Table Of Content·O
James Foley
Michael Haese
Robert Haese
Sandra Haese
Mark Humphries
Anna Rijken
Gregory Olesinski
0
Haese Mathematics
MATHEMATICS FOR THE INTERNATIONAL STUDENT
Mathematics HL (Core) third edition - WORKED SOLUTIONS
IB Diploma Programme
James Foley B.Ma.Comp.Sc.(Hons.)
Michael Haese B.Sc.(Hons.), Ph.D.
Robert Haese B.Sc.
Sandra Haese B.Sc.
Mark Humphries B.Sc.(Hons.)
AnnaRijken B.MaSc.
Gregory Olesinski B.Sc.
Haese Mathematics
152 Richmond Road, Marleston, SA 5033, AUSTRALIA
Telephone: +61 8 8210 4666, Fax: +61 8 8354 1238
En1ai l: [email protected]. at1
Web: www.haese1nathematics.con1.au
National Library ofA ustralia Card Number & ISBN 978-1-92 1972-12-6
© Haese & Harris Publications 2013
Published by Haese Mathematics
152 Richmond Road, Mar1eston, SA 5033, AUSTRALIA
First Edition 2005
Second Edition 2009
Third Edition 2013
Reprinted 2014
Artwork by Brian Houston and Gregory Olesinski.
Cover design by Piotr Poturaj.
Typeset in Australia by Charlotte Frost and Deanne Gallasch.
Typeset in Times Roman Si / 10
Printed in Singapore by Opus Grot1p.
The textbook, its acco1npanying CD and this book of fully worked solutions have been developed
independently of the International Baccalaureate Organization (IBO). These publications are in no way
coru1ected with, or endorsed by, the IBO.
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FOREWORD
This book gives you fully worked solutions for every question ( discussio·ns, investigations and projects
exce·pted) in each chapter of our textbook Mathematics HL (Core) third edition which is one of the
textbooks in our series Mathematics for the International Student intended for use with IB Diploma and
Middle Years courses.
Correct answers can sometimes be obtained by different methods. In this book, where applicable, each
worked solution is modelled on t11e worked exa1nple ii1 the textbook.
Be aware of the limitations of calculators and computer modelling packages. Understand that when your
calculator gives an answer that is different from the answer you fmd in the book, you have not necessarily
made a mistake, but tl1e book may not be wrong either.
We 11ave a list of errata for our books on our website. Please contact us if you notice any errors in this
book.
JMF PMH
RCH SHH MAH
AER GO
e-mail: [email protected]
web: www.haesemathematics.com.au
TABLE OF CONTENTS
Chapter I QUADRATICS 5
Chapter 2 FUNCTIONS 44
Chapter 3 EXPONENTIALS 89
Chapter 4 LOGARITHMS 110
Chapter 5 TRANSFORMING FUNCTIONS 140
Chapter 6 COMPLEX NUMBERS AND POLYNOMIALS 165
Chapter 7 SEQUENCES AND SERIES 209
Chapter 8 COUNTING AND THE BINOMIAL EXPANSION 245
Chapter 9 MATHEMATICAL INDUCTION 266
Chapter 10 THE UNIT CIRCLE AND RADIAN MEASURE 291
Chapter 11 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY 310
Chapter 12 TRIGONOMETRIC FUNCTIONS 327
Chapter 13 TRIGONOMETRIC EQUATIONS AND IDENTITIES 343
388
Chapter 14 VECTORS
Chapter 15 VECTOR APPLICATIONS 437
Chapter 16 COMPLEX NUMBERS 499
Chapter 17 INTRODUCTION TO DIFFERENTIAL CALCULUS 541
Chapter 18 RULES OF DIFFERENTIATION 559
598
Chapter 19 PROPERTIES OF CURVES
Chapter 20 APPLICATIONS OF DIFFERENTIAL CALCULUS 648
Chapter 21 INTEGRATION 690
Chapter 22 APPLICATIONS OF INTEGRATION 751
Chapter 23 DESCRIPTIVE STATISTICS 791
Chapter 24 PROBABILITY 810
Chapter 25 DISCRETE RANDOM VARIABLES 842
Chapter 26 CONTINUOUS RANDOM VARIABLES 872
Chapter 27 MISCELLANEOUS QUESTIONS 899
ChaP-ter
1
QUADRATICS
EXERCISE 1A. 1
1 a 4x2 + 7x = 0 b 3x2 - 7x = 0 c 2x2 - llx = 0
+
x(4x 7) = 0 x(3x - 7) = 0 x(2x - 11) = 0
+
x = 0 or 4x 7 = 0 x = 0 or 3x - 7 = 0 x = 0 or 2x - 11 = 0
{Null Factor law} {Null Factor law} {Null Factor law}
x = 0 or 7 x = 0 or ~ x - 0 or 11
- 2
4
= 2 2 + = 2 +
d 9x 6x e x 5x 6 0 f x 21 = lOx
-
6x2 9x = 0 (x - 2)(x - 3) = 0 x2 lOx + 21 = 0
- -
3x(2x - 3) = 0 x - 2 = 0 or x - 3 = 0 ( x - 3) ( x - 7) = 0
= = = =
{Null Factor law}
x 0 or 2x - 3 0 x - 3 0 or x - 7 0
{Null Factor law} x = 2 or 3 {Null Factor law}
=
x 3 or 7
x = 0 or ~
9 9 +x2 = 6x h x2 + x = 12 I• x2 +Bx= 33
+ +
2 2 + 2
x 6x 9 = 0 x x - 12 = 0 : . x 8x - 33 = 0
-
+
(x - 3)2 = 0 (x 4)(x - 3) = 0 (x+ ll)(x - 3) =0
+ = =
+
x-3=0 x 4 = 0 or x - 3 0 x 11 0 or x - 3 = 0
... x = 3 {Null Factor law} {Null Factor law}
x = -4 or 3 x = -11 or 3
a 9x2 12x + 4 = 0 b 2x2 l3x - 7 = 0 c 3x2 = l6x + 12
- -
= =
(3x - 2)2 = O (2x + l)(x - 7) 0 3x2 - l6x - 12 0
+
. x = x = _ l or 7 (3x 2)(x - 6) = 0
2
3 2
= -;
:. x or 6
d 3x2 + 5x = 2 e 2x2 + 3 = 5x f 3x2 + 8x + 4 = 0
3x2 + 5x - 2 = 0 2x2 5x + 3 = 0 ( 3x + 2) (x + 2) = 0
-
+
: . ( 3x - 1) ( x 2) = 0 (2x - 3)(x - 1) = 0 · x - - 2 or - 2
- 3
:. x = ~ or - 2 · x = ~ or 1
. . 2
g 3x2 = lOx + 8 h 4x2 + 4x = 3 i 4x2 = llx + 3
:. 3x2 l Ox - 8 = 0 4x2 + 4x - 3 = 0 : . 4x2 llx - 3 = 0
- -
+ + +
(3x 2) (x - 4) = 0 (2x 3) (2x - 1) = 0 ( 4x 1) ( x - 3) = 0
x = -~ or 4 x = - 3 or 1 · x = _l or 3
2 2 4
+ 2 2 + 2 + +
3 a (x 1) = 2x 5x 11 b 5 - 4x = 3(2x 1) 2
-
2 + + 2 + 2 + +
x 2x 1 = 2x 5x 11 5 - 4x = 6x 3 2
-
2 + 2 +
. . x 7x 10 = 0 :. 4x 6x = 0
-
= =
+
(x - 2)(x - 5) 0 2x(2x 3) 0
. ·. x = 2 or 5 . . x = 0 or 3
2
1 x+3 9
c 2x - - = - 1 d
x 1 - x x
=
2 +
2x - 1 -x x(x 3) = -9(1 - x)
2x2 + x - l = 0 2 + = +
x 3x - 9 9x
+
( 2x - 1) ( x 1) = 0 2 +
x 6x 9 = 0
-
x = .!. or -1 2
(x - 3) = 0
2
... x = 3
6 Mathematics HL (3rd edn), Chapter I - QUADRATICS
EXERCISE 1A.2
1 a (x + 5)2 = 2 b (x + 6)2 = - 11 c (x - 4)2 = 8
.'. x + 5 = ±v'2 11as no real solutions as ... x - 4 = ±v'S
(x + 6)2 cannot be negative
± v'2 = ± 2v'2
.'. x = - 5 .'. x 4
2 + 2 = 2
d 3(x - 2) = 18 e (2x 1) 3 f ( 1 - 3x) 7 = 0
-
= =
(x - 2)2 6 2x + 1 = ±v'3 (1 - 3x)2 7
±v6 v'3 ±v'?
x- 2 = .'. 2x = -1 ± 1 - 3x =
.'. x=2± v'6 .'. x=- ~ ±~\/'3 .·. 3x = 1 ± v7
.
x= l ± ~
. . 3 3
2 a x 2 - 4x + 1 = 0 b x2 + 6x + 2 = 0
+
2 2
.'. x 4x = - 1 .. x 6x = - 2
-
x2 4x + (-2)2 = - 1 + (-2)2 x 2 + 6x + 32 = -2 + 32
-
(x - 2)2 = 3 (x + 3)2 = 7
±v'3 ±v'?
+
.'. x - 2 = x 3 =
.'. x=2±\/'3 ± v'7
.'. x = - 3
c x2 - 14x + 46 = 0 d x 2 = 4x + 3
2 2 =
. . x 14x = - 46 x - 4x 3
-
=
x2 14x + (-7)2 = -46 + (-7)2 x 2 4x + (-2)2 3 + (-2)2
- -
2 2 =
(x - 7) = 3 (x - 2) 7
±V3
.'. x - 7 = x - 2 = ± v7
.'. x= 7± v'3
.'. x=2± v7
e x2 + 6x + 7 = 0 f x2 = 2x + 6
.'. x 2 +6x = -7 x 2 - 2x = 6
x 2 + 6x + 32 = - 7 + 32 x2 - 2x + (-1)2 = 6 + (-1)2
(x + 3)2 = 2 (x - 1)2 = 7
±v'2 ±v'?
+
.'. x 3 = .'. x - 1 =
± v'2 .. x= l ± v'?
.'. x = - 3
9 x 2 + 6x = 2 h x2 +10 = 8x
2 + + 2 + 2 2
x 6x 3 = 2 3 x - Bx= - 10
(x + 3)2 = 11 x 2 Bx+ (-4)2 = - 10 + (-4)2
-
x + 3 = ±vlil (x - 4)2 = 6
vli1 = ±v'6
.'. x = - 3 ± x - 4
.'. x=4± v'6
I• x2 +6x = - 11
x2 + 6x + 32 = - 11 + 32 x has no real solutions, since the
.'. (x + 3)2 = - 2 perfect square cannot be negative .
3 a 2x2 + 4x+ 1 = 0 b 2x2 lOx + 3 = 0
-
+;
2 + + 2
x 2x l = 0 .'. x 5x = 0
-
2
· x2 + 2x = _.! .'. x2 5x = - ;
-
. . 2
+ +
x2 + 2x + 12 = _ l + 12 x 2 - 5x (-~)2 = -~ (-~)2
2
! x _ = _ +
(x+ 1)2 = ( .§.) 2 3 25
2 2 4
¥
(x - ~)2 =
x+ l =±~
=
. x = - 1 ± . x - .§. ±~
. . _VL2
. . 2 2
. x=5± ~
. . 2 2
Mathematics HL (3rd edn), Chapter I - QUADRATICS 7
c 3x 2 + l2x + 5 = 0 d 3x2 = 6x+4
2 + + 5 = 2 + 4
x 4x 0 · x = 2x
. . 3
3
= - =
2 + 5 2 4
· x 4x x 2x
-
. . 3
3
2
2 2 5 2 + 2 + 2
x + 4x + 2 = - + 2 x 2x (-1) = ~ (-1)
-
3
(x + 2)2 = ~ (x - 1)2 = ;
±j i
=±[i
x+ 2 = x- 1
.. v!1 .. vfz
=
. x - 2 ± · x=l±
:3' ~
e 5x2 15x + 2 = 0 f 4x2 + 4x = 5
-
2 + = 2 +
:. x 3x ~ 0 :. x x = ~
-
· x 2 3x = - 2 x2 + x + ( l )2 = 5 + (l )2
-
. . 5
2 4 2
:. x2 - 3x + (-~)2 = -~ + (-~)2 (x + ~ )2 = ~
+
(x _ 3 )2 = _ 2 _g = 37
x +l =± ~
2 5 4 20
2 2
G
x-3=±y 20 . x = _ l ± ~
. . 2 2
2
vI E
. X= ~ ±
2o
. . 2
EXERCISE 1A .3
,
a x2 -4x - 3 = 0 b x2 + 6x + 7 = 0
has a = 1, b = -4, c = - 3 has a = 1, b = 6, c = 7
J6
- (-4) ± J (-4)2 - 4(1)(- 3) -6 ± 2 4(1)(7)
-
= - - - - - - - - - - x= - - - - - - -
x
2(1) 2(1)
v'28
4 ± - 6±vB
2 2
4±2V7 -6 ± 2v'2
2 2
= 2 ±v'7 = -3± \1'2
2 2
c x + 1 =4x d x +4x = 1
2 2
x -4x+ 1 = 0 x + 4x - 1 = 0
which has a = 1, b = -4, c = 1 which has a = 1, b = 4, c = - 1
J
-(-4) ± J(-4)2 - 4(1)(1) - 4 ± 42 - 4(1)(-1)
x = - - - - - - - - - - x= - - - - - - - -
2(1) 2(1)
v1i2 v'20
4 ± - 4 ±
2 2
4 ± 2y'3 - 4 ± 2v'5
2
2
v'5
=2±v'3 = -2 ±
8 Mathematics HL (3rd edn), Chapter I - QUADRATICS
e x2 - 4x + 2 = 0 f 2x2 - 2x - 3 = 0
has a = 1, b = -4, c = 2 has a= 2, b = -2, c = - 3
2
-(-4) ± JC-4) - 4(1)(2) -(-2) ± .j(-2)2 - 4(2)(-3)
x = - - - - - - - - - . x = - - - - - - - - - -
. . 2(1)
2(2)
v'28
±
4 ± v'8 2
2 4
±
4 2y'2 2±2V7
2 4
l±
=2±v'2 = v9
2 2
+ 2 + + =
g (3x 1) = -2x h ( x 3) ( 2x 1) 9
2 + + 2 + + +
.'. 9x 6x 1 = - 2x · 2x x 6x 3 = 9
2 + + = 2 + =
.'. 9x 8x 1 0 .'. 2x 7x - 6 0
wh:ich has a= 9, b = 8, c = 1 wh:ich has a= 2, b = 7, c = - 6
J1
-8 ± J82 4(9)(1) - 7 ± 2 4(2) (-6)
- -
x= - - - - - - - . x= - - - - - - -
. . 2(9)
2(2)
v'28
-8 ± - 7± J49+48
18 4
.m
- 8±2V7 .~ = _ 1 ±
±
or _4 tl 4 4
18
9 9
+
2
i x 2v'2x 2 = 0
-
= =
has a = 1, b - 2.J2, c 2
J
±
-(-2.J2) (-2.J2)2 - 4(1) (2)
x = - - - - - - - - - - -
. . 2(1)
± J8 -
2y'2 8
2
2y'2 ± 0
2
= v'2
+
a (x 2) ( x - 1) = 2 - 3x b ( 2x + 1) 2 = 3 - x
2 + 2 + +
. . x x 2x - 2 = 2 - 3x · 4x 4x 1 = 3 - x
-
2 + 2 +
.'. x 4x - 4 = 0 .'. 4x 5x - 2 = 0
which has a = 1, b = 4, c = -4 whicl1 has a = 4, b = 5, c = -2
·J
. x = - 4 ± 42 4(1)(- 4) . x = - 5 ± J52 4(4)(- 2)
- -
. . 2(4)
2(1)
J32
-4± -5 ± J25+32
2
8
- 4 ± 4v'2 ±
= .Y§2
_.§.
8 8
2
= -2 ± 2v'2
Mathematics HL (3rd edn), Chapter 1 - QUADRATICS 9
x- l
c (x - 2)2 = 1 + x d - - =2x+l
2 -x
2 + +
x 4x 4 = 1 x
-
+
x - 1 = ( 2x 1) ( 2 - x)
2 +
x - 5x 3 = 0
+
2
x - 1 = 4x - 2x 2 - x
which has a = 1, b = -5, c = 3
2x2 2x - 3 = 0
-
- (- 5) ± j(- 5)2 4(1)(3)
- which has a = 2, b = -2, c = - 3
x = - - - - - - - - - -
2(1)
-(-2) ± j(-2)2 - 4(2)(-3)
x = ---------------
5 ± v2s- i2
2(2)
2 v'28
±
2
=li±~ 4
2 2
2 ± 2y'7 -;;;
±
or l ::x..1.
2 2
4
1 1
e x - - = 1 f 2x - - =3
x x
x 2 1 = x 2x2 1=3x
- -
x 2 x - 1 = 0 2x2 3x - 1 = 0
- -
which has a = 1, b= - l c=- 1 which has a = 2, b = - 3, c = - 1
'
- (-1) ± j(-1)2 - 4(1)(-1) -(-3) ± j(-3)2 4(2) (-1)
-
= - - - - - - - - - - -
x = - - - - - - - - - - - x
2(1) 2(2)
±vi +4 3± v 9 +8
i
2 4
vs
= 1± =3 ±.:&
2 2 4 4
EXERCISE 18
,
=
a x2 + 7x - 3 0 b x 2 3x + 2 = 0
-
has a = 1, b = 7, c = -3 has a = 1, b = - 3, c = 2
2 = 2
~ = b - 4ac ~ b -4ac
= =
72 4(1)(- 3) (-3)2 4(1)(2)
- -
= 61 =1
> >
Since 0, there are two distinct Since 0 and is a square,
~ ~ ~
real solutions. there are two distinct rational solutions.
+
c 3x2 + 2x - 1 = 0 d 5x2 4x - 3 = 0
= = = = =
has a = 3, b 2, c - 1 has a 5, b 4, c - 3
2 = 2
~ = b 4ac ~ b 4ac
- -
= 22 4(3)(-1) = 42 4(5)(-3)
- -
=
16 = 76
> >
Since 0 and is a square, Since 0, there are two distinct
~ ~ ~
there are two distinct rational solutions. real solutions.
e x 2 + x +5 = 0 f 16x2 Bx + 1 = 0
-
= = =
has a 1, b 1, c 5 has a = 16, b = -8, c = 1
2 2
~ = b - 4ac ~ = b - 4ac
2 = 2
= 1 4(1)(5) (-8) 4(16)(1)
- -
=
- 19 = 0
<
Since 0, there are no real solutions. there is one repeated real solution.
~
