CAPE PURE
MATHΣMATICS
UNIT 1
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Table of Contents
MODULE ONE: ALGEBRA, RELATIONS and FUNCTIONS............................................................................................ 5
CHAPTER 1: INEQUALITIES ..............................................................................................................5
QUADRATIC INEQUALITIES..................................................................................................................................... 5
RATIONAL INEQUALITIES........................................................................................................................................ 7
CHAPTER 2: THE DISCRIMINANT ......................................................................................................9
CALCULATING THE DISCRIMINANT ........................................................................................................................ 9
THE DISCRIMINANT AND DISTINCT REAL ROOTS................................................................................................... 9
THE DISCRIMINANT AND EQUAL ROOTS ............................................................................................................. 10
THE DISCRIMINANT AND NO REAL ROOTS .......................................................................................................... 10
CHAPTER 3: SURDS ....................................................................................................................... 12
PROPERTIES OF SURDS ......................................................................................................................................... 12
SURD SIMPLIFICATION ......................................................................................................................................... 12
FRACTIONS INVOLVING SURDS ............................................................................................................................ 13
CHAPTER 4: LAWS OF INDICES ....................................................................................................... 15
MULTIPLICATION .................................................................................................................................................. 15
DIVISION ............................................................................................................................................................... 15
RAISING A POWER TO A POWER .......................................................................................................................... 15
NEGATIVE INDICES ............................................................................................................................................... 15
ZERO INDEX .......................................................................................................................................................... 16
FRACTIONAL INDICES ........................................................................................................................................... 16
FURTHER LESSONS ............................................................................................................................................... 16
CHAPTER 5: DISGUISED QUADRATIC EQUATIONS........................................................................... 18
CHAPTER 6: FUNCTIONS ................................................................................................................ 20
DEFINITION OF A FUNCTION ................................................................................................................................ 20
REPRESENTING A FUNCTION................................................................................................................................ 20
MAPPING DIAGRAMS ....................................................................................................................................... 20
ORDERED PAIRS................................................................................................................................................ 21
EQUATIONS ...................................................................................................................................................... 21
GRAPHS OF FUNCTIONS ................................................................................................................................... 21
THE VERTICAL LINE TEST ...................................................................................................................................... 22
THE RANGE OF A FUNCTION ................................................................................................................................ 22
POLYNOMIAL FUNCTIONS ................................................................................................................................ 22
SQUARE ROOT FUNCTIONS .............................................................................................................................. 23
RATIONAL FUNCTIONS ..................................................................................................................................... 23
CLASSES OF FUNCTIONS....................................................................................................................................... 24
INJECTIONS ....................................................................................................................................................... 24
DETERMINING WHETHER A FUNCTION IS ONE –TO – ONE ............................................................................. 24
THE HORIZONTAL LINE TEST ............................................................................................................................ 25
SURJECTIONS .................................................................................................................................................... 26
DETERMINING WHETHER A FUNCTION IS ONTO............................................................................................. 26
BIJECTIONS ....................................................................................................................................................... 26
FUNCTIONS AND THEIR INVERSES ....................................................................................................................... 28
ONE – TO – ONE QUADRATIC FUNCTIONS........................................................................................................... 29
PIECE-WISE FUNCTIONS ....................................................................................................................................... 31
1
CHAPTER 7: POLYNOMIALS ........................................................................................................... 34
SYNTHETIC DIVISION ............................................................................................................................................ 34
REMAINDER AND FACTOR THEOREM .................................................................................................................. 35
ROOTS OF POLYNOMIALS .................................................................................................................................... 37
CUBIC POLYNOMIALS ....................................................................................................................................... 39
FACTORING POLYNOMIALS .................................................................................................................................. 42
CHAPTER 8: LOGARITHMS ............................................................................................................. 43
PROPERTIES OF LOGARITHMS.............................................................................................................................. 44
LOGARITHMIC EQUATIONS .................................................................................................................................. 44
THE EXPONENTIAL FUNCTION ............................................................................................................................. 46
THE NATURAL LOGARITHM .................................................................................................................................. 47
LOGARITHMIC AND EXPONENTIAL EQUATIONS .................................................................................................. 47
UNKNOWN INDICES ............................................................................................................................................. 49
CHAPTER 9: MODULUS/ ABSOLUTE VALUE FUNCTION ................................................................... 51
MODULUS EQUATIONS ........................................................................................................................................ 51
MODULUS INEQUALITIES ..................................................................................................................................... 52
GRAPHS OF MODULUS FUNCTIONS ................................................................................................................. 54
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION............................................... 56
SEQUENCES .......................................................................................................................................................... 56
SERIES ................................................................................................................................................................. 57
SPECIAL SUMMATION FORMULAE............................................................................................................... 59
MATHEMATICAL INDUCTION ........................................................................................................................... 61
PROOF OF SUMMATION................................................................................................................................. 61
PROOF OF DIVISIBILITY ................................................................................................................................ 63
CHAPTER 11: THE REAL NUMBER SYSTEM ................................................................................... 65
BINARY OPERATIONS............................................................................................................................................ 65
CAYLEY TABLE....................................................................................................................................................... 65
CLOSURE ............................................................................................................................................................... 65
COMMUTATIVITY ................................................................................................................................................. 65
ASSOCIATIVITY...................................................................................................................................................... 66
IDENTITY AND INVERSE ........................................................................................................................................ 67
DIRECT PROOFS .................................................................................................................................................... 68
CHAPTER 12: REASONING AND LOGIC .......................................................................................... 69
TRUTH TABLES................................................................................................................................................... 69
CONVERSE, INVERSE AND CONTRAPOSITIVE................................................................................................ 70
LOGICAL EQUIVALENCE .................................................................................................................................... 70
LAWS OF BOOLEAN ALGEBRA .......................................................................................................................... 71
MODULE TWO: VECTORS, TRIGONOMETRY AND COORDINATE GEOMETRY ....................................................... 73
CHAPTER 13: TWO DIMENSIONAL VECTORS .................................................................................. 73
𝒊, 𝒋 REPRESENTATION ........................................................................................................................................... 73
𝒊, 𝒋, 𝒌 REPRESENTATION ....................................................................................................................................... 74
UNIT VECTORS...................................................................................................................................................... 74
SCALAR (DOT) PRODUCT ...................................................................................................................................... 74
ANGLE BETWEEN TWO VECTORS......................................................................................................................... 75
THE VECTOR EQUATION OF A LINE ...................................................................................................................... 78
VECTOR EQUATION OF A PLANE .......................................................................................................................... 79
CHAPTER 14: THE EQUATION OF A CIRCLE ..................................................................................... 82
DETERMINING THE EQUATION OF A CIRCLE........................................................................................................ 82
2
FINDING THE CENTRE AND RADIUS OF A CIRCLE ................................................................................................. 82
TANGENTS AND NORMALS .................................................................................................................................. 83
THE INTERSECTION OF TWO CIRCLES .................................................................................................................. 85
EQUATION OF A CIRCLE GIVEN 3 POINTS ............................................................................................................ 85
CHAPTER 15: THE LOCUS OF A POINT ............................................................................................ 87
CHAPTER 16: RADIAN MEASURE.................................................................................................... 89
CONVERTING RADIANS TO DEGREES ................................................................................................................... 89
CONVERTING DEGREES TO RADIANS ................................................................................................................... 89
ARC LENGTH ......................................................................................................................................................... 89
AREA OF SECTOR .................................................................................................................................................. 89
CHAPTER 17: TRIGONOMETRY ...................................................................................................... 92
TRIGONOMETRIC IDENTITIES ............................................................................................................................... 92
PROVING TRIGONOMETRIC IDENTITIES........................................................................................................... 93
GRAPHS OF TRIGONOMETRIC FUNCTIONS .......................................................................................................... 94
GRAPHS OF RECIPROCAL FUNCTIONS.............................................................................................................. 95
SKETCHING TRIGONOMETRIC GRAPHS ............................................................................................................ 95
TRIGONOMETRIC EQUATIONS ............................................................................................................................. 96
GENERAL SOLUTIONS ....................................................................................................................................... 96
SOLVING TRIGONOMETRIC EQUATIONS.......................................................................................................... 98
HARMONIC FORM............................................................................................................................................. 101
COMPOUND ANGLE FORMULAE ........................................................................................................................ 103
DOUBLE–ANGLE FORMULAE.............................................................................................................................. 105
HALF–ANGLE FORMULAE .................................................................................................................................. 106
FACTOR FORMULAE......................................................................................................................................... 107
CHAPTER 18: PARAMETRIC EQUATIONS..................................................................................... 112
MODULE THREE: CALCULUS.................................................................................................................................... 115
CHAPTER 19: LIMITS................................................................................................................... 115
LIMITS BY DIRECT SUBSTITUTION ................................................................................................................ 116
LIMIT PROPERTIES .......................................................................................................................................... 117
LIMITS AS 𝒙 APPROACHES INFINITY ............................................................................................................. 117
LIMITS AS 𝒙 APPROACHES 0 ........................................................................................................................... 118
Limit of 𝐬𝐢𝐧𝒙𝒙 ................................................................................................................................................ 118
LIMITS AND PIECEWISE FUNCTIONS............................................................................................................. 119
PIECEWISE FUNCTIONS AND CONTINUITY .............................................................................................. 120
DIFFERENTIATION FROM FIRST PRINCIPLES ...................................................................................................... 123
CHAPTER 20: DIFFERENTIATION .................................................................................................. 125
HOW TO DIFFERENTIATE.................................................................................................................................... 125
THE DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ..................................................................................... 126
PROPERTIES OF DERIVATIVES ........................................................................................................................ 126
DIFFERENTIATION RULES ............................................................................................................................... 126
THE SECOND DERIVATIVE............................................................................................................................... 128
PARAMETRIC DIFFERENTIATION .............................................................................................................. 129
APPLICATIONS OF DIFFERENTIATION................................................................................................................. 133
GRADIENTS AND DIFFERENTIATION............................................................................................................... 133
INCREASING AND DECREASING FUNCTIONS.................................................................................................. 134
POINTS OF INFLECTION .................................................................................................................................. 136
CURVE SKETCHING ...................................................................................................................................... 137
RATE OF CHANGE ........................................................................................................................................... 138
3
CHAPTER 21: INTEGRATION ........................................................................................................ 145
HOW TO INTEGRATE .......................................................................................................................................... 145
INTEGRATION NOTATION............................................................................................................................... 145
PROPERTIES OF INTEGRALS............................................................................................................................ 145
TRIGONOMETRIC INTEGRATION .................................................................................................................... 146
DIFFERENTIAL EQUATIONS............................................................................................................................. 147
DEFINITE INTEGRALS ...................................................................................................................................... 147
INTEGRATION BY SUBSTITUTION.............................................................................................................. 148
APPLICATIONS OF INTEGRATION .................................................................................................................. 151
THE EQUATION OF A CURVE .......................................................................................................................... 151
THE AREA UNDER A GRAPH ........................................................................................................................... 151
VOLUME OF REVOLUTION ABOUT THE 𝒙 AXIS .............................................................................................. 156
VOLUMES OF REVOLUTION ABOUT THE 𝒚 AXIS ...................................................................................... 156
CAPE 2012 .................................................................................................................................. 168
CAPE 2013 .................................................................................................................................. 171
CAPE 2014 .................................................................................................................................. 175
CAPE 2015 .................................................................................................................................. 178
CAPE 2016 .................................................................................................................................. 181
CAPE 2017 .................................................................................................................................. 184
ANSWERS FOR CAPE PAST PAPERS .............................................................................................. 188
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4
CHAPTER 1: INEQUALITIES
MODULE ONE: ALGEBRA, RELATIONS
and FUNCTIONS
CHAPTER 1: INEQUALITIES
At the end of this section, students should be able
to:
 find the solution set s of quadratics
inequalities using algebraic and graphical
methods
 find the solution sets of inequalities of the
𝑎𝑥+𝑏
form 𝑐𝑥+𝑑 > 0; ≥ 0; < 0; ≤ 0 using
algebraic and graphical methods.
QUADRATIC INEQUALITIES
Quadratic inequalities deal with determining the
range of values of 𝑥 which satisfy
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 > 0, 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 < 0 as well as
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ≥ 0 and 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ≤ 0.
If 𝑥1 and 𝑥2 are the roots of a quadratic equation,
with 𝑥1 < 𝑥2, then we have the following possible
ranges as solutions to the varying quadratic
inequalities.
NB: WE ideally want the coefficient of 𝑥 2 to be
positive so that the graphs have the shape below.
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 > 0
{𝑥 < 𝑥1 } ∪ {𝑥 > 𝑥2 }
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ≥ 0
{𝑥 ≤ 𝑥1 } ∪ {𝑥 ≥ 𝑥2 }
LESSON 1
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 < 0
Solution: 𝑥1 < 𝑥 < 𝑥2
Solve the inequality
𝑥 2 − 7𝑥 < 0
SOLUTION
𝑥 2 − 7𝑥 < 0
𝑥(𝑥 − 7) < 0
Roots: 𝑥 = 0, 7
0<𝑥<7
LESSON 2
Solve the inequality
𝑥 2 + 2𝑥 − 8 < 0
SOLUTION
𝑥 2 + 2𝑥 − 8 < 0
(𝑥 + 4)(𝑥 − 2) < 0
Roots 𝑥 = −4, 2
{−4 < 𝑥 < 2}
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ≤ 0
Solution: 𝑥1 ≤ 𝑥 ≤ 𝑥2
5
CHAPTER 1: INEQUALITIES
LESSON 3
Determine the range of values of
𝑥 for which 3𝑥 2 − 5𝑥 ≥ 2.
SOLUTION
3𝑥 2 − 5𝑥 ≥ 2
3𝑥 2 − 5𝑥 − 2 ≥ 0
(3𝑥 + 1)(𝑥 − 2) ≥ 0
1
Roots 𝑥 = − 3 , 2
1
{𝑥 ≤ − } ∪ {𝑥 ≥ 2}
3
(r) 𝑥 2 + 8𝑥 − 33 < 0
LESSON 4
Solve the inequality
15 − 7𝑥 − 2𝑥 2 > 0
SOLUTION
15 − 7𝑥 − 2𝑥 2 > 0
2𝑥 2 + 7𝑥 − 15 < 0
(2𝑥 − 3)(𝑥 + 5) < 0
3
Roots 𝑥 = −5, 2
3
{−5 < 𝑥 < }
2
(y) 20 − 8 𝑥 − 9 𝑥 2 < 0
NB: When we divide or multiply by a negative
number the inequality sign MUST be REVERSED.
…………………………………………………………………………..
EXERCISE 1.1
Solve each of the following inequalities.
(a) 𝑥 2 + 12𝑥 − 28 ≤ 0
(b) 𝑥 2 + 3 𝑥 − 70 ≥ 0
(c) 𝑥 2 + 15𝑥 − 16 < 0
(d) 𝑥 2 + 2𝑥 − 24 > 0
(e) 𝑥 2 + 12𝑥 − 64 < 0
(f) 𝑥 2 + 2𝑥 − 80 ≤ 0
(g) 𝑥 2 + 4𝑥 − 21 ≥ 0
(h) 𝑥 2 + 4𝑥 − 5 ≤ 0
2
(i) 𝑥 + 11𝑥 − 60 ≤ 0
(j) 𝑥 2 + 9𝑥 − 70 < 0
(k) 𝑥 2 + 11𝑥 − 80 ≥ 0
(l) 𝑥 2 + 8𝑥 − 48 > 0
(s) 𝑥 2 + 2𝑥 − 80 > 0
(t) 𝑥 2 + 6𝑥 − 7 > 0
(u) 2 − 2 𝑥 − 12 𝑥 2 > 0
(v) 6 − 7 𝑥 − 5 𝑥 2 ≥ 0
(w) 12 − 8 𝑥 − 15 𝑥 2 ≤ 0
(x) 16 − 12 𝑥 − 4 𝑥 2 > 0
SOLUTIONS
1.
(a) −14 ≤ 𝑥 ≤ 2
(b) {𝑥 ≤ −10} ∪ {𝑥 ≥ 7}
(c) −16 < 𝑥 < 1
(d) {𝑥 < −6} ∪ {𝑥 > 4}
(e) −16 < 𝑥 < 4
(f) −10 ≤ 𝑥 ≤ 8
(g) {𝑥 ≤ −7} ∪ {𝑥 ≥ 3}
(h) −5 ≤ 𝑥 ≤ 1
(i) −15 ≤ 𝑥 ≤ 4
(j) −14 < 𝑥 < 5
(k) {𝑥 ≤ −16} ∪ {𝑥 ≥ 5}
(l) {𝑥 < −12} ∪ { 𝑥 > 4}
(m) −4 < 𝑥 < 1
(n) −4 ≤ 𝑥 ≤ 1
(o) {𝑥 < −6} ∪ {𝑥 > 1}
(p) {𝑥 ≤ −8} ∪ {𝑥 ≥ 10}
(q) −5 ≤ 𝑥 ≤ 16
(r) −11 < 𝑥 < 3
(s) {𝑥 < −10} ∪ {𝑥 > 8}
(t) {𝑥 < −7} ∪ {𝑥 > 1}
1
1
(u) − 2 < 𝑥 < 3
3
(v) −2 ≤ 𝑥 ≤ 5
6
2
(w) {𝑥 ≤ − 5} ∪ {𝑥 ≥ 3 }
(x) −4 < 𝑥 < 1
10
(y) {𝑥 < −2} ∪ {𝑥 > 9 }
…………………………………………………………………………..
(m) 𝑥 2 + 3𝑥 − 4 < 0
(n) 𝑥 2 + 3𝑥 − 18 < 0
(o) 𝑥 2 + 5𝑥 − 6 > 0
(p) 𝑥 2 − 2𝑥 − 80 ≥ 0
(q) 𝑥 2 − 11𝑥 − 80 ≤ 0
6
CHAPTER 1: INEQUALITIES
RATIONAL INEQUALITIES
LESSON 5
Solve the inequality
𝑥 −3
≥0
𝑥 +2
SOLUTION
NB: We have to multiply
throughout by the square of the denominator.
𝑥 −3
≥ 0 × (𝑥 + 2)2
𝑥 +2
(𝑥 − 3)(𝑥 + 2) ≥ 0
Roots 𝑥 = −2, 3
{𝑥 < −2} ∪ {𝑥 ≥ 3}
NB: 𝑥 cannot equal −2 since that would lead to
division by zero.
LESSON 6
𝑥 for which
Determine the range of values of
2𝑥 − 3
≤1
𝑥 +1
SOLUTION
2𝑥 − 3
≤ 1 × (𝑥 + 1)2
𝑥+1
(2𝑥 − 3)(𝑥 + 1) ≤ (𝑥 + 1)2
2𝑥 2 − 𝑥 − 3 ≤ 𝑥 2 + 2𝑥 + 1
𝑥 2 − 3𝑥 − 4 ≤ 0
(𝑥 − 4)(𝑥 + 1) ≤ 0
Roots 𝑥 = −1, 4
{−1 < 𝑥 ≤ 4}
NB: 𝑥 ≠ −1 since that would lead to division by
zero.
…………………………………………………………………………..
EXERCISE 1.2
1. Determine 𝑥 ∈ ℝ for each of the following.
𝑥+10
(a) 𝑥−7 > 0
𝑥+6
(h) 𝑥−4 > 0
𝑥−3
(i) 𝑥−2 < 0
(j)
(c)
𝑥+10
𝑥−4
(l)
𝑥−3
≥3
𝑥−1
𝑥−2
(n) 𝑥+3 ≤ 4
(o)
𝑥+3
𝑥−3
≤2
𝑥+2
(p) 𝑥−5 − 1 ≥ 0
𝑥+3
(q) 𝑥−5 + 3 ≥ 1
𝑥−1
(r) 𝑥−4 − 2 ≥ 1
𝑥+2
(s) 𝑥−3 − 4 ≤ 0
𝑥+5
(t) 𝑥−1 + 1 ≤ 0
3𝑥+1
(u) 𝑥+4 ≥ 1
(v)
2.
3.
4.
𝑥−5
(e) 𝑥+8 > 0
𝑥+9
(f) 𝑥−10 < 0
𝑥+9
𝑥+1
(m) 𝑥−4 ≥ 2
(d) 𝑥+5 > 0
(g) 𝑥+2 > 0
<0
𝑥+1
<0
𝑥+1
𝑥−4
(k) 𝑥−2 ≥ 1
𝑥−3
(b) 𝑥−6 > 0
𝑥+2
2𝑥−3
𝑥+1
≤1
2𝑥+3
Solve for 𝑥 ∈ ℝ the inequality 3𝑥+4 < 1.
[10]
CAPE 2002
Find the range of values of 𝑥 ∈ ℝ for which
𝑥−2
> 0, 𝑥 ≠ 3.
[6]
𝑥+3
CAPE 2004
(i) Find 𝑎, 𝑏 ∈ ℝ such that
3𝑥
𝑎𝑥 + 𝑏
−2=
𝑥+1
𝑥+1
where 𝑥 ≠ −1.
[2]
(ii)
Hence, find the range of values of
3𝑥
𝑥 ∈ ℝ for which 𝑥+1 > 2.
[4]
CAPE 2006
5.
Solve, for 𝑥 ∈ ℝ, the inequality
2𝑥 − 3
−5 > 0
𝑥 +1
[5]
7
CHAPTER 1: INEQUALITIES
6.
Solve 3𝑥 2 + 4𝑥 + 1 ≤ 5.
CAPE 2010
[4]
CAPE 2013
SOLUTIONS
1. (a) {𝑥 < −10} ∪ {𝑥 > 7}
(b) {𝑥 < 3} ∪ {𝑥 > 6}
(c) −10 < 𝑥 < 4
(d) {𝑥 < −5} ∪ {𝑥 > −1}
(e) {𝑥 < −8} ∪ {𝑥 > 5}
(f) −9 < 𝑥 < 10
(g) {𝑥 < −9} ∪ {𝑥 > −2}
(h) {𝑥 < −6} ∪ {𝑥 > 4}
(i) 2 < 𝑥 < 3
(j) −2 < 𝑥 < 4
(k) 𝑥 > 2
(l) 3 < 𝑥 ≤ 5
(m) 4 < 𝑥 ≤ 7
14
(n) {𝑥 ≤ − 3 } ∪ {𝑥 > −3}
(o) {𝑥 < 3} ∪ {𝑥 ≥ 9}
(p) 𝑥 > 5
7
(q) {𝑥 ≤ 3} ∪ {𝑥 > 5}
11
(r) 4 < 𝑥 ≤ 2
(s) {𝑥 < 3} ∪ {𝑥 ≥
14
3
}
(t) −2 ≤ 𝑥 < 1
3
(u) {𝑥 < −4} ∪ {𝑥 ≥ }
2
(v) −1 < 𝑥 ≤ 4
4
2.
{𝑥 < − } ∪ {𝑥 > −1}
3.
4.
{𝑥 < −3} ∪ {𝑥 > 2}
(i) 𝑎 = 1, 𝑏 = −2 (ii) {𝑥 < −1} ∪ {𝑥 > 2}
5.
{𝑥 < − } ∪ {𝑥 > −1}
6.
−2 ≤ 𝑥 ≤ 3
3
8
3
2
…………………………………………………………………………..
8
CHAPTER 2: THE DISCRIMINANT
CHAPTER 2: THE DISCRIMINANT
At the end of this section, students should be able
to:
 determine the nature of roots of a
quadratic equation
___________________________________________________________
INTRODUCTION
For the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 we
have that
−𝑏 ± √𝑏2 − 4𝑎𝑐
𝑥=
2𝑎
The Discriminant is 𝑏2 − 4𝑎𝑐 from the above
formula and this gives us useful information about
the corresponding roots as shown below
Discriminant and Roots
Discriminant Roots of 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
𝑎, 𝑏 and 𝑐 real numbers, 𝑎 ≠ 0
𝒃𝟐 − 𝟒𝒂𝒄
Positive
Two distinct roots
0
Equal roots. One real root
(a double/repeated root)
Perfect Squares
Negative
No real roots.
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 > 0, 𝑎 > 0
This is explained graphically.
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 > 0
CALCULATING THE DISCRIMINANT
LESSON 1
For each of the following
quadratic equations determine the discriminant.
(a) 2𝑥 2 − 3𝑥 − 4 = 0
(b) 4𝑥 2 − 4𝑥 + 1 = 0
(c) 2𝑥 2 − 3𝑥 + 4 = 0
SOLUTION
(a) 2𝑥 2 − 3𝑥 − 4 = 0
𝑎 = 2, 𝑏 = −3 and 𝑐 = −4
Discriminant is 𝑏2 − 4𝑎𝑐
(−3) 2 − 4(2)(−4) = 41
Since 𝑏2 − 4𝑎𝑐 > 0 the equation 2𝑥 2 − 3𝑥 − 4
has 2 real and distinct roots
(b) 4𝑥 2 − 4𝑥 + 1 = 0
𝑎 = 4, 𝑏 = −4 and 𝑐 = 1
Discriminant is 𝑏2 − 4𝑎𝑐
(−4) 2 − 4(4)(1) = 0
Since 𝑏2 − 4𝑎𝑐 = 0 the equation
4𝑥 2 − 4𝑥 + 1 = 0 has a repeated (double)
root.
(c) 2𝑥 2 − 3𝑥 + 4 = 0
𝑎 = 2, 𝑏 = −3 and 𝑐 = 4
Discriminant is 𝑏2 − 4𝑎𝑐
(−3) 2 − 4(2)(4) = −23
Since 𝑏2 − 4𝑎𝑐 < 0 the equation
2𝑥 2 − 3𝑥 + 4 = 0 has no real roots.
THE DISCRIMINANT AND DISTINCT
REAL ROOTS
LESSON 2
Find the range of values of 𝑝 for
which the following equations has two distinct
real roots.
(a) 𝑥 2 + 2𝑝𝑥 − 5𝑝 = 0
(b) 𝑝(𝑥 2 − 1) = 3𝑥 + 3
9
CHAPTER 2: THE DISCRIMINANT
SOLUTION
(a) For 𝑥 2 + 2𝑝𝑥 − 5𝑝 = 0 to have 2 real and
distinct roots, 𝑏2 − 4𝑎𝑐 > 0
𝑎 = 1, 𝑏 = 2𝑝 and 𝑐 = −5𝑝
(2𝑝) 2 − 4(1)(−5𝑝) > 0
4𝑝2 + 20𝑝 > 0
4𝑝(𝑝 + 5) > 0
Roots: 𝑝 = 0, −5
{𝑥 < −5} ∪ {𝑥 > 0}
(b) 𝑝(𝑥 2 − 1) = 3𝑥 + 3
𝑝𝑥 2 − 𝑝 = 3𝑥 + 3
𝑝𝑥 2 − 3𝑥 − 𝑝 − 3 = 0
𝑎 = 𝑝, 𝑏 = −3 and 𝑐 = −𝑝 − 3
𝑏2 − 4𝑎𝑐 > 0
(−3) 2 − 4(𝑝)(−𝑝 − 3) > 0
9 + 4𝑝2 + 12𝑝 > 0
4𝑝2 + 12𝑝 + 9 > 0
(2𝑝 + 3)2 > 0
3
Root 𝑝 = − 2
3
𝑝≠−
2
THE DISCRIMINANT AND EQUAL
ROOTS
LESSON 3
Find the values of 𝑝 for which the
expression 𝑥 2 + (𝑝 + 3)𝑥 + 2𝑝 + 3 is a perfect
square.
SOLUTION
For perfect squares 𝑏2 − 4𝑎𝑐 = 0
𝑎 = 1, 𝑏 = 𝑝 + 3 and 𝑐 = 2𝑝 + 3
(𝑝 + 3)2 − 4(1)(2𝑝 + 3) = 0
𝑝2 + 6𝑝 + 9 − 8𝑝 − 12 = 0
𝑝2 − 2𝑝 − 3 = 0
(𝑝 − 3)(𝑝 + 1) = 0
𝑝 = 3, −1
THE DISCRIMINANT AND NO REAL
ROOTS
LESSON 4
Find the range of values of 𝑘 for
which the equation 𝑥 2 + 𝑘𝑥 + 25 = 0 has no real
roots.
SOLUTION
For no real roots 𝑏2 − 4𝑎𝑐 < 0
𝑘 2 − 4(1)(25) < 0
𝑘 2 − 100 < 0
(𝑘 + 10)(𝑘 − 10) < 0
Roots: 𝑘 = −10, 10
−10 < 𝑘 < 10
EXERCISE 2
1. If a quadratic equation has two distinct roots,
the value of the discriminant is
(A) Positive
(B) Negative
(C) Zero
2. If a quadratic equation has no real roots, the
value of the discriminant must be
(A) Positive
(B) Negative
(C) zero
3. For a quadratic equation to have a repeated
root the value of the discriminant
must be
(A) Positive
(B) Negative
(C) Zero
4. For a quadratic equation to have a repeated
root the quadratic expression must be
(a) A perfect square
(b) Factorisable
(c) Unfactorisable
5. Calculate the discriminant of
(i) 𝑥 2 − 3𝑥 + 5.
[−11]
[−39]
(ii) 3𝑥 2 + 9𝑥 + 10
6. (i) Calculate the discriminant of
[−52]
2𝑥 2 − 6𝑥 + 11.
(ii) State the number of real roots of the
equation 2𝑥 2 − 6𝑥 + 11 = 0.
7. (i) Calculate the discriminant of
[560]
5𝑥 2 + 20𝑥 − 8.
(ii) State the number of real roots of the
equation 5𝑥 2 + 20𝑥 − 8 = 0.
8. Find the set of values of 𝑘 for which the
equation 𝑥 2 + (𝑘 − 2)𝑥 + (2𝑘 − 4) = 0 has
{𝑘 < 2} ∪ {𝑘 > 10}
real roots.
2
(𝑘
9. The equation 𝑥 + 𝑘𝑥 + + 3) = 0, where 𝑘
is a constant, has different real roots.
(A) Show that 𝑘 2 − 4𝑘 − 12 > 0.
(B) Find the set of possible values of 𝑘.
{𝑘 < 2} ∪ {𝑘 > 6}
10. Find the set of values of 𝑝 for which the
equation 𝑝𝑥 2 + 4𝑥 + (5 − 𝑝) = 0 has 2
{𝑘 < 1} ∪ {𝑘 > 4}
distinct real roots.
11. Find the possible values of 𝑘 for which 𝑥 2 +
(𝑘 − 3)𝑥 + (3 − 2𝑘) = 0 has two distinct real
{𝑘 < −3} ∪ {𝑘 > 1}
roots.
12. The equation (𝑘 + 3)𝑥 2 + 6𝑥 + 𝑘 = 5 has two
distinct real roots. Determine the set of
[−4 < 𝑘 < 6]
possible values of 𝑘.
13. Given that the equation 𝑘𝑥 2 + 12𝑥 + 𝑘 = 0,
where 𝑘 is a positive constant, has equal
[𝑘 = 6]
roots, find the value of 𝑘.
10
CHAPTER 2: THE DISCRIMINANT
14. The equation 𝑥 2 + 3𝑝𝑥 + 𝑝 = 0, where 𝑝 is a
non – zero constant, has equal roots. Find the
4
value of 𝑝.
𝑝=9
15. The equation 2𝑥 2 − 3𝑥 − (𝑘 + 1) = 0, where
𝑘 is a constant, has no real roots. Find the set
17
of possible values of 𝑘.
[𝑘 < − 8 ]
16. The equation 𝑥 2 + 𝑘𝑥 + 8 = 𝑘 has no real
solutions for 𝑥.
(A) Show that 𝑘 satisfies 𝑘 2 + 4𝑘 − 32 < 0.
(B) Find the set of possible values of 𝑘.
[−8 < 𝑘 < 4]
11
CHAPTER 3: SURDS
CHAPTER 3: SURDS
At the end of this section, students should be able
to:
 perform operations involving surds
__________________________________________________________
INTRODUCTION
Surds are numbers left in 'square root form'. They
are therefore irrational numbers. The reason we
leave them as surds is because in decimal form
they would go on forever and so this is a very
clumsy way of writing them.
Note: √9 = 3 which is a rational number.
Roots such as √2, √3, √5 ……. are SURDS
PROPERTIES OF SURDS
1.
2.
3.
√𝑎 × √𝑏 = √𝑎𝑏 → √𝑎 × √𝑎 = √𝑎2 = 𝑎
𝑎
√𝑎
=√
𝑏
√𝑏
𝑎 √𝑐 + 𝑏 √𝑐 = (𝑎 + 𝑏)√𝑐 By factorization
SURD SIMPLIFICATION
A surd is simplified when the number under the
square root sign does not have a perfect square as
one of its factors.
LESSON 1
Simplify each of the following
(i) √50
(ii) √80
SOLUTION
(a) √50 = √25 × 2
= √25 × √2
= 5√2
(b) √80 = √16 × 5
= √16 × √5
= 4√5
LESSON 2
Simplify each of the following.
(a) 5√20 + 2√45
(b) 7√5 + 3√20 − √80
(c) 2(√3 + √12)
SOLUTION
(a) 5√20 + 2√45 = 5√5 × 4 + 2√9 × 5
= 5√4√5 + 2√9√5
= (5)(2) √5 + (2)(3)√5
= 10√5 + 6√5
= (10 + 6) √5
= 16√5
(b) 7√5 + 3√20 − √80
=7√5 + 3√5 × 4 − √16 × 5
= 7√5 + 3√5√4 − √16√5
= 7√5 + (3)(2)√5 − 4√5
= 7√5 + 6√5 − 4√5
= (7 + 6 − 4)√5
= 9√5
(c) 2(√3 + √12)
= 2(√3 + 2√3)
= 2(3√3)
= 6√3
LESSON 3
Simplify
(a) (2 + √5)(2 − √5)
(b) (√2 − 3)(√2 + 3)
(c) (3√3 + 4)(3√3 − 4)
SOLUTION
(a) (2 + √5)(2 − √5)
= 4 − 2√5 + 2√5 − √5√5
= 4 − √25
= 4 − 5 = −1
(b) (√2 − 3)(√2 + 3)
= √2√2 + 3√2 − 3√2 − 9
= √4 − 9
=2−9
= −7
(c) (3√3 + 4)(3√3 − 4)
= 9√3√3 − 12√3 + 12√3 − 16
= 9√9 − 16
= 27 − 16
= 11
…………………………………………………………………………..
EXERCISE 3.1
1.
Simplify each of the following surds.
a) √27
b) √63
c) √32
d) √44
e) √54
f) √72
g) √80
h) √96
i) √112
j) √108
k) √147
l) √192
m) 3√52 + 2√117
n) 4√18 + √98 − 2√8
o) √180 − 3√125 + √45
12
CHAPTER 3: SURDS
2.
3.
4.
Express √18 − √2 in simplified surd form.
Express √300 − √48 in the form 𝑘√3, where 𝑘
is an integer.
Express each of the following in the form 𝑘√2,
where 𝑘is an integer.
(i)
√200
(ii)
5√8 − 3√2
SOLUTIONS
1. (a) 3√3
(d) 2√11
(d)
=
(b) 3√7
(e) 3√6
2.
(g) 4√5
(h) 4√6
(j) 6√3
(k) 7√3
(m) 12√13 (n) 15√2
2√2
3.
6√3
4.
(i) 10√2
(c) 4√2
(f) 6√2
=
(i) 4√7
(l) 8√3
(o) −6√5
=
=
=
(ii) 7√2
FRACTIONS INVOLVING SURDS
When surds appear in the denominator of a
fraction, it is usual to eliminate them from. This is
called RATIONALISING THE DENOMINATOR i.e.
converting it from a SURD to a RATIONAL
Number. To do this we use two (2) facts
1.
2.
√𝑎 × √𝑎 = 𝑎
Difference of two Squares:
(𝑎2 − 𝑏2 ) = (𝑎 + 𝑏)(𝑎 − 𝑏)
where (𝑎2 − 𝑏2 ) is ALWAYS a Rational
Number
LESSON 4
Express each of the following in
𝑎√𝑐
the form 𝑏 ; 𝑎, 𝑏, 𝑐 ∈ ℤ
(b)
(c)
3
√2
5
√3
10√7
√5
4√45
(d) 5√8
SOLUTION
3
3 √2 3√2
(a)
=
×
=
2
√2 √2 √2
5
5 √3 5√3
(b)
=
×
=
3
√3 √3 √3
10√7
(c)
√5
10√7 √5
=
×
√5
√5
5√8
4√9 × 5
5√4 × 2
4√9√5
5√4√2
12√5
10√2
6√5
5√2
6√5(√2)
5√2(√2)
6√10
=
10
3√10
=
5
…………………………………………………………………………..
(a)
10√7√5
5
10√7 × 5
=
5
= 2√35
4√45
=
LESSON 5
the form
Express each of the following in
𝑎+𝑏√𝑐
1
𝑑
; 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℤ
(a) 2+√3
(b)
3
√3−6
2+√2
(c) 2−√2
(d)
3+√24
√6+2
SOLUTION
1
(a)
2 + √3
1
2 − √3
=
×
2 + √3 2 − √3
2 − √3
=
(2 + √3)(2 − √3)
2 − √3
=
2
22 − (√3)
2 − √3
=
4−3
= 2 − √3
(b)
3
√3 − 6
3
√3 + 6
=
×
√3 − 6 √3 + 6
13
CHAPTER 3: SURDS
3(√3 + 6)
(a) √108 −
(√3 − 6)(√3 + 6)
3(√3 + 6)
=
3 − 36
3(√3 + 6)
=
−33
(b) √45 +
=
=
(c)
=
=
2 − √2
2 + √2
2 − √2
=
𝑘√3
,
𝑘√5
,
𝑎√5 + 𝑏√2
Express each of the following in the indicated
form.
4
(a) 3−√7 ,
𝑎 + 𝑏√7
12
(c)
×
2 + √2
(2 + √2)(2 + √2)
√6 + 2
3 + √24
(d)
2 + √2
×
,
𝑎 + 𝑏√2
4+3√2
√5+3
,
𝑎√5 + 𝑏
(e) 3−√3 ,
𝑎 + 𝑏√3
√5−2
15+√3
SOLUTIONS
1. (a) 2√3
2.
𝑎 + 𝑏√7
(b) 7√5
(c) 3√5 + 2√2
(a) 6 + 2√7
(b) 9 − 3√5
(c) −25 + 18√2
(e) 8 + 3√3
(d) 11 + 5√5
(f) −3 + 2√7
EXAM QUESTIONS
1.
Express
5−√3
2+√3
in the form 𝑥 + 𝑦√3 where
𝑥, 𝑦 ∈ ℤ.
[5]
CAPE 2007
Without using calculators or tables, show that
(i)
√6 + 2 √6 − 2
(3 + √24)(√6 − 2)
Express each of the following in the indicated
form.
8−3√2
(f) 2+√7 ,
√6 − 2
EXERCISE 3.2
𝑎 − 𝑏√5
8+√7
2.
(√6 + 2)(√6 − 2)
3√6 − 6 − 2√24 − √24√6
=
6−4
3√6 − 6 − 2√6√4 − √144
=
2
3√6 − 6 − (2)(2)√6 − 12
=
2
3√6 − 6 − 4√6 − 12
=
2
−18 − √6
=
2
10√35
=
5
= 2√35
…………………………………………………………………………..
1.
√5
√5
,
(b) 3+√5 ,
3 + √24
=
2.
√3 + 6
−11
2 + √2
(2 − √2)(2 + √2)
4 + 2√2 + 2√2 + 2
=
4−2
4 + 4√2 + 2
=
2
6 + 4√2
=
2
= 3 + 2√2
(d)
(c)
15+√40
12
√3
20
(ii)
3.
4.
√6+√2
√6−√2
√6+√2
√6−√2
= 2 + √3
[5]
√6−√2
[5]
+
√6+√2
=4
CAPE 2008
Without the use of the tables or a calculator,
simplify √28 + √343 in the form 𝑘√7, where
𝑘 is an integer.
[5]
CAPE 2009
Without using calculators, find the exact value
2
of (√75 + √12) − (√75 − √12)
2
[3]
CAPE 2011
5. Without the use of a calculator, show that
√3 − 1 √3 + 1 √2 − 1 √2 + 1
+
+
+
= 10
√3 + 1 √3 − 1 √2 + 1 √2 − 1
[5]
CAPE 2014
SOLUTIONS
1. 13 − 7√3
2.
3. 9√7
4. 120
5.
…………………………………………………………………………..
14
CHAPTER 4: INDICES
CHAPTER 4: LAWS OF INDICES
MULTIPLICATION
LESSON 1
Simplify each of the following.
(i)
𝑏2 × 𝑏3
(ii)
𝑥4 × 𝑥
(iii)
2𝑚7 × 3𝑚5
(iv)
3𝑥 2 𝑦 × 4𝑥 5 𝑦 7
(v)
𝑎𝑛 × 𝑎𝑚
SOLUTION
(i)
𝑏2 × 𝑏3 = 𝑏 × 𝑏 × 𝑏 × 𝑏 × 𝑏 = 𝑏5
(ii)
𝑥4 × 𝑥 = 𝑥 × 𝑥 × 𝑥 × 𝑥 × 𝑥 = 𝑥5
(iii)
2𝑚7 × 3𝑚5 = 2 × 𝑚 × 𝑚 × 𝑚 × 𝑚 × 𝑚 ×
𝑚 ×𝑚 ×3×𝑚×𝑚 ×𝑚 ×𝑚×𝑚
= 6𝑚12
(iv)
3𝑥 2 𝑦 × 4𝑥 5 𝑦 7
= 3×𝑥 ×𝑥×𝑦 ×4×𝑥 ×𝑥 ×𝑥 ×𝑥 ×𝑥×
𝑦×𝑦 ×𝑦×𝑦×𝑦×𝑦×𝑦
= 12𝑥 7 𝑦 8
(v)
𝑎𝑛 × 𝑎𝑚 = 𝑎𝑛+𝑚
DIVISION
LESSON 2
ℎ3
(i)
ℎ3
(ii)
(iii)
(iv)
(v)
Simplify each of the following.
𝑘5
𝑚6 𝑟4
𝑎𝑚
SOLUTION
(i)
(ii)
(iii)
(iv)
(v)
ℎ3
𝑘9
𝑎−𝑛 =
=
𝑘×𝑘×𝑘×𝑘×𝑘×𝑘×𝑘×𝑘×𝑘
=
6𝑥𝑦3 𝑧4
= 𝑘4
𝑘×𝑘×𝑘×𝑘×𝑘
𝑚×𝑚×𝑚×𝑚×𝑚×𝑚×𝑟×𝑟×𝑟×𝑟
=
= 𝑚6 𝑟 2
𝑟 ×𝑟
2×𝑥×𝑥×𝑥×𝑦×𝑦×𝑦×𝑦×𝑦×𝑦×𝑦×𝑦×𝑧×𝑧
6×𝑥×𝑦×𝑦×𝑦×𝑧×𝑧×𝑧×𝑧
1𝑥 2 𝑦 5
=
3𝑧 2
= 𝑎𝑛−𝑚
RAISING A POWER TO A POWER
1
1
= 𝑎𝑛
𝑛
𝑎
𝑎−𝑛
Write each of the following using
LESSON 4
positive indices.
(i)
2−3
(ii)
𝑥 −4
(iii)
1
2
𝑎 −5
SOLUTION
1
(i)
2−3 = 23
= ℎ×ℎ×ℎ = 1
𝑟2
2𝑥3 𝑦8 𝑧2
𝑎𝑛
NEGATIVE INDICES
(ii)
ℎ×ℎ×ℎ
𝑘5
𝑚6 𝑟4
𝑎𝑚
(iv)
22 𝑥 2
52 𝑦 4
4𝑥 2
=
25𝑦 4
(𝑎𝑛 )𝑚 = 𝑎𝑛𝑚
2
=
1
𝑥 −4 = 𝑥4
1
2
(iii)
(iv)
1
1
1
2
𝑎
2𝑎5
𝑎 −5 = × 5 =
LESSON 4
form 𝑎𝑥 𝑛 .
1
(i)
𝑥3
6𝑥𝑦3
𝑎𝑛
2𝑥
( 5𝑥𝑦6 ) = (5𝑦2 )
(iii)
𝑟2
2𝑥3 𝑦8
2
2𝑥2 𝑦4
(iii)
(ii)
𝑘9
ℎ3
SOLUTION
(𝑏2 )3 = 𝑏2 × 𝑏2 × 𝑏2
(i)
= 𝑏2+2+2
= 𝑏6
(2𝑎3 )4 = 2𝑎3 × 2𝑎3 × 2𝑎3 × 2𝑎3
(ii)
= 24 𝑎3+3+3+3
= 24 𝑎12
Write each of the following in the
2
𝑥7
1
3𝑥2
5
7𝑥4
SOLUTION
1
−3
(i)
3 = 𝑥
(ii)
(iii)
(iv)
𝑥
2
𝑥7
= 2𝑥 −7
1
3𝑥2
5
7𝑥4
1
= 3 𝑥 −2
5
= 7 𝑥 −4
LESSON 3
Simplify each of the following.
(𝑏2 )3
(i)
(2𝑎3 )4
(ii)
2𝑥2 𝑦4
(iii)
( 5𝑥𝑦6 )
(iv)
(𝑎𝑛 )𝑚
2
15
CHAPTER 4: INDICES
ZERO INDEX
4𝑥 = 2
1
𝑥=
2
0
𝑎 =1
Any quantity, except zero, raised to the power
zero is 1
PROOF:
𝑎1 × 𝑎 −1 = 𝑎1+(−1) = 𝑎0
1
𝑎1 × 𝑎 −1 = 𝑎 × = 1
𝑎
∴ 𝑎0 = 1
LESSON 8
Without the use of a calculator
find the exact value of
1
1
3
1
(ii)
125−3
(iii)
643
(iv)
𝑥5
3
1
3
3
3 3
1
1
= 34 × 34 × 32
LESSON 8
show that
= 34+4++2
= 32
=9
Without the use of a calculator,
42
4
1 = 2 (√2)
−3
√2 × 8
Write each of the following in the
1
92
1
274 × 98 × 81 8 = (33 )4 × (32 )8 × (34 ) 8
1
(i)
1
SOLUTION
FRACTIONAL INDICES
𝑎 ⁄𝑛 = 𝑛√𝑎
1
1
𝑎 −𝑛 = 𝑛
√𝑎
𝑛
𝑛
𝑎 𝑚 = ( 𝑚√𝑎 )
𝑛
1
𝑎 −𝑚 = 𝑚 𝑛
( √𝑎 )
LESSON 6
form 𝑛√𝑎𝑚 .
3
274 × 98 × 818
1
2
2
3
(v)
2𝑥 −4
SOLUTION
1
(i)
92 = √9 = 3
(ii)
125−3 =
(iii)
𝑥 5 = ( 5√𝑥 )
(iv)
2𝑥 −4 =
1
2
3
1
1
125 3
2
3
𝑥4
=3
1
√125
SOLUTION
LHS
(22 )2
42
=
1
1
1
−
−
√2 × 8 3 22 × (23 ) 3
24
= 1
22 × 2−1
24
= 1
2−2
1
= 24 (22 )
= 24 √2
2
= 4
RHS
2
3
( √𝑥 )
FURTHER LESSONS
𝒂𝒏 = 𝒂𝒎 → 𝒏 = 𝒎
LESSON 7
Solve for 𝑥 the equations
(i) (3𝑥 )2 = 27𝑥−3
GRAPHS OF EXPONENTIAL FUNCTIONS APH OF
EXPONENTIAL FUNCTIONS
Any function of the form 𝑓(𝑥) = 𝑎𝑥 , 𝑎 > 0 is an
exponential function and they have the shape as
shown in the diagram.
1 𝑥
(ii) 3𝑥−2 = ( )
27
SOLUTION
(i) (3𝑥 ) 2 = 27𝑥−3
32𝑥 = 33(𝑥−3)
2𝑥 = 3(𝑥 − 3)
2𝑥 = 3𝑥 − 9
𝑥=9
1 𝑥
(ii) 3𝑥−2 = ( )
27
1 𝑥
𝑥−2
3
= ( 3)
3
3𝑥−2 = (3−3 )𝑥
3𝑥−2 = 3−3𝑥
𝑥 − 2 = −3𝑥
It is important to note that the graphs all cross the
𝑦 axis at the same point and that the graphs will
NEVER touch or cross the 𝑥 axis since 𝑎𝑥 > 0.
16
CHAPTER 4: INDICES
EXERCISE 4
4
1.
Solve the equation 163𝑥−2 = 82𝑥 .
[𝑥 = 3 ]
2.
Solve the equation 92𝑥−1 = 27𝑥 .
[𝑥 = 2]
3.
Solve the equation (23−4𝑥 )(4𝑥+4 ) = 2. [𝑥 = 5]
4.
(i)
Express
(ii)
Express (64) 𝑥 as a power of 2.
(iii)
Hence solve the equation 2𝑥 =
1
√32
5
[2−2 ]
as a power of 2.
1
6
[2 𝑥 ]
1
(64)𝑥
1
.
√32
3
[𝑥 = − 2]
Solve, for 𝑥 and 𝑦, the simulatneous equations
125𝑥 = 25(5𝑦 )
7𝑥 ÷ 49𝑦 = 1
4
2
[𝑥 = , 𝑦 = ]
5
5
EXAM QUESTIONS
5.
1 𝑥−1
6.
Solve for 𝑥, the equation 23−5𝑥 = ( )
7.
CAPE 2000
[𝑥 = 3]
Without using a calculator, find the exact
64
1
3
1
value of 274 × 98 × 818 .
. [3]
[3]
CAPE 2011
[Ans: 9]
17
CHAPTER 5: DISGUISED QUADRATIC EQUATIONS
CHAPTER 5: DISGUISED QUADRATIC EQUATIONS
At the end of this section, students should be able
to:
 solve equations in 𝑥 reducible to a
quadratic equation;
__________________________________________________________
It is often possible to solve equations by the use of
a substitution. In this section we look at equations
which can be solved by being transformed into
quadratic equations.
LESSON 1
Solve the equation
𝑥 4 − 5𝑥 2 − 6 = 0
SOLUTION
Let 𝑦 = 𝑥 2 since 𝑦 2 = (𝑥 2 )2 = 𝑥 4
2
𝑦 − 5𝑦 − 6 = 0
(𝑦 − 6)(𝑦 + 1) = 0
𝑦 = −1, 6
Re – substitute 𝑦 = 𝑥 2
𝑥 2 = −1 → NO REAL ROOTS
𝑥 2 = 6 → 𝑥 = ±√6
LESSON 2
Solve the equation
𝑥 6 − 9𝑥 3 + 8 = 0.
SOLUTION
Let 𝑦 = 𝑥 3 since 𝑦 2 = (𝑥 3 )2 = 𝑥 6
2
𝑦 − 9𝑦 + 8 = 0
(𝑦 − 1)(𝑦 − 8) = 0
𝑦 = 1, 8
Re – substituting 𝑦 = 𝑥 3
𝑥3 = 1 → 𝑥 = 1
𝑥3 = 8 → 𝑥 = 2
LESSON 3
Solve the equation
𝑥 − 3√𝑥 + 2 = 0.
2
SOLUTION
Let 𝑦 = √𝑥 since 𝑦 2 = (√𝑥) = 𝑥
𝑦 2 − 3𝑦 + 2 = 0
(𝑦 − 1)(𝑦 − 2) = 0
𝑦 = 1, 2
Re – substituting 𝑦 = √𝑥
√𝑥 = 1 → 𝑥 = 1
√𝑥 = 2 → 𝑥 = 4
LESSON 4
2
3
Solve the equation
1
3
2𝑥 − 5𝑥 = 12.
1
1 2
2
SOLUTION
Let 𝑦 = 𝑥 3 since 𝑦 2 = (𝑥 3 ) = 𝑥 3
2𝑦 2 − 5𝑦 − 12 = 0
(2𝑦 + 3)(𝑦 − 4) = 0
3
𝑦 = − ,4
2
1
Re – substituting 𝑦 = 𝑥 3
1
3
27
𝑥3 = − → 𝑥 = −
2
8
1
𝑥 3 = 4 → 𝑥 = 64
LESSON 5
Solve the equation
4𝑥 − 2𝑥+1 − 8 = 0.
SOLUTION
Rewrite expressions in 𝑥 using a
common base
(22 ) 𝑥 − 2(2𝑥 ) − 8 = 0
(2𝑥 ) 2 − 2(2𝑥 ) − 8 = 0
Let 𝑦 = 2𝑥
𝑦 2 − 2𝑦 − 8 = 0
(𝑦 + 2)(𝑦 − 4) = 0
𝑦 = −2, 4
Re – substituting 𝑦 = 2𝑥
2𝑥 = −2 → NO REAL ROOTS
2𝑥 = 4 → 𝑥 = 2
LESSON 6
Determine the values of 𝑥 ∈ ℝ for
which 3𝑥 + 32−𝑥 = 10.
SOLUTION
3𝑥 + 32−𝑥 − 10 = 0
32
3𝑥 + 𝑥 − 10 = 0
3
Let 𝑦 = 3𝑥
9
𝑦 + − 10 = 0
×𝑦
𝑦
2
𝑦 + 9 − 10𝑦 = 0
𝑦 2 − 10𝑦 + 9 = 0
(𝑦 − 1)(𝑦 − 9) = 0
𝑦 = 1, 9
Re – substituting 𝑦 = 3𝑥
3𝑥 = 1 → 𝑥 = 0
3𝑥 = 9 → 𝑥 = 2
LESSON 7
Solve the equation
23𝑥 − 8(2−3𝑥 ) = 7
SOLUTION
8
23𝑥 − 3𝑥 = 7
2
8
𝑦− =7
𝑦
𝑦 2 − 8 = 7𝑦
𝑦 2 − 7𝑦 − 8 = 0
(𝑦 − 8)(𝑦 + 1) = 0
𝑦 = −1, 8
Re – substituting 𝑦 = 23𝑥
23𝑥 = −1 → INVALID
23𝑥 = 8 → 𝑥 = 1
18
CHAPTER 5: DISGUISED QUADRATIC EQUATIONS
EXERCISE 5
1. Solve the equations
[𝑥 = ±1, ±3]
(i)
𝑥 4 − 10𝑥 2 + 9 = 0
[𝑥 = ±2, ±3]
(ii)
𝑥 4 + 36 = 13𝑥 2
2. Find the real roots of the equations
[𝑥 = ±1]
(i) 2𝑥 4 = 𝑥 2 + 1
1
(ii) 4𝑥 4 + 3𝑥 2 − 1 = 0.
[𝑥 = ± 2]
3. Solve the equations
[𝑥 = −2, 1]
(i) 𝑥 6 + 7𝑥 3 = 8
1
6
3
(ii) 8𝑥 + 7𝑥 − 1 = 0.
[𝑥 = −1, 2]
4. Solve the equation
[𝑥 = 9]
(i) 𝑥 = 2√𝑥 + 3
[𝑥 = 4, 36]
(ii) 𝑥 − 8√𝑥 + 12 = 0.
[𝑥 = 1, 9]
(iii) 𝑥 + 3 = 4√𝑥
25
(iv) 2𝑥 + 15 = 11 √𝑥
[𝑥 = 4 , 9]
5.
1
2.
3.
1
1
Solve for 𝑥 the equation 𝑥 3 − 4𝑥 −3 = 3.
[5]
[𝑥 = 64]
By using the substitution 𝑦 = 2𝑥 , or
otherwise, solve 4𝑥 − 3(2𝑥 +1 ) + 8 = 0.
[7]
CAPE 2010
[𝑥 = 1, 2]
7
6
Solve the equation 3 − 𝑥 − 𝑥 = 0.
[5]
9
81
1
[𝑥 = ]
2
1
2
[𝑥 = 4 , 9]
1
[𝑥 = 8]
(ii) 𝑥 3 + 3𝑥 3 − 10 = 0
2
3
1
3
8
(iii) 3𝑥 + 𝑥 − 2 = 0
1
1
2
1
[𝑥 = 27 ]
16
(iv) 3𝑥 2 − 8𝑥 4 + 4 = 0
7.
1.
Solve the following equations.
(i) 2𝑥 − 7𝑥 2 + 3 = 0
6.
EXAM QUESTIONS
[𝑥 = 81 , 16]
(v) 6𝑥 3 + 5𝑥 3 = 4
Solve the equations
(i) 22𝑥 − 5(2𝑥 ) + 4 = 0
(ii) 3(9𝑥 ) − 10(3𝑥 ) + 3 = 0
(iii) 4𝑥 − 12(2𝑥 ) + 32 = 0
(iv) 4𝑥 + 8 = 9(2𝑥 )
(v) 22𝑥 + 1 = 2𝑥+1
(vi) 22𝑥 + 128 = 3(2𝑥+3 )
64 1
[𝑥 = − 27 , 8]
[𝑥 = 0, 2]
[𝑥 = ±1]
[𝑥 = 2, 3]
[𝑥 = 0, 3]
[𝑥 = 0]
[𝑥 = 3, 4]
Find the real roots of the equation
3 10
−
−8 =0
𝑦4 𝑦2
1
8.
1
[𝑦 = ± 2]
Solve the equation 𝑥 − 6𝑥 2 + 2 = 0, giving
your answer in the form 𝑝 ± 𝑞 √𝑟, where 𝑝, 𝑞
and 𝑟 are integers.
[𝑥 = 16 ± 6√7]
9. Solve the equation 𝑥 − 8√𝑥 + 13 = 0, giving
your answer in the form 𝑝 ± 𝑞 √𝑟, where 𝑝, 𝑞
and 𝑟 are integers.
[𝑥 = 19 ± 8√3]
10. By using the substitution 𝑦 = (𝑥 + 2)2 , find
the real roots of the equation
(𝑥 + 2) 4 + 5(𝑥 + 2)2 − 6 = 0.
[𝑥 = −1, −3]
11. By using the substitution 𝑢 = (3𝑥 − 2)2 , find
the roots of the equation
(3𝑥 − 2) 4 − 5(3𝑥 − 2)2 + 4 = 0.
1 4
[𝑥 = 0, , 1, ]
3 3
19
CHAPTER 6: FUNCTIONS
CHAPTER 6: FUNCTIONS
At the end of this section, students should be able
to:








use terms to functions;
determine the range of a function given
its domain;
determine whether a given function is
many – to – one or one – to – one;
determine the inverse of a given function,
(if it exists);
plot and sketch functions and their
inverse, (if they exist);
state the geometrical relationship
between 𝑦 = 𝑓(𝑥) and its inverse 𝑓 −1 (𝑥);
find the composition of two functions;
recognise that, if 𝑔 is the inverse of 𝑓,
then 𝑓[𝑔(𝑥)] = 𝑥, for all 𝑥, in the domain
of 𝑔.
Figure 1
DEFINITION OF A FUNCTION
A function is a rule that produces a
correspondence/relation between two sets of
elements, say 𝐴 and 𝐵, such that to each element
in the first set, 𝐴, there corresponds one and only
one element in the second set, 𝐵.
The first set is called the domain and the second
set is called the co-domain. The set of all
corresponding elements in the second set is called
the range of the function.
Figure 2
REPRESENTING A FUNCTION
Functions can be represented using
- mapping diagrams
- ordered pairs
- equations
- graphs
Figure 3
MAPPING DIAGRAMS
LESSON 1
State giving reasons for your
answer whether the following mapping diagrams
represent functions. If the mapping diagram
represents a function, state its domain, co –
domain and range.
Figure 4
SOLUTION
𝑓: 𝐴 → 𝐵 in Figure 1 represents a function since
each element in 𝐴 is mapped to one and only one
element in 𝐵.
20
CHAPTER 6: FUNCTIONS
Domain = {0, 2, 3, 5}
Co – Domain = {1, 4, 6, 8}
Range = {1, 4, 6, 8}
This type of function is a one – to – one function
since each element of the domain is mapped to
only one element of the co – domain. Furthermore,
since the range and the co – domain are identical
the function is said to be onto.
𝑔: 𝐴 → 𝐵 in Figure 2 represents a function because
each element in 𝐴 is mapped to one and only one
element in 𝐵.
Domain = {𝑎, 𝑏, 𝑐, 𝑑}
Co – domain = {0, 1, 2, 3}
Range = {1, 2, 3}
This type of function is called a many – to – one
function since 1 ∈ 𝐵 is mapped onto by more than
one (many) element from the domain.
𝑓: 𝐶 → 𝐷 in Figure 3 does not represent a function
since ℎ which is an element of the domain is not
mapped to any element in 𝐵.
ℎ: 𝐸 → 𝐹 in Figure 4 does not represent a function
because 1 which is an element of the domain is
mapped to more than 1 element in the co –
domain.
ORDERED PAIRS
A function is a set of ordered pairs with the
property that no two ordered pairs have the same
first component and different second components.
The set of all first components in a function is
called the domain of the function, and the set of all
second components is called the range.
SOLUTION
(a) 𝑓 is a function with domain = {2, 3, 4, 5} and
range = {4, 6, 8 , 10}
(b) 𝑔 is a function with domain = {−1, 0, 1, 2} and
range = {4, 3, 2, 1}
(c) ℎ is not a function since the first number 10 is
mapped to 2 elements, as well as 5.
(d) 𝑓1 is a function with domain = {−10, −5, 0, 5, 10}
and range = {10, 5, 0}
(e) 𝑔1 is a function with domain = {0, 1, 2, 3, 4, 5} and
range = {1, 2}
(f) ℎ1 is not a function since the first numbers 1, 2
and 3 are each mapped to more than 1 element.
EQUATIONS
Functions can be of the form 𝑦 = 𝑓(𝑥). This is read
as “𝑦 equals 𝑓 of 𝑥. 𝑓 is the function, 𝑥 represents
the members of the domain and 𝑦 represents the
members of the range. These functions can be
represented graphically.
GRAPHS OF FUNCTIONS
Since the coordinates on a graph are of the form
(𝑥, 𝑦) and functions can be written in the form
𝑦 = 𝑓(𝑥), corresponding 𝑥 (domain) and 𝑦
(range) values can be illustrated as coordinates
and plotted to form a graph.
LESSON 3
−3 ≤ 𝑥 ≤ 3.
Graph the function 𝑓(𝑥) = 𝑥 2 for
SOLUTION
𝑓(𝑥) = 𝑥 2
𝑓(−3) = (−3)2 = 9 → (−3, 9)
LESSON 2
Write the functions from Figure 1
and Figure 2 as sets of ordered pairs.
SOLUTION
𝑓: 𝐴 → 𝐵 = {(0, 1), (2, 4), (3, 6), (5, 8)}
𝑓: 𝐶 → 𝐷 = {(𝑎, 1) , (𝑏, 1) , (𝑐, 2), (𝑑, 3)}
𝑓(−2) = (−2)2 = 4
→ (−2, 4)
𝑓(−1) = (−1)2 = 1
→ (−1, 1)
𝑓(0) = 02 = 0
→ (0, 0)
LESSON 2
Indicate whether each set of
ordered pairs defines a function. Find the domain
and range of each function.
𝑓(1) = 12 = 1
→ (1, 1)
𝑓(2) = 22 = 4
→ (2, 4)
𝑓(3) = 32 = 9
→ (3, 9)
(a) 𝑓 = {(2, 4), (3, 6), (4, 8), (5, 10)}
(b) 𝑔 = {(−1, 4), (0, 3), (1, 2), (2, 1)}
(c) ℎ = {(10, −10) , (5, −5), (0, 0) , (5, 5), (10, 10)}
(d) 𝑓1 = {(−10, 10), (−5, 5), (0, 0) , (5, 5), (10, 10)}
(e) 𝑔1 = {(0, 1), (1, 1) , (2, 1), (3, 2), (4, 2) , (5, 2)}
(f) ℎ1 = {(1, 1), (2, 1), (3, 1), (1, 2), (2, 2), (3, 2)}
21
CHAPTER 6: FUNCTIONS
Figure 2 represents a function since any vertical
line drawn through the graph does not cut the
graph in more than one place.
Figure 3 does not represent a function since any
vertical line drawn through the graph of the
function will cut the graph in two places.
THE RANGE OF A FUNCTION
POLYNOMIAL FUNCTIONS
The domain is assumed to be 𝑥 ∈ ℝ.
For polynomials of odd degree the range is 𝑦 ∈ ℝ.
LESSON 5
State the range of the function,
𝑓(𝑥) = 2𝑥 3 − 6𝑥 2 + 5𝑥 − 4.
THE VERTICAL LINE TEST
The vertical line test states that if any vertical line
drawn through a graph cuts the graph in only one
place then the graph represents a function.
SOLUTION
Since the function is of degree 3
the range of 𝑓 is 𝑦 ∈ ℝ.
LESSON 4
State which of the following graphs
represent functions.
Figure 1
For polynomials of even degree the range can be
determined via a sketch obtained by substituting
values for 𝑥. However, for quadratics written in
the form 𝑎(𝑥 + ℎ) 2 + 𝑘 the range is 𝑦 ≥ 𝑘 if 𝑎 > 0.
If 𝑎 < 0 the range is 𝑦 ≤ 𝑘.
LESSON 6
(i) Write 𝑓(𝑥) = 2𝑥 2 − 10𝑥 + 3 in the form
𝑎(𝑥 + ℎ) 2 + 𝑘 where 𝑎, ℎ and 𝑘 are real
numbers.
(ii) Hence, state the range of 𝑓.
Figure 2
SOLUTION
(i)
Figure 3
SOLUTION
Figure 1 represents a function since any vertical
line drawn through the graph will cut the graph in
only one place.
2𝑥 2 − 10𝑥 + 3
3
= 2 [𝑥 2 − 5𝑥 + ]
2
3 2
3
5 2
2
= 2 [𝑥 − 5𝑥 + ( )
+ −( ) ]
2
2
2
2
5
3 25
= 2 [(𝑥 − ) + − ]
2
2 4
2
5
19
= 2 [(𝑥 − ) − ]
2
4
22
CHAPTER 6: FUNCTIONS
5 2 19
= 2 (𝑥 − ) −
2
2
(b) 𝑓(𝑥) has a maximum value since the
coefficient of √1 − 2𝑥 is negative.
Range: 𝑦 ≤ −8
RATIONAL FUNCTIONS
LESSON 8
Determine the range of
4
(a) 𝑓(𝑥) = − 𝑥+3
(ii)
19
Range of 𝑓: 𝑦 ≥ − 2
SQUARE ROOT FUNCTIONS
For square root functions of the form
𝑓(𝑥) = 𝑎 √𝑥 + ℎ + 𝑘 we have the following:
 𝑓(𝑥) has a minimum value if 𝑎 > 0 with
range 𝑦 ≥ 𝑘
 𝑓(𝑥) has a maximum value if 𝑎 < 0 with
range 𝑦 ≤ 𝑘
LESSON 7
Determine the range of
(a) 𝑓(𝑥) = 2√𝑥 − 1 + 4
(b) 𝑓(𝑥) = −8 − 3√1 − 2𝑥.
SOLUTION
(a) 𝑓(𝑥) has a minimum value since the
coefficient of √𝑥 − 1 is positive.
Range: 𝑦 ≥ 4
2𝑥+5
(b) 𝑓(𝑥) = 3𝑥−1
SOLUTION
(a) We know that 𝑓(𝑥) is undefined when
𝑥 = −3. Therefore, for the range we simply
need to determine the value of 𝑦 which would
have corresponded to this undefined value.
4
Let 𝑦 = − 𝑥+3
𝑦(𝑥 + 3) = −4
𝑥𝑦 + 3𝑦 = −4
𝑥𝑦 = −3𝑦 − 4
−3𝑦 − 4
𝑥=
𝑦
Range: 𝑦 ∈ ℝ , 𝑦 ≠ 0
2𝑥+5
(b) 𝑦 = 3𝑥−1
𝑦(3𝑥 − 1) = 2𝑥 + 5
3𝑥𝑦 − 𝑦 = 2𝑥 + 5
3𝑥𝑦 − 2𝑥 = 𝑦 + 5
𝑥(3𝑦 − 2) = 𝑦 + 5
𝑦+5
𝑥=
3𝑦 − 2
Range: 𝑦 ∈ ℝ, 𝑦 ≠ 0
23
CHAPTER 6: FUNCTIONS
…………………………………………………………………………..
EXERCISE 6.1
1.
The diagram below shows an assignment 𝑔
from 𝐴 → 𝐵.
5.
Determine the range of each of the following
functions.
2𝑥−10
(a) 𝑓(𝑥) = −3𝑥−4
2𝑥+4
(b) 𝑓(𝑥) = 𝑥−7
2𝑥−3
(c) 𝑓(𝑥) = −𝑥−3
−4𝑥+7
(d) 𝑓(𝑥) = −2(𝑥+3)
4𝑥+9
(e) 𝑓(𝑥) = 𝑥+10
5𝑥−3
(f) 𝑓(𝑥) = −𝑥−7
2
a.
b.
c.
d.
Write 𝑔 as a set of ordered pairs
State 2 reasons why 𝑔 is not a function.
The assignment 𝑔 can be transformed
into a function, 𝑓: 𝐴 → 𝐵, by moving a
single arrow. Draw the function 𝑓.
Under 𝑓, what is
(i)
the pre – image of 4
(ii)
the image of 𝑥
(a) {(𝑥, 1), (𝑦, 3), (𝑧, 2), (𝑧, 5), (𝑢, 5)}
(b) 𝑡 is not mapped to and 𝑧 is mapped to 2
elements
(c) (d) (i) 𝑡
(ii) 1
2.
(a) 𝑦 ≠ − 3
(b) 𝑦 ≠ 2
(c) 𝑦 ≠ −2
(d) 𝑦 ≠ 2
(e) 𝑦 ≠ 4
(f) 𝑦 ≠ −5
…………………………………………………………………………..
CLASSES OF FUNCTIONS
Functions are classified by three classes:
(i) Injections (ii) Surjections and (iii) Bijections.
INJECTIONS
A function 𝑓: 𝐴 → 𝐵 is injective, or one–to–one, if
𝑦 ∈ 𝐵 is the image of only one 𝑥 ∈ 𝐴.
LESSON 9
Determine whether the following
functions are injective
State the range of the function
𝑓(𝑥) = 𝑥 5 − 4𝑥 3 + 2𝑥 − 1
𝑦∈ℝ
3.
State the range of each of the following
functions.
(a) 𝑓(𝑥) = 𝑥 2 − 4𝑥 − 6
(b) 𝑓(𝑥) = 𝑥 2 − 5𝑥 − 5
(c) 𝑓(𝑥) = 2𝑥 2 − 7𝑥 + 3
(d) 𝑓 (𝑥 ) = −3𝑥 2 − 6𝑥 + 5
(e) 𝑓(𝑥) = 2𝑥 2 + 3𝑥 + 10
(f) 𝑓(𝑥) = −3𝑥 2 − 2𝑥 − 6
45
25
(a) 𝑦 ≥ −10 (b) 𝑦 ≤ 4 (c) 𝑦 ≥ − 8
(d) 𝑦 ≤ 8
4.
(e) 𝑦 ≥
71
8
(f) 𝑦 ≤ −
17
3
Determine the range of each of the following
functions.
(a) 𝑓(𝑥) = −3√𝑥 − 2 − 6
(b) 𝑓(𝑥) = √2𝑥 + 8 + 4
(c) 𝑔(𝑥) = 3 √1 + 𝑥 − 4
(d) 𝑔(𝑥) = 5 √𝑥 + 3 − 7
(e) 𝑔(𝑥) = −2√4𝑥 + 10
(f) 𝑔(𝑥) = −3√𝑥 − 8 − 8
(a) 𝑦 ≤ −6
(b) 𝑦 ≥ 4
(c) 𝑦 ≥ 4
(d) 𝑦 ≥ −7
(e) 𝑦 ≤ 0
(f) 𝑦 ≤ −8
SOLUTION
In the diagrams above 𝑓: 𝐴 → 𝐵 is
injective since every element in 𝐵 is the image of
only one element in 𝐴. However, 𝑔: 𝐴 → 𝐵 is not
injective since, for example, 1 ∈ 𝐵 is the image of
𝑏, 𝑑 ∈ 𝐴.
DETERMINING WHETHER A FUNCTION IS ONE
–TO – ONE
𝑓: 𝐴 → 𝐵 is one-to-one if and only if for all 𝑎, 𝑏 ∈ 𝐴
the following exists
𝑓(𝑎) = 𝑓(𝑏) → 𝑎 = 𝑏
24
CHAPTER 6: FUNCTIONS
LESSON 10
The function, 𝑓, is defined on ℝ
by 𝑓: 𝑥 → 4𝑥 − 1.
This is made possible by the restriction on the
domain.
(i)
(ii)
THE HORIZONTAL LINE TEST
This test states that any if any horizontal line drawn
through the graph of a function cuts the graph in only
one place then the graph represents a one – to – one
function.
Show that 𝑓 is one to one.
Hence find the value of 𝑥 ∈ ℝ for which
𝑓(𝑓(𝑥 + 2)) = 𝑓(𝑥 − 5).
SOLUTION
(i)
(ii)
𝑓(𝑎) = 𝑓(𝑏)
4𝑎 − 1 = 4𝑏 − 1
4𝑎 = 4𝑏
𝑎=𝑏
Therefore 𝑓 is one to one.
Since 𝑓 is one to one
𝑓(𝑥 + 2) = 𝑥 − 5
4(𝑥 + 2) − 1 = 𝑥 − 5
4𝑥 + 8 − 1 = 𝑥 − 5
3𝑥 = −12
𝑥 = −4
LESSON 11
Determine if 𝑓(𝑥) =
LESSON 14
State which of the following graphs
represent one – to – one functions.
2𝑥+1
1−5𝑥
Figure 1
is
injective.
SOLUTION
𝑓(𝑎) = 𝑓(𝑏)
2𝑎 + 1 2𝑏 + 1
=
1 − 5𝑎 1 − 5𝑏
(2𝑎 + 1)(1 − 5𝑏) = (2𝑏 + 1)(1 − 5𝑎)
2𝑎 − 10𝑎𝑏 + 1 − 5𝑏 = 2𝑏 − 10𝑎𝑏 + 1 − 5𝑎
2𝑎 − 5𝑏 = 2𝑏 − 5𝑎
7𝑎 = 7𝑏
𝑎=𝑏
Therefore 𝑓 is injective.
LESSON 12
Determine if 𝑓 (𝑥 ) = 𝑥 2
represents a one - to – one function.
SOLUTION
𝑓(𝑎) = 𝑓(𝑏)
𝑎2 = 𝑏2
±𝑎 = ±𝑏
either 𝑎 = 𝑏 or 𝑎 = −𝑏 or −𝑎 = 𝑏 or 𝑎 = −𝑏
Therefore 𝑓 is not one to one.
Figure 2
SOLUTION
Figure 1 does not represent a one – to – one function
since any horizontal line drawn through the graph of
the function will cut the function in more than one
place. The function is many – to – one.
Figure 2 represents a one – to – one function since
any horizontal line drawn through the graph of the
function will cut the function in only one place.
LESSON 13
Let 𝐴 = {𝑥: 𝑥 ∈ ℝ, 𝑥 ≥ 1}
A function 𝑓: 𝐴 → ℝ is defined as
𝑓(𝑥) = 𝑥 2 − 2𝑥. Show that 𝑓 is one – to – one.
SOLUTION
𝑓(𝑎) = 𝑓(𝑏)
𝑎2 − 2𝑎 = 𝑏2 − 2𝑏
(𝑎 − 1)2 − 1 = (𝑏 − 1) 2 − 1
(𝑎 − 1)2 = (𝑏 − 1) 2
𝑎−1 = 𝑏−1
𝑎=𝑏
25
CHAPTER 6: FUNCTIONS
SURJECTIONS
A function 𝑓: 𝐴 → 𝐵 is surjective or onto if every
𝑦 ∈ 𝐵 is the image of at least one 𝑥 ∈ 𝐴.
𝑦−1
=𝑥
2
𝑦−1
Therefore, the real number 2 is the pre𝑦−1
LESSON 15
Determine whether the
following functions are surjective.
image of 𝑦 by 𝑓 and 2 is in ℝ for
any 𝑦 in ℝ so 𝑓 is onto.
However, for 𝑓: ℕ → ℕ, 𝑓(𝑥) = 2𝑥 + 1 is not
𝑦−1
onto since 2 is not in ℕ for every 𝑦 in ℕ. For
2−1
1
example, when 𝑦 = 2, 2 = 2 ∉ ℕ
(b) Let 𝑦 = 𝑔(𝑥)
𝑦 = 𝑥2 − 3
√𝑦 + 3 = 𝑥
Therefore, the real number √𝑦 + 3 is the preimage of 𝑦 by 𝑓, but √𝑦 + 3 is a real number
only when 𝑦 > −3, so 𝑓 is not onto.
LESSON 17
Determine if 𝑓(𝑥) =
2𝑥−3
4𝑥+1
,𝑥 ≠ −
1
4
is surjective.
SOLUTION
2
Since there is a horizontal asymptote at 𝑦 = , 𝑓 is
4
NOT onto.
SOLUTION
𝑓: 𝐴 → 𝐵 is surjective since every 𝑦 ∈ 𝐵 is the
image of one 𝑥 ∈ 𝐴
𝑔: 𝐴 → 𝐵 is surjective because every 𝑦 ∈ 𝐵 is the
image of at least one 𝑥 ∈ 𝐴
ℎ: 𝑋 → 𝑌 is not surjective because 3 ∈ 𝑌 is not the
image of a 𝑥 ∈ 𝑋
Consequently, for a function to be a surjection the
range of the function must be equal to the codomain of the function.
NB: 𝑓: 𝐴 → 𝐵 illustrates that an injection can be a
surjection.
BIJECTIONS
A function that is both injective and surjective is a
bijective function (one–to–one and onto).
LESSON18
Determine whether the
following functions are bijective.
DETERMINING WHETHER A FUNCTION IS
ONTO
A function 𝑓: 𝐴 → 𝐵 is onto if and only if for every
𝑦 ∈ 𝐵 there exist at least one 𝑥 ∈ 𝐴 such that
𝑦 = 𝑓(𝑥).
LESSON 16
Prove whether the following
functions, 𝑅 → 𝑅, are surjective.
(a) 𝑓(𝑥) = 2𝑥 + 1
(b) 𝑔(𝑥) = 𝑥 2 − 3
SOLUTION
𝑓: 𝐴 → 𝐵 is bijective since it is injective and
surjective
𝑔: 𝐴 → 𝐵 is surjective but not injective therefore it
is not bijective.
SOLUTION
(a) Let 𝑦 = 𝑓 (𝑥 )
𝑦 = 2𝑥 + 1
26
CHAPTER 6: FUNCTIONS
The four possible combinations of injective and
surjective features are illustrated in the following
diagrams.
Injective and Surjective (Bijective)
Does 𝑔−1 exist? Give a reason for your answer.
2.
3.
4.
Injective and Non – Surjective
5.
6.
Non – Injective and Surjective
7.
8.
Non – Injective and Non-Surjective
For surjective functions, the codomain of the
function is equal to its
(A) Domain
(B) Range
(C) Inverse
(D) Set of pre - images
If 𝐴 = {1, 2, 3} and 𝐵 = {𝑎, 𝑏, 𝑐, 𝑑} explain why
𝑓: 𝐴 → 𝐵 cannot be surjective.
Using 𝐴 = {𝑒, 𝑓, 𝑔, ℎ} and 𝐵 = {0, 4, 9},
construct a surjective function. Does the
function you created have an inverse? Give a
reason for your answer.
For a function to be bijective the domain of
the function must contain the same number of
elements as its
(A) Codomain
(B) Range
For a function to have an inverse it must be
(A) Injective
(B) Surjective
(C) Bijective
(D) None of the above
Show that each of the following functions are
1 – 1.
(i) 𝑓(𝑥) = 1 − 6𝑥
𝑥
(ii) 𝑔(𝑥) = 3 + 4
𝑥2
Show that the function 𝑓(𝑥 ) = 𝑥2 −1 , 𝑥 ≠ −1, 1
is not one – to – one.
9. The functions, 𝑓 and 𝑔, are defined by
𝑥2
1
𝑓: 𝑥 →
, 𝑥2 ≠ 1
𝑔: 𝑥 → 𝑥 − 3
1 − 𝑥2
2
(a) Explain clearly why 𝑓 is not one-to-one
(b) Find, and express in its simplest form,
𝑔𝑓(𝑥)
10. Show that the function 𝑓(𝑥) = 2𝑥 − 5 is one –
to – one.
11. Show that 𝑓: ℝ → ℝ, 𝑓(𝑥) = 3𝑥 − 2 is onto.
1
12. Show that 𝑓: ℝ → ℕ, 𝑓(𝑥) = 𝑥 + 1 is not
2
…………………………………………………………………………..
EXERCISE 6.2
1.
Given that 𝐴 = {𝑎, 𝑏, 𝑐, 𝑑} and 𝐵 = {1, 5, 7},
(i)
explain why 𝑓: 𝐴 → 𝐵 cannot be
injective.
(ii)
construct the injective function
𝑔: 𝐵 → 𝐴.
onto.
13. Consider the following sets 𝐴 = {1, 2,3, 4} and
𝐵 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒} and the set of ordered pairs
𝑓: {(1, 𝑏), (3, 𝑑),(2, 𝑏), (4, 𝑒)}. Draw the arrow
diagram to represent 𝑓 and answer the
following questions.
a. Is 𝑓 a function?
b. Is it injective? Is it surjective?
c. What is the image of 2?
d. What is the pre-image of 𝑏?
e. What is the pre-image of 𝑐?
27
CHAPTER 6: FUNCTIONS
14. If 𝑓: 𝐴 → 𝐵 and 𝐴 has 5 elements while 𝐵 has
4 elements, can 𝑓 be injective?
15. If 𝑓: 𝐴 → 𝐵 and 𝐴 has 5 elements while 𝐵 has
6 elements, can 𝑓 be surjective?
16. Construct a surjective function that is not
injective between {𝑎, 𝑏, 𝑐, 𝑑} and {𝑒, 𝑓, 𝑔}
SOLUTION
(i)
𝑓𝑔(𝑥)
𝑥+2
𝑥 +2
𝑓(
) = 3(
)−2
3
3
=𝑥
SOLUTIONS
(ii)
𝑔𝑓(𝑥)
1.
2.
3.
4.
5.
6.
7.
8.
9.
(i) A domain element would have to be
omitted for one – to – oneness.
(B)
At least one element in 𝐴 would have to be
mapped to more than one element in 𝐵.
No, since an element in co – domain would not
be mapped to.
(B)
(C)
(i)
(ii)
(a) Counter – Example
7𝑥2 −6
(b) 2(1−𝑥2 )
10.
11.
12.
13. (a) Yes (b) Non Injective, Non Surjective
(c) 𝑏 (d) 1, 2 (e) Does not exist
14. No
15. No
16.
…………………………………………………………………………..
3𝑥 − 2 + 2
3
=𝑥
Generally, 𝑓𝑓 −1 (𝑥) = 𝑥 and 𝑓 −1 𝑓(𝑥) = 𝑥. Since
𝑓𝑔(𝑥) = 𝑥, 𝑔 is the inverse of 𝑥. Furthermore, (as
expected) since 𝑔𝑓(𝑥) = 𝑥, 𝑓 is the inverse of 𝑔.
𝑔(3𝑥 − 2) =
The functions 𝑓 and 𝑔 written as sets of ordered
pairs are as follows:
𝑓 = {(−1, −1), (0, 1), (1, 3), (2, 5), (3, 7), (4, 9)}
𝑔 = {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
We realise that 𝑓 is a one – to – one function while
𝑔 is a many – to – one function.
Now let’s consider the inverses of each function by
simply switching the first and second elements of
each pair.
𝑓 −1 = {(−1, −1), (1, 0), (3, 1), (5, 2), (7, 3), (9, 4)}
𝑔−1 = {(9, −3), (4, −2), (1, −1), (0, 0), (1, 1), (4, 2), (9, 3)}
The inverse of 𝑓 is clearly a function. However,
𝑔 −1 is not a function since an element in the
domain, for example the element 9, is mapped to
more than one element in the range.
FUNCTIONS AND THEIR INVERSES
LESSON 19
it has to be
For a function to have an inverse
(A) Injective
(B) Surjective
(C) Bijective
(D) None of the above.
VIEW SOLUTION
LESSON 20
Given that 𝑓(𝑥) = 3𝑥 − 2 and
𝑥+2
𝑔(𝑥) =
, determine
3
(i)
𝑓𝑔(𝑥)
(ii)
𝑔𝑓 (𝑥 )
State the relationship between 𝑓 and 𝑔.
Generally, if 𝒇 is one – to – one and onto, then 𝒇 −𝟏
exists.
Here are the graphs of 𝑓 and 𝑓 −1, along with the
line 𝑦 = 𝑥. In the diagram we realise that the
graph of 𝑓 −1 is a reflection of the graph of 𝑓 in the
28
CHAPTER 6: FUNCTIONS
line 𝑦 = 𝑥. This is the geometric relation between
a function and its inverse. That is to say, if 𝑓 has an
inverse then this inverse, 𝑓 −1, is a reflection of 𝑓 in
the line 𝑦 = 𝑥.
LESSON 21
Diagram 1
The functions 𝑓 is defined by
𝑓: 𝑥 → 2𝑥 + 3,
𝑥∈ℝ
Sketch, in a single diagram, the graphs of 𝑦 = 𝑓(𝑥)
and 𝑦 = 𝑓 −1 (𝑥), making clearly the relationship
between the two graphs.
SOLUTION
Diagram 2
We will construct a table for corresponding pairs
of 𝑥 and 𝑦 values for 𝑓(𝑥)
𝑥
−2
0
2
𝑦
−1
3
7
We will plot these values to graph 𝑓(𝑥) and plot
the reverse values for 𝑓 −1 (𝑥)
𝑓 = {(−2, −1), (0, 3), (2, 7)}
𝑓 −1 = {−1, −2), (3, 0), (7, 2)}
Diagram 3
In diagram 1 it is clear that the graph does not
pass the horizontal line test and therefore it is not
one – to –one. However, by restricting the domain,
we can create a one – to – one relation. The largest
one – to – one relation will be created by using
line of symmetry as seen in the latter 2 diagrams
above.
ONE – TO – ONE QUADRATIC
FUNCTIONS
Diagram 1 shows the graph of
𝑓 (𝑥 ) = 3𝑥 2 − 12𝑥 + 4 with its line of symmetry.
Diagram 2 and diagram 3 show the right half and
the left half of the graph respectively.
LESSON 22
Given that 𝑓(𝑥) = 3𝑥 2 − 12𝑥 + 4
(a) Express 𝑓(𝑥 ) in the form 𝑎(𝑥 + 𝑏)2 + 𝑐, where
𝑎, 𝑏 and 𝑐 are real numbers.
(b) State the least value of 𝑥 for which 𝑓(𝑥) is one
– to – one.
(c) State the range of 𝑓(𝑥) for which 𝑓 −1 (𝑥)
exists.
(d) Determine the inverse of 𝑓(𝑥).
SOLUTION
(a) 𝑓(𝑥) = 3𝑥 2 − 12𝑥 + 4
4
𝑓(𝑥) = 3 [𝑥 2 − 4𝑥 + ]
3
𝑓(𝑥) = 3 [𝑥 2 − 4𝑥 + 𝑐 +
𝑐=(
−4 2
) =4
2
𝑓(𝑥) = 3 [𝑥 2 − 4𝑥 + 4 +
4
− 𝑐]
3
4
− 4]
3
29
CHAPTER 6: FUNCTIONS
8
𝑓(𝑥) = 3 [(𝑥 − 2)2 − ]
3
𝑓(𝑥) = 3(𝑥 − 2)2 − 8
(b) The equation for the axis of symmetry is
𝑥 = 2, therefore the least value of 𝑥 for which
𝑓(𝑥) is one – to – one is 𝑥 = 2.
(c) The minimum value of 𝑓(𝑥) is −8, therefore
the corresponding range is 𝑦 ≥ −8.
(d) Let 𝑦 = 𝑓 (𝑥 )
𝑦 = 3(𝑥 − 2)2 − 8
𝑦 + 8 = 3(𝑥 − 2)2
𝑦+8
= (𝑥 − 2)2
3
(ii) Express each of 𝑓 −1 (𝑥) and 𝑔−1 (𝑥) in
terms of 𝑥.
(iii) Show that the equation
𝑓 −1 (𝑥) = 𝑔 −1 (𝑥) has no real roots.
(iv)
Sketch, on a single diagram, the
graphs of 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑓 −1 (𝑥),
making clear the relationship
between these two graphs.
4.
𝑓(𝑥) = 2𝑥 + 1, 𝑥 ∈ ℝ, 𝑥 > 0
2𝑥 − 1
𝑔(𝑥) =
, 𝑥 ∈ ℝ, 𝑥 ≠ 0
𝑥 +3
(i) Solve the equation 𝑔𝑓(𝑥) = 𝑥.
(ii) Express 𝑓 −1 (𝑥) and 𝑔 −1 (𝑥) in terms of 𝑥.
(iii) State the range of 𝑔.
(iv) Show that the equation 𝑔 −1 (𝑥) = 𝑥 has no
solutions.
(v) Sketch, in a single diagram the graphs of
𝑦 = 𝑓(𝑥) and 𝑦 = 𝑓 −1 (𝑥), making clear
the relationship between the graphs.
𝑦+8
√
= 𝑥−2
3
𝑦+8
2+√
=𝑥
3
Replacing 𝑦 with 𝑥
𝑥+8
𝑓 −1 (𝑥) = 2 + √
3
NB: ± is not used since the function is one – to
– one. Since the least value of 𝑥 was required
we want the right half of the graph and
consequently + was used. – would have been
used if the left half of the graph was required.
…………………………………………………………………………..
EXERCISE 6.3
1. The function 𝑓 is defined as 𝑓(𝑥) = 3𝑥 − 1.
(i) Determine 𝑓 −1 (𝑥)
(ii) On a single graph, sketch the graphs of
𝑓(𝑥) and 𝑓 −1 (𝑥). Clearly show the
relationship between 𝑓 and its inverse.
2. The functions 𝑓 and 𝑔 are defined by
5.
(i) Find the set of values for which
𝑓(𝑥) > 15.
(ii) Show that 𝑔𝑓(𝑥) = 0 has no real solution.
(iii) Sketch in a single diagram the graphs of
𝑦 = 𝑔(𝑥) and 𝑦 = 𝑔 −1 (𝑥), clearly showing
the relationship between the two graphs.
3.
Functions 𝑓 and 𝑔 are defined by
𝑓(𝑥) = 2𝑥 − 5,
4
𝑔(𝑥) =
, 𝑥 ∈ ℝ, 𝑥 ≠ 2
2−𝑥
(i) Find the value of 𝑥 for which 𝑓𝑔(𝑥) = 7.
(i) Express 3𝑥 2 − 12𝑥 + 8 in the form
𝑎 (𝑥 + 𝑏) 2 + 𝑐, where 𝑎, 𝑏 and 𝑐 are
integers.
A function 𝑓 is defined by
𝑓(𝑥) = 3𝑥 2 − 12𝑥 + 8, 𝑥 ∈ ℝ.
(ii) Find the coordinates of the minimum
point on the graph of 𝑦 = 𝑓(𝑥).
A function is defined by
𝑔(𝑥) = 3𝑥 2 − 12𝑥 + 8, 𝑥 ∈ ℝ, where 𝑥 ≤ 𝑁.
(iii) State the greatest value of 𝑁 for which 𝑔
has an inverse.
(iv) Using the result obtained in part (i), find
an expression for 𝑔 −1 .
𝑓: 𝑥 → 𝑥 2 − 2𝑥, 𝑥 ∈ ℝ
𝑔: 𝑥 → 3𝑥 + 5, 𝑥 ∈ ℝ
Functions 𝑓 and 𝑔 are defined by
6.
(i) Express 2𝑥 2 − 8𝑥 + 3 in the form
𝑎(𝑥 + 𝑏) 2 + 𝑐, where 𝑎, 𝑏 and 𝑐 are
integers.
A function 𝑓 is defined by
𝑓(𝑥) = 2𝑥 2 − 8𝑥 + 3, 𝑥 ∈ ℝ.
(ii) Find the coordinates of the minimum
point on the graph of 𝑦 = 𝑓(𝑥).
(iii) Find the value of 𝑓 2 (0).
A function is defined by
𝑔(𝑥) = 2𝑥 2 − 8𝑥 + 3, 𝑥 ∈ ℝ, where 𝑥 ≤ 𝑁.
(iv) State the greatest value of 𝑁 for which 𝑔
has an inverse.
30
CHAPTER 6: FUNCTIONS
(v) Using the result obtained in part (i), find
an expression for 𝑔 −1 .
SOLUTIONS
𝑥+1
1. (i) 3
(ii)
2. (i) {𝑥 < −3} ∪ {𝑥 > 5}
(ii) Discriminant is negative (iii)
4
𝑥+5
2𝑥−4
3. (i) 3 (ii) 𝑓 −1 (𝑥) = 2 , 𝑔−1 (𝑥) = 𝑥
(iii) Discriminant is negative
4.
1
(i) 𝑥 = ±√2 (ii) 𝑓 −1 (𝑥) =
𝑥−1
2
3𝑥+1
, 𝑔 −1 (𝑥) = 2−𝑥
(iii) 𝑦 ≠ 2 (iv) Discriminant is negative
5.
LESSON 24
The function 𝑓 is defined by
𝑥 − 3; −5 ≤ 𝑥 < −2
−2 < 𝑥 < 0
𝑓(𝑥) = { 2;
0≤𝑥<9
√𝑥;
1.
2.
State the domain of 𝑓.
Determine the values of the following.
i. 𝑓𝑓(4)
ii. 𝑓𝑓(−1)
iii. 𝑓𝑓(−4)
SOLUTION
1.
(i) 3(𝑥 − 2)2 − 4 (ii) (2, −4) (iii) 2
Placing the intervals on a number line we
have
𝑥+4
(iv) 2 + √ 3
6.
(i) 2(𝑥 − 2)2 − 5 (ii) (2, −5) (iii) −3
𝑥+5
(iv) 2 (v) 2 + √
2.
2
…………………………………………………………………………..
PIECE-WISE FUNCTIONS
A Piece-wise function is a function whose formula
changes depending upon the input values used.
LESSON 23
The function 𝑓(𝑥) is defined as
𝑥 + 5;
𝑥≤0
𝑓(𝑥) = { 2
𝑥 ;
𝑥>0
Evaluate:
1.
2.
3.
4.
𝑓(0)
𝑓(5)
𝑓(−2)
𝑓(−1) + 𝑓(1)
SOLUTION
i. 𝑓(0)
0 is in the interval 𝑥 ≤ 0 therefore we use the
formula 𝑥 + 5
So 𝑓(0) = 0 + 5 = 5
ii. 𝑓(5)
5 is in the interval 𝑥 > 0, therefore we use the
formula 𝑥 2 .
So 𝑓(5) = 52 = 25
iii. 𝑓(−2) = −2 + 5 = 3 since −2 is in the
interval 𝑥 ≤ 0
iv. 𝑓(−1) = −1 + 5 = 4 and 𝑓(1) = 12 = 1
𝑓 (−1) + 𝑓 (1) = 4 + 1 = 5
The domain of 𝑓: {𝑥: −5 ≤ 𝑥 < 9, 𝑥 ≠ −2}
i. 𝑓𝑓(4)
using √𝑥
= 𝑓[√4]
= 𝑓(2)
using √𝑥
= √2
ii. 𝑓𝑓(−1)
using 2
= 𝑓 (2)
using √𝑥
= √2
iv. 𝑓𝑓(−4)
using 𝑥 − 3
= 𝑓(−7)
Undefined: −7 is outside domain of 𝑓
NB: For compositions of piece-wise functions DO
NOT create a formula for the composition.
…………………………………………………………………………..
EXERCISE 6.4
1.
𝑥 + 7, 𝑥 ≤ 2
For 𝑓 (𝑥 ) = { 𝑥 , 𝑥 > 2 , evaluate
3
(a) 𝑓(2)
(b) 𝑓(−4)
(c) 𝑓(15)
(d) 𝑓(63)
2.
𝑥 − 5, − 9 ≤ 𝑥 < 3
For 𝑓(𝑥) = { 𝑥 , 3 ≤ 𝑥 < 8 , determine
2
(i) the domain of 𝑓
(ii) 𝑓𝑓(0)
(iii) 𝑓𝑓(4)
SOLUTIONS
1. (a) 9
(b) 3
(c) 5
2. (i) −9 ≤ 𝑥 < 8 (ii) −10
(d) 21
(iii) −3
…………………………………………………………………………..
31
CHAPTER 6: FUNCTIONS
7.
EXAM QUESTIONS
1.
(i) The function 𝑓 is defined by
𝑓: 𝑥 → 1 − 𝑥 2 , 𝑥 ∈ ℝ.
𝑥−1
𝑔(𝑥) = √ 2 where 1 ≤ 𝑥 < ∞, 𝑥 ∈ ℝ .
(i)
Determine, in terms of 𝑥,
(a) 𝑓 2 (𝑥)
(b) 𝑓[𝑔(𝑥)]
(ii) Hence, or otherwise, state the
relationship between 𝑓 and 𝑔.
Show that 𝑓 is NOT one – to – one.
(ii) The function 𝑔 is defined by
1
𝑔: 𝑥 → 𝑥 − 3, 𝑥 ∈ ℝ.
2
(a) Find 𝑓𝑔(𝑥), and clearly state its
domain.
(b) Determine the inverse, 𝑔 −1 , of 𝑔 and
sketch on same pair of axes, the
graphs of 𝑔 and 𝑔−1 .
ADDITIONAL MATHEMATICS 2014
2.
3.
4.
5.
The functions 𝑓 and 𝑔 are defined on 𝑅 by
𝑓: 𝑥 → 2𝑥
𝑔: 𝑥 → 𝑥 2 − 3
Determine the set of values of 𝑥 for which
𝑓(𝑓(𝑥)) = 𝑔(𝑓(𝑥)).
[7]
CAPE 2000
The function 𝑓 and 𝑔 are defined on ℝ by
𝑓: 𝑥 → −3𝑥 + 6,
𝑔: 𝑥 → 𝑥 + 7
Solve, for 𝑥, the equation 𝑓(𝑔(2𝑥 + 1)) = 30
[5]
CAPE 2004
The function 𝑓(𝑥) has the property that
𝑓 (2𝑥 + 3) = 2𝑓 (𝑥 ) + 3, 𝑥 ∈ ℝ.
If 𝑓(0) = 6, find the value of
(i)
𝑓(3)
[4]
(ii)
𝑓(9)
[2]
(iii)
𝑓(−3)
[3]
CAPE 2011
The function 𝑓 and 𝑔 are defined by
𝑓(𝑥) = 𝑥 3 + 1,
0≤𝑥≤3
𝑔(𝑥) = 𝑥 + 5, 𝑥 ∈ ℝ
where ℝ is the set of real numbers.
(i) Determine the composition function
𝑔(𝑓(𝑥)).
[1]
(ii) State the range of 𝑔(𝑓(𝑥)).
[1]
(iii) Determine the inverse of 𝑔(𝑓(𝑥)).
[2]
ADDITIONAL MATHEMATICS 2012
6.
The functions 𝑓 and 𝑔 are defined as follows:
𝑓(𝑥) = 2𝑥 2 + 1
2𝑥−1
A function 𝑓(𝑥) is given by 𝑓(𝑥) = 𝑥+2 .
(i) Find an expression for the inverse
function 𝑓 −1 (𝑥).
[3]
(ii) The function 𝑔 is given by 𝑔(𝑥) = 𝑥 + 1.
Write an expression for the composite
function, 𝑓𝑔(𝑥). Simplify your answer. [2]
ADDITIONAL MATHEMATICS 2013
8.
9.
[3]
[3]
[1]
CAPE 2014
The diagram below (not drawn to scale)
represents the graph of the function
𝑓(𝑥) = 𝑥 2 + 1, −1 ≤ 𝑥 ≤ 1 and 𝑝, 𝑞 ∈ ℝ.
(a) Find
(i) the value of 𝑝 and 𝑞.
[2]
(ii) the range of the function 𝑓(𝑥 ) for the
given domain.
[1]
(b) determine whether 𝑓(𝑥)
(i) is surjective (onto)
[1]
(ii) is injective (one – to – one)
[1]
(iii) has an inverse.
[1]
CAPE 2006
The function 𝑓 on 𝑅 is defined by
𝑥 −3
if 𝑥 ≤ 3
𝑓(𝑥) = { 𝑥
if 𝑥 > 3
4
Find the value of
(i) 𝑓[𝑓(20)]
(ii) 𝑓[𝑓(8)]
(iii) 𝑓[𝑓(3)]
[3]
[2]
[2]
CAPE 2009
10. Let 𝐴 = {𝑥: 𝑥 ∈ ℝ, 𝑥 ≥ 1}.
A function 𝑓: 𝐴 → ℝ is defined as
𝑓(𝑥) = 𝑥 2 − 𝑥. Show that 𝑓 is one to one. [7]
CAPE 2013
11. The relations 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 are
functions which are both one – to – one and
onto. Show that (𝑔 ∘ 𝑓) is
(i)
one – to – one
[4]
(ii)
onto
[4]
CAPE 2015
32
CHAPTER 6: FUNCTIONS
SOLUTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
𝑥
2
(ii) (a) 1 − (2 − 3) , 𝑥 ∈ ℝ
(b) 𝑔−1 (𝑥) = 2(𝑥 + 3)
1 3
𝑥 = − 2,2
𝑥 = −8
(i) 15 (ii) 33 (iii) −3
(i) 𝑥 3 + 6 (ii) 𝑦 ∈ ℝ (iii) 3√𝑥 − 6
−1−2𝑥
2𝑥+1
(i) 𝑥−2 (ii) 𝑥+3
(i) (a) 8𝑥 4 + 8𝑥 2 + 3 (b) 𝑥
(a) (i) 𝑝 = 2, 𝑞 = 1
(ii) 1 ≤ 𝑦 ≤ 2
(b) (i) Yes (ii) No (iii) No
5
(i) 4
(ii) −1
(iii) −3
…………………………………………………………………………..
33
CHAPTER 7: POLYNOMIALS
CHAPTER 7: POLYNOMIALS
At the end of this section, students should be able
to:
 factorise polynomial expressions of
degree less than or equal to 4, leading to
real linear factors;
 apply the Remainder Theorem;
 use the Factor Theorem to find factors
and to evaluate unknown coefficients;
 use the relationship between the sum of
the roots, the product of the roots, the
sum of the product of the roots of the
roots pair – wise and the coefficients of
𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 0
 extract all factors of 𝑎𝑛 − 𝑏𝑛 for positive
integers 𝑛 ≤ 6;
__________________________________________________________
INTRODUCTION
𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 𝑎𝑛 ≠ 0
is called an 𝑛th degree polynomial function. 𝑎 is a
real number and 𝑛 is a non – negative integer.
 𝑃(𝑥) = 𝑎 is a constant function. A zero
degree polynoimial.
 𝑃(𝑥) = 𝑎𝑥 + 𝑏 is a linear function. A 1st
degree polynomial
 𝑃(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 is a quadratic
function. A 2nd degree polynomial.
 𝑃 (𝑥 ) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 is a cubic
polynomial. A 3rd degree polynomial.
SYNTHETIC DIVISION
LESSON 1
Divide 𝑥 2 + 5𝑥 + 6 by 𝑥 + 2.
SOLUTION
NB: 𝑥 2 + 5𝑥 + 6 is called the
dividend whereas 𝑥 + 2 is the divisor.
∴
𝑥 2 + 5𝑥 + 6
= 𝑥+3
𝑥 +2
LESSON 2
If 𝑥 − 1 is a factor of
𝑃(𝑥) = 𝑥 3 − 7𝑥 + 6 find the remaining factor(s).
SOLUTION
= (𝑥 − 1)(𝑥 2 + 𝑥 − 6)
= (𝑥 − 1)(𝑥 + 3)(𝑥 − 2)
LESSON 3
If 𝑥 + 3 is a factor of
𝑃(𝑥) = 2𝑥 3 + 9𝑥 2 + 6𝑥 − 9 find the remaining
factor(s).
SOLUTION
𝑃(𝑥) =
LESSON 4
Find the quotient and remainder
when 𝑃(𝑥) = 2𝑥 4 + 3𝑥 3 − 3𝑥 + 5 is divided by
2𝑥 − 1.
SOLUTION
Dividing 2𝑥 − 1 by 2 we
1
have 𝑥 − 2 which is then our divisor.
Then we have
1
𝑃(𝑥) ≡ (𝑥 − ) (2𝑥 3 + 4𝑥 2 + 2𝑥 − 2) + 4
2
1
≡ 2 (𝑥 − ) (𝑥 3 + 2𝑥 2 + 𝑥 − 1) + 4
2
≡ (2𝑥 − 1)(𝑥 3 + 2𝑥 2 + 𝑥 − 1) + 4
…………………………………………………………………………
EXERCISE 7.1
1. Divide
(a) 𝑥 3 − 11𝑥 2 + 38𝑥 − 40 by 𝑥 − 2
(b) 𝑥 3 − 𝑥 2 − 𝑥 + 1 by 𝑥 + 1
(c) 𝑥 3 − 2𝑥 2 − 25𝑥 + 50 by 𝑥 − 2
(d) 2𝑥 3 + 11𝑥 2 + 18𝑥 + 9 by 𝑥 + 3
(e) 2𝑥 3 − 5𝑥 2 − 11𝑥 − 4 by 𝑥 − 4
(f) 9𝑥 3 − 27𝑥 2 + 23𝑥 − 5 by 𝑥 − 1
(g) 𝑥 3 − 3𝑥 + 2 by 𝑥 + 2
(h) 4𝑥 3 + 20𝑥 2 + 29𝑥 + 10 by 𝑥 + 2
(i) 6𝑥 3 + 13𝑥 2 − 10𝑥 − 24 by 𝑥 + 2
(j) 3𝑥 3 − 10𝑥 2 − 16𝑥 + 32 by 𝑥 − 4
2.
𝑃(𝑥) = 𝑥 3 − 7𝑥 + 6
Divide
(a) 2𝑥 3 − 11𝑥 2 + 13𝑥 − 4 by 2𝑥 − 1
(b) 2𝑥 3 − 11𝑥 2 + 19𝑥 − 10 by 2𝑥 − 5
(c) 3𝑥 3 + 20𝑥 2 + 27𝑥 + 10 by 3𝑥 + 2
(d) 4𝑥 3 + 16𝑥 2 − 3𝑥 − 45 by 2𝑥 + 5
34
CHAPTER 7: POLYNOMIALS
SOLUTIONS
1. (a) (𝑥 − 4)(𝑥 − 5)
(c) (𝑥 + 5)(𝑥 − 5)
(e) (2𝑥 + 1)(𝑥 + 1)
(g) (𝑥 − 1)2
(i) (3𝑥 − 4)(2𝑥 + 3)
2. (a) (𝑥 − 1)(𝑥 − 4)
(c) (𝑥 + 1)(𝑥 + 5)
(b) (𝑥 − 1)2
(d) (2𝑥 + 3)(𝑥 + 1)
(f) (3𝑥 − 1)(3𝑥 − 5)
(h) (2𝑥 + 1)(2𝑥 + 5)
(j) (3𝑥 − 4)(𝑥 + 2)
(b) (𝑥 − 1)(𝑥 − 2)
(d) (2𝑥 − 3)(𝑥 + 3)
…………………………………………………………………………
REMAINDER AND FACTOR
THEOREM
INTRODUCTION
If 𝑅 is the remainder after dividing the polynomial
𝑃(𝑥) by 𝑥 + 𝑟, then
𝑃(−𝑟) = 𝑅
It is clear that if 𝑥 + 𝑟 is a factor of the polynomial
𝑃(𝑥) then the remainder (𝑅) is zero. Therefore
from (*) above we would have that 𝑃(−𝑟) = 0.
This leads us directly to the factor theorem.
If −𝑟 is a root of the polynomial 𝑃(𝑥), then 𝑥 + 𝑟 is
a factor of 𝑃(𝑥). Conversely, if 𝑥 + 𝑟 is a factor of
𝑃(𝑥 ), then – 𝑟 is a root of 𝑃(𝑥).
LESSON 5
The expression
𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 7𝑥 + 𝑏 is divisible by 𝑥 − 4,
determine
(i) the value of 𝑏,
(ii) the remainder when 𝑓 (𝑥 ) is divided by 2𝑥 − 1
SOLUTION
𝑥 − 4 is a factor → 𝑥 = 4 is a root
𝑃(4) = 0
by Factor Theorem
𝑃(4) = 2(4)3 − 3(4) 2 − 7(4) + 𝑏
2(4) 3 − 3(4)2 − 7(4) + 𝑏 = 0
128 − 48 − 28 + 𝑏 = 0
𝑏 = −52
2𝑥 3 − 3𝑥 2 − 7𝑥 − 52
1
2𝑥 − 1 → corresponding 𝑥 is 𝑥 =
2
1
1 3
1 2
1
𝑃 ( ) = 2 ( ) − 3 ( ) − 7 ( ) − 52 = −56
2
2
2
2
LESSON 6
Let 𝑓(𝑥) = 2𝑥 3 + 𝑎𝑥 2 − 4𝑥 + 𝑏
(i) Given that 𝑥 − 1 is a factor of 𝑓(𝑥) and that
there is a remainder of 9 when 𝑓(𝑥 ) is divided
by 𝑥 + 2, find the values of 𝑎 and 𝑏.
(ii) Hence
(a) factor 𝑓(𝑥)
(b) solve the equation 𝑓(𝑥) = 0.
SOLUTION
By the Factor Theorem
𝑃(−1) = 0
2(−1) 3 + 𝑎(−1) 2 + 𝑏 = 0
𝑎+𝑏 = 2
By the Remainder Theorem
𝑃(3) = 16
2(3) 3 + 𝑎(3) 2 + 𝑏 = 16
9𝑎 + 𝑏 = −38
Solving (*) and (**) simultaneously
𝑎+𝑏 = 2
9𝑎 + 𝑏 = −38
𝑎 = −5 and 𝑏 = 7
LESSON 7
Given that
𝑓 (𝑥 ) = 2𝑥 3 + 𝑝𝑥 2 − 12𝑥 + 𝑞 is divisible by
𝑥 2 − 2𝑥 − 3 show that 𝑝 = −1 and 𝑞 = −9.
SOLUTION
𝑥 2 − 2𝑥 − 3 = 0
(𝑥 − 3)(𝑥 + 1) = 0
𝑥 = −1, 3
The roots of 𝑥 2 − 2𝑥 − 3 have to be roots of
2𝑥 3 + 𝑝𝑥 2 − 12𝑥 + 𝑞
By the Factor Theorem
𝑃 (−1) = 0
2(−1) 3 + 𝑝(−1)2 − 12(−1) + 𝑞 = 0
𝑝 + 𝑞 = −10
(*)
𝑃(3) = 0
2(3) 3 + 𝑝(3) 2 − 12(3) + 𝑞 = 0
9𝑝 + 𝑞 = −18
(**)
Solving (*) and (**) simultaneously
𝑝 + 𝑞 = −10
9𝑝 + 𝑞 = −18
𝑝 = −1 and 𝑞 = −9
LESSON 8
Factorise 𝑥 3 + 2𝑥 2 − 𝑥 − 2.
SOLUTION
Let 𝑃(𝑥) = 𝑥 3 + 2𝑥 2 − 𝑥 − 2
We now need to find a root of 𝑃(𝑥), therefore we
try various values of 𝑥 until we get 𝑃(𝑥) = 0. By
inspection we see that 𝑃(1) = 13 + 2(1) 2 − 1 −
2 = 0, therefore (𝑥 − 1) is a factor of 𝑃(𝑥). We can
continue by guessing the other 2 roots (a cubic
equation has at most three roots) or we can use
polynomial long division
35
CHAPTER 7: POLYNOMIALS
SOLUTIONS
1. 5
2. −21
3. 𝑝 = −11, 𝑞 = 6
4. (i) 𝑎 = −3, 𝑏 = −2 (ii) (𝑥 − 2)(𝑥 + 1) 2
5. (2𝑥 − 1)(𝑥 − 3)(𝑥 + 5)
5
6. 𝑥 = −2, 2 , 3
7.
1
(i) −3, , 2
3
or synthetic division.
𝑃(𝑥) = (𝑥 − 1)(𝑥 2 + 3𝑥 + 2)
= (𝑥 − 1)(𝑥 + 1)(𝑥 + 2)
…………………………………………………………………………
EXERCISE 7.2
1. Use the remainder theorem to find the
remainder when 2𝑥 3 + 3𝑥 2 + 9 is divided by
𝑥 + 2.
2. The expression 𝑎𝑥 3 − 4𝑥 2 + 𝑏𝑥 − 5 has a
factor of 𝑥 + 1 and leaves a remainder of 4
when divided by 𝑥 − 3. Calculate the
remainder when the expression is divided by
𝑥 − 2.
3. Find the value of 𝑝 and 𝑞 for which
𝑥 2 − 5𝑥 − 6 is a factor of 2𝑥 3 + 𝑝𝑥 2 − 7𝑥 + 𝑞.
4. The cubic polynomial 𝑓(𝑥) is given by
𝑓(𝑥) = 𝑥 3 + 𝑎𝑥 + 𝑏
where 𝑎 and 𝑏 are constants. It is given that
(𝑥 + 1) is a factor of 𝑓(𝑥) and that the
remainder when 𝑓(𝑥) is divided by (𝑥 − 3) is
16.
(i) Find the values of 𝑎 and 𝑏.
(ii) Hence verify that 𝑓(2) = 0, and factorise
𝑓(𝑥) completely.
5. Factorise 2𝑥 3 + 3𝑥 2 − 32𝑥 + 15.
6. Factorise 2𝑥 3 − 7𝑥 2 − 7𝑥 + 30, hence solve
the equation 2𝑥 3 − 7𝑥 2 − 7𝑥 = −30.
7. Solve the following equations
(i)
2𝑥 3 + 𝑥 2 − 13𝑥 + 6 = 0
(ii)
3𝑥(𝑥 2 + 6) = 8 − 17𝑥 2
8. Factorise completely the expression
4𝑥 3 − 13𝑥 − 6 and hence solve the equation
3
2 (2𝑥 2 − ) = 13
𝑥
2
1
(ii) −4, −2,
1
3
8. − 2 , − 2 , 2
…………………………………………………………………………
EXAM QUESTIONS
1. The function 𝑓(𝑥) = 𝑥 3 − 𝑝2 𝑥 2 + 2𝑥 − 𝑝 has
remainder −5 when it is divided by 𝑥 + 1.
Find the possible values of 𝑝.
[6]
CAPE 2004
2. The function 𝑓(𝑥) is given by
𝑓(𝑥) = 𝑥 4 − (𝑝 + 1)𝑥 2 + 𝑝, 𝑝 ∈ ℕ.
(i)
Show that (𝑥 − 1) is a factor of 𝑓(𝑥)
for all values of 𝑝.
[2]
(ii)
If (𝑥 − 2) is a factor of 𝑓(𝑥), find the
value of 𝑝.
[2]
CAPE 2006
3. If 𝑥 − 1 is a factor of the function
𝑓 (𝑥 ) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 − 2, 𝑝 ∈ ℝ, find
(a) the value of 𝑝
[2]
(b) the remaining factors
[4]
CAPE 2007
4. The roots of the cubic equation
𝑥 3 + 3𝑝𝑥 2 + 𝑞𝑥 + 𝑟 = 0 are 1, −1 and 3. Find
the values of the real constants 𝑝, 𝑞 and 𝑟. [7]
CAPE 2008
5. Find the values of the constant 𝑝 such that
𝑥 − 𝑝 is a factor of
𝑓 (𝑥 ) = 4𝑥 3 − (3𝑝 + 2)𝑥 2 − (𝑝2 − 1) 𝑥 + 3.
[5]
6. Let 𝑓(𝑥) = 𝑥 3 − 9𝑥 2 + 𝑝𝑥 + 16
(i)
Given that (𝑥 + 1) is a factor of 𝑓(𝑥),
show that 𝑝 = 6.
[2]
(ii)
Factorise 𝑓(𝑥) completely.
[4]
(iii)
Hence, or otherwise, solve 𝑓(𝑥) = 0.
[3]
CAPE 2013
7. The polynomial 𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 + 𝑞 has
a factor (𝑥 − 5) and a remainder of 24 when
divided by (𝑥 − 1).
(i)
Find the values of 𝑝 and 𝑞.
[4]
(ii)
Hence, factorise
𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 + 𝑞 completely.
[5]
CAPE 2015
36
CHAPTER 7: POLYNOMIALS
SOLUTIONS
1. 𝑝 = −2, 1
2. (ii) 𝑝 = 4
3. (i) 𝑝 = 2
(ii) (𝑥 − 1)(𝑥 + 2)(𝑥 + 1)
4. 𝑝 = −1, 𝑞 = −1, 𝑟 = 3
3
5.
𝑝 = −1, 2
6.
7.
(iii) −1, 2, 8
(i) 𝑝 = −6, 𝑞 = 30 (ii) (𝑥 − 5)(𝑥 + 2)(𝑥 − 3)
(−𝑝)
𝑏
𝛼+𝛽=− =−
=𝑝
𝑎
1
𝑐
𝛼𝛽 = = 𝑝
𝑎
LESSON 10
If 𝛼 and 𝛽 are the roots of the
equation 𝑥 2 − 4𝑥 − 2 = 0, find the values of
a) 𝛼 + 𝛽
b) 𝛼𝛽
c) 𝛼 2 + 𝛽 2
SOLUTION
…………………………………………………………………………
𝑏
−4
=−
=4
𝑎
1
𝑐
2
b) 𝛼𝛽 = 𝑎 = − 1 = −2
a)
ROOTS OF POLYNOMIALS
INTRODUCTION
A number 𝑎 is called a root of the polynomial 𝑃(𝑥)
if 𝑃(𝑎) = 0. For a quadratic equation we have
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
𝑏
𝑐
→ 𝑥2 + 𝑥 + = 0
𝑎
𝑎
Also if 𝛼 and 𝛽 are the roots of a quadratic
equation then
(𝑥 − 𝛽 )(𝑥 − 𝛼 ) = 0
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 = 0
Equating coefficients of 𝑥 we then have:
𝑏
−(𝛼 + 𝛽) =
𝑎
𝑏
∴ 𝛼+𝛽 =−
𝑎
i.e. the sum of the roots of a quadratic equation =
𝑏
−
𝑎
Equating constants we have:
𝑐
𝛼𝛽 =
𝑎
i.e. the product of the roots of a quadratic
𝑐
equation =
𝑎
Thus, we know that given the roots of a quadratic
equation the corresponding equation is
𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
LESSON 9
Find the sums and products of the
roots of the following equations.
a) 3𝑥 2 − 5𝑥 − 2 = 0
b) 𝑥 2 − 𝑝𝑥 + 𝑝 = 0
SOLUTION
Let 𝛼 and 𝛽 be the roots of the equations
a) 3𝑥 2 − 5𝑥 − 2 = 0
𝑎 = 3, 𝑏 = −5, 𝑐 = −2
𝑏
−5 5
(sum of roots): 𝛼 + 𝛽 = − = −
=
𝑎
3
3
𝑐
2
(product of roots): 𝛼𝛽 = = −
𝑎
3
b) 𝑥 2 − 𝑝𝑥 + 𝑝 = 0
𝑎 = 1, 𝑏 = −𝑝, 𝑐 = 𝑝
𝛼+𝛽=−
c) (𝛼 + 𝛽)2 = 42
𝛼 2 + 2𝛼𝛽 + 𝛽 2 = 16
𝛼 2 + 𝛽 2 = 16 − 2𝛼𝛽
= 16 − 2(−2) = 20
LESSON 11
If 𝛼 and 𝛽 are the roots of the
equation 𝑥 2 + 𝑥 − 2 = 0, find equations whose
roots are
a) – 𝛼, −𝛽
b) 𝛼 2 , 𝛽 2
c) 𝛼 − 1, 𝛽 − 1
SOLUTION
𝑏
1
= − = −1
𝑎
1
𝑐 −2
𝛼𝛽 = =
= −2
𝑎
1
a) (– 𝛼) + (−𝛽) = −(𝛼 + 𝛽) = −(−1) = 1
(−𝛼)(−𝛽) = 𝛼𝛽 = −2
𝛼+𝛽 =−
𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
𝑥 2 − (1)𝑥 + (−2) = 0
𝑥2 − 𝑥 − 2 = 0
b) 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽 )2 − 2𝛼𝛽
= (−1)2 − 2(−2)
=5
(𝛼 2 )(𝛽 2 ) = (𝛼𝛽)2 = (−2) 2 = 4
𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
c)
𝑥 2 − 5𝑥 + 4 = 0
(𝛼 − 1) + (𝛽 − 1) = 𝛼 + 𝛽 − 2
= −1 − 2 = −3
(𝛼 − 1)(𝛽 − 1) = 𝛼𝛽 − 𝛼 − 𝛽 + 1
= 𝛼𝛽 − (𝛼 + 𝛽) + 1
= −2 − (−1) + 1
=0
𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
𝑥 2 − (−3)𝑥 + 0 = 0
𝑥 2 + 3𝑥 = 0
37
CHAPTER 7: POLYNOMIALS
LESSON 12
The roots of the equation
3𝑥 2 − 4𝑥 + 1 = 0 are 𝛼 and 𝛽.
Without solving the equation
(i)
write down the values of 𝛼 + 𝛽 and 𝛼𝛽
(ii)
find the value of 𝛼2 + 𝛽2
(iii)
obtain a quadratic equation whose roots are
1
α2
a.
b.
c.
d.
e.
1
and β2
f.
𝛼2 + 𝛽2
(𝛼 − 𝛽)2
𝛼3 + 𝛽3
1
1
+𝛽
𝛼
1
𝛼2
𝛼
𝛽2
1
+ 𝛽2
𝛽
+ 𝛼2
SOLUTION
4
(i) 𝛼 + 𝛽 = 3
𝛼𝛽
(ii) 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
4 2
1
= ( ) − 2( )
3
3
10
=
9
1
1
1
1
(iii) 𝑥 2 − (𝛼2 + 𝛽2 ) 𝑥 + (𝛼 ) (𝛽) = 0
7.
𝑥 2 + 10𝑥 + 9 = 0
…………………………………………………………………………
EXERCISE 7.3
1. For 𝑥 2 − 5𝑥 − 9 = 0 the sum of the roots is
(A) 5
(B) −5
1
(C)
4.
(a) 1, 125
5.
(a) −3
6.
(a) 4
5
2.
3.
3
(B) − 25
25
(C) 3
4.
5.
6.
SOLUTIONS
1. A
2. B
3. C
25
(D) − 3
Find the sum and product of the roots of the
following equations.
a. 5 + 𝑥 − 𝑥 2 = 0
b. 𝑥(𝑥 − 3) = 5(𝑥 − 2)
c. 3𝑥 2 − 7 = 0
1
d. 3𝑥 − 𝑥 = 4
If 𝛼 and 𝛽 are the roots of the equation
2𝑥 2 + 6𝑥 − 3 = 0, determine the values of
a. 𝛼 + 𝛽
b. 𝛼𝛽
c. 𝛼 2 + 𝛽 2
d. 𝛼 3 + 𝛽 3
If 𝛼 and 𝛽 are the roots of the equation
2𝑥 2 − 𝑥 − 4 = 0, find the values of
17
(f)
7.
1
(D) − 5
What is the value of the sum of the roots of
the equation 2𝑥 2 − 8𝑥 + 3 = 0?
(A) −4
(B) 4
(C) 8
(D) −8
What is the value of the product of the roots of
the equation 3𝑥 2 − 8𝑥 + 25 = 0?
3
(A) 25
If 𝛼 and 𝛽 are the roots of the equation
𝑥 2 + 𝑥 − 4 = 0, find equations whose roots are
a) – 𝛼, −𝛽
b) 𝛼 2 , 𝛽 2
c) 𝛼 − 1, 𝛽 − 1
(b) 8, 10
3
(b) − 2
33
(b) 4
7
(c) 0, − 3
(c) 12
25
(c) 8
4
1
(d) 3 , − 3
(d) −40.5
1
17
(d) − 4 (e) 16
25
32
(a) 𝑥 2 − 𝑥 − 4 = 0 (b) 𝑥 2 − 9𝑥 + 16 = 0
(c) 𝑥 2 + 3𝑥 − 2 = 0
…………………………………………………………………………
EXAM QUESTIONS
1. Let 𝛼 and 𝛽 be the roots of the equation
4𝑥 2 − 3𝑥 + 1 = 0
(a) Without solving the equation, write down
the values of
(i)
𝛼+𝛽
(ii)
𝛼𝛽
[1]
(b) Find the value of 𝛼 2 + 𝛽 2.
[2]
(c) Find the equation whose roots are
2
2
and 2.
[3]
2
𝛼
2.
3.
𝛽
CAPE 2000
Given that 𝛼 and 𝛽 are the roots of the
equation 𝑥 2 − 3𝑥 − 1 = 0, find the equation
whose roots are 1 + 𝛼 and 1 + 𝛽.
[5]
CAPE 2004
The roots of the quadratic equation
2𝑥 2 + 4𝑥 + 5 = 0 are 𝛼 and 𝛽. Without
solving the equation
(i)
write down the values of 𝛼 + 𝛽 and
𝛼𝛽
[2]
(ii)
calculate
(a) 𝛼 2 + 𝛽 2
[2]
(b) 𝛼 3 + 𝛽 3
[4]
(iii)
find a quadratic equation whose roots
are 𝛼 3 and 𝛽 3 .
[4]
38
CHAPTER 7: POLYNOMIALS
4.
CAPE 2008
The roots of the quadratic equation
2𝑥 2 + 4𝑥 + 5 = 0 are 𝛼 and 𝛽. Without
solving the equation, find a quadratic
2
2
equation with roots and .
[6]
𝛼
5.
CAPE 2009
The quadratic equation 𝑥 2 − 𝑝𝑥 + 24 = 0,
𝑝 ∈ ℝ, has roots 𝛼 and 𝛽.
(i) Express in terms of 𝑝
(a) 𝛼 + 𝛽
[1]
(b) 𝛼 2 + 𝛽 2
[4]
(ii) Given that 𝛼 2 + 𝛽 2 = 33, find the possible
values of 𝑝.
[3]
CAPE 2010
SOLUTIONS
3
1. (a) (i) 4
2.
3.
4.
5.
𝛽
1
(ii) 4
1
(b) 16
(c) 𝑥 2 − 2𝑥 + 64 = 0
𝑥 2 − 5𝑥 + 3 = 0
5
(a) (i) −2, 2 (ii) (a) −1
(b) 7
5𝑥 2 + 8𝑥 + 8 = 0
(i) (a) 𝑝
(b) 24
(ii) 𝑝 = ±9
…………………………………………………………………………
CUBIC POLYNOMIALS
INTRODUCTION
For a cubic polynomial we have
𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 0
𝑏
𝑐
𝑑
→ 𝑥3 + 𝑥2 + 𝑥 + = 0
𝑎
𝑎
𝑎
Now, if 𝛼, 𝛽 and 𝛾 are the roots of the polynomial
then
(𝑥 − 𝛼 )(𝑥 − 𝛽 )(𝑥 − 𝛾 ) = 0
→ 𝑥 3 − (𝛼 + 𝛽 + 𝛾)𝑥 2 + (𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)𝑥 − 𝛼𝛽𝛾 = 0
Equating coefficients of 𝑥 2 :
𝑏
−(𝛼 + 𝛽 + 𝛾) =
𝑎
𝑏
→ 𝛼+𝛽+𝛾 =−
𝑎
Equating coefficients of 𝑥:
𝑐
𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 =
𝑎
Equating constants:
𝑑
−𝛼𝛽𝛾 =
𝑎
𝑑
→ 𝛼𝛽𝛾 = −
𝑎
Therefore, for a cubic polynomial
𝑏
𝑑
(sum of roots) = − and (product of roots) = −
𝑎
𝑎
LESSON 13
For the following equations
determine the values of 𝛼 + 𝛽 + 𝛾, 𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾
and 𝛼𝛽𝛾.
(i)
𝑥 3 + 2𝑥 2 − 5𝑥 − 6 = 0
(ii)
4𝑥 3 + 𝑥 2 + 11𝑥 + 6 = 0
SOLUTION
(i)
𝑎 = 1, 𝑏 = 2, 𝑐 = −5, 𝑑 = −6
𝑏
2
𝛼 + 𝛽 + 𝛾 = − = − = −2
𝑎
1
𝑐 −5
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = =
= −5
𝑎
1
𝑑
−6
𝛼𝛽𝛾 = − = −
=6
𝑎
1
(ii)
𝑎 = 4, 𝑏 = 1, 𝑐 = 11, 𝑑 = 6
𝑏
1
𝛼+𝛽 +𝛾 =− =−
𝑎
4
𝑐 11
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = =
𝑎
4
𝑑
6
3
𝛼𝛽𝛾 = − = − = −
𝑎
4
2
LESSON 14
Find the cubic equation with
roots, 𝛼, 𝛽 and 𝛾 given that 𝛼 + 𝛽 + 𝛾 = −6,
𝛼 2 + 𝛽 2 + 𝛾 2 = 38 and 𝛼𝛽𝛾 = 30.
SOLUTION
NB: 𝛼 2 + 𝛽 2 + 𝛾 2 = (𝛼 + 𝛽 + 𝛾)2 − 2
(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 )
Required equation is
𝑥 3 − (𝛼 + 𝛽 + 𝛾)𝑥 2 + (𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)𝑥 − 𝛼𝛽𝛾 = 0
Using equation
38 = (−6)2 − 2(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)
38 − 36
= 𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾
−2
−1 = 𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾
𝑥 3 + 6𝑥 2 − 𝑥 − 30 = 0
LESSON 15
The cubic equation
𝑥 3 + 4𝑥 2 − 1 = 0 has roots 𝛼, 𝛽 and 𝛾. Determine
the cubic equations with roots
(a) 2𝛼, 2𝛽, 2𝛾
1 1 1
(b) 𝛼 , 𝛽 , 𝛾
SOLUTION
(a) We know that for the given equation
39
CHAPTER 7: POLYNOMIALS
𝑥 = 𝛼, 𝛽, 𝛾 and for the required equation we
will let 𝑋 = 2𝛼, 2𝛽, 2𝛾. Therefore we see that
𝑋
𝑋 = 2𝑥 → 𝑥 = 2
Substituting we get,
𝑋 3
𝑋 2
( ) + 4( ) − 1 = 0
2
2
𝑋3
𝑋2
+ 4( )− 1 = 0
8
4
𝑋3 + 8𝑋2 − 8 = 0
(b) Let 𝑋 =
1
𝑥
→
𝑥=
1
1 2
+
4
(
) −1 =0
𝑋3
𝑋
1
4
+
−1 =0
𝑋3 𝑋2
1 + 4𝑋 − 𝑋3 = 0
1
= −19
(𝛼 2 𝛽 2 𝛾 2) = (𝛼𝛽𝛾)2 = (−2)2 = 4
Required equation is
𝑥 3 − 27𝑥 2 − 19𝑥 − 4 = 0
LESSON 17
Two of the roots of the cubic
equation 2𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 + 3 are −1 and 3. Find
(i)
(ii)
the values of 𝑎 and 𝑏
the third root
SOLUTION
𝑋
(i)
𝑋3 − 4𝑋 − 1 = 0
LESSON 16
The cubic equation
𝑥 3 + 5𝑥 2 − 𝑥 + 2 = 0 has roots 𝛼, 𝛽 and 𝛾.
Determine the cubic equations with roots
(a) 𝛼𝛽, 𝛼𝛾, 𝛽𝛾
(b) 𝛼 2 , 𝛽 2, 𝛾2
(ii)
SOLUTION
𝑏
= −5
𝑎
𝑐
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = = −1
𝑎
𝑑
𝛼𝛽𝛾 = − = −2
𝑎
(a) Equation with roots 𝛼, 𝛽 and 𝛾 is
𝑥 3 − (𝛼 + 𝛽 + 𝛾)𝑥2 + (𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)𝑥 − 𝛼𝛽𝛾
=0
Therefore with the roots 𝛼𝛽, 𝛼𝛾 and 𝛽𝛾 the
equation is
𝑥 3 − (𝛼𝛽 + αγ + 𝛽𝛾)𝑥 2
+ [(𝛼𝛽)(𝛼𝛾) + (𝛽𝛾)(𝛼𝛽)
+ (𝛼𝛾)(𝛽𝛾)]𝑥 − (𝛼𝛽)(𝛼𝛾)(𝛽𝛾)
=0
𝑥 3 − (−1)𝑥 2 + [𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾)]𝑥 − (𝛼𝛽𝛾)2 = 0
𝑥 3 + 𝑥 2 + [−2(−5)]𝑥 − (−2)2 = 0
𝑥 3 + 𝑥 2 + 10𝑥 − 4 = 0
(b) Required equation is
𝛼+𝛽 +𝛾 =−
𝑥 3 − (𝛼 2 + 𝛽 2 + 𝛾 2 )𝑥 2 + [(𝛼 2 𝛽2) + (𝛼 2 𝛾 2) + (𝛽 2 𝛾 2)]𝑥
− (𝛼 2 𝛽2 𝛾 2) = 0
Now,
𝛼 2 + 𝛽 2 + 𝛾 2 = (𝛼 + 𝛽 + 𝛾)2 − 2(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)
= (−5)2 − 2(−1)
= 27
(𝛼 2 𝛽 2 ) + (𝛼 2 𝛾 2 ) + (𝛽 2 𝛾 2 )
= (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)2 − 2𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾)
= ( −1)2 − 2(−2)(−5)
By the Factor Theorem
𝑃(−1) = 0
2(−1) 3 + 𝑎(−1) 2 + 𝑏(−1) + 3 = 0
𝑎 + 𝑏 = −5
(*)
𝑃(3) = 0
2(3) 3 + 𝑎(3) 2 + 𝑏(3) + 3 = 0
3𝑎 − 𝑏 = 17
(**)
Solving (*) and (**) simultaneously
𝑎 + 𝑏 = −5
3𝑎 − 𝑏 = 17
𝑎 = 3 and 𝑏 = −8
𝑃(𝑥) = 2𝑥 3 + 3𝑥 2 − 8𝑥 + 3
Let the roots be 𝛼, 𝛽 and 𝛾 where 𝛾 is the
third root
𝛼𝛽𝛾 = −3
(−1)(3)𝛾 = −3
𝛾=1
…………………………………………………………………………
EXERCISE 7.4
1. Write down the sums and products of the
roots of the following equations.
(a) 2𝑥 3 + 3𝑥 2 − 8𝑥 − 12 = 0
(b) 2𝑥 3 + 5𝑥 2 − 3𝑥 = 0
(c) 3𝑥 3 + 4𝑥 2 − 5𝑥 − 2 = 0
(d) 𝑥 3 − 11𝑥 − 6 = 0
2. Given that
𝛼 + 𝛽 + 𝛾 = 0,
𝛼 2 + 𝛽 2 + 𝛾 2 = 14,
𝛼𝛽𝛾 = −18
find a cubic equation whose roots are 𝛼, 𝛽, 𝛾.
3. A cubic equation has roots 𝛼, 𝛽, and 𝛾 such
that
𝛼 + 𝛽 + 𝛾 = 4,
𝛼 2 + 𝛽 2 + 𝛾 2 = 14,
𝛼 3 + 𝛽 3 + 𝛾 3 = 34
Find the value of 𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾.
Show that the cubic equation is
𝑥 3 − 4𝑥 2 + 𝑥 + 6 = 0 and solve this equation.
4.
Given that 𝛼, 𝛽 and 𝛾 are the root of the
equation 𝑥 3 − 2𝑥 − 5 = 0, find the value of
𝛼 3 + 𝛽 3 + 𝛾 3.
40
CHAPTER 7: POLYNOMIALS
5.
6.
7.
8.
9.
Given that 𝛼, 𝛽 and 𝛾 are the roots of the
equation 𝑥 3 − 3𝑥 2 − 4𝑥 − 1 = 0, find the
exact values of 𝛼 2 + 𝛽 2 + 𝛾 2 and
𝛼 3 + 𝛽 3 + 𝛾 3.
The roots of the cubic equation
𝑥 3 − 7𝑥 2 + 2𝑥 − 3 = 0 are 𝛼, 𝛽 and 𝛾. Find the
values of
(i)
𝛼2 + 𝛽2 + 𝛾 2
(ii)
𝛼3 + 𝛽3 + 𝛾 3
The equation 8𝑥 3 + 12𝑥 2 + 4𝑥 − 1 = 0 has
roots 𝛼, 𝛽, 𝛾. Show that the equation with root
2𝛼 + 1, 2𝛽 + 1 and 2𝛾 + 1 is 𝑦 3 − 𝑦 − 1 = 0.
The cubic equation 𝑥 3 − 2𝑥 2 − 3𝑥 + 4 = 0 has
roots 𝛼, 𝛽 and 𝛾, find a cubic equation whose
roots are 𝛼 + 𝛽, 𝛽 + 𝛾, 𝛾 + 𝛼.
If the roots of the equation 𝑥 3 + 5𝑥 2 + ℎ𝑥 +
𝑘 = 0 are 𝛼, 2𝛼 and 𝛼 + 3, find the values of
𝛼, ℎ and 𝑘.
NB:
𝛼 2 + 𝛽 2 + 𝛾 2 = (𝛼 + 𝛽 + 𝛾)2 − 2(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)
𝛼 3 + 𝛽 3 + 𝛾 3 = (𝛼 + 𝛽 + 𝛾)3
− 3(𝛼 + 𝛽 + 𝛾)(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)
+ 3𝛼𝛽𝛾
(𝛼 + 𝛽)(𝛼 + 𝛾)(𝛽 + 𝛾) = (𝛼 + 𝛽 + 𝛾)(𝛼𝛽 + 𝛽𝛾 +
𝛼𝛾) − 𝛼𝛽𝛾.
SOLUTIONS
3
1. (a) − 2 , 6
2.
3.
4.
5.
6.
7.
8.
9.
5
(b) − 2 , 0
4 2
(c) − 3 , 3
(d) 0, 6
3
𝑥 − 7𝑥 + 18 = 0
𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 =1, 𝑥 = −1, 2, 3
66
17, 66
(i) 45 (ii) 310
𝑥 3 − 4𝑥 2 + 𝑥 + 2 = 0
𝛼 = −2, ℎ = 2, 𝑘 = −8
…………………………………………………………………………
EXERCISE 7.4.1
1. The cubic equation 𝑧 3 + 4𝑧 2 − 3𝑧 + 1 = 0 has
roots 𝛼, 𝛽 and 𝛾.
(i)
Write down the values of 𝛼 + 𝛽 + 𝛾,
𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 and 𝛼𝛽𝛾.
(ii)
Show that 𝛼 2 + 𝛽 2 + 𝛾 2 = 22.
(i) −4, −3, −1
2. The equation 𝑥 3 + 𝑝𝑥 2 + 𝑞𝑥 + 3 = 0 has roots
𝛼, 𝛽 and 𝛾, where
𝛼+𝛽+𝛾 = 4
𝛼2 + 𝛽2 + 𝛾 2 = 6
Find 𝑝 and 𝑞.
3.
4.
5.
𝑝 = −4, 𝑞 = 5
The equation 𝑥 3 + 𝑝𝑥 2 + 𝑞𝑥 + 𝑟 = 0 has roots
𝛼, 𝛽 and 𝛾, where
𝛼+𝛽+𝛾 = 3
𝛼𝛽𝛾 = −7
𝛼 2 + 𝛽 2 + 𝛾 2 = 13
(i)
Write down the values of 𝑝 and 𝑟.
(ii)
Find the value of 𝑞.
(i) 𝑝 = −3, 𝑟 = 7 (ii) 𝑞 = −2
The cubic equation 𝑥 3 − 5𝑥 2 + 𝑝𝑥 + 𝑞 = 0 has
roots 𝛼, −3𝛼 and 𝛼 + 3. Find the values of 𝛼, 𝑝
and 𝑞.
𝑝 = −8, 𝑞 = 12
The cubic equation 3𝑥 3 + 8𝑥 2 + 𝑝𝑥 + 𝑞 = 0
𝛼
has roots 𝛼, 6 and 𝛼 − 7. Find the values of
𝛼, 𝑝 and 𝑞.
6.
𝛼 = 2, 𝑝 = −33, 𝑞 = 10
The roots of the cubic equation
2𝑥 3 + 𝑥 2 + 𝑝𝑥 + 𝑞 = 0 are 2𝑤, −6𝑤 and 3𝑤.
Find the values of the roots and the values of
𝑝 and 𝑞.
3
Roots: 1, −3, 2 𝑝 = −12, 𝑞 = 9
The cubic equation 𝑥 3 + 3𝑥 2 − 7𝑥 + 1 = 0 has
roots 𝛼, 𝛽 and 𝛾.
(i)
Write down the values of 𝛼 + 𝛽 + 𝛾,
𝛼𝛽 + 𝛽𝛾 + 𝛾𝛼 and 𝛼𝛽𝛾.
(ii)
Find the cubic equation with roots
2𝛼, 2𝛽 and 2𝛾, simplifying your
answers as far as possible.
(i) −3, −7, −1 (ii) 𝑥 3 + 6𝑥 2 − 28𝑥 + 8 = 0
8. The roots of the cubic equation
2𝑥 3 + 𝑥 2 − 3𝑥 + 1 = 0 are 𝛼, 𝛽 and 𝛾.
Find the cubic equation whose roots are
2𝛼, 2𝛽 and 2𝛾, expressing your answer in a
form with integer coefficients.
𝑥 3 + 𝑥 2 − 6𝑥 + 4 = 0
9. The roots of the cubic equation
𝑥 3 + 3𝑥 2 − 7𝑥 + 1 = 0 are 𝛼, 𝛽 and 𝛾.
Find the cubic equation whose roots are
3𝛼, 3𝛽 and 3𝛾, expressing your answer in a
form with integer coefficients.
𝑥 3 + 9𝑥 2 − 63𝑥 + 27 = 0
10. The roots of the cubic equation 𝑥 3 − 2𝑥 2 −
8𝑥 + 11 = 0 are 𝛼, 𝛽 and 𝛾. Find the cubic
equation with roots 𝛼 + 1, 𝛽 + 1 and 𝛾 + 1.
𝑥 3 − 5𝑥 2 − 𝑥 + 16 = 0
11. The roots of the cubic equation
2𝑥 3 − 3𝑥 2 + 𝑥 − 4 = 0 are 𝛼, 𝛽 and 𝛾.
7.
41
CHAPTER 7: POLYNOMIALS
Find the cubic equation whose roots are
2𝛼 + 1, 2𝛽 + 1 and 2𝛾 + 1, expressing your
answer in a form with integer coefficients.
𝑥 3 − 6𝑥 2 + 11𝑥 − 22 = 0
12. The roots of the cubic equation
𝑥 3 − 4𝑥 2 + 8𝑥 − 3 = 0 are 𝛼, 𝛽 and 𝛾.
Find the cubic equation whose roots are
2𝛼 − 1, 2𝛽 − 1 and 2𝛾 − 1, expressing your
answer in a form with integer coefficients.
𝑥 3 − 5𝑥 2 + 19𝑥 + 1 = 0
13. The cubic equation 𝑥 3 − 5𝑥 2 + 3𝑥 − 6 = 0 has
roots 𝛼, 𝛽 and 𝛾. Find a cubic equation with
𝛼
𝛽
𝛾
roots 3 + 1, 3 + 1 and 3 + 1, simplifying your
answer as far as possible.
9𝑥 3 − 42𝑥 2 + 60𝑥 − 29 = 0
…………………………………………………………………………
FACTORING POLYNOMIALS
(i) 𝑢3 − 𝑣 3 = (𝑢 − 𝑣)(𝑢2 + 𝑢𝑣 + 𝑣 2 )
Difference of Cubes
(ii) 𝑢3 + 𝑣 3 = (𝑢 + 𝑣 )(𝑢2 − 𝑢𝑣 + 𝑣 2 )
Sum of Cubes
(iii) 𝑢4 − 𝑣 4 = (𝑢 − 𝑣 )(𝑢3 + 𝑢2 𝑣 + 𝑢𝑣 2 + 𝑣 3 )
= (𝑢 − 𝑣)(𝑢 + 𝑣)(𝑢2 + 𝑣 2 )
In general,
𝑢𝑛 − 𝑣 𝑛
= (𝑢 − 𝑣)(𝑢𝑛−1 + 𝑢𝑛−2 𝑣 1 + ⋯ + 𝑢1 𝑣 𝑛−2 + 𝑣 𝑛−1 )
LESSON 1
following.
Factorise, completely, each of the
(a) 8𝑚3 − 1
(b) 𝑥 3 + 27𝑧 3
(c) 𝑥 4 − 16
SOLUTION
(a) 8𝑚3 − 1
= (2𝑚) 3 − 13
= (2𝑚 − 1)[(2𝑚) 2 + (1)(2𝑚) + 12 ]
= (2𝑚 − 1)(4𝑚2 + 2𝑚 + 1)
…………………………………………………………………………
EXERCISE 7.5
1. Factorise each of the following completely.
(a) 9𝑥 2 − 4
(b) 4𝑎2 − 9𝑏2
(c) 𝑎2 𝑏2 − 𝑐 2
(d) 𝑥 3 𝑦 − 9𝑥𝑦 3
(e) 4𝑢3 𝑣 − 𝑢𝑣 3
(f) 𝑟 3 − 𝑡 3
(g) 𝑚3 + 𝑛3
(h) 𝑦 4 − 14
(i) 2𝑥 5 − 162𝑥
(j) 𝑎5 − 32
(k) 32 − 𝑏5
(l) 𝑎6 − 𝑏6
SOLUTIONS
1.
(a) (3𝑥 + 2)(3𝑥 − 2)
(b) (2𝑎 + 3𝑏)(2𝑎 − 3𝑏)
(c) (𝑎𝑏 − 𝑐)(𝑎𝑏 + 𝑐)
(d) 𝑥𝑦(𝑥 + 3𝑦)(𝑥 − 3𝑦)
(e) 𝑢𝑣(2𝑢 + 𝑣)(2𝑢 − 𝑣)
(f) (𝑟 − 𝑡)(𝑟 2 + 𝑟 2 𝑡 + 𝑡 2 )
(g) (𝑚 + 𝑛)(𝑚2 − 𝑚𝑛 + 𝑛2 )
(h) (𝑦 − 1)(𝑦 3 + 𝑦 2 + 𝑦 + 1)
(i) 2𝑥(𝑥 − 3)(𝑥 3 + 3𝑥 2 + 9𝑥 + 27)
(j) (𝑎 − 2)(𝑎4 + 2𝑎3 + 4𝑎2 + 8𝑎 + 16)
(k) (2 − 𝑏)(16 + 8𝑏 + 4𝑏2 + 2𝑏3 + 𝑏4 )
(l) (𝑎 − 𝑏)(𝑎5 + 𝑎4 𝑏 + 𝑎3 𝑏2 + 𝑎2 𝑏3 + 𝑎𝑏4 + 𝑏5 )
…………………………………………………………………………
EXAM QUESTIONS
12. Let 𝑥 and 𝑦 be positive real numbers such that
𝑥 ≠ 𝑦.
𝑥4 −𝑦4
(i)
Simplify 𝑥−𝑦 .
(ii)
Hence, or otherwise, show that
[6]
(𝑦 + 1)4 − 𝑦 4 = (𝑦 + 1)3 + (𝑦 + 1)2 𝑦 + (𝑦 + 1)𝑦 2 + 𝑦 3
(iii)
[4]
Deduce that
(𝑦 + 1)4 − 𝑦 4 < 4(𝑦 + 1)3 .
[2]
CAPE 2009
(i) 𝑥 3 + 𝑥 2 𝑦 + 𝑥𝑦 2 + 𝑥 3
(b) 𝑥 3 + 27𝑧 3
= 𝑥 3 + (3𝑧) 3
= (𝑥 + 3𝑧)[𝑥 2 − (𝑥)(3𝑧) + (3𝑧) 2 ]
= (𝑥 + 3𝑧)(𝑥 2 − 3𝑥𝑧 + 9𝑧 2 )
(c) 𝑥 4 − 16
= 𝑥 4 − 24
= (𝑥 − 2)(𝑥 3 + 𝑥 2 (2) + 𝑥(2) 2 + 23 )
= (𝑥 − 2)(𝑥 3 + 2𝑥 2 + 4𝑥 + 8)
42
CHAPTER 8: LOGARITHMS
CHAPTER 8: LOGARITHMS
At the end of this section, students should be able
to;


use the fact that log 𝑎 𝑏 = 𝑐 ↔ 𝑎𝑐 = 𝑏
where 𝑎 is any positive whole number;
simplify expressions by using the laws:
(a) log 𝑎 (𝑃𝑄) = log 𝑎 𝑃 + log 𝑎 𝑄 ;
𝑃
(b) log 𝑎 (𝑄) = log 𝑎 𝑃 − log 𝑎 𝑄;
(c) log 𝑎 𝑃 𝑏 = 𝑏 log 𝑎 𝑃;
(d) log 𝑎 𝑎 = 1
(e) log 𝑎 1 = 0
1
(f) log 𝑎 𝑥 = log 𝑎
𝑥
LESSON 1
Write the following in
logarithmic form
1.
3.
52 = 25
1
2−2 =
2.
1
643 = 4
4
SOLUTION
1. 52 = 25
3. 2−2 = 4
→ log 5 25 = 2
1
→ log 64 4 =
3
→ log 1 2 = −2
LESSON 2
Find 𝑥 or 𝑦 as indicated below
2.
1
643 = 4
1
4
log𝑏 𝑥
(g) log 𝑎 𝑥 = log 𝑎









𝑏
solve logarithmic equations;
use logarithms to solve equations of the
form 𝑎𝑥 = 𝑏;
apply logarithms to problems involving
the transformation of a given relationship
to linear form.
solve logarithmic equations
use logarithms to solve equations of the
form 𝑎𝑥 = 𝑏;
define an exponential function 𝑦 = 𝑎 𝑥 for
𝑎∈ℝ
sketch the graph of 𝑦 = 𝑎𝑥 ;
define the exponential function 𝑦 = 𝑒 𝑥
and its inverse 𝑦 = ln 𝑥, where
ln 𝑥 = log 𝑒 𝑥;
solve problems involving changing of the
base of a logarithm
Logarithm is another word for index of power.
Any statement in index (exponent) form has an
equivalent Logarithmic Form
Now, 23 = 8
(index form)
i.e.
3 is the power to which the base 2 must
be raised to obtain 8 or log base 2 of 8 is 3
log 2 8 = 3
(logarithmic form)
Similarly, 32 = 9
(index form)
i.e. log base 3 of 9 is 2 or log 3 9 = 2
(logarithmic form)
1 −2
Also ( ) = 25
(index form)
5
1
i. e. log base of 25 is − 2 or log 1 25 = −2
5
5
(logarithmic form)
1. log 2 𝑥 = 3
3. log 9 27 = 𝑦
5. log 7 0 = 𝑦
2. log 3 𝑥 = 2
1
4. log 5 25 = 𝑦
6. log 5(−5) = 𝑦
SOLUTION
1.
2.
3.
4.
5.
6.
log 2 𝑥 = 3
𝑥 = 23
𝑥=8
log 3 𝑥 = 2
𝑥 = 32
𝑥=9
log 9 27 = 𝑦
9𝑦 = 27
(32 ) 𝑦 = 33
32𝑦 = 33
2𝑦 = 3
3
𝑦=
2
1
log 5
=𝑦
25
1
5𝑦 =
25
1
𝑦
5 = 2 = 5−2
5
𝑦 = −2
log 7 0 = 𝑦
7𝑦 = 0
𝑦 is undefined
log 5 (−5) = 𝑦
5𝑦 = −5
𝑦 is undefined
NB: We can only find logarithms of positive
numbers, since 𝑎𝑛 > 0
43
CHAPTER 8: LOGARITHMS
LESSON 3
Find the logarithms to base 4 of
1
a) 16
b) 2
c)
4
d) 4
e) 1
f) 8
SOLUTION
a) 16 = 42 ↔ log 4 16 = 2
1
1
b) 2 = 42 ↔ log 4 2 =
2
1
1
−1
c) = 4 ↔ log 4 = −1
4
4
d) 4 = 41 ↔ log 4 4 = 1
e) 1 = 40 ↔ log 4 1 = 0
3
3
f) 8 = 42 ↔ log 4 8 =
2
= lg 9 − lg 𝑥 3 + lg 10
9
= lg 3 + lg 10
𝑥
90
= lg 3
𝑥
LESSON 5
Given that 𝑢 = log 9 𝑥, find in
terms of 𝑢
(i) 𝑥
(ii) log 9 (27𝑥)
(iii) log 3 𝑥
(iv) log 𝑥 81
SOLUTION
(i) log 9 𝑥 = 𝑢 → 𝑥 = 9𝑢
(ii) log 9(27𝑥)
= log 9 27 + log 9 𝑥
3
PROPERTIES OF LOGARITHMS
(a) log 𝑎 (𝑃𝑄) = log 𝑎 𝑃 + log 𝑎 𝑄 ;
𝑃
(b) log 𝑎 ( ) = log 𝑎 𝑃 − log 𝑎 𝑄;
𝑄
(c) log 𝑎 𝑃 𝑏 = 𝑏 log 𝑎 𝑃;
(d) log 𝑎 𝑎 = 1
(e) log 𝑎 1 = 0
1
(f) log 𝑎 𝑥 =
log𝑥 𝑎
log𝑏 𝑥
(g) log 𝑎 𝑥 = log 𝑎
𝑏
(h) lg 𝑥 = log 10 𝑥
LESSON 4
Write each of the following as a
single logarithm.
1. log 2 + log 7
2. log 3 − log 8
3. 2 log 𝑥 − 3 log 𝑦
log 6
4.
3
5. 3 log 2 + log 4 − log 16
6. 2 lg 3 − 3 lg 𝑥 + 1
SOLUTION
1. log 2 + log 7 = log(2 × 7) = log 14
3
2. log 3 − log 8 = log 8
3. 2 log 𝑥 − 𝑦 log 2
= log 𝑥 2 − log 2𝑦
= log 𝑥 2 − log 2𝑦
𝑥2
= log 𝑦
2
1
log 6
1
4.
=
log 6 = log 63
3
3
5.
6.
3 log 2 + log 4 − log 16
= log 23 + log 4 − log 16
= log 8 + log 4 − log 16
8×4
= log (
) = log 2
16
2 lg 3 − 3 lg 𝑥 + 1
= lg 32 − lg 𝑥 3 + lg 10
= log 9 92 + 𝑢
3
= +𝑢
2
(iii) log 3 𝑥
log 9 𝑥
=
log 9 3
𝑢
=
1
log 9 92
𝑢
=
= 2𝑢
1⁄
2
(iv) log 𝑥 81
log 9 81
=
log 9 𝑥
log 9 92 2
=
=
𝑢
𝑢
LOGARITHMIC EQUATIONS
Solving equations involving logarithms generally
require the application of the properties of
logarithms. In most cases the use of the property –
log 𝑎 𝑥 = log 𝑎 𝑦 → 𝑥 = 𝑦 is needed to obtain an
equation (linear, quadratic or otherwise) which
can then be easily solved.
LESSON 6
Solve the equation
3 lg(𝑥 − 1) = lg 27
SOLUTION
3 lg(𝑥 − 1) = lg 27
lg(𝑥 − 1)3 = lg 33
∴ (𝑥 − 1) 3 = 33
𝑥 −1 =3
𝑥=4
LESSON 7
Find the value(s) of 𝑥 ∈ ℝ which
satisfy 2 log 3 𝑥 = log 3 (𝑥 + 12)
SOLUTION
2 log 3 𝑥 = log 3(𝑥 + 12)
log 3 𝑥 2 = log 3(𝑥 + 12)
44
CHAPTER 8: LOGARITHMS
𝑥 2 = 𝑥 + 12
𝑥 2 − 𝑥 − 12 = 0
(𝑥 − 4)(𝑥 + 3) = 0
𝑥 = −3, 4
𝑥 = −3 is INVALID since substituting this value
into the original equation will result in us having
to find logarithms of a negative number.
LESSON 8a
Solve the equations
log 3 (4𝑥) + log 3 (𝑥 − 1) = 1
SOLUTION
log 3 (4𝑥) + log 3 (𝑥 − 1) = 1
log 3 [(4𝑥)(𝑥 − 1)] = log 3 3
4𝑥 2 − 4𝑥 = 3
4𝑥 2 − 4𝑥 − 3 = 0
(2𝑥 − 3)(2𝑥 + 1) = 0
3 1
𝑥 = ,−
2 2
1
𝑥 = − 2 is INVALID since substituting this value
into the original equation will result in us having
to find logarithms of a negative number.
LESSON 8b
Solve the equation
lg(10𝑥 ) − lg(𝑥 − 9) = 2
SOLUTION
lg(10𝑥) − lg(𝑥 − 9) = 2
lg(10𝑥) − lg(𝑥 − 9) = lg(102 )
10𝑥
lg (
) = lg 100
𝑥−9
10𝑥
∴
= 100
𝑥 −9
10𝑥 = 100𝑥 − 900
900 = 90𝑥
10 = 𝑥
LESSON 9
Solve the equation
log 2 𝑥 + 4 log 𝑥 2 = 5
SOLUTION
log 2 𝑥 + 4 log 𝑥 2 = 5
4
log 2 𝑥 +
=5
log 2 𝑥
Letting 𝑦 = log 2 𝑥 we have
4
𝑦+ =5
(× 𝑦)
𝑦
2
𝑦 + 4 = 5𝑦
𝑦 2 − 5𝑦 + 4 = 0
(𝑦 − 1)(𝑦 − 4) = 0
𝑦 = 1, 4
log 2 𝑥 = 1 → 𝑥 = 21 = 2
log 2 𝑥 = 4 → 𝑥 = 24 = 16
LESSON 10
Solve the equation
log 9 𝑥 = 1 + log 3 3𝑥, 𝑥 > 0.
SOLUTION
log 3 𝑥
log 9 𝑥 =
log 3 9
log 3 𝑥
=
2
log 9 𝑥 = 1 + log 3 3𝑥
1
log 𝑥 = log 3 3 + log 3 3𝑥
2 3
log 3 𝑥 = 2 log 3 3 + 2 log 3 3𝑥
log 3 𝑥 = log 3 9 + log 3 9𝑥 2
log 3 𝑥 = log 3 81𝑥 2
𝑥 = 81𝑥 2
81𝑥 2 − 𝑥 = 0
𝑥(81𝑥 − 1) = 0
1
𝑥 = since 𝑥 > 0
81
LESSON 11
Given that
𝑎3 + 𝑏3 + 3𝑎2 𝑏 = 24𝑎𝑏2 show that
𝑎+ 𝑏
3 log [
] = log 𝑎 + 2 log 𝑏
3
SOLUTION
𝑎+ 𝑏
3 log [
] = log 𝑎 + 2 log 𝑏
3
3
𝑎+𝑏
log [
] = log 𝑎 + log 𝑏2
3
(𝑎 + 𝑏) 3
log
= log 𝑎𝑏2
33
(𝑎 + 𝑏)3
= 𝑎𝑏2
27
𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏2 + 𝑏3 = 27𝑎𝑏2
𝑎3 + 𝑏3 + 3𝑎2 𝑏 = 24𝑎𝑏2
…………………………………………………………………………..
EXERCISE 8.1
1.
Write the following in logarithmic form.
(a) 34 = 81
1
(b) 3433 = 7
1
(c) 5−4 = 625
2
(d) 4 = 83
3
(e) 64 = 162
1
1
2
1
(f) 81−4 = 3
(g) 32−5 = 4
2.
(h) 5 = 3√125
Find 𝑥 or 𝑦 as indicated below
(a) log 5 125 = 𝑦
(b) log 3 𝑥 = 5
(c) log 2 (−8) = 𝑦
(d) log 4 1 = 𝑦
(e) log 3 𝑥 = −3
45
CHAPTER 8: LOGARITHMS
3.
4.
5.
Find the logarithm to base 2 of
(a) 1
(b) 2
(c) 8
1
(d) 2
Find the logarithms to base 10 of
(a) 1
(b) 10
(c) 100
(d) 0.001
Evaluate each of the following
(a) log 4 16
(b) log 2 16
(c) log 6 216
1
(d) log 5
125
(e) log 1 81
3
(f) log 3
8.
Express lg 𝑎 + 3 lg 𝑏 − 3 as a single logarithm.
Express each of the following as a single
logarithm:
(i) log 𝑎 2 + log 𝑎 3
(ii) 2 log 10 𝑥 − 3 log 10 𝑦
Given that log 𝑝 𝑋 = 9 and log 𝑝 𝑌 = 6, find
(i)
log 𝑝 √𝑋
(ii)
(iii)
(iv)
9.
1.
1
log 𝑝 (𝑋),
log 𝑝 (𝑋𝑌) ,
log 𝑌 𝑋
Given that log 8 𝑝 = 𝑥 and log 8 𝑞 = 𝑦, express
in terms of 𝑥 and/or 𝑦
(i)
log 8 √𝑝 + log 8 𝑞 2,
𝑞
(ii)
log 8 ( ),
8
10. Given that 𝑢 = log 4 𝑥, find, in its simplest form
in terms of 𝑢,
(i)
𝑥,
16
(ii)
log 4 ( 𝑥 )
11. Solve the equation 4 lg(𝑥 + 2) = lg 81.
12. Solve the equation 2 log 2(𝑥 − 1) = 4 log 2 3.
13. Solve the following equations.
(a) log 4 𝑥 + log 4 (2𝑥 − 4) = log 4(𝑥 + 3)
(b) log 5 (4𝑥) + log 5 (𝑥 + 2) = 1
(c) lg(8𝑥) − lg(𝑥 − 2) = lg 9
14. Solve for 𝑥
a) log 2 𝑥 − log 𝑥 8 − 2 = 0
b) log 2 𝑥 = log 𝑥 16
c) log 3 𝑥 = 4 log 𝑥 3
d) 3 log 8 𝑥 = 2 log 𝑥 8 + 5
log 𝑥
15. Find the value of log3 𝑥 .
9
1
(a) log 3 81 = 4
(b) log 343 7 = 3
1
2
(c) log 5 (625) = −4 (d) log 8 4 = 3
3
(e) log 16 64 = 2
1
2.
3.
4.
5.
27
2 8
6.
7.
SOLUTIONS
2
(g) log 32 (4) = − 5
(a) 𝑦 = 3
(c) INVALID
1
(e) 𝑥 = 27
(a) 0
(d) −1
(a) 0
(d) −3
(a) 2
(d) −3
𝑎𝑏3
6.
lg 1000
7.
(i) log 𝑎 6
8.
(i)
1
1
(h) log 125 5 = 3
(b) 𝑥 = 243
(d) 𝑦 = 0
(b) 1
(c) 3
(b) 1
(c)2
(b) 4
(e) −4
(c) 3
(f) 3
𝑥2
(ii) log 10 𝑦3
9
(ii) −9
2
(iv)
9.
1
(f) log 81 3 = − 4
(iii) 15
3
2
1
(i) 2 𝑥 + 2𝑦
(ii) 𝑦 − 1
10. (i) 4𝑢
11. 𝑥 = 1
12. 𝑥 = 10
(ii) 2 − 𝑢
13. (a) 𝑥 = 3
(b) 𝑥 = 2
1
1
(c) 𝑥 = 18
1
14. (a) 𝑥 = 2 , 8
(b) 𝑥 = 4 , 4
(c) 𝑥 = 9 , 9
15. 2
(d) 𝑥 = 2 , 64
1
1
…………………………………………………………………………..
THE EXPONENTIAL FUNCTION
INTRODUCTION
The function 𝑓(𝑥) = 𝑎𝑥 where 𝑎 > 0, 𝑎 ≠ 1 is
called an exponential
function. 2𝑥 , 3𝑥 , 7.56𝑥
are LESSONs of
exponential
functions. The
function 𝑓(𝑥) = 𝑒 𝑥 is
known as The
Exponential Function
where 𝑒 =
2.71828 …. The
graph of 𝑦 = 𝑒 𝑥 is
shown below.
From the graph we see, as for all exponential
functions, that 𝑦 = 𝑒 𝑥 does not cross the 𝑥-axis.
46
CHAPTER 8: LOGARITHMS
This indicates that exponential functions can
never be negative. It should also be noted that ALL
exponential functions cross the 𝑦-axis at 1, i.e.
they all pass through the point (0, 1).
(iii)
𝑦 > 0 is the range of 𝑓
when 𝑥 = 0, 𝑦 = 32(0) + 4 = 7
THE NATURAL LOGARITHM
The Natural Logarithm is a logarithm taken of
base 𝑒, log e 𝑥. Generally, logarithms of base 𝑒 are
referred to simply as ln 𝑥. The diagram below
shows the graphs of 𝑦 = 𝑒 𝑥 , 𝑦 = 𝑥 and 𝑦 = ln 𝑥.
What is the relationship between 𝑦 = 𝑒 𝑥 and
𝑦 = ln 𝑥?
𝑦 = 𝑒 𝑥 is the inverse of 𝑦 = ln 𝑥, therefore
𝑒 ln 𝑥 = 𝑥 and ln 𝑒 𝑥 = 𝑥.
LESSON 13a
Solve the equation
𝑒 2𝑥 − 7𝑒 𝑥 − 8 = 0.
SOLUTION
𝑒 2𝑥 − 7𝑒 𝑥 − 8 = 0
(𝑒 𝑥 ) 2 − 7𝑒 𝑥 − 8 = 0
Let 𝑦 = 𝑒 𝑥
𝑦 2 − 7𝑦 − 8 = 0
(𝑦 + 1)(𝑦 − 8) = 0
𝑦 = −1,
8
𝑒 𝑥 = −1
INVALID
𝑒𝑥 = 8
ln 𝑒 𝑥 = ln 8
𝑥 = ln 8
LESSON 13b
LOGARITHMIC AND EXPONENTIAL
EQUATIONS
LESSON 12
(i)
(ii)
(iii)
Given that 𝑓(𝑥) = 3𝑒 2𝑥 + 4, 𝑥 ∈ ℝ.
Determine 𝑓 −1 (𝑥).
State the range of 𝑓.
Sketch the graph of 𝑓.
SOLUTION
(i)
(ii)
Let 𝑦 = 𝑓(𝑥)
𝑦 = 3𝑒 2𝑥 + 4
𝑦−4
= 𝑒 2𝑥
3
𝑦−4
ln (
) = ln 𝑒 2𝑥
3
𝑦−4
ln (
) = 2𝑥
3
1
𝑦−4
ln (
)=𝑥
2
3
1
𝑥−4
𝑓 −1 (𝑥 ) = ln (
)
2
3
Since log 𝑎 is defined ONLY when 𝑎 > 0
𝑦−4
>0
3
Solve the equation
𝑒 3𝑥 + 9𝑒 −3𝑥 = 6
SOLUTION
𝑒 3𝑥 + 9𝑒 −3𝑥 = 6
9
𝑒 3𝑥 + 3𝑥 − 6 = 0
𝑒
Let 𝑦 = 𝑒 3𝑥
9
𝑦+ −6=0
𝑦
𝑦 2 + 9 − 6𝑦 = 0
𝑦 2 − 6𝑦 + 9 = 0
(𝑦 − 3)2 = 0
𝑦=3
𝑒 3𝑥 = 3
ln 𝑒 3𝑥 = ln 3
3𝑥 = ln 3
ln 3
𝑥=
3
LESSON 14
Solve for 𝑥 and 𝑦 the pair of
simultaneous equations
𝑒𝑥 𝑒𝑦 = 𝑒5
ln(2𝑥 + 𝑦) = ln 3 + ln 4
SOLUTION
𝑒𝑥 𝑒𝑦 = 𝑒5
→𝑥 +𝑦 = 5
ln(2𝑥 + 𝑦) = ln 3 + ln 4 → 2𝑥 + 𝑦 = 12
47
CHAPTER 8: LOGARITHMS
𝑥 +𝑦 = 5
(3)
2𝑥 + 𝑦 = 12
(4)
Subtracting (4) from (3)
−𝑥 = −7
𝑥=7
∴ 𝑦 = −2
LESSON 15
Solve the equation
15
2 ln 𝑥 + ln 𝑥 = 11, giving your answers as exact
values of 𝑥.
SOLUTION
15
2 ln 𝑥 +
= 11
ln 𝑥
Let 𝑦 = ln 𝑥
15
2𝑦 +
= 11
𝑦
2
2𝑦 + 15 = 11𝑦
2𝑦 2 − 11𝑦 + 15 = 0
(2𝑦 − 5)(𝑦 − 3) = 0
5
𝑦 = ,3
2
5
ln 𝑥 =
2
𝑁=
(ii)
5
𝑒 ln 𝑥 = 𝑒 2
5
𝑥 = 𝑒2
ln 𝑥 = 3
𝑒 ln 𝑥 = 𝑒 3
𝑥 = 𝑒3
(iii)
800
1
1 + 𝑘 ( 𝑟𝑡 )
𝑒
800
𝑁=
1
1 + 𝑘 ( ∞)
𝑒
𝑁 = 800
when 𝑡 = 0, 𝑁 = 50
800
50 =
1 + 𝑘𝑒 −𝑟(0)
800
50 =
1+𝑘
1 + 𝑘 = 16
𝑘 = 15
when 𝑡 = 1, 𝑁 = 200
800
200 =
1 + 15𝑒 −𝑟(1)
1 + 15𝑒 −𝑟 = 4
15𝑒 −𝑟 = 3
1
𝑒 −𝑟 =
5
1
−𝑟
ln 𝑒 = ln ( )
5
1
−𝑟 = ln ( )
5
1 −1
𝑟 = ln ( )
5
𝑟 = ln 5
800
𝑁=
1
1+15𝑒
LESSON 16
The enrolment pattern of
membership of a country club follows an
exponential logistic function 𝑁,
𝑁=
800
, 𝑘 ∈ ℝ, 𝑟 ∈ ℝ
1 + 𝑘𝑒 −𝑟𝑡
where 𝑁 is the number of members enrolled 𝑡
years after the formation of the club. The initial
membership was 50 persons and after one year,
there are 200 persons enrolled in the club.
(ln5 )𝑡
when 𝑡 = 3
800
𝑁=
1
1 + 15𝑒 3 ln5
800
𝑁=
1
1 + 15𝑒 ln125
800
𝑁=
1
1 + 15 (125)
𝑁 ≈ 715 members
(i) What is the LARGEST number reached by the
membership of the club?
(ii) Calculate the exact value of 𝑘 and 𝑟.
(iii) How many members will there be in the club
3 years after its formation?
CAPE 2006
SOLUTION
(i)
The largest membership will occur when
time goes on indefinitely. As 𝑡 → ∞
48
CHAPTER 8: LOGARITHMS
UNKNOWN INDICES
LESSON 17a
Solve, correct to 2 decimal places,
the equation 3𝑥 = 7
SOLUTION
3𝑥 = 7
ln 3𝑥 = ln 7
𝑥 ln 3 = ln 7
ln 7
𝑥=
ln 3
𝑥 = 1.77
LESSON 17b
Determine 𝑥 ∈ ℝ such that
2𝑥+1 = 15.
SOLUTION
2𝑥+1 = 15
ln 2𝑥+1 = ln 15
(𝑥 + 1) ln 2 = ln 15
ln 15
𝑥 +1 =
ln 2
ln 15
𝑥=
−1
ln 2
𝑥 = 2.91
LESSON 17c
Determine 𝑥 ∈ ℝ such that
4𝑦−2 = 32𝑦+1 .
SOLUTION
4𝑦−2 = 32𝑦+1
ln 4𝑦−2 = ln 32𝑦+1
(𝑦 − 2) ln 4 = (2𝑦 + 1) ln 3
𝑦 ln 4 − 2 ln 4 = 2𝑦 ln 3 + ln 3
𝑦 ln 4 − 2𝑦 ln 3 = 2 ln 4 + ln 3
𝑦(ln 4 − 2 ln 3) = 2 ln 4 + ln 3
2 ln 3 + ln 3
𝑦=
ln 4 − 2 ln 3
𝑦 = −4.77
…………………………………………………………………………
EXERCISE 8.2
1.
2.
3.
4.
Given that 𝑓(𝑥) = 𝑒 2𝑥 ,
(i)
Sketch the graph of 𝑓 and state the
coordinates of the point where the
graph crosses the 𝑦-axis.
(ii)
Determine 𝑓 −1.
A function 𝑓 is defined by 𝑓(𝑥) = 2𝑒 3𝑥 − 1 for
all real values of 𝑥.
(a) Find the range of 𝑓
(b) Find an expression for 𝑓 −1 (𝑥).
By using the substitution 𝑦 = 𝑒 𝑥 , find the
value of 𝑥 such that 𝑒 2𝑥 = 𝑒 𝑥 + 12.
(a) Given that 3𝑒 𝑥 = 4, find the exact value of
𝑥.
(b) (i) By substituting 𝑦 = 𝑒 𝑥 , show that the
equation 3𝑒 𝑥 + 20𝑒 −𝑥 = 19 can be
written as 3𝑦 2 − 19𝑦 + 20 = 0
7.
(ii) Hence solve the equation
3𝑒 𝑥 + 20𝑒 −𝑥 = 19, giving your
answers as exact values.
8
Solve the following equation 3 ln 𝑥 + ln 𝑥 = 14.
A curve has equation 𝑦 = ln(𝑥 2 + 5), show
that this equation can be written in the form
𝑥 2 = 𝑒 𝑦 − 5.
Express 𝑥 2 = 𝑒 𝑥−2 in the form ln 𝑥 = 𝑎𝑥 + 𝑏.
8.
Given that 𝑒 −2𝑥 = 3, find the exact value of 𝑥.
5.
6.
9.
The functions 𝑓 and 𝑔 are defined with their
respective domains by
𝑓(𝑥) = 𝑒 2𝑥 − 3, or all values of 𝑥
1
4
𝑔(𝑥) = 3𝑥+4 , for all real values of 𝑥, 𝑥 ≠ − 3
(a) Find the range of 𝑓.
(b) The inverse of 𝑓 is 𝑓 −1.
i.
Find 𝑓 −1
ii.
Solve the equation 𝑓 −1 (𝑥) = 0
(c) (i) Find an expression for 𝑔𝑓(𝑥).
(ii) Solve the equation 𝑔𝑓(𝑥) = 1,
giving your answer in an exact form.
10. The graph 𝑦 = 𝑒 2𝑥 − 9 cuts the 𝑥-axis at 𝐴 and
the 𝑦-axis at 𝐵.
(i) Determine the coordinates of 𝐴 and 𝐵.
(ii) Show that 𝑦 2 = 𝑒 4𝑥 − 18𝑒 2𝑥 + 81.
11. The curve 𝑦 = 𝑒 𝑥+1 and 𝑦 = 𝑒 4−2𝑥 meet at 𝑃.
Find the coordinates of 𝑃.
12. Given that 2𝑥 4𝑦 = 128 and that
ln(4𝑥 − 𝑦) = ln 2 + ln 5, calculate the value of
𝑥 and of 𝑦.
13. The temperature of water, 𝑥℃, in an insulated
tank at time, 𝑡 hours, may be modelled by the
equation
𝑥 = 65 + 8𝑒 −0.02𝑡
Determine the
(i) initial temperature of the water in the
tank
[2]
(ii) temperature at which the water in the
tank will eventually stabilize
[2]
(iii) time when the temperature of the water
in the tank is 70℃
[4]
14. Solve correct to 2 decimal places the following
equations
(a) 2𝑥 = 5
(b) 5𝑥 = 15
(c) 3𝑥+2 = 12
(d) 2𝑥−3 = 7
(e) 3𝑥+1 = 22𝑥−5
49
CHAPTER 8: LOGARITHMS
15. Solve the equation 24𝑥−1 = 35−2𝑥 , giving your
lg 𝑎
answer in the form .
lg 𝑏
16. Solve the equation 7𝑤−3 − 4 = 180, giving
your answer correct to 3 significant figures.
17. Solve the following equations, giving your
answer correct to 3 significant figures.
(i) 5𝑥−1 = 120
(ii) 7𝑥 = 2𝑥+1
(iii) 53𝑤−1 = 4250
SOLUTIONS
4.
5.
ln 𝑥
1.
(i) (0, 1)
2.
3.
(a) 𝑦 > −1 (b) ln ( )
3
2
𝑥 = ln 4
4.
(b) (i)
5.
6.
7.
𝑥 = 𝑒 3, 𝑒4
8.
− 2
9.
3.
(ii)
2
1
𝑥+1
4
(ii) 𝑥 = ln 5 , ln (3)
2
6.
1
ln 𝑥 = 𝑥 − 1
ln 3
2
7.
(a) 𝑦 > −3 (b) (i)
1
ln(𝑥+3)
2
(c) (i) 3𝑒2𝑥 −5
(ii) 𝑥 = 𝑒2
10. (i) 𝐴(ln 3 , 0) , 𝐵 (0, −8)
11. 𝑃(1, 𝑒 2 )
12. (3, 2)
13. (i) 73℃
(ii) 65℃
14. (a) 2.32
(b) 1.68
(d) 5.81
lg 486
15. lg 144
16. 5.68
17. (i) 3.97
(ii) 𝑥 = −2
2
(ii) −0.80
(iii) 23.5 hours
(c) 0.26
(iii)72.1
8.
(ii) By substituting 𝑦 = log 2 𝑥, or otherwise,
solve, for 𝑥, the equation
√log 2 𝑥 = log 2 √𝑥.
[6]
CAPE 2011
Solve the equation
log 2 (𝑥 + 3) = 3 − log 2 (𝑥 + 2)
[5]
CAPE 2013
Solve the equation
log 2(𝑥 + 1) − log 2(3𝑥 + 1) = 2
[4]
CAPE 2014
Let 𝑓(𝑥) = 3𝑥 + 2 and 𝑔(𝑥) = 𝑒 2𝑥 .
(i) Find
a) 𝑓 −1 (𝑥) and 𝑔 −1 (𝑥)
[4]
b) 𝑓[𝑔(𝑥)] (or 𝑓 ∘ 𝑔(𝑥))
[1]
(ii) Show that (𝑓 ∘ 𝑔) −1 (𝑥) = 𝑔−1 (𝑥) ∘
𝑓 −1 (𝑥).
[5]
CAPE 2013
Given that 𝑎3 + 𝑏3 + 3𝑎2 𝑏 = 5𝑎𝑏2 , show that
𝑎+𝑏
3 log ( 2 ) = log 𝑎 + 2 log 𝑏.
[5]
CAPE 2014
The population growth of bacteria present in
a river after time, 𝑡 hours, is given by
𝑁 = 300 + 5𝑡
Determine
(ii) the number of bacteria present at 𝑡 = 0.
[1]
(ii) the time required to triple the number of
bacteria.
[4]
CAPE 2015
Solve the equation
4
4
3 − 𝑥 − 𝑥 = 0.
9
81
SOLUTIONS
EXAM QUESTIONS
1.
(a) 𝑥 = 9, 27 (b) 𝑥 =
1.
2.
3.
(i) 𝑥 = 2
(i) (2, 3)
4.
𝑥=
5.
𝑥=−
6.
(i) (a) 3
7.
8.
9.
(i) 301 (ii) 4 hours
0.315
(i) Find 𝑥 such that
log 5 (𝑥 + 3) + log 5 (𝑥 − 1) = 1
[5]
(ii) Without the use of calculators, or tables,
evaluate
1
2
3
8
9
log10 ( ) + log10 ( ) + log10 ( ) + ⋯ + log10 ( ) + log10 ( )
2
3
4
9
10
2.
[3]
CAPE 2008
(i) Solve, for 𝑥 and 𝑦, the simultaneous
equations
log(𝑥 − 1) + 2 log 𝑦 = 2 log 3
log 𝑥 + log 𝑦 = log 6
[8]
1
256
,2
(ii) −1
(ii) 𝑥 = 1, 16
−5+√33
2
3
11
𝑥−2
(b) 3𝑒 2𝑥 + 2
…………………………………………………………………………
50
CHAPTER 9: THE MODULUS FUNCTION
CHAPTER 9: MODULUS/ ABSOLUTE VALUE FUNCTION
At the end of this section, students should be able
to:




define the modulus function;
use the properties:
(a) |𝑥| is the positive square root of 𝑥 2 ;
(b) |𝑥| < |𝑦| if, and only if 𝑥 2 < 𝑦 2
(c) |𝑥| < |𝑦| iff −𝑦 < 𝑥 < 𝑦
(d) |𝑥 + 𝑦| ≤ |𝑥| + |𝑦|
solve equations and inequalities involving
the modulus function, using algebraic or
graphical methods.
Illustrate by means of graphs, the
relationship between the function
𝑦 = 𝑓(𝑥) given in graphical form and
𝑦 = |𝑓(𝑥)|.
__________________________________________________________
Consider a number line, what is the distance from
the origin, 0, to 5 and −5? The answer in both
instances is 5. If 𝑎 is the ordinate of a point on a
real number line, then the distance from the origin
to 𝑎 is represented by |𝑎| and is referred to as the
modulus of 𝑎. Thus, |2| = 2 since the point with
ordinate 2 is two units from the origin,
9
9
9
|2 | = 2 since the point with ordinate 2 is 4.5 units
from the origin and |−3| = 3, since the point
with ordinate −3 is three units from the origin
(Fig 1). Thus, this provides us with a geometric
definition of modulus.
Symbolically, and more formally, we define
modulus as follows:
Modulus
−𝑥; 𝑥 < 0
|𝑥| = {
𝑥; 𝑥 ≥ 0
|4| = 4 |−7| = −(−7) = 7
[Note: −𝑥 is positive if 𝑥 is negative]
Both the geometric and non-geometric definitions
of modulus are useful.
NB: The modulus is never negative.
MODULUS EQUATIONS
LESSON 1
(i)
(ii)
Solve the equations
|𝑥| = 4
|3𝑥| = 9
(iii)
|2𝑥 − 5| = 11
SOLUTION
(i)
(ii)
It is clear that 4 and −4 both have a
distance of 4 units from the origin.
Therefore, 𝑥 = 4 or 𝑥 = −4
Note: |𝑥| is the positive square root of 𝑥 2 .
|3𝑥| = 9
3𝑥 = 9 or 3𝑥 = −9
𝑥 = 3 or 𝑥 = −3
Alternatively
|3𝑥|2 = 92
9𝑥 2 = 81
𝑥2 = 9
𝑥 = ±3
(iii)
2𝑥 − 5 = 11 or 2𝑥 − 5 = −11
2𝑥 = 16 or
2𝑥 = −6
𝑥 = 8 or
𝑥 = −3
Alternatively,
|2𝑥 − 5| = 11
|2𝑥 − 5| 2 = 112
4𝑥 2 − 20𝑥 + 25 = 121
4𝑥 2 − 20𝑥 − 96 = 0
𝑥 2 − 5𝑥 − 24 = 0
(𝑥 + 3)(𝑥 − 8) = 0
either 𝑥 + 3 = 0 or 𝑥 − 8 = 0
𝑥 = −3 or
𝑥=8
From the previous examples we see that squaring
both sides of the equation and solving the
resulting quadratic is an alternative method.
However, linear equations are generally easier to
solve than quadratics, thus we will be using this
method.
LESSON 2
Solve the equations
(i)
(ii)
|𝑥 − 2| = 3𝑥 + 1
|𝑥 − 2| = 𝑥 − 2
SOLUTION
(i)
𝑥 − 2 = 3𝑥 + 1 or 𝑥 − 2 = −(3𝑥 + 1)
= −3𝑥 − 1
−3 = 2𝑥
or 4𝑥 = 1
3
1
𝑥=−
or
𝑥=
2
4
Caution: We cannot have a negative absolute
value; therefore, we now need to check our
solutions to verify that the right hand side of the
equation is not negative.
51
CHAPTER 9: THE MODULUS FUNCTION
(ii)
3
7
1
7
3 (− ) + 1 = −
and 3 ( ) + 1 =
2
2
4
4
1
Therefore, 𝑥 = 4 is the only answer.
|𝑥 − 2| = 𝑥 − 2
𝑥 − 2 = 𝑥 − 2 or 𝑥 − 2 = −(𝑥 − 2)
It is quite clear that 𝑥 − 2 = 𝑥 − 2 for all
values of 𝑥. But since |𝑥 − 2| is positive,
𝑥 − 2 ≥ 0 → 𝑥 ≥ 2. This solution covers
both of the above equations.
LESSON 3
(i)
(ii)
(iii)
3.
Solve the equations
|2𝑥 − 1| = |4𝑥 + 9|
|𝑥 + 3| 2 − |𝑥 + 3| − 2 = 0
𝑥 2 − 4|𝑥| + 3 = 0
SOLUTION
(i)
2𝑥 − 1 = 4𝑥 + 9 or 2𝑥 − 1 = −(4𝑥 +
9) = −4𝑥 − 9
2𝑥 = −10
or
6𝑥 = −8
4
𝑥 = −5
or
𝑥=−
3
Both sides of the equations contain
absolute values ensuring that both sides
are positive. Consequently, there is no
need to check. But it is still a good
practice to do so.
|𝑥 + 3|2 − |𝑥 + 3| − 2 = 0
(ii)
Let 𝑦 = |𝑥 + 3|
𝑦2 − 𝑦 − 2 = 0
(𝑦 − 2)(𝑦 + 1) = 0
either 𝑦 − 2 = 0 or 𝑦 + 1 = 0
𝑦 = 2 or
𝑦 = −1
∴ |𝑥 + 3| = 2 or |𝑥 + 3| = −1 INVALID
𝑥 + 3 = 2 → 𝑥 = −1
𝑥 + 3 = −2 → 𝑥 = −5
(iii)
𝑥 2 − 4|𝑥| + 3 = 0
𝑥 2 − 4|𝑥| + 3 = 0
(|𝑥| − 3)(|𝑥| − 1) = 0
either (|𝑥| − 3) = 0 or (|𝑥| − 1) = 0
|𝑥| = 3
|𝑥| = 1
𝑥 = ±3
𝑥 = ±1
…………………………………………………………………………..
EXERCISE 9.1
1. Solve the following equations
a. |6𝑚| = 42
b. |−6𝑥| = 30
c. |𝑘 − 10| = 3
|𝑥|
d. 7 = 5
b. 2|3𝑥| = 4𝑥 + 10
c. |2𝑥 − 1| − 3 = 𝑥
d. |3𝑥 − 9| = 3𝑥 − 9
e. |2𝑥 − 3| = 3 − 2𝑥
f. |𝑥 + 4| = −(𝑥 + 4)
Solve the following equations
a. |3𝑥 + 4| = |2𝑥 − 3|
b. |𝑥 − 3| = |𝑥 + 2|
c. |3𝑥 + 1| = |3 − 2𝑥|
d. |2𝑥 − 3| 2 + 5|2𝑥 − 3| + 6 = 0
e. 3|5𝑥 − 1|2 − 16|5𝑥 − 1| + 5 = 0
f. 6|𝑥 + 7| 2 − 14|𝑥 + 7| = −4
g. 𝑥 2 − 5|𝑥| + 6 = 0
h. 𝑥 2 − 7|𝑥| − 44 = 0
i. 3𝑥 2 − 5|𝑥| − 2 = 0
SOLUTIONS
1. (a) 𝑚 = ±7
(c) 𝑘 = 7, 13
(e) 𝑥 = ±21
5
2. (a) 𝑥 = 3
(b) 𝑥 = ±5
(d) 𝑥 = ±35
(f) No Solution
(b) 𝑥 = −1, 5
2
(c) 𝑥 = − 3 , 4
(d) 𝑥 ≥ 3
(e) 𝑥 ≤ 2
(f) 𝑥 ≤ −4
3
3.
1
1
(a) 𝑥 = − 5 , −7
(b) 𝑥 = 2
2
(c) 𝑥 = −4, 5
4
(e) 𝑥 = − ,
2
(d) No Solution
,
4
,
6
5 15 15 5
(f) 𝑥 = −9, −
22
3
,−
20
3
, −5
(g) 𝑥 = ±2, ±3
(h) 𝑥 = ±11
(i) 𝑥 = ±2
…………………………………………………………………………..
MODULUS INEQUALITIES
Before we examine inequality questions dealing
with absolute value; let’s take a look at the
number line to help us understand the concept of
modulus inequalities.
The red line on the graph above represents the
interval (−3, 3). If 𝑥 is any number within this
interval then −3 < 𝑥 < 3. More precisely, |𝑥| < 3.
In general, if |𝑥| < 𝑝, 𝑝 ∈ ℝ then – 𝑝 < 𝑥 < 𝑝.
What if |𝑥| > 3?
𝑥
e. | | = 3
7
2.
f. −10|𝑣 + 2| = 70
Solve the following equations
a. |5 − 𝑥| = 2𝑥
From the graph our solution would be {𝑥 < −3} ∪
{𝑥 > 3}. Therefore, if |𝑥| > 𝑝, then {𝑥 < −𝑝} ∪
{𝑥 > 𝑝}
52
CHAPTER 9: THE MODULUS FUNCTION
LESSON 4
(i)
(ii)
(iii)
(iv)
(v)
Solve the inequalities
|𝑥| < 7
|2𝑥 − 5| < 9
|9𝑚 + 2| ≤ 1
|8𝑥 − 3| > 9
3 < |𝑥| < 7
SOLUTION
(i)
(ii)
(iii)
(iv)
(v)
|𝑥| < 7
−7 < 𝑥 < 7
|2𝑥 − 5| < 9
−9 < 2𝑥 − 5 < 9
−4 < 2𝑥 < 14
−2 < 𝑥 < 7
|9𝑚 + 2| ≤ 1
−1 ≤ 9𝑚 + 2 ≤ 1
−3 ≤ 9𝑚 ≤ −1
|8𝑥 − 3| > 9
8𝑥 − 3 > 9 8𝑥 − 3 < −9
8𝑥 > 12
8𝑥 < −6
3
3
{𝑥 > } ∪ {𝑥 < − }
2
4
3 < |𝑥| < 7
Before we attempt to solve as a double
inequality we will split up our inequality
into two parts so that we get some
valuable insight into how to solve this as a
double inequality.
|𝑥| < 7 → {−7 < 𝑥 < 7}
|𝑥 | > 3 → {𝑥 < −3} ∪ {𝑥 > 3}
These intervals are represented on the
number lines below. Combining our
intervals we get {−7 < 𝑥 < −3} ∪
{3 < 𝑥 < 7}.
If we try the double inequality approach
we should get 3 < 𝑥 < 7 and −3 < 𝑥 <
−7. But the signs of the second inequality
are reversed. How do we correct this? As
follows:
3 < 𝑥 < 7 and 3 < −𝑥 < 7
−3 > 𝑥 > −7
−7 < 𝑥 < −3
With this approach the intervals are more
easily identifiable
LESSON 5
which
(i)
(ii)
SOLUTION
Find the range of values of 𝑥 for
|𝑥 − 5| < |3𝑥 + 2|
𝑥 2 − 7|𝑥| + 10 < 0
(i)
At first glance we assume that we have
the following inequality to solve
– (3𝑥 + 2) < 𝑥 − 5 < 3𝑥 + 2 but what
about the inequality
3𝑥 + 2 < 𝑥 − 5 < −(3𝑥 + 2)? If we take
some time to think about it; since we do
not know the value of 𝑥 we do not know
whether 3𝑥 + 2 is positive or negative.
Hence, we have the before mentioned
possibilities. To counteract this potential
problem we proceed as follows:
|𝑥 − 5| < |3𝑥 + 2|
(𝑥 − 5) 2 < (3𝑥 + 2) 2 We square both
sides to avoid negativity
𝑥 2 − 10𝑥 + 25 < 9𝑥 2 + 12𝑥 + 4
2
−8𝑥 − 22𝑥 + 21 < 0
8𝑥 2 + 22𝑥 − 21 > 0
(4𝑥 − 3)(2𝑥 + 7) > 0
3
7
{𝑥 > } ∪ {𝑥 < − }
4
2
(ii)
𝑥 2 − 7|𝑥| + 10 < 0
(|𝑥| − 5)(|𝑥| − 2) < 0
Roots are |𝑥 | = 2, 5
From graph 𝑥 2 − 7|𝑥| + 10 < 0 when
2 < |𝑥| < 5
We thus have the following inequalities
(a) 2 < 𝑥 < 5 and
(b) 2 < −𝑥 < 5
−2 > 𝑥 > −5
−5 < 𝑥 < −2
Therefore the solution set is:
{−5 < 𝑥 < −2} ∪ {2 < 𝑥 < 5}
When is 𝑥 2 − 7|𝑥| + 10 > 0? Let’s see.
Roots are 𝑥 = 2, 5
𝑥 2 − 7|𝑥| + 10 > 0 when
|𝑥| > 5 and |𝑥| < 2
|𝑥| > 5 → {𝑥 < −5} ∪ {𝑥 > 5}
|𝑥 | < 2 → {−2 < 𝑥 < 2}
Therefore, we have
{𝑥 < −5} ∪ {𝑥 > 5} ∪ {−2 < 𝑥 < 2}
53
CHAPTER 9: THE MODULUS FUNCTION
…………………………………………………………………………..
EXERCISE 9.2
1. Solve the following inequalities.
a. |7𝑥| ≤ 42
b. |1 + 𝑥| < 8
c. 10|10𝑛 − 8| ≤ 80
2. Solve the following inequalities
a. |−5𝑝| > 20
b. |4 + 5𝑏| ≥ 24
𝑥
c. |2 | + 9 > 13
d. (|𝑥| − 4)(|𝑥| + 1) < 0
e. |𝑥|2 − 2|𝑥| − 3 < 0
f. 2|𝑥|2 + 2 > 5|𝑥|
g. |𝑥 − 1| < |2𝑥 + 1|
SOLUTIONS
1. (a) −6 ≤ 𝑥 ≤ 6
(b) −9 < 𝑥 < 7
8
(c) 0 ≤ 𝑛 ≤ 5
2. (a) {𝑝 < −4} ∪ {𝑝 > 4}
28
(b) {𝑏 ≤ − 5 } ∪ {𝑏 ≥ 4}
(c) {𝑥 < −8} ∪ {𝑥 > 8}
(d) −4 < 𝑥 < 4
(e) −3 < 𝑥 < 3
1
1
(f) {− < 𝑥 < } ∪ {𝑥 < −2} ∪ {𝑥 > 2}
2
2
LESSON 7
Draw the graph of
𝑓(𝑥) = |𝑥 2 + 𝑥 − 12|.
SOLUTION
𝑓(𝑥) = |𝑥 2 + 𝑥 − 12|
…………………………………………………………………………..
(g) {𝑥 < −2} ∪ {𝑥 > 0}
…………………………………………………………………………..
EXERCISE 9.3
GRAPHS OF MODULUS FUNCTIONS
1.
LESSON 6
Draw the graph of 𝑓(𝑥) = |𝑥 + 2|
for −7 ≤ 𝑥 ≤ 3.
SOLUTION
We have the following
𝑓(𝑥) = |𝑥 + 2|
𝑥
−7
−6
−5
−4
−3
−2
−1
0
1
2
3
|−7 + 2| = |−5| = 5
|−6 + 2| = |−4| = 4
|−5 + 2| = |−3| = 3
|−4 + 2| = |−2| = 2
|−3 + 2| = |−1| = 1
|−2 + 2| = |0| = 0
|−1 + 2| = |1| = 1
|0 + 2| = |2| = 2
|1 + 2| = |3| = 3
|2 + 2| = |4| = 4
|3 + 2| = |5| = 5
2.
Draw the graph of 𝑓(𝑥 ) = |3𝑥 − 1| for
−2 ≤ 𝑥 ≤ 3.
1
Draw the graph of 𝑓(𝑥) = |1 − 2 𝑥| for
−2 ≤ 𝑥 ≤ 4.
Draw the graph of 𝑓(𝑥) = |𝑥 2 − 2𝑥 − 3| for
−2 ≤ 𝑥 ≤ 4.
4. Draw the graph of 𝑓(𝑥) = |2𝑥 2 + 5𝑥 − 12| for
−5 ≤ 𝑥 ≤ 2.
…………………………………………………………………………..
3.
EXAM QUESTIONS
1. Find the real values of 𝑥 which satisfy the
equation |2𝑥 + 3| = 5.
[3]
CAPE 2000
2.
Solve, for 𝑥 ∈ ℝ, the equation
𝑥 2 − 6 |𝑥 | + 8 = 0
[4]
CAPE 2004
54
CHAPTER 9: THE MODULUS FUNCTION
3.
Let 𝐴 = {𝑥: 2 ≤ 𝑥 ≤ 7} and
𝐵 = {𝑥: |𝑥 − 4| ≤ ℎ}, ℎ ∈ ℝ.
Find the LARGEST value of ℎ for which 𝐵 ⊂ 𝐴.
[6]
CAPE 2006
4.
Solve |𝑥 − 4| − 6 > 0 for all 𝑥 ∈ ℝ.
5.
Solve, for real values of 𝑥, the inequality
𝑥 2 − |𝑥| − 12 < 0.
[5]
CAPE 2011
6.
Solve the following: |𝑥 + 2| = 3𝑥 + 5.
[4]
CAPE 2013
7.
On the same axes, sketch the graphs of
𝑓(𝑥) = 2𝑥 + 3 and 𝑔(𝑥) = |2𝑥 + 3|.
Show clearly ALL intercepts that may be
present.
[5]
CAPE 2007
[5]
CAPE 2013
SOLUTIONS
1.
2.
3.
4.
5.
6.
𝑥 = −4, 1
𝑥 = ±2, 𝑥 = ±4
ℎ=2
{𝑥 < −2} ∪ {𝑥 > 10}
−4 < 𝑥 < 4
7
3
𝑥 = − ,−
4
2
7.
…………………………………………………………………………
55
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
At the end of this section, students should be able
to:
 define a sequence of terms, 𝑎𝑛 , where 𝑛 is
a positive integer;
 write a specific term from the formula for
the 𝑛th term of a sequence;
 use the summation (∑) notation;
 establish simple proofs by using the
principle of mathematical induction.
_________________________________________________________
SEQUENCES
INTRODUCTION
A sequence is a list of numbers which obey a
particular pattern. Each number in the sequence is
called a term of the sequence. These are usually
denoted 𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛−1 , 𝑢𝑛 where 𝑢1 is the first
term, 𝑢2 is the second term and 𝑢𝑛 is the 𝑛th term.
In some cases the sequence can be defined by a
formula – an expression for the 𝑛th term.
LESSON 1
Write down the first 5 terms of
the following sequences:
(a) 𝑢𝑛 = 4𝑛 − 1
(b) 𝑢𝑛 =
(c) 𝑢𝑛 =
𝑛+1
𝑛
1
2𝑛
(d) 𝑢𝑛 = (−1)𝑛+1 (
𝑛
)
𝑛+1
SOLUTION
(a) 𝑢𝑛 = 4𝑛 − 1
𝑢1 = 4(1) − 1 = 3
𝑢2 = 4(2) − 1 = 7
𝑢3 = 4(3) − 1 = 11
𝑢4 = 4(4) − 1 = 15
𝑢5 = 4(5) − 1 = 19
3, 7, 11, 15, 19,
𝑛+1
(b) 𝑢𝑛 = 𝑛
1+1
𝑢1 =
=2
1
2+1 3
𝑢2 =
=
2
2
3+1 4
𝑢3 =
=
3
3
4+1 5
𝑢4 =
=
4
4
5+1 6
𝑢5 =
=
5
5
3 4 5 6
,
,
,
, …
2 3 4 5
1
(c) 𝑢𝑛 = 2𝑛
2,
1
1
=
21 2
1
1
𝑢2 = 2 =
2
4
1
1
𝑢3 = 3 =
2
8
1
1
𝑢4 = 4 =
2
16
1
1
𝑢5 = 5 =
2
32
1 1 1 1
1
,
, ,
,
, …
2 4 8 16 32
𝑛
(d) 𝑢𝑛 = (−1)𝑛+1 (𝑛+1 )
𝑢1 =
1
1
)=
1+1
2
2
2
3
𝑢2 = (−1) (
)=−
2+1
3
3
3
4
𝑢3 = (−1) (
)=
3+1
4
4
4
5
(
)
𝑢4 = −1 (
)=−
4+1
5
5
5
6
𝑢5 = (−1) (
)=
5+1
6
1
2 3
4 5
, − ,
, − ,
, …
2
3 4
5 6
…………………………………………………………………………..
EXERCISE 10.1
1. Write down the first 5 terms of the following
sequences:
(a) 𝑢𝑛 = 𝑛 + 3
(b) 𝑢𝑛 = 5 − 𝑛
(c) 𝑢𝑛 = 4𝑛 − 1
(d) 𝑢𝑛 = 2 − 5𝑛
(e) 𝑢𝑛 = 𝑛2 − 3𝑛
𝑢1 = (−1) 2 (
(f) 𝑢𝑛 =
….
𝑛+1
𝑛
𝑛+1
(g) 𝑢𝑛 = 𝑛2
1
(h) 𝑢𝑛 = 2𝑛
𝑛
(i) 𝑢𝑛 = 3𝑛
(j) 𝑢𝑛 = (−1)𝑛
(k) 𝑢𝑛 = (−1)𝑛+1
1
(l) 𝑢𝑛 = (−1)𝑛 𝑛
𝑛
(m) 𝑢𝑛 = (−1)𝑛+1 (𝑛+1)
SOLUTIONS
(a) 4, 7, 10, 13, 16
(b) 4, 3, 2, 1, 0
56
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
(c) 3, 7, 11, 15, 19
(d) −3, −8, −13, −18, −23
(e) −2, −2, 0, 4, 10
(f) Each numerator is 1 and the first number of
the denominator is 𝑛 and the second is 𝑛 + 1.
1
Therefore 𝑢𝑛 = 𝑛(𝑛+1)
3 4 5 6
(f) 2, 2 , 3 , 4 , 5
3 4
5
6
1 1 1
1
1
1 2
4
(g) The numerators are the natural numbers but
they begin with 2, i.e. 𝑛 + 1 and the
denominators are the square numbers.
𝑛+1
Therefore 𝑢𝑛 = 2
(g) 2, 4 , 9 , 16 , 25
(h) 2 , 4 , 8 , 16 , 32
3
5
𝑛
(i) 3 , 9 , 27 , 81 , 243
(j) −1, 1, −1, 1, −1
(k) 1, −1, 1, −1, 1
1
1 1
1
(l) −1, 2 , − 3 , 4 , − 5
1 1
1 1
(m) 1, − 2 , 3 , − 4 , 5
LESSON 2
For each of the following
sequences determine an expression for the 𝑛𝑡ℎ
term, 𝑢𝑛 .
(c)
3
,
(d) 1,
(e) 1,
2
4
3
,
1
,
1
− ,
2
1
2
,
4
1
(f) 1×2 ,
3
4
,
3
1
1
(g) 2,
5
4
,
,
2×3
4
9
,
6
5
,
7
1
8
1
1
4
5
3×4
5
16
,
,
4×5
4
3
,
5
…
,
6
1
− 2,
1
2
(f) 1×2 ,
,
4
,
1
(e) 1,
…
(g) 2,
, …
16
1
2
3
1
,
1
1
1
,
−2,
, …
, − ,
1
(c) 3 ,
(d) 1,
(a) 5, 8, 11, 14, ….
(b) 8, 6, 4, 2, 0,
1
…………………………………………………………………………..
EXERCISE 10.2
1. (i) For each of the following sequences
determine an expression for the 𝑛𝑡ℎ term,
𝑢𝑛 .
(ii) Determine the 50th term of sequence.
(a) 5, 8, 11, 14, ….
(b) 8, 6, 4, 2, 0, −2, …
4
1
,
,
2×3
3
4
4
,
9
,
5
,
7
, …
1
1
, − 4,
1
8
1
,
3×4
5
16
5
1
,
,
,
…
, …
16
1
4×5
,
1
5×6
,
….
…
SOLUTIONS
1
5×6
,
….
(a) 𝑢𝑛 = 3𝑛 + 2, 𝑢50 = 77
(b) 𝑢𝑛 = 10 − 2𝑛, 𝑢50 = −90
𝑛
50
(c) 𝑢𝑛 = 𝑛+3 , 𝑢50 = 53
…
(−1)𝑛+1
1
SOLUTION
(d) 𝑢𝑛 =
(a) Consecutive terms differ by 3 therefore we try
3𝑛. To create the right formula we add 2 i.e.
𝑢𝑛 = 3𝑛 + 2
(b) Consecutive terms differ by −2 therefore we
try −2𝑛. To create the correct expression we
need to add 10 i.e. 𝑢𝑛 = 10 − 2𝑛
(c) The numerators are the natural numbers 𝑛
and the denominators are two more than the
𝑛
numerator i.e. 𝑢𝑛 = 𝑛+2
(e) 𝑢𝑛 = 2𝑛−1 , 𝑢50 = 249
(d) Ignoring the signs, each numerator is 1 and
the denominators are the natural numbers 𝑛.
Since the signs alternate between positive and
negative, starting with positive, we use
(−1) 𝑛+1 . Therefore 𝑢𝑛 = (−1) 𝑛+1 (
𝑛
)
𝑛+1
(e) Each numerator is 1 and the denominators
1
are powers of 2 i.e. 𝑢𝑛 = 2𝑛−1
1
(f) 𝑢𝑛 =
𝑛
1
𝑛(𝑛+1)
𝑛+1
, 𝑢50 = − 50
1
, 𝑢50 =
1
50(51)
51
(g) 𝑢𝑛 = 𝑛2 , 𝑢50 = 2500
…………………………………………………………………………..
SERIES
INTRODUCTION
If 𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 , … is a sequence, then the
expression 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 + ⋯ is called a
series. If a sequence is finite, the corresponding
series is a finite series. If the sequence is infinite,
the corresponding series is an infinite series. For
example,
1, 2, 4, 8, 16
Finite sequence
1 + 2 + 4 + 8 + 16
Finite series
Series are often represented in a compact form
called summation notation using the symbol Σ,
which is a stylized version of the Greek letter
sigma, meaning ‘the sum of.’ Consider the
following examples:
57
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
∞
4
∑ 3𝑟 + 2
∑ 𝑎𝑘 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4
𝑟=1
𝑘=1
7
∑ 𝑏𝑘 = 𝑏3 + 𝑏4 + 𝑏5 + 𝑏6 + 𝑏7
(ii) Consecutive terms differ by −2 therefore we
try −2𝑛. To create the correct expression we
need to add 10 i.e. 10 − 2𝑛
∞
𝑘=3
𝑛
∑ 10 − 2𝑛
∑ 𝑐𝑘 = 𝑐0 + 𝑐1 + 𝑐2 + ⋯ + 𝑐𝑛
𝑘=0
The terms on the right are obtained from the
expression on the left by successively replacing
the summing index, 𝑘 with integers, starting with
the first number indicated below Σ and ending
with the number that appears above Σ. Thus, for
example, if we are given the sequence
1 1 1
1
, , ,… , 𝑛
2 4 8
2
the corresponding series is
1 1 1
1
+ + + ⋯+ 𝑛
2 4 8
2
or, more compactly,
𝑛
1
∑ 𝑘
2
𝑘=1
The letter 𝑘 as used here is called the index of
summation. It is a ‘dummy variable,’ so called
because it does not appear in the final result, and
this result would be unchanged if some other
letter was used. In fact any other letter can be
used. For LESSON,
𝑛
∑ 𝑎𝑟 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛
𝑟=1
(iii) The numerators are the natural numbers 𝑛
and the denominators are two more than the
𝑛
numerator i.e. 𝑛+2
∞
∑
𝑟=1
(iv) Ignoring the signs, each numerator is 1 and
the denominators are the natural numbers 𝑛.
Since the signs alternate between positive and
negative, starting with positive, we use
(−1) 𝑛+1 . Therefore
∞
∑(−1) 𝑛+1 (
𝑟=1
1
are powers of 2 i.e. 2𝑛−1
∞
1
2
3
4
5
𝑟=1
1
1
1
1
1
4
5
1
∞
𝑟=1
1
𝑛(𝑛 + 1)
(vii)
The numerators are the natural numbers
but they begin with 2, i.e. 𝑛 + 1 and the
denominators are the square numbers.
Therefore
∞
∑
1
(d) 1 + (− 2) + 3 + (− 4) + 5 + ⋯
1
2𝑛−1
(vi) Each numerator is 1 and the first number of
the denominator is 𝑛 and the second is 𝑛 + 1.
Therefore
(c) 3 + 4 + 5 + 6 + 7 + ⋯
1
1
∑
∑
(a) 5 + 8 + 11 + 14 + ⋯
(b) 8 + 6 + 4 + 2 + 0 + (−2) + ⋯
𝑛
)
𝑛+1
(v) Each numerator is 1 and the denominators
𝑟=1
LESSON 3
Write each of the following series
using sigma notation.
𝑛
𝑛+2
𝑟=1
1
𝑛+1
𝑛2
(e) 1 + 2 + 4 + 8 + 16 + ⋯
1
1
1
(f) 1×2 + 2×3 + 3×4 + 4×5 + 5×6 + ⋯
3
(g) 2 + 4 + 9 + 16 + ⋯
Some basic rules for manipulating expressions
involving Σ can be established as follows:
𝑛
∑ 𝑎 = 𝑎 + 𝑎 + ⋯ + 𝑎 = 𝑛𝑎
SOLUTION
𝑟=1
𝑛
(i) Consecutive terms differ by 3 therefore we try
3𝑛. To create the right formula we add 2 i.e.
3𝑛 + 2
∑(𝑘𝑢𝑟 ) = 𝑘𝑢1 + 𝑘𝑢2 + ⋯ + 𝑘𝑢𝑛
𝑟=1
= 𝑘(𝑢1 + 𝑢2 + ⋯ + 𝑢𝑛 )
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CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
𝑛
= 40
3
(b) ∑16
𝑟=10 𝑟
= 𝑘 ∑ 𝑢𝑟
16
𝑟=1
𝑛
∑(𝑢𝑟 + 𝑣𝑟 ) = (𝑢1 + 𝑣1 ) + (𝑢2 + 𝑣2 ) + ⋯ + (𝑢𝑛 + 𝑣𝑛 )
𝑟=1
= (𝑢1 + 𝑢2 + ⋯ + 𝑢𝑛 ) + (𝑣1 + 𝑣2 + ⋯ + 𝑣𝑛 )
𝑛
=
𝑛
𝑟=1
(16 + 1) 2 −
𝑛
𝑛
LESSON 6
Express in its simplest form
2𝑛
∑(𝑢𝑟 − 𝑣𝑟) = ∑ 𝑢𝑟 − ∑ 𝑣𝑟
𝑟=1
92
(9 + 1)
4
𝑟=1
This also holds for subtraction. That is,
𝑛
𝑟 =1
162
4
= 16 471
= ∑ 𝑢𝑟 + ∑ 𝑣𝑟
𝑟=1
9
= ∑ 𝑟3 − ∑ 𝑟3
𝑟=1
∑ 𝑟
𝑟=1
𝑟=𝑛+1
SOLUTION
LESSON 4
2𝑛
Evaluate
∑(2𝑟 + 3)
𝑟=𝑛+1
SOLUTION
4
4
∑(2𝑟 + 3) = 2 ∑ 𝑟 + ∑ 3
𝑟=1
𝑟=1
𝑟=1
= 2[1 + 2 + 3 + 4] + 4(3)
= 32
SPECIAL SUMMATION FORMULAE
The following standard results can be used to find
the sum of various series.
𝑛
∑𝑟 =
𝑟=1
𝑛
(𝑛 + 1),
2
𝑛
∑ 𝑟3 =
𝑟=1
𝑛
∑ 𝑟2 =
𝑟=1
𝑛
(𝑛 + 1)(2𝑛 + 1),
6
𝑛2
(𝑛 + 1) 2
4
𝑟=1
Find each of the following sums
(a) ∑4𝑟=1 𝑟(𝑟 + 1)
3
(b) ∑16
10 𝑟
SOLUTION
(a) ∑4𝑟=1 𝑟(𝑟 + 1)
4
= ∑(𝑟 2 + 𝑟)
𝑟 =1
4
4
= ∑ 𝑟2 + ∑ 𝑟
𝑟 =1
=
𝑟=1
4
4
(4 + 1)(2(4) + 1) + (4 + 1)
6
2
𝑟=1
2𝑛
𝑛
(2𝑛 + 1) − (𝑛 + 1)
2
2
𝑛
= [2(2𝑛 + 1) − (𝑛 + 1)]
2
𝑛
= [(4𝑛 + 2) − (𝑛 + 1)]
2
𝑛
= (3𝑛 + 1)
2
LESSON 7
Express each of the following in a
factorized form.
(a) ∑𝑛𝑟=1(𝑟 + 1)(𝑟 − 1)
(b) ∑𝑛𝑟=1 𝑟 2 (𝑟 + 2)
SOLUTION
(a) ∑𝑛𝑟=1(𝑟 + 1)(𝑟 − 1)
𝑛
= ∑(𝑟 2 − 1)
𝑟 =1
𝑛
𝑛
= ∑ 𝑟2 − ∑ 1
𝑟 =1
LESSON 5
𝑛
∑ 𝑟 = ∑ 𝑟 − ∑𝑟 =
𝑟 =1
4
2𝑛
4
𝑟=1
𝑛
= (𝑛 + 1)(2𝑛 + 1) − 𝑛
6
𝑛(𝑛 + 1)(2𝑛 + 1) − 6𝑛
=
6
𝑛[(𝑛 + 1)(2𝑛 + 1) − 6]
=
6
𝑛[2𝑛2 + 3𝑛 + 1 − 6]
=
6
𝑛[2𝑛2 + 3𝑛 − 5]
=
6
𝑛(2𝑛 + 5)(𝑛 − 1)
=
6
(b) ∑𝑛𝑟=1 𝑟 2 (𝑟 + 2)
𝑛
= ∑(𝑟 3 + 2𝑟 2 )
𝑟 =1
59
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
𝑛
𝑛
= ∑ 𝑟3 + 2 ∑ 𝑟2
𝑟 =1
𝑛2
𝑟=1
𝑛
(𝑛 + 1)2 + 2 [ (𝑛 + 1)(2𝑛 + 1)]
4
6
3𝑛2 (𝑛 + 1) 2 + 4𝑛(𝑛 + 1)(2𝑛 + 1)
=
12
𝑛(𝑛 + 1)[3𝑛(𝑛 + 1) + 4(2𝑛 + 1)]
=
12
𝑛(𝑛 + 1)[3𝑛2 + 3𝑛 + 8𝑛 + 4]
=
12
=
4.
5.
…………………………………………………………………………..
EXERCISE 10.3
1.
6.
3
State the first 5 terms, the 𝑛th term and the
(𝑛 + 1)st of
1
(b) ∑𝑛𝑟=1(𝑟 + 2)(𝑟 − 3) = 3 𝑛(𝑛2 − 19)
1
𝑛
(c) ∑𝑛𝑟=1 𝑟2 (𝑟 + 1) = 12 𝑛(𝑛 + 1)(𝑛 + 2)(3𝑛 + 1)
a. ∑(3𝑟 − 2)
1
(d) ∑𝑛𝑟=1 𝑟(𝑟2 − 3) = 4 𝑛(𝑛 + 1)(𝑛 + 3)(𝑛 − 2)
𝑟=1
𝑛
1
𝑟=1
𝑛
c. ∑(2𝑟 2 + 5)
𝑟=3
𝑛
d. ∑ 𝑟 3
𝑟=2
2.
(e) ∑𝑛𝑟=1 𝑟 2 (3 − 4𝑟) = 2 𝑛(𝑛 + 1)(1 − 2𝑛2 )
∑ 2𝑟 2
b.
Evaluate each of the following.
7
(a)
∑(7 − 𝑟)
𝑟=1
5
(b)
SOLUTIONS
1. (a)
1 + 4 + 7 + 10 + 13 + ⋯ + (3𝑛 − 2) +
(3𝑛 + 1)
(b) 2 + 8 + 18 + 32 + 50 + ⋯ + 2𝑛2 + 2(𝑛 + 1) 2
(c) 23 + 37 + 55 + 77 + 103 + ⋯ + (2𝑛2 + 5) +
(2(𝑛 + 1) 2 + 5)
(d) 8 + 27 + 64 + 125 + 216 + ⋯ + 𝑛3 + (𝑛 + 1) 3
73
2. (a) 21 (b) 45 (c) 405 (d) 12
3.
(e) ∑𝑛𝑟=1 2𝑟
4.
8
∑(2𝑟 2 + 5)
𝑟=4
5
(d)
∑
𝑟=2
3.
𝑟
𝑟 −1
Write the following series using sigma
notation [assume that there are 𝑛 terms in the
series]
(a) 8 + 13 + 18 + 23 + ⋯
(b) 2 + 6 + 18 + 54 + 162 + ⋯
(c) 64 + 32 + 16 + 8 + ⋯
(d)
4 + 9 + 16 + 25 + ⋯
1 1 1 1
(e)
+ + +
+⋯
2 4 8 16
1 1 1 1
(f) − + − +
−⋯
2 4 8 16
(a) ∑𝑛𝑟=1(5𝑟 + 3)
(c) ∑𝑛𝑟=1(27−𝑟 )
1
∑(20 − 𝑟 2 )
𝑟 =1
(c)
Evaluate
(a) ∑20
𝑟=1 𝑟
(b) ∑40
𝑟=1 3𝑟
(c) ∑24
𝑟=1(𝑟 + 3)
(d) (d) ∑6𝑟=1(𝑟 2 + 2𝑟 + 7)
(e) ∑100
𝑟=51 𝑟
Express each of the following in a factorised
form.
(c) ∑𝑛𝑟=1(𝑟 + 1)(𝑟 − 1)
(d) ∑𝑛𝑟=1 𝑟(𝑟 2 + 1)
(e) ∑𝑛𝑟=1 𝑟 2 (𝑟 − 1)
(f) ∑𝑛𝑟=1 𝑟 2 (𝑟 + 2)
Use standard series formulae to show that
1
(a) ∑𝑛𝑟=1[(𝑟 + 1)(𝑟 − 2)] = 𝑛(𝑛2 − 7)
(a) 210
(d) 5668
(b) ∑𝑛𝑟=1 2(3𝑟−1 )
(d) ∑𝑛𝑟=1(𝑟 + 1) 2
−1 𝑟
(f) ∑𝑛𝑟=1 ( 2 )
(b) 2460
(e) 3775
(c) 372
EXAM QUESTIONS
1.
Given that
𝑛
∑𝑟 =
𝑟=1
𝑛
(𝑛 + 1),
2
show that
𝑛
∑(3𝑟 + 1) =
𝑟=1
1
𝑛(3𝑛 + 5)
2
[4]
CAPE 2006
60
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
2.
Let
MATHEMATICAL INDUCTION
𝑛
INTRODUCTION
𝑆𝑛 = ∑ 𝑟 for 𝑛 ∈ ℕ
𝑟=1
LESSON 1 – The (𝑘 + 1)st Term
Find the value of 𝑛 for which 3𝑆2𝑛 = 11𝑆𝑛.
[4]
CAPE 2007
3.
(i) Show that
𝑛
∑ 𝑟(𝑟 + 1) =
𝑟=1
1
𝑛(𝑛 + 1)(𝑛 + 2), 𝑛 ∈ ℕ
3
[5]
(ii) Hence, or otherwise, evaluate
50
∑ 𝑟(𝑟 + 1)
𝑟=31
𝑛
Let 𝑃𝑛 : ∑(2𝑟 − 1) = 𝑛2
𝑟=1
𝑃1 : 2(1) − 1 = 12
1=1
Therefore, 𝑃1 is true
Assume 𝑃𝑛 is true for 𝑛 = 𝑘
𝑘
[3]
CAPE 2008
SOLUTIONS
1.
2.
3.
PROOF OF SUMMATION
LESSON 2
Prove by Mathematical Induction
that the sum of the first 𝑛 odd integers is 𝑛2 .
PROOF
𝑃𝑘 : ∑ 2𝑟 − 1 = 𝑘 2
𝑟=1
𝑘+1
𝑃𝑘+1 : ∑ 2𝑟 − 1 = (𝑘 + 1) 2
𝑟 =1
𝑛=5
34 280
…………………………………………………………………………..
Now,
𝑃𝑘+1 = 𝑃𝑘 + (𝑘 + 1)st term
𝑘+1
𝑘
∑ 2𝑟 − 1 = ∑ 2𝑟 − 1 + (𝑘 + 1)st term
𝑟=1
𝑟 =1
= 𝑘 2 + 2(𝑘 + 1) − 1
= 𝑘 2 + 2𝑘 + 2 − 1
= 𝑘 2 + 2𝑘 + 1
= (𝑘 + 1) 2
Therefore, 𝑃𝑘+1 is true whenever 𝑃𝑘 is true.
Hence, by Mathematical Induction
𝑛
∑ 2𝑟 − 1 = 𝑛2 for all 𝑛 ∈ ℕ
𝑟=1
LESSON 3
that
Prove by Mathematical Induction
𝑛
∑(6𝑟 + 5) = 3𝑛2 + 8𝑛
𝑟=1
for all positive integers 𝑛.
PROOF
𝑛
Let 𝑃𝑛 : ∑(6𝑟 + 5) = 3𝑛2 + 8𝑛
𝑟=1
6(1) + 5 = 3(1) 2 + 8(1)
6+5 =3+8
11 = 11
Thus, 𝑃1 is true.
Assume 𝑃𝑛 is true for 𝑛 = 𝑘
𝑃1 :
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CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
𝑘
𝑃𝑘 : ∑(6𝑟 + 5) = 3𝑘 2 + 8𝑘
𝑟=1
𝑘+1
𝑃𝑘+1 : ∑(6𝑟 + 5) = 3(𝑘 + 1)2 + 8(𝑘 + 1)
𝑟=1
= 3(𝑘 2 + 2𝑘 + 1) + 8𝑘 + 8
= 3𝑘 2 + 6𝑘 + 3 + 8𝑘 + 8
= 3𝑘 2 + 14𝑘 + 11
Now,
𝑃𝑘+1 = 𝑃𝑘 + (𝑘 + 1)st term
𝑘+1
1
𝑘(𝑘 + 1)(𝑘 + 2) + (𝑘 + 1)(𝑘 + 2)
3
1
= (𝑘 + 1)(𝑘 + 2)[𝑘 + 3]
3
Thus, 𝑃𝑘 +1 is true whenever 𝑃𝑘 is true.
=
Hence, by Mathematical Induction
𝑛
1
∑ 𝑟(𝑟 + 1) = 𝑛(𝑛 + 1)(𝑛 + 2) for all 𝑛 ∈ ℕ
3
𝑟=1
LESSON 5
𝑘
𝑃𝑘+1 : ∑(6𝑟 + 5) = ∑(6𝑟 + 5) + (𝑘 + 1)st term
𝑟=1
𝑟=1
𝑟=1
2
= 3𝑘 + 8𝑘 + 6(𝑘 + 1) + 5
= 3𝑘 2 + 8𝑘 + 6𝑘 + 6 + 5
= 3𝑘 2 + 14𝑘 + 11
Thus, 𝑃𝑘 +1 is true whenever 𝑃𝑘 is true.
Hence, by Mathematical Induction
𝑛
∑(6𝑟 + 5) = 3𝑛2 + 8𝑛 for all 𝑛 ∈ ℕ
𝑟=1
LESSON 4
Prove by Mathematical Induction
1
∑ 𝑟(𝑟 + 1) = 𝑛(𝑛 + 1)(𝑛 + 2)
3
𝑛
𝑟=1
for all positive integers 𝑛.
PROOF:
𝑛
1
Let 𝑃𝑛 : ∑ 𝑟(𝑟 + 1) = 𝑛(𝑛 + 1)(𝑛 + 2)
3
𝑟=1
1
𝑃1 : 1(1 + 1) = (1)(1 + 1)(1 + 2)
3
2=2
Therefore, 𝑃1 is true.
Assume 𝑃𝑛 is true for 𝑛 = 𝑘.
𝑘
𝑃𝑘 : ∑ 𝑟(𝑟 + 1) =
𝑟=1
𝑘+1
𝑃𝑘+1 : ∑ 𝑟(𝑟 + 1) =
𝑟=1
1
𝑘(𝑘 + 1)(𝑘 + 2)
3
1
(𝑘 + 1)(𝑘 + 1 + 1)(𝑘 + 1 + 2)
3
1
= (𝑘 + 1)(𝑘 + 2)(𝑘 + 3)
3
Now,
𝑃𝑘+1 = 𝑃𝑘 + (𝑘 + 1)st term
𝑘+1
𝑘
𝑃𝑘+1 : ∑ 𝑟(𝑟 + 1) = ∑ 𝑟(𝑟 + 1) + (𝑘 + 1)st term
𝑟=1
𝑟=1
Prove by Mathematical Induction
1
𝑛
∑
=
(𝑟 + 1)(𝑟 + 2) 2(𝑛 + 2)
𝑛
1
= 𝑘(𝑘 + 1)(𝑘 + 2) + (𝑘 + 1)(𝑘 + 1 + 1)
3
for all positive integers 𝑛.
PROOF
𝑛
1
𝑛
=
(𝑟 + 1)(𝑟 + 2) 2(𝑛 + 2)
𝑟=1
1
1
𝑃1 :
=
(1 + 1)(1 + 2) 2(1 + 2)
1
1
=
6
6
Therefore, 𝑃1 is true.
Assume 𝑃𝑛 is true for 𝑛 = 𝑘.
Let 𝑃𝑛 : ∑
𝑘
𝑃𝑘 : ∑
𝑟=1
1
=
(𝑟 + 1)(𝑟 + 2)
𝑘+1
𝑃𝑘+1 : ∑
𝑟=1
1
𝑘+1
=
(𝑟 + 1)(𝑟 + 2) 2(𝑘 + 3)
Now,
𝑃𝑘+1 = 𝑃𝑘 + (𝑘 + 1)st term
𝑘+1
1
𝑃𝑘+1 =
+
2(𝑘 + 3) (𝑘 + 2)(𝑘 + 3)
𝑘(𝑘 + 3)
1(2)
=
+
2(𝑘 + 2)(𝑘 + 3) 2(𝑘 + 2)(𝑘 + 3)
𝑘 2 + 3𝑘 + 2
2(𝑘 + 1)(𝑘 + 2)
(𝑘 + 1)(𝑘 + 2)
=
2(𝑘 + 2)(𝑘 + 3)
𝑘+1
=
2(𝑘 + 3)
Thus, 𝑃𝑘 +1 is true whenever 𝑃𝑘 is true.
=
Hence, by Mathematical Induction
𝑛
1
𝑛
∑
=
for all 𝑛 ∈ ℕ
(𝑟 + 1)(𝑟 + 2) 2(𝑛 + 2)
𝑟=1
62
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
LESSON 6
that
Prove by mathematical induction
Hence by mathematical induction
𝑛
∑ 𝑟 2 (𝑟 − 1) =
𝑛
1
∑ 𝑟 2 (𝑟 − 1) =
𝑛(𝑛2 − 1)(3𝑛 + 2)
12
𝑟=1
𝑟=1
1
𝑛(𝑛2 − 1)(3𝑛 + 2)
12
for all positive integers 𝑛.
for all positive integers 𝑛.
PROOF
…………………………………………………………………………..
EXERCISE 10. 4
𝑛
𝑃𝑛 : ∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
𝑃1 : 12 (1 − 1) =
1
𝑛(𝑛2 − 1)(3𝑛 + 2)
12
1.
1
(1)(12 − 1)(3(1) + 2)
12
2.
0=0
(i) Find the 𝑛th term of the series
1(2) + 2(5) + 3(8) + ⋯
(ii) Prove, by Mathematical Induction, that
the sum to 𝑛 terms of the series in (i)
above is 𝑛2 (𝑛 + 1).
Prove, by mathematical induction, that
𝑛
∑(4𝑟 + 1) = 𝑛(2𝑛 + 3)
Therefore, 𝑃1 is true.
𝑟=1
Assume 𝑃𝑛 is true for 𝑛 = 𝑘
𝑘
𝑃𝑘 : ∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
𝑘+1
𝑃𝑘+1 : ∑ 𝑟2 (𝑟 − 1) =
𝑟 =1
1
𝑘(𝑘 2 − 1)(3𝑘 + 2)
12
1
(𝑘 + 1)((𝑘 + 1)2 − 1)(3(𝑘 + 1) + 2)
12
for all positive integers 𝑛.
3. Prove the following by Mathematical
Induction for all 𝑛 ∈ ℤ+
𝑛
𝑛
a. ∑ 𝑟 2 = (𝑛 + 1)(2𝑛 + 1)
6
𝑟=1
𝑛
1
(𝑘 + 1)(𝑘 2 + 2𝑘 )(3𝑘 + 5)
=
12
=
1
𝑘(𝑘 + 1)(𝑘 + 2)(3𝑘 + 5)
12
Now, 𝑃𝑘+1 = 𝑃𝑘 + (𝑘 + 1) term
1
𝑃𝑘+1 =
𝑘(𝑘 2 − 1)(3𝑘 + 2) + (𝑘 + 1) 2 (𝑘 + 1 − 1)
12
1
12(𝑘 + 1)2 𝑘
=
𝑘(𝑘 + 1)(𝑘 − 1)(3𝑘 + 2) +
12
12
1
=
𝑘(𝑘 + 1)[𝑘 − 1)(3𝑘 + 2) + 12(𝑘 + 1)]
12
=
1
𝑘(𝑘 + 1)(3𝑘 2 − 𝑘 − 2 + 12𝑘 + 12)
12
=
1
𝑘(𝑘 + 1)(3𝑘 2 + 11𝑘 + 10)
12
=
1
𝑘(𝑘 + 1)(𝑘 + 2)(3𝑘 + 5)
12
Therefore 𝑃𝑘+1 is true whenever 𝑃𝑘 is true.
b.
∑ 𝑟3 =
𝑟=1
𝑛
c. ∑
𝑟 =1
𝑛
d. ∑
𝑟=1
𝑛
e. ∑
𝑟=1
𝑛2
(𝑛 + 1)2
4
1
𝑛
=
𝑟(𝑟 + 1) 𝑛 + 1
1
𝑛
=
(𝑟 + 1)(𝑟 + 2) 2(𝑛 + 2)
1
𝑛
=
(𝑟 + 3)(𝑟 + 4) 4(𝑛 + 4)
…………………………………………………………………………..
PROOF OF DIVISIBILITY
LESSON 7
Use Mathematical Induction to
show that 4𝑛3 − 𝑛 is divisible by 3, whenever 𝑛 is
a positive integer.
PROOF
State the proposition
Let 𝑃𝑛 be the proposition that 4𝑛3 − 𝑛 is divisible by
3.
Using the definition of divisibility we have
𝑃𝑛 : 4𝑛3 − 𝑛 = 3𝑟 for some integer 𝑟
63
CHAPTER 10: SEQUENCES, SERIES and MATHEMATICAL INDUCTION
Part 1
Show that 𝑃1 is true
𝑃1 : 4(1)3 − 1 = 3 = 3(1)
Thus, 𝑃1 is true.
Part 2 Assume that 𝑃𝑛 is true for 𝑛 = 𝑘.
This is called the inductive step.
𝑃𝑘 : 4𝑘 3 − 𝑘 = 3𝑟
Assume 𝑃𝑘 is true
𝑃𝑘+1 : 4(𝑘 + 1)3 − (𝑘 + 1) = 3𝑚 𝑚 ∈ ℤ
Show that 𝑃𝑘+1 must follow
Part 3 Show that 𝑃𝑘+1 is true given that 𝑃𝑘 is
true.
Now,
4(𝑘 + 1)3 − 𝑘 + 1
= 4(𝑘 3 + 3𝑘 2 + 3𝑘 + 1) − (𝑘 + 1)
= 4𝑘 3 + 12𝑘 2 + 12𝑘 + 4 − 𝑘 − 1
= 4𝑘 3 + 12𝑘 2 + 11𝑘 + 3
= (4𝑘 3 − 𝑘) + 12𝑘 2 + 12𝑘 + 3
= (4𝑘 3 − 𝑘) + 3(4𝑘 2 + 4𝑘 + 1)
= 3𝑟 + 3(4𝑘 2 + 4𝑘 + 1)
= 3(𝑟 + 4𝑘 2 + 4𝑘 + 1)
Thus, if 𝑃𝑘 is true, then 𝑃𝑘 +1 is true.
Conclusion:
Hence by Mathematical Induction 𝑃𝑛 is true for all
positive integers 𝑛.
LESSON 8
Prove that 42𝑛 − 1 is divisible by
5 for all positive integers 𝑛.
PROOF
Let 𝑃𝑛 : 42𝑛 − 1 = 5𝑟 𝑟 𝜖 ℤ
𝑃1 : 42(1) − 1 = 16 − 1
= 15 = 5(3)
Therefore, 𝑃1 is true.
Assume that 𝑃𝑛 is true for 𝑛 = 𝑘.
𝑃𝑘 : 42𝑘 − 1
∴ 42𝑘 = 5𝑟 + 1
𝑃𝑘+1 : 42(𝑘+1) − 1 = 5𝑚 𝑚 𝜖 ℤ
Now,
42(𝑘+1) − 1 = 42𝑘+2 − 1
= 42𝑘 . 42 − 1
= (5𝑟 + 1)16 − 1
= 80𝑟 + 16 − 1
= 80𝑟 + 15
= 5(16𝑟 + 3)
Thus, 42(𝑘+1) − 1 = 5𝑚 where 𝑚 = 16𝑟 + 3 is an
integer. Therefore 𝑃𝑘 +1 is true whenever 𝑃𝑘 is
true.
Hence, by Mathematical Induction, 42𝑛 − 1 is
divisible by 5 for all positive integers 𝑛.
…………………………………………………………………………..
EXERCISE 10.5
1. Prove the following by Mathematical Induction
for all 𝑛 ∈ ℤ+
(a) 72𝑛 − 1 is divisible by 48
(b) 4𝑛 + 6𝑛 − 1 is divisible by 9 for 𝑛 ≥ 1, 𝑛 ∈
ℤ+ .
EXAM QUESTIONS
1.
Prove, by mathematical induction, that
𝑛(𝑛2 + 5) is divisible by 6 for all positive
integers 𝑛.
[6]
CAPE 2011
2.
Use mathematical induction to prove that
5𝑛 + 3 is divisible by 2 for all values of 𝑛 ∈ ℕ.
[8]
CAPE 2013
3.
Use mathematical induction to prove that
𝑛
12 + 32 + 52 + ⋯ + (2𝑛 − 1)2 = (4𝑛2 − 1)
3
for 𝑛 ∈ ℕ.
[10]
CAPE 2014
4.
Given that 𝑆(𝑛) = 5 + 52 + 53 + 54 + ⋯ + 5𝑛 ,
use mathematical induction to prove that
4𝑆(𝑛) = 5𝑛+1 − 5 for 𝑛 ∈ ℕ.
[8]
CAPE 2015
…………………………………………………………………………..
64
CHAPTER 11: THE REAL NUMBER SYSTEM
CHAPTER 11: THE REAL NUMBER SYSTEM
At the end of this section, students should be able
to:


perform binary operations
use the concepts of identity, closure,
inverse, commutavity, associativity,
distributivity, addition, multiplication and
other binary operations.
BINARY OPERATIONS
The four basic operations are addition,
subtraction, multiplication and division. Their
operators are +, −,×, and ÷ respectively. A
binary operator is any operator other than these
four which combines two elements of a set to
produce a third element.
be CLOSED on the set, 𝐴, if 𝑎⨁𝑏 is ALWAYS a
member of the set, 𝐴.
LESSON 2a
An operation is defined on the set
𝐴 = {1, 2, 3, 4} as shown in the table below.
∗
1
2
3
4
1
1
3
5
7
2
2
4
6
8
3
3
5
7
9
4
4
6
8
10
Determine if the operation is Closed on the set 𝐴.
SOLUTION
Since the table contains elements
not found in the set 𝐴 the operation is NOT Closed
on the set 𝐴.
LESSON 2b
An operation is defined by
𝑎 ∗ 𝑏 = 3𝑎 − 𝑎𝑏 where 𝑎 and 𝑏 are real numbers.
Show that ∗ is closed on ℝ.
CAYLEY TABLE
A binary operation on a finite set can be
represented in the form of a table, sometimes
called a Cayley table.
LESSON 1
The operation ∗ is defined on the
set 𝐴 = {1, 2, 3, 4} by 𝑎 ∗ 𝑏 = 𝑎 + 2𝑏 − 2,
determine the corresponding table.
SOLUTION
The first column represents 𝑎
and the first row represents 𝑏. Here are the
calculations for the first column.
𝑎 ∗ 𝑏 = 𝑎 + 2𝑏 − 2
SOLUTION
If ∗ is closed on ℝ, then
𝑎 ∗ 𝑏 = 3𝑎 − 𝑎𝑏 is ALWAYS a real number.
3𝑎 + (−𝑎𝑏) ∈ ℝ since the sum of two real
numbers 3𝑎 and −𝑎𝑏 is ALWAYS real.
LESSON 2c
An operation ⨁ is defined by
𝑥+𝑦
𝑥⨁𝑦 =
where 𝑥 and 𝑦 are real numbers.
2
Determine if ⨁ is closed on ℤ.
𝑥+𝑦
SOLUTION
If ⨁ is closed on ℤ, 𝑥⨁𝑦 = 2 is
ALWAYS an Integer. We will use a counter –
example.
1 ∗ 1 = 1 + 2(1) − 2 = 1
1⨁2 =
2 ∗ 1 = 2 + 2(1) − 2 = 2
3
Since 2 is NOT an Integer, 𝑥⨁𝑦 is NOT ALWAYS a
member of ℤ. Therefore ⨁ is NOT closed.
3 ∗ 1 = 3 + 2(1) − 2 = 3
4 ∗ 1 = 4 + 2(1) − 2 = 4
∗
1
2
3
4
1
1
2
3
4
2
3
4
5
6
1+ 2 3
=
2
2
3
5
6
7
8
4
7
8
9
10
CLOSURE
Let 𝑎 and 𝑏 be members of a set, 𝐴, and ⨁ be a
binary operation on 𝐴. The operation ⨁ is said to
LESSON 2d
The operation ⊕ is defined on
the set of real numbers by 𝑚 ⊕ 𝑛 = √3𝑚 − 2𝑛.
Show that ⊕ is not Closed on 𝑅.
SOLUTION
⊕ is not closed on 𝑅 since the square root of a
negative number is not real.
COMMUTATIVITY
Given that 𝑎 and 𝑏 are members of the set 𝐴 and
the operation ⨁ is defined on the set 𝐴, then ⨁ is
COMMUTATIVE if 𝑎⨁𝑏 = 𝑏⨁𝑎.
65
CHAPTER 11: THE REAL NUMBER SYSTEM
LESSON 3
An operation is defined by
𝑥 ⋆ 𝑦 = 2𝑥 − 3𝑦 where 𝑥 and 𝑦 are real numbers.
State, with a reason, whether
i) ⋆ is Closed under the set of Real numbers,
ii) ⋆ is Commutative under the set of Real
numbers,
SOLUTION
(i)
2𝑥 is a real number and −3𝑦 is a real
number.
The sum of 2 real numbers is real ∴ ⋆ is
closed on ℝ.
(ii)
If ⋆ is commutative 𝑥 ⋆ 𝑦 = 𝑦 ⋆ 𝑥
𝑥 ⋆ 𝑦 = 2𝑥 − 3𝑦
𝑦 ⋆ 𝑥 = 2𝑦 − 3𝑥
2𝑥 − 3𝑦 ≠ 2𝑦 − 3𝑥 ∴ ⋆ is not
commutative
LESSON 3
The following tables represent
binary operations.
(a)
⨁
1
2
3
4
1
0
1
2
3
2
1
2
3
4
3
2
3
4
5
4
3
4
5
6
∗
1
2
3
4
1
1
3
5
7
2
2
4
6
8
3
3
5
7
9
4
4
6
8
10
⊠
𝒂
𝒃
𝒄
𝒅
𝒂
𝑐
𝑑
𝑎
𝑏
𝒃
𝑑
𝑎
𝑏
𝑐
𝒄
𝑎
𝑏
𝑐
𝑑
𝒅
𝑏
𝑐
𝑑
𝑎
(b)
(c)
State, giving a reason for your answers, which
tables represent operations which are
commutative.
SOLUTION
A binary operation is commutative if its
corresponding table is symmetric about the
leading diagonal. Therefore, ⨁ and ⊠ are
commutative.
ASSOCIATIVITY
Given that 𝑎, 𝑏 and c are members of the set 𝐴 and
the operation ⨁ is defined on the set 𝐴, ⨁ is
associative if 𝑎 ⊕ (𝑏 ⊕ 𝑐) = (𝑎 ⊕ 𝑏) ⊕ 𝑐.
LESSON 4
An operation is defined by
𝑥 ⋆ 𝑦 = 2𝑥 − 3𝑦 where 𝑥 and 𝑦 are real numbers.
State, with a reason, whether ⋆ is Associative
under the set of Real numbers.
SOLUTION
If ⋆ is associative (𝑥 ⋆ 𝑦) ⋆ 𝑧 = 𝑥 ⋆ (𝑦 ⋆ 𝑧) where
𝑧∈ℝ
(𝑥 ⋆ 𝑦) ⋆ 𝑧 = (2𝑥 − 3𝑦) ⋆ 𝑧
= 2(2𝑥 − 3𝑦) + 3𝑧
= 4𝑥 − 6𝑦 + 3𝑧
𝑥 ⋆ (𝑦 ⋆ 𝑧) = 𝑥 ⋆ (2𝑦 − 3𝑧)
= 2𝑥 − 3(2𝑦 − 3𝑧)
= 2𝑥 − 6𝑦 + 9𝑧
4𝑥 − 6𝑦 + 3𝑧 ≠ 2𝑥 − 6𝑦 + 9𝑧 ∴ ⋆ is not
associative
LESSON 4
Given that the operation ∗ is
defined on the set of real numbers such that
𝑎 ∗ 𝑏 = 5 + 𝑎 + 𝑏, prove that ∗ is associative on ℝ.
SOLUTION
If ∗ is associative on ℝ then
𝑎 ∗ (𝑏 ∗ 𝑐) = (𝑎 ∗ 𝑏) ∗ 𝑐 for all 𝑎, 𝑏, 𝑐 ∈ ℝ.
𝑎∗𝑏 =5+𝑎+𝑏
(𝑎 ∗ 𝑏) ∗ 𝑐 = 𝑎 ∗ (𝑏 ∗ 𝑐)
(5 + 𝑎 + 𝑏) ∗ 𝑐 = 𝑎 ∗ (5 + 𝑏 + 𝑐 )
5+5+𝑎+𝑏 +𝑐 =5+𝑎+5+𝑏 +𝑐
10 + 𝑎 + 𝑏 + 𝑐 = 10 + 𝑎 + 𝑏 + 𝑐
66
CHAPTER 11: THE REAL NUMBER SYSTEM
IDENTITY AND INVERSE
If 𝑒 is the identity element for an operation ∗ then
𝑎∗𝑒 =𝑒∗𝑎 = 𝑎
If 𝑦 is the inverse of an element 𝑎, then
𝑎∗𝑦 = 𝑦∗𝑎 = 𝑒
LESSON 5
The operation ∗ is defined on the
set of real numbers by 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 2.
Determine
(i) the identity element of ∗.
(ii) the inverse of 𝑎, 𝑎−1
SOLUTION
(i) Let 𝑒 be the identity element.
𝑎∗𝑒 = 𝑒∗ 𝑎 =𝑎
𝑎+𝑒−2= 𝑎
𝑒−2 =0
𝑒=2
The identity element is 2.
(ii) 𝑎 ∗ 𝑎 −1 = 𝑎 −1 ∗ 𝑎 = 𝑒
𝑎 ∗ 𝑎 −1 = 2
𝑎 + 𝑎 −1 − 2 = 2
𝑎 + 𝑎 −1 = 4
𝑎 −1 = 4 − 𝑎
LESSON 5
An operation ⨁ is defined on the
set 𝐴 = {1, 2, 3, 4} as shown in the following table.
⨁
1
2
3
4
1
0
1
2
3
2
1
2
3
4
3
2
3
4
5
4
3
4
5
6
Determine
(a) the identity element of ⨁
(b) the inverse of the element 3
SOLUTION
(a) Since the column headed by the 2 and the row
headed by the 2 correspond exactly to the set
𝐴, 2 is the identity element.
(b) Let 𝑦 represent the inverse of 3
3 ∗ 𝑦 = 𝑒 where 𝑒 is the identity element
EXERCISE 11
1. For all real numbers 𝑥 and 𝑦, let ⊺ be defined
as 𝑥 ⊺ 𝑦 = 3𝑥 2 + 𝑦 2 .
(i) Prove that ⊺ is Closed on the set of Real
numbers.
(ii) Show that ⊺ is not Commutative on the set
of Real numbers.
2. The operation △ is defined on 𝑅 by
𝑎 △ 𝑏 = 𝑎3 − 2𝑎𝑏 − 𝑏2 .
(i) Show that △ is Closed on 𝑅.
(ii) Show that △ is not Commutative on 𝑅.
3.
4.
5.
6.
7.
The operation ∗ is defined by
𝑥 ∗ 𝑦 = 𝑥 + 𝑦 − 𝑧 where 𝑥 and 𝑦 are real
numbers and 𝑧 is a real number.
State, with a reason, whether
(a) ∗ is closed in 𝑅
(b) ∗ is commutative in 𝑅
(c) ∗ is associative in 𝑅
The operation ∘ on real numbers is defined by
𝑝 ∘ 𝑞 = 𝑝|𝑞|.
(a) Show that ∘ is not commutative.
(b) Prove that ∘ is associative.
For all real numbers 𝑥 and 𝑦, let 𝑥 ∆ 𝑦 be
defined as 𝑥 ∆ 𝑦 = 𝑥 2 + 𝑥𝑦 + 𝑦 2 .
State, with a reason, whether
(a) ∆ is closed in 𝑅
(b) ∆ is commutative in 𝑅
(c) ∆ is associative in 𝑅
The operation 𝑥⨀𝑦 = 𝑥𝑦 + 𝑥 + 𝑦 for all real
numbers 𝑥 and 𝑦. Determine which of the
following is/are true
(a) 𝑥⨀𝑦 = 𝑦⨀𝑥
(b) (𝑥 − 1)⨀(𝑥 + 1) = (𝑥⨀𝑥) + 1
(c) 𝑥⨀(𝑦 + 𝑧) = (𝑥⨀𝑦) + (𝑥⨀𝑧)
A binary operation ∗ is defined on real
numbers 𝑥 and 𝑦 by
𝑥 ∗ 𝑦 = 2𝑥𝑦 + 𝑥 + 𝑦
You may assume that the operation ∗ is
commutative and associative.
(i) Explain briefly the meanings of the terms
‘commutative’ and ‘associative’
1
1
1
(ii) Show that 𝑥 ∗ 𝑦 = 2 (𝑥 + 2) (𝑦 + 2) − 2
𝑥 ∘ 𝑦 = is the remainder when 𝑥 ∗ 𝑦 is divided
by 7
(iii) Show that 4 ∘ 6 = 2.
3∗𝑦 =2
From the table 3 ∗ 1 = 2, therefore the inverse
of 3 is 1.
67
CHAPTER 11: THE REAL NUMBER SYSTEM
8.
The binary operation ∗ is defined on the set
{0, 1, 2, 3, 4} as shown in the table below.
*
0
1
2
3
4
0
0
1
2
3
4
1
1
2
3
4
0
2
2
3
4
0
1
3
3
4
0
1
2
4
4
0
1
2
3
(a) Is * closed on the set? Give a reason for
your answer.
(b) Is * commutative? Give a reason for your
answer.
(c) State the identity element of *.
(d) State the inverse element of 3.
(e) Determine if (3 ∗ 1) ∗ 4 = 3 ∗ (1 ∗ 4).
Hence, state if * is associative.
SOLUTIONS
1.
2.
3.
4.
5.
6.
7.
8.
(a) Yes
(b) Yes
(c) Yes
(a) Yes
(a) True
(b) Yes
(b) False
(c) No
(c) True
(a) Yes (b) Yes (c) 0 (d) 2 (e) Yes
EXAM QUESTIONS
1.
2.
A binary operator ⊕ is defined on a set of
positive real numbers by
𝑦 ⊕ 𝑥 = 𝑦 2 + 𝑥 2 + 2𝑦 + 𝑥 − 5𝑥𝑦
Solve the equation 2 ⊕ 𝑥 = 0.
[5]
CAPE 2013
A binary operator ⊕ is defined on a set of
positive real numbers by
𝑦 ⊕ 𝑥 = 𝑦 3 + 𝑥 3 + 𝑎𝑦 2 + 𝑎𝑥 2 − 5𝑦 − 5𝑥 + 16
where 𝑎 is a real number.
(i) State, giving a reason for your answer, if
⊕ is commutative in 𝑅.
[3]
(ii) Given that 𝑓(𝑥) = 2 ⊕ 𝑥 and (𝑥 − 1) is a
factor of 𝑓(𝑥),
a) find the value of 𝑎
[4]
b) factorise completely.
[3]
CAPE 2014
SOLUTIONS
1.
2.
𝑥 = 1, 8
(i) Yes, 𝑥 ⊕ 𝑦 = 𝑦 ⊕ 𝑥
(i) (a) 𝑎 = −2
(b) (𝑥 − 1)(𝑥 − 3)(𝑥 + 2)
…………………………………………………………………………
DIRECT PROOFS
Other mathematical statements can be proven by
using deduction; this is usually the case. In proof
by deduction, each step is deduced from the
previous one or is justified by quoting an accepted
fact or a result previously proven, to arrive at the
required result.
LESSON 1
Prove that for any two positive
numbers 𝑥 and 𝑦
𝑥+𝑦
√𝑥𝑦 ≤
2
SOLUTION
2
(√𝑥 − √𝑦) ≥ 0
𝑥 + 𝑦 − 2 √𝑥𝑦 ≥ 0
𝑥 + 𝑦 ≥ 2√𝑥𝑦
2√𝑥𝑦 ≤ 𝑥 + 2
𝑥+𝑦
√𝑥𝑦 ≤
2
LESSON 2
Prove that for 𝑛 ∈ ℕ, 𝑛(𝑛 + 1) is
even.
SOLUTION
Suppose 𝑛 is even, then 𝑛 + 1 is odd and the
product of an even number and an odd number is
even.
Suppose 𝑛 is odd, then 𝑛 + 1 is even and the
product of an odd number and an even number is
even. Therefore, 𝑛(𝑛 + 1) is even.
…………………………………………………………………………
1. Prove that if 𝑎 and 𝑏 are real numbers then
𝑎2 + 𝑏2 ≥ 2𝑎𝑏.
EXAM QUESTIONS
1.
2.
3.
The smallest of three consecutive integers is
𝑛.
Write down the other two integers.
Prove that the sum of any three consecutive
integers is divisible by 3.
Given that 𝑥 > 𝑦, and 𝑘 < 0 for the real
numbers 𝑥, 𝑦 and 𝑘, show that 𝑘𝑥 < 𝑘𝑦. [4]
CAPE 2004
Prove that the product of any two consecutive
integers 𝑘 and 𝑘 + 1 is an even integer.
[2]
CAPE 2011
…………………………………………………………………………
68
CHAPTER 12: REASONING AND LOGIC
CHAPTER 12: REASONING AND LOGIC
At the end of this section, students should be able
to:




identify simple and compound
propositions;
establish the truth value of compound
statements using truth tables;
state the converse, contrapositive and
inverse of a conditional (implication)
statement;
determine whether two statements are
logically equivalent.
__________________________________________________________
A proposition is a statement which is either true
or false. Propositions are usually denoted by
letters: 𝑝, 𝑞, 𝑟, … Connectors are used to form
compound propositions from two or more
propositions. Here is a list of connectors
∧
conjunction (i.e. and)
∨
inclusive (i.e. injunction / or)
~
negation (i.e. not)
→
implication (i.e. if…then)
↔
equivalence (i.e. if and only if)
The statement “I brush my teeth and I floss” is a
Compound Proposition since it is the combination
of the simple propositions – I brush my teeth and I
floss. With the use of propositional notation, also
known as mathematical logic, this can be written
as
I brush my teeth ∧ I floss
Furthermore, if we let
𝑎 = I brush my teeth
𝑏 = I floss
we can then simplify our statement to 𝑎 ∧ 𝑏. Thus,
we have created a Boolean expression.
LESSON 1
Given that
𝑝 = dogs are lovable
𝑞 = cats are friendly
𝑟 = I study a lot
𝑠 = I pass my exams
𝑡 = I score more than 80%
𝑣 = I receive a grade 𝐴
use the symbols of propositional notation to write
down Boolean expressions to represent the
statements:
(i)
(ii)
(iii)
Dogs are lovable and cats are friendly.
Dogs are lovable or cats are friendly.
Dogs are not lovable.
(iv)
(v)
(vi)
Dogs are lovable and cats are unfriendly.
If I study a lot then I will pass my exams.
If I score more than 80%, I will receive a
grade A, and if I receive a grade A, my
score will be more than 80%.
SOLUTION
(i)
(ii)
(iii)
(iv)
(v)
(vi)
𝑝∧𝑞
𝑝∨𝑞
∼𝑝
𝑝 ∧∼ 𝑞
𝑠→𝑡
(𝑡 → 𝑣) ∧ (𝑣 → 𝑡) or (𝑡 ↔ 𝑣)
TRUTH TABLES
A truth table shows the truth or falsity of a
statement by listing all of the possible outcomes.
A tautology is a statement that is always true.
A contradiction is a statement that is always false.
A contingency is a statement that is sometimes
true and sometimes false.
LESSON 2
Construct truth tables for
a) ∼ 𝑝
b) 𝑝 ∧∼ 𝑝
c) 𝑝 ∨∼ 𝑝
d) 𝑝 ∧ 𝑞
e) 𝑝 ∨∼ 𝑞
and state whether the proposition is a tautology,
contradiction or a contingency.
SOLUTION
a)
𝑝
∼𝑝
0
1
1
0
This is a contingency.
b)
𝑝
∼𝑝
𝑝 ∧∼ 𝑝
0
1
0
1
0
0
This is a contradiction. Furthermore,
𝑝 ∧∼ 𝑝 = 0
c)
𝑝
∼𝑝
𝑝 ∨∼ 𝑝
0
1
1
1
0
1
This is a tautology. Furthermore, 𝑝 ∨∼ 𝑝 = 1
69
CHAPTER 12: REASONING AND LOGIC
d)
SOLUTION
𝑝
𝑞
𝑝∧𝑞
0
0
0
0
1
0
1
0
0
1
1
1
This is a contingency.
(i)
e)
𝑝 ∨∼ 𝑞
1
0
1
1
(ii)
Construct truth tables for
(iii)
𝑝
𝑞
∼𝑞
0
0
1
0
1
0
1
0
1
1
1
0
This is a contingency.
LESSON 3
(i)
(ii)
SOLUTION
(i)
𝑞
0
0
1
1
0
0
1
1
𝑟
0
1
0
1
0
1
0
1
𝑝∧𝑞
0
0
0
0
0
0
1
1
∼𝑝
∼𝑞
(𝑝 ∧ 𝑞) ∨ 𝑟
0
1
0
1
0
1
1
1
(ii)
𝑝
0
0
0
0
1
1
1
1
LESSON 4
(i)
(ii)
(iii)
𝑞
𝑟
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
∼𝑞
∨
∼𝑝
1
1
1
1
1
1
0
0
Construct truth tables for
𝑞
0
1
0
1
𝑝
0
𝑞
0
1
0
1
1
1
(𝑝 ∧ 𝑞) ∨ 𝑟
(∼ 𝑞 ∨∼ 𝑝) ∧ 𝑟
𝑝
0
0
0
0
1
1
1
1
𝑝
0
0
1
1
(∼ 𝑞
∨∼ 𝑝)
∧𝑟
0
1
0
1
0
1
0
0
𝑝→𝑞
1
1
0
1
∼𝑝
1
1
0
0
∼𝑞
1
0
1
0
∼ 𝑝 →∼ 𝑞
1
0
1
1
NB: 𝑝 ↔ 𝑞 means (𝑝 → 𝑞) ∧ (𝑞 → 𝑝)
𝑝
𝑞
𝑝 → 𝑞 𝑞 → 𝑝 (𝑝 → 𝑞) ∧ (𝑞 → 𝑝)
0
0
1
1
1
0
1
1
0
0
1
0
0
1
0
1
1
1
1
1
CONVERSE, INVERSE AND
CONTRAPOSITIVE
LESSON 5
Consider the proposition 𝑝 → 𝑞
Its converse is 𝑞 → 𝑝.
Interchange propositions.
Its inverse is ∼ 𝑝 →∼ 𝑞.
Negate both propositions.
Its contrapositive is ∼ 𝑞 →∼ 𝑝.
Interchange propositions and negate.
LOGICAL EQUIVALENCE
LESSON 6
Construct a single truth table for
𝑝 → 𝑞 and ∼ 𝑝 ∨ 𝑞.
SOLUTION
𝑝
𝑞
∼𝑝
𝑝→𝑞
∼ 𝑝∨𝑞
0
0
1
1
1
0
1
1
1
1
1
0
0
0
0
1
1
0
1
1
Since the two columns for 𝑝 → 𝑞 and ∼ 𝑝 ∨ 𝑞 are
identical these two propositions are logically
equivalent.
𝑝→𝑞
∼ 𝑝 →∼ 𝑞
𝑝↔𝑞
70
CHAPTER 12: REASONING AND LOGIC
LAWS OF BOOLEAN ALGEBRA
Identity:
𝑝∨𝑝 ≡ 𝑝
𝑝∧𝑝 ≡ 𝑝
𝑝∨1 ≡1
𝑝∧1 ≡𝑝
𝑝∨0 ≡𝑝
𝑝∧0 ≡0
Commutative:
𝑝∨𝑞 ≡ 𝑞∨𝑝
𝑝∧𝑞 ≡ 𝑞∧𝑝
Complement:
𝑝 ∨∼ 𝑝 ≡ 1
𝑝 ∧∼ 𝑝 ≡ 0
Double Negation:
∼ (∼ 𝑝) ≡ 𝑝
Associative:
𝑝 ∨ (𝑞 ∨ 𝑟) ≡ (𝑝 ∨ 𝑞) ∨ 𝑟
𝑝 ∧ (𝑞 ∧ 𝑟) ≡ (𝑝 ∧ 𝑞) ∧ 𝑟
Distributive:
𝑝 ∨ (𝑞 ∧ 𝑟) ≡ (𝑝 ∨ 𝑞 ) ∧ (𝑝 ∨ 𝑟)
𝑝 ∧ (𝑞 ∨ 𝑟) ≡ (𝑝 ∧ 𝑞) ∨ (𝑝 ∧ 𝑟)
Absorbtion:
𝑝 ∨ (𝑝 ∧ 𝑞) ≡ 𝑝
𝑝 ∧ (𝑝 ∨ 𝑞) ≡ 𝑝
De Morgan’s:
∼ (𝑝 ∨ 𝑞) ≡∼ 𝑝 ∧∼ 𝑞
∼ (𝑝 ∧ 𝑞) ≡∼ 𝑝 ∨∼ 𝑞
LESSON 7
Use the laws of algebra of
propositions to show that
𝑝 ∧ (𝑝 → 𝑞) ≡ 𝑝 ∧ 𝑞.
SOLUTION
LHS
𝑝 ∧ (𝑝 → 𝑞)
= 𝑝 ∧ (∼ 𝑝 ∨ 𝑞)
Since 𝑝 → 𝑞 and ∼ 𝑝 ∨ 𝑞 are logically equivalent.
= (𝑝 ∧∼ 𝑝) ∨ (𝑝 ∧ 𝑞)
= 0 ∨ (𝑝 ∧ 𝑞)
=𝑝∧𝑞
RHS
LESSON 8
Show that
{𝑝 ∧ (∼ 𝑝 ∨ 𝑞)} ∨ {𝑞 ∧∼ (𝑝 ∧ 𝑞)} ≡ 𝑞.
SOLUTION
LHS
{𝑝 ∧ (∼ 𝑝 ∨ 𝑞)} ∨ {𝑞 ∧∼ (𝑝 ∧ 𝑞)}
= {𝑝 ∧∼ 𝑝) ∨ (𝑝 ∧ 𝑞)} ∨ {𝑞 ∧ (∼ 𝑝 ∨∼ 𝑞)}
= {0 ∨ (𝑝 ∧ 𝑞)} ∨ (𝑞 ∧∼ 𝑝) ∨ (𝑞 ∧∼ 𝑞)}
= (𝑝 ∧ 𝑞) ∨ {(𝑞 ∧∼ 𝑝) ∨ 0}
= (𝑝 ∧ 𝑞) ∨ (𝑞 ∧∼ 𝑝)
= (𝑞 ∧ 𝑝) ∨ (𝑞 ∧∼ 𝑝)
= 𝑞 ∧ (𝑝 ∨∼ 𝑝)
=𝑞∧1
=𝑞
RHS
…………………………………………………………………………
EXERCISE 12
1. Given that 𝑝 is the statement, “I will go to the
beach” and 𝑞 is the statement “It is extremely
hot” represent the following as Boolean
expressions.
(i) I will not go to the beach.
(ii) It is extremely hot and I will go to the
beach.
(iii) If it is extremely hot then I will go to the
beach.
2. Construct truth tables for each of the
following.
(a) ~𝑎
(b) ~𝑎 ∨ 𝑏
(c) 𝑎 ∨ 𝑏
(d) ∼ 𝑎 ∧ 𝑏
and state whether the proposition is a
tautology, contradiction or a contingency.
3.
4.
5.
6.
7.
8.
Construct truth tables for
(a) (𝑎 ∧ 𝑏) ∨ 𝑐
(b) (∼ 𝑎 ∧∼ 𝑏) ∨ 𝑐
Construct truth tables for
(a) ∼ 𝑎 → 𝑏
(b) (𝑎 ∧ 𝑏) → 𝑐
(c) (𝑎 ∨ 𝑏) → (𝑎 ∧ 𝑐)
State the converse, inverse and contrapositive
of (𝑝 ∨ 𝑞) → (𝑞 ∧ 𝑝).
Show that ∼ (𝑝 ∧ 𝑞) and ∼ 𝑝 ∨∼ 𝑞 are
logically equivalent.
Simplify
(a) 𝑎 ∨ (∼ 𝑎 ∧ 𝑏)
(b) 𝑎 ∧ [𝑏 ∨ (𝑎 ∧ 𝑏)] ∧ [𝑎 ∨ (∼ 𝑎 ∧ 𝑏)]
Prove the following statements, using the
laws of Boolean Algebra
(i) (𝑎 ∧ 𝑏) ∨ (𝑎 ∧ 𝑐 ) = 𝑎 ∧ (𝑏 ∨ 𝑐 )
(ii) (𝑎 ∨ 𝑏) ∧ (𝑎 ∧ 𝑏) = (𝑎 ∧ 𝑏)
(iii) (𝑎 ∧ 𝑏) ∨ (𝑎 ∧∼ 𝑏) = 𝑎
(iv) 𝑎 ∧ [(𝑏 ∧ 𝑐 ) ∨ (𝑏 ∧∼ 𝑐 )] = 𝑎 ∧ 𝑏
(v) ∼ 𝑎 ∧ (∼ 𝑏 ∨ 𝑎) =∼ (𝑎 ∨ 𝑏)
(vi) [𝑝 ∧ (∼ 𝑝 ∨ 𝑞)] ∨ [𝑞 ∧∼ (𝑝 ∧ 𝑞)] = 𝑞
71
CHAPTER 12: REASONING AND LOGIC
(vii) [(𝑝 ∨∼ 𝑞) ∧ (∼ 𝑝 ∨∼ 𝑞)] ∨ 𝑞 = 1
(viii) ∼ (𝑝 ∨∼ (𝑝 ∧ 𝑞)) = 0
EXAM QUESTIONS
9.
Let 𝑝 and 𝑞 be two propositions.
(i) State the converse of (𝑝 ∧ 𝑞) → (𝑞 ∨∼ 𝑝).
[1]
(ii) Show that the contrapositive of the
inverse of (𝑝 ∧ 𝑞) → (𝑞 ∨∼ 𝑝) is the
converse of (𝑝 ∧ 𝑞) → (𝑞 ∨∼ 𝑝).
[3]
CAPE 2013
10. Let 𝑝, 𝑞 and 𝑟 be three propositions. Construct
a truth table for the statement
(𝑝 → 𝑞) ∧ (𝑟 → 𝑞).
[5]
CAPE 2014
11. Let 𝑝 and 𝑞 be any two propositions.
(i) State the inverse and the contrapositive of
the statement 𝑝 → 𝑞.
[2]
(ii) Copy and complete the table below to
show the truth table for 𝑝 → 𝑞 and
∼ 𝑞 →∼ 𝑝.
𝒑
𝒒
∼𝒑 ∼𝒒
𝒑→𝒒
∼𝒒→
∼𝒑
T
T
T
F
F
T
F
F
[4]
(iii) Hence, state whether the compound
statement 𝑝 → 𝑞 and ∼ 𝑞 →∼ 𝑝 are
logically equivalent. Justify your response.
[2]
CAPE 2015
12. State the converse, inverse and contrapositive
of 𝑝 →∼ 𝑞.
[5]
APPLIED MATHEMATICS 2015
SOLUTIONS
SEE PAGE 166
…………………………………………………………………………
72
CHAPTER 12: REASONING AND LOGIC
MODULE TWO: VECTORS,
TRIGONOMETRY AND COORDINATE
GEOMETRY
CHAPTER 13: TWO DIMENSIONAL VECTORS
At the end of this section, students should be able
to:



















𝑥
express a vector in the form (𝑦 ) or
𝑥𝒊 + 𝑦𝒋; 𝑥, 𝑦 ∈ ℝ;
define equal vectors;
add and subtract vectors;
multiply a vector by a scalar quantity;
derive and use unit vectors;
find displacement vectors;
find the magnitude and direction of a
vector;
define the scalar product of two vectors:
(i) in terms of their components;
(ii) in terms of their magnitudes and the
angle between them;
find the angle between two given vectors;
apply properties of parallel and
perpendicular vectors.
𝑥
express a vector in the form (𝑦 ) or
𝑧
𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 where 𝒊, 𝒋 and 𝒌 are unit
vectors in the direction of 𝑥−, 𝑦 − and 𝑧 −
axis respectively;
define equality of two vectors;
add and subtract vectors;
multiply a vector by a scalar quantity;
derive and use unit vectors, position
vectors and displacement vectors;
find the magnitude and direction of a
vector;
find the angle between two vectors using
scalar product;
find the equation of a line in vector form,
parametric form, Cartesian form, given a
point on the line and a vector parallel to
the line;
determine whether two lines are parallel,
intersecting or skewed;

find the equation of the plane, in the form
𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 = 𝑑, 𝒓. 𝒏 = 𝑑, given a point
in the plane and the normal to the plane.
_________________________________________________________
𝒊, 𝒋 REPRESENTATION
INTRODUCTION
⃗⃗⃗⃗⃗ which is a vector such that
The diagram shows 𝐴𝐵
⃗⃗⃗⃗⃗ = (3)
𝐴𝐵
4
1
Alternately the base unit vectors 𝒊 = ( ) and
0
0
𝒋 = ( ) can be used to express vectors.
1
⃗⃗⃗⃗⃗ = 3 (1) + 4 (0) = (3) + (0) = (3)
𝐴𝐵
0
1
0
4
4
⃗⃗⃗⃗⃗ = 3𝒊 + 4𝒋
𝐴𝐵
73
CHAPTER 13: VECTORS
𝒊, 𝒋, 𝒌 REPRESENTATION
INTRODUCTION
In three dimensions, the base vectors are
1
0
0
𝒊 = (0) ,
𝒋 = (1) ,
𝒌 = (0)
0
0
1
that are along the 𝑥, 𝑦 and z coordinate directions,
respectively, as shown in the figure.
−4
For example, 𝒗 = ( 3 ) can be written as
2
−4𝒊 + 3𝒋 + 2𝒌.
LESSON 2
The three points 𝑂, 𝑃 and 𝑄 are
𝑞
2
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ = ( ). Given that
such that 𝑂𝑃 = ( ) and 𝑂𝑄
2𝑞
3
⃗⃗⃗⃗⃗ is a unit vector, calculate the possible values of
𝑃𝑄
𝑞.
SOLUTION
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑃𝑄 = ⃗⃗⃗⃗⃗
𝑃𝑂 + 𝑂𝑄
𝑞
2
= − ( ) + (2𝑞 )
3
−2 + 𝑞
=(
)
−3 + 2𝑞
Since ⃗⃗⃗⃗⃗
𝑃𝑄 is a unit vector the length of ⃗⃗⃗⃗⃗
𝑃𝑄 is 1,
therefore
√(𝑞 − 2) 2 + (2𝑞 − 3)2 = √1
𝑞 2 − 4𝑞 + 4 + 4𝑞 2 − 12𝑞 + 9 = 1
5𝑞 2 − 16𝑞 + 12 = 0
(5𝑞 − 6)(𝑞 − 2) = 0
6
𝑞= ,
2
5
LESSON 3
The magnitude of 𝒗:
|𝒗| = √(−4)2 + 32 + 22 = √29
Consequently,
−4
3
2
𝒗
̂=
𝒊+
𝒋+
𝒌
√29
√29
√29
UNIT VECTORS
LESSON 1
Given the position vectors of the
points 𝐴 and 𝐵 relative to a fixed point 𝑂 are
⃗⃗⃗⃗⃗ = (2) and 𝑂𝐵
⃗⃗⃗⃗⃗ = (−1), determine
𝑂𝐴
5
4
(i)
(ii)
(iii)
⃗⃗⃗⃗⃗ in the form 𝑥𝒊 + 𝑦𝒋
𝐴𝐵
⃗⃗⃗⃗⃗ |
|𝐴𝐵
⃗⃗⃗⃗⃗ .
the unit vector in the direction of 𝐴𝐵
SOLUTION
(i)
(ii)
(iii)
⃗⃗⃗⃗⃗ = 2𝒊 + 5𝒋 and 𝑂𝐵
⃗⃗⃗⃗⃗ = −𝒊 + 4𝒋
𝑂𝐴
⃗⃗⃗⃗⃗ = 𝐴𝑂
⃗⃗⃗⃗⃗ + 𝑂𝐵
⃗⃗⃗⃗⃗
𝐴𝐵
= −(2𝒊 + 5𝒋) + (−𝒊 + 4𝒋)
= −3𝒊 − 𝒋
⃗⃗⃗⃗⃗ | = √(−3) 3 + ( −1)2 = √10
|𝐴𝐵
⃗⃗⃗⃗⃗
Unit vector in the direction of 𝐴𝐵
1
(−3𝒊 − 𝒋)
√10
Determine the unit vector in the
−4
⃗⃗⃗⃗⃗ = ( 3 ).
direction of the vector 𝑂𝐴
2
SOLUTION
⃗⃗⃗⃗⃗ | = √(−4) 2 + (3) 2 + (2)2 = √29
|𝑂𝐴
−4
1
Unit Vector=
(3)
√29
2
SCALAR (DOT) PRODUCT
We define the scalar (dot) product of 2 vectors
𝑣 = 𝑎𝒊 + 𝑏𝒋 and 𝑤 = 𝑐𝒊 + 𝑑𝒋 as
𝑣. 𝑤 = (𝑎𝒊 + 𝑏𝒋 ). (𝑐𝒊 + 𝑑𝒋) = 𝑎𝑐 + 𝑏𝑑
LESSON 4
If 𝒗 = −2𝒊 + 𝒋 and 𝒘 = 3𝒊 − 4𝒋
determine 𝒗. 𝒘
SOLUTION
𝒗. 𝒘 = (−2𝒊 + 𝒋). (3𝒊 − 4𝒋)
= (−2)(3) + (1)(−4)
= −10
Given 3 dimensional vectors we define their dot
product as follows;
If 𝒗 = 𝑎𝒊 + 𝑏𝒋 + 𝑐𝒌 and 𝒘 = 𝑑𝒊 + 𝑒𝒋 + 𝑓𝒌 then
𝒗. 𝒘 = (𝑎𝒊 + 𝑏𝒋 + 𝑐𝒌 ). (𝑑𝒊 + 𝑒𝒋 + 𝑓𝒌)
= 𝑎𝑑 + 𝑏𝑒 + 𝑐𝑓
LESSON 5
If 𝒕 = −𝒊 + 3𝒋 − 2𝒌 and
𝒓 = 2𝒊 − 4𝒌 determine 𝒓. 𝒕
SOLUTION
𝒓. 𝒕 = (−𝒊 + 3𝒋 − 2𝒌). (2𝒊 − 4𝒌)
= ( −1)(2) + (3)(0) + (−2)( −4)
=6
74
CHAPTER 13: VECTORS
ANGLE BETWEEN TWO VECTORS
(−𝒊 + 5𝒋). (−2𝒊 − 3𝒋)
We can use the scalar product to find the angle
between two vectors, thanks to the following
formula:
𝒂. 𝒃 = |𝒂||𝒃| cos 𝜃
where 𝜃 is the angle between the vectors.
√(−1)2 + 52 √(−2) 2 + (−3)2
(−1)(−2) + (5)(−3)
= cos 𝜃
√26√13
2 − 15
13
=−
= cos 𝜃
√26√13
√26√13
13
cos −1 (−
)=𝜃
√26√13
135° = 𝜃
Notice that the vectors 𝒂 and 𝒃 are going away
from the angle, 𝜃.
An important fact is that two vectors are
perpendicular (orthogonal) if and only if their dot
product is zero. This is because if 𝜃 = 90°, then
𝒂. 𝒃 = 0 (Recall: cos 90° = 0)
LESSON 8
LESSON 6
The position vectors of points 𝐴
and 𝐵 with respect to the origin 𝑂 are given by
⃗⃗⃗⃗⃗ = 3𝒊 + 2𝒋 and 𝑂𝐵
⃗⃗⃗⃗⃗ = 2𝒊 − 4𝒋.
𝑂𝐴
Find
⃗⃗⃗⃗⃗ in terms of 𝒊 and 𝒋
(a) (i)
𝐴𝐵
⃗⃗⃗⃗⃗
(ii)
The magnitude of 𝐴𝐵
⃗⃗⃗⃗⃗
(b) Determine whether 𝑂𝐴 is perpendicular to ⃗⃗⃗⃗⃗
𝑂𝐵
SOLUTION
⃗⃗⃗⃗⃗ = 𝐴𝑂
⃗⃗⃗⃗⃗ + 𝑂𝐵
⃗⃗⃗⃗⃗
(a) (i) 𝐴𝐵
= (−3𝒊 − 2𝒋) + (2𝒊 − 4𝒋)
= −𝒊 − 6𝒋
⃗⃗⃗⃗⃗ | = √(−1) 2 + (−6)2 = √37
(ii) |𝐴𝐵
⃗⃗⃗⃗⃗ is perpendicular to 𝑂𝐵
⃗⃗⃗⃗⃗ then 𝑂𝐴
⃗⃗⃗⃗⃗ . 𝑂𝐵
⃗⃗⃗⃗⃗ = 0
(b) If 𝑂𝐴
⃗⃗⃗⃗⃗ . 𝑂𝐵
⃗⃗⃗⃗⃗ = (3𝒊 + 2𝒋). (2𝒊 − 4𝒋)
𝑂𝐴
= (3)(2) + (2)( −4)
= −2
Therefore, ⃗⃗⃗⃗⃗
𝑂𝐴 and ⃗⃗⃗⃗⃗
𝑂𝐵 are not perpendicular.
LESSON 7
The position vectors of 𝐴 and
𝐶 relative to an origin O are 𝒊 + 8𝒋 and 2𝒊 + 3𝒋. Use
a vector method to find ∠𝐴𝐶𝑂
SOLUTION
NB: Since we are finding ∠𝐴𝐶𝑂, 𝜃 is located at 𝐶.
Thus, we need the 2 vectors which are going away
⃗⃗⃗⃗⃗ and 𝐶𝑂
⃗⃗⃗⃗⃗
from C i.e. 𝐶𝐴
⃗⃗⃗⃗⃗ = 𝐶𝑂
⃗⃗⃗⃗⃗ + 𝑂𝐴
⃗⃗⃗⃗⃗
𝐶𝐴
⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
= −𝑂𝐶
𝑂𝐴
(
= −2𝒊 − 3𝒋) + (𝒊 + 8𝒋)
= −𝒊 + 5𝒋
⃗⃗⃗⃗⃗ . 𝐶𝑂
⃗⃗⃗⃗⃗
𝐶𝐴
= cos 𝜃
⃗⃗⃗⃗⃗ |. |𝐶𝑂
⃗⃗⃗⃗⃗ |
|𝐶𝐴
= cos 𝜃
2
2
Given that 𝒂 = ( −2) , 𝒃 = (6)
1
3
𝑝
and 𝒄 = ( 𝑝 ), find
𝑝+1
(i) The angle between the directions of 𝒂 and 𝒃.
(ii) The value of 𝑝 for which 𝒃 and c are
perpendicular
SOLUTION
(i)
𝒂. 𝒃 = |𝒂||𝒃| cos 𝜃
2
2
(−2) . (6)
1
3
= cos 𝜃
√22 + (−2)2 + 1√22 + 62 + 32
(2)(2) + (−2)(6) + (1)(3)
= cos 𝜃
√9√49
5
−
= cos 𝜃
21
5
𝜃 = cos −1 (− ) = 103.8°
21
(ii)
If 𝒃 and 𝒄 are perpendicular then 𝒃. 𝒄 = 0
𝑝
2
(6) . ( 𝑝 ) = 0
𝑝+1
3
2𝑝 + 6𝑝 + 3(𝑝 + 1) = 0
2𝑝 + 6𝑝 + 3𝑝 + 3 = 0
11𝑝 = −3
3
𝑝=−
11
…………………………………………………………………………
EXERCISE 13.1
1.
2.
Three points 𝐴, 𝐵 and 𝐶 have coordinates
(1, 2), (2, 5) and (0, −4) respectively, relative
to the origin 𝑂.
(i) Express the position vector of 𝐴, 𝐵 and 𝐶
in terms of 𝒊 and 𝒋
⃗⃗⃗⃗⃗ = 𝐶𝐷
⃗⃗⃗⃗⃗ , find the position vector of 𝐷
(ii) If 𝐴𝐵
in terms of 𝒊 and 𝒋.
The position vectors of points 𝐴 and 𝐵,
relative to an origin 𝑂, are 6𝑖 − 3𝑗 and
15𝑖 + 9𝑗 respectively.
75
CHAPTER 13: VECTORS
3.
4.
5.
6.
⃗⃗⃗⃗⃗ .
(i) Find the unit vector parallel to 𝐴𝐵
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ .
The point 𝐶 lies on 𝐴𝐵 such that 𝐴𝐶 = 2𝐶𝐵
(ii) Find the position vector of 𝐶.
The position vectors of points 𝐴 and 𝐵 relative
to an origin 𝑂 are −3𝑖 − 𝑗 and 𝑖 + 2𝑗
respectively. The point 𝐶 lies on 𝐴𝐵 and is
⃗⃗⃗⃗⃗ = 3 𝐴𝐵
⃗⃗⃗⃗⃗ . Find the position vector
such that 𝐴𝐶
5
of 𝐶 and show that it is a unit vector.
−17
4
Given that ⃗⃗⃗⃗⃗
𝑂𝐴 = (
) and ⃗⃗⃗⃗⃗
𝑂𝐵 = ( ), find
25
5
⃗⃗⃗⃗⃗ ,
(i)
the unit vector parallel to 𝐴𝐵
⃗⃗⃗⃗⃗ , such that 𝐴𝐶
⃗⃗⃗⃗⃗ = 3𝐴𝐵
⃗⃗⃗⃗⃗ .
(ii)
the vector 𝑂𝐶
The position vectors of the points 𝐴 and 𝐵,
relative to an origin 𝑂, are 𝑖 − 7𝑗 and 4𝑖 + 𝑘𝑗
respectively, where 𝑘 is a scalar. The unit
⃗⃗⃗⃗⃗ is 0.6𝑖 + 0.8𝑗,
vector in the direction of 𝐴𝐵
find the value of 𝑘.
The points 𝐴 and 𝐵 are such that the unit
⃗⃗⃗⃗⃗ is 0.28𝑖 + 𝑝𝑗,
vector in the direction of 𝐴𝐵
where 𝑝 is a positive constant.
(i)
Find the value of 𝑝.
The position vectors of 𝐴 and 𝐵, relative to an
origin 𝑂, are 𝑞𝑖 − 7𝑗 and 12𝑖 + 17𝑗
respectively.
(ii)
Find the value of the constant 𝑞.
7. The position vectors, relative to an origin 𝑂, of
three points 𝑃, 𝑄 and 𝑅 are 𝑖 + 3𝑗, 5𝑖 + 11𝑗
and 9𝑖 + 9𝑗 respectively.
(i)
By finding the magnitude of the
vectors ⃗⃗⃗⃗⃗
𝑃𝑅 , ⃗⃗⃗⃗⃗
𝑅𝑄 and ⃗⃗⃗⃗⃗
𝑄𝑃, show that
angle 𝑃𝑄𝑅 is 90°.
(ii)
Find the unit vector parallel to ⃗⃗⃗⃗⃗
𝑃𝑅 .
⃗⃗⃗⃗⃗⃗ = 𝑚𝑂𝑃
⃗⃗⃗⃗⃗ + 𝑛𝑂𝑅
⃗⃗⃗⃗⃗ , where
(iii)
Given that 𝑂𝑄
𝑚 and 𝑛 are constants, find the value
of 𝑚 and 𝑛.
8. Given that 𝑶𝑨 = 2𝒊 + 𝒋 and 𝑶𝑩 = 𝒊 + 3𝒋,
show that 𝑨𝑩 is perpendicular to 𝑶𝑨.
9. In a triangle 𝐴𝐵𝐶, the position vectors of 𝐴, 𝐵
and 𝐶 are respectively
𝒊 + 𝒋,
3𝒊 + 4𝒋,
and
4𝒊 − 𝒋
⃗⃗⃗⃗⃗ and 𝐴𝐶
⃗⃗⃗⃗⃗
(i) Find 𝐵𝐴
(ii) Show that ∠𝐵𝐴𝐶 = 90°
10. (a) The position vectors of points 𝐴 and 𝐵
with respect to the origin 𝑂 are given by
⃗⃗⃗⃗⃗ = 3𝒊 + 2𝒋 and 𝑂𝐵
⃗⃗⃗⃗⃗ = 2𝒊 − 4𝒋.
𝑂𝐴
Find
⃗⃗⃗⃗⃗ in terms of 𝒊 and 𝒋
(i) 𝐴𝐵
⃗⃗⃗⃗⃗
(ii) The magnitude of 𝐴𝐵
⃗⃗⃗⃗⃗ is perpendicular
(b) Determine whether 𝑂𝐴
to ⃗⃗⃗⃗⃗
𝑂𝐵.
11. The position vectors of three points 𝐴, 𝐵 and 𝐶
with respect to a fixed origin 𝑂 are
2𝑖 − 2𝑗 + 𝑘, 4𝑖 + 2𝑗 + 𝑘 and 𝑖 + 𝑗 + 3𝑘
respectively. Find the unit vectors in the
⃗⃗⃗⃗⃗ and 𝐶𝐵
⃗⃗⃗⃗⃗ .
directions of 𝐶𝐴
Calculate angle 𝐴𝐶𝐵 in degrees, correct to 1
decimal place.
12. Relative to an origin 𝑂, the position vectors of
points 𝐴 and 𝐵 are 2𝑖 + 𝑗 + 2𝑘 and
3𝑖 − 2𝑗 + 𝑝𝑘 respectively.
(i) Find the value of 𝑝 for which 𝑂𝐴 and 𝑂𝐵
are perpendicular.
(ii) In the case where 𝑝 = 6, use a scalar
product to find angle 𝐴𝑂𝐵, correct to the
nearest degree.
⃗⃗⃗⃗⃗ in terms of 𝑝 and
(iii) Express the vector 𝐴𝐵
hence find the values of 𝑝 for which the
length of 𝐴𝐵 is 3.5 units.
13. Relative to an origin 𝑂, the position vectors of
the points 𝐴 and 𝐵 are given by
⃗⃗⃗⃗⃗
𝑂𝐴 = 2𝑖 − 8𝑗 + 4𝑘 and ⃗⃗⃗⃗⃗
𝑂𝐵 = 7𝑖 + 2𝑗 − 𝑘
⃗⃗⃗⃗⃗ . 𝑂𝐵
⃗⃗⃗⃗⃗ and hence
(i)
Find the value of 𝑂𝐴
state whether angle 𝐴𝑂𝐵 is acute,
obtuse or a right angle.
⃗⃗⃗⃗⃗ = 2 𝐴𝐵
⃗⃗⃗⃗⃗ .
(ii)
The point 𝑋 is such that 𝐴𝑋
5
Find the unit vector in the direction
of 𝑂𝑋.
SOLUTIONS
2.
⃗⃗⃗⃗⃗ = 𝑖 + 2𝑗, 𝑂𝐵
⃗⃗⃗⃗⃗ = 2𝑖 + 5𝑗, 𝑂𝐶
⃗⃗⃗⃗⃗ = −4𝑗
(i) 𝑂𝐴
⃗⃗⃗⃗⃗⃗
(ii) 𝑂𝐷 = 𝑖 − 𝑗
1
⃗⃗⃗⃗⃗ = 12𝑖 + 5𝑗
(i) 5 (3𝑖 + 4𝑗)
(ii) 𝑂𝐶
3.
⃗⃗⃗⃗⃗ = − 3 𝑖 + 4 𝑗
𝑂𝐶
4.
(i) Unit vector: 29 (
5.
𝑘 = −3
6.
(i) 𝑝 =
7.
1 8
2
(i) (ii) Unit vector: 10 ( ) (iii) 𝑚 = 3, 𝑛 = 9
6
1.
5
5
1
24
25
21
)
−20
⃗⃗⃗⃗⃗ = ( 46 )
(ii) 𝑂𝐶
−35
(ii) 𝑞 = 5
8.
9.
⃗⃗⃗⃗⃗ = −2𝑖 − 3𝑗, ⃗⃗⃗⃗⃗
𝐵𝐴
𝐴𝐶 = 3𝑖 − 2𝑗
⃗⃗⃗⃗⃗
10. (a) (i) 𝐴𝐵 = −𝑖 − 6𝑗 (ii) √37
11.
(b) not perpendicular
1
1
(𝑖 − 3𝑗 − 2𝑘),
(3𝑖 + 𝑗 − 2𝑘) 𝜃 = 73.4°
√14
√14
12. (i) 𝑝 = −2 (ii) 𝜃 = 40° (iii) 𝑝 = 0.5, 3.5
4
1
13. (i) Obtuse (ii) ⃗⃗⃗⃗⃗
𝑂𝑋 = 6 (−4)
2
76
CHAPTER 13: VECTORS
EXAM QUESTIONS
1.
In a triangle, the position vectors of 𝐴, 𝐵 and 𝐶
are respectively 𝑖 + 𝑗, 3𝑖 + 4𝑗 and 4𝑖 − 𝑗
(i)
(ii)
2.
3.
4.
5.
(b) 𝑅, where 𝑅 is such that 𝑃𝑄𝑅𝑂,
labelled clockwise, forms a
parallelogram.
[3]
CAPE 2009
⃗⃗⃗⃗⃗ and 𝐴𝐶
⃗⃗⃗⃗⃗ .
Find 𝐵𝐴
Show that ∠𝐵𝐴𝐶 = 90°.
[2]
[2]
CAPE 2000
If the position vector of the point 𝐴 is 𝑖 − 3𝑗
and the position vector of the point 𝐵 is
2𝑖 + 5𝑗, find
⃗⃗⃗⃗⃗ |
(i)
|𝐴𝐵
[4]
(ii)
the position vector of the mid – point
⃗⃗⃗⃗⃗ .
of 𝐴𝐵
[3]
CAPE 2002
7.
The vectors 𝑝 and 𝑞 are given by𝑝 = 6𝑖 + 4𝑗
and 𝑞 = −8𝑖 − 9𝑗
(i)
Calculate, in degrees, the angle
between 𝑝 and 𝑞.
[5]
(ii)
(a) Find a non – zero vector 𝑣 such
that 𝑝. 𝑣 = 0.
(b) State the relationship between 𝑝
and 𝑣.
[5]
CAPE 2010
SOLUTIONS
1.
⃗⃗⃗⃗⃗ = −2𝑖 − 3𝑗, 𝐴𝐶
⃗⃗⃗⃗⃗ = 3𝑖 − 2𝑗
(i) 𝐵𝐴
⃗⃗⃗⃗⃗ | = √65, (ii) 𝑂𝑀
⃗⃗⃗⃗⃗⃗ = 3 𝑖 + 𝑗
(i) |𝐴𝐵
The position vector of a point 𝑃 is 𝑖 + 3𝑗. Find
⃗⃗⃗⃗⃗ . [2]
(a) the unit vector in the direction of 𝑂𝑃
⃗⃗⃗⃗⃗
(b) the position vector of a point 𝑄 on 𝑂𝑃
⃗⃗⃗⃗⃗⃗
produced such that |𝑂𝑄 | = 5.
[2]
(c) the value of 𝑡 such that the vector 3𝑡𝑖 + 4𝑗
⃗⃗⃗⃗⃗ .
is perpendicular to the vector 𝑂𝑃
[2]
CAPE 2004
2.
Given the vectors 𝑝 = 2𝑖 + 3𝑗 and 𝑞 = 3𝑖 − 2𝑗,
(a) Find 𝑥, 𝑦 ∈ ℝ such that
𝑥𝑝 + 𝑦𝑞 = −3𝑖 − 11𝑗.
[7]
(b) Show that 𝑝 and 𝑞 are perpendicular. [2]
CAPE 2006
6.
The position vectors of points 𝐴 and 𝐵 with
respect to an origin 𝑂 are given by
⃗⃗⃗⃗⃗ = 3𝑖 + 2𝑗 and 𝑂𝐵
⃗⃗⃗⃗⃗ = 2𝑖 − 4𝑗. Find
𝑂𝐴
…………………………………………………………………………
2
1
3.
⃗⃗⃗⃗⃗⃗ = √10 (𝑖 + 3𝑗)
(𝑖 + 3𝑗) (b) 𝑂𝑄
(a)
2
√10
4.
(c) 𝑡 = −4
(a) 𝑥 = −3, 𝑦 = 1
⃗⃗⃗⃗⃗ = −𝑖 − 6𝑗
(i) 𝐴𝐵
5.
7.
⃗⃗⃗⃗⃗ | = √37 not perpendicular
(ii) |𝐴𝐵
(i) (a) 30.01°,
(b) (i) 13 units 2
⃗⃗⃗⃗⃗⃗ = 𝑖 + 7𝑗 (b) 𝑂𝑅
⃗⃗⃗⃗⃗ = 4𝑖 + 2𝑗
(ii) (a) 𝑂𝑀
(i) 165.32°
(ii) (a) 𝑣 = 2𝑖 − 3𝑗 (1 possibility)
(b)⊥
⃗⃗⃗⃗⃗ in terms of 𝑖 and 𝑗
(i)
𝐴𝐵
[2]
⃗⃗⃗⃗⃗
(ii)
the magnitude of 𝐴𝐵
[2]
⃗⃗⃗⃗⃗
Determine whether 𝑂𝐴 is perpendicular to
⃗⃗⃗⃗⃗ .
𝑂𝐵
[3]
CAPE 2007
6.
The points 𝑃 and 𝑄 have position vectors
relative to the origin 𝑂 given respectively by
𝑝 = −𝑖 + 6𝑗 and 𝑞 = 3𝑖 + 8𝑗.
(i) (a) Calculate, in degrees, the size of the
acute angle 𝜃 between 𝑝 and 𝑞.
[5]
(b) Hence, calculate the area of triangle
𝑃𝑂𝑄.
[2]
(ii) Find, in terms of 𝑖 and 𝑗, the position
vector of
(a) 𝑀, where 𝑀 is the midpoint of 𝑃𝑄
[2]
77
CHAPTER 13: VECTORS
THE VECTOR EQUATION OF A LINE
INTRODUCTION
The equation of a line can be written in the form
𝑟 = 𝑎 + 𝜆𝑏
where 𝑎 is the position vector of any point on the
line and 𝑏 is any vector parallel to the line. 𝑏 is
referred to as the direction vector.
LESSON 9a
Determine the vector equation of
the line which passes through the point 𝐴(2, 9)
⃗⃗⃗⃗⃗ = (−1).
and is parallel to the vector 𝐴𝐵
−2
SOLUTION
2
⃗⃗⃗⃗⃗
𝑂𝐴 = 𝑎 = ( )
9
−1
𝑏=( )
−2
2
−1
𝑟 = ( )+ 𝜆( )
9
−2
𝑟 = 2𝒊 + 9𝒋 + 𝜆(−𝒊 − 2𝒋)
LESSON 9b
Determine the vector equation of
the line which is parallel to the vector
3𝒊 + 2𝒋 − 4𝒌 and passes through the point with
position vector 𝒊 − 2𝒋 + 𝒌.
SOLUTION
𝑎 = 𝒊 − 2𝒋 + 𝒌
𝑏 = 3𝒊 + 2𝒋 − 4𝒌
𝑟 = 𝒊 − 2𝒋 + 𝒌 + 𝜆(3𝒊 + 2𝒋 − 4𝒌)
LESSON 9C
The line 𝑙 passes through the
points 𝐴(2, −3, 1) and 𝐵(4, 0, −5). Determine the
vector equation of 𝑙.
SOLUTION
2
𝑎 = (−3)
1
⃗⃗⃗⃗⃗
𝑏 = 𝐴𝐵
⃗⃗⃗⃗⃗ = 𝐴𝑂 + 𝑂𝐵
𝐴𝐵
4
2
= − (−3) + ( 0 )
1
−5
2
=( 3 )
−6
2
2
𝑙 = (−3) + 𝜆 ( 3 )
1
−6
NB: 𝐵 could have been used as the needed
⃗⃗⃗⃗⃗ as the required parallel
position vector and 𝐵𝐴
vector.
LESSON 10
Determine the vector equation of
the line which passes through (2, −3, 1) and is
parallel to the vector 𝒊 − 𝒋 − 2𝒌 in
(i)
Vector form
(ii)
Parametric form and
(iii)
Cartesian form.
SOLUTION
Vector Form
2
1
𝑟 = (−3) + 𝜆 (−1)
1
−2
Parametric Form
𝑥
Let 𝑟 = (𝑦 )
𝑧
𝑥
2
1
(𝑦 ) = (−3) + 𝜆 (−1)
𝑧
1
−2
2+𝜆
= (−3 − 𝜆)
1 + 2𝜆
𝑥 =2+𝜆
(1)
𝑦 = −3 − 𝜆
(2)
𝑧 = 1 − 2𝜆
(3)
Cartesian Form
We need to eliminate the parameter.
From (1): 𝑥 − 2 = 𝜆
(2): −3 − 𝑦 = 𝜆
𝑧−1
(3): −2 = 𝜆
Since 𝜆 = 𝜆 = 𝜆
(𝑧 − 1)
𝑥 − 2 = −3 − 𝑦 =
−2
LESSON 11
Show that the following pair of
lines is parallel.
𝐿: 5𝒊 + 3𝒋 + 4𝒌 + 𝜆(−𝒊 + 2𝒋 + 3𝒌)
𝑁: 𝑟 = 4𝒊 − 2𝒋 + 𝒌 + 𝜇(3𝒊 − 6𝒋 − 9𝒌)
SOLUTION
We simply need to show that the
two direction vectors are parallel.
Since 3𝒊 − 6𝒋 − 9𝒌 = 3(−𝒊 + 2𝒋 + 3𝒌) the two
lines are parallel.
78
CHAPTER 13: VECTORS
LESSON 12
Show that the following pair of
lines intersect and determine the point of
intersection.
𝐿: 𝑟 = 4𝒊 − 3𝒋 + 𝒌 + 𝜆(𝒊 + 2𝒋 − 𝒌)
𝑁: 𝑟 = 2𝒊 + 6𝒋 − 𝒌 + 𝜇(−5𝒊 + 3𝒋 + 𝒌)
SOLUTION
There must exist 𝜆 and 𝜇 such
that
4𝒊 − 3𝒋 + 𝒌 + 𝜆(𝒊 + 2𝒋 − 𝒌)
= 2𝒊 + 6𝒋 − 𝒌 + 𝜇(−5𝒊 + 3𝒋 + 𝒌)
Equating coefficients of 𝒊:
4 + 𝜆 = 2 − 5𝜇
𝜆 + 5𝜇 = −2
(1)
Equating coefficients of 𝒋:
−3 + 2𝜆 = 6 + 3𝜇
2𝜆 − 3𝜇 = 9
(2)
Solving (1) and (2)
(1) × 2: 2𝜆 + 10𝜇 = −4
2𝜆 − 3𝜇 = 9
13𝜇 = −13
𝜇 = −1
𝜆=3
We will now equate the coefficients of 𝑘 to
determine if the values of 𝜆 and 𝜇 are consistent.
Equating coefficients of 𝒌:
1 − 𝜆 = −1 + 𝜇
1 − 3 = −1 + (−1)
−2 = −2
The values are consistent therefore 𝐿 and 𝑁
intersect.
We simply substitute 𝜆 = 3 into 𝐿 or 𝜇 = −1 into
𝑁 to determine the point of intersection.
4𝒊 − 3𝒋 + 𝒌 + 3(𝒊 + 2𝒋 − 𝒌) = 7𝒊 + 3𝒋 − 2𝒌
7
Point is ( 3 )
−2
NB: If the values for 𝜆 and 𝜇 are inconsistent and
the lines are not parallel they are referred to as
SKEWED.
VECTOR EQUATION OF A PLANE
The vector equation of a plane can be written as
𝑟 = 𝒂 + 𝜆𝒃 + 𝜇𝒄
where 𝑎 is a position vector of a point on the plane
and 𝑏 and 𝑐 are vectors parallel to the plane.
LESSON 13
Determine the equation of the
plane, in vector form and Cartesian form, which
contains the point (2, −3, 1) with normal
𝒊 − 2𝒋 + 3𝒌.
SOLUTION
The equation of a plane can be written in the form
𝑟. 𝑛 = 𝑎. 𝑛
where 𝑛 is a vector perpendicular to the plane and
𝑎 is a position vector of a point on the plane.
1
2
1
𝑟. ( −2) = (−3) . (−2)
3
1
3
1
𝑟. ( −2) = 2(1) + (−3)(−2) + 1(3)
3
1
𝑟. ( −2) = 11
3
Cartesian form
𝑥
Let 𝑟 = (𝑦 )
𝑧
𝑥
1
(𝑦 ) . (−2) = 11
𝑧
3
𝑥 − 2𝑦 + 3𝑧 = 11
…………………………………………………………………………
EXERCISE 13.2
1.
Determine the equation of the line which
passes through the point
(i) 𝐴(−1, 1, −3) and is parallel to the vector
2
( 0 ).
−1
(ii) 𝐵 (4, 3, −2) and is parallel to the vector
−9
(−2).
1
(iii) 𝐶(−3, 4, 3) and is parallel to the vector
8
(−3).
−7
(iv) 𝐷(5, 2, −3) and is parallel to the vector
−4
(−5).
7
(v) 𝐸(3, 5, −5) and is parallel to the vector
−8
(−8).
3
79
CHAPTER 13: VECTORS
2.
Find the equation of the line which passes
through the points 𝐴 and 𝐵 with position
vectors
(i)
5
0
(−4) and (−3) respectively.
−4
3
(ii)
5
−1
(2) and ( 0 ) respectively.
−4
2
(iii)
(iv)
(v)
−3
−2
( 0 ) and ( 0 ) respectively.
3
−2
4
3
( 0 ) and (3) respectively.
−5
0
−4
5
( 5 ) and ( 5 ) respectively.
0
−5
3.
The points 𝐴 and 𝐵 have position vectors
2𝑖 − 9𝑗 − 5𝑘 and 10𝑖 − 3𝑗 − 𝑘 respectively,
relative to an origin 𝑂. The line 𝑙 passes
through 𝐴 and 𝐵. Obtain a vector equation of 𝑙.
4.
Determine the vector equation of the line
which passes through (8, −3, 2) and is parallel
to the vector 𝒊 + 3𝒋 − 2𝒌 in
(i)
(ii)
(iii)
5.
6.
7.
Vector form
Parametric form and
Cartesian form.
The line 𝑙1 has equation
𝑟 = 2𝑖 + 3𝑗 − 4𝑘 + 𝜆(𝑖 + 2𝑗 + 𝑘) where 𝜆 is a
scalar parameter.
The line 𝑙2 has equation
𝑟 = 9𝑗 − 3𝑘 + 𝜇(5𝑖 + 2𝑘) where 𝜇 is a scalar
parameter.
Given that 𝑙1 and 𝑙2 meet at the point 𝐶, find
the coordinates of 𝐶.
The quadrilateral 𝐴𝐵𝐶𝐷 has vertices
𝐴(2, 1, 3), 𝐵(6, 5, 3), 𝐶(6, 1, −1) and
𝐷(2, −3, −1). The line 𝐿 has vector equation
6
1
𝑟 = ( 1 ) + 𝜆 (1 ).
−1
0
⃗⃗⃗⃗⃗ .
(a) (i) Find the vector 𝐴𝐵
(ii) Show that the line 𝐴𝐵 is parallel to 𝐿.
(iii) Verify that 𝐷 lies on 𝐿.
8.
9.
(b) The line 𝑀 passes through 𝐷(2, −3, −1)
and 𝐸(4, 1, 1).
(i)
Find the vector equation of 𝑀.
(ii)
Find the angle between 𝑀 and
𝐴𝐶.
The position vectors of the points 𝑃 and 𝑄
with respect to an origin 𝑂 are 5𝒊 + 2𝒋 − 9𝒌
and 4𝒊 + 4𝒋 − 6𝒌 respectively.
(i)
Find a vector equation for the line
𝑃𝑄.
The position vector of the point 𝑇 is
𝒊 + 2𝒋 − 𝒌.
(ii)
Write down a vector equation for the
line 𝑂𝑇 and show that 𝑂𝑇 is
perpendicular to 𝑃𝑄.
It is given that 𝑂𝑇 intersects 𝑃𝑄.
(iii)
Find the position vector of the point
of intersection of 𝑂𝑇 and 𝑃𝑄.
Lines 𝐿1 , 𝐿2 and 𝐿3 have vector equations
𝐿1 : 𝒓 = (5𝒊 − 𝒋 − 2𝒌) + 𝑠(−6𝒊 + 8𝒋 − 2𝒌)
𝐿2 : 𝒓 = (3𝒊 − 8𝒋) + 𝑡(𝒊 + 3𝒋 + 2𝒌)
𝐿3 : 𝒓 = (2𝒊 + 𝒋 + 3𝒌) + 𝜇(3𝒊 + 𝑐𝒋 + 𝒌)
(i) Calculate the acute angle between 𝐿1 and
𝐿2 .
(ii) Given that 𝐿1 and 𝐿3 are parallel, find the
value of 𝑐.
(iii) Given instead that 𝐿2 and 𝐿3 intersect,
find the value of 𝑐.
(i) Show that the straight line with equation
2
1
𝒓 = (−3) + 𝑡 ( 4 ) meets the line
5
−2
passing through (9, 7, 5) and (7, 8, 2), and
find the point of intersection of these
lines.
(ii) Find the acute angle between these lines.
10. Find the equation of the plane, in vector form
and Cartesian form, through the point
(−4, 3, 1) that is perpendicular to the vector
𝑎 = −4𝑖 + 7𝑗– 2𝑘.
11. Find an equation of the plane through the
point (6, 3, 2) and perpendicular to the vector
(−2, 1, 5). Check if (2, −1, 0) and (1, −2, 1) are
in that plane.
12. Find an equation of the plane through the
point (4, −2, 3) and parallel to the plane
3𝑥 − 7𝑧 = 12.
13. Find an equation for the plane through
𝑃(1, −1, 3) parallel to the plane
3𝑥 + 𝑦 + 𝑧 = 7.
14. Find an equation of the plane through the
points (0, 1, 1), (1, 0, 1) and (1, 1, 0).
80
CHAPTER 13: VECTORS
SOLUTIONS
1.
2.
3.
4.
(i) 68.5° (ii) 𝑐 = −4 (iii) 𝑐 = 5
−1
2
(i) 𝑟 = ( 1 ) + 𝜆 ( 0 )
3
−1
4
−9
(ii) 𝑟 = ( 3 ) + 𝜆 ( −2)
−2
1
−3
8
(iii) 𝑟 = ( 4 ) + 𝜆 (−3)
3
−7
5
−4
(iv) 𝑟 = ( 2 ) + 𝜆 (−5)
7
−3
3
−8
(v) 𝑟 = ( 5 ) + 𝜆 ( −8)
−5
3
5
−5
(i) 𝑟 = (−4) + 𝜆 ( 1 )
3
−7
5
−6
(ii) 𝑟 = (2) + 𝜆 ( −2)
−6
2
−3
1
(iii) 𝑟 = ( 0 ) + 𝜆 ( 0 )
3
−5
3
1
(iv) 𝑟 = ( 0 ) + 𝜆 (3)
−5
5
5
−9
(v) 𝑟 = ( 5 ) + 𝜆 ( 0 )
5
−5
𝑙 = 2𝑖 − 9𝑗 − 5𝑘 + 𝜆 (8𝑖 + 6𝑗 + 4𝑘 )
9.
(i)
8
1
(i) 𝑟 = (−3) + 𝜆 ( 3 ),
2
−2
2.
𝑥 = 8+𝜆
(ii) 𝑦 = −3 + 3𝜆
𝑧 = 2 − 2𝜆
5.
6.
7.
3
8.
(iii) 𝑥 − 8 =
5
𝐶 = (9)
5
4
(a) (i) ( 4)
0
𝑦+3
3
=
(ii)
(ii) 62.2°
−4
10. 𝑟. ( 7 ) = 35,
−4𝑥 + 7𝑦 − 2𝑧 = 35
−2
−2
11. 𝑟. ( 1 ) = 1, No, Yes
5
3
12. 𝑟. ( 0 ) = −9
−7
3
13. 𝑟. ( 1) = 7
1
1
14. 𝑟. ( 1) = 2
1
EXAM QUESTIONS
1. The points 𝐴(3, −1, 2), 𝐵(1, 2, −4) and
𝐶(−1, 1, −2) are three vertices of a
parallelogram 𝐴𝐵𝐶𝐷.
⃗⃗⃗⃗⃗ and 𝐵𝐶
⃗⃗⃗⃗⃗ in the
(i) Express the vectors 𝐴𝐵
form 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘.
[3]
(ii) Show that the vector 𝑟 = −16𝑗 − 8𝑘 is
perpendicular to the plane through 𝐴, 𝐵
and 𝐶.
[5]
(iii) Hence, find the Cartesian equation
through 𝐴, 𝐵 and 𝐶.
[4]
CAPE 2013
The points 𝑃 (3, −2, 1) , 𝑄( −1, 𝜆, 5) and
𝑅(2, 1, −4) are three vertices of a triangle
𝑃𝑄𝑅.
⃗⃗⃗⃗⃗ , 𝑄𝑅
⃗⃗⃗⃗⃗ and
Express EACH of the vectors 𝑃𝑄
⃗⃗⃗⃗⃗ in the form 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘.
𝑅𝑃
[4]
(ii) Hence, find the value of 𝜆, given that
𝑃𝑄𝑅 is right – angled with the side 𝑃𝑄 as
hypotenuse.
[6]
CAPE 2014
(i)
2−𝑧
2
(iii)
2
2
(b) (i) 𝑟 = (−3) + 𝜆 (4)
(ii) 90°
−1
2
(i) 𝑟 = 5𝑖 + 2𝑗 − 9𝑘 + 𝜆(−𝑖 + 2𝑗 + 3𝑘)
3
(ii) 𝑟 = 𝜇(𝑖 + 2𝑗 − 𝑘) (iii) ( 6 )
−3
SOLUTIONS
⃗⃗⃗⃗⃗ = −2𝑖 + 3𝑗 − 6𝑘,
1. (i) 𝐴𝐵
⃗⃗⃗⃗⃗ = −2𝑖 − 𝑗 + 2𝑘
𝐵𝐶
(ii)
(iii) 2𝑦 + 𝑧 = 0
⃗⃗⃗⃗⃗ = −4𝑖 + (2 + 𝜆)𝑗 + 4𝑘,
2. (i) 𝑃𝑄
⃗⃗⃗⃗⃗ = 3𝑖 + (1 − 𝜆) (ii) 𝜆 = 15
𝑄𝑅
…………………………………………………………………………
81
CHAPTER 14: THE EQUATION OF A CIRCLE
CHAPTER 14: THE EQUATION OF A CIRCLE
At the end of this section, students should be able
to:
 find the equations of tangents and normal
to circles;
 find the points of intersection of a curve
with a straight line;
 find the points of intersection of two
curves;
 determine the equation of a circle given
three points on the circle’s circumference
Therefore, the equation of a circle with radius 𝑟
and centre at (ℎ, 𝑘) is
(𝑥 − ℎ) 2 +(𝑦 − 𝑘)2 = 𝑟 2
𝑟>0
The equation of a circle with radius 𝑟 and centre at
(0, 0):
𝑥 2 + 𝑦2 = 𝑟 2
𝑟>0
The equation of a circle can also be written in the
form
𝑥 2 + 𝑦 2 + 2ℎ𝑥 + 2𝑘𝑦 + 𝑐 = 0 where
𝑐 = ℎ2 + 𝑘 2 − 𝑟 2
__________________________________________________________
For the line 𝐴𝐵 above we have the following
formulae:
DETERMINING THE EQUATION OF A
CIRCLE
𝑥1 + 𝑥2 𝑦1 + 𝑦2
,
)
2
2
Length of 𝐴𝐵 = √(𝑥1 − 𝑥2 )2 + (𝑦1 − 𝑦2 )2
𝑦1 − 𝑦2
Gradient of 𝐴𝐵 =
𝑥1 − 𝑥2
Generally, we have the equation of a line is of the
form
𝑦 = 𝑚𝑥 + 𝑐
Standard Form
where 𝑚 is the gradient of the line and 𝑐 is the 𝑦intercept i.e. where the line cuts the 𝑦-axis.
NEED TO KNOW
Two lines are parallel if they have the same
gradient.
Two lines are perpendicular if the product of their
gradients is −1.
Let’s find the equation of a
circle with radius 𝑟(𝑟 > 0)
and centre 𝐶(ℎ, 𝑘) in a
rectangular coordinate
system. The point 𝑃(𝑥, 𝑦) is
on the circle if and only if
the distance 𝑃𝐶 = 𝑟; that
is, if and only if
√(𝑥 − ℎ) 2 +(𝑦 − 𝑘)2 = 𝑟
𝑟>0
or equivalently,
(𝑥 − ℎ)2 +(𝑦 − 𝑘)2 = 𝑟 2
𝑟>0
Mid − point of 𝐴𝐵 = (
LESSON 1
Find the equation of the circle
with radius 4 and centre at:
(a) (−3, 6)
(b) (0, 0)
SOLUTION
(a) (ℎ, 𝑘) = (−3, 6) and 𝑟 = 4;
(𝑥 − ℎ) 2 +(𝑦 − 𝑘)2 = 𝑟 2
[𝑥 − (−3)]2 +(𝑦 − 6) 2 = 42
(𝑥 + 3) 2 +(𝑦 − 6) 2 = 16
(b) (ℎ, 𝑘) = (0, 0)and 𝑟 = 4
𝑥 2 +𝑦 2 = 𝑟 2
𝑥 2 + 𝑦 2 = 42
𝑥 2 + 𝑦 2 = 16
FINDING THE CENTRE AND RADIUS
OF A CIRCLE
LESSON 2
Find the centre and radius of the
circle with equation
i. 𝑥 2 + 𝑦 2 + 6𝑥 − 4𝑦 = 23
ii. −3𝑥 2 − 3𝑦 2 + 21 = 0
SOLUTION
We transform the equation into
the form (𝑥 − ℎ) 2 +(𝑦 − 𝑘)2 = 𝑟 2 by completing
the square relative to 𝑥 and relative to 𝑦. From
this standard form we can determine the centre
and radius.
i. 𝑥 2 + 𝑦 2 + 6𝑥 − 4𝑦 = 23
(𝑥 2 + 6𝑥 ) + (𝑦 2 − 4𝑦 ) = 23
(𝑥 2 + 6𝑥 + 9) + (𝑦 2 − 4𝑦 + 4) = 23 + 9 + 4
(𝑥 + 3) 2 + (𝑦 − 2) 2 = 36
[𝑥 − (−3)]2 + (𝑦 − 2) 2 = 62
Centre: 𝐶(ℎ, 𝑘) = (−3, 2) Radius: 𝑟 = 6
ii. −3𝑥 2 − 3𝑦 2 + 21 = 0
𝑥 2 + 𝑦2 − 7 = 0
𝑥 2 + 𝑦2 = 7
(𝑥 − 0) 2 + (𝑦 − 0) 2 = 7
82
CHAPTER 14: THE EQUATION OF A CIRCLE
2
(𝑥 − 0) 2 + (𝑦 − 0) 2 = (√7)
Centre:
𝐶(ℎ, 𝑘) = (0, 0)
Radius:
𝑟 = √7
TANGENTS AND NORMALS
LESSON 3
The circle 𝐶 has equation
(𝑥 − 4) 2 + (𝑦 − 3) 2 = 25.
(i) State the radius and the coordinates of the
centre of 𝐶.
(ii) Find the equation of the tangent at the point
(8, 6) on 𝐶.
(iii) Calculate the coordinates of the points of
intersection of 𝐶 with the straight line
𝑦 = −𝑥 + 8.
SOLUTION
(i)
Centre (4, 3) radius is 5
6−3
3
(ii)
Gradient of radius is 8−4 = 4 ∴ gradient of
need to show that there is only one point of
contact.
𝑦 = 𝑥−1
(1)
𝑥 2 + 𝑦 2 + 12𝑥 + 2𝑦 = −19
(2)
Solving (1) and (2) simultaneously
𝑥 2 + (𝑥 − 1)2 + 12𝑥 + 2(𝑥 − 1) + 19 = 0
2𝑥 2 + 12𝑥 + 18 = 0
𝑥 2 + 6𝑥 + 9 = 0
(𝑥 + 3) 2 = 0
𝑥 = −3
𝑦 = −4
Since (−3, −4) is the only point of contact, the line
𝑦 = 𝑥 − 1 is a tangent to the curve.
4
tangent is − 3
(iii)
4
𝑦 = 𝑚𝑥 + 𝑐
using (8, 6) and 𝑚 = −
3
4
6 = − (8) + 𝑐
3
50
=𝑐
3
4
50
𝑦=− 𝑥+
3
3
Solving the two equations simultaneously
(𝑥 − 4) 2 + (𝑦 − 3) 2 = 25
(1)
𝑦 = −𝑥 + 8
(2)
Subbing (2) into (1)
(𝑥 − 4) 2 + (−𝑥 + 8 − 3)2 = 25
𝑥 2 − 8𝑥 + 16 + 𝑥 2 − 10𝑥 + 25 = 25
2𝑥 2 − 18𝑥 + 16 = 0
𝑥 2 − 9𝑥 + 8 = 0
(𝑥 − 8)(𝑥 − 1) = 0
𝑥 = 1, 8
𝑦 = 7, 0
(1, 7) and (8, 0)
LESSON 5
A curve 𝐶 is defined by the
equation 𝑥 2 + 𝑦 2 − 6𝑥 − 2𝑦 + 1 = 0.
(i) Show that the centre and the radius of the
circle, 𝐶, are (3, 1) and 3 respectively.
(ii) (a) Find the equation of the normal to the
circle at the point (6, 1).
(b) Show that the tangent to the circle at the
point (6, 1) is parallel to the 𝑦-axis.
SOLUTION
(i)
𝑥 2 + 𝑦 2 − 6𝑥 − 2𝑦 = −1
𝑥 2 − 6𝑥 + 9 + 𝑦 2 − 2𝑦 + 1 = −1 + 9 + 1
(𝑥 − 3) 2 + (𝑦 − 1) 2 = 9
Centre (3, 1) radius is 3
1−1
(ii)
(a) 𝑚 =
=0
3−6
Gradient of normal is undefined
Equation of normal is 𝑥 = 6
(b) Gradient of the 𝑥-axis is 0 and the
gradient of the tangent is 0, therefore,
they are parallel.
LESSON 4
Prove that the line 𝑦 = 𝑥 − 1 is a
tangent to the curve 𝑥 2 + 𝑦 2 + 12𝑥 + 2𝑦 = −19.
SOLUTION
A tangent is a straight line that
touches a curve at only one point. Therefore we
83
CHAPTER 14: THE EQUATION OF A CIRCLE
LESSON 6
Find the length of the tangent
drawn from the point 𝐴(−4, 2) to the circle
𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 = −8.
SOLUTION
𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 = −8
(𝑥 2 − 6𝑥 ) + (𝑦 2 − 4𝑦 ) = −8
(𝑥 2 − 6𝑥 + 9) + (𝑦 2 − 4𝑦 + 4) = −8 + 9 + 4
(𝑥 − 3) 2 + (𝑦 − 2) 2 = 5
Centre (3, 2) and radius is √5
Distance between centre and 𝐴 is 7 and the radius
is √5, so by Pythagoras’ Theorem
4.
5.
6.
2
72 = (√5) + 𝑥 2
49 − 5 = 𝑥 2
44 = 𝑥 2
2√11 = 𝑥
7.
Prove that the line 𝑥 + 𝑦 = 9 is a tangent to
the curve 𝑥 2 + 𝑦 2 − 4𝑥 − 2𝑦 = 13.
A curve 𝐶 is defined by the equation
𝑥 2 + 𝑦 2 + 8𝑥 + 6𝑦 − 15 = 0.
(i)
Determine the coordinates of the
centre of 𝐶 and its radius.
(ii)
Find the equation of the normal to the
circle at the point (2, −5).
The circle 𝐶 has equation
𝑥 2 + 𝑦 2 + 2𝑥 − 4𝑦 = 15.
(i)
Determine the radius and the
coordinates of the centre of 𝐶.
(ii)
Find the equation of the tangent at
the point (−5, 4) on 𝐶.
(iii)
Calculate the coordinates of the
points of intersection of 𝐶 with the
straight line 𝑦 = −𝑥 + 3.
Find the length of the tangent drawn from the
point 𝐴(−2, 4) to the circle
𝑥 2 + 𝑦 2 − 10𝑥 + 4𝑦 = −11.
SOLUTIONS
1.
(a) 𝑥 2 + 𝑦 2 = 64
(b) 𝑥 2 + 𝑦 2 = 81
49
…………………………………………………………………………..
(c) 𝑥 2 + 𝑦 2 = 81
EXERCISE 14.1
(d) 𝑥 2 + 𝑦 2 = 107 + 20√7
(e) (𝑥 − 1)2 + (𝑦 − 3) 2 = 9
(f) (𝑥 − 1) 2 + (𝑦 + 3) 2 = 25
(ii) 10
(iii) (7, −2)
2
(𝑥
(iv) − 7) + (𝑦 + 2)2 = 25
(a) 𝐶(0, 0), 𝑟 = 9
(b) 𝐶(0, 0), 𝑟 = 7
1.
2.
3.
Find the equations of the circles with the
following radii and centres
(a) Centre (0, 0), radius =8
(b) Centre (0, 0), radius = 9
7
(c) Centre (0, 0), radius = 9
(d) 𝐶(0, 0),radius = 10 + √7
(e) Centre (1, 3), radius = 3
(f) Centre (1, −3), radius = 5
The points 𝐴, 𝐵 and 𝐶 have coordinates
(3, −5), (4, −6) and (11, 1) respectively.
(i)
Show that 𝐴𝐵 is perpendicular to 𝐵𝐶.
(ii)
Find the length of 𝐴𝐶.
(iii)
Find the coordinates of the mid –
point of 𝐴𝐶.
(iv)
Hence find the equation of the circle
which passes through the points 𝐴, 𝐵
and 𝐶.
Find the centre and radius of the circle with
equation
(a) 𝑥 2 + 𝑦 2 = 81
(b) 𝑥 2 + 𝑦 2 = 49
1
(c) 𝑥 2 + 𝑦 2 =
64
(d) (𝑥 − 2) 2 + (𝑦 + 1) 2 = 25
(e) 𝑥 2 + 𝑦 2 − 4𝑥 + 10𝑦 + 13 = 0
(f) 𝑥 2 + 𝑦 2 + 6𝑦 + 2 = 0
2.
3.
1
4.
5.
(c) 𝐶(0, 0), 𝑟 = 8
(d) 𝐶(2, −1), 𝑟 = 5
(e) 𝐶 (2, −5) , 𝑟 = 4
(f) 𝐶 (0, −3), 𝑟 = √7
(i) 𝐶(−4, −3), 𝑟 = √40
1
13
(ii) 𝑦 = − 3 𝑥 − 3
6.
7.
(i) 𝐶(−1, 2), 𝑟 = √20
(ii) 𝑦 = 2𝑥 + 14
(iii) (−3, 6) (3, 0)
√67
…………………………………………………………………………..
84
CHAPTER 14: THE EQUATION OF A CIRCLE
THE INTERSECTION OF TWO
CIRCLES
LESSON 7
Find the points of intersection of
the circles with equations
𝑥 2 + 𝑦 2 − 2𝑥 − 8𝑦 = −7 and
𝑥 2 + 𝑦 2 − 12𝑥 + 2𝑦 = −17.
SOLUTION
We need to solve both equations
simultaneously.
𝑥 2 + 𝑦 2 − 2𝑥 − 8𝑦 = −7
(1)
𝑥 2 + 𝑦 2 − 12𝑥 + 2𝑦 = −17
(2)
(1) – (2)
−2𝑥 − (−12𝑥) − 8𝑦 − 2𝑦 = −7 − (−17)
10𝑥 − 10𝑦 = 10
𝑥 −𝑦 = 1
𝑥 =𝑦+1
(3)
Sub. (3) into (1)
(𝑦 + 1)2 + 𝑦 2 − 2(𝑦 + 1) − 8𝑦 = −7
𝑦 2 + 2𝑦 + 1 + 𝑦 2 − 2𝑦 − 2 − 8𝑦 = −7
2𝑦 2 − 8𝑦 + 6 = 0
𝑦 2 − 4𝑦 + 3 = 0
(𝑦 − 1)(𝑦 − 3) = 0
𝑦 = 1, 3
𝑥 =𝑦+1
𝑥 =1+1
𝑥=2
(2, 1)
SOLUTION
The equation of a circle is (𝑥 − ℎ) 2 +(𝑦 − 𝑘)2 = 𝑟 2 ,
but for our purposes we will write it as
𝑥 2 − 2ℎ𝑥 + ℎ2 + 𝑦 2 − 2𝑦𝑘 + 𝑘 2 −𝑟 2 = 0 and let
ℎ2 + 𝑘 2 − 𝑟 2 = 𝑐
𝑥 2 − 2ℎ𝑥 + 𝑦 2 − 2𝑘𝑦 + 𝑐 = 0
Since the points line on the line they must satisfy the
equation of the line, therefore
(−1) 2 − 2(−1)ℎ + ℎ2 + 02 − 2(0)𝑘 + 𝑘 2 = 𝑟 2
1 + 2ℎ + ℎ2 + 𝑘 2 = 𝑟 2
1 + 2ℎ + 𝑐 = 0
2ℎ + 𝑐 = −1
(1)
12 − 2(1)(ℎ) + ℎ2 + 22 − 2(2)𝑘 + 𝑘 2 = 𝑟 2
1 − 2ℎ + ℎ2 + 4 − 4𝑘 + 𝑘 2 = 𝑟 2
1 − 2ℎ + 4 − 4𝑘 + 𝑐 = 0
−2ℎ − 4𝑘 + 𝑐 = −5
(2)
(−5)2 − 2(−5)ℎ + ℎ2 + 42 − 2(4)𝑘 + 𝑘 2 = 𝑟 2
25 + 10ℎ + ℎ2 + 16 − 8𝑘 + 𝑘 2 = 𝑟 2
10ℎ − 8𝑘 + 41 + 𝑐 = 0
10ℎ − 8𝑘 + 𝑐 = −41
(3)
From (1) we get 𝑐 = −1 − 2ℎ
Substituting 𝑐 in (2) we get: −ℎ − 𝑘 = −1 (4)
Substituting 𝑐 in (3)we get: ℎ − 𝑘 = −5 (5)
Solving (4) and (5) simultaneously we get 𝑘 = 3 and
ℎ = −2 ∴ 𝑐 = 3
Our equation is 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 + 3 = 0
…………………………………………………
𝑥 =𝑦+1
𝑥 =3+1
𝑥=4
(4, 3)
EXERCISE 14.2
1. Find the point(s) of intersection of the circles
with equations
(i) 𝑥 2 + 𝑦 2 + 10𝑥 − 4𝑦 = −9 and
𝑥 2 + 𝑦 2 + 6𝑥 − 2𝑦 = −5
(ii) 𝑥 2 + 𝑦 2 − 6𝑥 − 6𝑦 = 16 and
𝑥 2 + 𝑦 2 + 2𝑥 − 8𝑦 = 0
(iii) 𝑥 2 + 𝑦 2 − 6𝑥 = 59 and
𝑥 2 + 𝑦 2 + 10𝑦 = 9
(iv) 𝑥 2 + 𝑦 2 + 8𝑥 = 34 and
𝑥 2 + 𝑦 2 − 8𝑥 + 2𝑦 = 8
2. Find the equation of the circle which passes
through the points
(i) (−1, 2), (−3, 2) and (−3, 4)
(ii) (4, 1), (−4, 1) and (−2, −1)
(iii) (−4, 4), (1, 5) and (−5, 5)
(iv) (−1, −3), (−5, 5) and (−4, −2)
EQUATION OF A CIRCLE GIVEN 3
POINTS
SOLUTIONS
LESSON 8
Find the equation of the circle
which passes through the points (−1, 0), (1, 2) and
(−5, 4).
1.
2.
(i) (−1, 0) (ii) (0, 8) (−2, 0)
(iii) (5, 8) ( −5, −2) (iv) (1, −5) (2.08, 3.62)
(i) 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 = 11
(ii) 𝑥 2 + 𝑦 2 − 6𝑦 = 11
(iii) 𝑥 2 + 𝑦 2 + 4𝑥 − 14𝑦 = −40
(iv) 𝑥 2 + 𝑦 2 + 2𝑥 − 4𝑦 = 20
85
CHAPTER 14: THE EQUATION OF A CIRCLE
EXAM QUESTIONS
1.
2.
3.
The equation of the circle, 𝑄, with centre 𝑂 is
𝑥 2 + 𝑦 2 − 2𝑥 + 2𝑦 = 23.
(a) Express the equation of 𝑄 in the form
(𝑥 − 𝑎) 2 + (𝑦 − 𝑏)2 = 𝑐.
[5]
(b) Hence, or otherwise, state
a. the coordinates of the centre of 𝑄. [2]
b. the radius of 𝑄.
[1]
(c) Show that the point 𝐴(4, 3) lies on 𝑄. [3]
(d) Find the equation of the tangent to 𝑄 at
the point 𝐴.
[5]
(e) The centre of 𝑄 is the midpoint of its
diameter 𝐴𝐵. Find the coordinates of 𝐵.
[4]
CAPE 2005
The circle shown in the diagram below (not
drawn to scale) has
centre 𝐶 at (5, −4)
and touches the 𝑦 −
axis at the point 𝐷.
The circle cuts the
𝑥 −axis at points 𝐴
and 𝐵. The tangent at
𝐵 cuts the 𝑦 − axis at
the point 𝑃.
(a) Determine
(i) The length of the radius of the circle.
[2]
(ii) The equation of the circle.
[1]
(iii) The coordinates of the points 𝐴 and 𝐵,
at which the circle cuts the 𝑥 − axis.
[6]
(iv) The equation of the tangent at 𝐵. [4]
(v) The coordinates of 𝑃
[2]
(b) Show by calculation that 𝑃𝐷 = 𝑃𝐵
[5]
CAPE 2007
(a) In the Cartesian plane with origin 𝑂, the
coordinates of points 𝑃 and 𝑄 are
(−2, 0) and (8, 8) respectively. The
midpoint of 𝑃𝑄 is 𝑀.
(i) Find the equation of the line which
passes through 𝑀 and is
perpendicular to 𝑃𝑄.
[8]
(b) (i) Prove that the line 𝑦 = 𝑥 + 1 is a
tangent to the circle
𝑥 2 + 𝑦 2 + 10𝑥 − 12𝑦 + 11 = 0.
[6]
(ii) Find the coordinates of the point of
contact of this tangent to the circle.
[2]
CAPE 2008
4.
5.
6.
The circle 𝐶 has equation
(𝑥 − 3)2 + (𝑦 − 4) 2 = 25
(i) State the radius and the coordinates of
the centre of 𝐶.
[2]
(ii) Find the equation of the tangent at the
point (6, 8) on 𝐶.
[4]
(iii) Calculate the coordinates of the points of
intersection of 𝐶 with the straight line
𝑦 = 2𝑥 + 3.
[7]
CAPE 2009
The circle 𝐶1 has (−3, 4) and (1, 2) as
endpoints of a diameter.
Show that the equation of 𝐶1 is
𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5 = 0.
[6]
CAPE 2010
The circle 𝐶1 has (−3, 4) and (1, 2) as
endpoints of a diameter.
(i) Show that the equation of 𝐶1 is
𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5 = 0.
[6]
(ii) The circle 𝐶2 has equation
𝑥 2 + 𝑦 2 + 𝑥 − 5𝑦 = 0. Calculate the
coordinates of the points of intersection of
𝐶1 and 𝐶2 .
[9]
CAPE 2010
SOLUTIONS
2.
(a) (𝑥 − 1) 2 + (𝑦 + 1)2 = 25
(b) (i) 𝑂(1, −1)
(ii) 𝑟 = 5
3
(d) 𝑦 = − 4 𝑥 + 6
(e) (−2, −5)
(a) (i) 5
(ii) (𝑥 − 5)2 + (𝑦 + 4)2 = 25
3.
(iii) 𝐴(2, 0), 𝐵(8, 0) (iv) 𝑦 = − 4 𝑥 + 6
(v)(0, 6)
5
31
(a) (i) 𝑦 = − 𝑥 +
4.
(ii) 𝑥 2 + 𝑦 2 + 2𝑥 − 18𝑦 = 0
(b) (ii) (0, 1)
3
25
(i) 𝐶(3, 4) , 𝑟 = 5 (ii) 𝑦 = − 4 𝑥 + 2
1.
3
4
4
(iii) (−1, 1) (3, 9)
5.
6.
(i)
(ii) (−3, 2) (0, 5)
…………………………………………………………………………
86
CHAPTER 15: THE LOCUS OF A POINT
CHAPTER 15: THE LOCUS OF A POINT
At the end of this section, students should be able
to:

determine the loci of points satisfying
given properties.
__________________________________________________________
The locus of a point is the path which the point
follows as it obeys a particular rule.
√(−2 − 𝑥)2 + (𝑦 − 𝑦) 2 = √(𝑥 − 4)2 + (𝑦 − 0)2
√(−2 − 𝑥)2 + 02 = √𝑥 2 − 8𝑥 + 16 + 𝑦 2
√𝑥 2 + 4𝑥 + 4 = √𝑥 2 − 8𝑥 + 16 + 𝑦 2
𝑥 2 + 4𝑥 + 4 = 𝑥 2 − 8𝑥 + 16 + 𝑦 2
12𝑥 − 12 = 𝑦 2
This is the equation of a PARABOLA.
LESSON 1
Determine the equation of the
curve that is the locus of all points equidistant
from the two points (8, 3) and (2, 1).
SOLUTION
The first intuitive point which fits
the conditions would be the midpoint between the
two given points. From the diagram we see that
the locus of this point is the perpendicular
bisector of (8, 3) and (2, 1).
2+8 1+3
Midpoint = ( 2 , 2 ) = (5, 2)
1−3
1
Gradient = 2−8 = 3
⊥ gradient is −3
General form of equation of a line is
𝑦 = 𝑚𝑥 + 𝑐 (using 𝑚 = −3 and (5, 2))
2 = −3(5) + 𝑐
17 = 𝑐
𝑦 = −3𝑥 + 17
LESSON 2
Find the equation of the curve
which is the locus of the points equidistant from
the line 𝑥 = −2 and the point (4, 0).
SOLUTION
We require the distance of a
point (−2, 𝑦) on the line 𝑥 = −2 to an arbitrary
point (𝑥, 𝑦) to be the same as the distance from
(𝑥, 𝑦) to (4, 0).
LESSON 3
Determine the equation of the
curve which is the locus of the points 5 units from
the point (−2, 1).
SOLUTION
This locus is simply a circle with
centre (−2, 1) and radius 5.
2
(𝑥 − (−2)) + (𝑦 − 1)2 = 52
(𝑥 + 2) 2 + (𝑦 − 1) 2 = 25
…………………………………………………………………………..
EXERCISE 15
1.
Determine the equation of the curve that is
the locus of all points equidistant from the
two points (3, 0) and (−1, 4).
87
CHAPTER 15: THE LOCUS OF A POINT
2.
3.
4.
5.
6.
7.
8.
9.
Determine the equation of the curve that is
the locus of all points equidistant from the
two points (−4, −4) and (1, −2).
Determine the equation of the curve that is
the locus of all points equidistant from the
two points (1, −5) and (5, −4).
Find the equation of the curve which is the
locus of the points equidistant from the line
𝑥 = −1 and the point (3, 0).
Find the equation of the curve which is the
locus of the points equidistant from the line
𝑥 = −7 and the point (5, 0).
𝑃 is the point (𝑥, 𝑦) and 𝑆 is the point (6, 1).
The point 𝑃 moves in such a way that its
distance from 𝑆 is equal to its distance from
the line 𝑥 = −1. Show that the equation of the
parabola traced out by the point 𝑃 is
𝑦(𝑦 − 2) = 14𝑥 + 𝑘 where 𝑘 is a constant to
be found.
Determine the equation of the curve which is
the locus of the points 3 units from the point
(1, 2).
Determine the equation of the curve which is
the locus of the points 7 units from the point
(−4, 5).
Determine the equation of the curve which is
the locus of the points √10 units from the
point (0, 3).
SOLUTIONS
1.
(𝑥 − 2) 2 + (𝑦 − 3) 2 = 2
2.
(𝑥 − 3 ) + (𝑦 − 3) =
20 2
5 2
104
9
SOLUTIONS
1. 𝑦 = 𝑥 + 1
5
27
2. 𝑦 = − 2 𝑥 − 4
3.
4.
5.
6.
7.
8.
9.
15
𝑦 = −4𝑥 + 2
𝑦 2 = 8𝑥 − 8
𝑦 2 = 24𝑥 + 24
𝑘 = −36
(𝑥 − 1) 2 + (𝑦 − 2) 2 = 9
(𝑥 + 4) 2 + (𝑦 − 5) 2 = 49
𝑥 2 + (𝑦 − 3) 2 = 10
EXAM QUESTIONS
1.
A point, 𝑝, moves in the 𝑥 − 𝑦 plane such that
its distance from 𝐶(2, 3) is always √2 units.
Determine the locus of 𝑝.
[3]
CAPE 2014
2.
A point 𝑃(𝑥, 𝑦) moves so that its distance from
the fixed point (0, 3) is two times the distance
from the fixed point (5, 2). Show that the
equation of the locus of the point 𝑃(𝑥, 𝑦) is a
circle.
[12]
CAPE 2015
88
CHAPTER 16: RADIAN MEASURE
CHAPTER 16: RADIAN MEASURE
At the end of this section, students should be able
to:



define the radian;
convert degrees to radians and radians to
degrees;
use the formulae for arc length 𝑙 = 𝑟𝜃 and
1
sector area, 𝐴 = 2 𝑟 2 𝜃;
________________________________________________________
INTRODUCTION
In the diagram 𝑂𝐴, 𝑂𝐵
and 𝐴𝐵 all have lengths,
𝑟, where 𝑟 is the radius
of the circle. Therefore,
the angle 𝛼 which is the
angle subtended by the
sector 𝐴𝑂𝐵 has a size of
1 radian.
A radian is equivalent to how many degrees?
To answer this question we need to determine how
many radii are equivalent to the circumference of a
circle.
Now, we have
Radians
1 radian
𝟐𝝅 radians
Radians
𝟐𝝅 radians
𝝅 radians
1 radian
Arc Length
𝑟
2𝜋𝑟
Degrees
360
180
180
≅ 57.3
𝜋
CONVERTING RADIANS TO DEGREES
180
𝜋
Convert 1.75 radians to degrees.
Degrees = Radians ×
LESSON 1a
SOLUTION
180
Degrees = 1.75 ×
= 1.75 × 57.3 = 100.268°
𝜋
CONVERTING DEGREES TO RADIANS
𝜋
180
Convert 60° to radians.
Radians = Degrees ×
LESSON 1b
SOLUTION
Radians = 60 ×
Below is a list of the common radians and their
degree equivalent
𝜋
𝜋
𝜋
𝜋
= 30°
= 45°
= 60°
6
4
3
2
3𝜋
= 90°
= 135°
4
3𝜋
𝜋 = 180°
= 270° 2𝜋 = 360°
2
ARC LENGTH
𝜃
Recall: Arc Length = 360° × 2𝜋𝑟
Replacing 360° with the appropriate radian
measure we have
𝜃
× 2𝜋𝑟 = 𝑟𝜃
2𝜋
where 𝜃 is in radians.
𝜋
LESSON 2a
If 𝜃 = and 𝑟 = 2, what is the
2
length of the arc subtended?
SOLUTION
Arc Length = 𝑟𝜃
𝜋
=2( )
2
=𝜋
LESSON 2b
On a circle whose radius is 3,
what angle subtends an arc length of 4?
SOLUTION
Arc Length = 𝑟𝜃
4 = 3𝜃
4
=𝜃
3
AREA OF SECTOR
Recall: Area of Sector =
𝜃
360
× 𝜋𝑟 2
By replacing 360° with the appropriate radian
measure we have
𝜃
1
Arc Length =
× 𝜋𝑟 2 = 𝑟 2 𝜃
2𝜋
2
where 𝜃 is in radians.
LESSON 3 Find the area of the sector subtending
𝜋
an angle of 6 radians with radius 6.
SOLUTION
1
Area of Sector = 2 𝑟 2 𝜃
1
𝜋
= (6) 2 ( )
2
6
= 3𝜋
𝜋
𝜋
= radians
180 3
89
CHAPTER 16: RADIAN MEASURE
LESSON 4
The diagram shows part of a
circle centre 𝑂 and radius 5 cm. Given that ∠𝐴𝑂𝐶
is 1.2 radians, calculate
i.
the length of the arc 𝐴𝐵𝐶
ii.
the area of the shaded segment
1
(5) 2 (0.64)
2
1
− (4)(3)
2
= 2 cm2
2
= 3.35 cm
=
LESSON 6
SOLUTION
(i)
Arc length = 𝑟𝜃
= 5(1.2)
= 6 cm
(ii)
Area of segment = Area of sector 𝑂𝐴𝐵𝐶 –
Area of triangle 𝑂𝐴𝐶
1
1
= 𝑟 2 𝜃 − 𝑎𝑏 sin 𝐶
2
2
1
2
= (5) (1.2)
2
1
− (5)(5) sin 1.2
2
LESSON 5
The diagram below shows a
sector of a circle centre O, radius 5 cm. 𝐵𝑁 is
perpendicular to 𝑂𝐴. Given that 𝐵𝑁 = 3 cm,
The diagram shows a circle with centre 𝑂 and
radius 5 cm. The point 𝑃 lies on the circle, 𝑃𝑇 is a
tangent to the circle and 𝑃𝑇 = 12 cm. The line 𝑂𝑇
cuts the circle at the point 𝑄.
(i) Find the perimeter of the shaded region
(ii) Find the area of the shaded region.
SOLUTION
(i)
By Pythagoras’ Theorem 𝑂𝑇 = 13 cm
12
tan 𝑇𝑂̂𝑃 =
5
12
−1
𝑇𝑂̂𝑃 = tan ( ) = 1.176 radians
5
Perimeter = 𝑃𝑇 + 𝑄𝑇 + arc 𝑂𝑃
= 12 + (13 − 5) + 5(1.176)
calculate
(i)
(ii)
(iii)
angle 𝐵𝑂𝑁 in radians
the perimeter of the shaded region
the area of the shaded region
SOLUTION
(i)
(ii)
3
sin 𝐵𝑂̂𝑁 = 5
3
𝐵𝑂̂𝑁 = sin−1 ( ) = 0.64 radians
5
Perimeter = 𝐵𝑁 + 𝐴𝑁 + arc 𝐴𝐵
= 3 + (5 − 4) + 5(0.64)
= 7.2 cm
N.B. The value of 4 cm is obtained by the use of
Pythagoras’ Theorem
(iii)
Area of shaded region = Area of sector
𝐴𝑂𝐵 – Area of triangle 𝐴𝑂𝐵
= 25.88 cm
(ii)
Shaded area = Area of 𝑂𝑃𝑇 – Area of
sector 𝑂𝑃𝑄
1
1
= (12)(5) − (5)2 (1.176)
2
2
= 15.3 cm2
…………………………………………………………………………
EXERCISE 16
1. Convert the following angles measured in
radians to degrees, stating your answer to 1
decimal place where necessary.
5𝜋
12
d. 1.2
a.
2.
3𝜋
5
e. − 2
b.
c. −
5𝜋
4
Convert the following angles measured in
degrees to radians correct to 2 decimal places.
90
CHAPTER 16: RADIAN MEASURE
a. 72°
d. −60°
b. 54°
e. 202.5°
c. −45°
3.
Find the length of an arc of a circle with radius
𝜋
10 m that subtends a central angle of 6 .
4.
A central angle 𝜃 of a circle with radius 16 cm
subtends an arc of 19.36 cm. Find 𝜃.
Find the area of a sector of a circle with central
𝜋
angle 3 if the radius of the circle is 3 m.
5.
3.
SOLUTIONS
1. (a) 75° (b) 108° (c) −225°
(d) 68.8° (e) −114.6°
2𝜋
3𝜋
𝜋
2. (a) 5
(b) 10
(c) − 4
𝜋
(d) − 3
3.
4.
5.
5𝜋
Calculate
(a) the size, in radians, of the angle 𝛼.
[3]
(b) the length of the arc 𝐴𝐶𝐵.
[3]
CAPE 2004
The diagrams shown below, not drawn below,
represent
a. a sector, 𝑂𝐴𝐵𝐶, of a circle with centre at 𝑂
and a radius of 7 cm, where angle 𝐴𝑂𝐶
𝜋
measures 3 radians.
b. a right circular cone with vertex 𝑂 and a
circular base of radius 𝑟 cm which is
formed when the sector 𝑂𝐴𝐵𝐶 is folded
so that 𝑂𝐴 coincides with 𝑂𝐶.
9𝜋
(e) 8
3
1.21 radians
3𝜋
2
EXAM QUESTIONS
1.
The diagram, not drawn to scale, is a sketch of
a wedge in an electrical appliance in the form
of a sector of a circle, centre 𝑂 and radius 4
𝜋
cm. Angle 𝐴𝑂𝐵 measures 4 radians.
(i) Express the arc length in terms of 𝜋. [1]
(ii) Hence show that
7
(a) 𝑟 = 6
[3]
(b) if ℎ cm is the height of the cone, then
the exact value of ℎ is
4.
2.
(i) Show that the area of the shaded region is
2(𝜋 − 2√2).
[6]
(ii) Using the cosine rule, show that the
length of the chord 𝐴𝐵 is 4√2 − √2. [4]
CAPE 2003
The figure (not drawn to scale) represents a
cross – section through a tunnel. The cross
section is part of a circle with radius 5 metres
and centre 𝑂. The width 𝐴𝐵 of the floor of the
tunnel is 8 metres.
7√35
[2]
CAPE 2005
The circle in the diagram, not drawn to scale,
𝜋
has centre 𝑂 and the acute angle 𝐴𝑂𝐵 = 6
radians. 𝑂𝐴 = 6 cm and 𝐶 and 𝐷 are the
midpoints of 𝑂𝐴 and 𝑂𝐵 respectively.
6
.
Express in terms of 𝜋.
(a) the length of arc 𝐴𝐵
[2]
(b) the area of the shaded region 𝐴𝐵𝐷𝐶. [4]
CAPE 2007
SOLUTIONS
1.
2. (a) 𝛼 = 1.85 (b) 22.14
7𝜋
3.
(i)
4.
(a) 𝜋 (b) 3𝜋 − 4
3
9
…………………………………………………………………………
91
CHAPTER 17: TRIGONOMETRY
CHAPTER 17: TRIGONOMETRY
At the end of this section, students should be able
to:
 evaluate sine, cosine and tangent for
angles of any size given either in degrees
or radians;
 evaluate the exact values of sine, cosine
𝜋 𝜋 𝜋 𝜋
and tangent for 𝜃 = 0, 6 , 4 , 3 , 2 , … , 2𝜋













graph the functions sin𝑘𝑥 , cos 𝑘𝑥, tan 𝑘𝑥,
where 𝑘 is 1 or 2 and 0 ≤ 𝑥 ≤ 2𝜋;
derive the identity cos 2 𝜃 + sin2 𝜃 = 1
use the formula for sin(𝐴 ± 𝐵) , cos(𝐴 ±
𝐵) and tan(𝐴 ± 𝐵);
derive the multiple angle identities for
sin 2𝑥 , cos 2𝑥, tan 2𝑥;
solve simple identities;
find solutions of simple equations for a
given range, including those involving the
use of cos 2 𝜃 + sin2 𝜃 = 1
use compound – angle formulae;
use the reciprocal functions of
sec 𝑥 , csc 𝑥 and cot 𝑥;
derive identities for the following:
(a) sin 𝑘𝐴 , cos 𝑘𝐴 , tan 𝑘𝐴 for 𝑘 ∈ ℚ,
(b) tan2 𝑥 , cot 2 𝑥 , sec 2 𝑥 and csc 2 𝑥,
(c) sin 𝐴 ± sin 𝐵 , cos 𝐴 ± cos 𝐵,
express 𝑎 cos 𝜃 + 𝑏 sin𝜃 in the form
𝑟 cos(𝜃 ± 𝛼) and 𝑟 sin(𝜃 ± 𝛼), where 𝑟 is
𝜋
positive, 0 < 𝛼 < 2 ;
find the general solution of equations of
the form:
(a) sin 𝑘𝜃 = 𝑠
(b) cos 𝑘𝜃 = 𝑐
(c) tan 𝑘𝜃 = 𝑡
(d) 𝑎 cos 𝜃 + 𝑏 sin𝜃 = 𝑐
for 𝑎, 𝑏, 𝑐, 𝑘, 𝑠, 𝑡 ∈ ℝ;
find the solutions of the equations above
for a given range;
obtain the maximum or minimum of
(𝑎 cos 𝜃 + 𝑏 sin𝜃) for 0 ≤ 𝜃 ≤ 2𝜋
TRIGONOMETRIC IDENTITIES
INTRODUCTION
Recall the following formulae which apply to
triangles. 1 – 4 apply to right angled triangles only
whereas 5 – 6 are primarily applicable to nonright angled triangles.
1.
Pythagoras’ Theorem: 𝑎2 = 𝑏2 + 𝑐 2
2.
tan 𝜃 =
𝑜𝑝𝑝 𝑏
=
𝑎𝑑𝑗 𝑐
𝑜𝑝𝑝 𝑏
=
ℎ𝑦𝑝 𝑎
𝑎𝑑𝑗 𝑐
4. cos 𝜃 =
=
ℎ𝑦𝑝 𝑎
Quotient Identities
sin𝜃
1. tan 𝜃 =
cos 𝜃
Pythagorean Identities
1. 𝑏2 + 𝑐 2 = 𝑎2
2. sin2 𝜃 + cos 2 𝜃 = 1
Variations of (2)
sin2 𝜃 = 1 − cos 2 𝜃
= (1 − cos 𝜃)(1 + cos 𝜃)
cos 2 𝜃 = 1 − sin2 𝜃
= (1 − sin 𝜃)(1 + sin 𝜃)
3.
sin 𝜃 =
Reciprocal Identities
1
1.
= cot 𝜃
tan 𝜃
1
2.
= sec 𝜃
cos𝜃
1
3.
= csc 𝜃
sin𝜃
Further Quotient Identities
cos 𝜃
cot 𝜃 =
sin𝜃
Further Pythagorean Identities
1.
2.
1 + cot 2 𝜃 = cosec2 𝜃
tan2 𝜃 + 1 = sec 2 𝜃
92
CHAPTER 17: TRIGONOMETRY
PROVING TRIGONOMETRIC IDENTITIES
Many identities are derived from the fundamental
identities. In the following LESSONs we will learn
how to prove that an equation is an identity, thus
discovering new identities.
Hints for Proving Trigonometric Identities
1. Choose a side (LHS or RHS) and use known
identities to transform it into the other side.
Starting with the more complicated side is
usually the best option.
2. When dealing with these proofs it is generally
useful to rewrite the given functions in terms
of sine and cosine.
LESSON 1a Prove that tan 𝑥 +
1
tan 𝑥
≡
1
sin𝑥 cos 𝑥
SOLUTION
1
tan 𝑥
sin 𝑥 cos 𝑥
=
+
cos 𝑥 sin 𝑥
(sin 𝑥)(sin𝑥) + (cos 𝑥)(cos 𝑥)
=
sin 𝑥 cos 𝑥
(sin2 𝑥 + cos 2 𝑥)
=
sin 𝑥 cos 𝑥
1
=
sin 𝑥 cos 𝑥
= RHS
LESSON 1b
Prove that
( sin𝜃 + cos 𝜃 )2 ≡ 1 + 2 sin𝜃 cos 𝜃
SOLUTION
LHS: (sin𝜃 + cos 𝜃)(sin 𝜃 + cos 𝜃)
= sin2 𝜃 + 2 sin 𝜃 cos 𝜃 + cos 2 𝜃
= sin2 𝜃 + cos 2 𝜃 + 2 sin 𝜃 cos 𝜃
= 1 + 2 sin 𝜃 cos 𝜃
= RHS
LHS = tan 𝑥 +
LESSON 1c
Prove that
tan2 𝜃 − sin2 𝜃 ≡ tan2 𝜃 sin2 𝜃
SOLUTION
LHS: tan2 𝜃 − sin2 𝜃
sin2 𝜃
=
− sin2 𝜃
cos 2 𝜃
sin2 𝜃 − cos2 𝜃 sin2 𝜃
=
cos 2 𝜃
sin2 𝜃 (1 − cos 2 𝜃)
=
cos2 𝜃
sin2 𝜃
=
× sin2 𝜃
cos 2 𝜃
= tan2 𝜃 sin2 𝜃
= RHS
LESSON 2a
Prove that
sin 𝜃 tan 𝜃 + cos 𝜃 = sec 𝜃
SOLUTION
LHS: sin 𝜃 tan 𝜃 + cos 𝜃
sin 𝜃 sin 𝜃
+ cos 𝜃
cos 𝜃
sin2 𝜃 + cos2 𝜃
=
cos 𝜃
1
=
cos 𝜃
= sec 𝜃
= RHS
=
LESSON 2b
Prove that
1
1
+
≡ 2 csc 𝑥 cot 𝑥
sec 𝑥 + 1 sec 𝑥 − 1
SOLUTION
1
1
LHS:
+
sec 𝑥 + 1 sec 𝑥 − 1
sec 𝑥 − 1 + sec 𝑥 + 1
=
sec 2 𝑥 − 1
2 sec 𝑥
=
tan2 𝑥
cos 2 𝑥
= 2 sec 𝑥 ×
sin2 𝑥
2
cos 2 𝑥
=
×
cos 𝑥 sin2 𝑥
2
cos 𝑥
=
×
sin 𝑥 sin𝑥
= 2 csc 𝑥 cot 𝑥
= RHS
LESSON 2c
Prove that
cos 𝑥
1−sin𝑥
= sec 𝑥 + tan 𝑥
SOLUTION
cos 𝑥
LHS:
1 − sin 𝑥
cos 𝑥 1 + sin 𝑥
=
.
1 − sin 𝑥 1 + sin 𝑥
cos 𝑥 (1 + sin𝑥)
=
1 − sin2 𝑥
cos 𝑥 (1 + sin𝑥 )
=
cos 2 𝑥
1 + sin 𝑥
=
cos 𝑥
1
sin 𝑥
=
+
cos 𝑥 cos 𝑥
= sec 𝑥 + tan 𝑥
= RHS
…………………………………………………………………………
EXERCISE 17.1
1.
Prove the following identities.
sin 𝑥
cos 𝑥−1
(b)
+
=0
cos 𝑥+1
cos2 𝜃
sin𝑥
(c) 1+sin 𝜃 = 1 − sin 𝜃
(d) sin 𝜃 +
cos2 𝜃
1+sin𝜃
=1
(e) tan 𝜃 sin 𝜃 cos 𝜃 = sin2 𝜃
1
cos 𝜃
(f) sin 𝜃 cos 𝜃 − sin𝜃 = tan 𝜃
93
CHAPTER 17: TRIGONOMETRY
sin2 𝜃
(g) 1+cos 𝜃 = 1 − cos 𝜃
1
1
1
(h) cos2 𝑥 + sin2 𝑥 ≡ sin2 𝑥 cos2 𝑥
sin2 𝜃
(i) cos 𝜃 + 1+cos 𝜃 = 1
cos2 𝑥−sin2 𝑥
(j)
1−tan2 𝑥
1
= cos 2 𝑥
1
(k) cos 𝑥 sin𝑥 − tan 𝑥 ≡ tan 𝑥
cos 𝑥
(l)
2.
sin 𝑥
+
sin 𝑥
cos 𝑥
=
GRAPHS OF TRIGONOMETRIC
FUNCTIONS
INTRODUCTION
Here is a list of the graphs of the trigonometric
functions of 𝑦 = sin𝑥 , 𝑦 = cos 𝑥 and 𝑦 = tan 𝑥.
NB: For this section ALL angles will be assumed to
be in radians unless otherwise stated.
1
sin 𝑥 cos 𝑥
Prove the following identities.
(a) cos 𝑥 + sin 𝑥 tan 𝑥 = sec 𝑥
csc 𝑥−sin 𝑥
(b) sin𝑥 csc 𝑥 = csc 𝑥 − sin 𝑥
sec2 𝛽
1
(c) tan 𝛽 + tan 𝛽 = tan 𝛽
(d)
1+sin 𝜃
cos 𝜃
+ 1+sin 𝜃 = 2 sec 𝜃
cos 𝜃
(e) sec 𝑦 + tan 𝑦 =
cos 𝑦
1−sin𝑦
sin2 𝜃+cos2 𝜃+cot2 𝜃
(f)
= cot 2 𝜃
1+tan2 𝜃
(g) sin 𝜃 csc 𝜃 cos 𝜃 = cos 𝜃
(h) cot 𝜃 sin𝜃 cos 𝜃 = cos2 𝜃
(i) csc 𝜃 (sin 𝜃 + tan 𝜃) = 1 + sec 𝜃
sin 𝜃
(j) 1 − csc 𝜃 = cos 2 𝜃
(k) cos 𝜃 (sec 𝜃 − cos 𝜃 ) = sin2 𝜃
cot 𝜃
(l) csc 𝜃 = cos 𝜃
sec 𝜃
(m) csc 𝜃 = tan 𝜃
1
sin𝜃
(n) sin 𝜃 cos 𝜃 − cos 𝜃 = cot 𝜃
(o)
tan2 𝜃
sec2 𝜃−1
csc 𝜃
The maximum value of sin 𝑥 is 1
The minimum value of sin 𝑥 is −1
−1 ≤ sin𝑥 ≤ 1
……………………………………………………………………….
The maximum value of cos 𝑥 is 1
The minimum value of cos 𝑥 is −1
−1 ≤ cos 𝑥 ≤ 1
……………………………………………………………………….
− 1 = sec 𝜃
(p) sin 𝜃 − cot 2 𝜃 = 1
(q) sec 𝜃 (cos 𝜃 − cot 𝜃) = 1 − csc 𝜃
cos 𝜃
(r) 1 −
= sin2 𝜃
sec 𝜃
(s) sin 𝜃 (csc 𝜃 − sin 𝜃 ) = cos 2 𝜃
tan 𝜃
(t)
= sin𝜃
sec 𝜃
csc 𝜃
(u) sec 𝜃 = cot 𝜃
sin2 𝜃
(v) 1+cos 𝜃 = 1 − cos 𝜃
(w) sec 𝜃 csc 𝜃 = tan 𝜃 + cot 𝜃
sec 𝜃
(x) cos 𝜃 − tan2 𝜃 = 1
cos 𝜃
sin 𝜃
(y) sec 𝜃 + csc 𝜃 = 1
…………………………………………………………………………
The maximum value is ∞
The minimum value is −∞.
3𝜋
𝜋 𝜋 3𝜋
The graph is undefined at − 2 , − 2 , 2 , 2 .
NB: On most calculators undefined values are
indicated by MATH ERROR.
94
CHAPTER 17: TRIGONOMETRY
GRAPHS OF RECIPROCAL FUNCTIONS
SKETCHING TRIGONOMETRIC GRAPHS
LESSON 3
Sketch the graph of 𝑓(𝑥) = sin 𝑥
for 0 ≤ 𝑥 ≤ 2𝜋.
SOLUTION
……………………………………………………………………….
……………………………………………………………………….
LESSON 4
Sketch the graph of 𝑦 = cos 2𝑥
for 0 ≤ 𝑥 ≤ 2𝜋.
SOLUTION
We simply need to complete the
following table. When we graph draw the graph
we have to keep in mind that the graph will have
the same basic shape as that of 𝑦 = cos 𝑥.
𝜋
3𝜋
𝒙
0
𝜋
2𝜋
2
2
𝒚
1
−1
1
−1
1
LESSON 5
Sketch the graph of 𝑦 = tan 2𝑥
for 0 ≤ 𝑥 ≤ 2𝜋
SOLUTION
𝜋
3𝜋
𝒙
0
𝜋
2𝜋
2
2
𝒚
0
0
0
0
0
This table is clearly not very helpful so we will
include additional values.
𝜋
𝜋 3𝜋
3𝜋 7𝜋
𝒙 0
𝜋 5𝜋
2𝜋
4
2
4
4
4
2
𝒚 0 und 0 und 0 und 0 und 0
efin
efin
efin
efin
ed
ed
ed
ed
From the table we see that we have vertical
𝜋 3𝜋 5𝜋 7𝜋
asymptotes at 𝑥 = 4 , 4 , 4 , 4 .
95
CHAPTER 17: TRIGONOMETRY
TRIGONOMETRIC EQUATIONS
INTRODUCTION
Before we look at how to solve trigonometric
equations we need to develop what is called the
QUADRANT RULE.
The range 0 to 2𝜋 can be divided into quadrants as
seen above.
𝜋
QUADRANT I: 0 →
2
ALL ratios are positive
𝜋
QUADRANT II: 2 → 𝜋
SINE ONLY is positive
3𝜋
QUADRANT III: 𝜋 → 2
TANGENT ONLY is positive
3𝜋
QUADRANT IV: → 2𝜋
2
COSINE ONLY is positive
The acronym
All School Teachers Curse
is commonly used to remember the quadrant rule.
GENERAL SOLUTIONS
LESSON 6 Solve the following equations
1
(a) sin 𝜃 = 2 for
0 ≤ 𝜃 ≤ 2𝜋
(b) cos 𝜃 = 0.5 for 0 ≤ 𝜃 ≤ 2𝜋
(c) tan 𝑥 = 0.5 for 0 ≤ 𝑥 ≤ 2𝜋
Hence state the general solutions.
SOLUTION
(a) We need to find the angle in Quadrant I and
then determine the corresponding angles in
the appropriate quadrant(s) using the
information below. This initial angle we will
refer to as a Reference Angle, 𝑅𝐴.
𝐼: 𝜃 = 𝑅𝐴
96
CHAPTER 17: TRIGONOMETRY
𝐼𝐼: 𝜃 = 𝜋 − 𝑅𝐴
𝐼𝐼𝐼: 𝜃 = 𝜋 + 𝑅𝐴
𝐼𝑉: 𝜃 = 2𝜋 − 𝑅𝐴
1
sin 𝜃 =
2
1
𝜋
𝑅𝐴 = sin−1 ( ) =
2
6
Sine is postitive in I and II
𝜋
𝐼: 𝜃 =
6
𝜋 5𝜋
𝐼𝐼: 𝜃 = 𝜋 − =
6
6
Tangent is positive in I and III
𝐼: 𝑥 = 0.464𝑐
𝐼𝐼𝐼: 𝑥 = 𝜋 + 0.464𝑐 = 3.606𝑐
General Solutions:
0.464 + 𝑛𝜋
𝜃={
3.606 + 𝑛𝜋
General Solutions:
𝜋
+ 2𝑛𝜋
6
𝜃 = {5𝜋
+ 2𝑛𝜋
6
(b) cos 𝜃 =
𝑛∈ℤ
1
2
1
𝜋
𝑅𝐴 = cos−1 ( ) =
2
3
Cosine is positive in I and IV
𝜋
𝐼: 𝜃 =
3
𝜋 5𝜋
𝐼𝑉: 𝜃 = 2𝜋 − =
3
3
𝑛∈ℤ
LESSON 7
Solve the following equations
for 0 ≤ 𝜃 ≤ 2𝜋.
(a) 2 cos 𝜃 = √3
(b) 2 sin 𝜃 + 1 = 0
Hence determine the general solutions.
SOLUTION
(a) 2 cos 𝜃 = √3
√3
cos 𝜃 =
2
𝜋
√3
𝑅𝐴 = cos−1 ( ) =
2
6
Cosine is positive in I and IV
𝜋
𝐼: 𝜃 =
6
𝜋 11𝜋
𝐼𝑉: 𝜃 = 2𝜋 − =
6
6
General Solutions:
𝜋
+ 2𝑛𝜋
6
𝜃 = {11𝜋
+ 2𝑛𝜋
6
General Solutions:
𝜋
+ 2𝑛𝜋
3
𝜃 = {5𝜋
+ 2𝑛𝜋
3
(c) tan 𝑥 = 0.5
1
𝑅𝐴 = tan−1 ( ) = 0.464𝑐
2
𝑛∈ℤ
𝑛∈ℤ
(b) 2 sin 𝜃 + 1 = 0
2 sin 𝜃 = −1
1
sin 𝜃 = −
2
1
𝜋
−1
𝑅𝐴 = sin ( ) =
2
6
Sine is negative in 𝐼𝐼𝐼 and 𝐼𝑉.
97
CHAPTER 17: TRIGONOMETRY
𝜋 7𝜋
=
6
6
𝜋 11𝜋
𝐼𝑉: 𝜃 = 2𝜋 − =
6
6
𝐼𝐼𝐼: 𝜃 = 𝜋 +
General Solutions:
7𝜋
+ 2𝑛𝜋
6
𝜃 ={
11𝜋
+ 2𝑛𝜋
6
𝑛∈ℤ
SOLVING TRIGONOMETRIC EQUATIONS
LESSON 8
Solve the equation
5 tan(𝑥 + 40°) = −6 for 0 ≤ 𝑥 ≤ 360°, giving your
answer to 2 decimal places.
SOLUTION
5 tan(𝑥 + 40°) = −6
Let 𝜃 = 𝑥 + 40°
5 tan 𝜃 = −6
6
tan 𝜃 = −
5
6
−1
𝑅𝐴 = tan ( ) = 50.19°
5
Tan is negative in Quadrant 𝐼𝐼 and 𝐼𝑉.
𝐼𝐼: 𝜃 = 180° − 50.19° = 129.81°
𝜃 = 𝑥 + 40° = 129.81°
𝑥 = 89.81°
𝐼𝑉: 𝜃 = 360° − 50.19° = 309.81°
𝜃 = 𝑥 + 40° = 309.81°
𝑥 = 269.81°
LESSON 9
Solve the equation
6 sin2 𝜃 − 5 cos 𝜃 = 0 for 0° ≤ 𝜃 ≤ 360°.
Hence determine the general solutions.
SOLUTION
We have 2 trigonometric
functions in the equation therefore we need to
rewrite it in terms of a single function.
6 sin2 𝜃 − 5 cos 𝜃 = 0
6(1 − cos 2 𝜃) − 5 cos 𝜃 = 0 Using trig identities
6 − 6 cos 2 𝜃 − 5 cos 𝜃 = 0
× (−1)
6 cos 2 𝜃 + 5 cos 𝜃 − 6 = 0
(2 cos 𝜃 + 3)(3 cos 𝜃 − 2) = 0
either 3 cos 𝜃 − 2 = 0 or 2 cos 𝜃 + 3 = 0
2
3
cos 𝜃 =
or cos 𝜃 = −
3
2
2
𝜃 = cos −1 ( )
3
𝜃 = 48.19°
Cosine is positive in I and IV
𝐼: 𝜃 = 48.19°
𝐼𝑉: 𝜃 = 360° − 48.19° = 311.81°
3
𝜃 = cos −1 ( )
2
Invalid: − 1 ≤ cos θ ≤ 1
General Solutions:
48.19° + 360°𝑛
𝜃 ={
311.81° + 360°𝑛
𝑛∈ℤ
LESSON 10
Solve the following equations
5 tan2 𝜃 + 7 = 11 sec 𝜃 for 0 < 𝜃 < 2𝜋.
SOLUTION
5 tan2 𝜃 + 7 = 11 sec 𝜃
sin2 𝜃
1
5(
) + 7 = 11 (
)
cos 2 𝜃
cos 𝜃
5 sin2 𝜃 + 7 cos 2 𝜃 = 11 cos 𝜃
5(1 − cos 2 𝜃) + 7 cos2 𝜃 = 11 cos 𝜃
5 − 5 cos 2 𝜃 + 7 cos 2 𝜃 = 11 cos 𝜃
2 cos 2 𝜃 − 11 cos 𝜃 + 5 = 0
(2 cos 𝜃 − 1)(cos 𝜃 − 5) = 0
1
cos 𝜃 =
cos 𝜃 = 5 INVALID
2
1
𝜋
𝑅𝐴 = cos−1 ( ) =
2
3
Cosine is positive in I and IV
𝜋
𝐼: 𝜃 =
3
𝜋 5𝜋
𝐼𝑉: 𝜃 = 2𝜋 − =
3
3
ALTERNATELY
5 tan2 𝜃 + 7 = 11 sec 𝜃
98
CHAPTER 17: TRIGONOMETRY
5(sec 2 𝜃 − 1) + 7 = 11 sec 𝜃
5 sec 2 𝜃 − 5 + 7 = 11 sec 𝜃
5 sec 2 𝜃 − 11 sec 𝜃 + 2 = 0
(5 sec 𝜃 − 1)(sec 𝜃 − 2) = 0
1
sec 𝜃 =
sec 𝜃 = 2
5
1
1
1
=
=2
cos 𝜃 5
cos 𝜃
1
cos 𝜃 = 5
cos 𝜃 =
2
INVALID
𝑦
√3
cos ( ) =
2
2
𝜋
√3
𝑅𝐴 = cos−1 ( ) =
2
6
Cosine is positive in I and IV but IV is outside the
𝑦
range of 2
𝑦 𝜋
𝐼: =
2 6
𝜋
𝑦=
3
LESSON 11 Solve the equation
sin2 𝜃 + 2 sin𝜃 cos 𝜃 = 0 for 0 ≤ 𝜃 ≤ 180°
SOLUTION
sin2 𝜃 + 2 sin𝜃 cos 𝜃 = 0
sin 𝜃 (sin 𝜃 + 2 cos 𝜃) = 0
either sin𝜃 = 0
or sin𝜃 + 2 cos 𝜃 = 0
∴ 𝜃 = 0°, 180°
sin 𝜃 + 2 cos 𝜃 = 0
sin 𝜃 = −2 cos 𝜃
sin 𝜃
= tan 𝜃 = −2 (∗)
cos 𝜃
𝑅𝐴 = tan−1(2) = 63.43°
Tan is negative in II
𝐼𝐼: 𝜃 = 180° − 63.43° = 116.57°
𝜃 = 0°, 116.57°, 180°
NB:(∗)We are permitted to divide sin𝜃 by cos 𝜃
because it forms a trigonometric identity, thus, we
have not divided by zero.
LESSON 13
Solve the equation
(2 sin2 2𝜃 − 1) cos 2𝜃 = 0 for 0 ≤ 𝜃 ≤ 2𝜋.
SOLUTION
(2 sin2 2𝜃 − 1) cos 2𝜃 = 0
cos 2𝜃 = 0
2𝜃 =
𝜋 3𝜋 5𝜋 7𝜋
, , ,
2 2 2 2
From Graph
2 sin2 2𝜃 − 1 = 0
sin2 2𝜃 =
1
2
sin 2𝜃 = ±
1
√2
𝑅. 𝐴 = sin−1 (
𝐼1 : 2𝜃 =
1
√2
𝜋 3𝜋
=
4
4
𝐼𝐼𝐼3 : 2𝜃 = 𝜋 +
Find all the angles 0 ≤ 𝑦 ≤ 2𝜋
𝑦
𝜋 5𝜋
=
4
4
𝐼𝑉1 : 2𝜃 = 2𝜋 −
√3
which satisfy the equation cos ( ) =
2
2
SOLUTION
Since 0 ≤ 𝑦 ≤ 2𝜋 then
𝑦
𝑦
0 ≤ 2 ≤ 𝜋. We first solve for 2
𝐼2 : 2𝜃 =
𝜋
4
𝜋
4
𝐼𝐼1 : 2𝜃 = 𝜋 −
LESSON 12
)=
𝜋 7𝜋
=
4
4
𝜋
9𝜋
+ 2𝜋 =
4
4
99
CHAPTER 17: TRIGONOMETRY
𝐼𝐼2 : 2𝜃 =
3𝜋
11𝜋
+ 2𝜋 =
4
4
𝐼𝐼𝐼2 : 2𝜃 =
5𝜋
13𝜋
+ 2𝜋 =
4
4
𝐼𝑉2 : 2𝜃 =
7𝜋
15𝜋
+ 2𝜋 =
4
4
𝜃=
3.
4.
5.
𝜋 3𝜋 5𝜋 7𝜋 𝜋 3𝜋 5𝜋 7𝜋 9𝜋 11𝜋 13𝜋 15𝜋
,
,
, , ,
, ,
,
,
,
,
4 4 4 4 8 8 8 8 8 8
8
8
6.
LESSON 14
Find all the angles between
– 𝜋 ≤ 𝑥 ≤ 𝜋 which satisfy the equation
2(sin 𝑥 + cos 𝑥) = 3 sin 𝑥
SOLUTION
2(sin 𝑥 + cos 𝑥) = 3 sin 𝑥
2 sin 𝑥 + 2 cos 𝑥 = 3 sin𝑥
2 cos 𝑥 = 3 sin 𝑥 − 2 sin𝑥
2 cos 𝑥 = sin 𝑥
sin 𝑥
2=
= tan 𝑥
cos 𝑥
−1
𝑅𝐴 = tan 2 = 1.107 radians
The range – 𝜋 ≤ 𝑥 ≤ 𝜋 indicates that the required
values are bounded within I and II
(rotating anti-clockwise) and within III and
IV (rotating clockwise)
Tan is positive in I and III
𝐼: 𝑥 = 1.107 radians
(Quadrant I rotating anti-clockwise)
𝐼𝐼: 𝑥 = −(𝜋 − 1.107) = −2.03
(Quadrant III rotating clockwise)
𝑥 = 1.107 radians, −2.03 radians
.........................................................................................................
EXERCISE 17.2
√3
1.
Solve the equation sin 𝜃 = 2 for 0 ≤ 𝜃 ≤ 2𝜋.
2.
Solve the following equations for 0 ≤ 𝜃 ≤ 2𝜋.
√3
(i) sin 𝜃 = − 2
(ii) cos 𝜃 = −
1
2
(iii) 2 cos 𝜃 = −√3
Hence state the general solutions.
Solve the equation sin(𝑥 + 15°) = 0.5 for
0 ≤ 𝑥 ≤ 360°.
Solve the following equations for
0° < 𝜃 < 360°
(i)
5 cos 𝜃 + 2 sin2 𝜃 = 4
(ii)
3 sin 𝜃 tan 𝜃 = 8
Find the solutions of the following equations
for 0 ≤ 𝑦 ≤ 360°.
(i) 2 cos 𝑦 + 5 sin𝑦 cos 𝑦 = 0
(ii) tan 𝑦 sin 𝑦 + sin 𝑦 = 0
Solve the equation 2 cos2 𝜃 + 3 sin2 𝜃 − 3 = 0
for 0 ≤ 𝜃 ≤ 2𝜋.
7.
Find all the angles between 0° and 360° which
satisfy the equation
5 cos2 𝑥 − 8 sin 𝑥 cos 𝑥 = 0
8. Show that the equation sin2 𝜃 +
3 sin 𝜃 cos 𝜃 = 4 cos 2 𝜃 can be written as a
quadratic equation in tan 𝜃
Hence, or otherwise, solve the equation in
part sin2 𝜃 + 3 sin 𝜃 cos 𝜃 = 4 cos 2 𝜃 for
0 ≤ 𝜃 ≤ 𝜋.
9. Solve the following equations for 0 ≤ 𝜃 ≤ 2𝜋
(i) 2 cos 2𝜃 = 1
(ii) 1 + 2 sin 2𝜃 = 0
10. Solve the following equations for 0 ≤ 𝜃 ≤ 2𝜋.
𝜃
1
(i) sin (2 ) = 2
𝜃
1
(ii) cos (2 ) = − 2
𝑦
(iii) tan (2 ) = −√3
11. Solve the following equations for – 𝜋 ≤ 𝜃 ≤ 𝜋,
correct to 2 decimal places where necessary.
a) (sin 𝜃 − cos 𝜃) = 2 cos 𝜃
b) (1 + sin 𝑥)(2 + sin𝑥) = cos 2 𝑥
12. Solve the equation
5 tan2 𝜃 = 5 tan 𝜃 + 3 sec 2 𝜃
for 0 < 𝜃 < 360°
2𝜋
13. Solve the equations sec (2𝜃 + 9 ) = 2 for
0 ≤ 𝜃 ≤ 2𝜋.
14. Solve the following equations for – 𝜋 ≤ 𝜃 ≤ 𝜋,
correct to 2 decimal places where necessary.
a) 2 sec 𝜃 + 3 cos 𝜃 = 7
b) 2 cos 𝑥 + 3 sec 𝑥 = 7
15. Show that the equation 3 tan 𝜃 = 2 cos 𝜃 can
be expressed as
2 sin2 𝜃 + 3 sin 𝜃 − 2 = 0
Hence, find the general solutions of the
equation
3 tan 𝜃 = 2 cos 𝜃.
16. Find the general solutions for the equation
4 sin4 𝜃 + 5 = 7 cos 2 𝜃.
[Hint: Solve as a disguised quadratic]
100
CHAPTER 17: TRIGONOMETRY
17. Find the general solutions of the equation
4 cos 2 𝜃 − 4 sin 𝜃 = 1.
18. Solve the equation 3𝑥 3 − 𝑥 = 3𝑥 2 − 1, hence
find the general solutions of the equation
3 tan3 𝑥 − 3 tan2 𝑥 − tan 𝑥 + 1 = 0
SOLUTIONS
𝜋 2𝜋
1. 𝜃 = 3 , 3
2.
4𝜋
5𝜋
2𝜋
4𝜋
5𝜋
7𝜋
(i) 𝜃 = 3 + 2𝑛𝜋, 3 + 2𝑛𝜋
(ii) 𝜃 = 3 + 2𝑛𝜋, 3 + 2𝑛𝜋
6.
(iii) 𝜃 = 6 + 2𝑛𝜋, 6 + 2𝑛𝜋
𝑥 = 15°, 135°
(i) 𝜃 = 60°, 300°
(ii) 𝜃 = 70.5°, 289.5°
(i) 𝑦 = 90°, 203.6°, 270°, 336.4°
(ii) 𝑦 = 0°, 135°, 180°, 315°, 360°
𝜋 3𝜋
𝜃= ,
7.
8.
𝑥 = 32°, 90°, 212°, 270°
𝜋
𝜃 = 4 , 1.82
9.
(i) 𝜃 = 6 , 6 , 6 , 6
3.
4.
5.
2
HARMONIC FORM
INTRODUCTION
In many instances it is essential to find the
solutions of equations of the form
𝑎 sin 𝜃 + 𝑏 cos 𝜃 = 𝑐 or 𝑎 cos 𝜃 + 𝑏 sin𝜃 = 𝑐
The previous methods for solving a trig equation
cannot be applied directly to these equations.
Therefore, we need to find an alternate form (of a
single trig. ratio) of expressing the equation. This
form is derived as follows:
From the diagram at left we see that
2
𝜋 5𝜋 7𝜋 11𝜋
𝜋 5𝜋
10. (i) 𝜃 = 3 , 3
7𝜋 11𝜋 19𝜋 23𝜋
(ii) 𝜃 = 12 , 12 , 12 , 12
4𝜋
(ii) 𝜃 = 3
𝑐 +𝑑
𝑟
𝑟 sin(𝜃 + 𝛼) = 𝑐 + 𝑑
sin(𝜃 + 𝛼) =
4𝜋
(iii) 𝜃 = 3
5𝜋
𝜋
𝜋
11. (a) 𝜃 = −2.11, 1.03 (b) 𝑥 = − 6 , − 2 , − 6
12. 𝜃 = 71.6°, 153.4°, 251.6°, 333.4°
𝜋 13𝜋 19𝜋 31𝜋
13. 𝜃 = 18 , 18 , 18 , 18
14. (a) 𝜃 = −1.23 (b) 𝑥 = −1.05
𝜋 5𝜋
15. 𝜃 = 6 , 6 + 2𝑛𝜋
𝜋 5𝜋
16. 𝜃 = 6 , 6 + 𝑛𝜋
𝜋 5𝜋
17. 6 , 6 + 2𝑛𝜋
𝑐
𝑎
𝑑
cos 𝜃 =
𝑏
sin 𝜃 =
→ 𝑐 = 𝑎 sin𝜃
→ 𝑑 = 𝑏 cos 𝜃
𝑟 sin(𝜃 + 𝛼) = 𝑎 sin 𝜃 + 𝑏 cos 𝜃
Which is the form of our equation.
Also,
𝑟 sin(𝜃 − 𝛼) = 𝑐 − 𝑑
= 𝑎 sin 𝜃 − 𝑏 cos 𝜃
∴ 𝑎 sin 𝜃 ± 𝑏 cos 𝜃 = 𝑟 sin(𝜃 ± 𝛼)
𝜋 𝜋 5𝜋 5𝜋 7𝜋 11𝜋
18. 6 , 4 , 4 , 6 , 6 , 6 + 𝑛𝜋
…………………………………………………………………………..
Similarly,
𝑐 = 𝑑 −𝑒
𝑐 𝑑 −𝑒
=
𝑟
𝑟
𝑟 cos(𝜃 + 𝛼) = 𝑑 − 𝑒
𝑑
cos 𝜃 =
→ 𝑑 = 𝑎 cos 𝜃
𝑎
𝑒
sin 𝜃 =
→ 𝑒 = 𝑏 sin𝜃
𝑏
cos(𝜃 + 𝛼) = 𝑎 cos 𝜃 − 𝑏 sin 𝜃
Also,
𝑟 cos(𝜃 − 𝛼) = 𝑑 + 𝑒
= 𝑎 cos 𝜃 + 𝑏 sin 𝜃
∴ 𝑎 cos 𝜃 ± 𝑏 sin𝜃 = 𝑟 cos(𝜃 ∓ 𝛼)
Furthermore, from the diagrams above we determine
that
𝑏
𝑟 = √𝑎2 + 𝑏2 and 𝛼 = tan−1 ( )
𝑎
cos(𝜃 + 𝛼) =
101
CHAPTER 17: TRIGONOMETRY
NB: The absolute values of 𝑎 and 𝑏 are to be used
in the above calculations.
Summarising we have
LESSON 17
LESSON 15
a) Express 𝑓(𝜃) = √2 cos 𝜃 + sin𝜃 in the form
𝑅 cos(𝜃 − 𝛼) where 𝑅 > 0 and 𝛼 is acute.
b) Hence, find the minimum value of
𝑓(𝜃), where 0 ≤ 𝜃 ≤ 2𝜋.
c) Determine the value of 𝜃, 0 ≤ 𝜃 ≤ 2𝜋, at
which the minimum value of 𝑓(𝜃) occurs.
SOLUTION
𝑏
𝑎 sin 𝜃 ± 𝑏 cos 𝜃 = √𝑎2 + 𝑏2 sin(𝜃 ± 𝛼) ; 𝛼 = tan−1 ( )
𝑎
𝑏
−1
2
2
𝑎 cos 𝜃 ± 𝑏 sin𝜃 = √𝑎 + 𝑏 cos(𝜃 ∓ 𝛼) ; 𝛼 = tan ( )
𝑎
Thus, we are now equipped to solve the required
equations
(i) Express 3 cos 𝜃 − 4 sin𝜃 in the form
𝑅 cos(𝜃 + 𝛼)
(ii) Hence, solve the equation
3 cos 𝜃 − 4 sin𝜃 = 1 for 0° ≤ 𝜃 ≤ 360°
SOLUTION
(i) 3 cos 𝜃 − 4 sin𝜃 = √32 + 42 cos(𝜃 + 𝛼)
4
𝛼 = tan−1 ( ) = 53.13°
3
5 cos(𝜃 + 53.1°)
(ii) 5 cos(𝜃 + 53.13°) = 1
1
cos(𝜃 + 53.13°) =
5
1
(𝜃 + 53.13°) = cos −1 ( ) = 78.5°
5
Reference angle is 78.5°
Cosine is positive in I and IV
𝐼: (𝜃 + 53.13°) = 78.46°
𝜃 = 78.46° − 53.13° = 25.33°
𝐼𝑉: (𝜃 + 53.1°) = 360° − 78.46° = 281.54°
𝜃 = 281.54° − 53.13° = 228.41°
𝜃 = 25.33°, 228.41°
LESSON 16
(i) Express 4 sin 𝜃 − 3 cos 𝜃 in the form
𝑅 sin(𝜃 − 𝛼).
(ii) Hence, solve the equation
4 sin 𝜃 − 3 cos 𝜃 = 1 for 0° ≤ 𝜃 ≤ 360°.
SOLUTION
(i) 4 sin 𝜃 − 3 cos 𝜃 = √42 + 32 sin(𝜃 − 𝛼)
3
𝛼 = tan−1 ( ) = 36.87°
4
5 sin(𝜃 − 36.87°)
(ii) 4 sin 𝜃 − 3 cos 𝜃 = 1
5 sin(𝜃 − 36.87°) = 1
sin(𝜃 − 36.87°) = 0.2
𝑅𝐴 = sin−1(0.2) = 11.54°
Sine is positive in Quadrants I and II
𝐼: 𝜃 − 36.87° = 11.54°
𝜃 = 36.87° + 11.54° = 48.41°
𝐼𝐼: 𝜃 − 36.87° = 180° − 11.54° = 168.46°
𝜃 = 168.46° + 36.87° = 205.33°
2
a.
√2 cos 𝜃 + sin 𝜃 = √(√2) + 12 cos(𝜃 − 𝛼)
1
𝛼 = tan−1 ( ) = 0.615
√2
√3 cos(𝜃 − 0.615)
b.
We know that the minimum value of the cosine
function is −1
Hence, the minimum value of
𝑓(𝜃) = √3(−1) = −√3
c.
To determine this minimum value we need
to solve the equation
cos(𝜃 − 0.615) = −1
cos(𝜃 − 0.615) = −1 when (𝜃 − 0.615) = 𝜋
𝜃 = 𝜋 + 0.615 = 3.76
LESSON 18
(i)
Express 𝑓(2𝜃) = 4 sin 2𝜃 + 3 cos 2𝜃 in
the form 𝑟 sin(2𝜃 + 𝛼) where 𝑟 > 0 and
𝜋
0 < 𝛼 < 2.
(ii)
Hence, or otherwise, find the maximum
and minimum values of
1
6 − 𝑓(𝜃)
SOLUTION
(i)
(ii)
4 sin 2𝜃 + 3 cos 2𝜃
𝑟 = √42 + 32 = 5
3
𝛼 = tan−1 ( ) = 0.644𝑐
4
𝑓(2𝜃) = 5 sin(2𝜃 + 0.644)
The minimum and maximum values of
𝑓(2𝜃) and 𝑓(𝜃) are the same.
Maximum value of 𝑓(𝜃) = 5
Minimum value of 𝑓(𝜃) = −5
1
1
1
1
Maximum value of 6−𝑓(𝜃) = 6−5 = 1
1
Minimum value of 6−𝑓(𝜃) = 6−(−5) = 11
102
CHAPTER 17: TRIGONOMETRY
…………………………………………………………………………
EXERCISE 17.3
1. Express 5 sin 𝜃 + 3 cos 𝜃 in the form
𝑅 sin(𝜃 + 𝛼) where 𝑅 > 0 and 𝛼 is acute.
Hence solve the equation
5 sin 𝜃 + 3 cos 𝜃 = 5.
2. Express cos 𝑥 + √3 sin 𝑥 in the form
𝑅 cos(𝑥 − 𝛼), giving the exact values of 𝑅 and
𝛼 such that 𝑅 > 0 and 0° < 𝛼 < 90°.
Hence find the general solution of the
equation cos 𝑥 + √3 sin 𝑥 = 2, giving your
answer exactly, in degrees.
3. Express 5 sin 𝜃 + 6 cos 𝜃 in the form
𝑅 sin(𝜃 + 𝛼), where 𝑅 is positive and 𝛼 is
acute.
Hence
(i) find the value of 𝜃, between 0° and 90°,
for which 5 sin 𝜃 + 6 cos 𝜃 is a maximum.
(ii) solve the equation 5 sin 𝜃 + 6 cos 𝜃 = 4,
for 0° < 𝜃 < 360°.
4. Given that 3 cos 𝑥 − 4 sin 𝑥 ≡ 𝑅 cos(𝑥 + 𝛼)
where 𝑅 > 0 and 0° < 𝛼 < 90°, find the
values of 𝑅 and 𝛼, giving the value of 𝛼 correct
to two decimal places.
Hence solve the equation
3 cos 2𝜃 − 4 sin 2𝜃 = 2, for 0° < 𝜃 < 360°,
giving your answers correct to 2 decimal
places.
SOLUTIONS
1. √34 sin(𝜃 + 0.54𝑐 ); 𝜃 = 0.49𝑐 , 1.57𝑐
2. 2 cos(𝜃 − 60°); 60° + 360°𝑛
3. √61 sin(𝜃 + 50.2°) (i) 𝜃 = 39.8°
(ii) 𝜃 = 160.6°
4. 5 cos(𝜃 + 53.13°),
𝜃 = 6.65°, 120.23°, 186.65°, 300.23°
…………………………………………………………………………
COMPOUND ANGLE FORMULAE
INTRODUCTION
The formulae for compound angles are:
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin𝐵
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin𝐵
cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
tan 𝐴 + tan 𝐵
tan(𝐴 + 𝐵) =
1 − tan 𝐴 tan 𝐵
tan 𝐴 − tan 𝐵
tan(𝐴 − 𝐵) =
1 + tan 𝐴 tan 𝐵
LESSON 19
Evaluate
i.
cos(75°)
ii. tan(−15°)
SOLUTION
We choose compounds of special angles
30°, 45°, 60°
(i)
cos(75°) = cos(30° + 45°)
= cos(30°) cos(45°) − sin(30°) sin(45°)
1 1
√3 1
=
.
− .
2 √2 2 √2
1
√3
=
−
2√2 2√2
√3 − 1
=
2√2
(ii)
tan(−15°) = tan(30° − 45°)
tan(30°) − tan(45°)
=
1 + tan(30°) tan(45°)
1
( − 1)
= √3
1
1+
.1
√3
1 − √3
3
= √
√3 + 1
√3
1 − √3
=
1 + √3
LESSON 20
4
12
Given that sin 𝐴 = 13 and
cos 𝐵 = 5 find the value of sin(𝐴 + 𝐵), cos(𝐴 + 𝐵)
and hence deduce that 𝐴 + 𝐵 is obtuse.
SOLUTION
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin𝐵
103
CHAPTER 17: TRIGONOMETRY
Therefore, we need to determine the cos 𝐴 and
sin 𝐵. To do this we use the ratios given and
complete the corresponding triangles using
Pythagoras' Theorem.
5
3
cos 𝐴 =
,
sin 𝐵 =
13
5
12 4 5 3
sin(𝐴 + 𝐵) =
. +
.
13 5 13 5
48 15
=
+
65 65
63
=
65
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
5 4 12 3
=
. −
.
13 5 13 5
20 36
=
−
65 65
16
=−
65
Thus, we have that sin(𝐴 + 𝐵) is positive and
cos(𝐴 + 𝐵) is negative. This implies that 𝐴 + 𝐵 is
in the Second Quadrant. Hence, 𝐴 + 𝐵 is obtuse.
LESSON 21
(i)
𝜋
Given that sin =
4
√2
, where 𝑥 is acute,
2
…………………………………………………………………………..
EXERCISE 17.4
1. Evaluate
(i) sin 75°
(ii) cos 15°
(iii) tan 105°
2. Simplify each of these expressions
a. sin 18° cos 27° + cos 18° sin27°
b. cos 10° cos 80° − sin 10° sin80°
3𝜋
2𝜋
3𝜋
2𝜋
c. cos cos
+ sin sin
7
21
7
21
𝜋
𝜋
tan + tan
18
9
d.
𝜋
𝜋
1 − tan 18 tan 9
tan 73° − tan 13°
e.
1 + tan 73° tan 13°
13𝜋
𝜋
13𝜋
𝜋
f. cos
cos (− ) − sin
sin(− )
15
5
15
5
3. Given that 𝐴 and 𝐵 are acute angles and that
12
3
cos 𝐶 = 13 and that cos 𝐷 = 5, find the value of
each of the following
a. cos(𝐶 + 𝐷)
b. cos(𝐶 − 𝐷)
4. Using the identity
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin𝐵,
show that
(ii)
5𝜋
𝜋
√2
(sin 𝑥 − cos 𝑥)
sin (𝑥 − ) =
4
2
𝜋
√3
Using the fact that sin 3 = 2 and
𝜋
1
𝜋
3
2
12
cos = , find the exact value of sin
showing ALL steps in your working.
SOLUTION
(i)
By the compound angle formula for
sin(𝐴 − 𝐵 )
𝜋
𝜋
𝜋
sin (𝑥 − ) = sin𝑥 cos − cos 𝑥 sin
4
4
4
√2
√2
= sin 𝑥 ( ) − cos 𝑥 ( )
2
2
√2
(sin 𝑥 − cos 𝑥)
=
2
𝜋
𝜋
𝜋
(ii)
=3−4
12
𝜋
𝜋 𝜋
∴ sin = sin ( − )
12
3 4
𝜋 𝜋
𝜋
𝜋
𝜋
𝜋
sin ( − ) = sin cos − cos sin
3 4
3
4
3
4
1 √2
√3 √2
=
( )− ( )
2 2
2 2
√6 √2
=
−
4
4
√6 − √2
=
4
5.
6.
calculate the value of sin( 12 ).
i. State the exact value of
𝜋
𝜋
(a) cos 6 and (b) sin 6
ii. Hence, show that
𝜋
1
cos (𝑥 + ) = (√3 cos 𝑥 − sin 𝑥)
6
2
1+tan 𝑥
𝜋
Prove that 1−tan 𝑥 ≡ tan ( 4 + 𝑥)
SOLUTIONS
√6+√2
1. (i) 4
(ii)
√2
√6+√2
4
1
(iii) −2 − √3
√3
1
2.
(a) 2 (b) 0 (c) 2 (d) 3 (e) √3 (f) − 2
3.
(a) 65 (b) 65
4.
√6+√2
5.
(i) (a) 2
16
56
4
√3
1
(b) 2
6.
…………………………………………………………………………..
104
CHAPTER 17: TRIGONOMETRY
DOUBLE–ANGLE FORMULAE
INTRODUCTION
sin 2𝐴 = 2 sin 𝐴 cos 𝐴
cos 2𝐴 = cos 2 𝐴 − sin2 𝐴
= cos 2 𝐴 − (1 − cos 2 𝐴)
= 2 cos2 𝐴 − 1
cos 2𝐴 = cos 2 𝐴 − sin2 𝐴
= (1 − sin2 𝐴) − sin2 𝐴
= 1 − 2 sin2 𝐴
2 tan 𝐴
tan 2𝐴 =
1 − tan2 𝐴
3
LESSON 22
If sin 𝛼 = 5 and 𝛼 is an acute
angle, use the double angle formulae to find the
exact value of sin2𝛼 , cos 2𝛼 and tan 2𝛼 and the
quadrant of angle 2𝛼.
SOLUTION
3
4
3
sin 𝛼 =
cos 𝛼 =
tan 𝛼 =
5
5
4
3 4
24
sin 2𝛼 = 2 sin𝛼 cos 𝛼 = 2 ( ) ( ) =
5 5
25
cos 2𝛼 = cos 2 𝛼 − sin2 𝛼
4 2
3 2 16 9
7
= ( ) −( ) =
−
=
5
5
25 25 25
3
3
2 (4)
2 tan 𝛼
2
tan 2𝛼 =
=
=
1 − tan2 𝛼
3 2 1− 9
1 − (4)
16
3 7
3 16 24
= ÷
= ×
=
2 16 2 7
7
Since 2𝛼 has a positive value for all three ratios it
must be in the first quadrant.
LESSON 23
Solve the equation
3 sin 2𝜃 = sin 𝜃 , for 0 ≤ 𝜃 ≤ 2𝜋
SOLUTION
One function is written in 𝜃 and the other in 2𝜃
therefore we cannot solve directly. Hence, we use
the double angle formula for sine.
3 sin 2𝜃 = sin 𝜃
3(2 sin𝜃 cos 𝜃) = sin 𝜃
6 sin 𝜃 cos 𝜃 = sin 𝜃
6 sin 𝜃 cos 𝜃 − sin 𝜃 = 0
sin 𝜃 (6 cos 𝜃 − 1) = 0
sin 𝜃 = 0
→ 𝜃 = 0, 𝜋, 2𝜋
6 cos 𝜃 − 1 = 0
1
cos 𝜃 =
6
1
−1
𝑅𝐴 = cos ( ) = 1.40
6
Cosine is positive in I and IV
𝜃 = 1.40 radians
𝜃 = 2𝜋 − 1.40 = 4.88 radians
𝜃 = 0,
1.40,
𝜋,
4.88,
2𝜋
LESSON 24
Prove the identity
1 + cos 𝜃 + cos 2𝜃
1
=
sin𝜃 + sin 2𝜃
tan 𝜃
SOLUTION
LHS
1 + cos 𝜃 + cos 2𝜃
1
=
sin 𝜃 + sin 2𝜃
tan 𝜃
1 + cos 𝜃 + 2 cos2 𝜃 − 1
=
sin𝜃 + 2 sin 𝜃 cos 𝜃
cos 𝜃 (1 + 2 cos 𝜃)
=
sin 𝜃 (1 + 2 cos 𝜃)
cos 𝜃
=
sin 𝜃
1
=
tan 𝜃
= RHS
…………………………………………………………………………..
EXERCISE 17.5
1. Given that 𝜃 is an acute angle and that
4
sin 𝜃 = 5 , find the value of each of
the following:
a. sin 2𝜃
b. cos 2𝜃
c. tan 2𝜃
2. Solve each of the following equations for
0 ≤ 𝜃 ≤ 2𝜋, giving your answers correct to 3
decimal places
a. sin 2𝜃 + cos 𝜃 = 0
b. 4 cos 𝜃 = 3 sin 2𝜃
c. 3 cos 2𝜃 − cos 𝜃 + 2 = 0
3. (i) Prove the identity
cos 4𝜃 + 4 cos 2𝜃 ≡ 8 cos 4 𝜃 − 3.
(ii) Hence, solve the equation
cos 4𝜃 + 4 cos 2𝜃 = 2.
4. Show that ( cos 𝑥 − sin 𝑥 )2 ≡ 1 − sin 2𝑥
5. Solve the equation sin 2𝜃 = sin𝜃 for
0 ≤ 𝜃 ≤ 2𝜋.
6. Given that tan 𝐴 = 2 tan 𝐵, show that
sin2𝐵
tan(𝐴 − 𝐵) =
3 − cos 2𝐵
.
SOLUTIONS
24
7
24
1. (a) 25
(b) − 25
(c) − 7
2.
𝜋 7𝜋 3𝜋 11𝜋
(a) 𝜃 = 2 , 6 , 2 , 6
𝜋 3𝜋
(b) 𝜃 = 2 , 2 , 0.0.73𝑐 , 2.41𝑐
𝜋
3.
4.
5.
5𝜋
(c) 𝜃 = , 1.911𝑐 , , 4.373𝑐
6
6
(ii) 𝜃 = 0.48, 2.67, 3.62, 5.81
𝜋
5𝜋
𝜃 = 3 , 𝜋, 3
6.
105
CHAPTER 17: TRIGONOMETRY
HALF–ANGLE FORMULAE
INTRODUCTION
𝜃
𝜃
sin 𝜃 = 2 sin ( ) cos ( )
2
2
𝜃
2
cos 𝜃 = 2 cos ( ) − 1
2
𝜃
1 + cos 𝜃
cos ( ) = ±√
2
2
2.
sin 15°
sin
30°
1 − cos 30°
=√
2
2
(1)
√3
√1 − 2
=
2
(2)
(3)
=
𝜃
1 − cos 𝜃
sin ( ) = ±√
2
2
tan 𝜃 =
𝜃
2 tan (2 )
𝜃
1 − tan2 ( )
2
𝜃
1 − cos 𝜃
tan ( ) = ±√
2
1 + cos 𝜃
𝜃
1 − cos 𝜃
tan ( ) =
2
sin𝜃
𝜃
sin𝜃
tan ( ) =
2
1 + cos 𝜃
(4)
(5)
(6)
(7)
(8)
LESSON 25
Evaluate the following by using the appropriate
half angle formula.
𝜋
i. cos 8
ii. sin 15°
3𝜋
iii. tan ( )
8
SOLUTION
1. Since 𝜋/8 is half of 𝜋/4, then according to the
half angle formula
𝜋
𝜋
cos ( ) = 2 cos2 ( ) − 1
4
8
1
𝜋
= 2 cos2 ( ) − 1
8
√2
1
1 + √2
𝜋
+1 =
= 2cos 2 ( )
8
√2
√2
1 + √2
𝜋
= cos 2 ( )
8
2√2
1 + √2
𝜋
√
= cos ( )
8
2√2
NB: We only consider the positive square root
value because 𝜋/8 is in the first quadrant
√
=√
3.
tan (
2 − √3
2
2
2 − √3 √2 − √3
=
4
2
3𝜋
)
8
3𝜋
2 tan ( 8 )
3𝜋
tan ( ) =
3𝜋
4
1 − tan2 ( )
8
3𝜋
2 tan ( 8 )
−1 =
3𝜋
1 − tan2 ( )
8
3𝜋
3𝜋
tan2 ( ) − 1 = 2 tan ( )
8
8
3𝜋
3𝜋
2
tan ( ) − 2 tan ( ) − 1 = 0
8
8
Using the quadratic formula:
3𝜋
2 ± √(−2) 2 − 4(1)(−1)
tan ( ) =
8
2(1)
2 ± √4 + 4 2 ± √8 2 ± 2√2
=
=
=
2
2
2
= 1 ± √2
3𝜋
tan ( ) = 1 + √2
8
since it is in the first quadrant
LESSON 26
Prove that
𝑥 𝜋
tan ( + ) = sec 𝑥 + tan 𝑥
2 4
SOLUTION
LHS
𝑥 𝜋
tan ( + )
2 4
𝑥
𝜋
tan ( ) + tan ( )
2
4
=
𝑥
𝜋
1 − tan ( ) tan ( )
2
4
𝑥
tan (2 ) + 1
=
𝑥
1 − tan (2 )
106
CHAPTER 17: TRIGONOMETRY
𝑥
𝑥
(tan (2 ) + 1) (tan (2 ) + 1)
=
𝑥
𝑥
(1 − tan (2 )) (1 + tan (2 ))
𝑥
𝑥
tan2 (2 ) + 2 tan (2 ) + 1
=
𝑥
1 − tan2 (2 )
𝑥
𝑥
tan2 (2 ) + 1
2 tan (2 )
=
𝑥 +
𝑥
1 − tan2 (2 ) 1 − tan2 (2 )
𝑥
sin2 (2 )
𝑥 +1
cos 2 (2 )
=
𝑥 + tan 𝑥
sin2 (2 )
1−
𝑥
cos 2 (2 )
𝑥
𝑥
sin2 (2 ) + cos2 (2 )
𝑥
cos2 ( )
2
=
𝑥
𝑥 + tan 𝑥
2
cos (2 ) − sin2 (2 )
𝑥
cos2 ( )
2
1
=
𝑥
𝑥 + tan 𝑥
cos 2 (2 ) − sin2 (2 )
1
=
+ tan 𝑥
cos 𝑥
= sec 𝑥 + tan 𝑥
FACTOR FORMULAE
INTRODUCTION
𝑋+𝑌
𝑋−𝑌
) cos (
)
2
2
𝑋+𝑌
𝑋−𝑌
sin 𝑋 − sin 𝑌 = 2 cos (
) sin(
)
2
2
𝑋+𝑌
𝑋−𝑌
cos 𝑋 + cos 𝑌 = 2 cos (
) cos (
)
2
2
𝑋+𝑌
𝑋−𝑌
cos 𝑋 − cos 𝑌 = −2 sin (
) sin (
)
2
2
sin 𝑋 + sin 𝑌 = 2 sin(
LESSON 27
a) sin 105° − sin 15°
b) sin 105° sin15°
SOLUTION
We rewrite the difference as a product
sin 105° − sin 15°
105° + 15°
105° − 15°
= 2 cos (
) sin(
)
2
2
= 2 cos 60° sin 45°
1 1
= 2( )( )
2 √2
1
=
√2
Comparing sin105° sin 15° with
cos 𝑋 − cos 𝑌 = −2 sin (
=RHS
…………………………………………………………………………..
EXERCISE 17.6
1. Use a half-angle formula to find
𝜋
a. sin ( )
b. cos 165°
8
c. sin 75°
d. tan 22.5°
𝑎
12
𝜋
2.
Find the value ofsin (2 ) if cos 𝑎 = 13 , 0 < 𝑎 < 2
3.
Prove that 2 sin2 (2 ) + cos 𝑥 = 1
4.
Prove that 2 cos2 (2 ) sec 𝜃 = sec 𝜃 + 1
5.
Prove that cos 2 (2 ) − cos θ = sin2 (2 )
𝑥
𝜃
𝜃
SOLUTIONS
1.
(a)
(b)
4
√6+√2
(c) 4
(d) √2 − 1
2.
3.
4.
5.
√26
26
𝜃
𝑋+𝑌
𝑋−𝑌
) sin (
)
2
2
We see that
1
𝑋+𝑌
𝑋−𝑌
− (cos 𝑋 − cos 𝑌) = sin (
) sin(
)
2
2
2
𝑋+𝑌
= 105° → 𝑋 + 𝑌 = 210° (1)
2
𝑋−𝑌
= 15°
→ 𝑋 − 𝑌 = 30° (2)
2
Solving (1) and (2) simultaneously we get
𝑋 = 120°, 𝑌 = 90°
1
sin 105° sin15° = − (cos 120° − cos 90°)
2
1
1
= − (− − 0)
2
2
1
=
4
√2−√2
2
−√6−√2
Find the exact value of
LESSON 28
sin 𝐴+sin 𝐵
Simplify cos 𝐴−cos 𝐵
SOLUTION
𝐴+𝐵
𝐴−𝐵
2 sin( 2 ) cos ( 2 )
sin𝐴 + sin 𝐵
=
cos 𝐴 − cos 𝐵 −2 sin (𝐴 + 𝐵 ) sin (𝐴 − 𝐵 )
2
2
𝐴−𝐵
cos ( 2 )
𝐴−𝐵
=−
= − cot (
)
𝐴−𝐵
2
sin ( 2 )
107
CHAPTER 17: TRIGONOMETRY
LESSON 29
Solve the equation
cos 𝑥 + cos 2𝑥 = sin 2𝑥 − sin 𝑥 for 0 ≤ 𝑥 ≤ 2𝜋.
SOLUTION
cos 𝑥 + cos 2𝑥 = sin 2𝑥 − sin 𝑥
We apply the appropriate compound angle
formulae
2 cos (
2𝑥 + 𝑥
2𝑥 − 𝑥
2𝑥 + 𝑥
2𝑥 − 𝑥
) cos (
) = 2 cos (
) sin (
)
2
2
2
2
3𝑥
𝑥
3𝑥
𝑥
cos = 2 cos sin
2
2
2
2
3𝑥
𝑥
3𝑥
𝑥
2 cos cos − 2 cos sin = 0
2
2
2
2
3𝑥
𝑥
𝑥
2 cos (cos − sin ) = 0
2
2
2
3𝑥
2 cos
=0
2
3𝑥 𝜋
3𝜋
5𝜋
= ,
,
2
2
2
2
3𝑥
NB: 0 ≤
≤ 3𝜋
2
𝜋
5𝜋
𝑥= ,
𝜋,
3
3
𝑥
𝑥
cos − sin = 0
2
2
𝑥
𝑥
cos = sin
2
2
𝑥
sin 2
𝑥
𝑥 = tan 2 = 1
cos
2
𝑥 𝜋
=
2 4
𝑥
NB: 0 ≤ ≤ 𝜋
2
𝜋
𝑥=
2
𝜋
𝜋
5𝜋
𝑥= ,
,
𝜋,
3
2
3
2 cos
LESSON 30
If 𝐴, 𝐵 and 𝐶 are the angles of a
triangle prove that
𝐴
𝐵
𝐶
sin 𝐴 + sin𝐵 + sin𝐶 = 4 cos cos cos
2
2
2
SOLUTION
sin 𝐴 + sin 𝐵 + sin 𝐶
𝐴+𝐵
𝐴−𝐵
= 2 sin (
) cos (
) + sin 𝐶
2
2
𝐴+𝐵
𝐴−𝐵
𝐶
𝐶
= 2 sin (
) cos (
) + 2 sin cos
2
2
2
2
𝐴+𝐵
𝐴−𝐵
𝐶
𝐶
= 2 sin (
) cos (
) + 2 sin cos
2
2
2
2
𝐶
𝐴−𝐵
𝐴+𝐵
𝐶
) + 2 sin (90° − (
)) cos
= 2 sin (90° − ) cos (
2
2
2
2
𝐶
𝐴−𝐵
𝐴+𝐵
𝐶
= 2 cos cos (
) + 2 cos
cos
2
2
2
2
𝐶
𝐴−𝐵
𝐴+𝐵
= 2 cos (cos (
) + cos (
))
2
2
2
𝑋+𝑌
𝑋−𝑌
Recall: cos 𝑋 + cos 𝑌 = 2 cos (
) cos (
)
2
2
𝐴 −𝐵
𝐴+𝐵
Letting 𝑋 =
and 𝑌 =
we get
2
2
𝐴 −𝐵
𝐴+𝐵
𝐴
𝐵
cos (
) + cos (
) = 2 cos cos
2
2
2
2
𝐶
𝐴
𝐵
= 2 cos 2 cos cos
2
2
2
𝐴
𝐵
𝐶
= 4 cos cos cos
2
2
2
…………………………………………………………………………..
EXERCISE 17.7
1. Write the sum as a product (Apply negative
identities whenever possible)
(a) sin 5𝑥 + sin3𝑥
(b) sin 𝑥 − sin4𝑥
(c) cos 4𝑥 − cos 6𝑥 cos 9𝑥 + cos 2𝑥
(d) sin 2𝑥 − sin7𝑥
(e) sin 3𝑥 + sin4𝑥
2. Simplify
sin 𝐴+sin 𝐵
(a)
cos 𝐴+cos 𝐵
cos 𝐴+cos 𝐵
(b) sin 𝐴−sin 𝐵
(c)
(d)
3.
4.
𝐴 + 𝐵 + 𝐶 = 180°
𝐴 𝐵 𝐶
+ + = 90°
2 2 2
𝐶
𝐴 +𝐵
= 90° − (
)
2
2
sin 3𝐴−sin 𝐴
cos 3𝐴−cos 𝐴
sin 2𝐴+sin 3𝐴
cos 2𝐴−cos 3𝐴
Solve the following equations for 0 ≤ 𝑥 ≤ 𝜋
a. sin 𝑥 + sin 5𝑥 = sin 3𝑥
b. cos 𝑥 + cos 2𝑥 + cos 3𝑥 = 0
c. sin 𝑥 − 2 sin 2𝑥 + sin 3𝑥 = 0
d. cos 𝑥 − sin 2𝑥 = cos 5𝑥
Prove that
(a) sin 𝐴 + sin 2𝐴 + sin3𝐴 =
sin 2𝐴 (2 cos 𝐴 + 1)
(b) cos 𝐴 + 2 cos 3𝐴 + cos 5𝐴 =
4 cos 2 𝐴 cos 3𝐴
(c) cos 𝐴 − 2 cos 3𝐴 + cos 5𝐴 =
2 sin 𝐴 (sin2𝐴 − sin 4𝐴)
sin 𝐴−sin 2𝐴+sin 3𝐴
(d) cos 𝐴−cos 2𝐴+cos 3𝐴 = tan 2𝐴
108
CHAPTER 17: TRIGONOMETRY
SOLUTIONS
1. (a) 2 sin 4𝑥 cos 𝑥
c.
d.
5𝑥
3𝑥
9𝑥
5𝑥
7𝑥
𝑥
2
𝐴+𝐵
2
(b) 2 cos ( 2 ) sin (− 2 )
(c) −2 sin 5𝑥 sin(−𝑥)
11𝑥
7𝑥
(d) 2 cos ( 2 ) cos ( 2 )
(e) 2 cos ( 2 ) sin(− 2 )
(f) 2 sin ( ) cos (− )
2.
(a) tan ( 2 )
(b) cot (
𝐴−𝐵
2
)
(c) − cot 2𝐴
𝐴
(d) − cot (− 2 )
3.
𝜋 𝜋 5𝜋 2𝜋 7𝜋
(a) 𝑥 = 0, 9 , 3 , 9 , 3 , 9 , 𝜋
𝜋 3𝜋
(b) 𝑥 = 4 , 4
𝜋
(c) 𝑥 = 0, , 𝜋
2
𝜋
5𝜋 𝜋 13𝜋 17𝜋
(d) 𝑥 = 0, 18 , 18 , 2 , 18 , 18 , 𝜋
4.
…………………………………………………………………………..
EXAM QUESTIONS
1.
(a) Copy and complete the following table for
the function 𝑓(𝑥) = sin 𝑥, 0 ≤ 𝑥 ≤ 2𝜋.
𝑥
𝜋
2
0
3𝜋
2
𝜋
𝑓(𝑥)
2.
3.
4.
5.
2𝜋
sin 𝑟
cos(𝑝 + 𝑡)
[3]
[4]
CAPE 2007
6. Given that 𝐴 and 𝐵 are acute angles such that
sin 𝐴 = 3/5 and cos 𝐵 = 5/13 , find, without
using tables or a calculator, the EXACT values
of
(i) sin(𝐴 + 𝐵)
[3]
(ii) cos(𝐴 − 𝐵)
[3]
(iii) cos 2𝐴
[2]
CAPE 2009
7. (i) Solve cos 2𝜃 − 3 cos 𝜃 = 1 for 0 ≤ 𝜃 ≤ 2𝜋.
[6]
(ii) Prove that cos 4 𝐴 − sin4 𝐴 + 1 = 2 cos2 𝐴.
[5]
CAPE 2003
8. Solve the equation cos 2𝜃 = 3 cos 𝜃 − 2 for
0 ≤ 𝜃 ≤ 2𝜋.
[4]
CAPE 2004
9. (i) Show that cos 3𝜃 = 4 cos 3 𝜃 − 3 cos 𝜃.
[5]
(ii) The position vectors of two points 𝐴 and
𝐵 relative to the origin 𝑂 are
𝒂 = 4 cos 2 𝜃 𝒊 + (6 cos 𝜃 − 1)𝒋
𝒃 = 2 cos 𝜃 𝒊 − 𝒋
By using the identity in (i) above, find the
𝜋
value of 𝜃, 0 ≤ 𝜃 ≤ 4 , such that 𝒂 and 𝒃
are perpendicular.
[5]
CAPE 2005
1−cos 2𝑥
10. Show that
= tan2 𝑥.
[3]
1+cos 2𝑥
(b) Sketch the graph of 𝑓.
[4]
Solve, for 0° ≤ 𝜃 ≤ 180°, the equation
6 cos 2 𝜃 + sin 𝜃 = 4.
[7]
CAPE 2004
Solve the equation 4 cos2 𝜃 − 4 sin𝜃 − 1 = 0
for 0 ≤ 𝜃 ≤ 𝜋.
[5]
CAPE 2006
12
4
Given that sin 𝐴 = and sin𝐵 = , where 𝐴
13
5
and 𝐵 are acute angles, find cos(𝐴 − 𝐵) and
sin(𝐴 + 𝐵).
[8]
CAPE 2003
In the triangle below, not drawn to
3
5
scale, sin 𝑞 = and cos 𝑝 = .
5
13
CAPE 2006
1−tan2 𝜃
11. (i) Prove that cos 2𝜃 ≡ 1+tan2 𝜃.
[4]
(ii) Hence, show, without using calculators,
1
that tan 67 2 ° = 1 + √2
[7]
CAPE 2007
12. (i) Prove that sin 2𝜃 − tan 𝜃 cos 2𝜃 = tan 𝜃
[3]
(ii) Express tan 𝜃 in terms ofsin 2𝜃 and
cos 2𝜃.
[2]
(iii) Hence show, without using tables or
calculators, that tan 22.5° = √2 − 1. [4]
CAPE 2008
13. Find the general solutions of the equation
cos 𝜃 = 2 sin2 𝜃 − 1.
[7]
CAPE 2006
Determine the exact values of
a. cos 𝑞
b. sin 𝑝
[1]
[1]
14. Express 𝑓(𝜃) = √2 cos 𝜃 − sin𝜃 in the form
𝑅 cos(𝜃 + 𝛼).
[5]
i. Hence, find the minimum value of 𝑓(𝜃),
where 0 ≤ 𝜃 ≤ 2𝜋.
[1]
109
CHAPTER 17: TRIGONOMETRY
ii. Determine the value of 𝜃, 0 ≤ 𝜃 ≤ 2𝜋, at
which the minimum value of 𝑓(𝜃) occurs.
[2]
CAPE 2005
15. Given that 4 sin 𝑥 − cos 𝑥 = 𝑅 sin(𝑥 − 𝛼) , 𝑅 >
0 and 0° < 𝛼 < 90°, find the values of 𝑅 and 𝛼
correct to 1 decimal place.
[7]
CAPE 2006
16. (a) Express cos 𝜃 − sin𝜃 in the form
𝑅 cos(𝜃 + 𝛼) where 𝑅, 𝛼 ∈ ℝ, 𝑅 > 0 and
𝜋
0<𝛼< .
[5]
2
(b) Hence, find the general solution of
cos 𝜃 − sin 𝜃 = 1.
[3]
CAPE 2007
17. The diagram, which is
not drawn to scale,
shows a quadrilateral
𝐴𝐵𝐶𝐷 in which
𝐴𝐵 = 4 cm, 𝐵𝐶 = 9 cm,
𝐴𝐷 = 𝑥 cm and
∠𝐵𝐴𝐷 = ∠𝐵𝐶𝐷 =
𝜃 and ∠𝐶𝐷𝐴 is a right
angle.
(i) Show that 𝑥 = 4 cos 𝜃 + 9 sin 𝜃
[4]
(ii) By expressing 𝑥 in the form 𝑟 cos(𝜃 − 𝛼),
𝜋
where 𝑟 is positive and 0 ≤ 𝛼 ≤ 2, find
the MAXIMUM possible value of 𝑥.
[6]
CAPE 2009
3
𝐴
18. If cos 𝐴 = 5, find tan 2 .
[6]
CAPE 2003
19. Solve, for 0 ≤ 𝑥 ≤ 𝜋, the equation
sin 𝑥 + sin3𝑥 = 0.
[6]
CAPE 2004
20. Given that 𝐴, 𝐵 and C are the angles of a
triangle, prove that
𝐴+𝐵
𝐶
(a) sin 2 = cos 2
[3]
(b) sin 𝐵 + sin 𝐶 = 2 cos 2 cos 2
[2]
𝐴
𝐵−𝐶
Hence, show that
𝐴
𝐵
𝐶
sin 𝐴 + sin 𝐵 + sin 𝐶 = 4 cos 2 cos 2 cos 2
[5]
CAPE 2008
21. (a) By using 𝑥 = cos2 𝜃, or otherwise, find all
values of the angle 𝜃 such that
8 cos 4 𝜃 − 10 cos 2 𝜃 + 3 = 0, 0 ≤ 𝜃 ≤ 𝜋.
[6]
(b) The diagram, not drawn to scale, shows a
rectangle 𝑃𝑄𝑅𝑆 with sides 6 cm and 8 cm
inscribed in another rectangle 𝐴𝐵𝐶𝐷.
(i) The angle that 𝑆𝑅 makes with 𝐷𝐶 is 𝜃.
Find, in terms of 𝜃, the length of the
side 𝐵𝐶.
[2]
(ii) Find the value of 𝜃 if |𝐵𝐶| =7 cm. [5]
(iii) Is 15 a possible value for |𝐵𝐶|? Give a
reason for your answer.
[2]
1−cos 2𝜃
(c) (i) Show that sin 2𝜃 = tan 𝜃.
[3]
(ii) Hence, show that
1−cos 4𝜃
(a)
= tan 2𝜃
(b)
[3]
sin 4𝜃
1−cos 6𝜃
sin 6𝜃
= tan 3𝜃
[2]
(iii) Using the results in (c) (i) and (ii) above,
evaluate
𝑛
∑(tan 𝑟𝜃 sin2𝑟𝜃 + cos 2𝑟𝜃)
𝑟=1
where 𝑛 is a positive integer.
[2]
CAPE 2011
22. (a) (i) Given that
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 and
cos 2𝜃 = 2 cos2 𝜃 − 1, prove that
1
cos 3𝜃 ≡ 2 cos 𝜃 [cos2 𝜃 − sin2 𝜃 − 2 ].
[7]
(ii) Using the appropriate formula, show
that
1
[sin 6𝜃 − sin2𝜃] = (2 cos 2 2𝜃 − 1) sin 2𝜃
2
[5]
(iii) Hence, or otherwise, solve
𝜋
sin 6𝜃 − sin 2𝜃 = 0 for 0 ≤ 𝜃 ≤ 2 .
[5]
(b) Find ALL possible values of cos 𝜃 such
that 2 cot 2 𝜃 + cos 𝜃 = 0.
[8]
CAPE 2012
23. (a) 𝐴 is an acute angle and 𝐵 is an obtuse
4
3
angle, where sin 𝐴 = 5 and cos 𝐵 = − 5.
Without finding the values of angles 𝐴 and
𝐵, calculate cos 3𝐴.
[5]
110
CHAPTER 17: TRIGONOMETRY
(b) Solve the equation 4 cos 2𝜃 − 14 sin 𝜃 = 7
for values of 𝜃 between 0 and 2𝜋 radians.
[8]
CAPE 2013
2 tan 𝜃
24. (a) (i) Show that sin 2𝜃 =
[4]
2 .
1+tan 𝜃
(ii) Hence, or otherwise, solve equation
sin 2𝜃 − tan 𝜃 = 0 for 0 ≤ 𝜃 ≤ 2𝜋.
[8]
(b) (i) Express 𝑓(𝜃) = 3 cos 𝜃 − 4 sin 𝜃 in the
form 𝑟 cos(𝜃 + 𝛼) where 𝑟 > 0
𝜋
and0 < 𝛼 < 2.
[4]
(ii) Hence, find
(a) the maximum value of 𝑓(𝜃) [2]
1
(b) the minimum value of 8+𝑓(𝜃). [2]
(iii) Given that the sum of the angles 𝐴, 𝐵
and 𝐶 of a triangle is 𝜋 radians, show
that
(a) sin 𝐴 = sin(𝐵 + 𝐶)
[3]
(b) sin 𝐴 + sin 𝐵 + sin 𝐶
= sin(𝐴 + 𝐵) + sin(𝐵 + 𝐶) +
sin(𝐴 + 𝐶).
[2]
CAPE 2013
cot 𝑦−cot 𝑥
sin(𝑥−𝑦)
25. (a) (i) Prove that cot 𝑥+cot 𝑦 = sin(𝑥+𝑦).
[4]
(ii) Hence, or otherwise, find the possible
values for 𝑦 in the trigonometric
equation
cot 𝑦 − cot 𝑥
= 1, 0 ≤ 𝑦 ≤ 2𝜋
cot 𝑥 + cot 𝑦
1
𝜋
when sin𝑥 = 2 , 0 ≤ 𝑥 ≤ 2 .
[8]
(b) (i) Express 𝑓(𝜃) = sin 2𝜃 + 4 cos 2𝜃 in
the form 𝑟 sin(2𝜃 + 𝛼) where 𝑟 > 0
𝜋
and 0 < 𝛼 < 2.
[4]
(ii) Hence, or otherwise, determine
(a) the value of 𝜃, between 0 and 2𝜋
radians, at which 𝑓(𝜃) is a
minimum.
[4]
(b) the minimum and maximum
1
values of 7−𝑓(𝜃).
[5]
CAPE 2014
26. (a) (i) Show that cos 3𝑥 = 4 cos 3 𝑥 − 3 cos 𝑥.
[6]
(ii) Hence, or otherwise, solve
cos 6𝑥 − cos 2𝑥 = 0 for 0 ≤ 𝑥 ≤ 2𝜋.
[9]
(b) (i) Express 𝑓(2𝜃) = 3 sin2𝜃 + 4 cos 2𝜃 in
the form 𝑟 sin(2𝜃 + 𝛼) where 𝑟 > 0
𝜋
and 0 < 𝛼 < 2.
[6]
(ii) Hence, or otherwise, find the
maximum and minimum values of
1
.
[4]
( )
CAPE 2015
SOLUTIONS
1.
2.
𝜃 = 41.8°, 138.1°
3.
𝜃 = 6, 6
4.
5.
𝜋 5𝜋
63 56
,
65 65
4
12
63
56
63
(a) 5 (b) 13 (c) 65 (d)
10
7
6.
(i) 65 (ii) 65 (iii) 65
7.
(i) 𝜃 = 3 , 3
8.
𝜃 = 0, ,
9.
3√3−4
2𝜋 4𝜋
𝜋 5𝜋
3
3
, 2𝜋
2𝜋
(ii) 9
10.
11.
sin 2𝜃
12. (ii) tan 𝜃 = 1+cos 2𝜃
13. 𝜃 =
𝜋
3
+ 2𝑛𝜋, 𝜃 = 𝜋 + 2𝑛𝜋,
5𝜋
3
3 cos(𝜃 + 0.615𝑐 )
+ 2𝑛𝜋
14. (a) 𝑓(𝜃) = √
(b) −√3 (c) 𝜃 = 2.53𝑐
15. √14 sin(𝑥 − 14°)
𝜋
3𝜋
16. (a) √2 cos (𝜃 + 4 ) (b) 2𝑛𝜋, 2 + 2𝑛
17. (i)
1
18. 2
(ii) √97
𝜋
19. 𝑥 = 0, , 𝜋
2
20.
𝜋 𝜋 3𝜋 5𝜋
21. (a) 𝑥 = 6 , 4 , 4 , 6
(b) (i) 𝐵𝐶 = 8 sin 𝜃 + 6 cos 𝜃
(ii) 0.131𝑐
(iii) No
(c)(i)
(ii)
(iii) 𝑛
𝜋 𝜋 3𝜋
22. (a) (i)
(ii)
(iii) 𝜃 = 0, , ,
4 2
1
4
(b) − 2 , 0, 1
117
23. (a) − 125
(b) 3.40𝑐 , 6.03𝑐
𝜋 3𝜋
5𝜋 7𝜋
24. (a) (i)
(ii) 𝜃 = 0, 4 , 4 , 𝜋, 4 , 4 , 2𝜋
(b) (i) 5 cos(𝜃 + 0.927𝑐 )
(ii) (a) 5
1
(b) 13
25. (a) (i)
(ii) 𝑦 = 0, 𝜋, 2𝜋
(b) (i) √17 sin(2𝜃 + 1.33𝑐 )
(ii) (a) 𝜃 = 1.69𝑐
1
1
(b) min: 7+√17 ; max: 7−√17
26. (a) (i)
𝜋 𝜋 3𝜋
(ii) 𝑥 = 0, , ,
4 2
(b) (i) 5 sin(𝜃 + 0.927𝑐 )
4
, 𝜋,
5𝜋 3𝜋 7𝜋
4
,
2
1
,
4
, 2𝜋
1
(ii) max: 2; min: 12
7−𝑓 𝜃
111
CHAPTER 18: PARAMETRIC EQUATIONS
CHAPTER 18: PARAMETRIC EQUATIONS
At the end of this section, students should be able
to:


obtain the Cartesian equation of a curve
given its parametric representation;
obtain the parametric representation of a
curve given its Cartesian equation.
__________________________________________________________
If a curve is given by parametric equations, we
often are interested in finding an equation for the
curve in standard form: 𝑦 = 𝑓(𝑥)
LESSON 1
Find the Cartesian equation for:
𝑥 = 𝑡 2 − 1,
𝑦 = 2𝑡
SOLUTION
Solve the first equation for 𝑡 and plug that value
into the second equation.
𝑥 = 𝑡2 − 1
𝑡2 = 𝑥 + 1
𝑡 = ±√𝑥 + 1
𝑦 = ±2√𝑥 + 1
LESSON 2
Find the Cartesian equation for:
1
𝑥 = √𝑡, 𝑦 = 2 𝑡 + 3.
SOLUTION
𝑥 = √𝑡,
𝑦=
𝑥 = √𝑡
𝑡 = 𝑥2
1
𝑦 = 2 𝑥2 + 3
LESSON 3
1
𝑡+3
2
or
2𝑦 = 𝑥 2 + 6
Find the Cartesian equation for:
1
3
𝑥=
,
𝑦=
2−𝑡
1 + 2𝑡
SOLUTION
1
3
𝑥=
,
𝑦=
2−𝑡
1 + 2𝑡
1
𝑥=
2−𝑡
𝑥(2 − 𝑡) = 1
2𝑥 − 𝑥𝑡 = 1
2𝑥 − 1 = 𝑥𝑡
2𝑥 − 1
=𝑡
𝑥
LESSON 4
Find the Cartesian equation for
the following
𝑥 = 3 sin𝑡 ,
𝑦 = 5 cos 𝑡
SOLUTION
𝑥
𝑦
sin 𝑡 =
cos 𝑡 =
3
5
𝑥 2
𝑦 2
2
2
sin 𝑡 + cos 𝑡 = ( ) + ( )
3
5
𝑥 2 𝑦2
1=
+
9 25
This is the equation of an ELLIPSE with centre
(0, 0), 𝑥 – intercepts ±3 and 𝑦 – intercepts ±5.
LESSON 5
A point moves so that at time 𝑡
the distances from the coordinate axes are given
by
𝑥 = 3 + 2 sin𝑡 and 𝑦 = 2 + 3 cos 𝑡
(i) Find the maximum and minimum values of 𝑥
and 𝑦
(ii) Find the Cartesian equation of the curve
traced by the point
SOLUTION
(i) The maximum value of 𝑥 occurs when
sin 𝑡 = 1, therefore 𝑥max = 3 + 2(1) = 5
The minimum value of 𝑥 occurs when
sin 𝑡 = −1, therefore 𝑥min = 3 + 2(−1) = 1
Likewise, for 𝑦 we have
𝑦max = 2 + 3(1) = 5 and
𝑦min = 2 + 3(−1) = −1
(ii) 𝑥 = 3 + 2 sin 𝑡 , 𝑦 = 2 + 3 sin𝑡
3
3
3𝑥
𝑥 + 4𝑥 − 2
𝑦=
=
=
2(2𝑥 − 1)
𝑥
5𝑥 − 2
1+
𝑥
112
CHAPTER 18: PARAMETRIC EQUATIONS
𝑥 −3
𝑦−2
,
cos 𝑡 =
2
3
2
𝑥
−
3
𝑦−2 2
sin2 𝑡 = (
)
cos2 𝑡 = (
)
2
3
sin 𝑡 =
(𝑥 − 3)2 (𝑦 − 2) 2
+
4
9
2
2
(𝑥 − 3)
(𝑦 − 2)
1=
+
4
9
This is an ELLIPSE with centre (3, 2). The
horizontal distance from the centre is 2 in either
direction and the vertical distance from the centre
in either direction is 3.
…………………………………………………………………………..
6.
sin2 𝑡 + cos2 𝑡 =
7.
Determine the coordinates of the centre of the
circle and the radius of the circle.
Show that the Cartesian equation represented
by the parametric equations
𝑥 = 3 + cos 𝜃 ,
𝑦 = sin𝜃 − 1
represents the equation of a circle.
State the coordinates of the centre of the circle
and the radius of the circle.
For each of the following eliminate the
parameter and find the corresponding
Cartesian equations
𝑥 = 2 cos 𝜃 ,
𝑦 = 4 sin 𝜃
𝑥 = 𝑡 + 2,
𝑦=𝑡
Hence, determine the points of intersection of
the 2 curves.
EXERCISE 18
SOLUTIONS
1.
Find the Cartesian equations for each of the
following pairs of parametric equations.
𝑡
a) 𝑥 = 𝑡 + 4, 𝑦 = 4
1.
3.
4.
Find the Cartesian equations for each of the
following pairs of parametric equations
a. 𝑥 = cos 𝑡 , 𝑦 = 3 sin 𝑡
b. 𝑥 = 3 sin𝑡 , 𝑦 = cos 𝑡
c. 𝑥 = −4 + cos 𝑡 , 𝑦 = −1 + sin𝑡
d. 𝑥 = 4 + cos 𝑡 , 𝑦 = 9 sin𝑡
e. 𝑥 = 2 cos 𝑡 , 𝑦 = cos2 𝑡
A point moves so that at time 𝑡 the distances
from the coordinate axes are given by
𝑥 = −3 + 2 cos 𝑡 and 𝑦 = 1 + 3 sin𝑡
(e) 𝑦 =
1−𝑥
3𝑥2
(d) 𝑦 = 𝑥2 +4
(f) 𝑦 =
𝑥2
1−𝑥
2
2.
𝑦 = 𝑥 2 − 6𝑥 + 10
3.
(a) 𝑥 2 + = 1
(b) + 𝑦 2 = 1
9
9
(c) (𝑥 + 4) 2 + (𝑦 + 1)2 = 1
𝑦2
𝑥2
(d) (𝑥 − 4) 2 + = 1
(e) + (1 − 𝑦) = 1
4.
(a) 𝑥max = −1,
𝑦max = 4,
𝑦2
𝑥2
81
𝑥+3 2
5.
6.
7.
𝑦−1 2
4
𝑥min = −5,
𝑦min = −2
(b) ( ) + ( ) = 1
2
3
(𝑥 − 1) 2 + (𝑦 − 2) 2 = 1
(𝑥 − 3) 3 + (𝑦 + 1) 2 = 1
𝑥2
𝑦2
+ 16 = 1
6 16
(− , − ) ; (2, 0)
5
5
4
𝐶(1, 2); 𝑟 = 1
𝐶(3, −1); 𝑟 = 1
𝑦 = 𝑥−2
EXAM QUESTIONS
1.
Obtain the Cartesian equation of the curve
whose parametric representation is
𝑥 = 2𝑡 2 + 3, 𝑦 = 3𝑡 4 + 2 in the form
𝑦 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶, where 𝐴, 𝐵 and 𝐶 are real
numbers.
[6]
CAPE 2004
2.
(i) Find the coordinates of the centre and
the radius of the circle
𝑥 2 + 𝑦 2 + 2𝑥 − 4𝑦 = 4.
[4]
(ii) By writing 𝑥 + 1 = 3 sin 𝜃, show that the
parametric equations of this circle are
𝑥 = −1 + 3 sin𝜃, 𝑦 = 2 + 3 cos 𝜃.
[5]
a.
5.
Find the maximum and minimum values
of 𝑥 and 𝑦.
b. Find the Cartesian equation of the curve
traced by the point.
By eliminating the parameter from
𝑥 = 1 + cos 𝜃 ,
𝑦 = 2 + sin 𝜃
show that the corresponding Cartesian
equation represents the equation of a circle.
(b) 𝑦 = 4𝑥 + 4
4
𝑥
b) 𝑥 = 4 , 𝑦 = 𝑡 + 4
in the form 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 where 𝑎, 𝑏 and 𝑐
are real numbers.
𝑥−4
(c) 𝑦 = ±√2 + 1
𝑡
c) 𝑥 = 2𝑡 2 , 𝑦 = 𝑡 + 1
2
3
(d) 𝑥 =
,𝑦 =
1
+
𝑡
√𝑡
1
(e) 𝑥 =
, 𝑦 = 𝑡(1 + 𝑡)
𝑡+1
1
𝑡
(f) 𝑥 =
,𝑦 =
1 + 2𝑡
1 + 2𝑡
2. Obtain the Cartesian equation of the curve
whose parametric representation is
𝑥 = 2𝑡 2 + 3 𝑦 = 4𝑡 4 + 1
(a) 𝑦 =
113
CHAPTER 18: PARAMETRIC EQUATIONS
(iii) Show that the 𝑥 – coordinate of the points
of intersection of this circle with the line
3
𝑥 + 𝑦 = 1 are 𝑥 = −1 ± 2 √2.
[4]
CAPE 2006
3. The line 𝐿 has equation 𝑥 − 𝑦 + 1 = 0 and the
circle 𝐶 has equation 𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0.
(i) Show that 𝐿 passes through the centre of
𝐶.
[2]
(ii) If 𝐿 intersects 𝐶 at 𝑃 and 𝑄, determine the
coordinates of 𝑃 and 𝑄.
[3]
(iii) Find the constants 𝑎, 𝑏 and 𝑐 such that
𝑥 = 𝑏 + 𝑎 cos 𝜃 and 𝑦 = 𝑐 + 𝑎 sin𝜃 are
parametric equations of 𝐶.
[3]
(iv) Another circle 𝐶2 , with the same radius as
𝐶, touches 𝐿 at the centre of 𝐶. Find the
possible equations of 𝐶2 .
[7]
CAPE 2011
4. (i) Determine the Cartesian equation of the
curve, 𝐶, defined by the parametric
equations 𝑦 = sec 𝜃 and 𝑥 = 3 tan 𝜃. [5]
(ii) Find the points of intersection of the curve
𝑦 = √10𝑥 with 𝐶.
[9]
CAPE 2012
5.
Show that the Cartesian equation of the
curve that has the parametric equations
𝑥 = 𝑡 2 + 𝑡, 𝑦 = 2𝑡 − 4 is 4𝑥 = 𝑦 2 + 10𝑦 + 24.
[4]
CAPE 2013
6.
The parametric equations of a curve, 𝑆, are
given by
1
𝑡
𝑥=
and 𝑦 =
1+𝑡
1 − 𝑡2
Determine the Cartesian equation of the
curve,𝑆.
[6]
CAPE 2014
7.
The circles 𝐶1 and 𝐶2 are defined by th
eparametric equations as follows:
𝐶1 : 𝑥 = √10 cos 𝜃 − 3; 𝑦 = √10 sin𝜃 + 2
𝐶2 : 𝑥 = 4 cos 𝜃 + 3; 𝑦 = 4 sin𝜃 + 2
3.
(i)
(ii) (2√2, 1 + 2√2); (−2√2, 1 − 2√2)
(iii) 𝑎 = 4, 𝑏 = 0, 𝑐 = 1
2
2
(iv) (𝑥 − 2√2) + (𝑦 − (1 − 2√2) = 16
4.
𝑥2
(i) 𝑦 2 − 9 = 1
(ii) (45 + 12√14, 29.98); (45 − 12√14, 1)
5.
6.
7.
𝑦=
𝑥(1−𝑥)
2𝑥−1
(i) 𝐶1 :(𝑥 + 3)2 + (𝑦 − 2)2 = 10,
𝐶2 : (𝑥 − 3)2 + (𝑦 − 2)2 = 16
1
√15
1
√15
(ii) (− 2 , 2 + 2 ) ; (− 2 , 2 − 2 )
…………………………………………………………………………..
(i) Determine the Cartesian equations of 𝐶1
and 𝐶2 in the form
(𝑥 − 𝑎) 2 + (𝑦 − 𝑏)2 = 𝑟 2 .
[4]
(ii) Hence or otherwise, find the points of
intersection of 𝐶1 and 𝐶2 .
[9]
CAPE 2015
SOLUTIONS
1.
2.
3
9
17
𝑦 = 4 𝑥2 − 2 𝑥 + 2
(i) 𝐶(−1, 2); 𝑟 = 3
114
CHAPTER 19: LIMITS
MODULE THREE: CALCULUS
CHAPTER 19: LIMITS
At the end of this section, students should be able
to:
 use graphs to determine the
continuity and discontinuity of
functions;
 describe the behaviour of a function,
𝑓(𝑥) as 𝑥 gets arbitrarily close to
some given fixed number, using a
descriptive approach;
 use the limit notation lim 𝑓(𝑥) = 𝐿,
If we refer to the polygon as an 𝑛-gon, where 𝑛 is
the number of sides, we can make some
equivalent mathematical statements.



𝑥→𝑎

𝑓(𝑥) → 𝐿 as 𝑥 → 𝑎;
use the simple limit theorems:
if lim 𝑓(𝑥) = 𝐹, lim 𝑔(𝑥) = 𝐺 and 𝑘 is
𝑥→𝑎
𝑥→𝑎
a constant, then lim 𝑘𝑓(𝑥) 𝑘𝐹,
𝑥→𝑎
lim 𝑓(𝑥 )𝑔 (𝑥 ) = 𝐹𝐺, lim {𝑓 (𝑥 ) +
𝑥→𝑎
𝑥→𝑎
𝑔 (𝑥 )} = 𝐹 + 𝐺 and, provided
𝑓 (𝑥)
𝐹
𝐺 ≠ 0, lim
= ;


𝑥→𝑎 𝑔(𝑥)
𝐺
use the limit theorems in simple
problems:
sin 𝑥
use the fact that lim
= 1,
𝑥→0 𝑥
demonstrated by a geometric
approach;
 identify the point(s) for which a
function is (un)defined;
 identify the points for which a
function is continuous;
 identify the point(s) where a function
is discontinuous;
 use the concept of left – handed or
right – handed
__________________________________________________________
INTRODUCTION
To understand what is really going on in
differential calculus, we need to have an
understanding of limits. In the study of calculus,
what happens to the value of a function as the
independent variable gets very close to a
particular value is very important.
Consider the following in order to develop an
understanding of what is a limit:
If we increase the number of sides of a polygon, in
relation to a circle what can we say about the
polygon?
As the number of sides increases the polygon
becomes closer and closer to being a circle.
As 𝑛 gets larger, the 𝑛-gon gets closer to
being the circle.
As 𝑛 approaches infinity, the 𝑛-gon
approaches the circle.
The limit of the 𝑛-gon, as 𝑛 goes to
infinity, is the circle!
lim (𝑛 − gon) = circle
𝑛→∞
The 𝑛-gon never really gets to be the circle, but it
will get very close! So close, in fact, that, for all
practical purposes, it may as well be the circle.
That's what limits are all about!
TABULAR APPROACH
Now let’s look at a numerical example.
Find the limit as 𝑥 approaches 10 of the function
𝑓(𝑥) = 3𝑥 + 5. Firstly we need to note that 𝑥 can
approach 10 from the left or the right. This idea
will be demonstrated using a table of values and
the graph of the function.
Approaching 10 from the left
9.9999
𝑥
9
9.5
9.9
9.99
9.999
𝑓(𝑥) 32
33.5 34.7 34.97 34.997 34.9997
It appears that as 𝑥 gets closer and closer to 10
from the left that 𝑓(𝑥) gets closer and closer to 35.
Approaching 10 from the right
𝑥
𝑓(𝑥)
10.0001
35.0003
10.001
35.003
10.01
35.03
10.1
35.3
10.5
36.5
As 𝑥 gets closer and closer to 10 from the right
𝑓(𝑥) approaches 35. Hence, is seems reasonable to
conclude that as 𝑥 gets closer and closer to 10,
𝑓(𝑥) approaches 35.
GRAPHICAL APPROACH
We will now to take a look at the graph of 𝑓(𝑥) to
help verify our answer.
115
11
38
CHAPTER 19: LIMITS
LESSON 2
Evaluate the following limits.
𝑥 2 − 2𝑥 − 3
1. lim
𝑥→3
𝑥 −3
𝑥2 − 4
2. lim
𝑥→2 𝑥 − 2
√𝑥 − 2
3. lim
𝑥→2 𝑥 − 4
SOLUTION
Thus, we have the following limit as 𝑥 approaches
10 from the left (left-hand limit)
lim− 3𝑥 + 5 = 35
(∗)
𝑥→10
and from the right (right-hand limit)
lim+ 3𝑥 + 5 = 35 (∗∗)
𝑥→10
Therefore, since 𝑥 converges to 10 from both the
left and right, 𝑓(𝑥) approaches 35 we conclude
lim 3𝑥 + 5 = 35 (∗∗∗)
𝑥→10
(∗) and (∗∗) are one – sided limits, whereas, (∗∗∗)
is a two sided limit.
LIMITS BY DIRECT SUBSTITUTION
LESSON 1
Determine
𝑥→10
(ii) lim 3𝑥 3 − 𝑥 2 + 2𝑥 − 1
𝑥→3
𝑥→−1
lim
𝑥2 −2𝑥−3
𝑥−3
𝑥→3
(𝑥 + 1)(𝑥 − 3)
𝑥 −3
= lim 𝑥 + 1
= lim
𝑥→3
𝑥→3
=3+1
=4
Substituting 𝑥 = 3 immediately will lead directly
to division by zero. Therefore, we factor and
cancel before substituting.
𝑥2 − 4
𝑥→2 𝑥 − 2
(𝑥 − 2)(𝑥 + 2)
= lim
𝑥→2
𝑥 −2
= lim 𝑥 + 2
2. lim
𝑥→2
(i) lim 3𝑥 + 5
(iii) lim
1.
=2+2
=4
𝑥2 +3𝑥+1
𝑥2 +1
3. lim
√𝑥 − 2
𝑥→2 𝑥 − 4
SOLUTION
(i) lim 3𝑥 + 5
= lim
= 3(10) + 5
= 35
(ii) lim 3𝑥 3 − 𝑥 2 + 2𝑥 − 1
= lim
𝑥→10
𝑥→3
= 3 (3)3 − 32 + 2(3) − 1
= 77
(iii) lim
𝑥→−1
√𝑥 − 2
𝑥→2 (√𝑥 − 2)(√𝑥 + 2)
𝑥2 +3𝑥+1
𝑥2 +1
1
𝑥→2 (√𝑥 + 2)
=
1
√4 + 2
1
=
4
(−1)2 + 3(−1) + 1
=
(−1)2 + 1
1
=
2
In general, if 𝑓(𝑥) is continuous then:
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
It is important to note that all polynomial
functions are continuous.
116
CHAPTER 19: LIMITS
LIMIT PROPERTIES
1. lim 𝑓(𝑥) = 𝐿
or
𝑥→𝑎
as 𝑥 → 𝑎 iff
𝑓(𝑥) → 𝐿
lim 𝑓(𝑥) = 𝐿
and
𝑥→𝑎−
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎+
small? Yup, the graph is again getting closer and
closer to the 𝑥-axis
(which is 0.) It's
just coming in from
below this time.
1
lim ( ) = 0
𝑥→−∞ 𝑥
2. 𝑘. lim 𝑓(𝑥) = lim 𝑘. 𝑓(𝑥)
𝑥→𝑎
𝑥→𝑎
3. lim[𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
4. lim[𝑓(𝑥)𝑔(𝑥)] = lim 𝑓(𝑥) × lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
𝑓(𝑥)
𝑓(𝑥) lim
5. lim
= 𝑥→𝑐
𝑥→𝑐 𝑔(𝑥)
lim 𝑔(𝑥)
𝑥→𝑐
APPLICATIONS OF LIMIT LAWS
lim 5𝑥 = 5. lim 𝑥 = 5 . 4 = 20
𝑥→4
𝑥→4
lim(3𝑥 + 1)(2𝑥 − 1)
𝑥→2
= lim (3𝑥 + 1) × lim(2𝑥 − 1)
𝑥→2
𝑥→2
= (3 (2) + 1)(2(2) − 1)
=7×3
= 21
𝑥−1
𝑥→3 𝑥 2 + 1
lim 𝑥 − 1
= 𝑥→3 2
lim 𝑥 + 1
𝑥→3
3−1
=
9−1
1
=
4
lim
LESSON 3
5−3𝑥
Find lim (6𝑥+1)
𝑥→∞
SOLUTION
In solving these questions please
5−3(∞)
do not write 6(∞)+1 , this does not make
mathematical sense. Instead we use the fact
1
thatlim = 0. Therefore, we divide throughout by
𝑥→∞ 𝑥
𝑥 to get an expression that can be evaluated
5 − 3𝑥
lim (
)
𝑥→∞ 6𝑥 + 1
5
−3
= lim (𝑥
)
1
𝑥→∞
6+
𝑥
0−3
=
6+0
3
=−
6
1
=−
2
3𝑥2 −𝑥−2
LESSON 4
Evaluate lim 5𝑥2 +4𝑥+1
𝑥→∞
LIMITS AS 𝒙 APPROACHES INFINITY
INTRODUCTION
1
Now let’s look at the graph of 𝑓(𝑥) = 𝑥 from
another perspective. We are now going to examine
the limit as 𝑥 approaches infinity (both positive
and negative infinity).
1
Now, let's look at the graph of 𝑓(𝑥) = 𝑥 and see
what happens!
As 𝑥 gets really, really big, the graph gets closer
and closer to the 𝑥-axis which has a height of 0. So,
as 𝑥 approaches +∞, 𝑓(𝑥) is approaching 0. This is
called a limit at infinity.
SOLUTION
Divide throughout by the highest
power of 𝑥
3𝑥 2 𝑥
2
2 − 𝑥2 − 𝑥2
𝑥
lim
𝑥→∞ 5𝑥 2
4𝑥 1
+ 2+ 2
𝑥2
𝑥
𝑥
1 2
3−𝑥 − 2
𝑥
= lim
4 1
𝑥→∞
5+ + 2
𝑥 𝑥
3−0−0
=
5+0+0
3
=
5
1
lim ( ) = 0
𝑥→+∞ 𝑥
Now let's look at the green line... What is
happening to the graph as 𝑥 gets really, really
117
CHAPTER 19: LIMITS
LIMITS AS 𝒙 APPROACHES 0
INTRODUCTION
Although we cannot divide by 0 there are some
interesting, and important, limits where there is a
limiting value as 𝑥 approaches 0 and where it
would appear that we have a 0 denominator.
LESSON 5
Evaluate lim
𝑥→0
sin 3𝑥
𝑥
SOLUTION
The same graphical process can be used to derive
that
sin3𝑥
lim
=3
𝑥→0
𝑥
To illustrate we take a look at the graph of
1
𝑓(𝑥) = to see what happens as 𝑥 approaches 0.
𝑥
We need to look at two separate cases:
(1) the left – hand limit as 𝑥 approaches 0 and
(2) the right – hand limit as 𝑥 approaches 0
1
1
lim ( ) = −∞ and
lim ( ) = +∞
𝑥→0− 𝑥
𝑥→0+ 𝑥
Since the left – hand limit does not equal the right
– hand limit
1
lim ( ) = DOES NOT EXIST
𝑥→0 𝑥
In this case the 𝑦-axis is a vertical asymptote.
INTRODUCTION
Limit of
𝐬𝐢𝐧 𝒙
𝒙
We now look at the limit as 𝑥 approaches 0 of
sin𝑥
𝑥
We can use a table of values sufficiently close to 0
in order to evaluate the limit. This process reveals
that
sin 𝑥
lim
=1
𝑥→0 𝑥
This is supported by the graph below.
.
However, we are these limits can also be derived
algebraically as follows.
sin3𝑥
lim
𝑥→0
𝑥
sin 3𝑥 3𝑥
= lim
×
𝑥→0 3𝑥
𝑥
sin 3𝑥
3𝑥
= lim
× lim
𝑥→0 3𝑥
𝑥→0 𝑥
=1×3
=3
LESSON 6
Determine
sin 4𝑥
lim
𝑥→0 3𝑥
SOLUTION
sin4𝑥
lim
𝑥→0 3𝑥
sin 4𝑥 4𝑥
= lim
×
𝑥→0 4𝑥
3𝑥
sin 4𝑥
4𝑥
= lim
× lim
𝑥→0 4𝑥
𝑥→0 3𝑥
4
=1×
3
4
=
3
118
CHAPTER 19: LIMITS
LESSON 7
Evaluate
7.
sin 5𝑥
lim
𝑥→0 sin 2𝑥
SOLUTION
sin5𝑥
lim
𝑥→0 sin2𝑥
5𝑥 sin5𝑥
2𝑥
= lim
×
×
𝑥→0 2𝑥
5𝑥
sin 2𝑥
5
= ×1×1
2
5
=
2
…………………………………………………………………………..
EXERCISE 19.1
1.
Determine the limits of each function
(a) lim (3𝑥 2 + 5)
𝑥→2
a. lim
(c)
2.
lim
𝑥 2 + 3𝑥 + 2
𝑥→−1
𝑥2 + 1
6
𝑥 −2
𝑥→2 𝑥 2 − 4
𝑥 2 − 5𝑥 − 6
b. lim
𝑥→−1
𝑥 +1
𝑥 +2
c. lim 3
𝑥→−2 𝑥 + 8
𝑥−2
d. lim
𝑥→2 |𝑥 − 2|
5
𝑥→∞ 𝑥 4
4𝑥 2 + 1
𝑥→∞ 2 + 3𝑥 2
2𝑥 + 1
𝑥→∞ 5𝑥 − 1
𝑥2 + 2
e. lim 3
𝑥→+∞ 𝑥 + 𝑥 + 1
c. lim
d. lim
SOLUTIONS
1.
2.
(a) 17 (b) −17 (c) 0
1
1
(a) 4 (b) −7 (c) 12 (d) DOES NOT EXIST
3.
(a) 39
4.
5.
−
4
(a) Does not exist (b) does not exist
3
(c) 5 (d) 3 (e)
3
1
(b) − 11
4
6.
7.
(a) 0 (b) 0 (c)
Find
a. lim
b. lim
𝑥→∞ 𝑥
(b) lim (−2𝑥 2 + 1)
𝑥→−3
Evaluate
2
5
(d)
4
3
(e) 0
…………………………………………………………………………..
LIMITS AND PIECEWISE FUNCTIONS
LESSON 8
follows:
The function 𝑓(𝑥) is defined as
𝑓 (𝑥 ) = {
Find lim 𝑓(𝑥)
𝑥2
𝑥 +2
𝑥<0
𝑥≥0
𝑥→0
3.
Evaluate each of the following by applying the
appropriate law of limits.
a. lim(2𝑥 2 − 3𝑥 + 4)
SOLUTION
𝑥→5
𝑥 3 + 2𝑥 2 − 1
𝑥→−2
5 − 3𝑥
1
If lim 𝑓(𝑥) = − 2 and
b. lim
4.
𝑥→𝑐
2
𝑓(𝑥)
, find lim
𝑥→𝑐 𝑔(𝑥)
3
Evaluate each limit
1
4
a. lim 5
b. lim 11
𝑥→0 𝑥
𝑥→0 𝑥
sin 5𝑥
sin6𝑥
c. lim
d. lim
𝑥→0
𝑥→0 2𝑥
𝑥
3𝑥
e. lim
𝑥→0 sin 4𝑥
Show that
sin 𝑥 2
lim𝜋
=
𝑥
𝜋
𝑥→
lim 𝑔(𝑥) =
𝑥→𝑐
5.
6.
2
We will use the graph of 𝑓(𝑥) to help us solve the
question
Firstly, we note that the graph is discontinuous,
with a break, at 𝑥 = 0.
Since 𝑓(𝑥 ) = 𝑥 2 for 𝑥 < 0, we have
lim− 𝑓(𝑥) = lim− 𝑥 2 = 02 = 0
𝑥→0
𝑥→0
Since 𝑓(𝑥) = 𝑥 + 2 for 𝑥 ≥ 0, we have
lim+ 𝑓(𝑥) = lim+ 𝑥 + 2 = 0 + 2 = 2
𝑥→0
𝑥→0
So we have 2 different limiting values for 𝑓(𝑥)as
119
CHAPTER 19: LIMITS
𝑥 → 0. As a result we are left to conclude that lim 𝑓(𝑥)
𝑥→0
LESSON 11
does not exist.
LESSON 9
lim 𝑓(𝑥) if
𝑥→1
Use the graph of 𝑓(𝑥) to find
3 − 𝑥,
𝑓(𝑥) = {
1,
𝑥≠1
𝑥=1
The function 𝑓 on ℝ is defined by
𝑥2 + 1 𝑥 ≥ 3
𝑓(𝑥) = {
1 + 𝑏𝑥 𝑥 < 3
Determine
(a) 𝑓(3)
(b) lim+ 𝑓(𝑥)
𝑥→3
(c) lim− 𝑓(𝑥) in terms of the constant 𝑏
𝑥→3
(d) the value of 𝑏 such that 𝑓 is continuous at
𝑥 = 3.
SOLUTION
(a) 𝑓(3) = 32 + 1 = 10
(b) lim+ 𝑓(𝑥) = lim+ 𝑥 2 + 1 = 32 + 1 = 10
𝑥→3
𝑥→3
(c) lim− 𝑓 (𝑥 ) = lim− 1 + 𝑏𝑥 = 1 + 3𝑏
𝑥→3
SOLUTION
From the graph we see that
lim− 𝑓(𝑥) = 1 and lim+ 𝑓(𝑥) = 1
𝑥→1
𝑥→3
(d) If 𝑓 is continuous at 𝑥 = 3 then
lim+ 𝑓 (𝑥 ) = lim− 𝑓 (𝑥 )
𝑥→3
𝑥→1
𝑥→3
10 = 1 + 3𝑏
3=𝑏
Therefore,
lim 𝑓(𝑥) = 1
𝑥→1
PIECEWISE FUNCTIONS AND CONTINUITY
LESSON 10
The function 𝑓 on ℝ is defined by
4 − 𝑥 if 𝑥 ≥ 1
𝑓(𝑥) = {
2 + 𝑥 if 𝑥 < 1
(i) Sketch the graph of 𝑓(𝑥) for the domain
−1 ≤ 𝑥 ≤ 2.
(ii) Find
(a) lim+ 𝑓(𝑥)
𝑥→1
(b) lim− 𝑓(𝑥)
𝑥→1
(iii) Deduce that 𝑓(𝑥) is continuous at 𝑥 = 1.
LESSON 12
Determine the values of 𝑥 for
2𝑥+1
which the function 𝑓(𝑥) = 2
is continuous.
𝑥 +𝑥−2
SOLUTION
𝑓(𝑥) will be discontinuous at
those values of 𝑥 for which the denominator
𝑥 2 + 𝑥 − 2 = 0. Therefore for all other values of
𝑥, 𝑓(𝑥) is continuous.
𝑥2 + 𝑥 − 2 = 0
→ (𝑥 + 2)(𝑥 − 1) = 0
→ 𝑥 = −2, 1
𝑓(𝑥) is continuous for all real values of 𝑥 except
−2 and 1
…………………………………………………………………………..
EXERCISE 19.2
SOLUTION
(i)
1.
(ii) (a) lim+ 𝑓(𝑥) = lim+ 4 − 𝑥 = 4 − 1 = 3
𝑥→1
𝑥→1
(b) lim− 𝑓(𝑥) = lim− 2 + 𝑥 = 2 + 1 = 3
𝑥→1
𝑥→1
(iii) 𝑓(1) = 4 − 1 = 3
Since lim+ 𝑓(𝑥) = lim− 𝑓(𝑥) = 𝑓(1) = 3, 𝑓(𝑥)
𝑥→1
is continuous.
𝑥→1
2.
Determine whether or not each of the
following functions are continuous.
2
a. 𝑓(𝑥) =
1+𝑥
b. 𝑓(𝑥) = 𝑥 2 + 5
4 − 𝑥2
c. 𝑓(𝑥) = 3
𝑥 +8
Let 𝑓 be the function defined by
2
if 𝑥 < 1
𝑓(𝑥) = {2𝑥
4 − 𝑥 if 𝑥 ≥ 1
Graph 𝑓, and use the graph to find the
following
a. lim− 𝑓(𝑥)
𝑥→1
b.
lim 𝑓(𝑥)
𝑥→1+
c. lim 𝑓(𝑥)
𝑥→1
120
CHAPTER 19: LIMITS
3.
Determine the values of 𝑥 for which the
𝑥−1
function 𝑓(𝑥) = 2
is continuous.
4.
Determine the values of 𝑥 for which the
1
function 𝑓(𝑥) = 2√𝑥−9 is discontinuous.
𝑥 +2𝑥−3
5.
2.
3.
4.
(a) 2 (b) 3 (c) DOES NOT EXIST
𝑥 ≠ −3, 1
81
𝑥= 4
5.
(i) 3 (ii) −3 (iii) 2 (iv) 5 (v) 2 (vi) 1
2
1
3
1
Find
(vii) DOES NOT EXIST (viii) 2
𝑥2 − 9
(i) lim 2
𝑥→3 3𝑥 − 9𝑥
𝑥2 + 𝑥 − 2
(ii) lim 2
𝑥→1 𝑥 − 3𝑥 + 2
𝑥 2 − 2𝑥 − 3
(iii) lim 2
𝑥→3 𝑥 − 4𝑥 + 3
𝑥2 + 𝑥 − 2
(iv) lim 2
𝑥→−2 𝑥 + 5𝑥 + 6
𝑥3 + 8
(v) lim 3
𝑥→−2 𝑥 − 4𝑥
𝑥+4
(vi) lim
𝑥→∞ 𝑥 + 2
1
(vii) lim 2
𝑥→0 𝑥
3 + 𝑥2
(viii) lim
𝑥→∞ 1 + 2𝑥 2
11
6.
7.
8.
−4
−72
9.
2, 5
5
2
2
10. 𝑎 = −2, 𝑏 = 1
7
7
11. (i) 𝑥 ≠ ± 2 (ii) 𝑥 = ± 2 (iii) 𝑥 = −3, 6
(iv) 𝑥 = ±3
EXAM QUESTIONS
1.
Given that lim {4𝑓(𝑥)} = 5, evaluate
𝑥→−2
lim {𝑓(𝑥) + 2𝑥}.
[5]
𝑥→−2
CAPE 2004
2.
6.
Given that lim {4𝑓(𝑥)} = 5, evaluate
7.
lim {𝑓(𝑥) + 2𝑥}
𝑥→−2
Given that lim {𝑓(𝑥) + 3𝑥} = 1,
𝑥2 −2𝑥−3
(a) Evaluate lim 𝑥2 −4𝑥+3.
(b) Determine the values of 𝑥 ∈ ℝ for which
𝑥+2
the function 𝑥(𝑥+1) is NOT continuous.
𝑥→−2
[3]
𝑥→3
evaluate lim 9𝑓(𝑥)
CAPE 2004
𝑥→3
8.
Given that lim(𝑎√𝑥 + 𝑥) = 9, calculate the
9.
value of 𝑎.
sin 𝑥
Given that lim 𝑥 = 1,
𝑥→4
𝑥→0
sin2𝑥
evaluate lim
𝑥→0
𝑥
3.
𝑓(𝑥) =
𝑥→0
11. Determine the real values of 𝑥 for which the
following functions are continuous.
𝑥
i. 𝑓(𝑥) =
2|𝑥| − 7
𝑥
ii. 𝑓(𝑥) =
|2𝑥| − 7
𝑥2 + 1
iii. 𝑓(𝑥) =
|2𝑥 − 3| − 9
|𝑥 |
iv. 𝑓(𝑥) = 2
|𝑥| − 9
SOLUTIONS
(a) Discontinuous (b) Continuous
(c) Discontinuous
(a) Find the real values of 𝑥 for which the
function
𝑥
𝑥 2 − 2𝑥 − 8
is discontinuous.
sin 2𝑥
and lim sin 5𝑥 .
10. The function 𝑓 is defined by
1
𝑓(𝑥 ) =
(𝑥 + 2)(𝑥 − 1)
and is continuous for all values of 𝑥 except 𝑎
and 𝑏, where 𝑎 < 𝑏. Find the values of 𝑎 and 𝑏.
1.
[4]
𝑥→3
[3]
𝑥2 +𝑥−2
(b) Find lim 𝑥2 −3𝑥+2
[3]
𝑥→1
(c) Find the values of 𝑥 ∈ ℝ such that the
9−𝑥2
function 𝑓(𝑥) = (𝑥2 −3)(|𝑥|−3) is
discontinuous.
4.
𝑥3 +8
(a) Determine lim 𝑥3 −4𝑥 .
𝑥→−2
[4]
CAPE 2006
[4]
(b) Obtain the real values of 𝑥 such that the
𝑥2 +1
function 𝑓(𝑥) = |2𝑥−3|−9 is continuous.
[4]
CAPE 2007
5.
𝑥3 −27
Find lim 𝑥2 +𝑥−12 .
𝑥→3
[4]
CAPE 2008
121
CHAPTER 19: LIMITS
6.
𝑥3 −8
(a) Find lim 𝑥3 −6𝑥+8.
(b) The function 𝑓 on ℝ is defined by
3−𝑥 𝑥 ≥1
𝑓(𝑥) = {
1+𝑥 𝑥 <1
(i) Sketch the graph of 𝑓(𝑥) for the
domain −1 ≤ 𝑥 ≤ 2.
(ii) Find
(a) lim+ 𝑓(𝑥)
(b) the value of the constant 𝑝 such
that lim 𝑓(𝑥) exists.
[4]
𝑥→1
[2]
CAPE 2012
[2]
𝑥→1
(iii) Deduce that 𝑓(𝑥) is continuous at
𝑥 = 1.
[3]
CAPE 2009
10. A function 𝑓(𝑥) is defined as
𝑥+2 𝑥 <2
𝑓(𝑥) = { 2
𝑥
𝑥>2
(i) Find lim 𝑓(𝑥).
[4]
𝑥→2
(a) Find
𝑥2 −9
(i) lim 𝑥3 −27
[4]
(ii) lim sin2𝑥−4𝑥 .
[5]
𝑥→3
tan 𝑥−5𝑥
𝑥→0
(b) The function 𝑓 on ℝ is defined by
3𝑥 − 7 𝑥 > 4
𝑓(𝑥) = {
1 + 2𝑥 𝑥 ≤ 4
(i) Find
(a) lim+ 𝑓(𝑥)
[2]
𝑥→4
(b) lim− 𝑓(𝑥)
[2]
𝑥→4
(ii) Deduce that 𝑓(𝑥) is discontinuous at
𝑥 = 4.
[2]
CAPE 2010
𝑥2 +5𝑥+6
(a) Find lim 𝑥2 −𝑥−6
[4]
𝑥→−2
(b) The function 𝑓on ℝ is defined by
2
𝑓(𝑥) = { 𝑥 + 1 𝑥 ≥ 2
1 + 𝑏𝑥 𝑥 < 2
Determine
(i) 𝑓(2)
(ii) lim+ 𝑓 (𝑥 )
[2]
[2]
𝑥→2
(iii) lim− 𝑓(𝑥) in terms of the constant 𝑏.
𝑥→2
[2]
(iv) the value of 𝑏 such that 𝑓 is
continuous at 𝑥 = 2.
[4]
CAPE 2011
9.
(ii) Hence, determine the value of 𝑓(𝑥)
for 𝑓 to be continuous at the point
𝑥 = 1.
[1]
[2]
𝑥→1
8.
[2]
𝑥→1
(b) lim− 𝑓(𝑥)
7.
(i) Find
(a) lim+ 𝑓(𝑥)
[5]
𝑥→2
𝑥3 +8
(a) (i) Find the values of 𝑥 for which 𝑥2 −4 is
discontinuous.
[2]
𝑥3 +8
(ii) Hence, or otherwise, find lim 𝑥2 −4 .
𝑥→−2
(iii) By using the fact that lim
otherwise, find, lim
sin𝑥
𝑥→0 𝑥
2𝑥3 +4𝑥
𝑥→0 sin2𝑥
.
(b) The function 𝑓 on ℝ is defined by
𝑥2 + 1 𝑥 > 1
𝑓(𝑥) = {
4 + 𝑝𝑥 𝑥 < 1
[3]
= 1, or
[5]
(ii) Determine whether 𝑓(𝑥) is continuous at
𝑥 = 2. Give a reason for your answer. [2]
CAPE 2013
11. Let 𝑓(𝑥) be a function defined as
𝑎𝑥 + 2 𝑥 < 3
𝑓(𝑥) = { 2
𝑎𝑥
𝑥≥3
(i) Find the value of 𝑎 if 𝑓(𝑥) is continuous at
𝑥 = 3.
[4]
𝑥2 +2
(ii) Let 𝑔(𝑥 ) = 𝑏𝑥2 +𝑥+4.
Given that lim 2𝑔(𝑥) = lim 𝑔 (𝑥 ), find the
𝑥→1
value of 𝑏.
𝑥→0
[5]
CAPE 2014
12. Let 𝑓 be the function defined as
sin(𝑎𝑥)
𝑥 ≠ 0,
𝑎 ≠0
𝑓(𝑥) = { 𝑥
4
𝑥=0
If 𝑓 continuous at 𝑥 = 0, determine the value
of 𝑎.
[4]
CAPE 2015
SOLUTIONS
1.
2.
3.
4.
5.
6.
7.
11
−4
(a) 2 (b) 𝑥 = −1, 0
(a) 𝑥 = −2, 4 (b) −3 (c) 𝑥 = ±√3, ±3
3
(a) 2 (b) 𝑥 ≠ −3, 6
27
7
(a) 0 (b) (i)
(ii) (a) 2
(b) 2 (iii)
2
(a) (i) 9
(ii) 2 (b) (i) (a) 5 (b) 9
1
8.
(a) − 5 (b) (i) 5
(ii) 5
(iv) 𝑏 = 2
9. (a) (i) 𝑥 = ±2
(ii) −3
(b) (i) (a) 5
(b) 1
10. (i) 4 (ii) Yes
1
11. (i) (ii) 7
(ii)
(iii) 1 + 2𝑏
(iii) 2
(ii) 5
3
12. 4
…………………………………………………………………………..
122
CHAPTER 19: LIMITS
DIFFERENTIATION FROM FIRST
PRINCIPLES
INTRODUCTION
In this section, we will differentiate a function
from "first principles". This means we will start
from scratch and use algebra to find a general
expression for the slope of a curve, at any value 𝑥.
We wish to find an algebraic method to find the
slope of 𝑦 = 𝑓(𝑥) at 𝑃.
Therefore, 2𝑥 is the first derivative of
𝑦. At this stage we now introduce notation used
𝑑
for derivatives. 𝑑𝑥 is the operator for
𝑑
differentiation. 𝑑𝑥 means that we are
differentiating with respect to 𝑥, consequently
𝑑𝑦
means that we are differentiating 𝑦 with
𝑑𝑥
respect to 𝑥. Since 𝑦 = 𝑓(𝑥) the first derivative
can also be denoted as 𝑓 ′ (𝑥). Therefore,
𝑑
𝑑𝑦
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑦=
= lim
𝑑𝑥
𝑑𝑥 ℎ→0
ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
′
or 𝑓 (𝑥) = lim
ℎ→0
ℎ
LESSON 14
Differentiate from first principles
𝑦=
SOLUTION
1
𝑓(𝑥) =
𝑥
𝑓(𝑥 + ℎ) =
1
𝑥+ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥) =
1
𝑥
1
1
−
𝑥 +ℎ 𝑥
𝑥 − (𝑥 + ℎ)
𝑥(𝑥 + ℎ)
ℎ
=−
ℎ𝑥(𝑥 + ℎ)
1
=−
𝑥(𝑥 + ℎ)
𝑑𝑦
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
= lim
𝑑𝑥 ℎ→0
ℎ
−ℎ
1
= lim 2
×
ℎ→0 𝑥 + ℎ𝑥
ℎ
−1
= lim 2
ℎ→0 𝑥 + ℎ𝑥
1
=− 2
𝑥
=
We choose an arbitrary point close to 𝑃 say
𝑄 (𝑥 + ℎ, 𝑓(𝑥 + ℎ)) and then move this point
closer and closer to 𝑃. Eventually the line 𝑃𝑄 will
become the tangent of 𝑓(𝑥) at 𝑃. As 𝑄 gets closer
and closer to 𝑃, ℎ, the distance between 𝑥 and 𝑥 +
ℎ gets closer and closer to 0. Remember, we are
interested in determining the rate of change of a
variable, 𝑦, compared to another 𝑥. Hence, we
need to determine
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
lim
ℎ→0
ℎ
where [𝑓(𝑥 + ℎ) − 𝑓(𝑥)] is the change in 𝑦 and ℎ
is the change in 𝑥
LESSON 13
Differentiate 𝑦 = 𝑥 2 from first
principles.
SOLUTION
𝑓(𝑥) = 𝑥 2
𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 2
𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (𝑥 + ℎ) 2 − 𝑥 2
= 𝑥 2 + 2ℎ𝑥 + ℎ2 − 𝑥 2
= 2ℎ𝑥 + ℎ2
= ℎ(2𝑥 + ℎ)
ℎ(2𝑥 + ℎ)
ℎ
= lim 2𝑥 + ℎ
lim
ℎ→0
ℎ→0
= 2𝑥
LESSON 15
Differentiate from first principles
𝑓(𝑥) = 𝑥 3 + 2𝑥
SOLUTION
𝑓(𝑥) = 𝑥 3 + 2𝑥
𝑓(𝑥 + ℎ) = (𝑥 + ℎ) 3 + 2(𝑥 + ℎ)
= 𝑥 3 + 3𝑥 2 ℎ + 3𝑥ℎ2 + ℎ3 + 2𝑥 + 2ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
= 𝑥 3 + 3𝑥 2 ℎ + 3𝑥ℎ2 + ℎ3 + 2𝑥 + 2ℎ − 𝑥 3 − 2𝑥
= 3𝑥 2 ℎ + 3𝑥ℎ2 + ℎ3 + 2ℎ
= ℎ(3𝑥 2 + 2𝑥ℎ + ℎ2 + 2)
ℎ(3𝑥 2 + 3ℎ𝑥 + ℎ2 + 2)
𝑓 ′ (𝑥) = lim
ℎ→0
ℎ
= lim 3𝑥 2 + 3ℎ𝑥 + ℎ2 + 2
ℎ→0
= 3𝑥 2 + 2
123
CHAPTER 19: LIMITS
LESSON 16
𝑓(𝑥) = √𝑥
SOLUTION
𝑓(𝑥) = √𝑥
𝑓(𝑥 + ℎ) = √𝑥 + ℎ
𝑓(𝑥 + ℎ) − 𝑓(𝑥) = √𝑥 + ℎ − √𝑥
𝑓 ′ (𝑥) = lim
√𝑥 + ℎ − √𝑥
ℎ
(√𝑥 + ℎ − √𝑥)(√𝑥 + ℎ + √𝑥)
ℎ→0
= lim
(f) 𝑦 = cos 𝑥
Differentiate from first principles
ℎ→0
= lim
SOLUTIONS
(a) 0 (b) −2
lim
ℎ→0
= lim
ℎ→0 (√𝑥 + ℎ + √𝑥)
=
1
2√𝑥
LESSON 17
(e) 2 cos 2𝑥
√𝑥+ℎ−√𝑥
ℎ
=
1
.
[5]
2√𝑥
(b) Deduce, from first principles, the
derivative with respect to 𝑥 of 𝑦 = √𝑥.
ℎ
1
1
√𝑥
EXAM QUESTIONS
1. (a) Use the result that
(√𝑥 + ℎ + √𝑥)(√𝑥 + ℎ − √𝑥) = ℎ to
show that
ℎ→0 ℎ(√𝑥 + ℎ + √𝑥)
ℎ→0 ℎ(√𝑥 + ℎ + √𝑥)
(d)
(f) − sin 𝑥
ℎ(√𝑥 + ℎ + √𝑥)
𝑥 +ℎ−𝑥
= lim
6
(c) − 𝑥4
2.
[1]
CAPE 2005
Differentiate from first principles, with
1
respect to 𝑥, the function 𝑦 = 2 .
[6]
𝑥
CAPE 2009
…………………………………………………………………………..
Differentiate from first principles
𝑓(𝑥) = sin 2𝑥
SOLUTION
𝑓(𝑥) = sin 2𝑥
𝑓(𝑥 + ℎ) = sin2(𝑥 + ℎ)
𝑓 (𝑥 + ℎ) − 𝑓(𝑥 )
= sin(2𝑥 + 2ℎ) − sin 2𝑥
2𝑥 + 2ℎ + 2𝑥
2𝑥 + 2ℎ − 2𝑥
= 2 cos (
) sin (
)
2
2
= 2 cos(2𝑥 + ℎ) sin ℎ
2 cos(2𝑥 + ℎ) sin(ℎ)
ℎ
2 cos(2𝑥 + ℎ) sin(ℎ)
= lim
ℎ→0
ℎ
sin ℎ
= lim 2 cos(2𝑥 + ℎ) × lim
ℎ→0
ℎ→0 ℎ
= 2 cos 2𝑥 × 1
= 2 cos 2𝑥
𝑓 ′ (𝑥 ) = lim
ℎ→0
…………………………………………………………………………..
EXERCISE 19.3
1. Differentiate each of the following from first
principles.
(a) 𝑦 = 3
(b) 𝑦 = −2𝑥
2
(c) 𝑦 = 𝑥3
(d) 𝑦 = 2√𝑥
(e) 𝑦 = sin2𝑥
124
CHAPTER 20: DIFFERENTIATION
CHAPTER 20: DIFFERENTIATION
At the end of this section, students should be able
to:
 use the concept of the derivative at a point
𝑥 = 𝑐 as the gradient of the tangent to the
graph at 𝑥 = 𝑐;
𝑑𝑦
 use the 𝑓 ′ (𝑥) and 𝑑𝑥 notation for the first
derivative of 𝑓(𝑥);
𝑑
 use 𝑥 𝑛 = 𝑛𝑥 𝑛−1 where 𝑛 is any real

number;
𝑑
𝑑
use 𝑑𝑥 sin𝑥 = cos 𝑥 and 𝑑𝑥 cos 𝑥 = − sin 𝑥





𝑑𝑥


















use simple rules of derivatives to find
derivatives of sums and multiples of
functions;
calculate derivatives of polynomials and
trigonometric functions;
apply the chain rule in the differentiation of
composite functions;
differentiate products and quotients of simple
polynomials and trigonometric functions;
use the concept of the derivative as a rate of
change;
use the concept of stationary points;
locate stationary points, maxima and minima,
by considering sign changes of the derivative;
calculate the second derivative, 𝑓 ′′ (𝑥);
interpret the significance of the sign of the
second derivatives;
use the sign of the second derivative to
determine the nature of stationary points;
obtain equations of tangents and normal to
curves.
derive the derivative of a function at a point as
a limit;
differentiate, from first principles, functions
such as:
(a) 𝑓(𝑥) = 𝑘 where 𝑘 ∈ ℝ,
(b) 𝑓(𝑥) = 𝑥 𝑛 , where 𝑛 ∈
1 1
{−3, −2, −, 1−, − 2 , 2 , 1, 2, 3}
(c) 𝑓(𝑥) = sin 𝑥,
(d) 𝑓(𝑥) = cos 𝑥;
use the sum, product and quotient rules for
differentiation;
differentiate sums, products and quotients of:
(b) polynomials,
(c) trigonometric functions;
apply the chain rule in the differentiation of
(b) composite functions (substitution),
(c) functions given by parametric equations;
solve problems involving rates of change;


use the sign of the derivatives to investigate
where a function is increasing or decreasing;
apply the concept of stationary (critical)
points;
calculate second derivatives;
interpret the significance of the sign of the
second derivative;
use the sign of the second derivative to
determine the nature of stationary points;
sketch graphs of polynomials, rational
functions and trigonometric functions using
the features of the function and its first and
second derivative (including horizontal and
vertical asymptotes);
describe the behaviour of such graphs for
large values of the independent variable;
obtain equations of tangents and normal to
curves.
HOW TO DIFFERENTIATE
INTRODUCTION
𝑑
is the differential operator which indicates that
we are differentiating with respect to 𝑥.
𝑑𝑦
means that we are differentiating 𝑦 with
𝑑𝑥
𝑑𝑥
respect to 𝑥.
𝑑𝐴
means that we are differentiating 𝐴 with
𝑑𝑡
respect to 𝑡.
𝑑𝑦
Alternately, for functions of the form 𝑦 = 𝑓(𝑥), 𝑑𝑥
can be written as 𝑓 ′ (𝑥) or 𝑦′.
𝑑
𝑎𝑥 𝑛 = 𝑛𝑎𝑥 𝑛−1
𝑑𝑥
𝑑𝑦
LESSON 1
Determine for each of the
𝑑𝑥
following
1. 𝑦 = 𝑥 2
2. 𝑦 = 3𝑥 5
3. 𝑦 = √𝑥
1
4. 𝑦 = 2𝑥3
5.
4
𝑦=5 3
√𝑥
SOLUTION
1. 𝑦 = 𝑥 2
𝑑𝑦
= 2𝑥
𝑑𝑥
2. 𝑦 = 3𝑥 5
𝑑𝑦
= 15𝑥 4
𝑑𝑥
125
CHAPTER 20: DIFFERENTIATION
3.
4.
5.
1
𝑦 = √𝑥 = 𝑥 2
𝑑𝑦 1 −(1)
= 𝑥 2
𝑑𝑥 2
1
1
𝑦 = 2𝑥3 = 2 𝑥 −3
𝑑𝑦
3
3
= − 𝑥 −4 = − 4
𝑑𝑥
2
2𝑥
4
𝑦 = 5 3 = 4𝑥
√𝑥
2.
𝑑𝑥
𝑑
𝑑𝑥
[𝑓(𝑥) ± 𝑔(𝑥)] =
𝑓(𝑥)]
𝑑
𝑑𝑥
𝑓(𝑥) ±
𝑔(𝑥)
𝑑
[𝑓(𝑥)]𝑛 = 𝑛[𝑓(𝑥)]𝑛−1 𝑓 ′ (𝑥)
𝑑𝑥
𝑑
[𝑓(𝑔(𝑥))] = 𝑓′(𝑔(𝑥)). 𝑔′ (𝑥)
𝑑𝑥
LESSON 3
Differentiate each of the
following.
1. 𝑦 = (3𝑥 + 5)5
2.
3.
𝑦 = 5𝑥 3 − 𝑥 2 + 2 sin 𝑥
𝑑𝑦
1 1
= 15𝑥 2 − 𝑥 −2 + 2 cos 𝑥
𝑑𝑥
2
2
𝑦 = 7𝑥 4 + 12𝑥 −1 − 2𝑥 −5
𝑑𝑦
4 7
= 28𝑥 3 − 12𝑥 −2 + 𝑥 −5
𝑑𝑥
5
1
𝑦 = (4𝑥 + 𝑥 −5 )3
𝑦 = √𝑥 2 + 5𝑥 − 8
SOLUTION
1. 𝑦 = (3𝑥 + 5)5
𝑑𝑦
= 5(3𝑥 + 5)4 (3) = 15(3𝑥 + 5) 4
𝑑𝑥
1
2.
𝑦 = (4𝑥 + 𝑥 −5 )3
2
𝑑𝑦 1
= (4𝑥 + 𝑥 −5 ) −3 (4 − 5𝑥 −6 )
𝑑𝑥 3
3.
𝑦 = √𝑥 2 + 5𝑥 − 8 = (𝑥 2 + 5𝑥 − 8) 2
1
𝑑𝑦 1 2
= (𝑥 + 5𝑥 − 8)− 2 (2𝑥 + 5)
𝑑𝑥 2
𝑑𝑦
1 2
1.
𝑥
2
2. 𝑦 = 5𝑥 3 − √𝑥 + 2 sin 𝑥
12
2
3. 𝑦 = 7𝑥 4 +
−5
𝑥
√𝑥 2
2
4. 𝑦 = (2𝑥 − 3) − 5 cos 𝑥
(2𝑥 − 3) 2
5. 𝑦 =
𝑥
SOLUTION
1
1. 𝑦 = 3𝑥 5 + 𝑥 2
2
𝑑𝑦
4
= 15𝑥 + 𝑥
𝑑𝑥
3.
𝑑
𝑑𝑥
The Chain Rule
Find 𝑑𝑥 in each of the following
1
𝑥
9
= 4𝑥 − 12 + 9𝑥 −1
𝑥
𝑑𝑦
9
= 4 − 9𝑥 −2 = 4 − 2
𝑑𝑥
𝑥
𝑦 = 3𝑥 5 +
2.
4𝑥2 −12𝑥+9
DIFFERENTIATION RULES
[𝑐 𝑓(𝑥)] = 𝑐 [
LESSON 2
cases.
𝑦=
= 4𝑥 − 12 +
PROPERTIES OF DERIVATIVES
𝑑
5.
3
5
THE DERIVATIVES OF TRIGONOMETRIC
FUNCTIONS
𝑑𝑦
𝑦 = sin𝑥
= cos 𝑥
𝑑𝑥
𝑑𝑦
𝑦 = cos 𝑥
= − sin𝑥
𝑑𝑥
𝑑𝑦
𝑦 = tan 𝑥
= sec 2 𝑥
𝑑𝑥
𝑑𝑦
𝑦 = sec 𝑥
= sec 𝑥 tan 𝑥
𝑑𝑥
𝑑𝑦
𝑦 = csc 𝑥
= − csc 𝑥 cot 𝑥
𝑑𝑥
𝑑𝑦
𝑦 = cot 𝑥
= − csc 2 𝑥
𝑑𝑥
𝑑𝑥
𝑑
𝑦 = 4𝑥 2 − 12𝑥 + 9 − 5 cos 𝑥
𝑑𝑦
= 8𝑥 − 12 − 5(− sin 𝑥)
𝑑𝑥
= 8𝑥 − 12 + 5 sin 𝑥
−
𝑑𝑦
12 8
= − 𝑥 −5
𝑑𝑥
5
1.
4.
LESSON 4
1
Differentiate
2
√5𝑥−7
SOLUTION
1
2
𝑦=
= 2(5𝑥 − 7)−2
√5𝑥 − 7
3
𝑑𝑦
= −(5𝑥 − 7)−2 (5)
𝑑𝑥
𝑑𝑦
5
=−
𝑑𝑥
√(5𝑥 − 7)3
LESSON 5
Determine the derivative of each
of the following
1. 𝑦 = sin2𝑥
2. 𝑦 = −3 sin(4𝑥 − 1)
3. 𝑦 = 5 cos(𝑥 2 + 4)
126
CHAPTER 20: DIFFERENTIATION
SOLUTION
1. 𝑦 = sin2𝑥
𝑑𝑦
= 2 cos 2𝑥
𝑑𝑥
2.
𝑦 = (𝑥 3 + 7𝑥 − 1)(5𝑥 + 2)
𝑑𝑦
= (3𝑥 2 + 7)(5𝑥 + 2) + (𝑥 3 + 7𝑥 − 1)(5)
𝑑𝑥
= (3𝑥 2 + 7)(5𝑥 + 2) + 5(𝑥 3 + 7𝑥 − 1)
2.
3.
𝑦 = 𝑥√𝑥 + 3 = 𝑥(𝑥 + 3)2
1
1
𝑑𝑦
1
= 1(𝑥 + 3)2 + 𝑥 [ (𝑥 + 3)−2 (1)]
𝑑𝑥
2
1
1
𝑥
= (𝑥 + 3) 2 + (𝑥 + 3) −2
2
1
𝑥
= (𝑥 + 3) 2 [1 + (𝑥 + 3)−1 ]
2
4.
𝑦 = (𝑥 2 + 3𝑥 + 5) sin 𝑥
𝑑𝑦
= (2𝑥 + 3) sin 𝑥 + (𝑥 2 + 3𝑥 + 5) cos 𝑥
𝑑𝑥
5.
𝑦 = cos 𝑥 sin𝑥
𝑑𝑦
= (− sin𝑥) sin 𝑥 + cos 𝑥 (cos 𝑥)
𝑑𝑥
= cos 2 𝑥 − sin2 𝑥
3.
𝑦 = −3 sin(4𝑥 − 1)
𝑑𝑦
= −3 cos(4𝑥 − 1) × 4
𝑑𝑥
𝑑𝑦
= −12 cos(4𝑥 − 1)
𝑑𝑥
𝑦 = 5 cos(𝑥 2 + 4)
𝑑𝑦
= 5(− sin(𝑥 2 + 4) × 2𝑥)
𝑑𝑥
𝑑𝑦
= −10𝑥 sin(𝑥 2 + 4)
𝑑𝑥
LESSON 6
Differentiate each of the
following
1. 𝑦 = sin2 𝑥
2. 𝑦 = −3 cos 2 2𝑥
3. 𝑦 = 2 cos 3(4𝑥 3 + 2)
SOLUTION
1. 𝑦 = sin2 𝑥 = ( sin𝑥 )2
𝑑𝑦
= 2 sin𝑥 cos 𝑥
𝑑𝑥
1
LESSON 8
Differentiate 𝑦 = −2(𝑥 + 3) cos 𝑥
SOLUTION
2.
3.
𝑦 = −3 cos 2 2𝑥 = −3(cos 2𝑥)2
𝑑𝑦
= −6(cos 2𝑥)(−2 sin 2𝑥)
𝑑𝑥
𝑑𝑦
12 sin 2𝑥 cos 2𝑥
𝑑𝑥
𝑦 = 2 cos 3(4𝑥 3 + 2) = 2[cos(4𝑥 3 + 2)]3
𝑑𝑦
= 6[cos(4𝑥 3 + 2)]2 (− sin(4𝑥 3 + 2) × (12𝑥 2)
𝑑𝑥
𝑑𝑦
= −72𝑥 2 sin(4𝑥 3 + 2) cos 2(4𝑥 3 + 2)
𝑑𝑥
𝑦 = −2(𝑥 + 3) cos 𝑥
𝑑𝑦
= −2[1( cos 𝑥 ) + (𝑥 + 3)(− sin𝑥 )]
𝑑𝑥
𝑑𝑦
= −2[cos 𝑥 − (𝑥 + 3) sin𝑥]
𝑑𝑥
The Quotient Rule
𝑑 𝑓(𝑥)
𝑓 ′ (𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′ (𝑥)
[
]=
[𝑔(𝑥)]2
𝑑𝑥 𝑔(𝑥)
The Product Rule
𝑑
𝑓(𝑥)𝑔(𝑥) = 𝑓 ′ (𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)
𝑑𝑥
LESSON 7
Find the derivative of the
following functions
1. (𝑥 + 5)(𝑥 − 6)
2. (𝑥 3 + 7𝑥 − 1)(5𝑥 + 2)
3. 𝑥√𝑥 + 3
4. (𝑥 2 + 3𝑥 + 5) sin𝑥
5. cos 𝑥 sin 𝑥
SOLUTION
1. 𝑦 = (𝑥 + 5)(𝑥 − 6)
𝑑𝑦
= 1(𝑥 − 6) + (𝑥 + 5)(1) = 2𝑥 − 1
𝑑𝑥
LESSON 9
Find the derivatives of the
following functions
1.
2𝑥3
4−𝑥
sin 𝑥
3. 2+cos 𝑥
SOLUTION
2𝑥3
1. 𝑦 = 4−𝑥
(𝑥−4)2
2. (𝑥+3)2
𝑥2
4. sin2 𝑥
𝑑𝑦 6𝑥 2 (4 − 𝑥) − 2𝑥 3 (−1)
=
(4 − 𝑥)2
𝑑𝑥
2
24𝑥 − 6𝑥 3 + 2𝑥 3
=
(4 − 𝑥)2
2
24𝑥 − 4𝑥 3
=
(4 − 𝑥)2
127
CHAPTER 20: DIFFERENTIATION
=
2.
4𝑥 2 (6 − 𝑥)
(4 − 𝑥)2
(𝑥−4)2
𝑦 = (𝑥+3)2
𝑑𝑦 [2(𝑥 − 4)(1)](𝑥 + 3)2 − (𝑥 − 4)2 [2(𝑥 + 3)(1)]
=
[(𝑥 + 3)2 ]2
𝑑𝑥
2(𝑥 − 4)(𝑥 + 3) 2 − 2(𝑥 + 3)(𝑥 − 4) 2
=
(𝑥 + 3)4
2(𝑥 − 4)(𝑥 + 3)[(𝑥 + 3) − (𝑥 − 4)]
=
(𝑥 + 3) 4
2(𝑥 − 4)(7)
=
(𝑥 + 3) 3
14(𝑥 − 4)
=
(𝑥 + 3)3
3.
4.
THE SECOND DERIVATIVE
The second derivative is the derivative of the first
derivative.
sin𝑥
𝑦 = 2+cos 𝑥
𝑑𝑦 cos 𝑥 (2 + cos 𝑥) − sin 𝑥 (− sin 𝑥)
=
(2 + cos 𝑥)2
𝑑𝑥
2 cos 𝑥 + cos2 𝑥 + sin2 𝑥
=
(2 + cos 𝑥)2
2 cos 𝑥 + 1
=
(2 + cos 𝑥)2
𝑥2
𝑦 = sin2 𝑥
𝑑𝑦 2𝑥(sin2 𝑥) − 𝑥 2 (2(sin 𝑥) cos 𝑥)
=
(sin2 𝑥 )2
𝑑𝑥
2𝑥 sin2 𝑥 − 2𝑥 2 sin 𝑥 cos 𝑥
=
sin4 𝑥
2𝑥 sin𝑥 (sin 𝑥 − 𝑥 cos 𝑥 )
=
sin4 𝑥
2𝑥(sin𝑥 − 𝑥 cos 𝑥)
=
sin3 𝑥
LESSON 10
Notation
Function
Derivative
Derivative
𝑑𝑦
𝑑𝑥
𝑑 2𝑦
𝑑𝑥 2
𝑦
𝑦′
𝑦′′
𝑓(𝑥)
𝑓 ′ (𝑥)
𝑓 ′′ (𝑥)
LESSON 11
Determine the second derivative
for each of the following.
2
(a) 𝑦 = 3𝑥 4 − 𝑥 + 3
(b) 𝑓 (𝑥 ) = 3 cos 2 𝑥
SOLUTION
2
(a) 𝑦 = 3𝑥 4 − 𝑥 + 3 = 3𝑥 4 − 2𝑥 −1 + 3
𝑑𝑦
= 12𝑥 3 + 2𝑥 −2
𝑑𝑥
𝑑 2𝑦
4
= 36𝑥 2 − −3
2
𝑑𝑥
𝑥
(b) 𝑓(𝑥) = 3 cos 2 𝑥 = 3(cos 𝑥)2
𝑓 ′ (𝑥) = 6 cos 𝑥 (− sin 𝑥) = −6 sin 𝑥 cos 𝑥
Find 𝑓 ′ (𝑥 ) for the function
𝑥3
𝑓(𝑥) = 2
𝑥 +2
3𝑥 2 (𝑥 2 + 2) − 𝑥 3 (2𝑥)
(𝑥 2 + 2)2
4
3𝑥 + 6𝑥 2 − 2𝑥 4
=
(𝑥 2 + 2) 2
𝑥 4 + 6𝑥 2
= 2
(𝑥 + 2) 2
𝑥 2 (𝑥 2 + 6)
=
(𝑥 2 + 2) 2
Second
𝑦
𝑓 ′′ (𝑥) = −6 cos 𝑥 cos 𝑥 − 6 sin 𝑥 (− sin 𝑥)
= −6 cos 2 𝑥 + 6 sin2 𝑥
SOLUTION
𝑥3
𝑓(𝑥) = 𝑥2 +2
𝑓 ′ (𝑥) =
First
= 6(sin2 𝑥 − cos2 𝑥)
LESSON 12
Given that 𝑦 = √7𝑥 2 + 3,
𝑑𝑦
(i) obtain 𝑑𝑥
𝑑𝑦
(ii) show that 𝑦 𝑑𝑥 = 7𝑥
𝑑2 𝑦
𝑑𝑦 2
(iii) hence show that 𝑦 𝑑𝑥2 + (𝑑𝑥 ) = 7.
SOLUTION
(i)
1
𝑦 = (7𝑥 2 + 3) 2
1
𝑑𝑦 1
= (7𝑥 2 + 3) −2 (14𝑥)
𝑑𝑥 2
128
CHAPTER 20: DIFFERENTIATION
1
= 7𝑥(7𝑥 2 + 3)−2
(ii)
(ii)
𝑑𝑦
𝑦 𝑑𝑥
= √7𝑥 2 + 3 (
𝑑𝑦
= −3(1 + 𝑥
𝑑𝑥 2
7𝑥
)
√7𝑥 2 + 3
5
(1 + 𝑥 2 )2
1
−2
(iii)
= 7𝑥(7𝑥 2 + 3)
𝑑𝑥
1
3
𝑑 2𝑦
1
= 7(7𝑥 2 + 3)−2 + 7𝑥 [− (7𝑥 2 + 3)− 2 (14𝑥)]
𝑑𝑥 2
2
7
49𝑥 2
=
−
√7𝑥 2 + 3 (√7𝑥 2 + 3)3
𝑑2 𝑦
3𝑦
+
𝑑𝑥 2 (1 + 𝑥 2 )2
2
𝑑2 𝑦
𝑑𝑦 2
7
49𝑥 2
7𝑥
)+ (
)
𝑦 2 + ( ) = √7𝑥 2 + 3 (
−
3
𝑑𝑥
𝑑𝑥
√7𝑥 2 + 3 (√7𝑥 2 + 3)
√7𝑥 2 + 3
5
(1 + 𝑥 2 )2
=7−
=7
LESSON 13
(i)
(ii)
2𝑥
show that
2
𝑦
𝑥 𝑑𝑥 = 1+𝑥2
𝑑2 𝑦
𝑑𝑥2
3𝑦
+ (1+𝑥2 )2 = 0
SOLUTION
2𝑥
(i)
𝑦=√ 2
1+𝑥
1
1
1
2
2 −
𝑑𝑦 2(1 + 𝑥 )2 − 2𝑥 [2 (1 + 𝑥 ) 2 (2𝑥)]
=
1 2
𝑑𝑥
[(1 + 𝑥 2 )2 ]
=
1
1
2(1 + 𝑥 2 )2 − 2𝑥 2 (1 + 𝑥 2 )−2
1 + 𝑥2
1
= (2(1 + 𝑥 2 )2 −
=
=
=
𝑥
2
2𝑥 2
1
1 ) (1 + 𝑥 2 )
(1 + 𝑥 2 )2
2𝑥 2
1−
3
(1 + 𝑥 2 )2 (1 + 𝑥 2 )2
2(1 + 𝑥 2 ) − 2𝑥 2
3
(1 + 𝑥 2 )2
2
3
(1 + 𝑥 2 )2
𝑑𝑦
2
= 𝑥(
)
3
𝑑𝑥
(1 + 𝑥 2 )2
2𝑥
=
3
(1 + 𝑥 2 )2
𝑦
2𝑥
1
=(
1 ) (1 + 𝑥 2 )
1 + 𝑥2
(1 + 𝑥 2 )2
2𝑥
=
3
(1 + 𝑥 2 )2
6𝑥
2𝑥
1
)
5+ 3(
1) (
2 2
(1 + 𝑥 2 )2 (1 + 𝑥 )
(1 + 𝑥 2 )2
=−
6𝑥
+
6𝑥
5
(1 + 𝑥 2 )2
=0
1+𝑥
𝑑𝑦
=−
49𝑥 2
49𝑥 2
+ 2
2
7𝑥 + 3 7𝑥 + 3
If 𝑦 = √
5
2 )−2 (2𝑥)
6𝑥
=−
= 7𝑥
𝑑𝑦
3
= 2(1 + 𝑥 2 )− 2
𝑑𝑥
𝑑 2𝑦
PARAMETRIC DIFFERENTIATION
LESSON 14
A curve is represented
parametrically by
2
𝑥 = 2,
𝑦 = 𝑡 3 − 2𝑡
𝑡
𝑑𝑦
Find 𝑑𝑥 in terms of 𝑡.
SOLUTION
2
𝑥 = 𝑡2 = 2𝑡 −2
𝑑𝑥
4
=− 3
𝑑𝑡
𝑡
𝑦 = 𝑡 3 − 2𝑡
𝑑𝑦
= 3𝑡 2 − 2
𝑑𝑡
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
= (3𝑡 2 − 2) × (−
=−
𝑡3
)
4
𝑡 3 (3𝑡 2 − 2)
4
LESSON 15
The parametric equations of a
curve are given by
𝑥 = sin 𝜃 , 𝑦 = cos 𝜃 , 0 ≤ 𝜃 ≤ 2𝜋
𝑑𝑦
Find 𝑑𝑥 in terms of 𝜃.
Simplify the answer as far as possible.
SOLUTION
𝑥 = sin 𝜃
𝑑𝑥
= cos 𝜃
𝑑𝜃
𝑦 = cos 𝜃
𝑑𝑦
= − sin𝜃
𝑑𝜃
129
CHAPTER 20: DIFFERENTIATION
𝑑𝑦 𝑑𝑦 𝑑𝜃
=
×
𝑑𝑥 𝑑𝜃 𝑑𝑥
= − sin 𝜃 ×
= − tan 𝜃
1
cos 𝜃
4.
…………………………………………………………………………
5.
EXERCISE 20.1
1.
𝑑𝑦
𝑑𝑦
Determine 𝑑𝑥 for each of the following.
6.
(b) 𝑦 = 𝑥 8
7.
(a) 𝑦 = 𝑥
𝑓 ′ (𝑥) = 0, 𝑥 = 8 .
Use the product rule to differentiate the
(c) 𝑦 = 𝑥 0
following functions with respect to 𝑥.
(d) 𝑦 = 1
(a) (𝑥 + 3)(𝑥 − 4)
(e) 𝑦 = 𝑥 −4
(b) (3𝑥 − 4)(2𝑥 + 5)
(f) 𝑦 = 𝑥 −7
(c) (6 + 𝑥)(5 − 𝑥)
1
(d) (3 − 2𝑥)(7 + 3𝑥)
1
(e) 𝑥 2 (𝑥 + 3)4
(h) 𝑦 = 𝑥 −7
(i) 𝑦 = 𝑥
(f) 𝑥 4 (3𝑥 − 1)3
3
2
(g) 3𝑥 2 (2𝑥 + 5) 2
1
(h) 𝑥 3 (4𝑥 2 − 1)3
(j) 𝑦 = 𝑥 3
(i) (𝑥 + 2) 2 (𝑥 − 5)3
2
(k) 𝑦 = 𝑥 −5
(j) (2𝑥 − 1) 3 (𝑥 + 4) 2
(l) 𝑦 = √𝑥
(k) (5𝑥 + 2) 4 (4𝑥 − 3) 3
(m) 𝑦 = √𝑥 3
(l) (2 − 𝑥)6 (5 + 2𝑥)4
1
(n) 𝑦 = 3 2
(m) (3 + 5𝑥)2 (4 − 7𝑥)7
√𝑥
Find 𝑑𝑥 in each of the following cases.
𝑑𝑦
(n) 𝑥 3 √7 − 2𝑥
(a) 𝑥 + 𝑥2
2
(o) (2𝑥 − 1) √𝑥 + 3
16
(p) (1 − 3𝑥)√2𝑥 + 5
(b)𝑥 2 +
𝑥
(q) √𝑥(5𝑥 − 4)3
6
(c) 4𝑥 + 2
𝑥
(r) (3𝑥 + 5) 2 √𝑥 − 2
27
(d)10𝑥 + 𝑥2
(e)
(s) √2𝑥 − 3√4𝑥 + 1
16𝑥3 +4𝑥2 +1
(t) √6 + 𝑥√3 − 2𝑥
2𝑥2
54
(f) 𝑥 + 8𝑥 2
3.
= sin3 𝑥.
Given that 𝑓(𝑥) = sin 𝑥 + sin 2𝑥, show that for
𝑑𝑥
−1+√33
7
(g) 𝑦 = 𝑥 −4
2.
f. 𝑦 = (6𝑥 − 1)3
g. 𝑦 = (3𝑥 2 + 1) 4
Differentiate, with respect to, 𝑥,
(i) 𝑓(𝑥) = sin2 𝑥
(ii) 𝑓(𝑥) = sin2 𝑥 2
(iii) 𝑓(𝑥) = sin2 (2𝑥 + 3)
(iv) 𝑓(𝑥) = cos 2 𝑥
1
Given that 𝑦 = 3 cos 3 𝑥 − cos 𝑥, show that
Differentiate each of the following.
a. 𝑦 = (4𝑥 5 − 2𝑥 + 5)4
b. 𝑦 = sin4𝑥
c. 𝑦 = 4 cos(3𝑥 − 1)
d. 𝑦 = (1 − 𝑥 2 )10
e. 𝑦 = sin √𝑥
(u) √(𝑥 − 1)(2𝑥 + 1)
8.
Show that, if 𝑦 = sin 𝑥 cos 𝑥 then
𝑑𝑦
𝑑𝑥
9.
= 2 cos2 𝑥 − 1
Use the quotient rule to differentiate the given
function with respect to 𝑥
𝑥
(a) 𝑥−2
130
CHAPTER 20: DIFFERENTIATION
𝑥+3
(b) 𝑥−1
(c) 25 + 4 sec 𝑥
(d) csc 6𝑥 2
(e) 𝑥 tan 𝑥
(f) 𝑥 2 tan 3𝑥
(g) (6 − sec 2𝑥)3
12. Differentiate √5𝑥 3 − 4, with respect to 𝑥.
13. Given that 𝑦 = √2𝑥 2 + 3,
𝑑𝑦
(i) obtain 𝑑𝑥
3−𝑥
(c) 4+𝑥
4𝑥−3
(d) 𝑥+2
(e)
(f)
2𝑥−5
𝑥+4
5𝑥
𝑥+2
𝑑𝑦
1+3𝑥
(ii) show that 𝑦 𝑑𝑥 = 2𝑥
4𝑥+3
(iii) hence show that 𝑦 𝑑𝑥2 + (𝑑𝑥 ) = 2
(g) 2−5𝑥
(h) 2𝑥−1
(i)
(j)
14. Find the value of 𝑘 for which
𝑑 2𝑥 + 3
𝑘
(
)=
(𝑥 − 4) 2
𝑑𝑥 𝑥 − 4
𝑥2
𝑥+3
15. Given that 𝑦 = ((𝑥+3)(𝑥+5)2 ) , show that
𝑥−4
𝑑𝑦 2𝑥(𝑥 − 3)5 (𝑥 3 + 27𝑥 2 + 69𝑥 − 45)
=
(𝑥 + 3) 3 (𝑥 + 5)5
𝑑𝑥
(k) 2𝑥−3
(m)
(n)
(o)
(p)
𝑥5
𝑑𝑦
16. Determine 𝑑𝑥 for each of the following
3−𝑥
(a) 𝑥 = 𝑡 + 4, 𝑦 =
(3𝑥−2)2
𝑡
√𝑥
(5𝑥+1)3
√𝑥
5
(𝑥2 −4)
√𝑥
√𝑥
2𝑥−1
3−√𝑥
(q) (2+𝑥) 2
5+2√𝑥
(r) (5−4𝑥) 3
(s)
(t)
(3𝑥2 +2)
4
√2𝑥−1
(b)
(c)
Show that
√1−𝑥2
to 𝑥
1.
(a) 6𝑥 −3 + 2 tan 𝑥
(b) −3 cot 𝑥 + 2√𝑥
2
(a) 7𝑥 6 (b) 8𝑥 7 (c) 0 (d) 0 (e) −4𝑥 −5
2
1
2
1
7
1
1
3
1
5
2
(n) − 3 𝑥 −3
2.
4
16
𝑥
𝑥
1
3.
12
(a) 1 − 3 (b) 2𝑥 − 2 (c) 4 − 3
54
sin 2𝑥
𝑑𝑦
1
= − (cot 𝜃 − 1)
(j) 3 𝑥 −3 (k) − 5 𝑥 −5 (l) 2 𝑥 −2 (m) 2 𝑥 2
6−𝑥
11. Find 𝑑𝑥 in each of the following cases.
𝑑𝜃
3
𝑥 √5−𝑥2
(d) 𝑥2 +1
𝑑𝑦
(f) −7𝑥 −8 (g) 4𝑥 3 (h) 7𝑥 6 (i) 2 𝑥 2
𝑥−3
5−√𝑥
4
SOLUTIONS
𝑥(𝑥−1)3
𝑥3 √4−𝑥2
𝑡
(b) 𝑥 = , 𝑦 = 𝑡 + 4
4
(c) 𝑥 = 2𝑡 2 , 𝑦 = 𝑡 + 1
𝑑𝑦
17. Determine 𝑑𝑥 for each of the following
a. 𝑥 = cos 𝜃 , 𝑦 = 3 sin 𝜃
b. 𝑥 = 3 sin𝜃 , 𝑦 = cos 𝜃
c. 𝑥 = −4 + cos 𝜃 , 𝑦 = −1 + sin 𝜃
d. 𝑥 = 4 + cos 𝜃 , 𝑦 = 9 sin 𝜃
e. 𝑥 = 2 cos 𝜃 , 𝑦 = cos 2 𝜃
f. 𝑥 = 2 sec 𝜃 , 𝑦 = 3 + 2 tan 𝜃
g. 𝑥 = csc 𝜃 − cot 𝜃 , 𝑦 = csc 𝜃 − 2 cot 𝜃
18. A curve has parametric equations
𝑥 = 2 cos 𝜃 ,
𝑦 = sin 𝜃 + cos 𝜃
𝜋
where 0 ≤ 𝜃 ≤ 2 .
(2−3𝑥) 2
10. Differentiate each of the following with respect
(a)
2
𝑥(𝑥−3)3
𝑥2
𝑥3
(l)
𝑑𝑦 2
𝑑2 𝑦
𝑥
54
(d) 10 − 𝑥3 (e) 8 − 𝑥3 (f) − 𝑥2 + 16𝑥
(a) 8(4𝑥 5 − 2𝑥 + 5)3 (10𝑥 4 − 1)
(b) 4 cos 4𝑥 (c) −12 sin(3𝑥 − 1)
cos √𝑥
(d) −20𝑥(1 − 𝑥 2 )9 (e)
2√𝑥
131
CHAPTER 20: DIFFERENTIATION
4.
5.
6.
7.
(f) 18(6𝑥 − 1)2 (g) 24𝑥(3𝑥 2 + 1) 3
(i) sin2𝑥 (ii) 4 sin 𝑥 2 cos 𝑥 2
(iii) 4 sin(2𝑥 + 3) cos(2𝑥 + 3) (iv) − sin2𝑥
(a) 2𝑥 − 1
(b) 12𝑥 + 7
(c) −2𝑥 − 1
(d) −12𝑥 − 5
(e) 6𝑥(𝑥 + 1)(𝑥 + 3) 3
(f) 𝑥 3 (3𝑥 − 1)2 (21𝑥 − 4)
(g) 6𝑥(8𝑥 2 + 30𝑥 + 25)
(h) 3𝑥 2 (4𝑥 2 − 1)2 (12𝑥 2 − 1)
(i) (𝑥 − 5)2 (𝑥 + 2)(5𝑥 − 4)
(j) 2(𝑥 + 4)(2𝑥 − 1)2 (5𝑥 + 11)
(k) 4(4𝑥 − 3)2 (5𝑥 + 2)3 (35𝑥 − 9)
(l) −2(2 − 𝑥)5 (2𝑥 + 5)3 (10𝑥 + 7)
(m) −(4 − 7𝑥)6 (5𝑥 + 3)(315𝑥 + 107)
7(𝑥−3)𝑥2
6𝑥+11
−9𝑥−14
(n) −
(o) 2√𝑥+3
(p)
(q)
(s)
√7−2𝑥
(5𝑥−4)2 (35𝑥−4)
2√𝑥
8𝑥−5
12.
15𝑥2
2√5𝑥3 −4
2𝑥
13. (i) √
(ii)
(iii)
14. 𝑘 = −11
15.
1
16. (a) 4
(b) 4
(c) 4𝑡
2𝑥2 +3
1
1
17. (a) −3 cot 𝜃
(b) − 3 tan 𝜃
(c) − cot 𝜃
(e) cos 𝜃
(d) −9 cot 𝜃
(f) csc 𝜃
2 csc2 𝜃−cot 𝜃 csc 𝜃
(g) csc2 𝜃−cot 𝜃 csc 𝜃
18.
…………………………………………………………………………..
√2𝑥+5
(3𝑥+5)(15𝑥−19)
(r)
2√𝑥−2
−4𝑥−9
(t) 2√3−2𝑥√𝑥+6
√2𝑥−3√4𝑥+1
4𝑥−1
(u) √
2 2𝑥2 −𝑥−1
8.
9.
2
4
(a) − (𝑥−2)2
(b) − (𝑥−1)2
7
11
(c) − (𝑥+4)2
(d) (𝑥+2)2
13
10
(e)(𝑥+4)2
(f) (𝑥+2)2
11
10
(g) (2−5𝑥) 2
(h) − (2𝑥−1)2
𝑥(𝑥+6)
𝑥(𝑥−8)
(i) (𝑥+3)2
(j) (𝑥−4)2
𝑥2 (4𝑥−9)
(k) (2𝑥−3)2
(m)
(o)
(l)
(3𝑥−2)(9𝑥+2)
3
2𝑥 2
4
(𝑥2 −4) (19𝑥2 +4)
3
2𝑥 2
3𝑥−12√𝑥−2
(q) 2√𝑥(𝑥+2)3
𝑥4 (15−4𝑥)
(3−𝑥) 2
(5𝑥+1)2 (25𝑥−1)
(n)
(p) 2√𝑥(2𝑥−1)2
(r)
3
(s)
10. (a)
3
2𝑥 2
−2𝑥−1
5(4𝑥+12√𝑥+1)
√𝑥 (5−4𝑥) 4
(3𝑥2 +2) (45𝑥2 −24𝑥−2)
3
(2𝑥−1)2
2
2
3(𝑥−1) (𝑥 −4𝑥+1)
(𝑥−3)2
(t)
−9𝑥3 +22𝑥−12
3
(1−𝑥2 )2
(b)
𝑥3 −12𝑥2 +30
(6−𝑥) 2 √5−𝑥2
5
𝑥2 (7𝑥 2 −40𝑥2 −20√𝑥+120)
(c)
11.
2
2(5−√𝑥) √4−𝑥2
2[(𝑥2 +1) cos 2𝑥−𝑥 sin2𝑥]
(d)
(𝑥2 +1)2
18
1
(a) − 𝑥4 + 2 sec 2 𝑥 (b) 3 csc 𝑥 + 𝑥
√
(c) 4 sec 𝑥 tan 𝑥
(d) −12𝑥 csc 6𝑥 2 cot 6𝑥 2
2
(e) tan 𝑥 + 𝑥 sec 𝑥
(f) 𝑥(2 tan 3𝑥 + 3𝑥 sec 2 3𝑥)
(g) 6(6 − sec 2𝑥)2 (sec 2𝑥 tan 2𝑥)
132
CHAPTER 20: DIFFERENTIATION
APPLICATIONS OF DIFFERENTIATION
GRADIENTS AND DIFFERENTIATION
𝑑𝑦
When you find 𝑑𝑥 , we get a formula for the
gradient of the tangent to the curve any point, 𝑥. If
you want to find the gradient at a specific point we
then substitute the 𝑥 value of that point.
LESSON 1
Find the gradient of the tangent
to the curve 𝑦 = 𝑥 2 at the point (3, 9).
SOLUTION
𝑦 = 𝑥2
𝑑𝑦
= 2𝑥
𝑑𝑥
when 𝑥 = 3
𝑑𝑦
= 2(3) = 6
𝑑𝑥
LESSON 2
The gradient of the curve
𝑦 = 3𝑥 2 + 5𝑥 − 12 is 23 at the point 𝑃.
(i) Calculate the coordinates of 𝑃.
The curve cuts the 𝑥-axis at 𝑄 and 𝑅.
(ii) Find the gradient of the curve at 𝑄 and 𝑅.
SOLUTION
i.
𝑦 = 3𝑥 2 + 5𝑥 − 12
𝑑𝑦
= 6𝑥 + 5
𝑑𝑥
At 𝑃 we have
6𝑥 + 5 = 23
𝑥=3
when 𝑥 = 3
𝑦 = 3(3)2 + 5(3) − 12
= 30
∴ 𝑃(3, 30)
ii.
At 𝑄 and 𝑅 𝑦 = 0
3𝑥 2 + 5𝑥 − 12 = 0
(3𝑥 − 4)(𝑥 + 3) = 0
4
𝑥=
𝑥 = −3
3
4
When 𝑥 =
3
𝑑𝑦
4
= 6 ( ) + 5 = 13
𝑑𝑥
3
When 𝑥 = −3
𝑑𝑦
= 6(−3) + 5 = −13
𝑑𝑥
LESSON 3
Find the equation of the normal
6
to the curve 𝑦 = 3𝑥 + 𝑥 at the point (3, 11).
SOLUTION
6
𝑦 = 3𝑥 + = 3𝑥 + 6𝑥 −1
𝑥
𝑑𝑦
6
= 3 − 6𝑥 −2 = 3 − 2
𝑑𝑥
𝑥
when 𝑥 = 3
𝑑𝑦
6
7
=3− 2 =
𝑑𝑥
3
3
3
Gradient of normal is − 7
3
𝑦 = 𝑚𝑥 + 𝑐 using 𝑚 = − and (3, 11)
7
3
11 = − (3) + 𝑐
7
86
=𝑐
7
3
86
𝑦=− 𝑥+
→ 7𝑦 + 3𝑥 = 86
7
7
133
CHAPTER 20: DIFFERENTIATION
INCREASING AND DECREASING FUNCTIONS
INTRODUCTION
LESSON 4
If the gradient of the point (3, 16)
on the curve 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 13 is 7, find
𝑎 and 𝑏.
SOLUTION
Since (3, 16) is a point on the
curve it must satisfy the equation of the curve.
16 = 𝑎(3) 2 + 𝑏(3) + 13
3 = 9𝑎 + 3𝑏
1 = 3𝑎 + 𝑏
(1)
Also,
𝑑𝑦
=7
𝑑𝑥
2𝑎𝑥 + 𝑏 = 7
when 𝑥 = 3
2𝑎(3) + 𝑏 = 7
6𝑎 + 𝑏 = 7
(2)
Solving (1) and (2) simultaneously
3𝑎 + 𝑏 = 1
6𝑎 + 𝑏 = 7
𝑎 = 2, 𝑏 = −5
The above graph shows the path of a tennis ball
after it is dropped. Below are some of the
observations from the graph.
1. The height of the ball decreases after it is
released. Therefore, for 𝑥 < 𝑎 the graph is
decreasing
2. The height of the ball then increases to
the point 𝐵 as indicated on the graph.
Therefore, for 𝑎 < 𝑥 < 𝑏 the graph is
increasing.
3. Between the points 𝑥 = 𝑏 and 𝑥 = 𝑐 the
ball’s height is again decreasing. Thus for
this section the graph is decreasing.
4. Finally, after the point 𝐶, that is for 𝑥 > 𝑐,
the graph is increasing as the height of the
ball is increasing.
5. At the points 𝐴, 𝐵 and 𝐶 the ball
instantaneously stops as it changes
direction from decreasing to increasing
and vice versa.
The points 𝐴, 𝐵 and 𝐶 due to their nature are
called stationary points, since the ball stops at
these points. Alternatively, they are also referred
to as turning points as the ball’s direction turns
from decreasing to increasing and from increasing
to decreasing.
The points 𝐴 and 𝐶 look like the bottom of a valley
and as such are termed minimum points, or
minima. It should be noted that to the left of these
points the graph is decreasing and increasing to
the right.
The point 𝐶, on the other hand, appears to be the
top of a hill, and as a result is termed a maximum
point, or maxima. In contrast to the points 𝐴 and
𝐶, to the left of 𝐵 the graph is increasing and
decreasing to the right.
We can conclude that:
134
CHAPTER 20: DIFFERENTIATION
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
> 0 indicates that the graph is increasing
= 0 indicates a stationary point
< 0 indicates that the graph is decreasing
LESSON 5
For 𝑦 = 𝑥 3 − 6𝑥 2 − 15𝑥 + 1
determine the stationary points and their nature.
SOLUTION
𝑦 = 𝑥 3 − 6𝑥 2 − 15𝑥 + 1
𝑑𝑦
= 3𝑥 2 − 12𝑥 − 15
𝑑𝑥
𝑑𝑦
At stationary points 𝑑𝑥 = 0
∴ 3𝑥 2 − 12𝑥 − 15 = 0
𝑥 2 − 4𝑥 − 5 = 0
(𝑥 − 5)(𝑥 + 1) = 0
𝑥 = 5 and 𝑥 = −1
when 𝑥 = 5
𝑦 = 53 − 6(5) 2 − 15(5) + 1
= −99
(5, −99)
when 𝑥 = −1
𝑦 = (−1)3 − 6(−1) 2 − 15 (−1) + 1
=9
(−1, 9)
𝑑2 𝑦
𝑑𝑥2
𝑑𝑦
is the derivative of 𝑑𝑥
𝑑2 𝑦
If for a stationary point 𝑑𝑥2 < 0 then the point is a
𝑑2 𝑦
maximum point and a minimum point if 𝑑𝑥2 > 0.
Hence for the LESSON above
𝑑𝑦
= 3𝑥 2 − 12𝑥 − 15
𝑑𝑥
𝑑 2𝑦
= 6𝑥 − 12
𝑑𝑥 2
when 𝑥 = 5
𝑑 2𝑦
= 6(5) − 12 = 18 → minimum point
𝑑𝑥 2
when 𝑥 = −1
𝑑 2𝑦
= 6(−1) − 12 = −18 → maximum point
𝑑𝑥 2
And these results coincide with the results for the
sign change method used previously.
LESSON 7
A rectangular garden 𝑃𝑄𝑅𝑆 is to
be laid out as shown in the diagram. The garden
consists of a rectangular lawn 𝑇𝑈𝑉𝑊 surrounded
by flower beds. The lawn has an area of 240 m2 .
The flower beds are 3 m wide along the sides 𝑈𝑉
and 𝑇𝑊. Given that the distance 𝑇𝑊 is 𝑥 m, show
that the total area, 𝐴 m2 , of the garden 𝑃𝑄𝑅𝑆 is
given by
1440
𝐴 = 10𝑥 + 300 +
𝑥
𝑑𝐴
Given that 𝑥 varies, find an expression for .
𝑑𝑥
Hence, determine the dimensions of the garden
for which 𝐴 is a minimum.
To determine the nature of the stationary points
we examine the sign of the derivative on either
side of each stationary point.
Therefore, (−1, 9) is a maximum point and
(5, −99) is a minimum point.
LESSON 6 – The Second Derivative Test
Alternately, the Second Derivative Test can be
used to determine the nature of stationary points.
The second derivative is simply the derivative of
the first derivative.
SOLUTION
Let 𝑦 be the length of 𝑉𝑊: ∴ 𝑥𝑦 = 240
240
𝑦=
𝑥
𝐴 = (6 + 𝑥)(10 + 𝑦)
= 60 + 10𝑥 + 6𝑦 + 𝑥𝑦
135
CHAPTER 20: DIFFERENTIATION
= 60 + 240 + 10𝑥 + 6 (
= 300 + 10𝑥 +
240
)
𝑥
Concave Downwards
1440
𝑥
𝑑𝐴
1440
= 10 − 2
𝑑𝑥
𝑥
1440
0 = 10 − 2
𝑥
1440
10 = 2
𝑥
𝑥 = ±12
Since 𝑥 cannot be negative 𝑥 = 12
Dimensions: 𝑥 = 18, 𝑦 = 30
LESSON 9
Determine the coordinates of the
point(s) of inflection on the curve
𝑦 = 𝑥 3 + 3𝑥 2 + 3.
POINTS OF INFLECTION
LESSON 8
Determine the nature of the
stationary points on the curve 𝑦 = (𝑥 − 1) 3.
SOLUTION
𝑦 = (𝑥 − 1)3
𝑑𝑦
= 3(𝑥 − 1)2
𝑑𝑥
𝑑𝑦
For stationary points 𝑑𝑥 = 0
3(𝑥 − 1) 2 = 0
𝑥=1
𝑦=0
(1, 0)
SOLUTION
𝑦 = 𝑥 3 + 3𝑥 2 + 3
𝑑𝑦
= 3𝑥 2 + 6𝑥
𝑑𝑥
𝑑 2𝑦
= 6𝑥 + 6
𝑑𝑥 2
𝑑2 𝑦
For inflection point, 𝑑𝑥2 = 0
6𝑥 + 6 = 0
6𝑥 = −6
𝑥 = −1
𝑦 = (−1)3 + 3(−1) 2 + 3
=5
(−1, 5)
𝑑 2𝑦
= 6(𝑥 − 1)
𝑑𝑥 2
when 𝑥 = 1
𝑑 2𝑦
= 6(1 − 1) = 0
𝑑𝑥 2
This value for the second derivative implies that
the stationary point is a point of inflection.
𝑑2 𝑦
While an inflection point occurs when 𝑑𝑥2 = 0 it
𝑑𝑦
does not have to occur when 𝑑𝑥 = 0. An inflection
point occurs when the concavity of the curve
changes.
Concave Upwards
LESSON 10
The point 𝑃(−1, 3) is an
inflection point on the curve
𝑦 = 𝑥 3 + 𝑏𝑥 2 + 𝑐. Determine
(a) the values of 𝑏 and 𝑐.
(b) the equation of the normal to the curve at 𝑃.
SOLUTION
(a) Since 𝑃(−1, 3) lies on the curve it must satisfy
the equation of the curve
3 = (−1) 3 + 𝑏(−1)2 + 𝑐
4 =𝑏+𝑐
(1)
136
CHAPTER 20: DIFFERENTIATION
𝑑2 𝑦
(iii) 𝑑𝑥2 = 2(𝑥 + 3) + 2(𝑥 + 3) + 2𝑥(1)
𝑦 = 𝑥 3 + 𝑏𝑥 2 + 𝑐
𝑑𝑦
= 3𝑥 2 + 2𝑏𝑥
𝑑𝑥
𝑑 2𝑦
= 6𝑥 + 2𝑏
𝑑𝑥 2
6𝑥 + 2𝑏 = 0 at 𝑃(−1, 3)
6(−1) + 2𝑏 = 0
𝑏=3
Sub 𝑏 = 3 into equation (1)
4 =𝑏 +𝑐
4 =3+𝑐
1=𝑐
(b) Gradient at 𝑃(−1, 3):
𝑑𝑦
= 3𝑥 2 + 2𝑏𝑥 = 3𝑥 2 + 6𝑥
𝑑𝑥
3(−1) 2 + 6(−1) = −3
1
Gradient of normal: 3
= 6𝑥 + 12
when 𝑥 = −1
𝑑 2𝑦
= 6(−1) + 12 = 6
Minimum
𝑑𝑥 2
when 𝑥 = −3
𝑑 2𝑦
= 6(−3) + 12 = −6
Maximum
𝑑𝑥 2
(iv) We need to determine where the graph
crosses the 𝑥 and 𝑦 axes.
When 𝑦 = 0
𝑥(𝑥 + 3)2 = 0
𝑥 = −3, 0
(0, 0) and (−3, 0)
When 𝑥 = 0
0(0 + 3) 2 = 0
(0, 0)
1
𝑦 = 𝑚𝑥 + 𝑐 using 𝑚 = 3 and (−1, 3)
1
3 = (−1) + 𝑐
3
10
=𝑐
3
1
10
𝑦= 𝑥+
3
3
CURVE SKETCHING
LESSON 11
The equation of a curve is given
by
𝑓(𝑥) = 𝑥(𝑥 + 3)2
(i) Obtain an expression for 𝑓 ′ (𝑥).
(ii) Find the stationary point(s) of 𝑓.
(iii) Determine the nature of the stationary
point(s) of 𝑓.
(iv) Sketch the curve.
NB: To sketch polynomial functions we need to
know
-
the roots of the polynomial
the 𝑦 – intercept
the stationary points and their nature
LESSON 12
SOLUTION
(i) 𝑓(𝑥) = 𝑥(𝑥 + 3)2
𝑓 ′ (𝑥) = 1(𝑥 + 3) 2 + 𝑥[2(𝑥 + 3)(1)]
= (𝑥 + 3) 2 + 2𝑥(𝑥 + 3)
(ii) For stationary points 𝑓 ′ (𝑥) = 0
(𝑥 + 3) 2 + 2𝑥(𝑥 + 3) = 0
(𝑥 + 3)[(𝑥 + 3) + 2𝑥] = 0
(𝑥 + 3)(3𝑥 + 3) = 0
𝑥 = −1, −3
when 𝑥 = −1
𝑦 = (−1)(−1 + 3) 2 = −4
(−1, −4)
when 𝑥 = −3
𝑦 = −3(−3 + 3)2 = 0
(−3, 0)
SOLUTION
determine


Draw a sketch for the function
3𝑥 − 1
𝑓(𝑥) =
𝑥+2
To draw our sketch we need to
the intercepts
asymptotes
The Intercepts
When 𝑥 = 0
3(0) − 1
1
𝑦=
=−
0+2
2
1
(0, − )
2
When 𝑦 = 0
137
CHAPTER 20: DIFFERENTIATION
3𝑥 − 1
𝑥+2
0 = 3𝑥 − 1
1
=𝑥
3
1
( , 0)
3
= 2𝑥(5)
When 𝑥 = 6
𝑑𝐴
= 2(6)(5) = 60 cms2
𝑑𝑥
0=
LESSON 14
A sector of a circle of radius 𝑟 has
𝜋
an angle of 6 radians. Given that 𝑟 is increasing at a
Asymptotes
Vertical asymptotes occur when the denominator
is equal to zero.
𝑥 +2 =0
𝑥 = −2
Horizontal asymptotes investigate the limit as 𝑥
tends to infinity
3𝑥 1
−𝑥 3−0
3𝑥 − 1
lim
= 𝑥
=
=3
𝑥 2
𝑥→∞ 𝑥 + 2
1+0
+
𝑥 𝑥
𝑦=3
constant rate of 5 cms−1, calculate, correct to two
decimal places, the rate of increase, when 𝑟 = 6
cm, of
(i)
the area of the sector,
(ii)
the perimeter of the sector.
SOLUTION
(i)
Since 𝑟 is increasing at a rate of 5 cms −1,
𝑑𝑟
𝑑𝐴
= 5. We are therefore trying to determine ,
𝑑𝑡
𝑑𝑡
the rate of increase of 𝐴 (the area of the sector)
when 𝑟 = 6. Hence we begin with the formula for
the area of a sector.
1
𝐴 = 𝑟 2𝜃
2
𝑑𝐴
= 𝑟𝜃
Differentiating 𝐴 with
𝑑𝑟
respect to 𝑟. 𝜃 is a constant.
𝑑𝐴
𝑑𝐴 = 𝑟𝜃 𝑑𝑟
Treating like a
𝑑𝑟
fraction and cross multiplying
𝑑𝐴
𝑑𝑟
= 𝑟𝜃 𝑑𝑡
Dividing throughout by
𝑑𝑡
𝑑𝑡
𝑑𝐴
𝜋
= 6 ( ) (5)
𝑑𝑡
6
= 5𝜋
= 15.71 cm2 𝑠 −1
(ii)
The perimeter of a sector is 2𝑟 + 𝑟𝜃. We
𝑑𝑃
need to determine 𝑑𝑡 .
RATE OF CHANGE
The rate of change of 𝑥 =
change in 𝑥
𝑑𝑥
=
change in time 𝑑𝑡
LESSON 13
The length of the side of a square
is increasing at the rate of 5 cms −1. Find the rate
of increase of the area when the length is 6 cm.
SOLUTION
𝑑𝑥
=5
𝑑𝑡
𝐴 = 𝑥 2 where 𝑥 is the length of the side of the
square
𝑑𝐴
= 2𝑥
𝑑𝑥
𝑑𝐴 𝑑𝐴 𝑑𝑥
=
×
𝑑𝑡 𝑑𝑥 𝑑𝑡
𝑃 = 2𝑟 + 𝑟𝜃
𝑑𝑃
= 2+𝜃
𝑑𝑟
𝑑𝑃 = (2 + 𝜃)𝑑𝑟
𝑑𝑃
𝑑𝑟
= (2 + 𝜃)
𝑑𝑡
𝑑𝑡
𝑑𝑃
𝜋
= (2 + ) (5)
𝑑𝑡
6
5𝜋
= 10 +
6
= 12.62 cms−1
LESSON 15
A spherical balloon is released
from rest and expands as it rises. After rising for 𝑡
seconds its radius is 𝑟 cm, and its surface area is
𝐴 cm2 , where 𝐴 = 4𝜋𝑟 2 . The initial radius of the
balloon is 16 cm. Given that the rate of increase of
the radius is constant and has value 0.8 cms−1,
find the rate of increase of 𝐴 when 𝑡 = 5.
138
CHAPTER 20: DIFFERENTIATION
SOLUTION
𝑑𝑟
We know that 𝑑𝑡 = 0.8 and we are
…………………………………………………………………………..
𝑑𝐴
trying to determine 𝑑𝑡 when 𝑡 = 5.
𝐴 = 4𝜋𝑟 2
𝑑𝐴
= 8𝜋𝑟
𝑑𝑟
𝑑𝐴 = 8𝜋𝑟 𝑑𝑟
𝑑𝐴
𝑑𝑟
= 8𝜋𝑟
𝑑𝑡
𝑑𝑡
𝑑𝑟
At this stage we know the value for 𝑑𝑡 but not for
𝑟. To determine 𝑟 we use the fact that the initial
value for 𝑟 is 16 cm and 𝑟 increases 0.8 cms −1.
Therefore, when 𝑡 = 5, 𝑟 = 16 + (0.8)5 = 20 cm.
𝑑𝐴
= 8𝜋(20)(0.8) = 128𝜋 cm2 𝑠 −1
𝑑𝑡
LESSON 16
Suppose a water tank has the
shape of an inverted cone with base radius 2 m
and height 4 m. If the water is being pumped into
the tank at a rate of 2 m3 /min, find the rate at
which the water level is rising when the water is 3
m deep.
1
SOLUTION
Volume of Cone = 𝜋𝑟 2 ℎ
3
𝑑𝑉
=2
𝑑𝑡
1
𝑉 = 𝜋𝑟 2 ℎ
3
We have a slight problem since we do not know
the rate of change of 𝑟. The truth is, we are not
interested in it. From the diagram we see that
𝑟 2
ℎ
=
∴ 𝑟=
ℎ 4
2
1 ℎ 2
𝑉 = 𝜋( ) ℎ
3 2
𝜋 3
=
ℎ
12
𝑑𝑉 𝜋 2
= ℎ
𝑑ℎ 4
𝜋
𝑑𝑉 = ℎ2 𝑑ℎ
4
𝑑𝑉 𝜋 2 𝑑ℎ
= ℎ
𝑑𝑡 4 𝑑𝑡
𝜋
𝑑ℎ
2 = (3)2
4
𝑑𝑡
𝑑ℎ
8
=
m3 /min
𝑑𝑡 9𝜋
EXERCISE 20.2
1. Find the gradient of the curve 𝑦 = 8√𝑥 + 𝑥 at
the point whose 𝑥 coordinate is 9.
2.
5
Find the gradient of the curve 𝑦 = 𝑥 2 at the
point where 𝑥 = 4.
3. A curve has the equation 𝑦 = 5(1 − cos 2𝑥)
𝜋
and is defined for 0 ≤ 𝑥 ≤ 2 radians. Find
(i) the value of 𝑦 when 𝑥 = 1
(ii) the value of 𝑥 when 𝑦 = 3
𝜋
(iii) the gradient of the curve when 𝑥 = 4
4. Find the coordinates of the points on the
1
9
curve 𝑦 = 𝑥 3 + at which the tangent is
3
𝑥
parallel to the line 𝑦 = 8𝑥 + 3.
5. Find the equation of the normal to the curve
𝑦 = 𝑥 3 − 4𝑥 2 + 7 at the point (2, −1).
6. A curve has equation 𝑦 = 𝑥 2 + 𝑥.
(i)
Find the gradient of the curve at the
point for which 𝑥 = 2.
(ii)
Find the equation of the normal at the
point for which 𝑥 = 2.
7. Find the equation of the normal to the curve
6
𝑦 = 𝑥2 − 5 at the point on the curve where
𝑥 = 2.
8. Find the equation of the normal to the curve
2𝑥+4
𝑦 = 𝑥−2 at the point where 𝑥 = 4.
9. The equation of a curve is 𝑦 = 𝑥 3 − 8. Find the
equation of the normal to the curve at the
point where the curve crosses the 𝑥 axis.
𝑘
10. The curve 𝑦 = ℎ𝑥 2 + 𝑥 passes through point
𝑃(1, 1)and has a gradient of 5 at 𝑃.
Find
i. the values of the constants ℎ and 𝑘
ii. the equation of the tangent to the curve at
1
the point where 𝑥 = 2.
11. (i) Find the coordinates of the stationary
points on the curve 𝑦 = 𝑥 3 + 𝑥 2 − 𝑥 + 3.
(ii) Determine whether each stationary point
is a maximum or minimum point.
(iii) For what values of 𝑥 does 𝑥 3 + 𝑥 2 − 𝑥 + 3
decrease as 𝑥 increases?
12. (i) Find the coordinates of the stationary
points of the curve 𝑦 = 2𝑥 3 + 5𝑥 2 − 4𝑥.
(ii) State the set of values for 𝑥 for which
2𝑥 3 + 5𝑥 2 − 4𝑥 is a decreasing function,
139
CHAPTER 20: DIFFERENTIATION
(iii) Show that the equation of the tangent to
1
the curve at the point where 𝑥 = is
2
10𝑥 − 4𝑦 − 7 = 0.
13. (i) Find the coordinates of the stationary
6
point on the curve 𝑦 = 3𝑥 2 − 𝑥 − 2.
(ii) Determine whether the stationary point is
a maximum or a minimum point.
14. A curve has the equation 𝑦 = √𝑥 +
(i) Find expression for
𝑑𝑦
𝑑𝑥
and
𝑑2 𝑦
𝑑𝑥2
9
√𝑥
.
.
(ii) Show that the curve has a stationary
value when 𝑥 = 9.
(iii) Find the nature of this stationary value.
2
15. A curve has equation 𝑦 = 3𝑥 3 − 7𝑥 + 𝑥 .
(i) Verify that the curve has a stationary
point when 𝑥 = 1.
(ii) Determine the nature of this stationary
point.
(iii) The tangent to the curve at this stationary
point meets the 𝑦 axis at the point 𝑄. Find
the coordinates of 𝑄.
16. The curve 𝑦 = (1 − 𝑥)(𝑥 2 + 4𝑥 + 𝑘) has a
stationary point when 𝑥 = −3.
(i) Find the value of the constant 𝑘.
(ii) Determine whether the stationary point is
a maximum or minimum point.
(iii) Given that 𝑦 = 9𝑥 − 9 is the equation of
the tangent to the curve at the point 𝐴,
find the coordinates of 𝐴.
17. (i) Find the coordinates of the stationary
point on the curve 𝑦 = 𝑥 4 + 32𝑥.
(ii) Determine whether this stationary point
is a maximum or a minimum.
(iii) For what values of 𝑥 does 𝑥 4 + 32𝑥
increase as 𝑥 increases?
18. The curve 𝑦 = 𝑥 3 + 𝑝𝑥 2 + 2 has a stationary
point when 𝑥 = 4. Find the value of the
constant 𝑝 and determine whether the
stationary point is a maximum or minimum
point.
19. Find the point(s) of inflection on the following
curves.
(a) 𝑦 = 𝑥 4 − 2𝑥 3
(b) 𝑦 = 𝑥(15 − 4𝑥 2 )
(c) 𝑦 = 𝑥 3 − 3𝑥 2 + 3𝑥 − 1
(d) 𝑦 = 𝑥 3 − 3𝑥 2 + 3𝑥 − 4
(e) 𝑦 = 2𝑥 3 − 9𝑥 2 + 12𝑥 + 1
(f) 𝑦 = 𝑥 5 − 5
20. For the curve 𝑦 = 𝑥(𝑥 2 − 12), determine
(i) the coordinates of the stationary points,
(ii) the coordinates of the inflection point
(iii) the equation of the normal to the curve at
the origin.
21. The equation of a curve is given by
𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 4
(i) Obtain an expression for 𝑓 ′ (𝑥).
(ii) Find the stationary point(s) of 𝑓.
(iii) Determine the nature of the stationary
point(s) of 𝑓.
(iv) Sketch the curve.
𝑥
22. Draw a sketch of the function 𝑓(𝑥) = 𝑥−2.
23.
The diagram shows the curve with equation
𝑎𝑥 + 𝑏
𝑦=
𝑥+𝑐
where 𝑎, 𝑏 and 𝑐 are constants.
Given that the asymptotes of the curve are
𝑥 = −1 and 𝑦 = −2 and that the curve
passes through (3, 0), find the values of 𝑎, 𝑏
and 𝑐.
24. The diagram shows a rectangular field 𝐴𝐵𝐶𝐷
with 𝐴𝐵 = 50 m and 𝐴𝐷 = 80 m. The field is
partitioned by three fences 𝐴𝑃, 𝐴𝑄 and 𝑃𝑄.
The distance of 𝑃 from 𝐶 is twice the distance
of 𝑄 from 𝐷.
140
CHAPTER 20: DIFFERENTIATION
Given that 𝐷𝑄 = 𝑥 m, show that the area,
𝐴 m2 , of triangle 𝐴𝑃𝑄 is given by
𝐴 = 𝑥 2 − 40𝑥 + 2000.
Given that 𝑥 varies, find the stationary value
of 𝐴 and determine whether this is a
maximum or a minimum.
25. The circumference of a circle is increasing at
the rate of 3 cms −1. Find
i. the rate of increase of the radius
ii. the rate of increase of the area, at the
instant when the radius is 100 cm
26. At the instant when the radius is 5 cm. The
radius, 𝑟 cm, of a circle is increasing in such a
way that the rate of increase of the area of the
circle is 𝜋𝑟 cm2 s−1 . Calculate the rate of
increase of the radius.
27. The surface area of a sphere is increasing at a
constant rate of 6 cm2 s−1. Given that the
surface area of a sphere of radius 𝑟 is 4𝜋𝑟 2
4
and that the volume is 3 𝜋𝑟 3 , find the rate of
increase of
i.
the radius
ii.
the volume
28. The radius 𝑟 cm of a sphere is increasing at a
constant of 2 cms−1. Given that the volume of
4
a sphere of radius 𝑟 is 3 𝜋𝑟 3, find, in terms of
𝜋, the rate at which the volume is increasing
at the instant when the volume is 36𝜋 cm3.
29. A spherical balloon is released from rest and
expands as it rises. After rising for 𝑡 seconds
its radius is 𝑟 cm, and its surface area is 𝐴 cm2 ,
where 𝐴 = 4𝜋𝑟 2 .
The initial radius of the balloon is 16 cm.
Given that the rate of increase of the radius is
constant and has value 0.8 cms −1, find the
rate of increase of 𝐴 when 𝑡 = 5.
2𝑥
30. A curve has equation 𝑦 = 2 .
𝑥 +9
(i) Find the 𝑥 coordinate of each of the
stationary points of the curve.
(ii) Given that 𝑥 is increasing at the rate of 2
units per second, find the rate of
increase of 𝑦 when 𝑥 = 1.
31. A trough has the shape of the prism shown in
the diagram. The vertical ends 𝐴𝐵𝐶 and 𝐷𝐸𝐹
are identical isosceles triangles of height 20
cm with 𝐴𝐵 = 16 cm and 𝐴𝐶 = 𝐵𝐶. The open
top 𝐴𝐵𝐸𝐷 is horizontal and rectangular in
shape with 𝐴𝐷 = 30 cm.
The trough is being filled with water, 𝑋𝑌
indicating the level reached. At time 𝑡 seconds
after filling starts, the depth of water is 𝑥 cm
and 𝑋𝑌 is 𝑦 cm, as shown.
By using similar triangles, express 𝑦 in terms of
𝑥 and hence show that the volume, 𝑉 cm3 , of
water in the tank at time 𝑡 seconds is given by
𝑉 = 12𝑥 2 .
Given that the water is flowing into the trough
at the rate of 60 cm3 s −1, find the rate at which
𝑥 is increasing when 𝑥 = 10.
32. The parametric equations of a curve are
𝑥 = 3 + 2 sin 𝜃 ,
𝑦 = 4 − 2 cos 𝜃
where 0 ≤ 𝜃 < 2𝜋.
𝑑𝑥
𝑑𝑦
(i) Write down and , and hence express
𝑑𝜃
𝑑𝑦
𝑑𝜃
in terms of 𝜃.
𝑑𝑥
(ii) It is given that the curve is a circle. Use
the identity sin2 𝜃 + cos 2 𝜃 = 1 to find the
Cartesian equation of this circle, and state
the centre and radius.
33. A curve is represented parametrically by
𝑥 = 𝑡 2 + 3𝑡,
𝑦 = 𝑡 2 − 2𝑡
Find
𝑑𝑦
(i) an expression for 𝑑𝑥 in terms of 𝑡
(ii) the coordinates of the stationary point of
the curve.
SOLUTIONS
1.
7
3
2.
3.
4.
20
(i) 7.08 (ii) 0.58 (iii) 5
(3, 12) (−3, −12)
5.
𝑦= 𝑥−
6.
(i) 5 (ii) 𝑦 = − 5 𝑥 + 5
7.
𝑦 = 3𝑥 − 6
8.
𝑦 = 2𝑥 + 4
1
3
4
2
1
2
32
29
1
141
CHAPTER 20: DIFFERENTIATION
9.
1
1
𝑦 = 12 𝑥 + 6
2.
𝑥2 −1
(a) Given that 𝑦 = 𝑥2 +1,
𝑑𝑦
10. (i) ℎ = 2, 𝑘 = −1 (ii) 6
(i) find 𝑑𝑥 in terms of 𝑥
1 76
11. (i) 𝐴(−1, 4), 𝐵 (3 , 27)
(ii) 𝐴 − max,𝐵 − min (iii) −1 < 𝑥 <
1
19
1
3
12. (i) (−2, 12), (3 , − 27)
1
(ii) −2 < 𝑥 <
(iii)
3
13. (i) (−1, 7) (ii) minimum
𝑑𝑦
14. (i) 𝑑𝑥 =
𝑥−9 𝑑2 𝑦
27−𝑥
3
2𝑥 2
5
4𝑥 2
, 𝑑𝑥2 =
3.
(iii) minimum
15. (ii) minimum (iii) 𝑄(0, −2)
16. (i) 𝑘 = −5 (ii) minimum (iii) (−2, −27)
17. (i) (−2, −48) (ii) minimum (iii) 𝑥 > −2
18. 𝑝 = −6, minimum
19. (a) (0, 0), (1, −1) (b) (0, 0) (c) (1, 0)
3 11
(d) (1, −3) (e) ( , ) (f) (0, −5)
2
2
4.
1
21. (i) 3𝑥 2 − 6𝑥 (ii) 𝐴(0, 4), 𝐵(2, 0)
(iii) 𝐴 − max, 𝐵 − min
(iv)
22.
23. 𝑎 = 2, 𝑏 = −2, 𝑐 = 1
24. 1600, minimum
3
25. (i) 2𝜋 (ii) 300
5.
𝑥
2
3
4𝜋𝑟
(ii) 3𝑟
6.
28. 72𝜋
29. 128𝜋
8
30. (i) 𝑥 = ±3 (ii) 25
4
1
31. 𝑦 = 5 𝑥, 4 cms −1
32. (i)
𝑑𝑥
= 2 cos 𝜃 ,
𝑑𝜃
2𝑡−2
33. (i) 2𝑡+3
𝑑𝑦
𝑑𝜃
= 2 sin 𝜃 (ii) 𝐶(3, 4), 𝑟 = 2
(ii) (4, −1)
7.
EXAM QUESTIONS
1.
(b) the value of 𝑑𝑥2 at 𝑃
[2]
(c) the equation of the normal to the curve at
𝑃.
[4]
CAPE 2005
(a) Find the coordinates of the stationary
points of the function
𝑓: 𝑥 → 𝑥 3 − 3𝑥 2 − 9𝑥 + 6.
[6]
(b) Determine the nature of the stationary
points of 𝑓.
[3]
CAPE 2005
(a) The function 𝑓(𝑥) is defined by
2−𝑥
𝑓(𝑥) = 2 for 𝑥 ∈ ℝ, 𝑥 ≠ 0.
Determine the nature of the critical
value(s) of 𝑓(𝑥).
(b) Differentiate, with respect to 𝑥,
𝑓(𝑥) = sin2 (𝑥 2 )
1
27. (i)
[5]
(b) By investigating the sign of 𝑓 ′ (𝑥),
determine the range of real values of 𝑥 for
which 𝑥 5 − 5𝑥 + 3 is decreasing.
[8]
CAPE 2004
𝑃 is the point on the curve 𝑦 = 2𝑥 3 + 𝑘𝑥 − 5
where 𝑥 = 1 and the gradient is −2. Find
(a) the value of the constant 𝑘
[3]
𝑑2 𝑦
20. (i) (−2, 16), (2, 16) (ii) (0, 0) (iii) 𝑦 = 12 𝑥
26.
[5]
𝑑𝑦
4
(ii) show that 𝑥(𝑥 2 + 1) 𝑑𝑥 − 4𝑦 = 𝑥2 +1 .
(a) Given that 𝑓(𝑥) = 𝑥 3 − 5𝑥 2 + 3𝑥, find
(i) the coordinates of the stationary
points of 𝑓(𝑥).
[6]
(ii) the second derivative of 𝑓(𝑥), and
hence, determine which stationary
point is a local maximum and which
is a local minimum.
[5]
1
𝑑2 𝑦
2
(b)If 𝑦 = 𝑥2 +2 , show that 𝑑𝑥2 = 2(3𝑥 − 2)𝑦 3 .
[7]
CAPE 2003
8.
[6]
[3]
CAPE 2006
The curve 𝑦 = 𝑝𝑥 3 + 𝑞𝑥 + 𝑟 passes through
the origin 𝑂 and the point 𝑃(1, 2). The
gradient of the curve at 𝑃 is equal to 8.
(a) Calculate the values of the constants 𝑝, 𝑞
and 𝑟.
[6]
(b) Obtain the equation of the normal to the
curve at 𝑃.
[3]
CAPE 2007
For the function 𝑓: 𝑥 → 12𝑥 − 𝑥 3 , determine
(a) the stationary points
[4]
(b) the nature of EACH of the stationary
points.
[3]
CAPE 2007
A chemical process is controlled by the
𝑢
function 𝑃 = 𝑡 + 𝑣𝑡 2 , where 𝑢 and 𝑣 are
constants. Given that 𝑃 = −1 when 𝑡 = 1 and
the rate of change of 𝑃 with respect to 𝑡 is −5
1
when 𝑡 = 2 , find the values of 𝑢 and 𝑣.
[6]
9.
CAPE 2008
Given that 𝑦 = sin2𝑥 + cos 2𝑥, show that
𝑑2𝑦
𝑑𝑥2
+ 4𝑦 = 0.
[6]
142
CHAPTER 20: DIFFERENTIATION
CAPE 2009
10. The diagram represents a piece of thin
cardboard 16 cm by 10 cm. Shaded squares,
each of side 𝑥 cm, are removed from each
corner. The remainder is folded to form a tray.
(i) Show that the volume, 𝑉 cm3,
of the tray is given by
𝑉 = 4(𝑥 3 − 13𝑥 2 + 40𝑥).
[5]
(ii) Hence, find a possible value of 𝑥 such that
𝑉 is a maximum.
[8]
CAPE 2009
11. The curve 𝑦 = 𝑝𝑥 3 + 𝑞𝑥 2 + 2 passes through
the point 𝑇(1, 2) and its gradient at 𝑇 is 7. The
line 𝑥 = 1 cuts the 𝑥 −axis at 𝑀, and the
normal to the curve at 𝑇 cuts the 𝑥 −axis at 𝑁.
Find
(i) the values of the constants 𝑝 and 𝑞.
[6]
(ii) the equation of the normal to the curve at
𝑇.
[3]
(iii) the length of 𝑀𝑁.
[2]
CAPE 2011
12. A chemical process in a manufacturing plant is
𝑣
controlled by the function 𝑀 = 𝑢𝑡 2 + 𝑡2
where 𝑢 and 𝑣 are constants.
Given that 𝑀 = −1 when 𝑡 = 1 and that the
35
rate of change of 𝑀 with respect to 𝑡 is 4
when 𝑡 = 2, find the values of 𝑢 and 𝑣.
[8]
CAPE 2012
13. (i) Differentiate from first principles the
function 𝑓(𝑥 ) = 𝑥 3 with respect to 𝑥. [6]
(ii) Given that 𝑓(𝑥) = 𝑟𝑥 2 + 𝑠𝑥 + 𝑡, 𝑟 ≠ 0,
(a) find
(i) 𝑓 ′ (𝑥)
(ii) 𝑓 ′′ (𝑥)
[2]
(b) find, in terms of 𝑟 and 𝑠, the
conditions under which 𝑓(𝑥) will
have a maximum.
[3]
(c) find the maximum.
[3]
(iii) The curve 𝑦 = 𝑝𝑥 3 + 𝑞𝑥 2 + 3𝑥 + 2 passes
through the point 𝑇(1, 2) and its gradient
at 𝑇 is 7. Find the values of the constants
𝑝 and 𝑞.
[5]
CAPE 2004
14. (a) Given that 𝑦 = √5𝑥 2 + 3,
𝑑𝑦
(i) obtain 𝑑𝑥
[4]
𝑑𝑦
(ii) show that 𝑦 𝑑𝑥 = 5𝑥
(iii) hence, or otherwise, show that
𝑑2 𝑦
[2]
𝑑𝑦 2
𝑦 𝑑𝑥2 + (𝑑𝑥 ) = 5.
[4]
(b) At a certain port, high tides and low tides
occur daily. Suppose 𝑡 minutes after high
tide, the height, ℎ metres, of the tide
above a fixed point is given by
𝜋𝑡
ℎ = 2 (1 + cos
),0 ≤ 𝑡
450
[Note: High tide occurs when 𝒉 has its
maximum value and low tide when 𝒉 has
its minimum value.]
Determine
(i) the height of the tide when high tide
occurs for the first time
[2]
(ii) the length of time which elapses
between the first high tide and the
first low tide.
[3]
(iii) the rate, in metres per minute, at
which the tide is falling 75 minutes
after high tide.
[5]
CAPE 2007
15. Differentiate with respect to 𝑥
(i) 𝑦 = sin(3𝑥 + 2) + tan 5𝑥
[3]
𝑥2 +1
(ii) 𝑦 = 𝑥3 −1
[4]
CAPE 2010
16. (i) Given that 𝑦 = √4𝑥 2 + 7, show that
𝑑𝑦
𝑦 𝑑𝑥 = 4𝑥.
[3]
(ii) Hence, or otherwise, show that
𝑑𝑦
𝑑𝑦 2
𝑦
+( ) =4
𝑑𝑥
𝑑𝑥
[3]
CAPE 2012
𝑥2 +2𝑥+3
17. (a) Let 𝑦 = (𝑥2 +2)3 . Show that
𝑑𝑦 −4𝑥 3 − 10𝑥 2 − 14𝑥 + 4
=
(𝑥 2 + 2)4
𝑑𝑥
[5]
(b) The equation of an ellipse is given by
𝑥 = 1 − cos 𝜃 , 𝑦 = 2 sin 𝜃 , 0 ≤ 𝜃 ≤ 2𝜋
𝑑𝑦
Find 𝑑𝑥 in terms of 𝜃.
[5]
CAPE 2013
18. (a) (i) Let 𝑦 =
𝑑𝑦
𝑑𝑥
.
1
. Using first principles, find
√𝑥
[8]
143
CHAPTER 20: DIFFERENTIATION
(ii) If 𝑦 =
𝑑𝑦
𝑥
√1+𝑥
, determine an expression
for 𝑑𝑥 . Simplify your answer FULLY.
[4]
(b) The parametric equations of a curve are
given by 𝑥 = cos 𝜃 , 𝑦 = sin 𝜃,
𝑑𝑦
0 ≤ 𝜃 ≤ 2𝜋. Find 𝑑𝑥 in terms of 𝜃. Simplify
your answer as far as possible.
[4]
CAPE 2014
19. (a) Using first principles, determine the
derivative of 𝑓(𝑥) = sin(2𝑥).
[6]
2𝑥
(b) If 𝑦 = √ 2 , show that
1+𝑥
𝑑𝑦
𝑦
(i) 𝑥 𝑑𝑥 = 1+𝑥2
𝑑2𝑦
18. (a) (i) −
1
3
2𝑥 2
𝑥+2
(ii)
3
(b) − cot 𝜃
2(𝑥+1)2
19. (a) 2 cos(2𝑥) (b) (i)
(ii)
…………………………………………………………………………..
[7]
3𝑦
(ii) 𝑑𝑥2 + (1+𝑥2 )2 = 0
[8]
CAPE 2015
SOLUTIONS
1 13
1.
(a) (i) 𝐴 (3 , 27) , 𝐵(3, −9)
2.
(ii) 𝐴-maximum, 𝐵- minimum
4𝑥
(a) (i) (𝑥2 +1)2
3.
(b) −1 < 𝑥 < 1
1
23
(a) 𝑘 = −8 (b) 12 (c) 𝑦 = 2 𝑥 − 2
4.
5.
6.
(a) 𝐴(−1, 11), 𝐵(3, −21)
(b) 𝐴-maximum, 𝐵-minimum
(a) minimum
(b) 2𝑥 sin(2𝑥 2 )
(a) 𝑝 = 3, 𝑞 = −1, 𝑟 = 0
1
7.
8.
17
(b) 𝑦 = − 8 𝑥 + 8
(a) 𝐴(−2, −16), 𝐵(2, 16)
(b) 𝐴-minimum, 𝐵-maximum
4
9
𝑢 = 5, 𝑣 = − 5
9.
10. (ii) 𝑥 = 2
1
15
11. (i) 𝑝 = 7, 𝑞 = −7 (ii) 𝑦 = − 7 𝑥 + 7
(iii) 14
12. 𝑢 = 2, 𝑣 = −3
13. (i) 3𝑥 2 (ii) (a) (i) 2𝑟𝑥 + 𝑠
(ii) 2𝑟
𝑠2
(b) 2𝑟𝑥 + 𝑠 = 0 and 𝑟 < 0 (c) 𝑡 − 4𝑟
(iii) 𝑝 = 10, 𝑞 = −13
5𝑥
14. (a) (i)
(ii)
(iii)
2
√5𝑥 +3
(b)(i) 4
𝜋
(iii) −
450
15. (i) 3 cos(3𝑥 + 2) + 5 sec 2(5𝑥)
(ii) −
16. (i)
17. (a)
(ii) 450
𝑥(𝑥3 +3𝑥+2)
(𝑥3 −1)2
(ii)
(b) 2 cot 𝜃
144
CHAPTER 21: INTEGRATION
CHAPTER 21: INTEGRATION
At the end of this section, students should be able
to:
 recognise integration as the reverse
process of differentiation;
 use the notation ∫ 𝑓(𝑥) 𝑑𝑥;
 show that the indefinite integral
represents a family of functions which
differ by constants;
 use simple integration rules;
 integrate functions of the form (𝑎𝑥 + 𝑏)𝑛
where 𝑎, 𝑏, 𝑛 are real numbers and 𝑛 ≠
−1;
 find the indefinite integrals using
formulae and integration theorems;
 integrate simple trigonometric functions;
 compute definite integrals;
 formulate the equation of a curve given
its gradient function and a point on the
curve;



demonstrate use of the following
integration theorems;
(a) ∫ 𝑐𝑓(𝑥) 𝑑𝑥 = 𝑐 ∫ 𝑓(𝑥) 𝑑𝑥, where 𝑐 is a
constant,
(b) ∫{𝑓(𝑥) ± 𝑔(𝑥)} 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ±
∫ 𝑔(𝑥) 𝑑𝑥;
integrate using substitution;
use the results:
𝑏
𝑏
(a) ∫𝑎 𝑓(𝑥 ) 𝑑𝑥 = ∫𝑎 𝑓 (𝑡) 𝑑𝑡
𝑎


𝑎
(b) ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑥 − 𝑎) 𝑑𝑥 for
𝑎 > 0,
𝑏
(c) ∫𝑎 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎), where
𝐹 ′ (𝑥) = 𝑓(𝑥);
apply integration to:
(a) finding areas under the curve;
(b) finding areas between two curves;
(c) finding volumes of revolution by
rotating regions about both the 𝑥and 𝑦- axes;
given a rate of change with or without
initial boundary conditions:
(a) formulate a differential equation of
the form 𝑦 ′ = 𝑓(𝑥) or 𝑦 ′′ = 𝑓(𝑥)
where 𝑓 is a polynomial or a
trigonometric function,
(b) solve the resulting differential
equation in (a) above and interpret
the solution where applicable.
__________________________________________________________
HOW TO INTEGRATE
INTRODUCTION
Integration is referred to as anti – differentiation.
Therefore, the process of integration is the reverse
of differentiation.
Before we look at integration let’s review
differentiation by differentiating each of the
following
1. 𝑦 = 5𝑥 3 + 2𝑥
2. 𝑦 = 5𝑥 3 + 2𝑥 − 7
3. 𝑦 = 5𝑥 3 + 2𝑥 + 99
The derivative of each of the above functions is
15𝑥 2 + 2 though each function is differs in the
value of their constant.
INTEGRATION NOTATION
∫ is the symbol for integration
∫ 2𝑥 5 𝑑𝑥 means that we are integrating 2𝑥 5 with
respect to 𝑥. This is read “the integral of 2𝑥 5 with
respect to 𝑥.”
In general,
𝑎𝑥 𝑛+1
∫ 𝑎𝑥 𝑛 𝑑𝑥 =
+𝑐
𝑛 +1
where 𝑐 is known as the constant of integration.
The constant of integration (arbitrary constant)
compensates for the fact that the integral could
have an unknown constant.
PROPERTIES OF INTEGRALS
∫ 𝑘 𝑓(𝑥) 𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥) 𝑑𝑥 where 𝑘 is a constant
∫[𝑓 (𝑥 ) ± 𝑔 (𝑥 )] 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 ± ∫ 𝑔 (𝑥 ) 𝑑𝑥
LESSON 1
1. ∫ 4 𝑑𝑥
3
2. ∫ 𝑥2 𝑑𝑥
Evaluate each of the following.
4
3.
∫ (5√𝑥 − 2𝑥 3 + 3√ 5 ) 𝑑𝑥
4.
5.
∫ 𝑥 2 (3 − 𝑥) 𝑑𝑥
∫(2𝑥 − 1) 3 𝑑𝑥
𝑥
SOLUTION
1. ∫ 4 𝑑𝑥 = ∫ 4𝑥 0 𝑑𝑥
4𝑥 0+1
=
+𝑐
0+1
= 4𝑥 + 𝑐
2.
3
∫ 𝑥2 𝑑𝑥 = ∫ 3𝑥 −2 𝑑𝑥
3𝑥 −2+1
=
+𝑐
−2 + 1
145
CHAPTER 21: INTEGRATION
3𝑥 −1
+𝑐
−1
3
=− +𝑐
𝑥
ALTERNATELY,
∫ cos 𝑎𝑥 𝑑𝑥 =
=
∫ sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
∫ sec 2 𝑎𝑥 𝑑𝑥 =
𝑥 −2+1
]+𝑐
−2 + 1
𝑥 −1
= 3[
]+𝑐
−1
3
=− +𝑐
𝑥
∫ csc 2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝑐
1
∫ csc 2 𝑎𝑥 𝑑𝑥 = − cot 𝑎𝑥 + 𝑐
𝑎
∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐
∫ sec 𝑎𝑥 tan 𝑎𝑥 𝑑𝑥 =
1
4
∫ (5√𝑥 − 2𝑥 3 + 3√ 5 ) 𝑑𝑥 = ∫ (5𝑥 2 − 2𝑥 3 +
𝑥
5
−
3
4𝑥 ) 𝑑𝑥
1
=
3
5𝑥 2
5
2𝑥 4
+
4
+𝑐
3
2
−
2
3
2
10 3 1 4
=
𝑥 2 − 𝑥 − 6𝑥 −3 + 𝑐
3
2
5.
−
2
4𝑥 −3
LESSON 2
1 (𝑎𝑥+𝑏) 𝑛+1
N.B ∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 = 𝑎 [ (𝑛+1) ] + 𝑐
1 (2𝑥 − 1)3+1
= [
]+𝑐
(3 + 1)
2
(2𝑥 − 1)4
=
+𝑐
8
These are referred to as Indefinite Integrals since
they contain the arbitrary constant 𝑐.
TRIGONOMETRIC INTEGRATION
1
csc 𝑎𝑥 + 𝑐
𝑎
Evaluate each of the following.
1.
∫(1 + sin 2𝑥) 𝑑𝑥
2.
∫ (2 sec 2(−5𝑥) + 2 𝑥 2 ) 𝑑𝑥
1
3
−
SOLUTION
1. ∫(1 + sin 2𝑥) 𝑑𝑥
∫ 𝑥 2 (3 − 𝑥) 𝑑𝑥 = ∫(3𝑥 2 − 𝑥 3 ) 𝑑𝑥
3𝑥 2+1 𝑥 3+1
=
−
+𝑐
2+1 3+1
𝑥4
= 𝑥3 − + 𝑐
4
∫(2𝑥 − 1) 3 𝑑𝑥
1
sec 𝑎𝑥 + 𝑐
𝑎
∫ csc 𝑥 cot 𝑥 𝑑𝑥 = csc 𝑥 + 𝑐
∫ csc 𝑎𝑥 cot 𝑎𝑥 𝑑𝑥 =
5𝑥 2+1 2𝑥 3+1 4𝑥 −3+1
=
−
+
+𝑐
1
5
3+1
+
1
−
+
1
2
3
4.
1
tan 𝑎𝑥 + 𝑐
𝑎
∫ 3𝑥 −2 𝑑𝑥 = 3 ∫ 𝑥 −2 𝑑𝑥
= 3[
3.
1
sin𝑎𝑥 + 𝑐
𝑎
= ∫ 1 𝑑𝑥 + ∫ sin2𝑥 𝑑𝑥
2.
𝑥 0+1
1
=1(
) + (− cos 2𝑥) + 𝑐
0+1
2
1
= 𝑥 − cos 2𝑥 + 𝑐
2
3
1
∫ (2 sec 2(−5𝑥) + 2 𝑥 −2 ) 𝑑𝑥
3
1
= 2 ∫ sec 2(−5𝑥) 𝑑𝑥 + ∫ 𝑥 −2 𝑑𝑥
2
3
1
1 𝑥 −2+1
= 2 [− tan(−5𝑥)] + (
)+𝑐
5
2 −3 + 1
2
−1
2
1 𝑥2
= − tan(−5𝑥) + (
)+𝑐
5
2 −1
2
−1
2
2
= − tan(−5𝑥) − 𝑥 + 𝑐
5
∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐
1
∫ sin 𝑎𝑥 𝑑𝑥 = − cos 𝑎𝑥 + 𝑐
𝑎
∫ cos 𝑥 𝑑𝑥 = sin𝑥 + 𝑐
146
CHAPTER 21: INTEGRATION
DIFFERENTIAL EQUATIONS
A Differential Equation contains derivatives or
differentials
LESSON 3
1.
2.
3.
𝑑𝑦
𝑑𝑥
Solve the differential equation
= 6𝑥 2 − 5
𝑦′ = 𝑥 3 − 5
𝑑2 𝑦
𝑑𝑥2
1
= 6𝑥 3 + 𝑥3
SOLUTION
1.
𝑏
∫ 𝑓(𝑥) 𝑑𝑥
𝑎
To determine the definite integral we
a. integrate the function, excluding 𝑐
b. substitute the upper limit, 𝑥 = 𝑏, into the
integral
c. substitute the lower limit, 𝑥 = 𝑎, into the
integral
d. subtract the value from step (3) from the
value from step (2)
The answer which will be a number is the definite
integral.
LESSON 4a
𝑦′ = 𝑥 3 − 5
Evaluate the following
3
∫ 𝑥 3 𝑑𝑥
𝑦 = ∫ 𝑥 3 − 5 𝑑𝑥
0
=
2.
𝑑𝑦
𝑑𝑥
𝑥4
− 5𝑥 + 𝑐
4
= 6𝑥 2 − 5
𝑑𝑦 = (6𝑥 2 − 5) 𝑑𝑥
∫ 1 𝑑𝑦 = ∫(6𝑥 2 − 5) 𝑑𝑥
SOLUTION
a. We integrate the function, excluding the
constant 𝑐
𝑥 3+1 3
𝑥4 3
= [(
)] = [ ]
3+1 0
4 0
b. Substituting the upper limit into the integral we
34
c.
𝑦 = 2𝑥 3 − 5𝑥 + 𝑐
3.
𝑑2 𝑦
𝑑𝑥2
∫
1
= 6𝑥 3 + 𝑥3
𝑑 2𝑦
𝑑𝑦
𝑑𝑥 =
2
𝑑𝑥
𝑑𝑥
= ∫ 6𝑥 3 + 𝑥 −3 𝑑𝑥
3𝑥 4 𝑥 −2
−
+𝐴
2
2
4
−2
3𝑥
𝑥
∫ 𝑑𝑦 = ∫
−
+ 𝐴 𝑑𝑥
2
2
5
3𝑥
1
𝑦=
+
+ 𝐴𝑥 + 𝑐
10 2𝑥
=
DEFINITE INTEGRALS
If 𝐹(𝑥) is the integral of 𝑓(𝑥) then the definite
integral is defined as
𝑥=𝑏
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = [𝐹(𝑥)] = 𝐹(𝑏) − 𝐹(𝑎)
𝑎
𝑥=𝑎
where 𝑏 ∈ ℝ and 𝑏 is called the upper limit of 𝑓(𝑥)
𝑎 ∈ ℝ is the lower limit of 𝑓(𝑥)
𝐹(𝑏) is the value of the integral at the upper limit
𝑥=𝑏
𝐹(𝑎) is the value of the integral at the lower limit
𝑥=𝑎
It is accepted to simply write
get
d.
81
get [ 4 ] = 4
Substituting the lower limit into the integral we
04
4
=0
Subtracting the value from (c) from the value
81
81
from (b) 4 − 0 = 4 which is the definite
integral
The next LESSON illustrates the general way in
which a question involving definite integrals is
solved.
LESSON 4b
Evaluate
4
∫(𝑥 2 − 4 + 4𝑥 −2 )𝑑𝑥
1
SOLUTION
4
∫(𝑥 2 − 4 + 4𝑥 −2 )𝑑𝑥
1
𝑥 2+1
𝑥 0+1
𝑥 −2+1 4
− 4(
)+4(
)]
2+1
0+ 1
−2 + 1 1
𝑥3
4
𝑥3
4 4
= [ − 4𝑥 − 4𝑥 −1 ] = [ − 4𝑥 − ]
3
1
3
𝑥 1
43
4
13
4
= [ − 4(4) − ] − [ − 4(1) − ]
3
4
3
1
64
1
= [ − 16 − 1] − [ − 4 − 4]
3
3
= 12
=[
147
CHAPTER 21: INTEGRATION
LESSON 4c
LESSON 6
Determine ∫ 𝑥 3 (1 − 5𝑥 4 )2 𝑑𝑥
SOLUTION
Let 𝑢 = 1 − 5𝑥 4
𝑑𝑢
= −20𝑥 3
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
−20𝑥 3
Find
2
∫
1
1
√3𝑥 − 2
𝑑𝑥
SOLUTION
2
∫
1
1
√3𝑥 − 2
𝑑𝑥
∫ 𝑥 3 (1 − 5𝑥 4 )2 𝑑𝑥
2
1)
= ∫(3𝑥 − 2)−(2
1
= ∫ 𝑥 3 𝑢2 (
1
1
1 (3𝑥 − 2)2 2
2(3𝑥 − 2) 2 2
= [
] =[
]
1
3
1
3
1
⁄2
1
1
2(3(2) − 2)2
2(3(1) − 2)2
=[
]−[
]
3
3
4 2
−
3 3
2
=
3
1
∫ 𝑢2 𝑑𝑢
20
1 𝑢3
=− ( )+𝑐
20 3
=−
−
=
LESSON 4d
Find
𝜋
∫ 1 + sin 𝑥 𝑑𝑥
–𝜋
𝑑𝑢
)
−20𝑥 3
1
(1 − 5𝑥 4 )3 + 𝑐
60
LESSON 7
SOLUTION
Let 𝑢 = sin2𝑥
𝑑𝑢
= 2 cos 2𝑥
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
2 cos 2𝑥
Find ∫ sin 2𝑥 cos 2𝑥 𝑑𝑥
SOLUTION
∫ sin 2𝑥 cos 2𝑥 𝑑𝑥
∫ 1 + sin𝑥 𝑑𝑥
= ∫ 𝑢 cos 2𝑥 (
𝜋
–𝜋
𝜋
= [𝑥 − cos 𝑥]
−𝜋
= [𝜋 − cos(𝜋)] − [−𝜋 − cos(−𝜋)]
= [𝜋 − (−1)] − [−𝜋 − (−1)]
= 2𝜋
INTEGRATION BY SUBSTITUTION
LESSON 5
Find ∫(2𝑥 3 + 5) 4 (6𝑥 2 )𝑑𝑥
SOLUTION
𝑑𝑢
𝑑𝑢
= 6𝑥 2 → 2 = 𝑑𝑥
𝑑𝑥
6𝑥
So our integral can be rewritten as
𝑑𝑢
∫(𝑢4 )(6𝑥 2 ) 2 = ∫ 𝑢4 𝑑𝑢
6𝑥
4+1
𝑢
∫ 𝑢4 𝑑𝑢 =
+𝑐
4+1
5
𝑢
=
+𝑐
5
Resubstituting we get that
(2𝑥 3 + 5)5
∫(2𝑥 3 + 5)4 (6𝑥 2 ) 𝑑𝑥 =
+𝑐
5
𝑢 = 2𝑥 3 + 5 then
1
∫ 𝑢 𝑑𝑢
2
1 𝑢2
= ( )+𝑐
2 2
𝑢2
=
+𝑐
4
sin2 2𝑥
=
+𝑐
4
𝑑𝑢
)
2 cos 2𝑥
=
LESSON 8
SOLUTION
0
Find ∫−1 𝑥 2 (1 − 2𝑥 3 )4 𝑑𝑥
𝑑𝑢
𝑑𝑢
= −6𝑥 2 → − 2 = 𝑑𝑥
𝑑𝑥
6𝑥
Since the limits are in terms of 𝑥 an we will be
integrating in terms of 𝑢 we need to determine the
limits in terms of 𝑢, as follows
When 𝑥 = 0; 𝑢 = 1 − 2(0)3 = 1 (Upper Limit)
When 𝑥 = −1; 𝑢 = 1 − 2(−1)3 = 3 (Lower Limit)
Therefore,
Let 𝑢 = 1 − 2𝑥 3 then
0
∫ 𝑥 2 (1 − 2𝑥 3 )4 𝑑𝑥
−1
148
CHAPTER 21: INTEGRATION
1
= ∫ 𝑥 2 (𝑢4 )
3
6
1
𝑑𝑢
𝑑𝑢
=
∫
𝑢4
−6𝑥 2
−6
3
(e) ∫ 𝑥3 𝑑𝑥
(f) ∫(5𝑥 3 − 6𝑥 + 1) 𝑑𝑥
(g) ∫(𝑥 3 + 8𝑥 − 5) 𝑑𝑥
(h) ∫(𝑥 2 + 4)(𝑥 − 6) 𝑑𝑥
1 1
= − ∫ 𝑢4 𝑑𝑢
6 3
1 𝑢4+1 1
=− [
]
6 4+ 1 3
1 𝑢5 1
=− [ ]
6 5 3
1 15
35
= − ([ ] − [ ])
6 5
5
1 1 243
=− ( −
)
6 5
5
121
=
15
LESSON 9
evaluate
6
(i) ∫ (4𝑥 + 𝑥2 ) 𝑑𝑥
(j) ∫(2 − 6√𝑥) 𝑑𝑥
(k) ∫ 12√𝑥 𝑑𝑥
(l) ∫ 𝑥(𝑥 2 − 4) 𝑑𝑥
(m) ∫ 𝑥 3 (𝑥 2 − 𝑥 + 5) 𝑑𝑥
(n) ∫ 𝑥(𝑥 − 1)2 𝑑𝑥
1
(o) ∫
∫ sin4𝑥 cos 4𝑥 𝑑𝑥
0
SOLUTION
Instead of adjusting the limits we
will integrate firstly and then use the original
limits.
𝑢 = sin 4𝑥
𝑑𝑢
= 4 cos 4𝑥
𝑑𝑥
𝑑𝑢
= 𝑑𝑥
4 cos 4𝑥
∫ sin 4𝑥 cos 4𝑥 𝑑𝑥
1
2.
Evaluate
(a) ∫ cos 2𝑥 𝑑𝑥
(b) ∫ 𝑥 + 2 sin(−3𝑥) 𝑑𝑥
3. Find the general solution of each differential
equation.
𝑑𝑦
(a)
= 2𝑥 + 1
𝑑𝑥
′
(b) 𝑦 = 1 − 3𝑥
1
(c) 𝑓 ′ (𝑥) = − 𝑥3
(d) 𝑦 ′′ = cos 2𝑥
4. Evaluate each of the following.
6
(a) ∫1 𝑥( 𝑥 2 − 4) 𝑑𝑥
1
(b) ∫0 √3𝑥 + 1 𝑑𝑥
4
(c) ∫2 6𝑥 2 − 4𝑥 + 5 𝑑𝑥
3
(d) ∫0 5𝑥 2 − 9 𝑑𝑥
0 𝑥3 +2𝑥2
(e) ∫−2
4
10
8
2
5.
Express ∫1 (3√𝑥 +
6.
𝑎 + 𝑏√2, where 𝑎 and 𝑏 are integers.
The positive constant 𝑎 is such that
0
EXERCISE 21.1
1. Evaluate each of the following
(a) ∫ 6𝑥 −4 𝑑𝑥
(b) ∫ 24𝑥 −3 𝑑𝑥
(c) ∫ 18𝑥 −4 𝑑𝑥
𝑑𝑥
𝑥
(f) ∫2 (𝑥 3 − 𝑥2 ) 𝑑𝑥
8
1
𝜋
1
= [ sin2 (4 × )] − [ sin2(4 × 0)]
8
8
8
1
=
8
…………………………………………………………………………
𝑑𝑥
(q) ∫(3𝑥 + 1) −2 𝑑𝑥
𝑑𝑢
= ∫ 𝑢 cos 4𝑥 (
)
4 cos 4𝑥
1
= ∫ 𝑢 𝑑𝑢
4
1
= 𝑢2
8
𝜋
1
𝜋/8
∫ sin 4𝑥 cos 4𝑥 𝑑𝑥 = [ sin2 4𝑥]
8
0
𝑥
(p) ∫(2𝑥 − 1) 7 𝑑𝑥
Using the substitution 𝑢 = sin 4𝑥
𝜋
8
𝑥3 +3𝑥 2
√𝑥
) 𝑑𝑥 in the form
2𝑎 2𝑥3 −5𝑥2 +4
∫𝑎 (
𝑥2
) 𝑑𝑥 = 0.
(i) Show that 3𝑎3 − 5𝑎2 + 2 = 0.
(ii) Show that 𝑎 = 1 is a root of
3𝑎3 − 5𝑎2 + 2 = 0, and hence find the
other possible value of 𝑎, giving your
answer in simplified surd form.
7.
1
(i) Prove that sin2 𝜃 cos 2 𝜃 = 8 (1 − cos 4𝜃).
8
(d) ∫ 𝑥3 𝑑𝑥
149
CHAPTER 21: INTEGRATION
(ii) Hence, find the exact value of
𝜋
3
8.
3.
(a) 𝑥 2 + 𝑥 + 𝑐
3𝑥2
∫0 sin2 𝜃 cos 2 𝜃 𝑑𝜃
Evaluate
(a) ∫ cos 2𝑥 𝑑𝑥
(b) ∫ sec(−3𝑥)tan(−3𝑥) 𝑑𝑥
(c) ∫ csc 5𝑥 cot 5𝑥 𝑑𝑥
(d) ∫ csc 2(−𝑥) 𝑑𝑥
(b) 𝑥 − 2 + 𝑐
1
(c) 2𝑥2 + 𝑐
1
(d) − 4 cos(2𝑥 ) + 𝑐
4.
1015
(a)
(b)
9.
Evaluate each of the following
(i) ∫ 3𝑥 2 (𝑥 3 + 4)5 𝑑𝑥 𝑢 = 𝑥 3 + 4
(ii) ∫ √𝑥 3 + 𝑥 2 (3𝑥 2 + 2𝑥) 𝑑𝑥 𝑢 = 𝑥 3 + 𝑥 2
sin √𝑥
(vi) ∫
√𝑥
1 2
𝑑𝑥
𝑢 = √𝑥
√𝑥 3 + 1 𝑑𝑥
3
(vii) ∫−1 𝑥
𝑢=𝑥 +1
10. Using the substitution 𝑢 = 𝑥 − 2, determine
3
3
(a) ∫2 (𝑥 − 2) 2 𝑑𝑥
3
(b) ∫2 𝑥√𝑥 − 2 𝑑𝑥
SOLUTIONS
2
1. (a) − 3 + 𝑐
9
(c) 98
(d) 18
4
(e) − 3
√𝑥−1
(iii) ∫
𝑑𝑥 𝑢 = √𝑥 − 1
√𝑥
7
(iv) ∫ sin 𝑥 cos 𝑥 𝑑𝑥 𝑢 = sin 𝑥
sin𝑥
(v) ∫ cos5 𝑥 𝑑𝑥 𝑢 = cos 𝑥
4
14
115
(f) 2
5.
−6 + 40√2
6.
(ii) 3 ± 3
7.
(ii) 192 (3√3 + 8𝜋)
8.
(a) 2 sin 2𝑥 + 𝑐
(b) − 3 sec(−3𝑥) + 𝑐
(c) − 5 csc 5𝑥 + 𝑐
(d) − cot 𝑥 + 𝑐
1
√7
1
1
1
9.
(i)
6
(𝑥3 +4)
6
sec4 𝑥
(v)
6
(vii) 9
(c) − 3 + 𝑐
4
4√2
2
𝑥
4
3
2
+ 𝑐 (ii) (𝑥 2 (𝑥 + 1))2 + 𝑐
3
(iii) 𝑥 − 2√𝑥 + 𝑐 (iv)
𝑥
12
(b) − 𝑥2 + 𝑐
1
sin8 𝑥
8
+𝑐
+ 𝑐 (vi) −2 cos √𝑥 + 𝑐
26
10. (a) (b)
5
15
11.
…………………………………………………………………………
(d) − 2 + 𝑐
𝑥
3
(e) − 𝑥2 + 𝑐
5𝑥4
(f) 4 − 3𝑥 2 + 𝑥 + 𝑐
𝑥4
(g) 4 + 4𝑥 2 − 5𝑥 + 𝑐
𝑥4
(h) 4 − 2𝑥 3 + 2𝑥 2 − 24𝑥 + 𝑐
(i)
2(𝑥3 −3)
+𝑐
𝑥
3
(j) 2𝑥 − 4𝑥 2 + 𝑐
3
(k) 8𝑥 2 + 𝑐
𝑥4
(l)
− 2𝑥 2 + 𝑐
4
𝑥6
𝑥5
𝑥4
2𝑥3
5𝑥4
(m) 6 − 5 + 4 + 𝑐
(n)
4
𝑥3
−
3
+
𝑥2
2
+𝑐
(o) 3 + 6√𝑥 + 𝑐
1
(p) 16 (2𝑥 − 1)8 + 𝑐
1
(q)
2.
2(3𝑥+1)2
1
3
+𝑐
(a) sin 2𝑥 + 𝑐
2
𝑥2
2
(b) 2 − 3 cos(3𝑥) + 𝑐
150
CHAPTER 21: INTEGRATION
APPLICATIONS OF INTEGRATION
THE EQUATION OF A CURVE
LESSON 1
The gradient of a particular curve
is given by the formula 3𝑥 2 − 2𝑥. Given that this
curve passes through the point (2, 5), find the
equation of the curve.
SOLUTION
𝑑𝑦
= 3𝑥 2 − 2𝑥
𝑑𝑥
𝑑𝑦
∫
𝑑𝑥 = ∫(3𝑥 2 − 2𝑥) 𝑑𝑥
𝑑𝑥
3𝑥 2+1 2𝑥 2
𝑦=
−
+𝑐
2+1
2
3
2
𝑦 = 𝑥 −𝑥 +𝑐
Since the curve passes through (2, 5), this point
must satisfy the equation of the curve.
5 = 23 − 22 + 𝑐
1=𝑐
𝑦 = 𝑥3 − 𝑥2 + 1
LESSON 2
The gradient of the normal to a
1
curve at the point (𝑥, 𝑦) is
. Given that the
3−4𝑥
curve passes through the point (1, 3), find the
equation of the curve.
SOLUTION
Since the normal to a curve is ⊥
to the tangent of the curve, the gradient of the
curve is 4𝑥 − 3 because
1
(4𝑥 − 3) ×
= −1
3 − 4𝑥
𝑦 = ∫(4𝑥 − 3) 𝑑𝑥
𝑥 1+1
𝑥 0+1
)− 3 (
)+𝑐
1+1
0+1
𝑥2
𝑥1
= 4( ) −3( )+ 𝑐
2
1
𝑦 = 2𝑥 2 − 3𝑥 + 𝑐
The point (1, 3) must satisfy the equation of the
curve.
3 = 2(1) 2 − 3(1) + 𝑐
3 =2−3+𝑐
4=𝑐
So, 𝑦 = 2𝑥 2 − 3𝑥 + 4
= 4(
LESSON 3
−2 + 𝑎 = 0
𝑎=2
𝑑𝑦
Thus, for the curve 𝑑𝑥 = −𝑥 + 2
𝑦 = ∫(−𝑥 + 2)𝑑𝑥
𝑥 1+1
𝑥 0+1
)+2(
)+𝑐
1+1
0+1
𝑥2
𝑥1
= −( ) + 2( )+ 𝑐
2
1
𝑥2
𝑦 = − + 2𝑥 + 𝑐
2
Curve passes through (2, 3) so
22
3 = − + 2(2) + 𝑐
2
3 = −2 + 4 + 𝑐
1=𝑐
𝑥2
𝑦 = − + 2𝑥 + 1
2
= −(
THE AREA UNDER A GRAPH
In general,
𝑏
∫ 𝑓(𝑥)𝑑𝑥
𝑎
is the area of the region bounded by the curve
𝑦 = 𝑓(𝑥), the 𝑥-axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏.
LESSON 4a
Determine the area under the
curve 𝑓(𝑥) = 1, bounded by the 𝑥-axis and the
lines 𝑥 = 0 and 𝑥 = 2.
SOLUTION
We need to determine
2
∫ 1 𝑑𝑥
0
= [2] − [0]
= 2 units 2
2
= [𝑥]
0
𝑑𝑦
The curve for which 𝑑𝑥 = −𝑥 + 𝑎,
where 𝑎 is a constant, has a stationary point at
(2, 3). Find the equation of the curve.
SOLUTION
(2, 3) is a stationary point
therefore this point
𝑑𝑦
=0
𝑑𝑥
−𝑥 + 𝑎 = 0
LESSON 4b
What is the area region bounded
by the curve 𝑓(𝑥) = 2𝑥 + 1, the 𝑥-axis and the
lines 𝑥 = 1 and 𝑥 = 3?
151
CHAPTER 21: INTEGRATION
SOLUTION
SOLUTION
3
∫ 2𝑥 + 1 𝑑𝑥
1
2𝑥 1+1
3
+ 𝑥]
1+1
1
3
= [𝑥 2 + 𝑥]
1
= [32 + 3] − [11 + 1]
= 10 units 2
=[
The graph cuts the 𝑥-axis when 𝑦 = 0
9 − 𝑥2 = 0
𝑥2 = 9
𝑥 = ±3
3
∫(9 − 𝑥 2 ) 𝑑𝑥
−3
𝑥3 3
]
3 −3
(−3)3
33
= [9(3) − ] − [9(−3) −
]
3
3
= [27 − 9] − [−27 + 9]
= 18 − (−18)
= 36 units 2
= [9𝑥 −
LESSON 5
Find the area of the region
6
bounded by the curve 𝑦 = 𝑥2 , the 𝑥 −axis and the
lines 𝑥 = 1 and 𝑥 = 2.
SOLUTION
2
2
6
∫ 2 𝑑𝑥 = ∫ 6𝑥 −2 𝑑𝑥
𝑥
1
1
𝑥 −2+1 2
𝑥 −1 2
)] = [6 (
)]
−2 + 1 1
−1 1
6 2
2
= [−6𝑥 −1 ] = [− ]
1
𝑥 1
6
6
= [− ] — = (−3)— 6
2
1
= −3 + 6
= 3 units 2
= [6 (
LESSON 6
Find the area bounded by the
curve 𝑦 = 9 − 𝑥 2 and the 𝑥-axis and the points
where the graph cuts the 𝑥-axis
SUM FORMULA FOR THE AREA UNDER
A GRAPH
INTRODUCTION
While we have alreadu looked at how to
determine the area under a curve bounded by the
𝑥 – axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏 we now
look at it from a more theoretical aspect to see
how the formula wich you already know was
developed.
To determine the area under a curve we could
proceed as follows.
 Divide the shape into rectangles each of
equal width, (𝑥1 − 𝑥0 ) i.e. (𝑥𝑛+1 − 𝑥𝑛 )
 Determine the area of each individual
rectangle; 𝑓(𝑥𝑛 )(𝑥𝑛+1 − 𝑥𝑛 )
152
CHAPTER 21: INTEGRATION

Sum these areas to obtain an estimate of
the area under the curve
3
= ∫(−2𝑥 2 + 10𝑥 − 12) 𝑑𝑥
2
Area under curve :
𝑥=𝑥𝑛
∑ 𝑓(𝑥)(𝑥𝑛+1 − 𝑥𝑛 )
= [−
2𝑥 3 10𝑥 2
3
+
− 12𝑥]
3
2
2
𝑥=𝑥0
This procedure is not very efficient. As we can see
from the graph this approximation of the area
under the curve is well off of the actual value. To
improve the accuracy of our approximation we
can reduce the width of each rectangle until we
have an almost accurate approximation.
Therefore, we can say that as the width of the
rectangle gets closer and closer to zero our
approximation gets closer and closer to the
required area. If we denote the width (𝑥𝑛+1 − 𝑥𝑛 )
as 𝑑𝑥 and let 𝑥0 = 𝑎 and 𝑥𝑛 = 𝑏 we have
Area under a curve
2(3)3
2(2)3
+ 5(3)2 − 12(3)] − [−
+ 5(2)2 − 12(2)]
3
3
= [−
=
1
units 2
3
LESSON 8
region
Find the area of the shaded
𝑥=𝑏
lim ∑ 𝑓(𝑥)𝑑𝑥
𝑑𝑥→0
𝑥=𝑎
The standard notation for this area is
𝑏
∫ 𝑓(𝑥)𝑑𝑥 where ∫ is an elongated 𝑆 for sum
𝑎
In general,
𝑏
∫ 𝑓(𝑥)𝑑𝑥
𝑎
is the area of the region bounded by the curve
𝑦 = 𝑓(𝑥), the 𝑥-axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏.
AREA BETWEEN 2 CURVES
LESSON 7
region.
Find the area of the shaded
SOLUTION
To determine the limits of integration, along the 𝑥axis, we need to find, 𝑥-ordinates of the points of
intersection of the two curves. These will be the
limits of integration
𝑦 = 2𝑥
𝑦 2 = 6𝑥
∴ (2𝑥)2 = 6𝑥
4𝑥 2 = 6𝑥
4𝑥 2 − 6𝑥 = 0
2𝑥(2𝑥 − 3) = 0
3
𝑥=0
𝑥=
2
Limits of integration are 𝑥 = 0 and 𝑥 = 3/2
Area of the area region:
3/2
3/2
∫ √6𝑥 𝑑𝑥 − ∫ 2𝑥 𝑑𝑥
0
0
3/2
1
(6𝑥)2 𝑑𝑥 − ∫ 2𝑥 𝑑𝑥
3/2
=∫
0
3
1 (6𝑥)2
SOLUTION
3
3
AREA = ∫2 (6𝑥 − 𝑥 2 − 5) 𝑑𝑥 − ∫2 (𝑥 2 − 4𝑥 + 7) 𝑑𝑥
0
3
3
2𝑥 2
=[
]2 − [
]2
6 3⁄ 0
2 0
2
3 3
3
1(6𝑥)2
2
2
=[
] − [𝑥 ] 2
9
0
0
153
CHAPTER 21: INTEGRATION
3
3 2
(6 ( ))
2
=
9
3
6(0) 2
3 2
−
− [( ) − 02 ]
9
2
[
9
= [3 − 0] − [ ]
4
3
2
= units
4
]
AREA BELOW THE 𝒙 −AXIS
If the required area of 𝑓(𝑥) is below the 𝑥 −axis
then
𝑏
(−2)5
15
− 20(1)] − [
− 20(−2)])
5
5
99
32
= ([− ] − [− + 40])
5
5
267
=−
5
267
= |−
|
5
267
=
units 2
5
= ([
LESSON 11
region
Find the area of the shaded
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = |∫ 𝑓(𝑥) 𝑑𝑥 |
𝑎
𝑎
LESSON 9
Determine the area under the
curve 𝑦 = − sin 𝑥 between the 𝑥-axis and the lines
𝑥 = 0 and 𝑥 = 𝜋.
SOLUTION
𝜋
∫ − sin 𝑥 𝑑𝑥
0
𝜋
= [(−1)(− cos 𝑥)]
0
= [cos 𝜋] − [cos 0]
= (−1) − 1
= −2
= | −2|
= 2 units 2
LESSON 10
region.
Find the area of the shaded
SOLUTION
We have to separate the graphs into sections: (1)
above the 𝑥-axis and (2) below the 𝑥-axis.
1
0
∫ 𝑥3
𝑑𝑥 + | ∫ 𝑥 3 𝑑𝑥 |
0
−1
𝑥4 1
𝑥4 0
= [ ] + |[ ] |
4 0
4 −1
4
4
1
0
04 (−1)4
= [ − ] + |[ −
]|
4
4
4
4
1
1
= + |− |
4
4
1
=
2
SOLUTION
Since area is below the 𝑥-axis we
have to evaluate
1
| ∫(𝑥 4 − 20) 𝑑𝑥 |
−2
=[
𝑥5
1
− 20𝑥]
5
−2
154
CHAPTER 21: INTEGRATION
AREA BETWEEN A CURVE AND THE 𝒚AXIS
In general, the area bounded by the curve
𝑦 = 𝑓(𝑥) and the 𝑦-axis between the lines
𝑦 = 𝑓(𝑎) and 𝑦 = 𝑓(𝑏) is determined by
𝑓(𝑏)
SOLUTION
First of all we need to determine
the limits of integration. One limit is 𝑦 = 3 and the
other occurs when along the cuve 𝑦 = 4 − 𝑥 2 at
the point where 𝑥 = 0.
when 𝑥 = 0; 𝑦 = 4
Therefore, we need to evaluate
4
∫ 𝑥 𝑑𝑦
∫ 𝑓(𝑦) 𝑑𝑦
𝑓(𝑎)
LESSON 12
region 𝐴
3
Determine the area of the shaded
To find 𝑓(𝑦), we make 𝑥 the subject 𝑦 = 4 − 𝑥 2
∴ 𝑓(𝑦) = √4 − 𝑦
4
4
1
∫ √4 − 𝑦 𝑑𝑦 = ∫(4 − 𝑦) 2 𝑑𝑦
3
3
3
(4 − 𝑦) 2 4
= [−
]
3⁄
3
2
3
3
2(4 − 4) 2
2(4 − 3) 2
= [−
] − [−
]
3
3
2
= 0 − (− )
3
2
= 3 units 2
SOLUTION
Area of 𝐴:
Since 𝑦 = 𝑥 2 − 1
𝑥 = √𝑦 + 1
where 𝑥 = 𝑓(𝑦)
1
1
1
∫ √𝑦 + 1 𝑑𝑦 = ∫(𝑦 + 1)2 𝑑𝑦
0
=[
3
(𝑦 + 1) 2
3⁄
2
0
]
3
1
0
3
2(1 + 1)2 2(0 + 1) 2
=
−
3
3
4√2 2
=
−
3
3
4√2 − 2
=
units 2
3
LESSON 13
The diagram shows part of the
curve 𝑦 = 4 − 𝑥 2. The line 𝑦 = 3 meets this part of
the curve at the point (1, 3). Calculate the area of
the shaded region.
155
CHAPTER 21: INTEGRATION
VOLUME OF REVOLUTION ABOUT THE 𝒙 AXIS
The volume when the area bounded by the curve,
𝑦 = 𝑓(𝑥), the 𝑥-axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏
is rotated 360° about the 𝑥-axis is determined by
𝑏
VOLUMES OF REVOLUTION ABOUT THE 𝒚 AXIS
When the shaded region is rotated 360° about the
𝑦 – axis, the volume that is generated by the
formula
𝑏
𝑏
𝜋 ∫[𝑓(𝑥)]2 𝑑𝑥 OR 𝜋 ∫ 𝑦 2 𝑑𝑥
𝑎
𝑎
𝑎
LESSON 14
Find the volume obtained when
the region bounded by the line
𝑦 = 3𝑥, the 𝑥-axis and the lines 𝑥 = 0 and 𝑥 = 2 is
rotated 360° about the 𝑥-axis.
SOLUTION
where
𝑉 = 𝜋 ∫(3𝑥)2 𝑑𝑥



2
0
2
= 𝜋 ∫ 9𝑥 2 𝑑𝑥
0
𝑥3 2
]
3 0
23 03
= 9𝜋 [ − ]
3
3
= 24𝜋 units 3
= 9𝜋 [
LESSON 15
Find the volume obtained when
the region bounded by the curve 𝑦 = √𝑥 , the 𝑥axis and the lines 𝑥 = 3 and 𝑥 = 9 is rotated 360°
about the 𝑥-axis.
SOLUTION
𝑦 = √𝑥 → 𝑦 2 = 𝑥
9
𝑉 = 𝜋 ∫ 𝑥 𝑑𝑥
3
𝑥2 9
]
2 3
92 32
=𝜋[ − ]
2
2
= 36𝜋 units 3
=𝜋[
𝑏
𝑉 = 𝜋 ∫ 𝑥 2 𝑑𝑦 or 𝜋 ∫[𝑓(𝑦)]2 𝑑𝑦
𝑎
𝑥 = 𝑓(𝑦) is the equation of the curve
expressed in terms of 𝑦
𝑎 and 𝑏 are the upper and lower limits 𝑦 of the
area being rotated
𝑑𝑦 indicates that the area is being rotated
about the 𝑦 – axis
LESSON 16
Find the volume of the solid of
revolution when the area bounded between the
curve 𝑦 = 𝑥 3, the 𝑦 − axis and the lines 𝑦 = 0 and
𝑦 = 4 is rotated 360° about the 𝑦 −axis.
SOLUTION
1
𝑦 = 𝑥 3 → 𝑥 = 𝑦3
4
1 2
𝑉 = 𝜋 ∫ (𝑦 3 ) 𝑑𝑦
0
4
2
= 𝜋 ∫ 𝑦 3 𝑑𝑦
0
5
𝑦3 4
=𝜋[ ]
5⁄ 0
3
5
3𝑦 3 4
=𝜋[
]
5 0
5
5
3(4) 3 3(0)3
=𝜋[
−
]
5
5
= 𝜋[6.05 − 0]
= 6.05𝜋 units 3
156
CHAPTER 21: INTEGRATION
…………………………………………………………………………..
EXERCISE 21.2
1.
10. (i) Find the area of the region enclosed by the
12
curve 𝑦 = 2 , the 𝑥 −axis and the
𝑥
lines 𝑥 = 1 and 𝑥 = 3.
1
2
(i) Find ∫ (6𝑥 − 1) 𝑑𝑥.
(ii) Hence find the equation of the curve for
1
𝑑𝑦
which 𝑑𝑥 = 6𝑥 2 − 1 and which passes
2.
3.
through the point (4, 17).
𝑑𝑦
The gradient of a curve is given by 𝑑𝑥 = 12 √𝑥.
The curve passes through the point (4, 50).
Find the equation of the curve.
A curve has an equation which satisfies
𝑑𝑦
= 𝑘𝑥(2𝑥 − 1) for all values of 𝑥. The point
𝑑𝑥
𝑃(2, 7) lies on the curve and the gradient of
the curve at 𝑃 is 9.
(i)
Find the value of the constant 𝑘.
(ii)
Find the equation of the curve.
4.
5.
The gradient of a curve is given by
𝑑𝑦
= 3𝑥 2 + 𝑎, where 𝑎 is a constant. The curve
𝑑𝑥
passes through the points (−1, 2) and (2, 17).
Find the equation of the curve.
𝑑𝑦
𝜋
A curve is such that 𝑑𝑥 = 2 cos (2𝑥 − 2 ). The
(ii)
The area of the region enclosed by
12
the curve 𝑦 = 𝑥2 , the 𝑥 −axis and the
lines 𝑥 = 2 and 𝑥 = 𝑎, where 𝑎 > 2, is
3 units 2. Find the value of 𝑎.
11. The diagram shows the curve 𝑦 = 4𝑥 − 𝑥 2,
which crosses the 𝑥 axis at the origin 𝑂 and
the point 𝐴. The tangent to the curve at the
point (1, 3) crosses the 𝑥 axis at the point 𝐵.
(i)
Find the coordinates of 𝐴 and 𝐵.
(ii)
Find the area of the shaded region.
12. The graph shows part of the curve
𝜋
𝑦 = 3 sin 𝑥 + 4 cos 𝑥 for 0 ≤ 𝑥 ≤ .
2
𝜋
curve passes through the point ( 2 , 3).
(i) Find the equation of the curve.
(ii) Find the equation of the normal to the
3𝜋
curve at the point where 𝑥 = .
4
𝜋
NB: cos (𝜃 ± 2 ) = sin𝜃
6.
𝑑𝑦
𝜋
A curve is such that 𝑑𝑥 = 6 cos (2𝑥 + 2 ) for
𝜋
− ≤𝑥≤
4
𝜋
5𝜋
4
. The curve passes through the
point ( , 5). Find
4
(i)
(ii)
7.
8.
9.
the equation of the curve,
the 𝑥 coordinate of the stationary
points of the curve,
(iii)
the equation of the normal to the
curve at the point on the curve where
3𝜋
𝑥= 4.
Find the area enclosed between the curve
𝑦 = 3𝑥 2 − 3𝑥 − 6, the 𝑥 – axis from
𝑥 = 1 to 𝑥 = 3
Find the area enclosed between the curve
𝑦 = 4𝑥 − 𝑥 2, the 𝑥 – axis from 𝑥 = 2 to 𝑥 = 4
Find the area enclosed between the curve
𝑦 = (𝑥 + 2)2 + 3, the 𝑥 – axis from 𝑥 = −5 to
𝑥 = −2.
(i)
Find the coordinates of the maximum
point of the curve.
(ii)
Find the area of the shaded region.
13. Find the volume generated when the region
1
bounded by the curve 𝑦 =
, the 𝑥-axis, the
2𝑥+1
line 𝑥 = 2 and the line 𝑥 = 7 is rotated
through 360° about the 𝑥-axis
14. Find the volume generated when the region
9
bounded by the curve 𝑦 = 2𝑥 + 𝑥, the 𝑥-axis
and the lines 𝑥 = 1 and 𝑥 = 3 is rotated
through 360° about the 𝑥-axis
15. Show that (cos 𝑥 + sin 𝑥)2 ≡ 1 + sin 2𝑥.
Hence find the volume generated when the
region bounded by the curve
𝑦 = cos 𝑥 + sin 𝑥, the 𝑥-axis and the lines
𝜋
𝜋
𝑥 = 4 , 𝑥 = 2, is rotated through a complete
revolution about the 𝑥-axis.
157
CHAPTER 21: INTEGRATION
16. Find the areas of the shaded regions.
(a)
(b)
18. Determine the areas of the shaded regions.
(a)
(b)
(c)
(c)
17. The diagram shows part of the curve
𝑦 = 7 + 6𝑥 − 𝑥 2
(d)
Find
(i) the coordinates of the points 𝑃, 𝑄, 𝑅 and 𝑆
(ii) the area of the three shaded regions.
158
CHAPTER 21: INTEGRATION
19. The diagram shows the curve 𝑦 = 𝑥 4 + 3 and
the line 𝑦 = 19 which intersect at (−2, 19)
and (2, 19). Find the exact area of the shaded
region enclosed by the curve and the line.
20. The diagram shows part of the curve 𝑦 =
(𝑥 − 3) 2 intersected by a straight line at
𝐴(0, 9) and 𝐵(4, 1). Evaluate the area of the
shaded region.
21. The diagram shows part of the curve
𝑦 = 4 − 𝑥 2 . The line 𝑦 = 3 meets this part of
the curve at the point (1, 3). Calculate the
area of the shaded region.
22. The diagram shows part of the curves
𝑦 = 2𝑥 2 + 5 and
𝑦 = 3𝑥 2 + 1, intersecting at (2, 13). Find the
area of the shaded region.
23. The diagram shows part of the curve
3𝑦 = 5(𝑥 2 − 1). Calculate the volume
generated when the shaded region is rotated
through 360° about the 𝑦-axis.
24. The diagram shows part of the curve
𝑦 = 4 − 𝑥 2 . The line 𝑦 = 3 meets this part of
the curve at the point (1, 3).
Calculate the volume obtained when the
shaded region is rotated through 360° about
the 𝑦-axis.
25. The diagram shows part of the curves
𝑦 = 2𝑥 2 + 5 and 𝑦 = 3𝑥 2 + 1, intersecting at
(2, 13). Find the volume generated when the
shaded region is rotated through 360° about
the 𝑦-axis.
159
CHAPTER 21: INTEGRATION
EXAM QUESTIONS
1.
𝑥
4−𝑥2
(i) Show that for 𝑓(𝑥) = 𝑥2 +4, 𝑓 ′ (𝑥) = (𝑥2 +4)2 .
[4]
2 12−3𝑥2
(ii) Hence, evaluate ∫0 (𝑥2 +4)2 𝑑𝑥
2.
[4]
CAPE 2003
In the diagram (not drawn to scale), the line
𝑦 = 2𝑥 + 3 cuts the curve 𝑦 = 𝑥 2 at the points
𝑃 and 𝑄.
SOLUTIONS
3
𝑥2
3
𝑥2
1.
(i) 4𝑥 2 − 2 + 𝑐 (ii) 𝑦 = 4𝑥 2 − 2 − 7
2.
𝑦 = 8𝑥 2 − 14
3.
(i) 𝑘 = 2 (ii) 𝑦 = 𝑥 3 − 4 + 2
4.
5.
𝑦 = 𝑥 3 + 2𝑥 + 5
(i) 𝑦 = − cos(2𝑥) + 2
3
3𝑥2
3
1
6.
1
9.
[4]
(b) Calculate the area of the shaded portion
𝑃𝑂𝑄 shown in the diagram.
[5]
CAPE 2007
(ii) 𝑦 = 2 𝑥 + 8
(i) 𝑦 = 3 cos 2𝑥 + 5 (ii) 𝑥 = 0, 𝜋
(iii) 𝑦 = − 6 𝑥 +
7.
8.
(a) Determine the coordinates of 𝑃 and 𝑄.
16−3𝜋
40+𝜋
8
2
3.
16
3
116
𝑑𝑦
3
𝑑𝑥
10. (i) 8 (ii) 𝑎 = 4
1
2
2
3
11. (i) 𝐴(4, 0), 𝐵 (− , 0) (ii) 10
12. (i) (0.64, 5) (ii) 7
𝜋
13. 15
14.
482𝜋
3
𝜋+2
15. 𝜋 ( 4 )
4
16. (a) 3
3
(b) 15 4
125
4
11
18. (a) 3 (b) 3 (c) 6 (d) 0.29
256
5
20. 10
21.
2
4.
(c) 2
17. (i) 𝑃(7, 0), 𝑄(−1, 0), 𝑅(0, 7), 𝑆(6, 7)
19.
The curve 𝐶 passes through the point (−1, 0)
and its gradient at any point (𝑥, 𝑦) is given by
2
(ii) 44 3
= 3𝑥 2 − 6𝑥.
(i) Find the equation of 𝐶.
[3]
(ii) Find the coordinates of the stationary
points of 𝐶 and determine the nature of
EACH point.
[7]
(iii) Sketch the graph of 𝐶 and label the
𝑥 −intercepts.
[5]
CAPE 2008
The function 𝑓(𝑥) is such that
𝑓 ′ (𝑥) = 3𝑥 2 + 6𝑥 + 𝑘 where 𝑘 is a constant.
Given that 𝑓(0) = −6 and 𝑓(1) = −3, find the
function 𝑓(𝑥).
[5]
CAPE 2009
2
3
3
16
22. 3
23. 35𝜋
𝜋
24. 2
25. 8𝜋
160
CHAPTER 21: INTEGRATION
5.
6.
In the diagram m (not drawn to scale), the
line 𝑥 + 𝑦 = 2 intersects the curve 𝑦 = 𝑥 2 at
the points 𝑃 and 𝑄.
(i) Find the coordinates of the point 𝑃 and 𝑄.
[5]
(ii) Calculate the area of the shaded region of
the diagram bounded by the curve and
the straight line.
[5]
CAPE 2010
The curve, 𝐶, passes through the point (−1, 0)
and its gradient at the point (𝑥, 𝑦)is given by
𝑑𝑦
𝑑𝑥
= 3𝑥 2 − 6𝑥.
(i) Find the equation of 𝐶.
[4]
(ii) Find the coordinates of the stationary
points of 𝐶.
[3]
(iii) Determine the nature of EACH stationary
point.
[3]
(iv) Find the coordinates of the points 𝑃 and 𝑄
at which the curve 𝐶 meets the 𝑥 −axis.
[5]
(v) Hence, sketch the curve 𝐶, showing
(a) the stationary points
(b) the points 𝑃 and 𝑄
[4]
CAPE 2012
7.
8.
The diagram (not drawn to scale) shows the
curve 𝑦 = 𝑥 2 + 3 and the line 𝑦 = 4𝑥.
(i) Determine the coordinates of the points 𝑃
and 𝑄 at which the curve and the line
intersect.
[4]
(ii) Calculate the area of the shaded region.
[5]
CAPE 2013
The gradient of a curve which passes through
the point (−1, −4) is given by
𝑑𝑦
𝑑𝑥
= 3𝑥 2 − 4𝑥 + 1.
(i) Find
(a) the equation of the curve
[4]
(b) the coordinates of the stationary
points and determine their
nature.
[8]
(ii) Sketch the curve in (a) (i) (a) above,
clearly making ALL stationary points
and intercepts.
[4]
CAPE 2014
161
CHAPTER 21: INTEGRATION
9.
(a) In the diagram given, not drawn to scale,
the area under the curve
𝑦 = (1 + 𝑥)−1, 0 ≤ 𝑥 ≤ 1, is
approximated by a set of 𝑛 rectangular
1
strips of width 𝑛 units.
(c) Show that the curve 𝑓(𝑥) touches the
𝑥 −axis at 𝑥 = 1.
[4]
(d) Sketch the curve, 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2,
−2 ≤ 𝑥 ≤ 2.
[6]
(e) Find the area bounded by this curve and
the 𝑥 −axis for −2 ≤ 𝑥 ≤ 1.
[6]
CAPE 2004
12. The three points 𝑃, 𝑄 and 𝑅, on the curve
𝑦 = 𝑥 2 − 2𝑥 are shown in the diagram (not
drawn to scale).
Show that the sum, 𝑆𝑛 , of the areas of the
1
1
1
rectangular strips is 𝑛+1 + 𝑛+2 + ⋯ + 2𝑛.
[7]
(a) (i) Sketch the curve 𝑦 = 𝑥 2 + 1.
[3]
(ii) Find the volume obtained by rotating
the portion of the curve between
𝑥 = 0 and 𝑥 = 1 through 2𝜋 radians
about the 𝑦 axis.
[7]
CAPE 2003
10. The diagram below is a rough diagram of
𝑦 = |𝑥 − 2| for real values of 𝑥 from
𝑥 = 0 to 𝑥 = 4.
(a) Find the coordinates of EACH of the
points 𝑃, 𝑄 and 𝑅.
[4]
(b) Find the TOTAL area bounded by the
curve shown, the 𝑥 −axis and the lines
𝑥 = −1 and 𝑥 = 2.
[4]
CAPE 2005
13. The diagram (not drawn to scale) shows the
16
shaded area, 𝐴, bounded by the curve 𝑦 = 𝑥2
1
and lines 𝑦 = 2 𝑥 − 1, 𝑥 = 2 and 𝑥 = 3.
(a) Find the coordinates of the points 𝐴 and
𝐵.
[2]
(b) Find the volume generated by rotating the
triangle 𝑂𝐴𝐵 shown above through
360° about the 𝑥 −axis.
[4]
CAPE 2004
11. (a) Find the stationary point(s) of the curve,
𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2,
[6]
(b) Determine the nature of the stationary
point(s).
[3]
(a) Express the shaded area, 𝐴, as the
difference of two definite integrals.
[1]
(b) Hence, show that
3
3
1 3
𝐴 = 16 ∫2 𝑥 −2 𝑑𝑥 − 2 ∫2 𝑥 𝑑𝑥 + ∫2 𝑑𝑥.
[2]
(c) Find the value of 𝐴.
[3]
162
CHAPTER 21: INTEGRATION
CAPE 2006
14. Use the result
𝑎
𝑎
∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥, 𝑎 > 0, to show
that
𝜋
𝜋
(a) ∫0 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 ∫0 sin𝑥 𝑑𝑥 −
𝜋
∫0 𝑥 sin 𝑥 𝑑𝑥
𝜋
(b) ∫0 𝑥 sin 𝑥 𝑑𝑥 = 𝜋
[2]
[5]
CAPE 2006
15. (a) Differentiate, with respect to 𝑥, the
𝑥2 −4
function 𝑓(𝑥) = 𝑥3 +1.
(b) Using the substitution 𝑢 = sin 2𝑥, or
[4]
𝜋
otherwise, evaluate ∫04 sin2𝑥 cos 2𝑥 𝑑𝑥.
[4]
CAPE 2007
16. (a) (i) Use the result
𝑎
𝑎
∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥, 𝑎 > 0, to
𝜋
show that if 𝐼 = ∫02 sin2 𝑥 𝑑𝑥, then
𝜋
𝐼 = ∫02 cos 2 𝑥 𝑑𝑥.
[2]
𝜋
(ii) Hence, or otherwise, show that 𝐼 = 4.
[6]
(b) (i) Sketch the curve 𝑦 = 𝑥 2 + 4.
[4]
(ii) Calculate the volume created by
rotating the plane figure bounded by
𝑥 = 0, 𝑦 = 4, 𝑦 = 5 and the curve
𝑦 = 𝑥 2 + 4 through about the
𝑦 −axis.
[8]
CAPE 2007
17. (a) Differentiate with respect to 𝑥
(i) 𝑥√2𝑥 − 1
[3]
(ii) sin2(𝑥 3 + 4)
[4]
6
(b) (i) Given that ∫1 𝑓(𝑥) 𝑑𝑥 = 7,
6
evaluate ∫1 [2 − 𝑓(𝑥)] 𝑑𝑥
(ii) The area under the curve
𝑦 = 𝑥 2 + 𝑘𝑥 − 5, above the 𝑥-axis and
bounded by the lines 𝑥 = 1 and 𝑥 = 3
2
is 14 3 units 2.
Find the value of the constant 𝑘. [4]
(c) The diagram below not drawn to scale
represents a can in the shape of a closed
cylinder with a hemisphere at one end.
The can has volume of 45𝜋 units 3
(i) Taking 𝑟 units as the radius of the
cylinder and ℎ units as its height,
show that
45
2𝑟
(a) ℎ = 𝑟2 − 3
[3]
(b) 𝐴 =
5𝜋𝑟2
3
90𝜋
+ 𝑟 , where 𝐴 units is
the external surface area of the
can.
[3]
(ii) Hence, find the value of 𝑟 for which 𝐴
is a minimum and the corresponding
minimum value of 𝐴.
[5]
4 3
Volume of a sphere = 𝜋𝑟 ,
[
]
3
Surface area of a sphere = 4𝜋𝑟 2
Volume of a cylinder = 𝜋𝑟 2 ℎ,
[
]
Curved surface area of a cylinder = 2𝜋𝑟ℎ
CAPE 2008
𝑎
𝑎
18. Given that ∫0 (𝑥 + 1) 𝑑𝑥 = 3 ∫0 (𝑥 − 1) 𝑑𝑥,
𝑎 > 0, find the value of the constant 𝑎.
1
1 2
[6]
CAPE 2009
19. (i) Evaluate ∫−1 (𝑥 − 𝑥 ) 𝑑𝑥.
[6]
(ii) Using the substitution 𝑢 = 𝑥 2 + 4, or
otherwise, find ∫ 𝑥√𝑥 2 + 4 𝑑𝑥.
[4]
CAPE 2010
4
20. The function 𝑓(𝑥) satisfies ∫1 𝑓(𝑥) 𝑑𝑥 = 7.
4
(i) Find ∫1 [3𝑓(𝑥) + 4] 𝑑𝑥
[4]
(ii) Using the substitution 𝑢 = 𝑥 + 1, evaluate
3
[4]
∫0 2𝑓(𝑥 + 1) 𝑑𝑥.
CAPE 2010
21. (a) The diagram (not drawn to scale) is a
sketch of the section of the function
𝑓(𝑥) = 𝑥(𝑥 2 − 12) which passes through
the origin 𝑂. 𝐴 and 𝐵 are the
stationary points on the curve.
163
CHAPTER 21: INTEGRATION
Find
(i) the coordinates of each of the
stationary points, 𝐴 and 𝐵
[8]
(ii) the equation of the normal to the
curve 𝑓(𝑥) = 𝑥(𝑥 2 − 12) at the
origin.
[2]
(iii) the area between the curve and the
positive 𝑥 −axis.
[6]
(b) (i) Use the result
𝑎
𝑎
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥) 𝑑𝑥, 𝑎 > 0
0
0
to show that
𝜋
𝜋
∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥 𝑑𝑥
0
0
[2]
(ii) Hence show that
𝜋
(a) ∫0 𝑥 sin 𝑥 𝑑𝑥 =
𝜋
𝜋
𝜋 ∫0 sin 𝑥 𝑑𝑥 − ∫0 𝑥 sin 𝑥 𝑑𝑥
[2]
[5]
CAPE 2011
22. (a) (i) By using the substitution 𝑢 = 1 − 𝑥,
2
find ∫ 𝑥(1 − 𝑥) 𝑑𝑥.
[5]
(ii) Given that 𝑓(𝑡) = 2 cos 𝑡,
𝑔(𝑡) = 4 sin 5𝑡 + 3 cos 𝑡, show that
The perimeter of the track must be 600
metres.
600−2𝑥
(i) Show that 𝑟 = 2+𝜋 .
[2]
(ii) Hence, determine the length, 𝑥, that
maximises the area enclosed by the
track.
[6]
(c) (i) Let 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝐴𝑥 + 𝐵,
where 𝐴 and 𝐵 are constants. Show
that 𝑦 ′′ = 𝑥 sin𝑥.
[4]
(ii) Hence, determine the specific
solution of the differential equation
𝑦 ′′ = 𝑥 sin𝑥, given that when 𝑥 = 0,
𝑦 = 1 and when 𝑥 = 𝜋, 𝑦 = 6.
[4]
CAPE 2013
23. The equation of a curve is given by
𝑓 (𝑥 ) = 2𝑥√1 + 𝑥 2 .
3
(i) Evaluate ∫0 𝑓(𝑥) 𝑑𝑥
[5]
(ii) Find the volume generated by rotating the
area bounded by the curve in (i) above,
the 𝑥 −axis, and the lines 𝑥 = 0 and 𝑥 = 2
about the 𝑥 −axis.
[4]
CAPE 2014
24. (a) The diagram (not drawn to scale) shows
the region bounded by the lines
𝑦 = 3𝑥 − 7, 𝑦 + 𝑥 = 9 and 3𝑦 = 𝑥 + 3.
𝜋
(b) ∫0 𝑥 sin 𝑥 𝑑𝑥 = 𝜋
∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = ∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑔(𝑡) 𝑑𝑡
[4]
(b) A sports association is planning to
construct a running track in the shape of a
rectangle surmounted by a semicircle, as
shown in the diagram. The letter 𝑥
represents the length of the rectangular
section and 𝑟 represents the radius of the
semicircle.
(i) Show that the coordinates of 𝐴, 𝐵 and
𝐶 are (4, 5), (3, 2) and (6, 3)
respectively.
[5]
(ii) Hence, use integration to determine
the area bounded by the lines.
[6]
(b) The gradient function of a curve 𝑦 = 𝑓(𝑥)
which passes through the point (0, −6) is
given by 3𝑥 3 + 8𝑥 − 3.
(i) Determine the equation of the curve.
[3]
164
CHAPTER 21: INTEGRATION
(ii) Find the coordinates and natures of
the stationary points of the curve in
(b) (i) above.
[8]
(iii) Sketch the curve in (b)(i) by clearly
labelling the stationary points.
[3]
CAPE 2015
SOLUTIONS
3
1.
(ii) 4
2.
(a) 𝑃(−1, 1), 𝑄(3, 9) (b) 3
3.
4.
(i) 𝑦 = 𝑥 3 − 3𝑥 2 + 4
(ii) (0, 4) max, (2, 0) min
𝑓(𝑥) = 𝑥 3 + 3𝑥 2 − 𝑥 − 6
5.
(i) 𝑃(−2, 4), 𝑄(1, 1) (ii) 2
6.
7.
(i) 𝑦 = 𝑥 3 − 3𝑥 2 + 4 (ii) 𝐴(0, 4), 𝐵(2, 0)
(iii) 𝐴 – max, 𝐵 – min
4
(i) 𝑃(1, 4), 𝑄(3, 12) (ii) 3
8.
(i) (a) 𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥
9.
(a)
22
9
1 4
(b) ( , ) max,
3 27
5𝜋
(b) (i)
(ii) 3
10. (a) 𝐴(0, 2) , 𝐵 (2, 0) (b)
4𝜋
3
11. (a) 𝐴(−1, 4), 𝐵(1, 0)
(b) 𝐴-maximum, 𝐵-minimum (c)
3
(d) (e) 6
4
8
12. (a) 𝑃 (−1, 3) , 𝑄(1, −1), 𝑅(2, 0) (b) 3
3 16
3 1
13. (a) ∫2 𝑥2 𝑑𝑥 − ∫2 (2 𝑥 − 1) 𝑑𝑥 (b)
14. (a)
29
(c) 12
(b)
15. (a) −
𝑥(𝑥3 −12𝑥−2)
(𝑥3 +1)2
1
(b) 4
16. (a) (i)
(ii) (b) (i)
(ii)
3𝑥 −1
17. (a) (i)
(ii) 3𝑥 2 sin(2(𝑥 3 + 4))
√2𝑥−1
(b) (i) 3
(c) (i) (a)
18. 𝑎 = 4
16
(ii) 𝑘 = 4
(b)
(ii)𝑟 = 3, 𝐴 = 45𝜋
3
1
19. (i) − 3 (ii) 3 (𝑥 2 + 4) 2 + 𝑐
20. (i) 33 (ii) 14
1
21. (a) (i) (−2, 16), (2, −16) (ii) 𝑦 = 12 𝑥 (iii) 36
(b) (i) (ii) (a)
(b)
22. (a) (i)
(1−𝑥) 4
4
−
(1−𝑥)3
3
+𝑐
(ii)
2400
(b) (i)
(ii) 𝑥 = 16+4𝜋
(c) (i)
(ii) 𝑦 = −𝑥 sin𝑥 − 2 cos 𝑥 + 11 + 3
2
23. (i) 3 (10√10 − 1)
𝑥
544𝜋
(ii) 15
24. (a) (ii) 4 (b) (i) 𝑦 = 𝑥 3 + 4𝑥 2 − 3𝑥 − 6
1
176
(ii) (−3, 12) max, (3 , − 27 ) min (iii)
165
ANSWERS FOR REASONING AND LOGIC
1.
2.
(i) ~𝑝
(ii) 𝑞 ∧ 𝑝
(iii) 𝑞 → 𝑝
(a)
𝑎
∼𝑎
0
1
1
0
This is a contingency.
(b)
𝑎
𝑏
∼𝑎
0
0
1
0
1
1
1
0
0
1
1
0
This is a contingency.
(c)
𝑎
𝑏
𝑎∨𝑏
0
0
0
0
1
1
1
0
1
1
1
1
This is a contingency.
(b)
~𝑎 ∨ 𝑏
1
1
0
1
4.
(∼ 𝑎
∧∼ 𝑏)
∨𝑐
0
1
0
1
0
1
0
0
𝑎
𝑏
𝑐
∼𝑎
∼𝑏
∼𝑎
∧
∼𝑏
1
1
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
𝑎
0
0
1
1
𝑏
0
1
0
1
∼𝑎
1
1
0
0
𝑎
𝑏
c
𝑎∧𝑏
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
0
0
0
0
0
1
1
(𝑎 ∧ 𝑏)
→𝑐
1
1
1
1
1
1
0
1
𝑎
∧𝑐
0
0
0
0
0
1
0
1
(𝑎 ∨ 𝑏)
→ (𝑎 ∧ 𝑐)
1
1
0
0
0
1
0
1
(a)
∼𝑎→𝑏
0
1
1
1
(b)
(d)
𝑎
𝑏
~𝑎
0
0
1
0
1
1
1
0
0
1
1
0
This is a contingency.
3.
~𝑎 ∧ 𝑏
0
1
0
0
(a)
𝑎
0
0
0
0
1
1
1
1
𝑏
0
0
1
1
0
0
1
1
𝑐
0
1
0
1
0
1
0
1
𝑎∧𝑏
0
0
0
0
0
0
1
1
(𝑎 ∧ 𝑏) ∨ 𝑐
0
1
0
1
0
1
1
1
(c)
5.
6.
7.
𝑎
𝑏
𝑐
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
𝑎
∨𝑏
0
0
1
1
1
1
1
1
Converse: (𝑞 ∧ 𝑝) → (𝑝 ∨ 𝑞)
Inverse: ∼ (𝑝 ∨ 𝑞) → (𝑞 ∧ 𝑝)
Contrapositive: ∼ (𝑞 ∧ 𝑝) →∼ (𝑝 ∨ 𝑞)
(a) 𝑎 ∨ 𝑏 (b) 𝑎 ∧ 𝑏
166
8.
9. (i) (𝑞 ∨ ~𝑝) → (𝑝 ∧ 𝑞) (ii)
10.
𝑝
𝑞
𝑝
𝑟→𝑞
𝑟
→𝑞
0
0
0
1
1
0
0
1
1
0
0
1
0
1
1
0
1
1
1
1
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
1
1
1
1
1
11. (i) ∼ 𝑝 →∼ 𝑞 ∼ 𝑞 →∼ 𝑝
(ii)
𝒑
𝒒
∼𝒑 ∼𝒒
𝒑→𝒒
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
T
F
T
T
(𝑝 → 𝑞)
∧ (𝑟 → 𝑞)
1
0
1
1
0
0
1
1
∼𝒒→
∼𝒑
T
F
T
T
(iii) They are logically equivalent since they
have the same truth table values.
12. Converse: ~𝑞 → 𝑝
Inverse: ~𝑝 → 𝑞
Contrapositive: 𝑞 → ~𝑝
167
CAPE 2012 PAST PAPER
CAPE 2012
SECTION A
1.
(a) The expression 𝑓(𝑥) = 2𝑥 3 − 𝑝𝑥 2 + 𝑞𝑥 − 10 is divisible by 𝑥 − 1 and has a remainder −6 when
divided by 𝑥 + 1.
Find
(i) the values of the constants 𝑝 and 𝑞.
[7]
(ii) the factors of 𝑓(𝑥)
[3]
2
(b) Find positive integers 𝑥 and 𝑦 such that (√𝑥 + √𝑦) = 16 + √240.
[8]
(c) (i) Solve, for real values of 𝑥, the inequality |3𝑥 − 7| ≤ 5.
[5]
(ii) Show that no real solution, 𝑥, exists for the inequality |3𝑥 − 7| + 5 ≤ 0.
[2]
Total 25 marks
2.
(a) The function 𝑓 on ℝ is defined by 𝑓: 𝑥 → 𝑥 2 − 3.
(i) Find, in terms of 𝑥, 𝑓(𝑓(𝑥)).
[3]
(ii) Determine the values of 𝑥 for which 𝑓(𝑓(𝑥)) = 𝑓(𝑥 + 3).
[6]
(b) The roots of the equation 4𝑥 2 − 3𝑥 + 1 = 0 are 𝛼 and 𝛽.
Without solving the equation
(i) write down the values of 𝛼 + 𝛽 and 𝛼𝛽
[2]
(ii) find the value of 𝛼 2 + 𝛽 2
[2]
2
2
(iii) obtain a quadratic equation whose roots are 𝛼2 and 𝛽2 .
[5]
(c) Without the use of calculators or tables, evaluate
1
3
5
7
9
(i) log 10 (3) + log 10 (5) + log 10 (7) + log 10 (9) + log 10 (10)
𝑟
(ii) ∑99
𝑟=1 log 10 (𝑟+1)
[3]
[4]
Total 25 marks
168
CAPE 2012 PAST PAPER
SECTION B
3.
(a) (i) Given that cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin𝐴 sin 𝐵 and cos 2𝜃 = 2 cos 2 𝜃 − 1, prove that
1
cos 3𝜃 ≡ 2 cos 𝜃 [cos2 𝜃 − sin2 𝜃 − ]
2
[7]
(ii) Using the appropriate formula, show that
1
[sin 6𝜃 − sin2𝜃] ≡ (2 cos 2 2𝜃 − 1) sin 2𝜃
2
[5]
𝜋
(iii) Hence, or otherwise, solve sin 6𝜃 − sin2𝜃 = 0 for 0 ≤ 𝜃 ≤ .
[5]
(b) Find ALL possible values of cos 𝜃 such that 2 cot 2 𝜃 + cos 𝜃 = 0.
[8]
2
Total 25 marks
4.
(a) (i) Determine the Cartesian equation of the curve, 𝐶, defined by the parametric equations 𝑦 = 3 sec 𝜃
and 𝑥 = 3 tan 𝜃.
[5]
(ii) Find the points of intersection of the curve 𝑦 = √10𝑥 with 𝐶.
[9]
(b) Let 𝑝 and 𝑞 be two position vectors with endpoints (−3, 4) and (−1, 6) respectively.
(i) Express 𝑝 and 𝑞 in the form 𝑥𝑖 + 𝑦𝑗.
[2]
(ii) Obtain the vector 𝑝 − 𝑞.
[2]
(iii) Calculate 𝑝. 𝑞.
[2]
(iv) Let the angle between 𝑝 and 𝑞 be 𝜃. Use the result of (iii) above to calculate 𝜃 in degrees.
[5]
Total 25 marks
169
CAPE 2012 PAST PAPER
SECTION C
5.
𝑥3 +8
(a) (i) Find the values of 𝑥 for which 𝑥2 −4 is discontinuous.
[2]
𝑥3 +8
(ii) Hence, or otherwise, find lim 𝑥2 −4
[3]
𝑥→−2
(iii) By using the fact that lim
sin𝑥
𝑥→0 𝑥
2𝑥3 +4𝑥
= 1, or otherwise, find, lim sin2𝑥
[5]
𝑥→0
(b) The function 𝑓 on ℝ is defined by
𝑓(𝑥) = {
𝑥 2 + 1,
4 + 𝑝𝑥,
𝑥>1
𝑥<1
(i) Find
a)
lim 𝑓(𝑥)
[2]
𝑥→1+
b) the value of the constant 𝑝 such that lim 𝑓(𝑥) exists.
[4]
𝑥→1
(ii) Hence, determine the value of 𝑓(1) for 𝑓 to be continuous at the point 𝑥 = 1.
[1]
𝑣
(c) A chemical process in a manufacturing plant is controlled by the function 𝑀 = 𝑢𝑡 2 + 𝑡2 where 𝑢 and
𝑣 are constants. Given that 𝑀 = −1 when 𝑡 = 1 and that the rate of change of 𝑀 with respect to 𝑡 is
when 𝑡 = 2, find the values of 𝑢 and 𝑣.
35
4
[8]
Total 25 marks
6.
(a) (i) Given that 𝑦 = √4𝑥 2 − 7, show that 𝑦
𝑑2 𝑦
𝑑𝑦
𝑑𝑥
= 4𝑥.
[3]
𝑑𝑦 2
(ii) Hence, or otherwise, show that 𝑦 𝑑𝑥2 + (𝑑𝑥 ) = 4.
[3]
(b) The curve, 𝐶, passes through the point (−1, 0) and its gradient at the point (𝑥, 𝑦) is given by
𝑑𝑦
= 3𝑥 2 − 6𝑥.
𝑑𝑥
(i) Find the equation of 𝐶.
[4]
(ii) Find the coordinates of the stationary points of 𝐶.
[3]
(iii) Determine the nature of EACH stationary point.
[3]
(iv) Find the coordinates of the points 𝑃 and 𝑄 at which the curve 𝐶 meets the 𝑥-axis.
[5]
(v) Hence, sketch the curve 𝐶, showing
a) the stationary points
b) the points 𝑃 and 𝑄
[4]
Total 25 marks
170
CAPE 2013 PAST PAPER
CAPE 2013
SECTION A
1.
(a) Let 𝑝 and 𝑞 be two propositions. Construct a truth table for the statements
(i) 𝑝 → 𝑞
[1]
(ii) ~(𝑝 ∧ 𝑞)
[2]
(b) A binary operator ⨁ is defined on a set of positive real numbers by 𝑦⨁𝑥 = 𝑦 2 + 𝑥 2 + 2𝑦 + 𝑥 − 5𝑥𝑦.
Solve the equation 2⨁𝑥 = 0.
[5]
(c) Use mathematical induction to prove that 5𝑛 + 3 is divisible by 2 for all values of 𝑛 ∈ 𝑁.
[8]
(d) Let 𝑓(𝑥) = 𝑥 3 − 9𝑥 2 + 𝑝𝑥 + 16.
(i)
Given that (𝑥 + 1) is a factor of 𝑓(𝑥), show that 𝑝 = 6.
[2]
(ii)
Factorise 𝑓(𝑥) completely.
[4]
(iii)
Hence, or otherwise, solve 𝑓(𝑥) = 0.
[3]
Total 25 marks
2.
(a) Let 𝐴 = {𝑥: 𝑥 ∈ ℝ, 𝑥 ≥ 1}.
A function 𝑓: 𝐴 → ℝ is defined as 𝑓(𝑥) = 𝑥 2 − 𝑥. Show that 𝑓 is one to one.
[7]
(b) Let 𝑓(𝑥) = 3𝑥 + 2 and 𝑔(𝑥) = 𝑒 2𝑥 .
(i)
(ii)
Find
a) 𝑓 −1 (𝑥) and 𝑔 −1 (𝑥)
[4]
b) 𝑓 [𝑔(𝑥 )] (or 𝑓 ∘ 𝑔 (𝑥 )).
[1]
Show that (𝑓 ∘ 𝑔) −1 (𝑥) = 𝑔−1 (𝑥) ∘ 𝑓 −1 (𝑥).
[5]
(c) Solve the following:
(i)
3𝑥 2 + 4𝑥 + 1 ≤ 5
[4]
(ii)
|𝑥 + 2| = 3𝑥 + 5
[4]
Total 25 marks
171
CAPE 2013 PAST PAPER
SECTION B
3.
2 tan 𝜃
(a) (i) Show that sin 2𝜃 = 1+tan2 𝜃.
[4]
(ii) Hence, or otherwise, solve sin 2𝜃 − tan 𝜃 = 0 for 0 ≤ 𝜃 ≤ 2𝜃.
[8]
𝜋
(b) (i) Express 𝑓(𝜃) = 3 cos 𝜃 − 4 sin 𝜃 in the form 𝑟 cos(𝜃 + 𝛼) where 𝑟 > 0 and 0° ≤ 𝛼 ≤ 2 .
[4]
(ii) Hence, find
a) the maximum value of 𝑓(𝜃)
[2]
1
b) the minimum value of 8+𝑓(𝜃)
[2]
(iii) Given that the sum of the angles 𝐴, 𝐵 and 𝐶 of a triangle is 𝜋 radians, show that
a) sin 𝐴 = sin(𝐵 + 𝐶)
[3]
b) sin 𝐴 + sin 𝐵 + sin 𝐶 = sin(𝐴 + 𝐵) + sin(𝐵 + 𝐶) + sin(𝐴 + 𝐶)
[2]
Total 25 marks
4.
(a) A circle 𝐶 is defined by the equation 𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 + 4 = 0.
(i) Show that the centre and the radius of the circle, 𝐶, are (3, 2) and 3, respectively.
[3]
(ii) a) Find the equation of the normal to the circle 𝐶 at the point (6, 2).
[3]
b) Show that the tangent to the circle at the point (6, 2) is parallel to the 𝑦 – axis.
(b) Show that the Cartesian equation of the curve that has the parametric equations 𝑥 = 𝑡 2 + 𝑡,
𝑦 = 2𝑡 − 4 is 4𝑥 = 𝑦 2 + 10𝑦 + 24.
[3]
[4]
(c) The points 𝐴(3, −1, 2), 𝐵(1, 2, −4) and 𝐶(−1, 1, −2) are three vertices of a parallelogram 𝐴𝐵𝐶𝐷.
⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗
(i) Express the vectors 𝐴𝐵
𝐵𝐶 in the form 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘.
[3]
(ii) Show that the vector 𝑟 = −16𝑗 − 8𝑘 is perpendicular to the plane through 𝐴, 𝐵 and 𝐶.
[5]
(iii) Hence, find the Cartesian equation of the plane through 𝐴, 𝐵 and 𝐶.
[4]
Total 25 marks
172
CAPE 2013 PAST PAPER
SECTION C
5.
𝑥 + 2, 𝑥 < 2
(a) A function 𝑓(𝑥) is defined as 𝑓(𝑥) = { 2
𝑥 , 𝑥>2
(i) Find lim 𝑓(𝑥)
[4]
(ii) Determine whether 𝑓(𝑥) is continuous at 𝑥 = 2. Give a reason for your answer.
[2]
𝑥→2
𝑥2 +2𝑥+3
𝑑𝑦
(b) Let 𝑦 = (𝑥2 +2)3 . Show that 𝑑𝑥 =
−4𝑥3 −10𝑥2 −14𝑥+4
.
(𝑥2 +2)6
[5]
(c) The equation of an ellipse is given by 𝑥 = 1 − 3 cos 𝜃 , 𝑦 = 2 sin 𝜃 , 0 ≤ 𝜃 ≤ 2𝜋.
𝑑𝑦
Find 𝑑𝑥 in terms of 𝜃.
[5]
(d) The diagram below (not drawn to scale) shows the curve 𝑦 = 𝑥 2 + 3 and the line 𝑦 = 4𝑥.
(i) Determine the coordinates of the points 𝑃 and 𝑄 at which the curve and the line intersect.
[4]
(ii) Calculate the area of the shaded region.
[5]
Total 25 marks
173
CAPE 2013 PAST PAPER
6.
(a) (i) By using the substitution 𝑢 = 1 − 𝑥, find ∫ 𝑥(1 − 𝑥)2 𝑑𝑥.
[5]
(ii) Given that 𝑓(𝑡) = 2 cos 𝑡 , 𝑔(𝑡) = 4 sin 5𝑡 + 3 cos 𝑡, show that
∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = ∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑔(𝑡) 𝑑𝑡.
[4]
(b) A sports association is planning to construct a running track in the shape of a rectangle surmounted
by a semicircle, as shown in the diagram blow. The letter 𝑥 represents the length of the rectangular
section and 𝑟 represents the radius of the semicircle.
The perimeter of the track must be 600 metres.
(i) Show that 𝑟 =
600−2𝑥
2+𝜋
.
[2]
(ii) Hence, determine the length, 𝑥, that maximises the area enclosed by the track.
[6]
(c) (i) Let 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝐴𝑥 + 𝐵, where 𝐴 and 𝐵 are constants. Show that 𝑦 ′′ = 𝑥 sin 𝑥.
[4]
(ii) Hence, determine the specific solution of the differential equation 𝑦 ′′ = 𝑥 sin 𝑥 given that 𝑥 = 0,
𝑦 = 1 and when 𝑥 = 𝜋, 𝑦 = 6.
[4]
Total 25 marks
174
CAPE 2014 PAST PAPER
CAPE 2014
1.
SECTION A
(a) Let 𝑝, 𝑞 and 𝑟 be three propositions. Construct a truth table for the statement
(𝑝 → 𝑞) ∧ (𝑟 → 𝑞)
[5]
(b) A binary operator ⊕ is defined on a set of positive real numbers by
𝑦 ⊕ 𝑥 = 𝑦 3 + 𝑥 3 + 𝑎𝑦 2 + 𝑎𝑥 2 − 5𝑦 − 5𝑥 + 16 where 𝑎 is a real number.
(i)
State, giving a reason for your answer, if ⊕ is commutative in ℝ.
(ii)
Given that 𝑓(𝑥) = 2 ⊕ 𝑥 and (𝑥 − 1) is a factor of 𝑓(𝑥),
a)
find the value of 𝑎
b)
factorize 𝑓(𝑥) completely.
(c) Use mathematical induction to prove that
𝑛
12 + 32 + 52 + ⋯ + (2𝑛 − 1) 2 = 3 (4𝑛2 − 1) for 𝑛 ∈ ℕ.
[3]
[4]
[3]
[10]
Total 25 marks
2.
(a) The functions 𝑓 and 𝑔 are defined as follows:
𝑓(𝑥) = 2𝑥 2 + 1
𝑥−1
𝑔(𝑥) = √
2
where 1 ≤ 𝑥 ≤ ∞, 𝑥 ∈ ℝ
(i) Determine, in terms of 𝑥,
(a) 𝑓 2 (𝑥)
(b) 𝑓[𝑔(𝑥)]
(ii) Hence, or otherwise, state the relationship between 𝑓 and 𝑔.
(b) Given that 𝑎3 + 𝑏3 + 3𝑎2 𝑏 = 5𝑎𝑏2 , show that 3 log (
𝑎+𝑏
2
) = log 𝑎 + 2 log 𝑏.
(c) Solve EACH of the following equations
1
(i)
𝑒 𝑥 + 𝑒𝑥 − 2 = 0
(ii)
log 2 (𝑥 + 1) − log 2 (3𝑥 + 1) = 2
(d) Without the use of a calculator, show that
√3 − 1 √3 + 1 √2 − 1 √2 + 1
+
+
+
= 10
√3 + 1 √3 − 1 √2 + 1 √2 − 1
[3]
[3]
[1]
[5]
[4]
[4]
[5]
Total 25 marks
175
CAPE 2014 PAST PAPER
SECTION B
3.
(a) (i)
(ii)
(b) (i)
cot 𝑦−cot 𝑥
sin(𝑥−𝑦)
Prove that cot 𝑥+cot 𝑦 = sin(𝑥+𝑦).
[4]
Hence, or otherwise, find the possible values of 𝑦 in the trigonometric equation
cot 𝑦 − cot 𝑥
= 1, 0 ≤ 𝑦 ≤ 2𝜋,
cot 𝑥 + cot 𝑦
1
𝜋
when sin𝑥 = 2 , 0 ≤ 𝑥 ≤ 2 .
[8]
𝜋
Express 𝑓(𝜃) = 3 sin 2𝜃 + 4 cos 2𝜃 in the form 𝑟 sin(2𝜃 + 𝛼) where 𝑟 > 0 and 0 < 𝛼 < 2 .
[4]
(ii)
Hence, or otherwise, determine
(a) the value of 𝜃, between 0 and 2𝜋 radians, at which 𝑓(𝜃) is a minimum
1
(b) the minimum and maximum values of 7−𝑓(𝜃).
[4]
[5]
Total 25 marks
4.
(a) Let 𝐿1 and 𝐿2 be two diameters of a circle 𝐶. The equations of 𝐿1 and 𝐿2 are 𝑥 − 𝑦 + 1 = 0 and
𝑥 + 𝑦 = 5, respectively.
(i)
Show that the coordinates of the centre of the circle, 𝐶, where 𝐿1 and 𝐿2 intersect are (2, 3).
[3]
(ii)
𝐴 and 𝐵 are endpoints of the diameter 𝐿1. Given that the coordinates of 𝐴 are (1, 2) and that
the diameters of a circle bisect each other, determine the coordinates of 𝐵.
[3]
(iii)
A point, 𝑝, moves in the 𝑥 − 𝑦 plane such that its distance from 𝐶(2, 3) is always √2 units.
Determine the locus of 𝑝.
[3]
(b) The parametric equations of a curve, 𝑆, are given by
1
𝑡
𝑥=
and 𝑦 =
1+ 𝑡
1 − 𝑡2
Determine the Cartesian equation of the curve, 𝑆.
[6]
(c) The points 𝑃(3, −2, 1), 𝑄(−1, 𝜆, 5) and 𝑅(2, 1, 4) are three vertices of a triangle 𝑃𝑄𝑅.
⃗⃗⃗⃗⃗ , 𝑄𝑅
⃗⃗⃗⃗⃗ and 𝑅𝑃
⃗⃗⃗⃗⃗ in the form 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌.
(i)
Express EACH of the vectors 𝑃𝑄
[4]
(ii)
Hence, find the value of 𝜆, given that 𝑃𝑄𝑅 is right – angled with the side 𝑃𝑄 as hypotenuse.
[6]
Total 25 marks
176
CAPE 2014 PAST PAPER
SECTION C
5.
(a) Let 𝑓(𝑥) be a function defined as
𝑎𝑥 + 2, 𝑥 < 3
𝑎𝑥 2 , 𝑥 ≥ 3
(i) Find the value of 𝑎 if 𝑓(𝑥) is continuous at 𝑥 = 3.
𝑓(𝑥) = {
[4]
𝑥2 +2
(ii) Let 𝑔(𝑥) = 𝑏𝑥2 +𝑥+4.
Given that lim 2𝑔(𝑥) = lim 𝑔(𝑥), find the value of 𝑏.
(b) (i) Let 𝑦 =
(ii) If 𝑦 =
1
𝑥→1
𝑥→0
[5]
𝑑𝑦
. Using first principles, find 𝑑𝑥 .
√𝑥
𝑥
[8]
𝑑𝑦
, determine an expression for 𝑑𝑥 .
√1+𝑥
Simplify the answer FULLY.
(c) The parametric equations of a curve are given by
𝑥 = cos 𝜃 , 𝑦 = sin𝜃 , 0 ≤ 𝜃 ≤ 2𝜋
𝑑𝑦
Find 𝑑𝑥 in terms in 𝜃.
Simplify the answer as far as possible.
6.
[4]
[4]
Total 25 marks
𝑑𝑦
(a) The gradient of a curve which passes through the point (−1, −4) is given by 𝑑𝑥 = 3𝑥 2 − 4𝑥 + 1.
(i) Find
(a) the equation of the curve
(b) the coordinates of the stationary points and determine their nature.
(ii) sketch the curve in (a) (i) (a) above, clearly making ALL stationary points and intercepts.
(b) The equation of a curve is given by 𝑓(𝑥) = 2𝑥√1 + 𝑥 2 .
3
(i)
Evaluate ∫0 𝑓(𝑥) 𝑑𝑥.
(ii)
[4]
[8]
[4]
[5]
Find the volume generated by rotating the area bounded by the curve in (b) (i) above, the
𝑥 −axis, and the lines 𝑥 = 0 and 𝑥 = 2 about the 𝑥 − axis.
[4]
Total 25 marks
177
CAPE 2015 PAST PAPER
CAPE 2015
SECTION A
1.
(a) Let 𝑝 and 𝑞 be any two propositions.
(i) State the inverse and the contrapositive of the statement 𝑝 → 𝑞.
[2]
(ii) Copy and complete the table below to show the truth table for 𝑝 → 𝑞 and ~𝑞 → ~𝑝.
𝒑
𝒒
T
T
T
F
F
T
F
F
~𝒑
~𝒒
𝒑→𝒒
~𝒒 → ~𝒑
[4]
(iii) Hence, state whether the compound statements 𝑝 → 𝑞 and ~𝑞 → ~𝑝 are logically equivalent.
Justify your response.
[2]
(b) The polynomial 𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 + 𝑞 has a factor (𝑥 − 5) and a remainder of 24 when divided by
(𝑥 − 1).
(i) Find the values of 𝑝 and 𝑞.
[4]
(ii) Hence, factorize 𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 + 𝑞 completely.
2
3
4
[5]
𝑛
(c) Given that 𝑆(𝑛) = 5 + 5 + 5 + 5 + ⋯ + 5 , use mathematical induction to prove that
4𝑆(𝑛) = 5𝑛+1 − 5 for 𝑛 ∈ 𝑁.
[8]
Total 25 marks
2.
(a) The relations 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 are functions which are both one-to-one and onto.
Show that (𝑔 ∘ 𝑓) is
(i) one – to – one
[4]
(ii) onto
[4]
(b) Solve EACH of the following equations:
4
4
(i) 3 − 9𝑥 − (81)𝑥 = 0
[7]
(ii) |5𝑥 − 6| = 𝑥 + 5
[5]
(c) The population growth of bacteria present in a river after time, 𝑡 hours, is given by 𝑁 = 300 + 5𝑡 .
Determine
(i) the number of bacteria present at 𝑡 = 0.
[1]
(ii) the time required to triple the number of bacteria.
[4]
Total 25 marks
178
CAPE 2015 PAST PAPER
SECTION B
3.
3
(a) (i) Show that cos 3𝑥 = 4 cos 𝑥 − 3 cos 𝑥.
[6]
(ii) Hence, or otherwise, solve
cos 6𝑥 − cos 2𝑥 = 0 for 0 ≤ 𝑥 ≤ 2𝜋
(b)
[9]
𝜋
(i) Express𝑓(2𝜃) = 3 sin 2𝜃 + 4 cos 2𝜃 in the form 𝑟 sin(2𝜃 + 𝛼) where 𝑟 > 0 and 0 < 𝛼 < 2 .
[6]
1
(ii) Hence, or otherwise, find the maximum and minimum values of 7−𝑓(𝜃).
[4]
Total 25 marks
4.
(a) The circles 𝐶1 and 𝐶2 are defined by the parametric equations as follows:
𝐶1 : 𝑥 = √10 cos 𝜃 − 3;
𝐶2 : 𝑥 = 4 cos 𝜃 + 3 ;
𝑦 = √10 sin𝜃 + 2
𝑦 = 4 sin 𝜃 + 2
(i) Determine the Cartesian equations of 𝐶1 and 𝐶2 in the form (𝑥 − 𝑎) 2 + (𝑦 − 𝑏) 2 = 𝑟 2.
[4]
(ii) Hence or otherwise, find the points of intersection of 𝐶1 and 𝐶2 .
[9]
(b) A point 𝑃(𝑥, 𝑦) moves so that its distance from the fixed point (0, 3) is two times the distance from the
fixed point (5, 2). Show that the equation of the locus of the point 𝑃(𝑥, 𝑦) is a circle.
[12]
Total 25 marks
179
CAPE 2015 PAST PAPER
SECTION C
5.
(a) Let 𝑓 be a function defined as
𝑓(𝑥) = {
sin(𝑎𝑥)
𝑥
4
if 𝑥 ≠ 0, 𝑎 ≠ 0
if 𝑥 = 0
If 𝑓 is continuous at 𝑥 = 0, determine the value of 𝑎.
[4]
(b) Using first principles, determine the derivative of 𝑓(𝑥) = sin(2𝑥).
(c) If 𝑦 = √
2𝑥
1+𝑥2
[6]
show that
𝑑𝑦
𝑦
𝑑2 𝑦
3𝑦
(i) 𝑥 𝑑𝑥 = 1+𝑥2
[7]
(ii) 𝑑𝑥2 + (1+𝑥2 )2 = 0
[8]
Total 25 marks
6.
(a) The diagram below (not drawn to scale) shows the region bounded by the lines
𝑦 = 3𝑥 − 7, 𝑦 + 𝑥 = 9 and 3𝑦 = 𝑥 + 3.
(i) Show that the coordinates of 𝐴, 𝐵 and C are (4, 5), (3, 2), and (6, 3) respectively.
[5]
(ii) Hence, use integration to determine the area bounded by the lines.
[6]
(b) The gradient function of a curve 𝑦 = 𝑓(𝑥) which passes through the point (0, −6) is given by
3𝑥 2 + 8𝑥 − 3.
(i) Determine the equation of the curve.
[3]
(ii) Find the coordinates and nature of the stationary points of the curve in (b) (i) above.
[8]
(iii) Sketch the curve in (b) (i) by clearly labelling the stationary points.
[3]
Total 25 marks
180
CAPE 2016 PAST PAPER
CAPE 2016
SECTION A
1.
(a) Let 𝑓(𝑥) = 2𝑥 3 − 𝑥 2 + 𝑝𝑥 + 𝑞.
(i) Given that 𝑥 + 3 is a factor of 𝑓(𝑥) and that there is a remainder of 10, when 𝑓(𝑥) is divided by
𝑥 + 1 show that 𝑝 = −25 and 𝑞 = −12.
[7]
(ii) Hence, solve the equation 𝑓(𝑥) = 0.
[6]
(b) Use mathematical induction to prove that 6𝑛 − 1 is divisible by 5 for all natural numbers 𝑛.
(c) (i)
[6]
Given that 𝑝 and 𝑞 are two propositions, complete the truth table below:
𝒑
𝒒
T
T
T
F
F
T
F
F
𝒑→𝒒
𝒑∨𝒒
𝒑∧𝒒
(𝒑 ∨ 𝒒) → (𝒑 ∧ 𝒒)
[4]
(ii) State, giving a reason for your response, whether the following statements are logically
equivalent:
-
𝑝→𝑞
-
(𝑝 ∨ 𝑞 ) → (𝑝 ∧ 𝑞 )
[2]
Total 25 marks
2.
(a) Solve the following equation for 𝑥:
log 2(10 − 𝑥 ) + log 2 𝑥 = 4
[6]
𝑥+3
(b) A function 𝑓 is defined by 𝑓(𝑥) = 𝑥−1 , 𝑥 ≠ 1. Determine whether 𝑓 is bijective, that is, both one – to –
one and onto.
[8]
(c) Let the roots of the equation 2𝑥 3 − 5𝑥 2 + 4𝑥 + 6 = 0 be 𝛼, 𝛽 and 𝛾.
(i)
State the values of 𝛼 + 𝛽 + 𝛾, 𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 and 𝛼𝛽𝛾.
[3]
(ii)
Hence, or otherwise, determine an equation with integer coefficients which has roots 𝛼2 , 𝛽2
1
1
and 𝛾2 .
1
[8]
NOTE: (𝛼𝛽 )2 + (𝛼𝛾 )2 + (𝛽𝛾 )2 = (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 )2 − 2𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾 )
𝛼 2 + 𝛽 2 + 𝛾 2 = (𝛼 + 𝛽 + 𝛾)2 − 2(𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)
Total 25 marks
181
CAPE 2016 PAST PAPER
SECTION B
3.
(a) (i)
Show that
sec 2 𝜃 =
csc 𝜃
.
csc 𝜃 − sin 𝜃
[4]
(ii)
Hence, or otherwise, solve the equation
csc 𝜃
4
=
csc 𝜃 − sin𝜃 3
for 0 ≤ 𝜃 ≤ 2𝜋.
[5]
𝜋
(b) (i) Express the function 𝑓(𝜃) = sin 𝜃 + cos 𝜃 in the form 𝑟 sin(𝜃 + 𝛼), where 𝑟 > 0 and 0 ≤ 𝜃 ≤ .
2
[5]
(ii) Hence, find the maximum value of 𝑓 and the smallest non – negative value of 𝜃 at which it occurs.
[5]
(c) Prove that
tan(𝐴 + 𝐵 + 𝐶) =
tan 𝐴 + tan 𝐵 + tan 𝐶 − tan 𝐴 tan 𝐵 tan 𝐶
1 − tan 𝐴 tan 𝐵 − tan 𝐴 tan 𝐶 − tan 𝐵 tan 𝐶
[6]
Total 25 marks
4.
(a) (i)
(ii)
Given that sin 𝜃 = 𝑥, show that tan 𝜃 = √
𝑥
1−𝑥2
𝜋
, where 0 < 𝜃 < 2 .
[3]
Hence, or otherwise, determine the Cartesian equation of the curve defined parametrically
𝜋
by 𝑦 = tan 2𝑡 and 𝑥 = sin𝑡 for 0 < 𝑡 < 2.
[5]
1
2
(b) Let 𝑢 = (−3) and 𝑣 = ( 1) be two position vectors in ℝ3 .
2
5
(i)
Calculate the lengths of 𝑢 and 𝑣 respectively.
[3]
(ii)
Find cos 𝜃 where 𝜃 is the angle between 𝑢 and 𝑣 in ℝ3 .
[4]
(c) A point 𝑃(𝑥, 𝑦) moves such that its distance from the 𝑥 − axis is half its distance from the origin.
Determine the Cartesian equation of the locus of 𝑃.
[5]
(d) The line 𝐿 has the equation 2𝑥 + 𝑦 + 3 = 0 and the circle 𝐶 has the equation 𝑥 2 + 𝑦 2 = 9. Determine
the points of intersection of the circle 𝐶 and the line 𝐿.
[5]
Total 25 marks
182
CAPE 2016 PAST PAPER
SECTION C
5.
1
(a) Use an appropriate substitution to find ∫(𝑥 + 1)3 𝑑𝑥.
[4]
(b) The diagram below represents the finite region 𝑅 which is enclosed by the curve 𝑦 = 𝑥 3 − 1 and the
lines 𝑥 = 0 and 𝑦 = 0.
Calculate the volume of the solid that results from rotating 𝑅 about the 𝑦 − axis.
𝑎
[5]
𝑎
(c) Given that ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥 𝑎 > 0, show that
1
∫
0
𝑒𝑥
1
𝑑𝑥 =
𝑒 𝑥 + 𝑒 1−𝑥
2
[6]
Total 15 marks
Question 6d is NOT on C.A.P.E Unit 1 Pure Mathematics Syllabus
6.
(a) Find the equation of the tangent to the curve 𝑓(𝑥) = 2𝑥 3 + 5𝑥 2 − 𝑥 + 12 at the point where 𝑥 = 3.
[4]
(b) A function 𝑓 is defined on ℝ as
2
𝑓(𝑥) = {𝑥 + 2𝑥 + 3
𝑎𝑥 + 𝑏
𝑥≤0
𝑥>0
(i)
Calculate the lim− 𝑓(𝑥) and lim 𝑓(𝑥).
[4]
(ii)
Hence, determine the values of 𝑎 and 𝑏 such that 𝑓(𝑥) is continuous at 𝑥 = 0.
[5]
(iii)
If the value of 𝑏 = 3, determine 𝑎 such that
𝑥→0
𝑥→0
𝑓 ′ (0) = lim
𝑡→0
𝑓(0 + 𝑡) − 𝑓(0)
.
𝑡
[6]
(c) Use first principles to differentiate 𝑓(𝑥 ) = √𝑥 with respect to 𝑥.
[6]
Total 25 marks
183
CAPE 2017 PAST PAPER
CAPE 2017
SECTION A
1.
(a) Let 𝑝 and 𝑞 be two propositions
𝑝: It is raining
𝑞: John is sick
Write EACH of the statements below in terms of 𝑝 and 𝑞.
(i) It is not raining or John is sick.
[1]
(ii) If it is raining then John is sick.
[1]
(b) An operation * is defined on the set {1, 2, 3, 4} as shown in the following table.
(i)
*
1
1
2
2
4
3
1
4
3
Prove that * is commutative.
(ii)
Show that the identity element of * is 3.
2
4
3
2
1
3
1
2
3
4
4
3
1
4
2
[1]
[2]
(c) The polynomial 𝑓(𝑥) = 𝑎𝑥 3 + 9𝑥 2 − 11𝑥 + 𝑏 has a factor of (𝑥 − 2) and a remainder of 12 when
divided by (𝑥 + 2).
(i)
Show that 𝑎 = 2 and 𝑏 = −30.
[4]
(ii)
Hence, solve 𝑎𝑥 3 + 9𝑥 2 − 11𝑥 + 𝑏 = 0.
[9]
(d) Use mathematical induction to prove that
8 + 16 + 24 + 32 + ⋯ + 8𝑛 = 4𝑛(𝑛 + 1) for all 𝑛 ∈ ℕ.
[7]
Total 25 marks
184
CAPE 2017 PAST PAPER
2.
(a) (i)
(ii)
𝑎+𝑏
1
Given that 𝑎2 + 𝑏2 = 14𝑎𝑏, prove that ln ( 4 ) = 2 (ln 𝑎 + ln 𝑏) .
[5]
Solve the equation 2−𝑥 + 3(2𝑥 ) = 4.
[6]
[Your response may be expressed in terms of logarithms.]
3𝑥−4
(b) The following diagram shows the graph of the function 𝑓(𝑥) = 𝑥+4 .
On the diagram,
(i)
insert the asymptotes for the function 𝑓
[2]
(ii)
sketch the graph of 𝑓 −1, the inverse of 𝑓 showing the asymptotes for 𝑓 −1.
[4]
(c) Given that 𝛼, 𝛽 and 𝛾 are the roots of the equation 𝑥 3 + 3𝑥 + 2 = 0, form an equation whose roots are
𝛽𝛾, 𝛼𝛾 and 𝛼𝛽.
[8]
Total 25 marks
185
CAPE 2017 PAST PAPER
SECTION B
3.
(a) (i)
Prove the identity
tan(𝐴 + 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
[4]
(ii)
3
1
Given that sin 𝐴 = 5 and cos 𝐵 = − 2 where angle 𝐴 is acute and angle 𝐵 is obtuse, express
tan(𝐴 + 𝐵) in the form 𝑎 + 𝑏√3, where 𝑎 and 𝑏 are real numbers.
[6]
(b) Solve the equation sin2 𝜃 − 2 cos2 𝜃 + 3 cos 𝜃 + 5 = 0 for 0 ≤ 𝜃 ≤ 4𝜋.
[6]
(c) (i)
Express 𝑓(𝜃) = 6 cos 𝜃 + 8 sin𝜃 in the form 𝑟 sin(𝜃 + 𝛼) where 0 ≤ 𝛼 ≤ 90°.
[3]
Hence, or otherwise, find the general solution of 𝑓(𝜃) = 2.
[6]
(ii)
Total 25 marks
4.
(a) (i)
The circle 𝐶1 with equation 𝑥 2 + 𝑦 2 − 4𝑥 + 2𝑦 − 2 = 0 and the circle 𝐶2 have a common
centre. Given that 𝐶2 passes through the point (−1, −2), express the equation of 𝐶2 in the
form (𝑥 − ℎ) 2 + (𝑦 − 𝑘 )2 = 𝑘.
[3]
(ii)
The equation of the line 𝐿 is 𝑥 + 3𝑦 = 3. Determine whether 𝐿1 is a tangent to the circle 𝐶1.
[7]
(b) Let 𝑃(3, 1, 2) and 𝑄(1, −2, 4)
(i)
Express the vector ⃗⃗⃗⃗⃗
𝑃𝑄 in the form 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘.
[2]
(ii)
Determine the Cartesian equation of the plane which passes through the point 𝑄 and is
perpendicular to ⃗⃗⃗⃗⃗
𝑃𝑄 .
[6]
(c) The vector equations of two lines, 𝐿1 and 𝐿2 are:
𝐿1 = −𝑖 + 𝑗 − 2𝑘 + 𝛼(−2𝑖 + 𝑗 − 3𝑘)
𝐿2 = −2𝑖 + 𝑗 − 4𝑘 + 𝛽(𝑖 − 𝑗 + 𝑘)
(i)
Show that 𝐿1 and 𝐿2 intersect.
[5]
(ii)
Hence, determine the coordinates of the point of intersection of the two lines.
[2]
Total 25 marks
186
CAPE 2017 PAST PAPER
SECTION C
𝑥5 −1
5.
(a) Determine the value of 𝑘 for which 𝑓(𝑥) = { 𝑥−1 , 𝑥 ≠ 1 is continuous for all values of 𝑥.
𝑘,
𝑥=1
[4]
(b) A curve, 𝐶, is described parametrically by the equations 𝑥 = 5𝑡 + 3 and 𝑦 = 𝑡 3 − 𝑡 2 + 2.
𝑑𝑦
(i)
Find 𝑑𝑥 in terms of 𝑡.
[3]
(ii)
Hence, determine all points of 𝐶 such that 𝑑𝑥 = 0.
𝑑𝑦
[6]
Given that 𝑦 = √2 + 2𝑥 2, show that
(c) (i)
𝑑𝑦
(a) 𝑦 𝑑𝑥 − 2𝑥 = 0
𝑑2 𝑦
4
(b) 𝑑𝑥2 − 𝑦3 = 0
[9]
𝑑2 𝑦
(ii)
Hence, find the value of 𝑑𝑥2 when 𝑥 = 0.
[3]
Total 25 marks
6.
(a) Triangle 𝑃𝑄𝑅 has vertices 𝑃(0, 1), 𝑄(3, 3) and 𝑅(4, 2).
(i) On the axes below, sketch triangle 𝑃𝑄𝑅.
[1]
(ii) Determine the equations of EACH of the following:
- 𝑃𝑄
- 𝑄𝑅
- 𝑃𝑅
(iii) Hence use integration to determine the area of triangle 𝑃𝑄𝑅.
3
[7]
[7]
3
(c) Given that ∫−1[3𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 = 5 and ∫−1[5𝑓(𝑥) − 2𝑔(𝑥)] 𝑑𝑥 = 1, determine
3
-
∫−1 𝑓(𝑥) 𝑑𝑥
-
∫−1 𝑔 (𝑥 ) 𝑑𝑥
3
[5]
Total 20 marks
Question 6b is NOT on C.A.P.E Unit 1 Pure Mathematics Syllabus
187
ANSWERS TO CAPE PAST PAPERS
ANSWERS FOR CAPE PAST PAPERS
CAPE 2012
1. (a) (i) 𝑝 = −7, 𝑞 = 1 (ii) (𝑥 − 1)(2𝑥 + 5)(𝑥 + 2)
(b) (10, 6)or (6, 10)
3 1
1
2
(c) (i) 3 ≤ 𝑥 ≤ 4
2.
(a) (i) 𝑥 4 − 6𝑥 2 + 6 (ii) 𝑥 = −3, 0, 1, 2
3.
(a) (iii) 𝜃 = 0, 8 , 8 , 2
4.
(a) (i) 𝑦 2 = 𝑥 2 + 9 (ii) (1, √10) (9, 3√10)
(iv) 27.41°
5.
(a) (i) ±2 (ii) −3 (iii) 2
6.
(b) (i) 𝑦 = 𝑥 3 − 3𝑥 2 + 4 (ii) (0, 4) and (2, 0) (iii) (0, 4) Max (2, 0) Min (iv) (−1, 0) (2, 0)
𝜋 3𝜋 𝜋
(b) (i) 4 , 4 (ii) 16 (iii) 𝑥 2 − 2𝑥 + 64 = 0 (c) (i) 1 (ii) −2
(b) 0, 1 ± √3
(b) (i) 𝑝 = −3𝑖 + 4𝑗,
𝑞 = −𝑖 + 6𝑗 (ii) −2𝑖 − 2𝑗 (iii) 27
(b) (i) a) 2 b) 𝑝 = −2 (ii) 2 (c) 𝑢 = 2, 𝑣 = −3
(v) (a)
CAPE 2013
1.
(a) (i)
𝑝
0
0
1
1
𝑞
0
1
0
1
𝑝→𝑞
1
1
0
1
𝑝∧𝑞
0
0
0
1
(d) (ii) (𝑥 + 1)(𝑥 − 2)(𝑥 − 8)
(b) 𝑥 = 1, 8
𝑥−2
ln 𝑥
(iii) 𝑥 = −1, 2, 8
2
3
2.
(b) (i) 𝑓 −1 (𝑥) =
3.
(a) (ii) 𝜃 = 0, 𝜋, 2𝜋, 4 , 4 , 4 , 4
4.
⃗⃗⃗⃗⃗ = −2𝑖 + 3𝑗 − 6𝑘 𝐵𝐶
⃗⃗⃗⃗⃗ = −2𝑖 − 𝑗 + 2𝑘 (iii) 2𝑦 + 𝑧 = 0
(ii) a) 𝑦 = 2 (c) (i) 𝐴𝐵
5.
(a) (i) 4 (ii) Continuous (c) cot 𝜃 (d) (i) (3, 12) (1, 4) (ii)
6.
(a) (i)
3
, 𝑔 −1 (𝑥) =
~(𝑝 ∧ 𝑞)
1
1
1
0
𝜋 3𝜋 5𝜋 7𝜋
(1−𝑥) 4
4
−
(1−𝑥)3
3
2
(c) (i) {−2 ≤ 𝑥 ≤ 3} (ii) − 2 = 𝑥
1
(b) (i) 𝑓(𝜃) = 5 cos(𝜃 + 0.927) (ii) a) 5 b) 13
2
4
3
3
600
1
+ 𝑐 (b) (ii) 4+𝜋 = 𝑥 (c) (ii) 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝜋 𝑥 + 3
188
ANSWERS TO CAPE PAST PAPERS
CAPE 2014
1. (a)
𝑝
0
0
0
0
1
1
1
1
𝑞
0
0
1
1
0
0
1
1
𝑟
0
1
0
1
0
1
0
1
𝑝→𝑞
1
1
1
1
0
0
1
1
(𝑝 → 𝑞) ∧ (𝑟 → 𝑞)
1
0
1
1
0
0
1
1
𝑟→𝑞
1
0
1
1
1
0
1
1
2.
(b) (i) Commutative (ii) a) 𝑎 = −2 b) (𝑥 − 1)(𝑥 − 3)(𝑥 + 2)
3
(a) (i) (a) 8𝑥 4 + 8𝑥 2 + 3 (b) 𝑥 (ii) Inverse of each other (c) (i) 𝑥 = 0 (ii) 𝑥 = − 11
3.
(ii) 0, 𝜋, 2𝜋 (b) (i) 𝑓(𝜃) = 5 sin(2𝜃 + 0.927) (ii) (a) 𝜃 = 1.89 (b) 2 , 12
4.
(a) (ii) (3, 4) (iii) (𝑥 − 2)2 + (𝑦 − 3)2 = 2 (b) 𝑦 = 2𝑥−1
1
𝑥(1−𝑥)
1
(c) (i) ⃗⃗⃗⃗⃗
𝑃𝑄 = −4𝒊 + (2 + 𝜆)𝒋 + 4𝒌,
⃗⃗⃗⃗⃗ = 3𝒊 + (1 − 𝜆)𝒋 − 𝒌, 𝑅𝑃
⃗⃗⃗⃗⃗ = 𝒊 − 3𝒋 − 3𝒌 (ii) −1
𝑄𝑅
5.
6.
1
1
3
(a) (i) 𝑎 = 3 (ii) 7 (b)(i) − 2 𝑥 −2 (ii)
𝑥+2
3
(c) − cot 𝜃
2(1+𝑥)2
1 4
20√10−2
3 27
3
(a) (i) (a) 𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥 (b) Max (1, 0) Min ( , ) (b) (i)
(ii)
544𝜋
15
CAPE 2015
1. (a) (i) Inverse ~𝑝 → ~𝑞 Contrapositive ~𝑞 → ~𝑝
(ii)
𝒑
𝒒
~𝒑
~𝒒
𝒑→𝒒
~𝒒 → ~𝒑
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
(iii) Logically Equivalent (b) (i) 𝑝 = −6, 𝑞 = 30 (ii) (𝑥 − 5)(𝑥 − 3)(𝑥 + 2)
ln 2
11 1
2.
(b) (i) ln 9 (ii) 4 , 6 (c) (i) 301 (ii) 4 hours
3.
(a) (ii) 𝑥 = 2 , 2 , 4 , 4 , 4 , 4 (b) (i) 𝑓(2𝜃) = 5 sin(2𝜃 + 0.927) (ii) 2 , 12
4.
(a) (i) 𝐶1 : (𝑥 + 3)2 + (𝑦 − 2) 2 = 10 𝐶2 : (𝑥 − 3) 2 + (𝑦 − 2) 2 = 16 (ii) (− 2 ,
5.
(a) 4 (b) 2 cos 2𝑥
6.
(a) (ii) 4 (b) 𝑦 = 𝑥 3 + 4𝑥 2 − 3𝑥 − 6 (ii) Min (3 , − 27 ) Max (−3, 12)
𝜋 3𝜋 𝜋 3𝜋 5𝜋 7𝜋
1
1
1 4+√15
1
2
1 4−√15
) , (− 2 ,
2
)
176
189
ANSWERS TO CAPE PAST PAPERS
CAPE 2016
1
1. (a) (ii) 𝑥 = −3, − , 4 (c) (i)
2
𝒑
𝒒
𝒑→𝒒
𝒑∨𝒒
𝒑∧𝒒
(𝒑 ∨ 𝒒) → (𝒑 ∧ 𝒒)
T
T
T
T
T
T
T
F
F
T
F
T
F
T
T
T
F
T
F
F
T
F
F
T
(ii) Not Logically Equivalent
5
2.
(a) 2, 8 (b) Not Bijective (c) (i) , 2, −3 (ii) 36𝑥 3 − 76𝑥 2 + 9𝑥 − 4 = 0
3.
(a) (ii) 6 , 6 , 6 , 6
4.
(a) (ii) 𝑦 =
5.
(a)
6.
(a) 𝑦 = 83𝑥 − 141 (b) (i) 3, 𝑏 (ii) 𝑎 ∈ ℝ, 𝑏 = 3 (iii) 2 (c) 2 𝑥
2
𝜋 5𝜋 7𝜋 11𝜋
2𝑥 √1−𝑥2
4
(b) (i) √14, √30 (ii) 𝑦 =
1−2𝑥2
4
3(𝑥+1)3
𝜋
𝜋
(b) (i) 𝑓(𝜃) = √2 sin (𝜃 + 4 ) (ii) 4
2𝑥√1−𝑥2
1−2𝑥2
12 9
(c) 3𝑦 2 = 𝑥 2 (d) (0, −3) (− 5 , 5 )
3𝜋
+ 𝑐 (b) 5
1
√
CAPE 2017
3
1. (a) (i) ~𝑝 ∨ 𝑞 (ii) 𝑝 → ~𝑞 (c) (ii) 𝑥 = −5, − 2 , 2
ln 3
2.
(a) (ii) 0, −
3.
(a) (ii) 11 + 11
(c) 𝑥 3 − 3𝑥 2 − 4 = 0
ln 2
25√3
4.
𝜃 = −25.33 + 360°𝑛
𝑛∈ℤ
𝜃 = 131.69° + 360°𝑛
2
2
(a) (i) (𝑥 − 2) + (𝑦 + 1) = 10 (ii) Not a tangent (b) (i) −2𝑖 − 3𝑗 + 2𝑘 (ii) −2𝑥 − 3𝑦 + 2𝑧 = 12
(c) (ii) −3𝑖 + 2𝑗 − 5𝑘
5.
(a) 5 (b) (i)
6.
(a) (ii) 𝑦 = 𝑥 + 1 𝑦 = −𝑥 + 6
48
2
3
(b) 𝜋, 3𝜋 (c) (i) 𝑓(𝜃) = 10 sin(𝜃 + 36.87°) (ii) {
3𝑡2 −2𝑡
5
19 50
(ii) (3, 2) ( 3 , 27) (c) (ii) √2
1
5
4
2
𝑦 = 𝑥 + 1 (iii)
(c) 1, 2
190