, find the inverse of each matrix.
1 1 −12
2 1
0
3
3 −12 −34
4 2
4
6 4
5
3 3
(
)
(
(
)
(
(
)
2
5
1
1
)
)
In questions 6–10, (a) find the determinant of each matrix, (b) find the matrix of
cofactors and hence find the inverse of each matrix.
0
1 2 1
1 1
6
7 0 1 1
2 1 −1
3 1
2
0 0 2
5 −1
4
4 3 −1
8
9 2 4 4
3 −2
2
1
1 −1
3 2
3
1 0 5
10 4 1 0
2 3 4
(
(
(
)
)
(
(
)
)
)
In questions 11–15, find the inverse of each matrix by row reduction.
11
13
15
(
(
(
3
−1
1
1 0
4 1
2 3
−1
3
−2
1 2
1 0
3 1
5
0
4
2 1
1 1
1 7
)
)
)
(
14 (
12
: :
x
1
2
x3
1
8
16 Solve the equation x2 1 4 = 0.
370
1 4 1
2 1 1
−2 3 1
2 1 5
4 1 6
2 3 2
)
)
MODULE 3tCHAPTER 16
:
:
:
:
x2 1
1 = 0.
x3 1
x
17 Solve the equation x2 x
x3
x−1
18 Solve the equation −1
x+1
1
19 Solve the equation x + 1
x
1
1
1
:
x+1
1 = 0.
x−1
x
1
x+1
:
x+1
x = 0.
1
20 Find the values of a satisfying the equation
:
1
1
a
a+1
a−1
2a
:
1
a − 1 = 0.
a+1
21 Sanjeev pays TT$300 for 4 shirts and 2 pairs of trousers while Saleem pays
TT$700 for 2 shirts and 5 pairs of trousers. If x and y represent the price of a
shirt and a pair of trousers respectively, write a system of linear equation in
matrix form based on this information. Determine the price of a shirt and a
pair of trousers.
22 Michael feeds his dog Zezu with different mixtures of three types of food, A, B
and C. A scoop of each food contains the following nutrients.
Food A: 15 g of protein, 10 g carbohydrates and 20 g vitamins
Food B: 20 g of protein, 15 g carbohydrates and 10 g vitamins
Food C: 20 g of protein, 10 g carbohydrates and 20 g vitamins
Assume that dogs require 160 g of protein, 110 g of carbohydrates and 150 g of
vitamins. Find how many scoops of each food Michael should feed his dog daily
to satisfy their nutrient requirements.
23 Deanne has TT$50 000 and wishes to invest this for her retirement. She puts all
the money in a fixed deposit, trust fund and a money market fund. The amount
she puts in the money market fund is TT$10 000 more than that in the trust
fund. After one year, she receives a profit totalling TT$3000. The fixed deposit
pays 5% interest annually, the trust fund pays 6% annually and the money
market fund pays 7% annually.
By denoting the amount of money invested in the fixed deposit, trust fund
and the money market fund as x, y and z respectively, form a system of linear
equations based on the information given.
Write the system of linear equations in matrix form.
Find the amount of money invested in each category of the fund.
24 Show that the equations
x + 5y + 4z = 19
2x − 4y + z = −4
4x + 6y + 7z = 30
have a unique solution and hence find the solution by row reducing the
augmented matrix to echelon form.
371
M O DUL E 3
25 Write the augmented matrix for the following equations.
x − 2y + 3z = −3
6x + y + z = 12
3x − 2y + 4z = 0
Reduce the augmented matrix obtained to echelon form.
Hence, solve the system of equations.
26 Write the augmented matrix for the following equations.
3x + 4y + 2z = 1
x−y−z=2
5x + 2y = 3
Reduce the augmented matrix obtained to echelon form.
Hence, show that the system of equations has no solutions.
27 Write the augmented matrix for the following equations.
8y + 3z = 2
5x + 4y − z = 8
3x − y + 7z = 6
Reduce the augmented matrix obtained to echelon form.
Hence, solve the system of equations.
28 Write the augmented matrix for the following equations.
7x − 5y + 10z = 3
8x + 2y + z = 11
5x − 3y + 6z = 10
Reduce the augmented matrix obtained to echelon form.
Hence, solve the system of equations.
29 Write the augmented matrix for the following equations.
10x − 8y + 5z = 15
11x + 3y − 6z = 5
7x − 6y + 4z = 11
Reduce the augmented matrix obtained to echelon form.
Hence, solve the system of equations.
30 Write the following equations in the form AX = B.
3x − y + 2z = 7
x+y+z=2
4x + 5y + 3z = 1
Find (a) the determinant of A
(b) the matrix of cofactors of A.
Hence, write down the inverse of A and solve the system of equations.
372
MODULE 3tCHAPTER 16
(
2
31 Find the inverse of the matrix 5
equations
3
)
3
−1
4 −3 . Hence, or otherwise solve the
−2 −1
2x − y + 3z = −12
5x + 4y − 3z = 4
3x − 2y − z = 9
32 Write the following equations in the form AX = B.
2x + y + z = 4
x + 2y + z = 2
x + y + 2z = 6
Find the inverse of A and hence solve the equations.
33 Solve the equations
x − y + 5z = 4
4x + 3y + 3z = 13
5x − 4y − 2z = −5
34 Write the following set of equations in the form AX = B.
x − 3y + 5z = 2
x + 4y − z = 1
7y − 6z = a
Row reduce the augmented matrix to echelon form.
Hence, find the value of a for which the equations are consistent. Solve the
equations for this value of a.
35 Given the set of equations
x + 5y + az = 5
2x − 4y + z = 3
4x + 6y + 7z = 11
(a) write the equations in the form AX = B,
(b) find the set of values of a for which A is non-singular,
(c) solve the equations when a = 4.
36 Showing all working clearly, decide whether the following system of
equations is consistent.
x + 5y + 3z = −18
2x − 4y + z = 7
4x + 6y + 7z = −29
(
)
0 −1
α is non-singular.
3
8 −6 −2
(a) Find the set of possible values for α.
1
37 It is given that the matrix A = 0
(b) Find the inverse of A for these values of α.
373
M O DUL E 3
(c) Solve the equations
x − z = −2
3y + αz = 1
8x − 6y − 2z = 4
(
)
(
2 −1
1
3 and B = 2
3
2 1
1
−1
−1
(a) find A and B ,
1
38 Given that A = 0 1
−1
1
1
)
4
1,
2
(b) find (AB)−1 and (BA)−1.
39 A store has three outlets in Chaguanas, Grand Bazaar and Gulf City. Each store
sells toy cars, toy trains and toy buses at the same price. The following table
gives the numbers of each toy sold in one week in each of the outlets.
Chaguanas
Grand Bazaar
Gulf City
Trains
10
8
6
Buses
15
9
7
Cars
12
2
15
The total sales for the week for these three toys were TT$354 at Chaguanas,
TT$200 at Grand Bazaar and TT$247 at Gulf City. At what price did the stores
sell a train, a bus and a car?
374
MODULE 3tCHAPTER 16
SUMMARY
Matrices
A matrix is of order m × n where
m = no. of rows
and n = no. of columns.
A matrix A is singular iff |A| = 0
A matrix A is non-singular iff |A| ≠ 0
A matrix of order n × n is a
square matrix.
a b
1 d –b
If A = c d then A–1 = |A| –c a
Matrices are conformable for
multiplication if the number
of columns of the first is equal
to the number of rows of the second.
For a 3 × 3 matrix,
A–1 = 1 adj (A)
|A|
where adj (A) = (matrix of cofactors)T
Matrix multiplication is not
commutative.
Row reduction
( )
( )
Ri ↔ Rj
The transpose (AT) of a matrix
A is found by interchanging the
rows and columns of A.
Ri → c Ri
Ri → aRi + bRj
(AT)T = A
(kA)T = kAT
(AB)T = BTAT
Systems of linear equations
a11x + a12 y + a13z = b1
a21x + a22 y + a23z = b2
a31x + a32 y + a33z = b3
A square matrix is called
symmetric iff AT = A
AX = B
a11 a12 a13
where A = a21 a22 a23
a31 a32 a33
A square matrix is called
skew-symmetric iff AT = –A
a b
= ad – bc
The determinant
c d
| |
|
a11 a12 a13
a a
a21 a22 a23 = a11 22 23
a32 a33
a31 a32 a33
– a12
|
|
|
a21 a23
a a
+ a13 21 22
a31 a33
a31 a32
|
|
( )
() ()
b1
x
X = y , B = b2
b3
z
X = A–1B
|
|
If every element of a row or column
of a matrix A is 0 then |A| = 0
|AT| = |A|
If two rows or columns of a matrix
are interchanged, the sign of the value of the
determinant is changed.
System of linear equations in
two variables:
(i) Consistent and independent:
one solution, lines intersect at a point
(ii) Inconsistent: no solutions,
two lines are parallel
(iii) Consistent and dependent:
infinite solutions, lines coincide
System of linear equations in
three variables:
(i) Consistent and independent:
one solutions, all three planes
intersect at a point
(ii) Consistent and dependent:
(a) Infinite set of solutions, all
three planes intersect in a line –
one of the equations is a linear.
combination of the other two.
(b) All three planes are identical.
(iii) Inconsistent:
(a) All three planes are parallel
(b) Any two of the planes are
parallel and distinct
(c) One plane is parallel to the
line of intersection of the other
two.
A system of equations with three equations
and three unknowns can have
(i) a unique solution (|A| ≠ 0)
(ii) an infinite set of solutions (|A| = 0)
(iii) no solutions (|A| = 0)
To make a decision we can row reduce
the augmented matrix (A|B) to echelon form.
(A|B) has no row of zeros
⇒ unique solution.
In (A|B) the matrix A alone has a row of zeros
⇒ no solution.
The augmented matrix has a row of zeros
⇒ an infinite set of solutions.
|kA| = kn|A|, A is an n × n matrix.
|AB| = |A||B|
375
M O DUL E 3
Checklist
Can you do these?
■ Identify the order of a matrix.
■ Identify equal matrices.
■ Multiply a matrix by a scalar.
■ Identify matrices conformable to addition, subtraction, multiplication.
■ Add, subtract and multiply matrices.
■ Use the properties of matrix addition.
■ Use the properties of matrix multiplication.
■ Identify the identity matrix for an n × n matrix.
■ Find the transpose of a matrix.
■ Know and use the properties of the transpose.
■ Identify symmetric and skew-symmetric matrices.
■ Find the determinant of a matrix (2 × 2 and 3 × 3).
■ Recall and use the properties of determinants.
■ Determine singular matrices.
■ Determine non-singular matrices.
■ Solve simultaneous equations using determinants.
■ Find the inverse of a matrix (2 × 2 and 3 × 3).
■ Identify the matrix of cofactors.
■ Identify a system of linear equations.
■ Solve a system of linear equations using the inverse of a matrix.
■ Row reduce a matrix.
■ Find the inverse of a matrix using row reduction.
■ Write down an augmented matrix.
■ Solve simultaneous equations using row reduction.
■ Decide whether a system of equations has (i) one solution, (ii) an infinite set of
solutions or (iii) no solutions.
■ Solve application problems.
376
MODULE 3tCHAPTER 16
Review e x e r c i s e 1 6
1
(
)
8
1 1
(a) Find the inverse of the matrix A = 7 −2 9 .
4 −6 8
(b) Write the following set of equations in the form AX = B.
8p + q + r = 1
7p − 2q + 9r = −3
4p − 6q + 8r = −5
(c) Hence find the values of p, q, r.
2
3
Find the values of x satisfying the equation
x 1 2
x2 1 4 = 0.
x3 1 8
: :
Write the set of equations
9x + 3y − 4z = 13
2x − 5y + 2z = 14
7x + 3y − 2z = 3
in the form AX = B, where A is the coefficient matrix and X and B are column
vectors.
Write down the augmented matrix and row reduce this matrix to echelon form.
Hence solve the system of equations.
4
Two types of cars are rented by a rental company, which has a total of 54 cars.
One compact car rents for TT$900 per month and one mid-size car rents for
TT$1250 per month. If all cars are rented and the total rental income is TT$55 600
per month, use a matrix method to find how many cars of each type are rented.
5
Anslem invests TT$20 000, part in a bank at 5.00% interest, part in mutual
funds at 6.00% and part in a Unit trust at 6.50% interest. The total annual
interest is TT$1053. Three times the amount invested at 6.00% equals the
amount invested at 5.00% and 6.50% combined. Use a matrix method to find
how much is invested at each rate.
6
A company manufactures three types of goods, A, B and C, each of which is
made from three types of inputs X, Y and Z. Each unit of A requires 1 unit of X,
7 units of Y and 3 units of Z. Each unit of B requires 4 units of X, 3 units of Y
and 1 unit of Z. One unit of C requires 2 units of X, 4 units of Y and 2 units of Z.
On one day of production the company uses up 105 units of X, 135 units of Y
and 55 units of Z.
(a) Write three equations to represent the usage of X, Y and Z in the
production of a, b and c units of A, B and C respectively.
(b) Write the equations in matrix form.
(c) Use a matrix method to find a, b and c.
377
M O DUL E 3
7
The supply function for a commodity is given by q s(x) = ax 2 + bx + c, where
a, b and c are constants. When x = 1, the quantity supplied is 4; when x = 2, the
quantity supplied is 12; when x = 3, the quantity supplied is 26.
Use a matrix method to find the values of a, b and c.
:
:
:
x−1
1
−1
x−2
1
1 = 4 − 3x.
−1
x−3
1
0
2x − 1
1
9 Solve the equation 1
x−3
2 = 3 − 4x.
0
x+2
1
1 −6
1 2 4
10 Given the matrices A = −1 3 2 and B = −3 3
5
1 2 1
0
find AB.
8
Solve the equation
(
)
:
(
)
8
6 ,
−5
Hence deduce the inverse of A.
11 A system of equations is given by
2x + y + z = 6
x + 4y + 2z = 4
x−y−z=0
(a) Express the system of equations in the form AX = B, where A is a matrix
and X and B are column vectors.
(b) Find A−1.
(c) Hence, solve the system of equations.
12 A system of equations is given by
2x + y + z = 5
3y − 2z = 4
2x − 2y + 3z = 2
Write down the augmented matrix for the system of equations.
Use the method of row reduction to echelon form on the augmented matrix to
show that the system is inconsistent.
13 A system of equations is given by
4x + py = 6
2x + y = q
where p, q ∈ ℝ.
State the values of p and q for which the equations have an infinite set of solutions.
14 The matrices A and B are given by
A=
( 10
)
−1 2 ,
1 3
(
2
B= 4
3
1
−2
3
(a) Calculate (i) AB (ii) BTAT.
(b) Deduce that (AB)T = BTAT.
378
)
MODULE 3tCHAPTER 16
(
8
3
10
4
15 The matrix P is given by P = −5 −2
)
−12
8 .
−15
By performing elementary row operations on the matrix (P|I ), find the inverse
of P.
2
x
Hence solve the equations PX = 3 , where X = y .
z
1
()
()
16 Find the set of values of k for which this system of equations has a unique
solution.
x + 2y + 3z = 1
5x + 4y + kz = 4
−5x + ky + 11z = −3
Solve the system of equations when k = 4.
17 Find the value(s) of k for which the system of equations
x + 2y + z = 3
kx − y − 11z = 2
−2x + ky + 12z = 1
has (a) no solution, (b) a unique solution, (c) more than one solution.
Find the solution set for part (c).
379
M O DUL E 3
CHAPTER 17
Differential Equations and
Mathematical Modelling
At the end of this chapter you should be able to:
■ identify a first order differential equation
dy
+ Py = Q,
■ write a linear first order differential equation in the form ___
dx
where P and Q are functions of x only or constants
■ solve first order linear differential equations using the integrating factor
d2y
dy
+ b ___ + cy = 0
■ solve second order differential equations of the form a ___
dx
d x2
■ solve second order differential equations of the form
d2y
dy
a ___2 + b ___ + cy = f (x) where f (x) is a polynomial in x of maximum degree 2
dx
dx
■ solve second order differential equations of the form
d2y
dy
a ___2 + b ___ + cy = f (x) where f (x) is a trigonometric function
dx
dx
d2y
dy
+ b ___ + cy = λemx
■ solve differential equations of the form a ___
2
dx
dx
■ solve differential equations that can be reduced using a substitution
■ solve application problems
■ solve modelling problems from different areas of the syllabus.
KEYWORDS/TERMS
PSEJOBSZEJČFSFOUJBMFRVBUJPOtPSEFS
tEFHSFFtĕSTUPSEFSMJOFBSEJČFSFOUJBM
FRVBUJPOtTFDPOEPSEFSMJOFBSEJČFSFOUJBM
FRVBUJPOtJOUFHSBUJOHGBDUPStHFOFSBM
TPMVUJPOtIPNPHFOFPVTtSPPUTtBVYJMJBSZ
RVBESBUJDFRVBUJPOtDPNQMFNFOUBSZ
GVODUJPOtQBSUJDVMBSJOUFHSBMtQPMZOPNJBM
tUSJHPOPNFUSJDGVODUJPOtJOJUJBM
DPOEJUJPOT
380
MODULE 3tCHAPTER 17
dy d2y d3y
Any relation between the variables x, y and any or all of the derivatives ___, ____2 , ___3 ,
dx dx dx
. . . is called an ordinary differential equation.
Note
An equation of
degree 1 is linear.
The order of a differential equation is the highest derivative in the differential equation and the degree of the differential equation is the power to which the derivative
of the highest order is raised.
dy
For example ___ + y = x is a first order linear differential equation: first order since
dx
the highest derivative is the first derivative and linear (degree 1) since it is raised to the
power of 1.
dy
d2y ___
___
+
+ y = x is a second order linear differential equation: second order since
2
dx
dx
the highest derivative is the second derivative and linear since the highest power
of the highest derivative is 1. This chapter deals with solving first order linear differential equations and second order differential equations.
First order linear differential equations
d
___
[ ye∫Pdx ]
dx
u = y, v = e∫Pdx
dy
du = ___
___
dx dx
dv = Pe∫Pdx
___
dx
d
___
[ ye∫Pdx ] =
dx
dy
Pye∫Pdx + e∫Pdx ___
dx
dy
= e∫Pdx ___ + Py
dx
[
]
dy
The equation ___ + Py = Q where P and Q are functions of x alone or constants,
dx
dy
involving y and ___ to the first degree only, is known as a first order linear differential
dx
equation.
dy
We can find a solution to ___ + Py = Q in the following way:
dx
Multiplying both sides of the equation by e ∫Pdx gives
dy
e ∫Pdx ___ + Py = Qe ∫Pdx
dx
[
]
[
dy
d ye ∫Pdx = Qe∫Pdx, using the product rule ___
d ye∫Pdx = e ∫Pdx ___
⇒ ___
+ Py
[
]
[
]
dx
dx
dx
∫
⇒ ye ∫Pdx = Qe ∫Pdx d x
]
[A]
e ∫Pdx is called the integrating factor of the differential equation, and the solution for
the differential equation can be found from [A]. We can make y the subject of the
formula, which is the general solution to our differential equation.
dy
Let us use this result to solve differential equations of the form ___ + Py = Q. In this
dx
dy
___
equation, the coefficient of
must be 1, and P and Q are either functions of x only
dx
or constants.
EXAMPLE 1
SOLU TION
dy
Solve the differential equation ___ + y = x.
dx
By solving the differential equation we are to find a function y that satisfies the
differential equation. This can be done in the following way.
dy
dy
___
+ y = x is of the form ___ + Py = Q where P = 1 and Q = x.
dx
dx
The integrating factor of this equation is
e ∫1dx = ex
381
M O DUL E 3
Ask yourself
Is the equation of
the form
dy
___
+ Py = Q?
dx
If so, can you
identify P and Q?
∫
Using ye∫Pdx = Qe∫Pdx dx and substituting, we have
ye x
∫
=
xex d x
∫
To find xex d x we integrate by parts
dv = ex
u = x, ___
dx
du = 1, v = ex
___
dx
xex d x = xex − ex d x = xex − ex + c
∫
∴
∫
yex = xex − ex + c
Hence
xex − ex + c
y = __________
ex
y = x − 1 + ce−x
The general solution of the differential equation is y = x − 1 + ce−x.
EXAMPLE 2
SOLU TION
Is the equation in
the form
dy
___
+ Py = Q?
dx
dy
Solve the differential equation x ___ + y = x3 given that y = 2 when x = 1.
dx
dy
This equation can be written in the form ___ + Py = Q by dividing throughout by x.
dx
dy
1 y = x2
___
+ __
x
dx
1 , Q = x2
P = __
x
Find the integrating factor (IF):
( )
IF = e∫ x dx = eln x = x
1
__
∫__1x dx = ln x
∫
Using ye∫Pdx = Qe∫Pdx dx
e∫Pdx
= x, Q = x2, we have
1 x4 + c
xy = x3 d x = __
4
1 x3 + __c
y = __
x
4
where
∫
1 x3 + __c .
∴ The general solution of the equation is y = __
x
4
We can now use the condition y = 2 when x = 1 to find c as follows.
Substituting x = 1, y = 2 into
1 x3 + __c
y = __
x
4
1
2 = __ (1)3 + __c
4
1
1
__
2− =c
4
7
c = __
4
7
1 x3 + ___
∴ y = __
4
4x
7.
1 x3 + ___
The solution of the equation when x = 1, y = 2 is y = __
4
4x
382
MODULE 3tCHAPTER 17
EXAMPLE 3
dy
Solve the differential equation ___ + 3y = e2x.
dx
SOLU TION
dy
dy
The equation ___ + 3y = e2x is of the form ___ + Py = Q where P = 3, Q = e2x.
dx
dx
Find the integrating factor:
IF = e∫3 dx = e3x
∫
Using ye∫Pdx = Qe∫Pdx dx
where e∫Pdx = e3x, Q = e2x, we have
∫
ye3x = ∫e5x d x
y(e3x) = (e2x)(e3x) dx
1 e5x + c
ye3x = __
5
1 e5x + c
__
5
________
y=
e3x
1
__
y = e2x + ce−3x
5
1 e2x + ce−3x.
The general solution of the differential equation is y = __
5
EXAMPLE 4
SOLUTION
dy
Solve (x + 2) ___ + y = (x + 2)3.
dx
dy
We need to write the equation in the form ___ + Py = Q.
dx
dy
1 y = (x + 2)2
Divide by (x + 2) to get ___ + _____
dx x + 2
1 ,
P = _____
x+2
Q = (x + 2)2
1
_____
IF = e ∫ x + 2 dx = e ln (x+2) = (x + 2)
∫
Using ye∫Pdx = Qe∫Pdx dx
where
e∫Pdx
= x + 2, Q = (x + 2)2, we have
∫
y(x + 2) = (x + 2)2(x + 2) dx
∫
y (x + 2) = (x + 2)3 d x
(x + 2)4
y (x + 2) = _______ + c
4
(x
+
2)4
_______
+c
4
y = ___________
(x + 2)
c
1 (x + 2)3 + _____
= __
4
x+2
c .
1 (x + 2)3 + _____
The general solution of the differential equation is y = __
4
x+2
383
M O DUL E 3
EXAMPLE 5
dy
Solve the differential equation ___ + y tan x = sin x when x = 0, y = 1.
dx
SOLU TION
dy
___
+ y tan x = sin x
dx
dy
The equation is of the form ___ + Py = Q where P = tan x and Q = sin x.
dx
IF = e ∫ tan x dx = e ln sec x = sec x
∫
Using ye∫Pdx = Qe∫Pdx dx
where
e∫Pdx
= sec x, Q = sin x, we have
∫
y sec x = ∫ tan x d x
y(sec x) = (sec x)(sin x) dx
sec x sin x
1
= _____
cos x sin x
= tan x
y sec x = ln sec x + c
The general solution of the equation is y = (cos x)(ln sec x + c)
Using x = 0, y = 1 to find c, we get
1 = (cos 0)(ln sec 0 + c)
1=c
∴ y = cos x (ln sec x + 1)
Hence the solution of the equation when x = 0, y = 1 is y = cos x (ln sec x + 1).
EXAMPLE 6
SOLU TION
dy
Find the general solution of the differential equation x ___ + 4y = (x + 1).
dx
Hence, find the solution when x = 2, y = 1.
dy
x ___ + 4y = (x + 1)
dx
dy 4
x+1
y = _____
÷ x ⇒ ___ + __
x
dx x
4
__
IF = e ∫ x dx = e 4 ln x = e ln x = x4
4
∫
Using ye∫Pdx = Qe∫Pdx dx
x + 1 , we have
where e∫Pdx = x4, Q = _____
x
x
+
1
4
4
_____
y(x ) = x
dx
x
∫ ( )
yx4 = ∫(x + 1) x3 d x
= ∫(x4 + x3) d x
5
4
x + c, which gives the general solution
x + __
= __
5
4
25 + __
24 + c
When x = 2, y = 1, we have 1(24) = __
5
4
32
16 − ___ − 4 = c
5
28 = c
___
5
384
MODULE 3tCHAPTER 17
5
4
x + __
x + ___
28
∴ yx4 = __
5
5
4
5
4
x
x
28
1
y = __4 __ + __ + ___
5
4
x 5
28
x
1
= __ + __ + ___ x−4
5 4
5
[
]
Practical applications
EXAMPLE 7
Ohm’s law gives the drop in voltage due to the resistor as RI, where R is the resistance
dI , where L is the
and I is the current. The voltage drop due to the inductor is L __
dt
inductance and t is time. Kirchhoff ’s law states that the sum of the voltage drops is
dI + RI = E(t).
equal to the voltage supplied E(t). Therefore, we have L __
dt
Suppose that in the circuit shown the resistance is 6 ohms and the inductance is
2 henry. If the battery gives a voltage of 30 V and the switch is closed so that when
t = 0, I = 0, find (a) I in terms of t and (b) the limiting value of the current.
6Ω
30 V
SOLU TION
2H
dI + RI = E(t):
(a) Substitute into the equation L __
dt
dI + 6I = 30
2 __
dt
dI + 3I = 15
__
dt
Solve using the integrating factor method:
IF = e ∫3dt = e3t
The general solution is given by
∫
I(e3t) = 15e3t d t
Ie3t = 5e3t + c
c
÷ e3t ⇒ I = 5 + ___
e3t
I = 5 + ce−3t
When t = 0, I = 0
0 = 5 + ce0
0=5+c
c = −5
∴ I = 5 − 5e−3t
(b) As t → ∞, e−3t → 0 ∴ I → 5
Hence the limiting value of the current is 5 amperes.
385
M O DUL E 3
EXAMPLE 8
Given that the initial population of a colony of insects is 100 and the population
dP = 0.08P 1 − _____
P , solve this equation and find the time t in
growth equation is ___
1000
dt
seconds taken for the population to reach 1050.
(
P
dP = 0.08P 1 − ____
___
SOLU TION
Note
(
dt
k
is called the
model for
population
growth or the
logistic model.
)
1000
)
Separate variables:
1
___________
dP = 0.08 d t
P
P 1 − ____
1000
Integrate both sides:
1
___________
dP = ∫0.08 d t
P
P 1 − ____
1000
1
into partial fractions:
Separate ___________
P
P 1 − ____
1000
(
The differential
equation
dP = KP 1 − __
P
___
(
dt
)
)
∫ (
)
(
)
B
1
A + ________
___________
= __
P
P
1 − ____
1000
P + B (P)
1 = A 1 − ____
1000
When P = 1000
P
P 1 − ____
1000
(
)
(
)
1 = B (1000)
1
B = _____
1000
When P = 0
A=1
1
_____
1000 = __
1
1 + ________
1
1 + _________
= __
∴ ___________
P 1 − ____
P
P
P 1 − ____
1000
1000
So the integral becomes
1
1 + ________
__
dP = 0.08 d t
P 1000 − P
(
)
∫
P
1000 − P
∫
∴ ln |P| − ln |1000 − P| = 0.08t + c is the general solution.
When t = 0, P = 100
∴ ln 100 − ln (1000 − 100) = c
100 = ln __
1
c = ln 100 − ln 900 = ln ____
9
900
( )
()
1
∴ ln |P| − ln |1000 − P| = 0.08t + ln ( __
9)
is the solution to the equation.
When P = 1050
1
ln |1050| − ln |1000 − 1050| = 0.08t + ln __
9
1
0.08t = ln (1050) − ln (50) − ln __
9
1050 × 9 = 12.5 ln 189 = 65.5 sec (3 s.f.)
1 ln ________
t = ____
0.08
50
()
(
386
)
()
MODULE 3tCHAPTER 17
EXERCISE 17A
1
2
3
4
dy
1 ___
Find the general solution of the equation __
x d x + 2y = 1.
dy
3y .
Find y in terms of x given that (x + 3) ___ = 2 − __
x
dx
dy
Solve the differential equation (tan x) ___ + y = x2 tan x.
dx
dy
Find the general solution of x ___ − 4y = x 5ex.
dx
5
Solve the differential equation 2(y − 4x2) d x + x dy = 0.
6
Solve the differential equations
dy
dy
(a) ___ + 2y = x2
(b) ___ + y = 3x + 2
dx
dx
7
dy
Obtain the solution of the differential equation x ___ − y = x2 such that y = 1
dx
when x = 1.
8
Show that the solution to the differential equation
dy
1 (x sec x + sec x + sin x)
(cos x) ___ − y sin x = cos2 x is y = __
2
dx
1
__
when x = 0, y = .
2
9
Find the solution of the differential equation
dy
1
______
given that x = 1 and y = 2.
(1 + x2) ___ + xy = _______
dx
1
+
x2
√
10 Find the general solution of the differential equation
dy
___
+ y = 3x + 5.
dx
11 Find the solution of the differential equation
dy
π, y = 1.
(sin x) ___ + (cos x) y = sin x e cos x when x = __
2
dx
12 Show that the general solution of the differential equation
dy
ln c (x + 2)
1 can be written as y = _________
where c is a constant.
(1 + x) ___ + y = _____
x+1
x+2
dx
13 Find the solution of the differential equation
5
__
dy
(1 + x3) ___ + x2y = (1 + x3) 3 when x = 0 and y = 1.
dx
dI + 3 I = 18 sin 30 t, find I in terms of t.
14 Given the differential equation __
dt
15 The equation of motion of a train of mass m is given by
dV = m(k − ke−t − cV), where V is the speed of the train and c and k are
m ___
dt
constants. Find the speed of the train.
387
M O DUL E 3
Second order differential equations
A differential
equation of this
form is called a
homogeneous
differential
equation.
An equation is second order when the highest derivative in the equation is the
second derivative. Let us consider second order differential equations of the form
dy
d2y
a ___2 + b ___ + cy = 0
dx
dx
where a, b and c are constants.
To solve this equation we let y = emx where m is a constant.
Since y = emx
dy
___
= memx
dx
d2y
___
= m2emx
d x2
d2y
dy
Substituting into a ___2 + b ___ + cy = 0 we get
d
x
dx
am2 emx + bmemx + cemx = 0
emx (am2 + bm + c) = 0
am2 + bm + c = 0,
emx ≠ 0
The quadratic equation am2 + bm + c = 0 is called the auxiliary quadratic
equation (AQE).
d2y
The coefficient of ___2 is the coefficient of m2 in the AQE.
dx
dy
The coefficient of ___ is the coefficient of m in the AQE.
dx
The coefficient of y is the constant in the AQE.
The solution of the second order differential equation depends on the roots of the
auxiliary quadratic equation. Recall that there are three types of roots for a quadratic
equation:
(i) real and equal roots
(ii) real and distinct roots
(iii) complex roots.
When the roots of the AQE are real and equal
am2 + bm + c = 0 ⇒ m = m1; there is one root to the AQE.
The general solution of the differential equation is
y = (Ax + B) em1x.
EXAMPLE 9
d2y
dy
Solve the differential equation ___2 + 2 ___ + y = 0.
dx
dx
SOLU TION
Forming the auxiliary quadratic equation, we have
m2 + 2m + 1 = 0
388
MODULE 3tCHAPTER 17
Solving, (m + 1)2 = 0
m+1=0
m = −1
Since the roots are real and equal, the general solution is
y = (Ax + B) emx where m = −1
i.e. y = (Ax + B) e−x
EXAMPLE 10
d2y
dy
Solve the differential equation ____2 + 10 ___ + 25y = 0.
dx
dx
SOLU TION
dy
d2y
___
+ 10 ___ + 25y = 0
dx
d x2
Form the auxiliary quadratic equation:
m2 + 10m + 25 = 0
Solve (m + 5)(m + 5) = 0
m = −5
Since the roots are real and equal, the solution is of the form
y = (Ax + B) emx where m = −5
∴ y = (Ax + B) e−5x
Try these 17.1
Ask yourself
Are the roots real
and equal?
What is the
solution for these
roots?
Find the general solution of the following differential equations.
d2y
dy
(a) 9 ___2 − 6 ___ + y = 0
dx
dx
d2y
dx
dy
dx
(b) 16 ___2 − 24 ___ + 9y = 0
d2y
dx
dy
dx
(c) 9 ___2 + 12 ___ + 4y = 0
When the roots of the AQE are real and distinct
If the auxiliary quadratic equation has two distinct roots, i.e. m = m1, m2, then the
general solution of the differential equation is
y = Aem1x + Bem2x
EXAMPLE 11
d2y
dy
Find the general solution of the differential equation ___2 − 3 ___ + 2y = 0.
dx
dx
SOLU TION
dy
d2y
___
− 3 ___ + 2y = 0
2
dx
dx
Form the AQE:
m2 − 3m + 2 = 0
389
M O DUL E 3
Solve (m − 2)(m − 1) = 0
m = 2 or 1
Since the roots are real and distinct, the general solution is of the form
y = Aem1x + Bem2x.
Since m1 = 2, m2 = 1,
y = Ae2x + Bex
EXAMPLE 12
d2y
dy
Find the general solution of the differential equation 3 ___2 + ___ − 2y = 0.
dx
dx
SOLU TION
d2y
dy
3 ___2 + ___ − 2y = 0
dx
dx
Form the AQE:
3m2 + m − 2 = 0
Solve (3m − 2)(m + 1) = 0
3m − 2 = 0, m + 1 = 0
2 or
m = __
m = −1
3
Since the roots are real and distinct, the general solution is of the form
y = Aem1x + Bem2 x
2
__
∴ y = Ae 3 x + Be−x
Try these 17.2
Find the general solution of the following differential equations.
d2y
dy
(a) ___2 + 6 ___ + 8y = 0
dx
dx
2
dy
dy
(b) ___2 + 4 ___ − 12y = 0
dx
dx
d2y
dx
dy
dx
(c) 12 ___2 + 11 ___ − 5y = 0
When the roots of the AQE are complex
___
Complex roots are of the form m = m1 ± im2 where i = √−1 .
If the roots of the AQE are complex, i.e. m = m1 ± im2, then the solution is of the form
y = em1x [A cos m2 x + B sin m2 x]
EXAMPLE 13
d2y dy
Find the general solution of ___2 + ___ + y = 0.
dx
dx
SOLU TION
dy
d2y ___
___
+
+y=0
2
dx
dx
The AQE is m2 + m + 1 = 0
390
MODULE 3tCHAPTER 17
_____
± √1 − 4
__________
Solving: m = −1
2
___
± √−3
________
= −1
2
__ ___
√ 3 √ −1
1
__
________
=− ±
2
2
__
√3
1 ± ___
i
= − __
2
2
__
√3
1 , m = ___
∴ m1 = − __
2
2
2
The general solution is of the form
y = em1x [A cos m2x + B sin m2 x]
1
__
__
[
__
√3
√3
Hence y = e− 2 x A cos ___
x + B sin ___
x
2
2
]
EXAMPLE 14
d2y
dy
Find the general solution of ___2 + 2 ___ + 5y = 0.
dx
dx
SOLU TION
dy
d2y
___
+ 2 ___ + 5y = 0
2
dx
dx
The AQE is m2 + 2m + 5 = 0
__________
−2 ± √ 4 − 4(5)(1)
m = ________________
2
____
−2 ± √ −16
= ___________
2
___
___
√ 16 √ −1
−2 ± ________
= ___
2
2
4i
= −1 ± __
2
= −1 ± 2i
m1 = −1, m2 = 2
The solution is of the form y = em1x [A cos m2 x + B sin m2 x]
The general solution is y = e−x [A cos 2x + B sin 2x].
Try these 17.3
Ask yourself
Are the roots
complex?
What is the
solution for these
roots?
Find the general solution of the following differential equations.
d2y
dy
(a) ___2 − 2 ___ + 2y = 0
dx
dx
2
dy
dy
(b) ___2 − 4 ___ + 5y = 0
dx
d2y
dx
dx
dy
dx
(c) 8 ___2 − 12 ___ + 5y = 0
391
M O DUL E 3
Non-homogeneous second order
differential equations
A non-homogeneous second order differential equation is of the form
dy
d2y
a ___2 + b ___ + cy = f(x), where f(x) ≠ 0.
d
x
dx
We look at equations where f (x) is
(i) a polynomial in x of at most degree 2
(ii) a trigonometric function
(iii) an exponential function.
d2y
dy
Let h(x) be a solution of the equation and g (x) be a solution of a ___2 + b ___ + cy = 0.
dx
dx
Let y = h(x)
dy
___
= h′(x)
dx
d2y
___
= h″(x)
d x2
∴ ah ″(x) + bh′(x) + ch(x) = f(x)
[1]
Let y = g(x)
dy
___
= g′(x)
dx
d2y
___
= g ″(x)
d x2
∴ ag ″(x) + bg′(x) + cg(x) = 0
[2]
[1] + [2] gives
ah″(x) + ag ″(x) + bh′(x) + bg′(x) + ch(x) + cg (x) = f (x)
a[h″(x) + g ″(x)] + b[h′(x) + g′(x)] + c [h(x) + g(x)] = f (x)
Since h(x) + g(x) satisfies the differential equation,
y = h(x) + g(x) is a solution of the equation.
g(x) is called the complementary function (CF) and h(x) is called the particular
integral (PI) of the solution.
d2y
dy
∴ The general solution of a ___2 + b ___ + cy = f (x) is
d
x
dx
y = CF + PI
dy
d2y
To find the CF solve a ___2 + b ___ + cy = 0
d
x
dx
The PI depends on f (x). We will consider three cases for the PI:
(a) when f (x) is a polynomial
(b) when f (x) is a trigonometric function
(c) when f (x) is an exponential function.
392
MODULE 3tCHAPTER 17
When f(x) is a polynomial of degree n
In this case, the particular integral is also a polynomial of degree n.
EXAMPLE 15
SOLU TION
Ask yourself
Is the equation of
the form
d2y
dy
a___2 + b___ + cy
dx
dx
= f(x)?
Is f(x) a
polynomial?
What is the
degree of the
polynomial?
What is the form
of the particular
integral?
d2y
dy
Solve the differential equation ___2 − 6 ___ + 9y = 4x + 1.
d
x
dx
dy
d2y
___
− 6 ___ + 9y = 4x + 1
2
dx
dx
The general solution is y = CF + PI
dy
d2y
To find the CF solve ___2 − 6 ___ + 9y = 0
d
x
dx
Form the AQE:
m2 − 6m + 9 = 0
(m − 3)2 = 0
m=3
∴ The CF is y = (Ax + B)e3x, since the roots of the AQE are real and equal.
For the PI:
Since f (x) is a polynomial of degree 1, the PI must also be a polynomial of degree 1.
Let y = λ x + μ, where we need to find λ and μ.
dy
___
=λ
dx
d2y
___
=0
d x2
d2y
dy
Substituting into the differential equation ___2 − 6 ___ + 9y = 4x + 1
dx
dx
we get −6λ + 9(λx + μ) = 4x + 1
−6λ + 9 λx + 9μ = 4x + 1
9λx + 9μ − 6λ = 4x + 1
4
9λ = 4 ⇒ λ = __
9
9μ − 6 λ = 1
Equating coefficients of x:
Equating constants:
4 =1
9μ − 6 __
9
()
24
9μ = 1 + ___
9
33
___
9μ =
9
11 × __
1 = ___
11
μ = ___
3
9 27
11
4 x + ___
∴ PI is y = __
9
27
4 x + ___
11 .
The general solution is y = (Ax + B)e3x + __
9
27
393
M O DUL E 3
EXAMPLE 16
SOLU TION
d2y
dy
Find the general solution of the equation ___2 − 3 ___ + 2y = x2 − 1.
dx
dx
The general solution is y = CF + PI
dy
d2y
To find CF solve ___2 − 3 ___ + 2y = 0
dx
dx
AQE is m2 − 3m + 2 = 0
(m − 2)(m − 1) = 0
m = 2 or 1
∴ CF is y = Ae 2x + Be x.
To find the PI:
Remember
Do not leave out
any terms from
the form of the
PI. Even though
f(x) = x2 −1 our
PI contains the
term in x, i.e. ax2
+ bx + c.
Since the function f (x) is a quadratic the PI is also a quadratic.
Let y = ax2 + bx + c, where a, b and c are constants to be found.
dy
___
= 2ax + b
dx
d 2y
___
= 2a
d x2
dy
d2y
Substituting into ___2 − 3 ___ + 2y = x 2 − 1 gives
d
x
dx
2a − 3(2ax + b) + 2(ax2 +bx +c) = x2 − 1
⇒ 2a − 6ax − 3b + 2ax2 + 2bx + 2c = x2 − 1
∴ 2ax2 + x (−6a + 2b) + 2a − 3b + 2c = x2 − 1
Equating coefficients of x2:
1
2a = 1 ⇒ a = __
2
Equating coefficients of x:
−6a + 2b = 0
1 ⇒ −6 __
1 + 2b = 0
a = __
2
2
2b = 3
3
b = __
2
Equating constants:
( )
2a − 3b + 2c = −1
3 ⇒ 2 __
3 + 2c = −1
1 , b = __
1 − 3 __
a = __
2
2
2
2
9 + 2c = −1
1 − __
2
( ) ( )
9
2c = −1 − 1 + __
2
5
c = __
4
The general
solution is
y = CF + PI
394
3 x + __
5
1 x2 + __
∴ The PI is y = __
4
2
2
3 x + __
5.
1 x2 + __
The general solution is y = Ae2x + Bex + __
4
2
2
MODULE 3tCHAPTER 17
E X A M P L E 17
SOLU TION
Remember
Find the general
solution first, i.e.
CF + PI, then use
the boundary conditions to find the
constants in the
general solution.
d2y dy
Solve the differential equation ___2 − ___ − 6y = x − 1 given that x = 0 when y = 0
dx
dx
dy
___
= 2.
when
dx
dy
d2y ___
___
−
− 6y = x − 1
2
dx
dx
First we find the general solution, which is y = CF + PI.
To find the CF we solve
dy
d2y ___
___
−
− 6y = 0
2
d
x
dx
The AQE is
m2 − m − 6 = 0
(m − 3) (m + 2) = 0
m = 3 or −2
CF is y = Ae−2x + Be3x
To find the PI, since f (x) = x −1 is a polynomial of degree 1, our PI must also be a
polynomial of degree 1.
Let y = ax + b
dy
___
=a
dx
d2y
___
=0
d x2
Substituting into the differential equation
dy
d2y ___
___
−
− 6y = x − 1
d x2 d x
we get
−a − 6 (ax + b) = x − 1
⇒ −6ax − 6b − a = x − 1
Equating coefficients of x:
1
−6a = 1 ⇒ a = − __
6
Equating constants:
−6b − a = −1
1 = −1
−6b + __
6
7
b = ___
36
7
1 x + ___
The PI is y = − __
6
36
395
M O DUL E 3
The general solution is
7
1 x + ___
y = Ae−2x + Be3x − __
6
36
Remember
Do not use the
conditions
x = 0, y = 0,
dy
___
= 2 until the
dx
general solution
has been obtained.
dy
We need to find the constants A and B using x = 0, y = 0 when ___ = 2.
dx
Substitute x = 0, y = 0 into
7
1 x + ___
y = Ae−2x + Be3x − __
6
36
7
0 = A + B + ___
36
7
∴ A + B = −___
36
Differentiate
7
1 x + ___
y = Ae−2x + Be3x − __
6
36
dy
1
___
= −2Ae−2x + 3Be3x − __
6
dx
dy
Substituting x = 0, ___ = 2 gives
dx
1
2 = −2A + 3B − __
6
13
∴ 2A − 3B = −___
6
Solving simultaneously, we have
7
[1]
A + B = −___
36
13
2A − 3B = −___
6
[2]
[1] multiplied by 2 gives
7
2A + 2B = −___
18
[3]
[3] minus [2] gives
16
5B = ___
9
16
B = ___
45
16 = − ___
7
A + ___
45
36
11
A = − ___
20
The solution under the given conditions is
1
16 e3x − __
7
11 e−2x + ___
___
y = − ___
6 x + 36
45
20
396
MODULE 3tCHAPTER 17
E X A M P L E 18
SOLU TION
d2y
dy
Solve the differential equation ___2 − 8 ___ + 16y = 32x − 16 given that x = 0,
dx
dx
dy
___
= 2.
y = 0,
dx
dy
d2y
___
− 8 ___ + 16y = 32x − 16
2
dx
dx
We first find the general solution of the equation, y = CF + PI
To find the CF we solve
dy
d2y
___
− 8 ___ + 16y = 0
2
dx
dx
The AQE is m2 − 8m + 16 = 0
⇒ (m − 4)2 = 0
m=4
∴ The CF is y = (Ax + B) e4x
Since f(x) = 32x − 16, the PI is of the form
y = λx + μ
dy
___
=λ
dx
d 2y
____
=0
d x2
Substituting into the differential equation, we have
−8λ + 16(λx + μ) = 32x − 16
−8λ + 16λx + 16μ = 32x − 16
16λ = 32
(equating coefficients of x)
λ=2
−8λ + 16μ = −16
(equating constants)
λ = 2 ⇒ −16 + 16μ = −16
16μ = 0
μ=0
∴ y = 2x
(substituting λ = 2, μ = 0 into y = λx + μ)
The general solution of the equation is
y = (Ax + B) e4x + 2x
We now use the given values to find A and B.
When x = 0, y = 0, we have
0 = B + 2(0)
B=0
397
M O DUL E 3
y = (Ax) e4x + 2x
Differentiate
Ax e4x using the
product rule.
dy
___
= 4(Ax) e4x + Ae4x + 2
dx
dy
When x = 0, ___ = 2
dx
∴2=A+2
A=0
∴ The solution is y = 2x
Try these 17.4
d2y
dx
dy
dx
(a) Find the general solution of ___2 − 9 ___ + 20y = 7x + 2.
(b) Given that y = ax2 + bx + c is a solution of the differential equation
d2y
dy
___
−12 ___ + 36y = 3x2 − x + 1,
2
dx
dx
find a, b and c. Hence solve the differential equation.
When f(x) is a trigonometric function
Note
The angle
remains the same
in the PI and the
function f(x). The
coefficients of the
trigonometric
functions may
change.
d2y
dy
a ___2 + b ___ + cy = λ cos mx + μ sin mx.
dx
dx
The general solution is y = CF + PI.
dy
d2y
The CF is found by solving a ___2 + b ___ + cy = 0.
dx
dx
The PI is also a trigonometric function of the same form.
Let f(x) = λ cos mx + μ sin mx.
The PI is of the same form y = a cos mx + b sin mx.
E X A M P L E 19
d2y dy
Solve ___2 + ___ − 6y = cos x.
dx
dx
SOLU TION
The general solution is y = CF + PI
To find the CF:
d2y dy
Solve ___2 + ___ − 6y = 0
dx
dx
The AQE is m2 + m − 6 = 0
(m − 2)(m + 3) = 0
m = 2 or −3
The CF is
y = Ae2x + Be−3x
398
(since the roots are real and distinct)
MODULE 3tCHAPTER 17
Remember
The PI contains
both sine and
cosine, even
when f (x) contains only one of
sine or cosine.
Since f (x) = cos x + 0 sin x the PI is of the same form, i.e. y = a cos x + b sin x,
where a and b are constants to be found.
To find a and b, we find
dy
___
= −a sin x + b cos x
dx
d2y
___
= −a cos x − b sin x
d x2
Substituting into the differential equation
dy
d 2y ___
___
+
− 6y = cos x, we get
2
d
x
dx
−a cos x − b sin x + (−a sin x + b cos x) − 6(a cos x + b sin x) = cos x
⇒ (−a cos x + b cos x − 6 a cos x) + (−b sin x − a sin x − 6b sin x) = cos x
⇒ −7 a cos x + b cos x − 7b sin x − a sin x = cos x
Equating coefficients of cos x gives
−7a + b = 1
[1]
Equating coefficients of sin x gives
−7b − a = 0
[2]
Solving [1] and [2] simultaneously:
From [2]
a = −7b
Substitute into [1]
−7(−7b) + b = 1
50b = 1
1
b = ___
50
7
a = −___
50
7 cos x + ___
1 sin x
Therefore the PI is y = − ___
50
50
7 cos x + ___
1 sin x
The general solution is y = Ae2x + Be−3x − ___
50
50
E X A M P L E 20
d2y
Solve ___2 − y = 3 cos 2x + 4 sin 2x.
dx
SOLU TION
The general solution is y = CF + PI
d2y
To find the CF, solve ___2 − y = 0
dx
2
The AQE is m − 1 = 0, so
m2 = 1
m = ±1
399
M O DUL E 3
Since roots are real and distinct, the CF is y = Ae x + Be−x
To find the PI:
Since f(x) = 3 cos 2x + 4 sin 2x
The PI is y = λ cos 2x + μ sin 2x
where λ and μ are constants to be found
dy
___
= −2 λ sin 2x + 2μ cos 2x
dx
d2y
___
= −4 λ cos 2x − 4 μ sin 2x
d x2
d2y
Substituting into ___2 − y = 3 cos 2x + 4 sin 2x, we have
dx
−4λ cos 2x − 4μ sin 2x −(λcos 2x + μ sin 2x) = 3 cos 2x + 4 sin 2x
⇒ −4 λ cos 2x − λ cos 2x − 4μ sin 2x − μ sin 2x = 3 cos 2x + 4 sin 2x
⇒ −5λ cos 2x − 5μ sin 2x = 3 cos 2x + 4 sin 2x
Equating coefficients of cos 2x:
3
−5λ = 3 ⇒ λ = −__
5
Equating coefficients of sin 2x:
4
−5μ = 4 ⇒ μ = −__
5
3
4
__
∴ The PI is y = −__
5 cos 2x − 5 sin 2x
The general solution is
3 cos 2x − __
4 sin 2x
y = Aex + Be−x − __
5
5
Try these 17.5
Find the general solution of
d2y
dx
dy
dx
d2y
dx
dy
dx
(a) 2 ___2 − 5 ___ + 3y = cos x + sin x
(b) 4 ___2 − 4 ___ + y = 3 sin 2x
EXAMPLE 21
Show that y = ax sin x + bx cos x is a particular solution of the differential equation
d2y
d2y
___
___
+
y
=
sin
x.
Hence
solve
the
differential
equation
+ y = sin x
d x2
d x2
dy
given that when x = 0, y = 0 and ___ = 0.
dx
SOLU TION
Let y = ax sin x + bx cos x
dy
___
= ax cos x + a sin x − bx sin x + b cos x, using the product rule.
dx
d2y
___
= a cos x − ax sin x + a cos x − bx cos x − b sin x − b sin x
d x2
= 2a cos x − ax sin x − bx cos x − 2b sin x
400
MODULE 3tCHAPTER 17
Substituting into the differential equation, we have
2a cos x − ax sin x − bx cos x − 2b sin x + ax sin x + bx cos x = sin x
⇒ 2a cos x − 2b sin x = sin x
Equating coefficients of cos x, we have
2a = 0, a = 0
Equating coefficients of sin x, we have
−2b = 1
1
b = −__
2
1
Therefore the PI is y = −__
2 x cos x which is of the form y = ax sin x + bx cos x
1
with a = 0, b = −__
2
For the CF
d2y
___
+y=0
d x2
The AQE is
m2 + 1 = 0
___
m = ± √ −1 = ±i
Remember
The values x = 0,
dy
y = 0 and ___ = 0
dx
are used to find
the constants
in the general
solution of the
differential
equation.
Investigation
Can you identify
why the standard
PI y = a cos x +
b sin x does not
work in this
particular case?
Therefore the CF is y = A cos x + B sin x
The general solution is
1 x cos x
y = A cos x + B sin x − __
2
When x = 0, y = 0
Since the roots are complex,
the solution is of the form
y = em1x (A cos m2x + B sin m2x)
where m1 = 0, m2 = 1.
∴0=A
Differentiating y
dy
1 x sin x
1 cos x + __
___
= −A sin x + B cos x − __
2
2
dx
dy
When x = 0, ___ = 0
dx
1
∴ 0 = B − __
2
1
B = __
2
dy
Hence the solution when x = 0, y = 0 and ___ = 0 is
dx
1
1
__
__
y = sin x − x cos x
2
2
E X A M P L E 22
Find the general solution of the differential equation
d2x + 4 ___
d x + 3x = 2 cos 2t + 3 sin 2t. Hence show that, for large positive values of t,
___
dt
d t2
__
x ≈ 0.2 √5 sin (2t − α), where tan α = 2.
SOLU TION
The general solution is x = CF + PI
401
M O DUL E 3
To find the CF we solve
d x + 3x = 0
d2x + 4 ___
___
2
dt
dt
The AQE is m2 + 4m + 3 = 0
(m + 1) (m + 3) = 0
m = −1, −3
The CF is x = Ae−t + Be−3t
To find the PI:
let x = a cos 2t + b sin 2t
d x = −2a sin 2t + 2b cos 2t
___
dt
d2x = −4a cos 2t − 4b sin 2t
___
d t2
Substituting into the differential equation, we get
−4a cos 2t − 4b sin 2t + 4(−2a sin 2t + 2b cos 2t) + 3(a cos 2t + b sin 2t)
= 2 cos 2t + 3sin2t
Equating coefficients of cos 2t:
−4a + 8b + 3a = 2
Equating coefficients of sin 2t:
−4b − 8a + 3b = 3
Solve
−a + 8b = 2
[1]
−8a − b = 3
[2]
−8a + 64b = 16
13 = 0.2
[3] − [2] gives 65b = 13, b = ___
65
−a + 8 (0.2) = 2
[3]
[1] × 8 gives
a = −0.4
The PI is x = −0.4 cos 2t + 0.2 sin 2t
Therefore the general solution is
x = Ae−t + Be−3t − 0.4 cos 2t + 0.2 sin 2t
When t is large and positive, e−t → 0 and e−3t → 0
∴ Ae−t → 0 and Be−3t → 0
∴ y ≈ 0.2 sin 2t − 0.4 cos 2t
Let 0.2 sin 2t − 0.4 cos 2t = R sin(2t − α)
We have 0.2 sin 2t − 0.4 cos 2t = R sin 2t cos α − R cos 2t sin α
Equating coefficients, we have
R cos α = 0.2
R sin α = 0.4
402
MODULE 3tCHAPTER 17
Dividing gives tan α = 2
______________
__
Now R = √(0.2)2 + (−0.4)2 = 0.2 √5
__
Hence x ≈ 0.2 √ 5 sin (2t − α), where tan α = 2
When f (x) is an exponential function
d2y
dy
To solve an equation of the form a ___2 + b ___ + cy = f (x), where f (x) = λemx, we find
dx
dx
d2y
dy
(i) the complementary function by solving a ___2 + b ___ + cy = 0,
dx
dx
(ii) the particular integral. In this case, since f (x) = λemx, the PI is of the same
form, that is, y = μemx. (Notice the emx is the same in both f (x) and the PI.
What changes is the coefficient.)
(iii) the general solution, which is the sum of the particular integral and the
complementary function.
E X A M P L E 23
d2y dy
Find the general solution of the differential equation ___2 − ___ − 6y = 4e2x.
dx
dx
SOLUTION
The general solution is y = CF + PI
To find the CF we solve
d2y ___
dy
___
− 6y = 0
−
2
dx
dx
The AQE is
m2 − m − 6 = 0
(m − 3)(m + 2) = 0
m = 3 or −2
Since the roots are real and distinct, the CF is y = Ae3x + Be−2x
To find the PI:
Since f (x) = 4e2x, the PI is of the same form. Therefore y = ae2x
dy
___
= 2ae2x
dx
d2y
___
= 4ae2x
dx2
Substituting into the differential equation, we have
4ae2x − 2ae2x − 6ae2x = 4e2x
−4ae2x = 4e2x
−4a = 4
a = −1
Hence the PI is y = −e2x
The general solution is y = Ae3x + Be−2x − e2x
403
M O DUL E 3
E X A M P L E 24
SOLUTION
d2y
dy
Find the general solution of the differential equation ___2 − 4 ___ + 13y = 20ex.
dx
dx
dy
___
= 1.
Hence find the solution when x = 0, y = 0 and
dx
The general solution is y = CF + PI
To find the CF we solve
d 2y
dy
___
− 4 ___ + 13y = 0
dx
dx2
The AQE is
m2 − 4m + 13 = 0
Solving, we have ____________
−(−4) ± √ (−4)2 − 4(13)
m = ______________________
2
____
4 ± √ −36
m = _________
2
4
±
6i
m = ______ = 2 ± 3i
2
Since the roots are complex, the CF is y = e2x(A cos 3x + B sin 3x)
To find the PI:
Since f(x) = 20ex, the PI is of the same form.
Therefore y = aex
dy
___
= aex
dx
d2y
___
= aex
dx2
Substituting into the differential equation, we have
aex − 4aex + 13aex = 20ex
10aex = 20ex
10a = 20
a=2
Hence the PI is y = 2ex
The general solution is y = e2x (A cos 3x + B sin 3x) + 2ex
dy
We next need to find the solution when x = 0, y = 0, ___ = 1.
dx
Substituting x = 0, y = 0 into y = e2x (A cos 3x + B sin 3x) + 2ex gives
0 = e2(0) (A cos 3(0) + B sin 3(0)) + 2e0
0 = A + 2, so A = −2
Differentiating y with respect to x, we have
dy
___
= e2x (−3A sin 3x + 3B cos 3x) + 2e2x (A cos 3x + B sin 3x) + 2ex
dx
dy
Substituting x = 0, ___ = 1 gives:
dx
1 = 3B + 2A + 2
1 = 3B − 4 + 2
3B = 3, so B = 1
404
(substituting A = −2)
MODULE 3tCHAPTER 17
Substituting A = −2, B = 1 into y = e2x (A cos 3x + B sin 3x) + 2ex, we get
y = e2x (−2 cos 3x + sin 3x) + 2ex.
dy
Hence the solution to the equation when x = 0, y = 0 and ___ = 1 is
dx
y = e2x (−2 cos 3x + sin 3x) + 2ex.
Equations reducible to a recognisable form
dy
At this stage you should be able to solve differential equations of the form ___ + Py = Q
dx
d2y
dy
and second order differential equations of the form a ___2 + b ___ + cy = f (x). There
dx
dx
are some differential equations that can be reduced to one of these familiar forms
with a suitable substitution.
Let us start with first order differential equations.
E X A M P L E 25
dy
Solve the differential equation xy ___ = x2 + y2 using the substitution y = vx.
dx
SOLUTION
Let y = vx
Differentiating with respect to x, using the product rule on the right-hand side,
we have
dy
d [x] + x ___
d [v]
___
= v ___
dx
dx
dx
dy
dv
___
= v + x ___
dx
dx
dy
dy
dv into xy ___
= x2 + y2 gives
Substituting y = vx and ___ = v + x ___
dx
dx
dx
dv = x2 + (v x)2
x(vx) v + x ___
dx
dv = x2 + v 2 x2
vx2 v + x ___
dx
x2 (1 + v 2)
dv = _________
v + x ___
dx
vx2
1 + v2 − v
dv = ______
x ___
v
dx
1 + v2 − v2
dv = __________
x ___
v
dx
1
dv = __
x ___
dx v
Separating variables and integrating gives:
(
(
Notice that the
substitution
reduced the
differential
equation to a
standard form.
)
)
∫v dv = ∫ __1x dx
1 v 2 = ln x + c
__
2
y
Since y = vx, v = __
x
1 __y 2 = ln x + c
∴ __
2 x
( )
405
M O DUL E 3
( __xy )
2
= 2 ln x + 2c
_________
y__
=
2 ln x + 2c
√
x
_________
y = x√ 2 ln x + 2c
_________
Hence the general solution of the equation is y = x√2 ln x + 2c .
EXAMPLE 26
SOLUTION
dy
Using the substitution y = vx solve the differential equation (x + y) ___ = x − y
dx
Let y = vx
dy
dv
(using the product rule)
Then ___ = v + x ___
dx
dx
Substituting into the differential equation, we have
dv = x − vx
(x + vx) v + x ___
dx
x(1 − v)
dv = _______
v + x ___
dx x(1 + v)
(1 − v)
dv = ______
x ___
−v
(1
+ v)
dx
2
1 − v − v(1 + v) _____________
dv = ______________
x ___
=1−v−v−v
1+v
1+v
dx
(
)
dv = __________
1 − 2v − v
x ___
1+v
dx
2
We now have an equation that we can solve by separating the variables and integrating:
1+v
1 dx
dv = ∫ __
∫ __________
x
1 − 2v − v 2
− 2 − 2v dv = __
1 __________
1 dx
− __
x
2 1 − 2v − v2
∫
∫
1 ln |1 − 2v − v2| = ln x + c
− __
2
y
Since y = vx, v = __
x
y2
2y __
1 ln 1 − __
− 2 = −2 ln x − 2c
∴ −__
x
2
x
|
|
Notice by using the substitution the differential equation is converted to an equation
that is much simpler to solve. Let us now look at second order differential equations,
using a substitution to simplify.
E X A M P L E 27
406
d2y d2y dy
Show that if y depends on x and x = eu then x2 ___2 = ___2 − ___. Given that y satisfies
du
dx
du
the differential equation
d2y
dy
x2 ___2 − 2x ___ − 10y = 3x3
dx
dx
show that
dy
d2y
___
− 3 ___ − 10y = 3e3u
2
du
dx
Hence solve the differential equation, giving y as a function of x.
MODULE 3tCHAPTER 17
SOLUTION
By the chain rule:
dy ___
dy ___
___
=
× du
dx du dx
Since x = eu
dx = eu
___
du
du = __
1
___
dx
Note that the
dy
differential of ___
du
with respect to
d2y du
x is ____2 × ___
,
dx
du
using the chain
rule again.
eu
dy
dy
1 ___
∴ ___ = __
u
e
dx
du
dy
dy
1
___ = __ ___
dx x du
dy dy
x ___ = ___
dx du
Differentiating with respect to x again
d2y dy d2y du
x ___2 + ___ = ___2 × ___
dx du
dx
dx
2y
2y
d
d
dy
x2 ___2 + x ___ = ___2
dx du
dx
d2y d2y
dy
x2 ___2 = ___2 − x ___
dx
dx
du
d2y dy
= ___2 − ___
du
du
Substituting into the differential equation, we have
dy
dy
d2y ___
___
− 2 ___ − 10y = 3e3u
−
2
du
du
du
2y
d
dy
___ − 3 ___ − 10y = 3e3u
du
dx2
We now solve this differential equation to find y in terms of u.
The general solution of the equation is y = CF + PI
To find the CF we solve
d2y
dy
___
− 3 ___ − 10y = 0
du
dx2
The AQE is
m2 − 3m − 10 = 0
(m − 5)(m + 2) = 0
m = 5 or m = −2
∴ y = Ae5u + Be−2u
To find the PI, let y = ae3u
dy
___
= 3ae3u
du
d2y
___
= 9ae3u
du2
407
M O DUL E 3
Substituting into the differential equation, we have
9ae3u − 9ae3u − 10ae3u = 3e3u
−10a = 3
3
a = −___
10
3 3u
∴ The PI is y = −___
10 e
3 e3u
The general solution is y = Ae5u + Be−2u − ___
10
We now have y as a function of u and since x = eu, taking logs to base e, we get:
u = ln x
3 e3 ln x
∴ y = Ae5 ln x + Be−2 ln x − ___
10
3 e ln x3
y = Ae ln x5 + Be ln x−2 − ___
10
3 x3
B − ___
y = Ax5 + __
x2 10
3 x3.
B − ___
Hence our solution is y = Ax5 + __
x2 10
E X A M P L E 28
1 and the variable Z depends on x. Use this
The variables x and t are related by x = __
t
relationship to show that the differential equation
2
d Z + (2x4 − 5x3) ___
dZ + 4xZ = 6x + 3
x5 ____
dx
dx2
can be reduced to
2
d Z + 5 ___
dZ + 4Z = 3t + 6
____
dt
dt2
Hence solve the equation to find Z in terms of x.
SOLUTION
By the chain rule:
dZ ÷ ___
dx
dZ = ___
___
dx
dt
dt
1
Since x = __t
1
dx = −__
___
2
t
dt
1
dZ = ___
dZ ÷ −__
∴ ___
t2
dx
dt
( )
dZ
= −t2 ___
dt
1 ___
dZ = −__
dZ
∴ ___
x2 dt
dx
dZ
dZ = ___
−x2 ___
dx
dt
408
dZ makes use of the chain
Differentiate again with respect to x ; the differential of ___
dt
dZ is found by using the product rule.
rule and the differential of −x2 ___
dx
2
2
d Z ÷ ___
d Z − 2x ___
dZ = ____
dx
−x2 ____
dx
dt
dx2
dt2
MODULE 3tCHAPTER 17
d2Z ÷ − __
d2Z − 2x ___
dZ = ____
1
−x2 ____
2
dx
dx
dt2
t2
( )
2
dt
= −t2 ___
dt2
d2Z − 2x ___
dZ = − __
d2Z
1 ____
∴ −x2 ____
2
2
dx
dx
x dt2
d2Z
d2Z + 2x3 ___
dZ = ____
x4 ____
dx
dx2
dt2
Now
dZ
dZ = −x2 ___
___
dt
dx
and
2
2
dt2
dx2
d Z + 2x3 ___
dZ
d Z = x4 ____
____
dx
d2Z
dZ + 4xZ = 6x + 3
x5 ____2 + (2x4 − 5x3) ___
dx
dx
2
3
d Z + (2x3 − 5x2) ___
dZ + 4Z = 6 + __
(dividing by x throughout)
x4 ____
x
dx
dx2
3
d2Z + 2x3 ___
dZ − 5x2 ___
dZ + 4Z = 6 + __
x4 ____
x
dx
dx
dx2
dZ + 4Z = 6 + 3t
d2Z + 5 ___
1
____
since t = __
x
dt
dt2
dZ + 4Z = 3t + 6, we find the general solution, which is Z = CF + PI
d2Z + 5 ___
To solve ____
dx
dt2
To find the CF we solve
(
)
(
)
2
d Z + 5 ___
dZ + 4Z = 0
____
dt2
dt
The AQE is
m2 + 5m + 4 = 0
(m + 1)(m + 4) = 0
m = −1 or m = −4
∴Z=
Ae−t
+ Be−4t
To find the PI:
Z = at + b
dZ = a
___
dt
2
d Z=0
____
dt2
∴ 5a + 4(at + b) = 3t + 6
Equating coefficients of t, we have
3
4a = 3, a = __
4
Equating the constants, we have
5a + 4b = 6
409
M O DUL E 3
15 + 4b = 6
___
4
9
4b = __
4
9
b = ___
16
9
3 t + ___
∴ The PI is Z = __
4
16
9
3 t + ___
The general solution is Z = Ae−t + Be−4t + __
4
16
1 into the solution above.
We can now find Z as a function of x by substituting t = __
x
Hence our solution is
4
1
__
__
9
3 + ___
Z = Ae− x + Be−x + ___
4x 16
E X A M P L E 29
Use the substitution y = x2 to show that the differential equation
dx 2 + 5x ___
dx + 3x2 = sin 2t + 3 cos 2t
d2x + ___
x ___
2
dt
dt
dt
( )
( )
can be converted to:
d2y
dx + 6y = 2 sin 2t + 6 cos 2t
___
+ 5 ___
dt
dt2
Hence solve the equation to find x in terms of t.
SOLUTION
By the chain rule:
dy
dx
___
= 2x ___
dt
dt
Since y = x2, we have
d2y
d2x + 2 ___
dx ___
dx
___
___
=
2x
dt dt
dt2
dt2
( )( )
d x + ___
Substituting into: x ___
( dxdt ) + 5x( ___dxdt ) + 3x = sin 2t + 3 cos 2t, we have
dt
2
2
2
2
2
d y __
dy
1 ___
__
+ 5 ___ + 3y = sin 2t + 3 cos 2t
2 dt2 2 dt
d2y
dy
___
+ 5 ___ + 6y = 2 sin 2t + 6 cos 2t
2
dt
dt
To solve this equation, we find the general solution, y = CF + PI
dy
d2y
To find the CF: ___2 + 5 ___ + 6y = 0
dt
dt
2
The AQE is
m + 5m + 6 = 0
(m + 2)(m + 3) = 0
m = −2 or m = −3
∴y=
410
Ae−2t
+
Be−3t
To find the PI: y = a sin 2t + b cos 2t
dy
___
= 2a cos 2t − 2b sin 2t
dt
d2y
___
= −4a sin 2t − 4b cos 2t
dt2
MODULE 3tCHAPTER 17
Substituting into the differential equation, we have
−4a sin 2t − 4b cos 2t + 5 (2a cos 2t − 2b sin 2t) + 6 (a sin 2t + b cos 2t)
= 2 sin 2t + 6 cos 2t
Equating coefficients of sin 2t, we have
−4a − 10b + 6a = 2
2a − 10b = 2
[1]
Equating coefficients of cos 2t, we have
−4b + 10a + 6b = 6
10a + 2b = 6
[2]
Multiplying [1] by 5: 10a − 50b = 10
[3]
[2] − [3] gives: 52b = −4
4
b = −___
52
1
= −___
13
Substituting into equation [1], we have
1
2a − 10 −___
13 = 2
(
)
10
2a = 2 − ___
13
16
2a = ___
13
8
a = ___
13
8 sin 2t − ___
1 cos 2t
∴ The PI is y = ___
13
13
8 sin 2t − ___
1 cos 2t
The general solution is y = Ae−2t + Be−3t + ___
13
13
When y = x2
8 sin 2t − ___
1 cos 2t
x2 = Ae−2t + Be−3t + ___
13
13
EXERCISE 17B
Solve the differential equations in questions 1–12.
1
d2y
dy
___
+ 6 ___ + 9y = 0
2
dx
dx
2
d2y
dy
9 ___2 − 6 ___ + y = 0
dx
dx
3
d2y
dy
36 ___2 − 60 ___ + 25y = 0
dx
dx
4
d2y
dy
9 ___2 + 12 ___ + 4y = 0
dx
dx
5
d2y
dy
49 ___2 − 28 ___ + 4y = 0
dx
dx
6
dy
dy
___
+ 6 ___ + 8y = 0
7
d2y
dy
___ + 4 ___
− 12y = 0
2
dx
dx
2
8
d x2
dx
d2y
dy
12 ___2 + 11 ___ − 5y = 0
dx
dx
411
M O DUL E 3
9
d2y
dy
30 ___2 − 17 ___ − 35y = 0
d
x
dx
2
d2y
dx
dy
dy
11 ___2 − 4 ___ + 5y = 0
dx
d2y
dx
dy
dx
10 48 ___2 + 64 ___ − 35y = 0
dy
dx
12 2 ___2 − 2 ___ + y = 0
dx
Find the general solutions for the differential equations in questions 13–26.
d2y
dx
dy
dx
13 4 ___2 + 4 ___ + y = 3x − 2
d2y
dx
dy
dx
15 9 ___2 − 30 ___ + 25y = 5x
d2y
dy
17 ___2 + 3 ___ − 10y = 2e4x
dx
d2y
dx
d2y
dx
dy
dx
14 25 ___2 − 20 ___ + 4y = 4x2 − 8x − 70
2
dy
dy
16 ___2 − 14 ___ + 49y = 2x2 − 4
dx
d2y
dx
dy
dx
dx
18 6 ___2 + 7 ___ − 20y = 36e−2x
dy
dx
dy
dy
20 ___2 − 8 ___ + 17y = 10ex
19 6 ___2 − ___ − 40y = 22e3x
d2y
dx
dx
2
dx
dx
2
dy
dy
22 ___2 − 10 ___ + 29y = 7e2x
dy
dx
21 5 ___2 − 2 ___ + y = 3
dx
dx
2
dy
dy
23 ___2 + 3 ___ + 2y = cos x + sin x
dx
dx
d2y
dy
24 ___2 + ___ − 12y = 3 cos 2x + sin 2x
dx
d2y
dx
dx
dy
dx
25 2 ___2 − ___ − 6y = 4 cos x
d2y
dx
dy
dx
26 4 ___2 − 11 ___ + 6y = sin 4x
d2y
dx
dy
dx
27 Given that 64 ___2 + 48 ___ + 9y = 9x + 3, find y in terms of x when x = 0,
dy
y = 0 and ___ = 0.
dx
d2y
dx
dy
dx
28 Find the general solution of the differential equation ___2 − 4 ___ + 13y
dy
= 50e−2x. Hence find the solution when x = 0, y = 0 and ___ = 1.
dx
d2y
dy
dx
dx
is y = (Ax + B)e−5x + 2x − 1. Hence find the solution when x = 0, y = 1
dy
and ___ = 1.
dx
29 Show that the solution of the differential equation ___2 + 10 ___ + 25y = 50x − 5
30 Show that y = −2x + 3 is a particular integral of the differential equation
dy
d2y
12 ___2 − 23 ___ − 9y = 18x + 19. Hence find the general solution of the
dx
dx
differential equation.
d2y
dy
d x __ d x
5√ 2 cos (x − 81.9°).
positive y ≈ ____
69
31 Given that ___2 + 8 ___ + 12y = cos x + sin x, show that when x is large and
412
MODULE 3tCHAPTER 17
dy
dx
dy
33 Solve the differential equation ___ = (x + y + 4)2, using the substitution
dx
v = x + y + 4.
32 Use the substitution y = vx to solve the differential equation 2xy ___ = 4x2 + y2.
34 Use the substitution x = eu to find the general solution of the differential
d2y
dy
equation x2 ___2 + 10x___ + 20y = 0.
dx
dx
35 By using the substitution x = eu, show that the differential equation
d2y
d2y
dy
dy
x2 ___2 − 3x___ − 12y = 0 reduces to ___2 − 4___ − 12y = 0. Hence show that the
dx
dx
dx
dx
B.
6
general solution of the equation is y = Ax + __
x2
SUMMARY
Differential equations
Second order differential equation
First order linear differential equation
Write in the form dy + Py = Q
dx
dy
where the coefficient of
is 1,
dx
P and Q are constants or functions
of x only.
Find the general solution using
IF = e∫ pdx
pdx
∫
ye
= ∫ Qe∫ pdxdx
Given any conditions,
use the conditions to
find the constants in
the general solution.
a
d2y
dy
= b + cy = 0
dx2
dx
a
AQE am2 + bm + c = 0
d2y
dy
+ cy = f(x)
+b
dx2
dx
Equations reducible
to a recognisable
form by using a
substitution
General solution
y = CF + PI
Roots are real and equal
i.e. m = m1
y = (Ax + B)e m1x
For CF solve
d2y
dy
a 2 + b + cy = 0
dx
dx
Roots are real and distinct
i.e. m = m1, m2
y = Ae m1x + Be m2x
For PI
f(x) = a cos mx + b sin mx
f(x) is a polynomial
Roots are complex
i.e. m = m1 + im2
y = e m1x (A cos m2x + B sin m2x)
f(x) = xmx
y is a polynomial of
y = λ cos mx + μ sin mx
the same degree
PI: y = μemx
as f(x).
Given any conditions,
use the conditions to
find the constants in the
general solution.
413
M O DUL E 3
Checklist
Can you do these?
dy
+ Py = Q.
■ Write a first order differential equation in the form ___
dx
■ Find the integrating factor of first order differential equations.
■ Solve first order differential equations.
■ Find a solution of first order differential equations given conditions.
d2y
dy
+ b ___ + cy = 0.
■ Solve a differential equation of the form a ___
dx
d x2
d2y
dy
+ b ___ + cy = f (x).
■ Find the CF of an equation of the form a ___
dx
d x2
d2y
dy
d2y
dy
+ b ___ + cy = f (x) where f(x) is a
■ Find the PI of an equation of the form a ___
dx
d x2
polynomial.
+ b ___ + cy = f (x) where f (x) is a
■ Find the PI of an equation of the form a ___
dx
d x2
trigonometric function.
d2y
dy
+ b___ + cy = λemx.
■ Find the solution of an equation of the form a___
dx
dx2
■ Find the general solution of a second order differential equation of the form
d2y
dy
a ___2 + b ___ + cy = f (x).
d
x
dx
d2y
dy
+ b ___ + cy = f(x)
■ Find the solution of a differential equation of the form a ___
dx
d x2
dy
given a condition on x, y and ___.
dx
■ Find the solution of a second order differential equation given an initial condition.
■ Solve differential equations that can be reduced using a substitution.
Review exercise 17
414
1
dy
Find the solution of the differential equation (1 + x3) ___ = x 2y, given that x = 1
dx
when y = 2.
2
dy
Find the general solution of the differential equation ___ + y cot x = cos x.
dx
3
d x + 3x = t giving x
Find the general solution of the differential equation ___
dt
explicitly in terms of t.
4
Find the general solution of the differential equation
dy
d2y
___
− 8 ___ + 41y = 6 sin 2x + 4 cos 2x.
2
dx
dx
MODULE 3tCHAPTER 17
5
6
7
8
9
dy
Find the general solution of the differential equation x ___ − x3 + 2x = y.
dx
The angular speed of a flywheel (ω) at time t is given by
dω +12ω = 8 + 12 sin2 (2t).
4 ___
dt
If ω = 0 when t = 0, find the angular speed in terms of the time t.
A circuit has an equation for current i given by
d2i + 400 __
di + 80 000i = 0. Solve the equation for the current, given that
0.5 ___
dt
d t2
di = 50.
when t = 0, i = 0 and __
dt
d2y
dy
Solve the differential equation ___2 + 8 ___ + 32y = 32 sin 2x given that when
d
x
dx
dy
___
= 0.
x = 0, y = 0 and
dx
d2y
dy
Find the general solution of the differential equation ___2 − 7 ___ + 12y = 2.
dx
dx
dy
Hence find the solution given that when x = 0, y = 2 and ___ = 5.
dx
d2y
dy
d
x
dx
2
integral of the form y = ax + bx + c where a, b and c are constants. Find a, b
and c. Find the solution of the differential equation given that when x = 0,
dy
y = 2 and ___ = 1.
dx
10 The differential equation 49 ___2 + 42 ___ + 9y = 9x2 + 3x has a particular
11 In a galvanometer the deflection α satisfies the differential equation
dα + α = 4. Find α in terms of t, given that when t = 0, α = 0
d2α + 2 ___
___
dt
d t2
α
d
___
and
= 0.
dt
dy
12 Solve the differential equation (1 + x2)2 ___ + 2x(1 + x2)y = 1.
dx
13 The equation of motion when a particle moves in a resisting medium is given by
dv = −(av + bt), where a and b are constants. Solve the equation for v given
___
dt
that v = v0 when t = 0.
dy
14 Find the general solution of the curve ___ − 2y = x + 1.
dx
Sketch the solution curve when x = 0 and y = 0.
dy
dx
15 Find the general solution of x ___ − x3 + 2x = y.
16 The equation of motion of a particle moving in a straight line under damping is
2
d x + ___
d x + x = k sin 2t, where k is a constant. Find x as a function of t
given by ___
dt
d t2
d x = 0.
given that when t = 0, x = 0 and ___
dt
415
M O DUL E 3
17 Solve the following differential equations giving y as a function of x.
dy
(a) ___ + 2y = x2
dx
dy
(b) ___ + y = 3x + 2
dx
dy
(c) (sin x) ___ + (cos x) y = sin2 x
dx
18 Find the general solution of the differential equation
d2x + ___
d x − 6x = 12t + 16.
2 ___
dt
d t2
d2q
dq
dt C
dt
variation of capacitance charge in an alternating current circuit with inductance
L, resistance R, and capacitance C. Given that ω = 20 radians per second,
5 henry, C = ___
1 farads and V = 300 volts, find q in terms of
R = 10 ohms, L = __
0
3
30
dq
___
t, if when t = 0, q = 0 and
= 0.
dt
______
dy
Find the general solution of the differential equation x2___ = y + √x2 − y2 ,
dx
using the substitution y = vx.
dy
Solve the differential equation x2 ___ = xy + y2, using y = vx.
dx
Use the substitution x = eu, to find the general solution of the differential
d2y
dy
equation x2 ___2 − 7x___ + 10y = 0.
dx
dx
Use the substitution v = xt to solve the differential equation
d2t + 2(1 + 2x)___
dt + 4(1 + x)t = 64e2x.
x ___
dx
dx2
19 The differential equation L ___2 + R ___ + __1 q = V0 sin ω t describes the
20
21
22
23
416
MODULE 3tCHAPTER 17
Mathematical modelling
Many practical problems in business and science are too complicated to be described
by simple formulae. With the necessary assumptions and testing we can develop a
mathematical model to analyse data collected for our problem and make predictions.
The model can be developed in the following way:
(i) We can first formulate our problem into a mathematical model based on
assumptions, data collecting, analysing and using knowledge from different key
areas to identify variables and equations.
(ii) We next analyse our model using various tools of mathematics: calculus,
statistics numerical analysis and computer science.
(iii) After analysing the model any conclusions drawn are applied to the real-world
problem. This is to check the accuracy of the model and to make predictions.
(iv) We next test the model by looking at new data to check the accuracy of any
predictions using the model. If the predictions are not consistent with the
new data the model can be adjusted and the process of formulation, analysis,
interpretation and testing is repeated.
Let us look at some questions that make use of mathematical models:
MODELLING EXERCISE
1
In a certain chemical reaction the amount, y grams, of a substance present is
decreasing. The rate of decrease of y is proportional to the product of y and t
where t is the time in seconds after the start of the reaction.
dy
(a) Show that the differential equation formed is given by ___ =−kyt, where k
dt
is a positive constant.
(b) At the start of the reaction, when t = 0, y = 50.
1 2
__
Show that y = 50e− 2kt .
(c) 10 seconds after the start of the reaction the amount of substance present is
40 grams. Find the time after the start of the reaction at which the amount
of substance present is 25 grams.
2
The rate of cooling of a body is proportional to the excess of its temperature
above that of its surroundings. Given that θ is the temperature of the body at
time t and θ0 is the temperature of the surroundings, write the information in
the form of a differential equation.
A body cools from 90 °C to 70 °C in 3 minutes at a surrounding temperature of
15 °C. Determine how long it will take for the body to cool to 50 °C.
3
The charge Q on a capacitor is given by Q = Q1(1 − e−αt) where Q1 is the initial
charge, α is a constant and t is the time.
(a) Find an equation for t.
(b) Sketch the graph of Q.
1Q .
(c) Find t when Q = __
2 1
417
M O DUL E 3
4
A cup of hot chocolate, initially at boiling point, cools so that after t minutes the
temperature 0 °C is given by
__t
θ = 10 + 90e−8
Sketch the graph of θ against t. Find the value of t when the temperature
reaches 50 °C.
5
A survey was carried out on 200 students who owned either a Nokia, an iPhone
or a Motorola cell phone. 90 of the students were male. Out of the 100 students
who owned Nokia cell phones, 50 were female. 30 males owned iPhones and
40 females owned Motorola cell phones. If a student who owned a cell phone is
selected at random, find the probability that the student
(a) owned a Nokia or iPhone cell phone
(b) is a female or owned an iPhone cell phone
(c) owned an iPhone given that the student is a female
(d) is a male who owned an iPhone or a female who owned a Nokia.
6
Kirchhoff ’s second law states that the sum of the voltage drop across the inducdi and the voltage drop across the resistor (iR) is the same as the impressed
tor L __
dt
voltage (E(t)) on the circuit, i.e.
di + Ri = E(t), where L is the inductance and R the resistance.
L __
dt
Determine the current i if the initial current is zero in a 12-volt battery circuit
in which the inductance is 0.5 henry and the resistance is 10 ohms.
7
Sandra has a pond at the back of her house; she puts 2000 fish in it. She decides
to add an additional 20 fish to the pond each month. In addition, it is known
that the fish population is growing by 4% per month. The size of the population
after n months is given by the sequence
fn = 1.04 fn − 1 + 20, f0 = 2000.
How many fish are in the pond after three months?
8
Ryan went shopping and bought 2 pairs of jeans, 2 shirts, and 4 T-shirts for
TT$90.00. At the same store, Michael bought 1 pair of jeans and 3 T-shirts
for TT$42.50, while Rajeev bought 1 pair of jeans, 3 shirts and 2 T-shirts for
TT$62.00. Use a matrix method to determine the price of each clothing item.
9
An electrician applied Kirchhoff ’s law to a circuit and obtained the following
equations:
I1 − I2 − I3 = 0
6I1 + 3I3 = 0
6I1 + 6I2 = 36
Use a matrix method to find I1, I2, I3.
______
10 Find the area of the region bounded by the curve = √1 + x3 , the x-axis and the
line x = 0.5, using
(a) the trapezium rule with 6 equal subintervals
______
(b) the binomial expansion of √1 + x3 up to and including the term in x6.
418
MODULE 3tCHAPTER 17
11 Find the first three non-zero terms in the Maclaurin expansion of e−x2.
Hence evaluate
0.5
∫0
e−x dx.
2
12 A body moves in a straight line so that its distance x metres from the origin
d2x + ω2x = 0. Solve the equation for x given
after time t seconds is given by ___
dt2
dx
2π.
___
that x = λ and
= 0 when t = ___
ω
dt
13 The differential equation describing the variation of capacitor charge in an
alternating current circuit is given by
dq
d 2q
___
+ 2 ___ + 2q = 50 cos 2t
2
dt
dt
Find an expression for the charge q of the capacitor in the circuit at any time t
dq
seconds, given that t = 0, q = 0 and ___ = 0.
dt
14 A cricket coach has 20 players from which to choose a team of 11 for the first
one-day international match. His 20 players consist of 3 wicket-keepers,
8 batsmen and 9 bowlers. Two of the bowlers can bowl spin.
(a) In how many ways can the coach choose his team if the two spinners must
be on the team?
(b) In how many ways can the coach choose his team if there must be
four batsmen, one wicket-keeper and six bowlers?
(c) How many of the choices will contain at least one spin bowler?
15 A manufacturer estimates that the monthly output at a certain factory is
given by the Cobbs–Douglas function Q(K, L) = 100K0.4 L0.6 units, where K
is the capital investment measured in units of TT$1000 and L the size of the
labour force measured in worker-hours.
∂Q
(a) Find the marginal productivity of capital ___ and the marginal
∂K
∂Q
productivity of labour ___ when the capital expenditure is TT$600 000 and
∂L
the level of labour is 825 worker-hours.
(b) The manufacturer wishes to increase output. Should the manufacturer
consider adding capital or increasing the labour level?
16 A petroleum company currently employs 20 engineers and 80 technicians. The
weekly output of the company is given by Q(x, y) = 100x + 400y + x2y − x3 − y2
units, where x is the number of engineers employed and y is the number of
technicians.
(a) Estimate the change in the weekly output if the company adds one more
engineer to the workforce and the number of technicians remains unchanged.
(b) What is the change in the weekly output if the company adds one more
technician and the number of engineers remains unchanged?
17 The output of a certain factory is given by Q = 2s3 + s2u + u3 units, where
s is the number of hours of skilled labour and u is the number of hours of
unskilled labour. The existing labour force consists of 40 hours of skilled and
30 hours of unskilled labour. The factory has to maintain the output at its
current level but increase its skilled labour by 1 hour. Estimate the change in
unskilled labour that should be made.
419
M O DUL E 3
18 A vendor buys a machine to knead flour for the making of roti. When the
machine is t years old the revenue generated by using this machine is
y (t) = 4000 − 17t2 TTdollars per year and that the cost to operate and service
the machine is x (t) = 1500 + 8t2 TTdollars per year.
(a) What is the useful life of the machine?
(b) What is the net profit generated by the machine during its useful life?
19 The present value (PV) of an income stream that is deposited continuously at
a rate x(t) into an account that earns interest at an annual rate r compounded
continuously for a term of T years is PV =
T
∫0 x(t)e−rt dt
Cyd-Marie is trying to decide between two investments. The first investment
costs TT$2000 and is expected to generate a continuous income stream of
x1(t) = 4000e0.04t TTdollars per year. The second investment costs TT$5000
and is estimated to generate an income stream at a rate of
x2(t) = 6000 TTdollars per year. The annual interest rate on the money
deposited is fixed at 5% compounded continuously for the next 10 years.
Which investment is better for this time period?
20 (a) A soft drink can is a cylinder h cm tall and with radius r cm. Its volume
is given by V = πr2 h. A particular can is 15 cm tall with radius 4 cm.
Estimate the change in volume that results if the radius is increased by
1 cm and the height remains 15 cm.
(b) For the soft drink can the surface area is given by S = 2πr2 + 2πrh.
Estimate the change in surface area if
(i) the radius is increased from 4 cm to 5 cm while the height stays
at 15 cm
(ii) the height is decreased from 15 cm to 14 cm while the radius stays
at 4 cm.
420
MODULE 3tCHAPTER 17
Module 3 Tests
Module 3 Test 1
1
(a) How many three-letter words can be formed from the letters in the word
NUMBERS? How many of these three-letter words
(i) contain the letter S
(ii) do not contain any vowel?
[8]
(b) Find the general solution of the differential equation
d2x − 3 ___
d x − 4x = 50 sin 2t. Given that x = 0 when t = 0 and that x
___
dt
d t2
remains finite as t → ∞, find x in terms of t.
[11]
(c) A box contains 3 blue balls and 5 red balls. Two balls are randomly taken out
of the box one at a time and without replacement. Find the probability of
2
(i) drawing two red balls
[3]
(ii) drawing a blue ball and a red ball.
[3]
(a) (i) Find the number of different selections of five letters from the word
SELECTION.
[5]
(ii) Find the number of different arrangements of these five letters.
[4]
(b) An experiment is carried out with three coins. Two of the coins are biased
1, while the third coin is fair,
so that the probability of obtaining a head is __
3
1. The three
so that the probability of obtaining a head on any throw is __
2
coins are tossed and the events X and Y are defined as:
X occurs if all three coins show the same result;
Y occurs if the unbiased coin shows a tail.
Find (i) P(X)
(ii) P(X ∪ Y)
(iii) P(X ∩ Y′).
[7]
dy
(c) Solve the differential equation cos x___ + y sin x = e x cos2 x given that
dx
y = 2 when x = 0.
3
[9]
(a) A financial institution in Trinidad and Tobago distributes grants to three
groups of students: primary, secondary and tertiary students. The table
below shows the number of recipients for the years 2005, 2006 and 2007.
Year
Primary
Secondary
Tertiary
Total (TT$)
2005
10
100
50
85 000
2006
15
120
70
119 000
2007
18
105
100
136 250
421
M O DUL E 3
Assume that the values of each grant are x, y and z respectively per year.
(i) Obtain the system of linear equations based on the information given. [2]
(ii) Write the linear equations in the form Ax = b.
[2]
(iii) Find A−1.
[6]
(iv) Hence, solve the equations to find the value of each grant.
[4]
(b) Solve for x the equation
x+2
−4
x−3
1
−1
4
2
=0
2
x−1
|
|
[11]
Module 3 Test 2
1
(a) The digits of the number 1213348 are rearranged to form an odd number.
How many odd numbers can be formed? Explain your answer clearly. [6]
(b) Four letters of the word SELECTED are chosen.
(i) Find the number of different selections that can be made.
[5]
(ii) Find also the number of different arrangements of the four letters
chosen.
[4]
(c) A bag contains 4 red balls and 6 green balls. Three balls are drawn at
random from the bag, without replacement. Calculate the probability that
(i) all the balls are red
[2]
(ii) at least one ball of each colour is drawn
[4]
(iii) at least two red balls are drawn given that at least one of each colour is
drawn.
[4]
2
1
1
1
c = (a − b)(c − a)(b − c).
(a) Show that a
b
a2 b2 c2
(b) A system of linear equations is given by
|
|
[8]
2x + y − z = 1
3x + 4y + 2z = 7
9x + 7y − z = 10
(i) Write the equations in matrix form.
[2]
(ii) Show by row reduction that the system has an infinite set of
solutions.
[5]
(iii) Hence, solve the equations.
[4]
1
(c) Given that A = 0
3
(
Find (i) AB
(ii) A−1
422
1
1
−1
3
0
1 and B = 3
2
−3
)
(
−2
2
4
1
−1
1
)
[4]
[2]
MODULE 3tCHAPTER 17
3
dy
(a) Solve the differential equation ___ − 2y = e3x given that y = 1 when
dx
x = 0.
[7]
(b) Use the substitution x = eu to find the general solution of the differential
d2y
dy
equation x2 ___2 − x ___ − 15y = 0.
[8]
dx
dx
dy
d2y
(b) Solve the differential equation ___2 − ___ − 6y = 6x2 + 8x + 11, given that,
dx
dx
dy
___
= 1.
[10]
when x = 0, y = 0 and
dx
423
Unit 2 Multiple Choice Tests
Multiple Choice Test 1
__
1
The argument of the complex number −3 − (√3 )i is
5π
π
A __
B ___
6
6
5π
π
C −__
D − ___
6
6
2
The complex number z such that z2 = −5 + 12i is
I
3
4
5
6
IV −2 + 3i
A
I only
B II and III only
C
IV only
D I and IV only
The complex number (1 − i)5 is equivalent to
A
−4 + 4i
B −4 − 4i
C
−2 + 2i
D −2 − 2i
The locus of z where |z − 1 + 2i| = 2 is:
A
A circle centre (1, −2) and radius 2
B
A circle centre (−1, 2) and radius 2
C
A circle centre (1, −2) and radius √2
D
A circle centre (−1, 2) and radius √2
__
(sin 2x + 3) esin 2x+3
D 2 cos (2x + 3)esin (2x + 3)
A curve is defined parametrically by x = t2 + t, y = 2t − 1. The values of t
where the curve meets the line y = 2x − 3 are
1 and −1
2 and −2
B 0 and 1
D 3 and −1
dy
For x3y + 2xy = 4y, ___ at the point where x = 0 and y = 0 is equal to
dx
A 0
B −2
1
1
C __
D −__
2
2
C
424
__
The gradient of the curve y = (x + 1) ln (x + 2) at x = 0 is
1 + ln 2
A ln (2)
B __
2
1
C __
D ln 4
2
d [esin (2x + 3)] is
___
dx
A 2 cos 2x(esin 2x+3)
B (sin 2x + 3)e2 cos 2x
A
8
II 2 + 3i
III −2 − 3i
C
7
2 − 3i
9
dy
Given that y = x ln x, then ___ is
dx
A ln x
B ln x + 1
x2 ln x + ln x
C x + ln x
D __
2
10 The function f(x) is defined by f(x) = cos 2x then f ″(x) is
A
B −4 cos 2x
4 sin 2x
−4 sin 2x
D 4 cos 2x
2
11 __________
in partial fractions is
x2 + 4x + 3
1 + _____
1
1
1 − _____
B _____
A _____
x+1 x+3
x+1 x+3
1
1
1
1
_____
_____
D − _____
C − _____
x+1+x+3
x+1−x+3
C
12
13
1
dx is
∫___________
4 + (x − 1)2
A
1 ln(x − 1)2 + c
__
C
(
(
)
1
C
ln 2
π + __1 ln 2
__
4 2
1 ln 2
B __
2
π + ln 2
D __
4
ln x dx is equal to
∫____
x
A
C
15
)
1 tan−1(x − 1) + c
B __
2
x−1 +c
D tan−1 _____
2
1 + x dx is
∫0 ______
x2 + 1
A
14
2
x−1 +c
1 tan−1 _____
__
2
2
ln x + c
____
2x2
1 ln x + c
__
x2
1 (ln x)2 + c
B __
2
1 ln x + c
D ___
2x2
∫xex dx is
2
A
C
1 e x2 + c
__
2
1 e x2 + c
___
2x
B x2e x + c
2
D xe x + e x + c
2
2
50
16
∑ (3r − 2) =
r=1
A
3700
B 3823
C
3825
D 3725
17 The fourth term of (1 − 2x2)9 is
A
−672x6
B 1249x6
C
4x6
D 28x3
425
18
(n + 1)!
n! + _______
_______
=
(n + 1)! (n + 2)!
n!
A _______
(n + 2)!
2n + 3
C _____________
(n + 1)(n + 2)
n(2n + 3)
B _________
n+2
2n(n + 3)
D _____________
(n + 2)(n + 3)
19 In the expansion of (2 + 3x)n, the coefficients of x3 and x4 are in the ratio 8:15.
The value of n is
A
6
B 8
C
12
D 5
20 The first three terms in the expression of (1 − x)−1 are:
A
1 − x − x2
B 1 − x + x2
C
1 + x − x2
D 1 + x + x2
21 The expression (2 + 3x)−4 is valid for
A
−1 < x < 1
C
−1 < 3x < 1
22 The series
2
2
__
B −__
3<x<3
3
3
__
D −__
2<x<2
2 x is a Maclaurin series for which function?
∑ ( ____
n! )
∞
n n
n=0
A
ex
B e2x
C
2ex
D e4x
23 The second order Taylor polynomial for e4x about x = 1 is:
A
e4 + e4 (x − 1)
C
1+x
(x − 1)2
B e4 + e4 (x − 1) + e4 _______
2!
D 4 + x + x2
24 The equation ex = 25x − 10 has a root in the interval:
A
[−1, 0]
B [−2, −1]
C
[0, 1]
D [2, 3]
25 Linear interpolation is applied to the equation 2 sin x − x = 0 to find a first
approximation x to the root in the interval [1.5, 2]. What is the value of x?
A
1.8659
B 1.6500
C
1.8969
D 1.8955
Questions 26 and 27 refer to the following series:
2 − 6 + 18 − 54 − . . .
26 The series is:
A
finite
B infinite
C
convergent
D periodic
27 The rth term of the series is
426
A
2(3r)
B 2(−3)r − 1
C
2(3)−r + 1
D 3(2)r − 1
28 The next iteration value of the root of x3 + 3x2 + 5x + 9 = 0 using the
Newton–Raphson method, if the interval is −2.5, is:
A
−2.456
B −2.457
C
−1.750
D −2.458
Questions 29 and 30 refer to the following:
A sequence an satisfies a1 = 1 and an+1 = 2an + 4.
29 The first three terms of the sequence are
A
1, 6, 16
B 6, 16, 36
C
0, 1, 6
D 2, 6, 18
A
5(2n − 1) + 4
B 5(2n − 1) − 4
C
4(2n − 1) + 5
D 2(41(n − 1)) + 4
30 an is
31 If a fair coin is tossed three times, what is the probability of getting exactly two heads?
3
B __
4
8
1
C __
D 0
8
32 Using an ordinary deck of 52 playing cards, what is the probability of drawing
three black cards in a row if each drawn card is not returned to the deck?
4
1
A __
B ___
7
33
3
2
C ___
D ___
25
17
33 The number of permutations of the letters in the word HISTORY in which the
letters O and R are not together is
A
1
__
A
1440
B 3600
C
5040
D 720
34 Nine different books are to be arranged on a bookshelf. Four of these books are
written by Mohammed, two by Mungal and three by Miss Ramcharitar. The
number of possible permutations of the books if the books by Miss Ramcharitar
are separated from each other is
A
30 240
B 282 400
C
151 200
D 357 840
35 A committee of five people is to be selected from a group of six men and
nine women. If the selection is made randomly, what is the probability that the
committee consists of three men and two women?
240
1
B __
A ____
3
1001
1260
13
C ___
D ____
3003
18
36 If we toss a coin five times, what is the probability of obtaining exactly one head?
3
5
A ___
B ___
32
32
2
1
C ___
D ___
32
32
427
37 A bag contains six white balls, three red balls, and one blue ball. If one ball is
drawn from the bag, what is the probability it will be red?
A
0.1
B 0.3
1
C 0.6
D __
3
38 A and B are two independent events. Given that P(A) = 0.2 and P(B) = 0.5,
which of the following are true?
I
P(A ∪ B) = 0.7
II P(A ∩ B) = 0.10
III P(A B) = 0.2
IV P(A B′) = 0.5
A
I and II only
B II and III only
C
II and IV only
D III and IV only
39 A committee of five members is to be selected from six seniors and four juniors.
What is the number of ways in which this can be done if the committee has at
least one junior?
A
252
B 120
C
246
D 6
(
1
Questions 40–42 refer to the matrix 1
a
40 The cofactor of 3 is
−1
1
0
2
3
5
A
a
B −a
C
0
D 1
)
41 The determinant of the given matrix is
A
a−2
B 10 − 5a
C
5a − 10
D 5a + 10
42 Given that the matrix is non-singular, then
A
a=2
B a = −2
C
a≠2
D a ≠ −2
43 The particular integral of the differential equation
2
dy
dy
___
+ 3 ___ − 4y = cos 2x
dx
dx2
is of the form:
A
a sin 2x
B a cos 2x + b sin 2x
C
a cos x + b sin x
D a cos x
A
y = Ae2x + B
d2y
dx
B y = Ae2x + Bex
C
y = (Ax + B)e2x
D y = Ax + B
dy
dx
44 The general solution of the differential equation ___2 + 4 ___ − 4y = 0 is
428
dy
dx
45 The general solution of the differential equation x ___ − 2y = x5 is
A
C
1 x5 + c
y = __
3
y = x5 + x 2
1 x5 + cx 2
B y = __
3
1 x + cx 2
D y = __
3
Multiple Choice Test 2
1
i49 is
A
2
C i
3 + i in the form a + ib is
_____
2−i
A 1+i
C
3
4
5
6
7
B −1
1
1−i
D −i
6+i
B __
5
1 + __
1i
D __
5 5
−2 − 2i in the form re iθ is
__ __
π
B 2 √2 e− 4 i
__ ___
3π
D 2 √2 e− 4 i
A
2 √2 e 4 i
C
2 √2 e 4 i
__
π
__
__
3π
___
(2 + i) 4 is
A
7 − 24i
B 7 + 24i
C
−7 − 24i
D −7 + 24i
The equation of a curve is −x2 + y2 − 2xy − 2 = 0. The gradient function of
the curve is
A
x+y
_____
y−x
1 + 2y
B _______
2y − 2x
C
1 − 2y
_______
1 + 2y
D _______
2y + 2x
2y + 2x
The gradient function of y = ln (3x2 + 5) is
6
6x
A _______
B ______
x2 + 5
3x2 + 5
6x
1
C _______
D _______
3x2 + 5
3x2 + 5
dy
Given that y = sin−1(3x), ___ is
dx
3x
1
_______
_______
B ________
A ________
2
√1 − 9x
√1 − 9x2
______
C
√
1
______
1 − x2
__
9
3
______
D _______
√9 − x2
429
8
9
10
dy
Given that y = tan−1(x2 + 2), ___ is
dx
2x
A ___________
B
1 + (x2 + 2)2
1
C ___________
D
1 + (x2 + 2)2
C
x2
B + _____
Given that ____________
≡ A + _____
x+1 x+2
(x + 1)(x + 2)
A
A = 1, B = 1, C = −4
B A = 0, B = 1, C = −4
C
A = 1, B = −1, C = 4
D A = 1, B = 1, C = 4
1
________
dx is
∫ √_______
9 − 4x2
A
C
11
1 |n|x2 + 2
___
2x
2x
___________
1 + (x + 2)2
( )
( )
( )
( )
2x + c
sin−1 ___
3
2x + c
1 sin−1 ___
__
2
3
2x + c
B cos−1 ___
3
2x + c
1 sin−1 ___
D __
3
6
∫sin 4θ cos 4θ dθ is
1 sin2 4θ + c
1 sin2 4θ + c
__
B __
4
8
1
1 cos2 4θ + c
C −___
D __
16 cos 8θ + c
4
12 The parametric equations of a curve are y = 2t2 + 5, x = t + 3
The gradient of the curve is
1
1
B __
A __
4t
2t
1
C 4t
D __
2
dy
13 The parametric equations of a curve are x = 2 cos θ, y = 2 + 2 sin θ. Then ___ is
dx
1
3θ
A cosec2 θ
B −__
cosec
2
1
__
3
C
D −cosec2 θ
cosec θ
2
A
14
∫sin x ecos x dx is
A
−ecos x + c
B ecos x + c
C
sin2 x ecos x + c
D sin x esin x + c
dy
dx
15 Given that y = 2x + 1, ___ is
A
(x + 1)2x
C
2x + 1 ln 2
B 2x + 1
1 2x + 1
D ___
ln 2
16 The first three terms of the Maclaurin expansion of ex are
A
C
430
2
3
x + __
x
x + __
2
6
2
x
1 − x − __
2
x2
B 1 + x + ___
2
2
x3
x + __
D x − ___
6
2
17 Given that y(0) = 2, find the first three terms of the Maclaurian expansion of
dy
y where y2 + 2y ___ = 0.
dx
2
x3
x
A −x + __ − __
2
6
2
x
__
C 2+x+
2
n
18
2
x
B 2 − x + __
4
2
x
__
D 2−x−
4
∑ (5 − 4r) is
r=1
A
n(3 − 2n)
B 5 − 2n(n + 1)
C
n(2n + 3)
D n(2n − 3)
50
19
∑ (5 − 4r) is
r=1
A
4850
B −4850
C
5095
D −5095
dy
dx
from a second order Taylor polynomial about x = 0 is
20 Given that y(x) is the solution to ___ = 2y2 + 1, y(0) = 2, the value of y(0.2)
A
9.00
B 5.24
C
2.35
D 3.72
21 The root of the equation x3 = 1 − x is found by using the Newton–Raphson
method. The successive iteration values of the root are given in the table below:
Iteration number
Value of root
0
0.5
1
0.7143
2
0.6823
3
0.6823
4
0.6823
At which iteration number will you first trust at least two significant digits in
the answer?
A
1
B 2
C
3
D 4
Questions 22–24 refer to the following: A sequence of positive terms u1, u2, u3, . . . is
given by u1 = 2 and un+1 = 2un for n ≥ 1.
22 The first four terms of this sequence are
A
2, 4, 6, 8
B 2, 4, 8, 16
C
4, 8, 16, 32
D 1, 2, 4, 8
431
23 The sequence is
A
convergent
B decreasing
C
periodic
D divergent
24 un is
1
2n
B ___
2n
n
D 2n
C __
2
Questions 25 and 26 refer to (1 + 2x)−3 = f(x).
A
25 The first three terms of the expansion of f(x) are
A
1 − 6x + 24x2
B 1 − 6x + 12x2
C
1 − 3x + 12x2
D 1 − 3x − 24x2
26 The expansion of f(x) is valid for
A
−2 < x < 2
C
−1 < x < 1
1
1
__
B −__
4<x<4
1
1
__
D −__
2<x<2
Questions 27 and 28 refer to the following: The sum, Sn, of the first n terms of an
arithmetic progression is given by Sn = pn + qn2, where S3 = 6 and S5 = 11.
17
1 , q = ___
27 A p = ___
10
10
B p = 10, q = 17
17 , q = ___
1
C p = ___
10
10
7
1 , q = __
D p = __
5
5
28 The nth term of the series is
1 (2n + 8)
A 2n + 8
B ___
10
1 (2n + 16)
1 (2n + 4)
D ___
C ___
10
10
29 The equation 2 ln x + x = 2 has a root in the interval
A
[1, 2]
B [2, 3]
C
[3, 4]
D [4, 5]
30 The coefficient of the x6 term in the Maclaurin expression for cos (2x) is
A
C
1
___
720
4
−___
45
1
B −___
720
D 0
d2y
dy
Questions 31–33 refer to ___2 − 3__ + 2y = x + 1.
dx
dx
31 The CF of the equation is
432
A
y = (Ax + B)e2x
B y = Aex + Be2x
C
y = Ae−x + Be−2x
D y = ex[A cos 2x + B sin 2x]
32 The PI of the equation is
A
C
y = 2x + 5
1x
y = __
2
1x + 1
B y = __
2
1 x + __
5
D y = __
4
2
dy
dx
7
5
1
1
1 x + __
1
__
__
__
__
x
2x
x
B y = −2e + e2x + __
A y= e + e + x
4
4
2
2
2
2
1 e−x + __
1 e−2x + __
1x
1 e−x + __
1 e−2x + __
1x + 1
C y = __
D y = __
2
2
2
2
2
2
dy
Question 34 and 35 refer to the differential equation __ − 2y = 2e4x.
dx
34 The general solution of the differential equation is
33 The solution of the equation when x = 0, y = 1 and ___ = 2 is
A
y = e4x + c
B y = 1 + ce2x
C
y = 3 + ce4x
D y = 2 + ce2x
35 The solution of the differential equation when x = 0 and y = 4 is
A
y = e4x + 3
B y = 1 + 3e2x
C
y = 3 + e4x
D y = 2 + 2e4x
Questions 36–38 refer to the following: Three boys and five girls go to the movies.
36 The total number of arrangements for all the children sitting in a row is
A
40 320
B 4032
C
5040
D 80 640
37 The number of arrangements with all the boys sitting next to each other is
A
80 640
B 5040
C
720
D 4320
38 The number of arrangements with no two boys sitting next to each other is
A
14 400
B 36 000
C
720
D 3600
39 A set of 20 students is made up of 10 students from each of two different yeargroups. Five students are to be selected from the set and the order of selection is
unimportant. The number of selections in which there are at least two students
from each of the two year-groups is
A
15 504
B 5400
C
10 800
D 54 000
40 A bag contains three red balls and three green balls. Balls are drawn from the
bags at random, one by one and without replacement. The probability that the
first three balls drawn are red is
6
1
B ___
A __
30
2
1
2
D ___
C ____
20
120
433
Questions 41–43 refer to the equations
x + 5y + az = p
2x − 4y + z = q
4x + 6y + 7z = r
where a, p, q, r are constants.
41 The determinant of the coefficient matrix is
A
56
B 28a − 84
C
84 − 28a
D a−3
42 The set of equations have a unique solution when
a≠3
B a=3
1
1
C a = __
D a ≠ __
3
3
43 When a = 3, p = −18, q = 7, r = −29 the equations have
A
A
a unique solution
B no solution
C
an infinite set of solutions
D two solutions
(
1
Questions 44 and 45 refer to the matrix M = 2
3
44 |M| is
A
−1
B 1
C
2
D −3
)
3
5
8
4
−1 .
4
11
8
9
1
−1
−1
45 M −1 is
A
C
(
(
−28 11 −1
8 −1
−20
1
23 −9
−28
11
−1
−20
8
−1
)
23
−9
1
B
)
(
(
28
20
23
28 20 −23
D −11 −8
9
1
1 −1
Multiple Choice Test 3
1
2
434
2 + i can be written as
_____
3−i
5i
5 + __
A __
9 9
5i
7 + ___
C ___
10 10
__
The argument of −1 − √3 i is
π
A __
3
2π
C −___
3
)
1 + __
1i
B __
2 2
D 1+i
2π
B ___
3
π
D −__
3
)
3
π + i sin __
π is equivalent to:
2 cos __
4
4
__
__
1__ i
1__ + ___
A √2 + √2 i
B ___
2
2
√
√
__
__
__
__
2√2 i
2√2 + ____
√2
√2
i
D ___
C ____
− ___
4
4
2
2
π + i sin __
π 6 is
cos __
6
6
A −1 + i
B −e −πi
__
__
3
√
√
___
___
π
i
C −e
D
+ 3i
2
2
[ ( )
( )]
4
[ ( )
5
Given that |z − 1 + 4i = 3|, the locus of z is
6
7
8
9
( )]
A
A circle centre (−1, 4) and radius 3
B
A circle centre (1, −4) and radius 3
C
A circle centre (−1, 4) and radius √3
D
A circle centre (1, −4) and radius √3
__
__
The first derivative of e cos (2x) with respect to x is
A
esin (2x)
B −2 sin (2x) ecos 2x
C
ecos 2x
D cos 2x ecos (2x)−1
1 .
A curve is defined parametrically by the expression x = t3 + 2, y = _____
t+2
The gradient of the curve, in terms of t, is
3t2
1
B −_________
A −_______
(t + 2)2
3t2(t + 2)2
C
−3t2(t + 2)2
C
1
2 + _____
_____
x−3 x+2
D 3t2
3x − 4
In partial fractions the expression ____________
may be written as
(x − 3)(x + 2)
−1 + _____
2
2
1 + _____
B _____
A ______
x+2
x−3
x−3 x+2
1
−2 − _____
D _____
x−3 x+2
1
may be expressed as
∫___________
4x2 + 4x + 5
A
x+1 +c
1 tan−1 _____
__
B
1 ln | 4x2 + 4x + 5 | + c
______
4
(
)
2
8x + 4
1 tan−1 ______
2x + 1 + c
C __
2
2
2x + 1 + c
1 tan−1 ______
D __
4
2
dy
x + 1 , then ___
10 Given that y = ln _____
is
x+2
dx
x+2
B
A _____
x+1
3
2 − _____
D
C _____
x+1 x+2
(
(
)
)
(
)
(x + 2)2
_______
x+1
1
____________
(x + 1)(x + 2)
435
11
∫sin2 (2x) dx is
A
C
12
1 sin3 (2x) + c
__
6
1 sin 4x + c
1 x − __
__
8
2
1 x − __
1 sin 4x + c
B __
4
2
D x − sin 4x + c
∫xe3x dx may be expressed as
1 xe3x + __
1 e3x + c
__
B 3xe3x − 9e3x + c
9
3
1 e3x + c
1 x2e3x − __
1 e3x + c
1 xe3x − __
D __
C __
9
9
3
3
dy
13 The value of ___ when x = 1, y = 0 for the function xy + 2y2 + 3x = 3 is
dx
3
A __
B −3
5
1
C 3
D __
3
x+2
14 __________
dx is
x2 + 4x + 5
A
∫
A
(x + 2) ln | x2 + 4x + 5| + c
C
ln | x2 + 4x + 5| + c
B 2 ln | x2 + 4x + 5| + c
1 ln | x2 + 4x + 5| + c
D __
2
dy
dx
15 If y = tan−1 (x2) then ___ is
A
x2
______
1 + x4
2x
B ______
1 + x4
C
x2
______
1 + x2
2x
D ______
1 + x2
16
un
n
1
The term that best describes the behaviour of the sequence {un} shown above is
A
finite
B periodic
C
divergent
D convergent
17 The 6th term in the sequence that is defined by the relation un = (−1)n(n + 1)2 is
436
A
49
B −49
C
−36
D 36
n
18
∑(1 − r ) =
2
r=1
A
n(n + 1)(2n + 1)
1 − ______________
6
n 2
B −__
6 (2n + 3n − 5)
n(n + 1)(2n + 1)
n(n2 + 3n + 5)
_______________
D __
6
6
(n
+
1)!
19 The expression _______ can be simplified and written as
(n − 2)!
3
A n −1
B n2 + n
C
C
n3 − n
D n2 − n
9
20 The term independent of x in the expansion of ( x2 − __2x ) is
A
5376
B 84
C
−84
D −5376
Questions 21–23 refer to S as defined below:
S = 4 + 7 + 10 + 13 + 16 + 19 + 22
21 S is described as
A
C
a finite series
an infinite sequence
B an infinite series
D a finite sequence
22 The general term of S is
A
C
(3r − 1)
(r + 3)
B (3r + 1)
D (2r + 1)
23 S may be written as
7
A
7
∑(3r + 1)
B
C
∑
D
∑(3r − 1)
r=n
r=1
7
∞
(3r − 1)
r=1
∑(3r − 1)
r=1
1
__
24 The fourth term in the power series expansion of (1 + 2x)−2 , |x| < __21 is
A
3 x2
__
2
5 3
−__
2x
A
x e
e + xe + __
2!
x − 12
1 + (x − 1) + ______
2!
5 x3
B __
6
3 2
C
D −__
2x
25 The Maclaurin series of e2x up to and including the term in x3, ∀x ∈ ℝ is
1 2 __
1 x3 . . .
1 3...
1x2 + __
A 1 + x + __
B 1 + 2x + __
4x + 8 x
2
6
1 x3 . . .
4 x3 . . .
D 1 + 2x + 2x2 + __
C 1 + 2x − 2x2 + __
8
3
26 The second order Taylor’s expression for f (x) = ex about x = 1 is
C
2
[
(x − 1)2
B e 1 + (x − 1) + _______
2!
x2
D 1 + x + __
2!
]
437
27 The Maclaurin series for ln (x + 1) is valid for
A
−1 < x < 1
B −1 < x ≤ 1
C
−1 ≤ x ≤ 1
D x < −1
28 The function f (x) = x3 + 10x2 + 10x − 4 has a root in the closed interval
A
[−1, 0]
B [2, 3]
C
[5, 6]
D [0, 1]
29 An estimate of cos (0.1) using the first three terms of the Maclaurin series of
cos x is
A
0.995
B 0.996
C
0.9951
D 0.994
30 The sum of the first n terms of a series is [ 1 − ( __23 ) ]. The value of the second
n
term is
1
B __
3
5
2
C __
D __
9
9
31 The number of ways in which three boys and two girls can sit in a row if they
can sit in any position is
A
1
A
C
3!
3! × 2!
B 2!
D 5!
32 In how many ways can the letters of the word POWER be arranged so that the
O and the E are always together?
4! × 2
B 5!
5!
C __
D 4!
2!
33 How many six-digit odd numbers can be formed using the digits of the number
112 431?
A
A
480
B 20
C
80
D 160
34 A team of eleven players is to be chosen from 20 players. Given that the two
oldest players must be chosen, the number of ways in which the team can be
chosen is
A
20C
C
20C
9
9
×
20C
2
B
20C
D
18C
11
9
35 The number of ways of choosing four letters from the word ADVANCE is
438
A
15
B 30
C
45
D 25
36 In a recent survey, it was reported that 75 per cent of the population of a certain
county lived within ten miles of the largest city and that 40 per cent of those
who lived within ten miles of that city lived in condos. If a resident of this
county is selected at random, what is the probability that the person lives in a
condo within ten miles of the largest city?
A
0.15
C
0.30
B 0.10
D 0.53
37 Events A and B are such that P(A) = __31 and P(A′ ∩ B′) = __61. P(A ∪ B) is
1
1
A ___
B __
18
6
5
1
D __
C __
2
6
1 2 1
1 0 4
38 Given that A = −1 0 1 and B = −1 1 1 , AB is
3 2 2
2 1 0
3
5
6
6
1 3
B 1
A 1 1 −4
1 −4
5 4 14
5 −4 14
3
6
6
−1
1 3
D
C
1 −4
1
1 1 −4
5 −4 14
−5 4 14
(
(
(
)
(
)
)
(
)
(
(
1 0
0
)
)
)
39 The determinant of A where A = −1 2 1 is
A
−2
C
4
1 3 2
B −4
D −6
d2y
Questions 40–42 refer to the equation ___2 − y = 4x.
dx
40 The complementary function of the equation is:
A
y = A cos x + B sin x
B y = e x (A cos x + B sin x)
C
y = Aex + Be−x
D y = Ae x cos x
41 The particular integral of the differential equation is
A
y = 4x
B y = 4x + 2
C
y=x−3
D y = −4x
A
y = 3ex − e−x − 4x
dy
dx
B y = 3 cos x + sin x − 4x
C
y = ex + e−x − 4x
D y = e x (cos x + sin x) − 4x
42 The solution of the equation when x = 0, y = 2 and ___ = 0 is
(
1 2
1
3 3
a
)
43 The matrix A = −1 1 1 . For what value of a is A singular?
A
1
C
−3
B 3
D −1
439
dy 1
Questions 44 and 45 refer to the equation __ − _
y = x2
dx x
44 The general solution of the equation is
1 3
1 3
y = __
B y = __
2x
2 x + cx
1 x3 + cx2
C __
D y = x3 + cx2
2
45 The solution of the equation when x = 1, y = 2 is
3
1 3
1 3 __
A y = __
B y = __
2x
2x + 2x
3 2
1 3 __
D y = x3 + x 2
C y = __
2x + 2x
A
440
Index
A
addition
on Argand diagrams 15–16
complex numbers 5
matrices 323, 324
addition rule 271
alternating sequences 158
approximations, binomial
expansion use for 208–9
AQE see auxiliary quadratic
equations
area
trapezium 146
under curve 146
Argand diagrams
addition on 15–16
arguments of complex
numbers 17–18
modulus of complex numbers
16, 18
multiplication by i 16
representation of complex
numbers on 15
subtraction on 15–16
see also loci of complex
numbers
arguments, of complex
numbers 17–18
arithmetic progressions
(AP) 216–21
common differences 216
proving sequences are 220–1
sum of first n terms 218–20
summary 236
terms of 216
augmented matrices 348, 358
auxiliary quadratic equations
(AQE) 388–91
when roots are complex
390–1
when roots are real and
distinct 389–90
when roots are real and equal
388–9
B
binomial expansion (theorem)
approximations with 208–9
for integers 200–3
partial fractions and 209–12
proofs 200–1
for real numbers 205–7
region of validity 205
series properties 202–3
summary 213
term independent of x 203–4
binomials 197
box method 274
C
Cartesian form, complex
numbers 15, 35
CF 392
chain rule 43
circle, as locus of complex
number 27–8
Cobb–Douglas function 76–7
coefficient matrix 336
cofactors, matrices 339–41
coinciding lines 357
combinations 199–200, 285–8
distinct objects 285–7
formula 199
notation 199
permutations compared
with 291
and probability 311–14
with repetition 287–8
summary 291
common differences 216
common ratios 224
complement 298
complementary function (CF) 392
complex numbers 1972
addition 5
arguments 17–18
Cartesian form 15, 35
conjugate of 7
division 8
equality of 6
exponential form 21
extension from real
numbers 3
imaginary part 3
locus see loci of complex
numbers
modulus 16, 18
multiplication 5, 16
polar form 19
quadratic equations solution
3, 11–13
real part 3
representation on Argand
diagrams 15
square roots 9–10
subtraction 5
summary 37
trigonometric form 19
complex roots, polynomials
11, 13
conditional probability 299–302
conjugates, of complex
numbers 7
convergent sequences 157, 162
convergent series 180–3
cos x, integration of powers
of 123–4, 138–9
counting principles 271
Cramer’s rule 335–8
D
D’Alembert’s ratio test 182–3
De Moivre’s theorem 22–3
De Morgan’s laws 299
determinants 331–2
to solve simultaneous
equations 335–8
differential equations 381–411
degree of 381
first order linear see
first order linear differential
equations
homogeneous 388
441
INDEX
differential equations (Continued)
integrating factor 381
order of 381
second order linear see
second order linear
differential equations
summary 413
differentials
ex 43
ln x 42–3
standard 42
differentiation
chain rule 43
combinations of functions 50–3
composite functions 43
exponential functions 43–7
function of a function rule 43
implicit 58–62
inverse trigonometric
functions 62–4
logarithms 47–9
parametric 67–71
partial derivatives see partial
derivatives
product rule 50
quotient rule 50
second derivatives 65–6
summary 80
trigonometric functions 51–3
divergence test 182
divergent sequences 157, 162
divergent series 180–3
division, complex numbers 8
E
empty sets 298
equal matrices 323
events
exhaustive 271, 296
independent 299
mutually exclusive 271, 296
ex, differential of 43
exhaustive events 271, 296
explicit functions 58
exponential form, complex
numbers 21
exponential functions
differentiation 43–7
integration 101
442
F
factorial notation 197–8
factors, quadratic 11
first order linear differential
equations 381–6
applications 385–6
general solutions 381
integrating factor 381
summary 413
first (order) partial
derivatives 72–3
fractions
improper 85, 94–5
partial see partial fractions
proper see proper fractions
rational 85
function of a function rule 43
G
geometric progressions
(GP) 224–33
common ratios 224
convergence 231–3
proving series are 230–1
sum of first n terms 227–9
sum to infinity 229–30
summary 236
terms 224
graphical solutions, equations 242
H
half line, as locus of complex
number 29–30
harmonic series 182
homogeneous differential
equations 388
I
identity matrix 328
imaginary numbers 3
imaginary unit i 3
powers of 4
implicit differentiation 58–62
implicit functions 58
improper fractions 85, 94–5
IMVT 241, 253
independent events 299
inequality, as locus of complex
number 31–3
infinite series 168
integers 3
integral test 181–2
integrals
standard 100
trapezoidal (trapezium) rule
to estimate 146–50
trigonometric functions 121
integrating factor 381
integration 100
exponential functions 101
general forms 104, 105, 106
inverse trigonometric
functions 113–14
logarithmic functions 102–3
using partial fractions 100,
116–21
by parts 100, 112–15, 137
powers of cos x 123–4, 138–9
powers of sin x 123–4, 137–8,
140–1
powers of tan x 125, 139
products of sines and cosines
125–6
by recognition 100–7
reduction formulae 137–44
by substitution 100, 108–12,
126–32
summary 133
trigonometric functions 121–31
when numerator is differential
of denominator 104–5
intermediate value theorem
(IMVT) 241, 253
intersecting lines 355–6
intersection 298
interval bisection 242–3
intervals 146–50
inverse matrices 339, 341–3,
348–52
inverse trigonometric functions
differentiation 62–4
integration 113–14
iterative formulae 247
L
Laplace’s equation 76
L’Hôpital’s rule 164
limit laws, sequences 163–4
INDEX
limits, sequences 157, 162
linear interpolation 243–6
lines
coinciding 357
intersecting 355–6
parallel 356
ln x, differential of 42–3
loci of complex numbers
Cartesian form 35
circles 27–8
half lines 29–30
inequalities 31–3
intersecting 33–4
perpendicular bisector of line
segment 28–9
straight lines 30–1
logarithms (logs)
differentiation 47–9
integration 102–3
rules of 48
logistic model 386
M
Maclaurin expansion (series)
259–65
of common functions 265
conditions for 260
summary 266
validity 260
mathematical induction see PMI
mathematical models 417
matrices 322–70
addition 323, 324
applications 367–70
augmented 348, 358
coefficient 336
cofactors 339–41
conformability 323, 326, 328
determinants 331–2, 335–8
elements 322
equal 323
identity 328
inverse 339, 341–3, 348–52
multiplication 324, 325–30
negative of 324
non-singular 335
order 322
row echelon form 347–8
row equivalent 346
singular 334
skew-symmetric 330
square 322–3
subtraction 323–4
summary 375
symmetric 330
transpose of 330
zero 323, 324
see also row reduction; systems
of linear equations
method of differences 176–80
model for population growth
386
models, mathematical 417
modulus, complex numbers
16, 18
multiplication complex numbers
5, 16
matrices 324, 325–30
multiplication rule 271
mutually exclusive events
271, 296
N
natural numbers 3
Newton–Raphson (N–R)
method 247–52
accuracy 248
convergence failure 251–2
summary 253
non-singular matrices 335
normals 54–7
equations 56–7
gradients 54
number systems 3
O
ordinary differential
equations see
differential equations
oscillating sequences 157–8
outcomes 295
P
parallel lines 356
parametric differentiation
67–71
parametric equations 67
first derivatives of 67–70
line in three dimensions 365
second derivatives of 70–1
partial derivatives 72–9
applications 74–6
first (order) 72–3
functions of three variables
77–9
notation for 73
second (order) 73–4
partial fractions 85
and binomial expansion
209–12
integration of 100, 116–21
summary 97
partial sums 168
particular integral (PI) 392
when f(x) is exponential
function 403–5
when f(x) is polynomial
393–7
when f(x) is trigonometric
function 398–403
Pascal’s triangle 24, 197
patterns, sequences 160
periodic sequences 157–8
permutations 272–83
box method 274
combinations compared
with 291
n distinct objects 272–3
and probability 307–10
r out of n distinct
objects 274–5
repeated objects 275–7
with restrictions 277–80
with restrictions and
repetition 281–3
summary 291
perpendicular bisector of line
segment, as locus of
complex number 28–9
PI see particular integral (PI)
PMI 23, 186
and sequences 186–9
and series 190–3
summary 195
polar form, complex
numbers 19
polynomials, roots 11, 13
443
INDEX
power series 256–65
Maclaurin expansion see
Maclaurin expansion
Taylor expansion 256–9, 266
principle of mathematical
induction see PMI
probability 296–314
and combinations 311–14
conditional 299–302
definition 296–7
formulae 297
and permutations 307–10
rules 298
summary 317
tree diagrams 302–4
and Venn diagrams 298–9, 301
product rule 50
proper fractions 85
repeated linear factors 88–91
repeated quadratic factors 93
unrepeated linear factors 85–8
unrepeated quadratic factors
91–2
Q
quadratic equations, solution
with complex numbers
3, 11–13
quadratic factors 11
quotient rule 50
R
rational fractions 85
rational numbers 3
recurrence relations 161–2
recursive sequences 161, 165
reduction formulae 137–44
roots
approximations of 247–8
auxiliary quadratic equations
388–91
complex 11, 13
graphs to find 242
interval bisection 242–3
linear interpolation 243–6
summary 253
see also Newton–Raphson
(N–R) method
row echelon form 347–8
444
row reduction
simultaneous equations
solution by 352–4
to find inverse matrices 348–52
to row echelon form 346–7
S
sample points 295–6
sample spaces 295–6
scalar quantities 324
second derivatives 65–6
second order (linear) differential
equations 381, 388–411
auxiliary quadratic equations
(AQE) for see auxiliary
quadratic equations
complementary function
(CF) 392
general solutions 388, 389,
390, 392, 395
homogeneous 388–91
non-homogeneous 392–405
reducible to recognisable
form 405–11
summary 413
see also particular integral
second (order) partial derivatives
73–4
sequences 157–64
alternating 158
convergent 157, 162
divergent 157, 162
finite 157
general term 158, 160–1
infinite 157, 158
limit laws 163–4
limits 157, 162
oscillating 157–8
patterns 160
periodic 157–8
and PMI 186–9
recursive 161, 165
summary 166
terms 158–61
see also arithmetic progressions
series 168–83
binomial expansion properties
202–3
convergent 180–3
D’Alembert’s ratio test 182–3
divergence test 182
divergent 180–3
harmonic 182
infinite 168
integral test 181–2
partial sums 168
and PMI 190–3
in sigma notation 168–9
sum see sum of series
summary 184
see also geometric progressions;
power series
sets 298–9
sigma notation 168–9
simultaneous equations
solution by determinants
335–8
solution by row reduction
352–4
see also systems of linear
equations
sin x, integration of powers
of 123–4, 137–8, 140–1
singular matrices 334
skew-symmetric matrices 330
square matrices 322–3
square roots, complex numbers
9–10
standard differentials 42
standard integrals 100
statistical experiments 295
straight line, as locus of complex
number 30–1
subintervals 146–50
subtraction
on Argand diagrams 15–16
complex numbers 5
sum of series 169–80
method of differences use
176–80
standard results use 169–74
summation laws 169–70
symmetric matrices 330
systems of linear equations 355
consistent 355, 358
geometrical interpretation
365–6
inconsistent 358
INDEX
infinite solutions conditions
358, 362
no solutions conditions
358, 361
solution by matrices 343–6
three unknowns 358–65
two unknowns 355–7
unique solution conditions 358
see also simultaneous equations
T
tan x, integration of powers
of 125, 139
tangents 54–7
equations 56–7
gradients 54
Taylor expansion (series) 256–9,
266
trapezium, area 146
trapezoidal (trapezium) rule
146–52
tree diagrams 302–4
trigonometric form, complex
numbers 19
trigonometric functions
differentiation 51–3
integrals 121
integration 121–31
reduction formulae for 137–9
see also inverse trigonometric
functions
U
union 298
V
Venn diagrams, and
probability 298–9, 301
Z
zero matrix 323, 324
445