Mathematics (T) PELANGI BESTSELLER STPMText TERM 1 Lee Yoon Woh Tan Guan Hin Tey Kim Soon 1 COPIES SOLD MORE THAN MPRE-U
Mathematics (T) Lee Yoon Woh Tan Guan Hin Tey Kim Soon PRE-U TERM 1 © Penerbitan Pelangi Sdn. Bhd. 2022 All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of Penerbitan Pelangi Sdn. Bhd. ISBN: 978-967-2856-93-1 eISBN: 978-967-2878-08-7 (eBook) First Published 2022 Tel: 03-8922 3993 Fax: 03-8926 1223 / 8920 2366 E-mail: [email protected] Enquiry: [email protected] Printed in Malaysia by The Commercial Press Sdn. Bhd. Lot 8, Jalan P10/10, Kawasan Perusahaan Bangi, Bandar Baru Bangi, 43650 Bangi, Selangor Darul Ehsan, Malaysia. Please log on to https://plus.pelangibooks.com/errata/ for up-to-date adjustments to the contents of the book (where applicable). STPMText
Analysis of STPM Papers (2015 - 2018) Mathematics (T) Term 1 Chapter Year 2015 2016 2017 2018 Section Section Section Section A B A B A B A B 1. Functions Q1 Q7 Q1 Q7 Q1 Q1 Q7 2. Sequences and Series Q2 Q2 Q2 Q2 3. Matrices Q3 Q8 Q3 Q3 Q7 Q3 4. Complex Numbers Q4 Q4 Q4 Q4 5. Analytic Geometry Q5 Q5 Q8 Q5 Q5 6. Vectors Q6 Q6 Q6 Q8 Q6 Q8
ii Mathematics Term 1 STPM PREFACE This book ‘Pre-U STPM Text Mathematics (T) – Term 1’ is one in a series of three books, specially written to meet the requirements of the new revised Mathematics (T) syllabus in the STPM Examination, which will take effect from 2012. Under the new system introduced by the Malaysian Examinations Council (MEC), the new Form Six curriculum will be spread over three terms, with candidates sitting for an examination at the end of each term, and also coursework. This is to enhance the teaching and learning orientation of Form Six, so as to be in line with the orientation of teaching and learning in colleges and universities. The new syllabus fulfils the requirements of this new system, and this book covers all the topics specified in the syllabus of Mathematics (T) for Term 1. It is also suitable for use by students pursuing a foundation course in science or matriculation programmes at local universities. This book seeks to fulfil the aims and objectives as set out in the new mathematics syllabus, i.e. to develop the understanding of mathematical concepts and mathematical thinking, and acquire skills in problem solving and the applications of mathematics related to science and technology. This will prepare the students with an adequate foundation, before they proceed to programmes in the field of science and technology at institutions of higher learning. The contents of this book, in six chapters based on the syllabus, are organised and planned in a systematic manner so as to make learning more effective. Each new topic in a chapter is clearly presented via a simple and practical approach, which leads to a better overall picture and understanding of the topic concerned. The concepts presented in each subtopic include relevant explanations in detail, followed by the all-important worked examples, presented clearly in a step by step manner. This approach presumes that the student has only prior basic mathematical knowledge and skills up to the SPM level. The questions at the end of each subtopic are planned in such a way that the student will be able to test his understanding of, and apply, the concepts learned to solve basic problems. They are also suitable for the coursework. A summary of the concepts and other important formulae is given at the end of each chapter. A revision exercise covering all the subtopics in the chapter is also included. The questions chosen are all planned such that they are of equivalent standard as those in the STPM examination. To help the student in his revision and to assess his preparedness in facing the examination, a set of sample STPM examination paper is included at the end of the book, together with complete worked solutions. This book may be used as a textbook in the classroom, for coursework or by the student studying on his own. It is hoped that both teachers and students alike will benefit from using this book, and find the learning process both effective and enjoyable.
iii CONTENTS • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 1 FUNCTIONS 1 1.1 Functions 2 1.2 Polynomial and Rational Functions 21 1.3 Exponential and Logarithmic Functions 43 1.4 Trigonometric Functions 56 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 2 SEQUENCES AND SERIES 92 2.1 Sequences 93 2.2 Series 99 2.3 Binomial Expansions 119 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 3 MATRICES 136 3.1 Matrices 137 3.2 Systems of Linear Equations 160 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 4 COMPLEX NUMBERS 172 4.1 Complex Numbers 173 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 5 ANALYTIC GEOMETRY 192 5.1 Analytic Geometry 193 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Chapter 6 VECTORS 225 6.1 Vectors in Two and Three Dimensions 226 6.2 Vector Geometry 246 STPM Model Paper (954/1) 266 Answers 268
Mathematics Term 1 STPM Mathematical Notation Symbols = is equal to ≠ is not equal to is identical to or is congruent to ≈ is approximately equal to , is less than < is less than or equal to . is greater than > is greater than or equal to ∞ infinity \ therefore Set notation is an element of is not an element of ∅ empty set {x | …} set of x such that … N set of natural numbers, {0, 1, 2, 3, …} Z set of integers Z+ set of positive integers Q set of rational numbers R set of real numbers [ a, b ] closed interval {x | x R, a < x < b} ( a, b ) open interval {x | x R, a , x , b} [ a, b ) interval {x | x R, a < x , b} ( a, b ] interval {x | x R, a , x < b} union intersection Functions f a function f f(x) value of a function f at x f : A → B f is a function under which each element of set A has an image in set B f : x ↦ y f is a function which maps the element x to the element y f –1 inverse function of f f ° g composite function of f and g which is defined by f ° g(x) = f[g(x)] ex exponential function of x loga x logarithm to base a of x ln x natural logarithm of x, loge x sin, cos, tan, csc, sec, cot trigonometric functions sin–1, cos–1, tan–1, inverse trigonometric functions csc–1, sec–1, cot–1 Matrices A a matrix A 0 null matrix I identity matrix AT transpose of matrix A A–1 inverse of a non-singular square matrix A det A determinant of a square matrix A Complex numbers i square root of –1 z a complex number z |z| modulus z arg z argument of z z* complex conjugate of z Geometry AB length of the line segment with end points A and B ∠BAC angle between the line segments AB and AC ∆ABC triangle whose vertices are A, B and C // is parallel to is perpendicular to Vectors a a vector a |a| magnitude of a vector a i, j, k unit vectors in the directions of the Cartesian coordinates axes →AB vector represented in magnitude and direction by the directed line segment from point A to point B | →AB| magnitude of →AB a · b scalar product of vectors a and b a × b vector product of vectors a and b iv
1 Mathematics Term 1 STPM Chapter 1 Functions CHAPTER 1 FUNCTIONS Subtopic Learning Outcome 1.1 Functions (a) State the domain and range of a function, and find composite functions. (b) Determine whether a function is one-to-one, and find the inverse of a one-to-one function. (c) Sketch the graphs of simple functions, including piecewise-defined functions. 1.2 Polynomial and rational functions (a) Use the factor theorem and the remainder theorem. (b) Solve polynomial and rational equations and inequalities. (c) Solve equations and inequalities involving modulus signs in simple cases. (d) Decompose a rational expression into partial fractions in cases where the denominator has two distinct linear factors, or a linear factor and a prime quadratic factor. 1.3 Exponential and logarithmic functions (a) Relate exponential and logirithmic functions, algebraically and graphically. (b) Use the properties of exponents and logarithms. (c) Solve equations and inequalities involving exponential or logarithmic expressions. 1.4 Trigonometric functions (a) Relate the periodicity and symmetries of the sine, cosine and tangent functions to their graphs, and identify the inverse sine, inverse cosine and inverse tangent functions and their graphs. (b) Use basic trigonometric identities and the formulae for sin (A ± B), cos (A ± B) and tan (A ± B), including sin 2A, cos 2A and tan 2A. (c) Express a sin θ + b cos θ in the forms of r sin (θ ± ∝) and r cos (θ ± ∝). (d) Find the solutions, within specified intervals, of trigonometric equations and inequalities. composite – gubahan domain – domain equation – persamaan exponential function – fungsi eksponen factor theorem – teorem faktor function – fungsi inequality – ketaksamaan inverse – songsangan logarithmic function – fungsi logaritma partial fraction – pecahan separa polynomial function – fungsi polinomial range – julat rational function – fungsi nisbah remainder theorem – teorem baki trigonometric function – fungsi trigonometri Bilingual Keywords
2 1 Mathematics Term 1 STPM Chapter 1 Functions 1.1 Functions Functions Suppose we have set A = {1, 4, 9, 16, 25} and set B = {1, 2, 3, 4, 5}. Each element in set B is the square root of a corresponding element in set A. Figure 1.1 shows the relation ‘is the square root of ’ between the elements in set A and set B. 1 1 4 2 9 3 16 4 25 5 A 'is the square root of' B Figure 1.1 The relation between set A and set B can also be represented by the set of ordered pairs as follows: {(1, 1), (4, 2), (9, 3), (16, 4), (25, 5)} In the relation between set A and set B, each element in set A is connected to a unique element in set B. A relation such as this is known as a one-to-one relation. Now, suppose set P = {–2, –1, 1, 2, 3,} and set Q = {1, 4, 9, 16}. The relation between set P and set Q is as shown in the following diagram: –2 –1 1 1 4 2 9 3 16 P 'is the square of' Q Figure 1.2 Notice that in the relation between set P and set Q, the elements –2 and 2 of set P are connected to the element 4 of set Q, and the elements –1 and 1 of set P are connected to the element 1 in set Q. A relation such as this is known as a many-to-one relation. A relation whereby each element in a set X is connected to one and only one element in a set Y is known as a function. Hence, one-to-one relations and many-to-one relations are functions. If a relation is one-to-one, then the function is one-to-one. A function f from a set X to a set Y is defined as a rule that associates exactly one element of Y with each element of X. We say, f maps X into Y and we write f : X → Y. If x X and y Y such that y is assigned to x under f, we say f maps x to y and we write f : x ↦ y or y = f(x), where y is the image of x. In general, a function f is defined for certain values of x only. This set of values of x for which f is defined is called the domain of f. The set of values of f(x) for a given domain is called the range of f. In the previous example, let g be the function “is the square of ’ from set P to set Q. We write g : x ↦ x2 or g(x) = x2 .
3 1 Mathematics Term 1 STPM Chapter 1 Functions Set P is the domain and set Q is the codomain. The set of images {1, 4, 9} is the range of this function. –2 –1 1 1 4 2 9 3 16 P 'is the square of' Q Domain Codomain Range Figure 1.3 Example 1 –2 1 0 2 5 –1 0 1 2 X Y f (a) Show that the diagram above defines a function f from set X to set Y. (b) Find f(x). (c) State the domain of f and its range. Solution: (a) For each element x in set X, there exists a unique image y in set Y. Hence, f can be defined as a function from set X to set Y. (b) f(0) = 1 = 02 + 1 f(–1) = f(1) = 2 = 12 + 1 f(–2) = f(2) = 5 = 22 + 1 Hence, f(x) = x2 + 1. (c) Domain of f is {–2, –1, 0, 1, 2} Range of f is {1, 2, 5} Example 2 Determine whether each equation defines y as a function of x. If so, determine whether the function is one-to-one. (a) y = x + 2, (b) y = x2 + 3x + 2, (c) y2 = x – 2. Solution: (a) y = x + 2 For each value of x R, the value of y is unique. Thus y is a function of x. No two values of x R have the same image. Thus the function is one-to-one.
4 1 Mathematics Term 1 STPM Chapter 1 Functions (b) y = x2 + 3x + 2 For each value of x R, the value of y is unique. Thus y is a function of x. When x = 0, y = 0 + 0 + 2 = 2. When x = –3, y = 9 – 9 + 2 = 2. There exist two values of x R with the same image. Thus the function is not one-to-one. (c) y2 = x – 2 y = ± x – 2 For each value of x R with x . 2, there exist two possible values of y. For example, y = ±2 when x = 6. Thus y is not a function of x. Algebraic operations on functions Algebraic operations can be applied to functions. If f : x ↦ f(x) and g : x ↦ g(x), then (i) f + g : x ↦ f(x) + g(x), (ii) f – g : x ↦ f(x) – g(x), (iii) f · g : x ↦ f(x) · g(x), (iv) f g : x ↦ f(x) g(x) , g(x) ≠ 0, (v) af : x ↦ af(x), a R. For example, let f(x) = x + 1 and g(x) = x2 , where the domain is the set of real numbers, R. Then, f + g : x ↦ (x + 1) + x2 , x R f – g : x ↦ (x + 1) – x2 , x R f · g : x ↦ (x + 1)x2 , x R f g : x ↦ x + 1 x2 , x ≠ 0 2f : x ↦ 2(x + 1), x R Composite functions Suppose that the functions f and g are defined for the set of real numbers R, with f : x ↦ y and g : y ↦ z This means that we can create a new function which maps x directly to z. This function is called a composite function and is written as g º f, where g ° f(x) = g[f(x)]. Consider the diagram below, which shows the mapping of f, g and g ° f.
5 1 Mathematics Term 1 STPM Chapter 1 Functions y x z f g g ° f Figure 1.4 For example, if f : x ↦ x – 2 and g : x ↦ 3x, i.e. f(x) = x – 2 and g(x) = 3x, then g[f(x)] = 3f(x) = 3(x – 2) Hence, g ° f : x ↦ 3(x – 2) For example, if x = 5, f(5) = 5 – 2 = 3 g(3) = 3(3) = 9 i.e. g[f(5)] = g(3) = 9 3 5 9 f g g ° f Figure 1.5 Example 3 If f(x) = x2 and g(x) = x + 1, find (a) f ° g(x) (b) g ° f(x) (c) f ° f(x) (d) g ° g(x) Solution: (a) f ° g(x) = f[g(x)] (b) g ° f(x) = g[f(x)] = f(x + 1) = g(x2 ) = (x + 1)2 = x2 + 1 (c) f ° f(x) = f[f(x)] (d) g ° g(x) = g[g(x)] = f(x2 ) = g(x + 1) = (x2 ) 2 = (x + 1) + 1 = x4 = x + 2 Note: 1. f ° g ≠ g ° f, from (a) and (b). 2. f[g(x)] ≠ f(x) · g(x) since f[g(x)] = (x + 1)2 whereas f(x) · g(x) = x2 (x + 1).
6 1 Mathematics Term 1 STPM Chapter 1 Functions Example 4 Given that the function f : x ↦ 2x + 1, find the function g if (a) the composite function f ° g is f ° g : x ↦ 6x + 11, (b) the composite function g ° f is g ° f : x ↦ 1 x – 1 , x ≠ 1. Solution: (a) Given that f(x) = 2x + 1. If f ° g(x) = 6x + 11. f[g(x)] = 6x + 11. ∴ 2g(x) + 1 = 6x + 11 2g(x) = 6x + 10 g(x) = 3x + 5 Hence, g : x ↦ 3x + 5 (b) If g ° f(x) = 1 x – 1 . g[f(x)] = 1 x – 1 . ∴ g(2x + 1) = 1 x – 1 Let u = 2x + 1 then x = 1 2 (u – 1) Hence, g(u) = 1 1 2 (u – 1) – 1 = 2 u – 3 i.e. g : u ↦ 2 u – 3 , u ≠ 3 or g : x ↦ 2 x – 3 , x ≠ 3. Exercise 1.1 1. If f(x) = x2 – 9, x R, find (a) f(0), (b) f(4), (c) f(3), (d) f(–3). 2. Given that f(x) = x + 1 , x –1, (a) find the value of (i) f(8), (ii) f(0), (iii) f(–1), (b) find the value of x such that (i) f(x) = 5, (ii) f(x) = 10. 3. By drawing a suitable sketch graph, find the range of each of the following functions. (a) f(x) = 2x, x [–1, 3], (b) f(x) = 2 – x, x [–1, 3], (c) f(x) = x2 – x – 2, x [–2, 3].
7 1 Mathematics Term 1 STPM Chapter 1 Functions 4. Let X = {1, 2, 3, 4,}. Show whether each of the following relations, which define the set of ordered pairs, represent a function from X to X. (a) f : {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)} (b) g : {(3, 1), (4, 2), (1, 1)} (c) h : {(2, 1), (3, 4), (1, 4), (2, 3), (4, 4)} 5. The domain of a function h, where h : x ↦ 2x , is R. (a) Find the images of 0, 2, –1 and 6. (b) Find the element in the domain with image 16. (c) Find the range of the function h. 6. If f(x) = 2x – 3 and f ° g(x) = 2x + 1, find g(x). 7. If f(x) = x – 1 and g ° f(x)= 3 + 2x – x2 , find g(x). 8. If f(x) = cos x and g(x) = 1 + x, x R, find (a) f ° g(x) (b) f ° f(x) (c) g ° f(x) 9. If f(x) = x , x 0 and g(x) = 1 – x2 , x R, find (a) f ° g(x) (b) g ° f(x) (c) g ° g(x) 10. If f : R → R and g : R → R are defined by f(x) = x2 + 3x + 1 and g(x) = 2x – 3, find (a) f ° g (b) g ° f (c) g ° g (d) f ° f 11. If f : R → R and g : R → R are defined by f(x) = 2x – 3 and g(x) = x2 + 5, find (a) g ° f(2) (b) f ° g(3) (c) f ° g(a – 1) (d) g ° f(x) (e) f ° g(x + 1) (f) g ° g(x) Inverse functions Let f and g be two functions defined respectively by f : x ↦ 2x + 1 and g : x ↦ 1 2 (x – 1) 1 i.e. f(x) = 2x + 1 g(x) = —(x – 1) 2 123 1423 Then the composite function g ° f is defined as g ° f(x) = g[f(x)] = g(2x + 1) = 1 2 [(2x + 1) – 1] = x i.e. we get back the object for the function f, i.e. x. Similarly, f ° g(x) = f[g(x)] = f[ 1 2 (x – 1)] = 2[ 1 2 (x – 1)] + 1 = x Object of f Image of f Object of g Image of g
8 1 Mathematics Term 1 STPM Chapter 1 Functions Hence, g is the inverse function of f and is written as f–1. The inverse of a function f can be obtained by inverting the direction of the arrow in the diagram, such as the one shown below. a b f f –1 Domain of f (Range of f –1) Range of f (Domain of f –1) Figure 1.6 For the function f : x ↦ 2x + 1, the inverse of f is f –1 : x ↦ 1 2 (x – 1). –1 0 1 2 –1 1 3 5 A x f 2x +1 B f –1 –1 1 3 5 –1 0 1 2 B A x 2 – 1 (x–1) Figure 1.7(a) Figure 1.7(b) The diagram in Figure 1.7(a) shows the function f : x ↦ 2x + 1. The diagram in Figure 1.7(b) shows the inverse of the function f, which maps each element in the set B to only one element in the set A. This relation is a one-to-one relation. Hence, the inverse of the function f is also a function, and is called the inverse function of f. Now consider the function f below, i.e. f : x ↦ x2 , x R The inverse of the function f is f –1 : x2 ↦ x or f–1 : x ↦ ±x . – 2 –1 1 2 4 1 P Q x f x 2 4 1 –2 –1 1 2 Q P x f + –1 x – Figure 1.8(a) Figure 1.8(b) The diagram in Figure 1.8(a) shows the function f : x ↦ x2 . In the inverse of function f, as shown in Figure 1.8(b), the elements 4 and 1 in the set Q are each mapped to two elements in the set P. Each of the elements 4 and 1 does not have a unique image. Hence, the inverse of the function f is not a function. Now consider the function f again, where f : x ↦ x2 , x R+ The inverse of this function is f –1 : x ↦ x (positive square root only) Hence, the inverse of the function f is also a function, i.e. if f : x ↦ x2 , x R+ 0, then the inverse function of f is f–1 : x ↦ x, x R+ 0.
9 1 Mathematics Term 1 STPM Chapter 1 Functions From the above examples, we make the following conclusions: (a) The inverse of a function is not necessarily also a function. (b) The inverse of a given function is also a function if the given function is a one-to-one function. (c) The inverse of a many-to-one function can be a function if we restrict the domain. Example 5 Find the inverse of each of the following functions, indicating its domain. (a) f : x ↦ 3x + 5, x R (b) f : x ↦ 2 x – 3 , x ≠ 3 Solution: (a) f(x) = 3x + 5 Let f –1(x) = a ∴ f(a) = x 3a + 5 = x a = x – 5 3 Hence f–1(x) = x – 5 3 , x R or f –1 : x ↦ 1 3 (x – 5), x R (b) f(x) = 2 x – 3 , x ≠ 3. Let f –1(x) = b ∴ f(b) = x 2 b – 3 = x b = 2 + 3x x Hence, f –1(x) = 2 + 3x x , x ≠ 0 or f –1 : x ↦ 2 + 3x x , x ≠ 0. Exercise 1.2 1. 1 a b c d 2 3 4 5 X Y (a) Show that the elements in sets X and Y define a function f from set X to set Y. (b) Find f(1) and f(3). (c) State the domain of f. (d) State the range of f. (e) Is f a one-to-one function?
10 1 Mathematics Term 1 STPM Chapter 1 Functions 2. Given that g : x ↦ x2 + x + 1 and x X, where X = {–2, –1, 0, 1, 2}. (a) Sketch a diagram to represent g. (b) State the domain of g. (c) State the range Y of g. (d) Is g a one-to-one function? 3. Given that h : x ↦ 2x – 3, where x {1, 3, 5, 7, 9} and that the range is {–5, –3, –1, 3, 5, 7, 11, 15}. (a) Sketch a diagram to represent h. (b) Find h(1) and h(9). (c) Is h a one-to-one function? 4. The function f is defined by f : x ↦ x2 – x, x R. State whether f is one-to-one, giving a reason for the statement. If the domain of f is restricted to the subset of R for which x k, find the least value of k for which f is one-to-one. 5. The function f is defined by f : x ↦ x x 2 + 1 , x R 1 If a R and a ≠ 0, find the image of — under f. a Deduce that f is not one-to-one. Show that if a, b R with a . b 1, then f(b) . f(a). Deduce that, if the domain of f is restricted to the subset of R given by {x : x 1}, then f is one-to-one. State the range of f in this case. 6. Find the inverse of each of the following functions, stating its domain. (a) f : x ↦ x – 2, x R (b) f : x ↦ x2 + 1, x 0 (c) f : x ↦ (x – 1)2 , x 1 (d) g : x ↦ (x – 2)(x – 4), x 3 (e) g : x ↦ (x – 3)(x + 3), x R+ (f) h : x ↦ 2 x – 3 , x ≠ 3 (g) h : x ↦ x + 2 x – 2 , x ≠ 2 7. Given that f(x) = x 2 + 1 2x 2 + 1 and x 2, find f–1(x). Prove that (f–1 ° f)(x) = x. 8. The functions f and g are defined for x R (excluding –1, 0 and 1) by f : x ↦ 1 + x 1 – x and g : x ↦ 1 x Show that (f ° g)–1 = f ° g. 9. Sketch the graph of each of the following functions and its inverse. (a) f(x) = x + 3, x R (b) f(x) = 2x – 1, x R (c) f(x) = (x – 1)(x + 1), x R+ (d) f(x) = x2 + 3x – 4, x . – 3 2
11 1 Mathematics Term 1 STPM Chapter 1 Functions Graphs of linear, quadratic and cubic-functions The simplest algebraic function is the linear function, usually in the form f(x) = ax + b, where a and b are rational numbers, either positive or negative. The graph of a linear function is a straight line, and a is known as the gradient and b the intercept on the y-axis. The graph of y = ax + b for different values of a and b are as follows: Linear function f(x) = ax + b O x y O x y O x y O x y a . 0, b . 0 a . 0, b , 0 a , 0, b . 0 a , 0, b , 0 Figure 1.9 The shape of the graphs for a quadratic and a cubic function are as shown below. Quadratic function Cubic function f(x) = ax2 + bx + c f(x) = ax3 + bx2 + cx + d or or a . 0 a , 0 a . 0 a , 0 Figure 1.10 Notice that the graph of a quadratic function has one stationary point, whereas a cubic function has two stationary points. Example 6 Sketch the graphs of the following functions. (a) f(x) = x2 + 4x + 5, (b) f(x) = –2x2 + 3x + 1. Solution: (a) f(x) = x2 + 4x + 5 can be expressed as f(x) = (x + 2)2 – 4 + 5, by completing the square = (x + 2)2 + 1 The graph cuts the y-axis when x = 0, i.e. y = 5. As x → `, y → ` Similarly when x → –`, y → `. When x = –2, y = 1 is the minimum value of f(x). Hence the graph of y = x2 + 4x + 5 is as shown below. 0 –2 1 5 x y y = x 2 + 4x + 5
12 1 Mathematics Term 1 STPM Chapter 1 Functions (b) f(x) = –2x2 + 3x + 1 can be expressed as f(x) = –2(x2 – 3 2 x) + 1 = –2 1x – 3 4 2 2 – 9 16 + 1, by completing the square = –21x – 3 4 2 2 + 9 8 + 1 = –21x – 3 4 2 2 + 17 8 . The graph cuts the y-axis when x = 0, i.e. y = 1. As x → `, y → –`. Similarly when x → –`, y → –`. When x = 3 4 , y = 17 8 is the maximum value of f(x). Hence the graph of y = –2x2 + 3x + 1 is as shown below. 0 1 17 – 8 3 – 4 x y y = –2x 2 + 3x + 1 Example 7 Sketch the graphs of (a) y = x3 – 2x2 – 5x + 6 (b) y = x3 – x2 – x + 1 Solution: (a) Let y = x3 – 2x2 – 5x + 6 = (x + 2)(x – 1)(x – 3) The graph of y = x3 – 2x2 – 5x + 6 is in the form . The graph crosses the x-axis when y = 0, i.e. (x + 2)(x – 1)(x – 3) = 0. Hence, x + 2 = 0, x – 1 = 0 or x – 3 = 0. i.e. x = –2, 1 or 3. The graph crosses the y-axis when x = 0, i.e. y = 6 As x → ∞, y → ∞. As x → –∞, y →–∞. The maximum point must lie in the interval [–2, 1]. The minimum point lies in the interval [1, 3].
13 1 Mathematics Term 1 STPM Chapter 1 Functions With the information obtained above, the graph of y = x3 – 2x2 – 5x + 6 may be drawn as shown below. x y 3 6 y = x3 – 2x 2 – 5x + 6 –2 10 (b) Let y = x3 – x2 – x + 1 = (x + 1)(x2 – 2x + 1) = (x + 1)(x – 1)2 The graph of y = x3 – x2 – x + 1 is in the form . The graph crosses the x-axis when y = 0, i.e. (x + 1)(x – 1)2 = 0 Hence, x + 1 = 0 or (x – 1)2 = 0 i.e. x = –1, 1, 1 (repeated) The graph crosses the y-axis when x = 0, i.e. y = 1. As x → ∞, y → ∞. As x → –∞, y → –∞. The value x = 1 (repeated) shows that the curve touches the x-axis at the point x = 1. The minimum point is at (1, 0). The maximum point lies in the interval [–1, 1]. The graph of y = x3 – x2 – x + 1 is as shown below. x y y = x3 – x 2 – x + 1 1 1 –1 0 Graphs of power functions (a) If n is a positive integer (n Z+ ), the curve y = kxn meets the x-axis at y = 0, i.e. xn = 0. If n is even, the curve touches the x-axis at the origin (0, 0). The curve has a minimum point if k . 0 and a maximum point if k , 0.
14 1 Mathematics Term 1 STPM Chapter 1 Functions The graph of y = kxn (n even) is as follows: k fi 0 y x O minimum point k fi 0 y x O maximum point Figure 1.11(a) Figure 1.11(b) Notice that the graph is symmetrical about the y-axis. (b) If n is odd, the curve has a point of inflexion at the origin (0, 0). The graph of y = kxn (n odd) is as follows: k fi 0 y point of inflexion x O O k fi 0 y point of inflexion x Figure 1.12(a) Figure 1.12(b) Example 8 Sketch each of the following curves. (a) y = 5x4 (b) y = 3(x – 2)5 Solution: (a) The graph of y = 5x4 has a (b) The graph of y = 3x5 has a point of minimum point at the inflexion at the origin. The shape of origin and is symmetrical the curve y = 3(x – 2)5 is similar of about the y-axis. to that y = 3x5 but has a point of inflexion at (2, 0). O y x y = 5x 4 y = 3x 5 y = 3(x – 2) y 5 x 0 (2, 0)
15 1 Mathematics Term 1 STPM Chapter 1 Functions Graphs of rational functions A rational function is a function in the form f(x) g(x) , where f(x) and g(x) are algebraic functions with g(x) ≠ 0. An example of a rational function is xy = 1, i.e. y = 1 x or x = 1 y . Notice that when x → ±∞, y → 0 and when y → ±∞, x → 0. The lines y = 0 and x = 0 are known as the asymptotes for the curve xy = 1. In general, a curve which approaches a straight line when x → ±∞ or when y → ±∞ will have the straight line as its asymptote. y x xy = 1 O asymptotes Figure 1.13 Example 9 Sketch the curve with equation y = 2x – 1 x + 1 . State the asymptotes of the curve. Solution: y = 2x – 1 x + 1 1 x(2 – —) x Rewriting the equation as y = , 1 x(1 + —) x 1 We get 2 – —x y = . 1 1 + —x When x → ±∞, 1 x → 0 and y → 2. Hence, y = 2 is an asymptote to the curve y = 2x – 1 x + 1 . When x = –1, y = 2x – 1 x + 1 is undefined. Hence, x = –1 is also an asymptote to the curve. When x = 0, y = –1. When y = 0, x = 1 2 .
16 1 Mathematics Term 1 STPM Chapter 1 Functions The graph of y = 2x – 1 x + 1 is as shown below. The asymptotes to the curve are y = 2 and x = –1. y x 0 asymptote (y = 2) asymptote (x = –1) 2x – 1 x + 1 y = – 2 –1 –1 1 – 2 Graphs of functions involving modulus signs The absolute value or modulus of x is defined by |x| = x, x > 0, –x, x , 0. The graphs of y = x and y = |x| are as follows: 0 –1–2–3 1 2 3 y x y = x –1–2–3 0 1 2 3 y x y = |x | Figure 1.14(a) Figure 1.14(b) For x , 0, the graph of y = |x| is the reflection of the graph y = x about the x-axis, i.e. the graph of y = –x. Example 10 Sketch the graph of y = x2 – 3x + 2, noting the points of intersection of the graph with the x-axis. On a separate diagram, sketch the graph of y = |x2 – 3x + 2|. Solution: y = x2 – 3x + 2 At the point of intersection with the x-axis, y = 0. x2 – 3x + 2 = 0 (x – 1)(x – 2) = 0 x = 1 or 2. At the y-axis, x = 0 ⇒ y = 2. The graph of y = x2 – 3x + 2 is as shown on the right. 0 1 1 2 2 y x y = x 2 – 3x + 2
17 1 Mathematics Term 1 STPM Chapter 1 Functions The graph of y = |x2 – 3x + 2| can be drawn by reflecting the section of the graph that is below the x-axis, as shown below. 0 1 1 2 2 y x y = fix 2 – 3x + 2fi Example 11 Sketch the graph of y = |2x – 3|. By drawing the graph of y = x on the same diagram, determine the values of x such that |2x – 3| . x. Solution: |2x – 3| = 0 when x = 3 2 . The graph of y = |2x – 3| is symmetrical about x = 3 2 . When x = 0, y = 3. When x = 3, y = 3. The graph of y = |2x – 3| is as shown below. 1 3 3 y x A 0 2 B y = x y = fi2x – 3fi _3 2 Notice that the graph of y = x intersects the graph of y = |2x – 3| at two points, A and B. At A, 2x – 3 = x x = 3 At B,–(2x – 3) = x –2x + 3 = x 3x = 3 x = 1 Hence, from the graph, |2x – 3| . x if x , 1 or x . 3.
18 1 Mathematics Term 1 STPM Chapter 1 Functions Graphs of piecewise functions A function can be defined differently for different intervals in a given domain. We say that the function is defined piecewise on a domain, or that the function is a piecewise function. Consider the function f defined by x + 1 , –1 < x , 0. f(x) = x2 , 0 < x , 1, 2 , 1 < x < 2. We notice that the graph of f has 3 sections or pieces, as shown below. We say that the function f is piecewise for the domain [–1, 2]. 2 1 –1 0 1 2 y x Figure 1.15 Note: • indicates that this point is included; ° indicates that this point is excluded. Example 12 Sketch the graph of the function f defined by x2 , –3 < x , 2. f(x) = 5 , x = 2, 8 – x , 2 , x < 4. Solution: For –3 < x , 2, the graph is quadratic. For x = 2, y = 5. For 2 , x < 4, the graph is straight line. Hence, the graph of the function f is as shown below. 0 4 –3 2 4 x y y = 8 – x y = 5 y = x 2 5 6 9 Sketch Graphs INFO
19 1 Mathematics Term 1 STPM Chapter 1 Functions Example 13 For x R, [x] is defined as the greatest integer not exceeding x, for example, 3 5 2 4 = 2, [3] = 3 and [–1.2] = –2. Sketch the graph of f defined by f(x) = [x] for –2 < x , 3. Solution: From the definition of [x]. –2 for –2 < x , –1, –1 for –1 < x , 0, f(x) = 0 for 0 < x , 1, 1 for 1 < x , 2, 2 for 2 < x , 3. The sketch graph of f is as follows: y x 0 1 2 –2 –1 1 2 3 –1 –2 y = [ ] x Exercise 1.3 1. Sketch the graphs of (a) y = 4x + 3 (b) y = –3x + 8 (c) 2y = 5x + 6 2. Sketch the graphs of (a) y = x2 + 3 2 (b) y = 3x2 – 1 (c) y = –4x2 + 5x + 2 3. Sketch the graphs of (a) y = x3 – 1 (b) y = –2x3 + 1 (c) y = – 1 2 x3 + 5 4. Sketch the graph for each of the following curves, indicating the points of intersection of the curve with the axes. (a) y = x(x + 1)(x – 2) (b) y = (x + 1)2 (x – 2) (c) y = (x – 2)3 (d) y = x3 – 2x (e) y = x3 – 3x2 + 2x 5. Sketch the graph of (a) y = x(1 – x2 ) (b) y2 = x(1 – x2 )
20 1 Mathematics Term 1 STPM Chapter 1 Functions 6. Sketch the graph for each of the following curves, showing the asymptotes. (a) y = 2 1 + x (b) y = 1 – x 1 + 2x (c) y = x2 1 + x 7. Sketch the graph of (a) y = x 1 + x (b) y2 = x 1 + x 8. Sketch the graph of (a) y = |2x – 1| (b) y = |(x – 1)(x – 2)| (c) y = 3 + |x + 1| (d) y = |2x + 5| – 4 9. Sketch the graphs of y = |x + 1| and y = |3x – 5| on the same diagram. Find the set of values of x such that |x + 1| . |3x – 5|. 10. Sketch the graphs of y = |x| and y = |x2 – 7x + 6| on the same diagram. Determine the set of values of x such that |x| , |x2 – 7x + 6|. 11. The function f is defined on the domain [–4, 4] by f(x) = x + 5, –4 < x < –2; f(x) = x2 + 2, –2 , x < 2; and f(x) = –x + 7, 2 , x < 4. Sketch the graph of f. 12. The function f is defined on the domain [–2, 5] as follows: –2x2 + 3, –2 < x < 0, f(x) = x2 + 1, 0 , x < 2, –2x + 8, 2 , x < 5. Sketch the graph of f. 13. The function f is defined by |2x – 1|, – 3 2 < x , 3 2 , f(x) = 4, 3 2 < x < 3, – x2 3 + 6, 3 , x < 5. Sketch the graph of f. 14. The function f is defined by x + 1, –4 < x , –2, f(x) = x + 2, –2 < x , 0, x + 3, 0 < x , 2, x + 4, 2 < x < 4. Sketch the graph of f.
21 1 Mathematics Term 1 STPM Chapter 1 Functions 1.2 Polynomial and Rational Functions Polynomial functions Consider the algebraic expression f(x) = 3x3 + 4x2 – x + 5 This is an expression in terms of x, where the highest power of x is 3. An expression such as this is called a function of x of degree 3. Similarly, g(x) = 7x4 – 5x3 + 2x2 + x – 3 is called a function of x of degree 4 as the highest power of x is 4. In general, a function of x in the form P(x) ≡ a0 xn + a1 x n – 1 + … + ar xn – r + … + an – 1 x + an where ar R and a0 ≠ 0, n Z+ is called a polynomial function of degree n. Polynomials of degrees 1, 2, 3 and 4 are also known as linear, quadratic, cubic and quartic functions respectively. The value of a polynomial, P(x), when x = a is written as P(a). For example, if P(x) = 3x2 + 4x – 1 then when x = 1, P(1) = 3 + 4 – 1 = 6 and when x = 2, P(2) = 12 + 8 – 1 = 19 Algebraic operations on polynomials Consider two linear polynomials f(x) and g(x), which are defined respectively as f(x) = 5x + 4 and g(x) = 3x – 1 Notice that f(x) + g(x) = (5x + 4) + (3x – 1) = 5x + 4 + 3x – 1 = 8x + 3 f(x) – g(x) = (5x + 4) – (3x – 1) = 2x + 5 f(x) · g(x) = (5x + 4)(3x – 1) = 15x2 + 12x – 5x – 4 = 15x2 + 7x – 4 From the above results, we see that we can add, subtract or multiply two polynomials P(x) and Q(x) to obtain a new polynomial. The addition, subtraction and multiplication of two polynomials obey the commutative, associative and distributive laws respectively.
22 1 Mathematics Term 1 STPM Chapter 1 Functions Example 14 If P(x) = 2x3 + 4x2 – x + 3 and Q(x) = 3x2 + x – 1, find (a) P(x) + Q(x), (b) P(x) – Q(x), (c) P(x) · Q(x). Solution: (a) P(x) + Q(x) = 2x3 + 4x2 – x + 3 + 3x2 + x – 1 = 2x3 + 7x2 + 2 (b) P(x) – Q(x) = 2x3 + 4x2 – x + 3 – (3x2 + x – 1) = 2x3 + 4x2 – x + 3 – 3x2 – x + 1 = 2x3 + x2 – 2x + 4 (c) P(x) · Q(x) = (2x3 + 4x2 – x + 3) · (3x2 + x – 1) = 2x3 (3x2 + x – 1) + 4x2 (3x2 + x – 1) – x(3x2 + x – 1) + 3(3x2 + x – 1) = 6x5 + 2x4 – 2x3 + 12x4 + 4x3 – 4x2 – 3x3 – x2 + x + 9x2 + 3x – 3 = 6x5 + 14x4 – x3 + 4x2 + 4x – 3 Polynomial of degree 5. Note: If P(x) is a polynomial of degree m and Q(x) is a polynomial of degree n, then P(x) · Q(x) is a polynomial of degree (m + n). For the division of two polynomials, the long division method may be used, as shown in the following example. Example 15 Determine the quotient and remainder when 2x3 – 7x2 – 9x + 38 is divided by (x – 3). Solution: 2x2 – x – 12 x – 3 2x3 – 7x2 – 9x + 38 2x3 – 6x2 – x2 – 9x – x2 + 3x – 12x + 38 – 12x + 36 2 Using the long division method, the quotient is 2x2 – x – 12 and remainder 2. From Example 15 above, we know that a polynomial P(x), of degree m, may be divided by another polynomial, Q(x), of degree n, only if m n, m, n ∈ Z+ . If P(x) is divisible by Q(x) exactly, i.e. without any remainder, then the quotient is another polynomial of degree (m – n).
23 1 Mathematics Term 1 STPM Chapter 1 Functions For example, if P(x) = 2x5 – x4 – 5x2 – 2x – 3 and Q(x) = x2 + x + 1, then P(x) Q(x) may be found using long division as shown below. 2x3 – 3x2 + x – 3 x2 + x + 1 2x5 – x4 + 0x3 – 5x2 – 2x – 3 2x5 + 2x4 + 2x3 –3x4 – 2x3 – 5x2 –3x4 – 3x3 – 3x2 x3 – 2x2 – 2x x3 + x2 + x – 3x2 – 3x – 3 – 3x2 – 3x – 3 0 Add a term 0x 3 to avoid confusion during working. Hence, P(x) Q(x) = 2x3 – 3x2 + x – 3, i.e. a polynomial of degree 3. If it is known that the division of two polynomials is exact, the quotient may also be obtained by using the method as shown in Example 16 below. Example 16 Find the quotient if x3 – 4x2 + 5x – 2 can be divided by (x – 2) exactly. Solution: Let the quotient be the polynomial q(x). x3 – 4x2 + 5x – 2 Hence, ––––––––––––––– = q(x) x – 2 i.e. x3 – 4x2 + 5x – 2 ≡ q(x) · (x – 2) Multiply both sides by (x – 2) 1442443 Since q(x) is a polynomial of degree 2 (i.e. a quadratic function), it must be of the form ax2 + bx + c. Hence, x3 – 4x2 + 5x – 2 ≡ (ax2 + bx + c)(x – 2) = ax3 + bx2 + cx – 2ax2 – 2bx – 2c = ax3 + (b – 2a)x2 + (c – 2b)x – 2c Equating coefficients of x3 : 1 = a Equating coefficients of x2 : –4 = b – 2a –4 = b – 2(1) b = –2 Equating coefficients of x: 5 = c – 2b 5 = c – 2(–2) c = 1 Hence, a = 1, b = –2 and c = 1, and the quotient is x2 – 2x + 1. Polynomial of degree 3 Polynomial of degree 2
24 1 Mathematics Term 1 STPM Chapter 1 Functions Example 17 Find the constants A, B and C, such that x2 – 5x + 12 ≡ A (x – 1)(x – 2) + B(x + 1)(x – 1) + C(x – 2)(x + 3). Solution: Since the given equation is an identity, we can substitute suitable values of x into the identity to determine the values of A, B and C. When x = 1, 1 – 5(1) + 12 = A(0) + B(0) – 4C 8 = –4C C = –2 When x = 2, 22 – 5(2) + 12 = A(0) + 3B + C(0) 6 = 3B B = 2 When x = 0, 12 = 2A – B – 6C 12 = 2A – 2 – 6(–2) 12 = 2A – 2 + 12 A = 1 Hence, A = 1, B = 2 and C = –2. Note: We can also find the constants A, B and C in Example 17 by equating the coefficients of x2 , x and the constants on both sides of the identity. Exercise 1.4 1. If P(x) = 2x3 – 3x2 + 4x + 1, find the values of (a) P(0) (b) P1 1 2 2 (c) P(2) 2. If Q(x) = 2x4 – 5x2 + x – 3, find the values of (a) Q(–1) (b) Q1 3 2 2 (c) Q(2) 3. If F(x) = x2 + x – 1 and G(x) = 1 + 2x, find (a) F(x) + 2G(x) (b) F(x) – G(x) (c) 3F(x) + x · G(x) (d) (1 + x) · F(x) (e) G(x) · F(x) 4. By substituting suitable values of x into each of the identities below, find the values of the constants A, B and C. (a) 3x + 3 ≡ A(x – 1) + B(2 + x) (b) 7x + 6 ≡ A(x – 2) + B(x + 3) (c) 2x + 5 ≡ A(x + 1) + B(x – 2) (d) 2x2 – 5x + 7 ≡ A(x + 1)(x – 2) + B(x + 1)(x – 1) + C(x – 2)(x – 1) (e) x2 – 6x – 19 ≡ A(x + 5)(x – 1) + B(x2 – 1) + C(x + 5)(x + 1) 5. Find the product of (a) 2x2 – x + 7 and x + 2 (b) 3x3 – 2x2 + 5x – 1 and 2x – 3 (c) x2 + 3x – 2 and 4x2 – x + 1 (d) 5x3 – 2x + 3 and x2 – 1 (e) 3x4 – 2x3 + 6x – 4 and 3x + 2
25 1 Mathematics Term 1 STPM Chapter 1 Functions 6. Find, using long division, the quotient and remainder for (a) (3x2 – 2x + 5) ÷ (x + 1) (b) (4x3 – 4x2 + 5x + 1) ÷ (2x – 3) (c) (x3 – 3) ÷ (x + 3) (d) (3x5 – x4 – 6x3 + 11x2 – 1) ÷ (3x – 1) (e) (x4 – 2x3 + 6x – 5) ÷ (x2 – x – 1) 7. The expression ax3 – 8x2 + bx + 6 is divisible by x2 – 2x – 3. Find the values of a and b. 8. Find the values of A and B such that x3 + 4x2 + 4x + 1 ≡ (x + 1)(x2 + Ax + B) 9. Find J(x) if x3 + 3x2 + x – 2 ≡ (x + 2) · J(x) 10. Find numbers a, b and c, such that 2x4 + 2x3 + 5x2 + 3x + 3 ≡ (x2 + x + 1)(ax2 + bx + c) for all values of x. The remainder theorem When the polynomial 2x3 – 7x2 + 11x – 7 is divided by (x – 2), the quotient is 2x2 – 3x + 5 with remainder 3. We can write 2x3 – 7x2 + 11x – 7 3 ––––––––––––––––– = 2x2 – 3x + 5 + ––––– x – 2 14243 x – 2 Quotient or 2x3 – 7x2 + 11x – 7 = (2x2 – 3x + 5)(x – 2) + 3 Remainder (A constant) 14424443 14243123 The dividend (Polynomial degree 3) Quotient (Polynomial degree 2) Divisor (Polynomial degree 1) In general, if a polynomial of degree n, P(x), is divided by (x – a), the quotient, Q(x), is a polynomial of degree (n – 1), and the remainder, R, is a constant, i.e. P(x) x – a = Q(x) + R x – a or P(x) ≡ Q(x) · (x – a) + R By substituting x = a, we see that P(a) = R. When a polynomial P(x) is divided by (x – a), the remainder is P(a). Example 18 Find the remainder when the polynomial P(x) = 2x3 + 7x2 – 5x – 4 is divided by (x + 3). Solution: P(x) = 2x3 + 7x2 – 5x – 4 Since x + 3 is the divisor, choose a value of x such that x + 3 = 0, i.e. x = –3. P(–3) = 2(–3)3 + 7(–3)2 – 5(–3) – 4 = –54 + 63 + 15 – 4 = 20 Hence, the remainder when the polynomial P(x) is divided by (x + 3) is 20. Remainder
26 1 Mathematics Term 1 STPM Chapter 1 Functions Example 19 Find the remainder when the polynomial 2x4 – 5x3 + x2 – 7x + 1 is divided by (2x + 1). Solution: Let P(x) = 2x4 – 5x3 + x2 – 7x + 1 Since 2x + 1 is the divisor, we choose a value of x such that 2x + 1 = 0, i.e. x = – 1 2 . P1– 1 2 2 = 21– 1 2 2 4 – 51– 1 2 2 3 + 1– 1 2 2 2 – 71– 1 2 2 + 1 = 1 8 + 5 8 + 1 4 + 7 2 + 1 = 11 2 or 5 1 2 Hence, the remainder when the polynomial 2x4 – 5x3 + x2 – 7x + 1 is divided by (2x + 1) is 5 1 2 . When a polynomial P(x) is divided by (ax + b), the remainder is P1– b a 2. Example 20 The polynomial ax4 – 5x3 + bx2 – 7x + 1 leaves a remainder of –8 when it is divided by (x – 1), and a remainder of 11 2 when divided by (2x + 1). Determine the values of a and b. Solution: Let P(x) = ax4 – 5x3 + bx2 – 7x + 1 Hence, P(1) = –8 i.e. a – 5 + b – 7 + 1 = –8 a + b = 3 ………… Also, P1– 1 2 2 = 11 2 i.e. a1– 1 2 2 4 – 51– 1 2 2 3 + b1– 1 2 2 2 – 71– 1 2 2 + 1 = 11 2 a 16 + 5 8 + b 4 + 7 2 + 1 = 11 2 a + 10 + 4b + 56 + 16 = 88 a + 4b = 6 ………… – : 3b = 3 b = 1 Substitute b = 1 into : a + 1 = 3 a = 2 Example 21 When the polynomial P(x) is divided by (x – 1), its remainder is 5. When P(x) is divided by (x – 2), its remainder is 7. Given that P(x) may be written in the form (x – 1)(x – 2) · Q(x) + Ax + B, where Q(x) is a polynomial, A and B are constants, find the remainder when P(x) is divided by (x – 1)(x – 2).
27 1 Mathematics Term 1 STPM Chapter 1 Functions Solution: Given that P(x) ≡ (x – 1)(x – 2) · Q(x) + Ax + B When P(x) is divided by (x – 1), the remainder is 5. Hence, P(1) = 5 i.e. 0 + A + B = 5 A + B = 5 ………… When P(x) is divided by (x – 2), the remainder is 7. Hence, P(2) = 7 i.e. 0 + 2A + B = 7 2A + B = 7 ………… – : A = 2 Substituting A = 2 into : 2 + B = 5 B = 3 Hence, P(x) ≡ (x – 1)(x – 2) · Q(x) + 2x + 3. P(x) 2x + 3 and –––––––––––– = Q(x) + –––––––––––– (x – 1)(x – 2) (x – 1)(x – 2) Hence, when P(x) is divided by (x – 1)(x – 2), the remainder is (2x + 3). Note: Example 21 above shows that when a polynomial P(x) is divided by a quadratic expression (x – a)(x – b), the remainder is a linear function in the form Ax + B, where A and B are constants. The factor theorem From the remainder theorem, we have shown that when a polynomial P(x) is divided by (x – a), its remainder, R, is P(a). On the other hand, if (x – a) is a factor of P(x), then its remainder is zero, i.e. R = P(a) = 0. For a polynomial P(x), (x – a) is a factor of P(x) if and only if P(a) = 0. This theorem is very useful in the factorisation of polynomials of degrees higher than 2. By using this theorem, the linear factors of the polynomial may be obtained. Example 22 Factorise P(x) = x3 – 7x – 6. Solution: [Hints to obtain the factors: Notice that the constant term for P(x) = x3 – 7x – 6 is –6. Hence, if (x – a) were to be a factor of P(x), then a must be a factor of –6, i.e. a = ±1, ±2, ±3 or ±6. Try substituting these values into P(x) such that P(a) = 0.] If x = 1, P(1) = 1 – 7 – 6 = –12 ≠ 0 Hence, (x – 1) is not a factor of P(x). If x = –1, P(–1) = (–1)3 – 7(–1) – 6 = –1 + 7 – 6 = 0 Hence, (x + 1) is a factor of P(x).
28 1 Mathematics Term 1 STPM Chapter 1 Functions If x = 2, P(2) = 8 – 14 – 6 = –12 ≠ 0 Hence, (x – 2) is not a factor of P(x). If x = –2, P(–2) = (–2)3 – 7(–2) – 6 = –8 + 14 – 6 = 0 Hence, (x + 2) is a factor of P(x). If x = 3, P(3) = 33 – 7(3) – 6 = 27 – 21 – 6 = 0 Hence, (x – 3) is a factor of P(x). Since P(x) is of degree 3, it has only three linear factors. P(x) = x3 – 7x – 6 = (x + 1)(x + 2)(x – 3) Apart from the above method of using trial and error to obtain all the factors of a polynomial, we can also use long division method, as shown in Example 23 below. Example 23 Show that (x + 2) is a factor of f(x) = 6x3 + 13x2 – 4. Hence, factorise f(x) completely and find the values of x such that f(x) = 0. Solution: f(x) = 6x3 + 13x2 – 4 f(–2) = 6(–2)3 + 13(–2)2 – 4 = –48 + 52 – 4 = 0 Hence, by factor theorem, (x + 2) is a factor of f(x). Using long division, 6x2 + x – 2 x + 2 6x3 + 13x2 + 0x – 4 6x3 + 12x2 x2 + 0x x2 + 2x – 2x – 4 – 2x – 4 0 f(x) = (x + 2)(6x2 + x – 2) = (x + 2)(3x + 2)(2x – 1) When x = –2, f(–2) = 0 When x = – 2 3 , f1– 2 3 2 = 0 When x = 1 2 , f1 1 2 2 = 0 The values of x such that f(x) = 0 are –2, – 2 3 and 1 2 . Note that –2, – 2 3 and 1 2 are called the zeros of f.
29 1 Mathematics Term 1 STPM Chapter 1 Functions Example 24 Given that (x – 2) is a factor of f(x), where f(x) ≡ ax3 – 10x2 + bx – 2, a, b R. When f(x) is divided by (x – 3), its remainder is 16. Find the values of a and b. Solution: Given that (x – 2) is a factor of f(x) ≡ ax3 – 10x2 + bx – 2. By the factor theorem, f(2) = 0 a(2)3 – 10(2)2 + b(2) – 2= 0 8a – 40 + 2b – 2 = 0 8a + 2b = 42 4a + b = 21 ………… Given that when f(x) is divided by (x – 3), the remainder is 16. By the remainder theorem, f(3) = 16 a(3)3 – 10(3)2 + b(3) – 2= 16 27a – 90 + 3b – 2 = 16 27a + 3b = 108 ………… 9a + b = 36 – : 5a = 15 a = 3 Substituting a = 3 into , 12 + b = 21 b = 9 Hence, f(x) = 3x3 – 10x2 + 9x – 2. Exercise 1.5 1. Use the remainder theorem to find the remainder when each of the following polynomials is divided by the linear polynomial given. (a) x3 + 4x2 – 3x + 2; x – 2 (b) x4 – 3x3 + 2x – 1; x + 2 (c) x5 – x3 + 6; x + 1 (d) x3 + 2x + 1; 2x – 1 (e) x3 – 2x2 + x + 1; 2x – 3 (f) x3 + 15x – 1; 3x + 1 2. Determine whether each of the following linear polynomials is a factor of the polynomial given. (a) x + 2; x4 + 4x3 + 4x2 (b) x – 1; x5 + 3x2 – 6x + 3 (c) x + 1; x3 – 2x2 + 6x + 9 (d) x – 2; x3 – 4x2 + 3x + 2 (e) 2x + 1; 2x3 – 3x2 + 2x + 2 (f) 2x – 1; x3 + 4x2 – 3x – 1 3. Factorise each of the following polynomials. (a) 2x3 – 3x2 + 1 (b) 3x3 – 2x2 – 7x – 2 (c) x4 – x2 – 72 (d) x5 + x3 + x (e) 4x3 – 13x + 6 (f) 4x4 – 4x3 – 9x2 + x + 2 4. If (x – 2) is a factor of ax3 + 3x2 – 2x + a, find the value of a. 5. Show that (x – a) is a factor of x3 + (1 – a)x2 + (3 – a)x – 3a.
30 1 Mathematics Term 1 STPM Chapter 1 Functions 6. When the polynomial x2 + ax + b is divided by (x – 1) and (x + 2), its remainder is 4 and 5 respectively. Find the values of a and b. 7. When x3 + px2 + qx + 1 is divided by (x – 2), its remainder is 9; when it is divided by (x + 3), its remainder is 19. Find the values of p and q. 8. The expression 2x3 + ax2 + b can be divided exactly by (x + 1), and its remainder is 16 when divided by (x – 3). Find the values of a and b. 9. The expression ax3 – 8x2 + bx + 6 can be divided exactly by x2 – 2x – 3. Find the values of a and b. 10. When the polynomials x3 + 4x2 – 2x + 1 and x3 + 3x2 – x + 7 are divided by (x – p), the remainders are equal. Find the possible values of p. 11. The expressions x3 – 4x2 + x + 6 and x3 – 3x2 + 2x + k have a common factor. Find the possible values of k. 12. If (x – a) is a factor of the expression ax3 – 3x2 – 5ax – 9, find the possible values of a. Factorise the expression for each of the values of a. 13. Given that f(x) ≡ x3 + kx2 – 2x + 1 has a remainder k when it is divided by (x – k), find the possible values of k. 14. Find the value of k if (x + 1) is a factor of 2x3 + 7x2 + kx – 3. Using this value of k, solve the equation 2x3 + 7x2 + kx – 3 = 0. 15. Find the values of a and b if f(x) ≡ ax3 + bx2 + 12 can be divided exactly by both (x + 1) and (x – 2) respectively. With these values of a and b, solve the equation f(x) = 0. 16. The expression x3 + ax2 + bx – 8 is divisible by (x + 1). When it is divided by (x – 2), its remainder is 42. Find the values of a and b and the value of the expression when x = 1. Polynomial and rational inequalities The relations such as x . 0, 2x – 1 < 0 and 3x2 + 2x . 1 are known as inequalities. The basic rules governing inequalities involving real numbers are as follows: For any a, b R with a . b, (a) a + c . b + c, c R, (b) ac . bc, c R, c . 0, (c) ac , bc, c R, c , 0. Inequalities involving algebraic polynomials also obey the basic rules of inequalities for real numbers. The set of numbers which satisfy an inequality is called the solution set. For example, if 2x – 3 < 4x + 9, then the solution set is {x : x –6}. An inequality may be solved according to the type of function given, using the analytical method or graphical method, or by drawing the real number line.
31 1 Mathematics Term 1 STPM Chapter 1 Functions Example 25 If x . 0, find the range of values of x which satisfy each of the following inequalities. (a) 5x – 2 , 3 2 x + 5 8 (b) 3x 2 + 1 x Solution: (a) 5x – 2 , 3 2 x + 5 8 5x – 3 2 x , 5 8 + 2 7 2 x , 21 8 x , 3 4 Since x . 0, the range of values of x for 5x – 2 , 3 2 x + 5 8 is 0 , x , 3 4 . (b) 3x 2 + 1 x 3x – 2 – 1 x 0 Since x . 0, 13x – 2 – 1 x 2 · x 0 i.e. 3x2 – 2x – 1 0 (3x + 1)(x – 1) 0 Since x . 0, (3x + 1) is always positive. Hence, x – 1 0 or x 1 The range of values of x for the inequality 3x 2 + 1 x is x 1. Example 26 Show that the following inequalities are true for all x R. (a) 2x2 + 8x + 9 . 0 (b) –3x2 + 2x – 5 , 0 Solution: (a) Let h(x) = 2x2 + 8x + 9 = 21x2 + 4x + 9 2 2 = 21x2 + 4x + 22 – 22 + 9 2 2 = 23(x + 2)2 + 1 2 4 = 2(x + 2)2 + 1 Now, for all x R, (x + 2)2 0 Hence, 2x2 + 8x + 9 . 0 for all x R. (b) Let k(x) = –3x2 + 2x – 5 = –31x2 – 2 3 x + 5 3 2 = –33x2 – 2 3 x + 1– 1 3 2 2 – 1– 1 3 2 2 + 5 3 4 = –331x – 1 3 2 2 + 14 9 4 = –31x – 1 3 2 2 – 14 3
32 1 Mathematics Term 1 STPM Chapter 1 Functions For all x R, –31x – 1 3 2 2 < 0. Hence, –3x2 + 2x – 5 , 0 for all x R. Alternative Method: (a) h(x) = 2x2 + 8x + 9 with a = 2, b = 8, c = 9. b2 – 4ac = 82 – 4(2)(9) = 64 – 72 = –8 Hence, b2 – 4 ac , 0 and the graph of y = h(x) does not intersect the x-axis. Since a . 0, the graph of y = h(x) is always above the x-axis. Hence, 2x2 + 8x + 9 . 0 for all x R. (b) k(x) = –3x2 + 2x – 5 with a = –3, b = 2 , c = –5. b2 – 4ac = 22 – 4(–3)(–5) = 4 – 60 = –56 Hence, b2 – 4ac , 0 and the graph of y = k(x) does not intersect the x-axis. Since a , 0, the graph of y = k(x) is always below the x-axis. Hence, –3x2 + 2x – 5 , 0 for all x R. Example 27 Find the set of values of x which satisfy the inequality 2x2 + x . 3. Solution: Given 2x2 + x . 3 or 2x2 + x – 3 . 0 (2x + 3)(x – 1) . 0 Consider the graph of the function f(x) = (2x + 3)(x – 1) At the points of intersection with the x-axis, (2x + 3)(x – 1) = 0 Hence 2x + 3 = 0 or x – 1 = 0 x = – 3 2 or x = 1 From the graph of f(x), f(x) . 0 if x , – 3 2 or x . 1. Hence, the range of values of x which satisfy the inequality 2x2 + x . 3 is x , – 3 2 or x . 1. The solution set of the inequality is {x : x , – 3 2 or x . 1}. 0 y x y = h(x) 0 y x y = k(x) 3 0 1 – _ 2 x y f(x) = 2x 2 + x – 3
33 1 Mathematics Term 1 STPM Chapter 1 Functions Example 28 If f(x) = 2x3 – 9x2 + 3x + 14, factorise f(x). Sketch the graph of y = f(x). Hence, find the set of values of x such that f(x) 0. Solution: f(x) = 2x3 – 9x2 + 3x + 14 f(–1) = –2 – 9 – 3 + 14 = 0 Hence, (x + 1) is a factor of f(x). Using long division method, f(x) = (x + 1)(2x2 – 11x + 14) = (x + 1)(x – 2)(2x – 7) At the x-axis, f(x) = 0 i.e. (x + 1)(x – 2)(2x – 7) = 0 7 x = –1, 2 or — 2 So the graph of f(x) intersects the x-axis at x = –1, 2 and 7 2 . The graph of f(x) is as shown on the right. From the sketch graph of y = f(x), we see that f(x) 0 if –1 < x < 2 or x 7 2 . Hence, the set of values of x such that f(x) 0 is {x : –1 < x < 2 or x 7 2 }. Example 29 Find the set of values of x such that x . 6 x + 1. Solution: Given that x . 6 x + 1 x – 6 x – 1 . 0 x2 – x – 6 x . 0 (x – 3)(x + 2) x . 0 Let f(x) = (x – 3)(x + 2) x Notice that f(x) changes sign when x passes through x = 3, 0 and –2. We draw up a table for the sign of f(x) by considering the sign of each factor in each interval of x, as shown below. x , –2 –2 , x , 0 0 , x , 3 x . 3 (x + 2) – + + + x – – + + (x – 3) – – – + f(x) – + – + From the above table, we see that f(x) . 0 or x . 6 x + 1, when –2 , x , 0 or x . 3. Hence, the set of values of x is {x : –2 , x , 0 or x . 3}. –1 0 2 _7 2 x y y = f(x) The sign of f(x) is obtained by multiplying the signs of the 3 factors in each interval.
34 1 Mathematics Term 1 STPM Chapter 1 Functions Example 30 If x x + 8 < 1 x – 1 , find the set of values of x which satisfy the inequality. Solution: Given that x x + 8 < 1 x – 1 x x + 8 – 1 x – 1 < 0 x(x – 1) – (x + 8) (x + 8)(x – 1) < 0 x2 – 2x – 8 (x + 8)(x – 1) < 0 (x + 2)(x – 4) (x + 8)(x – 1) < 0 By considering each of the four factors as positive, we get x + 2 . 0, i.e. x . –2 x – 4 . 0, i.e. x . 4 x + 8 . 0, i.e. x . –8 x – 1 . 0, i.e. x . 1 These ranges can be represented on the real number line as follows. – 8 –2 0 1 4 x + 2 > 0 x – 4 > 0 x – 1 > 0 x x + 8 > 0 The numerator (x + 2)(x – 4) 0 in {x : x < –2 or x 4} The denominator (x + 8)(x – 1) . 0 in {x : x , –8 or x . 1} Hence, (x + 2)(x – 4) (x + 8)(x – 1) < 0 if the numerator and denominator are of opposite signs, i.e. when –8 , x < –2 or 1 , x < 4. Hence, the set of values of x such that x x + 8 < 1 x – 1 is {x : –8 , x < –2 or 1 , x < 4}. Inequalities involving modulus signs Inequalities involving absolute value or modulus signs may be solved using the analytical or graphical method. Example 31 Find the values of x such that |2x + 1| , 3. Solution: Given that |2x + 1| , 3 This means –3 , 2x + 1 , 3 –4 , 2x , 2 –2 , x , 1 Hence, the inequality is valid if –2 , x , 1, i.e. x (–2, 1).
35 1 Mathematics Term 1 STPM Chapter 1 Functions Note: Some inequalities involving moduli may be solved by squaring both sides of the inequality. However, it is only valid if both sides of the inequality are positive or zero for all x R. Example 32 Find the values of x such that 2|x – 1| < |x + 3|. Solution: Given that 2|x – 1| < |x + 3| 4(x – 1)2 < (x + 3)2 4x2 – 8x + 4 < x2 + 6x + 9 3x2 – 14x – 5 < 0 (3x + 1)(x – 5) < 0 From the sketch graph of f(x) = 3x2 – 14x – 5, we see that f(x) < 0 if – 1 3 < x < 5, i.e. in the closed interval 3– 1 3 , 54. Example 33 Using graphical method, find the range of values of x for which the inequality |2x – 1| < |x| + 3 2 is valid. Solution: We first sketch the graphs of y = |2x – 1| and y = |x| + 3 2 as follows: —5 2 1 – — 2 —1 2 y x O 1 —3 2 y = –x + — 3 2 y = –2x + 1 y = x + — 3 2 y = 2x – 1 A B The graphs of y = |2x – 1| and y = |x| + 3 2 intersect at A and B. For A, we solve y = 2x – 1 and y = x + 3 2 , i.e. 2x – 1 = x + 3 2 ⇒ x = 5 2 . For B, we solve y = –2x + 1 and y = –x + 3 2 , i.e. –2x + 1 = –x + 3 2 ⇒ x = – 1 2 . From the graphs, we notice that |2x – 1| < |x| + 3 2 is valid in the range – 1 2 < x < 5 2 . Both LHS and RHS are > 0. Squaring both sides of inequality. 1 0 5 – _ 3 x f(x ) f(x) = 3x 2 – 14x – 5
36 1 Mathematics Term 1 STPM Chapter 1 Functions Exercise 1.6 1. With the aid of a sketch graph, solve each of the following inequalities. (a) (x + 1)(x – 2) < 0 (b) (x – 3)(x + 5) . 0 (c) (2x – 3)(x + 4) , 0 (d) (2x + 1)(4x – 1) > 0 (e) ( 1 2 x + 5)(x – 3) < 0 (f) (x – 2)(5x + 2) . 0 2. Solve each of the following inequalities. (a) x2 > 9 (b) x2 + 2x + 1 . 0 (c) x(x + 1) < –2(2x + 3) (d) 5x2 < 3x + 2 (e) (x – 2)2 . 9x2 (f) 3x2 – 2x > x2 + 3x + 3 3. Find the range of values of x such that each of the following inequalities is valid. (a) (x + 2)(x – 1)(x + 3) , 0 (b) (x – 2)2 (x + 1) < 0 (c) x3 + 3x2 – 4 > 0 (d) 2x3 + 3x2 – 3x , 2 (e) x(5x2 + 8) < 1 2 (47x2 – 48) (f) 2x3 > 7x2 + 17x – 10 4. Find the range of values of x which satisfy each of the following inequalities. (a) 4 x + 3 . 2 – x (b) 4 – 5x 1 – 2x . 3 (c) 14 x – 2 > 2x – 1 (d) 13 – 4x x – 1 , 35 x – 3 (e) 9 4 – x < 7x + 5 x + 3 (f) x + 1 2x – 1 . 3 x – 2 5. Find the set of values of x which satisfy each of the following inequalities. (a) |x – 2| , 1 (b) |x – 3| > 5 (c) |3x + 4| . 5 (d) |2x – 5| < 11 (e) |x| > |x – 1| (f) 2|x – 2| , |x – 3| (g) 3|x + 2| < |x – 6| (h) 5|2x – 3| . 4|x – 5| (i) |2x + 1| , 3x + 2 (j) x x + 4 , 2 (k) x2 – 4 x < 3 (l) x + 1 x – 1 , 1 Partial fractions Let f(x) = 2x + 1 and g(x) = x2 + 3x + 2. When f(x) is divided by g(x), the resulting function 2x + 1 2x + 1 –––––––––– ≡ –––––––––––– x2 + 3x + 2 (x + 1)(x + 2) is known as a rational function. Notice that in the above case, f(x) is linear and g(x) is quadratic, i.e. the degree of f(x) is less than the degree of g(x). If f(x) is a polynomial function of degree m, and g(x) is a polynomial of degree n, f(x) where m , n, then h(x) =— is considered a proper rational function. g(x) x + 2 x2 + 1 2 For example, –––––––––––– , –––––––––– and –––––– are all proper rational functions. (x – 1)(x + 3) 2x3 + x + 1 3x + 1 f(x) However, if m n, then h(x) = —is considered an improper rational function. g(x) x2 + 2x + 3 x2 + 3x + 1 2x3 + x2 – 3x + 1 For example, –––––––––––, ––––––––––– and ––––––––––––––– are all improper rational functions. x + 1 x2 + 4x – 5 (x + 1)(x + 2)
37 1 Mathematics Term 1 STPM Chapter 1 Functions Denominator with linear factors Consider the addition of two proper rational functions as follows. f(x) = 2 x + 1 + 3 x + 2 The function f(x) can be expressed as a single rational function with a common denominator, i.e. f(x) = 2 x + 1 + 3 x + 2 = 2(x + 2) + 3(x + 1) (x + 1)(x + 2) = 5x + 7 (x + 1)(x + 2) Notice that the addition of two proper rational functions results in a single proper rational function. From the operation of addition as illustrated above, we see that the inverse process can also be carried out, i.e. expressing the proper rational function 5x + 7 (x + 1)(x + 2) as the sum of two rational functions with denominators (x + 1) and (x + 2) respectively. Let 5x + 7 (x + 1)(x + 2) ≡ A x + 1 + B x + 2 , where A and B are constants. Hence 5x + 7 (x + 1)(x + 2) ≡ A(x + 2) + B(x + 1) (x + 1)(x + 2) 5x + 7 ≡ A(x + 2) + B(x + 1) = (A + B)x + (2A + B) Equating coefficients of x: 5 = A + B ………… Equating constants: 7 = 2A + B ………… – : 2 = A Substituting A = 2 into , 5 = 2 + B B = 3 This process is known as expressing the function 5x + 7 (x + 1)(x + 2) in partial fractions with terms 2 x + 1 and 3 x + 2 . Notice that these two partial fractions are proper rational functions with the numerator as a constant and the denominator as a linear function. Example 34 Express 2x + 3 (x – 1)(x + 2) in partial fractions. Solution: Let 2x + 3 (x – 1)(x + 2) ≡ A x – 1 + B x + 2 ≡ A(x + 2) + B(x – 1) (x – 1)(x + 2) 2x + 3 ≡ A(x + 2) + B(x – 1) This is an identity and is valid for all x ∈ R
38 1 Mathematics Term 1 STPM Chapter 1 Functions When x = 1, 2 + 3 = A(3) + B(0) 5 = 3A A = 5 3 When x = –2, –4 + 3 = A(0) + B(–3) –1 = –3B B = 1 3 Hence, 2x + 3 (x – 1)(x + 2) ≡ 5 3(x – 1) + 1 3(x + 2) Denominator with prime quadratic factors We have learnt that for a partial fraction with a linear denominator, its numerator is a constant and the partial fraction is a proper rational function. In the same way, if the denominator of a partial fraction is a quadratic function, its numerator must be a constant or a linear function such that the partial fraction is a proper rational function. Hence, if the denominator of a proper rational function has a prime quadratic factor in the form ax2 + bx + c, where a, b and c are constants, the given partial fraction must be of the form Ax + B ax2 + bx + c , where A and B are suitable constants. Example 35 Express 3x + 4 (x + 2)(x2 – x + 1) in partial fractions. Solution: Let 3x + 4 (x + 2)(x2 – x + 1) ≡ A x + 2 + Bx + C x2 – x + 1 ≡ A(x2 – x + 1) + (Bx + C)(x + 2) (x + 2)(x2 – x + 1) 3x + 4 ≡ A (x2 – x + 1) + (Bx + C)(x + 2) When x = –2, –6 + 4 = A(4 + 2 + 1) + 0 –2 = 7A A = – 2 7 Equating coefficients of x2 : 0 = A + B B = 2 7 Equating the constants: 4 = A + 2C = – 2 7 + 2C 2C = 30 7 C = 15 7 Hence, 3x + 4 (x + 2)(x2 – x + 1) ≡ – 2 7(x + 2) + 2x + 15 7(x2 – x + 1) Notice that the two partial fractions are proper rational functions with the numerator of one of them as a linear function, i.e. 2x + 15.
39 1 Mathematics Term 1 STPM Chapter 1 Functions Denominator with repeated linear factors If one of the terms in the denominator of a proper rational function is a repeated linear factor such as (x + 1)2 , this factor can be considered to be a quadratic factor. So, the partial fraction concerned is of the form Ax + B (x + 1)2 . Suppose we want to express 3x2 + 6x – 1 (x – 1)(x + 1)2 in partial fractions. Then, 3x2 + 6x – 1 (x – 1)(x + 1)2 ≡ A x – 1 + Bx + C (x + 1)2 ≡ A(x + 1)2 + (Bx + C)(x – 1) (x – 1)(x + 1)2 3x2 + 6x – 1 ≡ A(x + 1)2 + (Bx + C)(x – 1) When x = 1, 3 + 6 – 1 = 4A 8 = 4A A = 2 Equating coefficients of x2 : 3 = A + B 3 = 2 + B B = 1 Equating constants: –1 = A – C –1 = 2 – C C = 3 Therefore, 3x2 + 6x – 1 (x – 1)(x + 1)2 ≡ 2 x – 1 + x + 3 (x + 1)2 Notice that the partial fractions obtained is correct, except that it is not in its simplest form, because x + 3 (x + 1)2 ≡ (x + 1) + 2 (x + 1)2 ≡ 1 x + 1 + 2 (x + 1)2 Hence, the partial fractions in its simplest form is 3x2 + 6x – 1 (x – 1)(x + 1)2 ≡ 2 x – 1 + 1 x + 1 + 2 (x + 1)2 This means that the function 3x2 + 6x – 1 (x – 1)(x + 1)2 can be expressed as partial fractions in the following way. Let 3x2 + 6x – 1 (x – 1)(x + 1)2 ≡ A x – 1 + B x + 1 + C (x + 1)2 ≡ A(x + 1)2 + B(x – 1)(x + 1) + C(x – 1) (x – 1)(x + 1)2 Hence, 3x2 + 6x – 1 ≡ A(x +1)2 + B(x – 1)(x + 1) + C(x – 1) When x = 1, 3 + 6 – 1 = 4A 8 = 4A A = 2 When x = –1, 3 – 6 – 1 = –2C –4 = –2C C = 2
40 1 Mathematics Term 1 STPM Chapter 1 Functions Equating the coefficients of x2 : 3 = A + B 3 = 2 + B B = 1 Therefore, 3x2 + 6x – 1 (x – 1)(x + 1)2 ≡ 2 x – 1 + 1 x + 1 + 2 (x + 1)2 In the same way, if the denominator of a proper rational function contains a repeated linear factor such as (x + 1)3 , the denominator of the partial fractions concerned are (x + 1), (x + 1)2 and (x + 1)3 respectively. Example 36 Express x2 + 1 (x – 1)(x + 1)3 in partial fractions. Solution: Since the denominator contains the repeated linear factor (x + 1)3 , the partial fractions consist of terms with denominators (x +1), (x + 1)2 and (x + 1)3 . Hence, x2 + 1 (x – 1)(x + 1)3 ≡ A x – 1 + B x + 1 + C (x + 1)2 + D (x + 1)3 Multiply both sides of the identity with (x – 1)(x + 1)3 , x2 + 1 ≡ A(x + 1)3 + B(x – 1)(x + 1)2 + C(x – 1)(x + 1) + D(x – 1) When x = 1, 2 = 8A A = 1 4 When x = –1, 2 = –2D D = –1 Equating the coefficients of x3 , 0 = A + B 0 = 1 4 + B B = – 1 4 When x = 0, 1 = A – B – C – D = 1 4 + 1 4 – C + 1 C = 1 2 Hence, x2 + 1 (x – 1)(x + 1)3 ≡ 1 4(x – 1) – 1 4(x + 1) + 1 2(x + 1)2 – 1 (x + 1)3 . Degree of denominator less than or equal to numerator The rational function f(x) g(x) , where the degrees of f(x) and g(x) are m and n respectively, with m n, is considered to be an improper rational function. In this case, f(x) must first be divided by g(x) before expressing in terms of partial fractions.
41 1 Mathematics Term 1 STPM Chapter 1 Functions Example 37 Express x2 + 2x + 3 (x + 2)(x – 1) in partial fractions. Solution: x2 + 2x + 3 (x + 2)(x – 1) ≡ x2 + 2x + 3 x2 + x – 2 ≡ (x2 + x – 2) + (x + 5) x2 + x – 2 = 1 + x + 5 x2 + x – 2 = 1 + x + 5 (x + 2)(x – 1) 14243 Let x + 5 (x + 2)(x – 1) ≡ A x + 2 + B x – 1 ≡ A(x – 1) + B(x + 2) (x + 2)(x – 1) x + 5 ≡ A(x – 1) + B(x + 2) When x = 1, 6 = 0 + 3B B = 2 When x = –2, –2 + 5 = –3A + 0 3 = –3A A = –1 x + 5 (x + 2)(x – 1) ≡ – 1 x + 2 + 2 x – 1 Hence, x2 + 2x + 3 (x + 2)(x – 1) ≡ 1 – 1 x + 2 + 2 x – 1 Example 38 Express x 3 + 1 (x – 1)(x + 2) in partial fractions. Solution: We have x3 + 1 (x – 1)(x + 2) ≡ x3 + 1 x2 + x – 2 By using long division, x – 1 x2 + x – 2 x3 + 0x2 + 0x + 1 x3 + x2 – 2x – x2 + 2x + 1 – x2 – x + 2 3x – 1 Write x 3 + 1 as x3 + 0x2 + 0x + 1. Hence, x3 + 1 (x – 1)(x + 2) ≡ (x – 1) + 3x – 1 (x – 1)(x + 2) ≡ (x – 1) + A x – 1 + B x + 2 x3 + 1 ≡ (x – 1)2 (x + 2) + A(x + 2) + B(x –1) x + 5 –––––––––– can be expressed (x +2)(x – 1) as partial fractions.
42 1 Mathematics Term 1 STPM Chapter 1 Functions When x = 1, 1 + 1 = 0 + 3A + B(0) 2 = 3A A = 2 3 When x = –2, –8 + 1 = 0 + A(0) + B(–3) –7 = –3B B = 7 3 Hence, x3 + 1 (x – 1)(x + 2) = x – 1 + 2 3(x – 1) + 7 3(x + 2) Exercise 1.7 Express each of the following expressions in partial fractions. 1. 2x + 4 (x + 1)(x + 2) 2. x – 4 (x + 2)(x – 1) 3. 3 x2 – 9 4. x x2 – 1 5. x + 1 x2 – 4 6. x + 3 x2 + 5x + 4 7. 4x + 10 (2 + x)(3 + x)(4 + x) 8. 4x + 2 (x + 2)(x2 – 1) 9. x2 – 2x + 4 2x(x – 3)(x + 1) 10. x2 – 4 x(x + 1)(x + 3) 11. x (1 + x) 2 12. 9 (x – 1)(x + 2)2 13. 3x2 + x – 2 (2x – 1)(x – 2)2 14. x2 – 1 x2 (2x + 1) 15. 1 x2 (x2 – 1) 16. x2 + 1 x2 – 1 17. x2 + x – 1 (x + 1)(x + 2) 18. 2x2 + 3x + 2 x2 + 3x + 2 19. x3 + x + 1 (x + 1)2 20. x3 + 3 (x + 1)(x – 1) 21. 2x – 1 (x + 2)(x2 + 1) 22. x + 1 x(x2 + x + 1) 23. 3x (x – 1)(x2 + x + 1) 24. 1 (x – 1)(x2 – x + 1) 25. x (x + 1)(x – 1)3
43 1 Mathematics Term 1 STPM Chapter 1 Functions 1.3 Exponential and Logarithmic Functions Exponential functions An exponential function f is defined by f(x) = ax , where x R and a 0, a ≠ 1. Examples of exponential functions are, for example, 2x , 3–2x , 10x – 1 and 53x + 1. For the exponential function f(x) = ax , x R and a 1, the basic properties are as follows: (a) f(x) 0 for x R. (b) As x increases, f(x) also increases at a quicker rate. (c) f(x) = 1 when x = 0, i.e. f(0) = 1. (d) As x decreases (in the direction of the negative x-axis), f(x) also decreases at a quicker rate until as x → –∞, f(x) → 0. (y = 0 is an asymptote.) Based on the above properties, the graph of y = ax , a 1, may be sketched. y x 0 y = ax (a > 1) 1 Figure 1.16 For the exponential function f(x) = ax , x R and 0 a 1, the basic properties are as follows: (a) f(x) 0 for x R. (b) As x increases, f(x) decreases at a quicker rate until as x → ∞, f(x) → 0. (y = 0 is an asymptote.) (c) f(x) = 1 when x = 0, i.e. f(0) = 1. (d) As x decreases (in the direction of the negative x-axis), f(x) increases at a quicker rate. Based on the above properties, the graph of y = ax , 0 a 1, may be sketched. y x 0 y = ax (0 < a < 1) 1 Figure 1.17
44 1 Mathematics Term 1 STPM Chapter 1 Functions Notice that when 0 a 1, 1 a . 1. Let 1 a = b, 0 a 1. Then 1 b = a. Hence, the function y= ax = 1 1 b 2 x = 1 bx = b–x, b 1 Therefore, the graph of y = ax , 0 a 1, is equivalent to the graph of y = a–x , a . 1. Hence, the graph of y = a–x can be obtained by reflecting the graph of y = ax about the y-axis, as shown in Figure 1.18. y x 0 y = ax y = a (a > 1) –x (a > 1) 1 Figure 1.18 Logarithmic functions Consider an exponential function f : x ↦ ax , x R, a 0, a ≠ 1. This function is a one-to-one function in the range R+ . Therefore, this function f possesses an inverse function. This inverse function of an exponential function is called a logarithmic function and is written as f–1 : x ↦ loga x, x R+ , a 0, a ≠ 1. or g(x) = loga x. The basic properties of a logarithmic function g(x) = loga x, a 1, are as follows: (a) The function is undefined for x 0. (b) When x = 1, g(x) = 0, i.e. g(1) = 0. (c) For x 1, g(x) 0 and when x → ∞, g(x) → ∞. (d) For 0 x 1, g(x) 0 and when x → 0, g(x) → –∞. Based on the above properties, the graph of y = loga x, a 1, can be drawn. Exponential Functions VIDEO