Answers [EXTENSION 1 MATHEMATICS HSC MATHS IN FOCUS]

562 Maths In Focus Mathematics Extension 1 HSC Course

Chapter 1: Geometry 2

Exercises 1.1

1. °°°°

( )

( )

( )

ABE ABDCBE CBD

ABDABE

ABD CBD

180180180

180straight angle

straight angle

given

+ ++ +

++

+ +

= −= −= −=

= —

2.

( )AFB is a straight angle+

° ( )°

° ( ° )

( )

( )

DFB xx

AFC xCFE x x

xAFC CFE

CD AFE

AFB180 180

180 180 2

bisects

is a straight angle

vertically opposite angles`

`

`

+

++

+ ++

+= − −=== − + −

==

3.

°WBC BCY x x2 115 65 2

180+ ++ = + + −

=

These are supplementary cointerior angles . VW XY` <

4. °°

( )x yA D

180180

given

` + ++ =

+ =

These are supplementary cointerior angles.

( )AB DC

A BAlso similarly`

+ +<

=

These are supplementary cointerior angles.

.AD BCABCD is a parallelogram

`

`

<

5. °

,

( )

( )

ADB CDBABD CBD

BD

ABD CBD

BD ABC

110

is commonby AAS

given

bisects

`

+ ++ + +

/∆ ∆

= ==

6.

,

( )

( )

( )

AB AEB E

BC DEABC AED

(a)

by SAS

given

base angles of isosceles

given

`

+ +

/∆ ∆

∆===

`

( )corresponding angles in congruent s∆

°°

( )

BCA EDA

ACD BCAEDA

ADCACD

BCD180180

(b)

since base angles are equal, is isosceles

is a straight angle

+ +

+ ++

+∆

=

= −= −=

7.

`

°

,

( )

( )

( )

( )

DC BCB D

DM AD

BN AB

DM BNMDC NBC

MC NCc

90

21

21

and

by SAS

given

given

given

orresponding sides in congruent s

`

`

+ +

/∆ ∆

∆

== =

=

=

=

=

8. °

( )

( )

( )

OCA OCBOA OB

OCOAC OBC

AC BC

OC AB

90

is commonby RHS,

bisects

given

equal radii

corresponding sides in congruent s

`

`

`

+ +

/∆ ∆

∆

= ==

=

9. ° ( )

( )

( )

CDB BECACB ABC

CBCDB BEC

CE BD

90

is commonby AAS,

altitudes given



`

`

+ ++ +

/∆ ∆

∆

∆

= ==

=

10.

`

( )

( )

( )

( )

AB ADBC DC

ACABC ADCDAC BAC

AC DABBCA DCA

AC DCB

is commonby SSS,

So bisectsAlso

bisects

given

given

corresponding angles in congruent s

corresponding angles in congruent s

`

`

+ +

++ +

+

/∆ ∆

∆

∆

==

=

=

11. ( )

( )

NMO MOPPMO MON

MO

MNO MPO

MN PO

PM ON

(a)

is commonby AAS,

alternate angles,

alternate angles,

`

+ ++ +

/

<

<

∆ ∆

==

( )

( )

( )

PMO MONMN NO

MON NMOPMO NMOPMQ NMQ

PM ON(b)

i.e.

alternate angles,

given


`

+ +

+ ++ ++ +

<

∆

=====

)(corresponding sides in congruent s∆

( )

( ( ))

MN NOPM NO

PM MNPMQ NMQ

MQPMQ NMQ

(c)

is commonby SAS,

given

from b

`

`

+ +

/∆ ∆

==

==

`

( )corresponding angles in congruent s∆

°°

( )

MQN MQP

MQN MQPMQN MQP

PQN18090

(d)

But straight angle

+ +

+ ++ +

+

=

+ == =

12.

Answers

563ANSWERS

Let ABCD be a parallelogram with diagonal AC .

,

( )

( )

DAC ACBBAC ACD

ACAAS ADC ABC

AD BC

AB DC

is commonby

alternate angles,

alternate angles,

`

+ ++ +

/

<

<

∆ ∆

==

13.

`

`

( )

( s)

BD

ADC ABCADC ABC

A C

12

Similarly, by using diagonal we canprove

opposite angles are equal

see question

corresponding angles in congruent+ +

+ +

/∆ ∆∆=

=

14.

`

.

( )

( )

( )

AB DCBM DN

AB BM DC DNAM NC

AM NC

CN

ABCD

i.e.AlsoSince one pair of sides is both parallel andequal, is a parallelogram

opposite sides in gram

given

is a gram

AM

<

<

<

==

− = −=

15.

`

( )AD BCBC FEAD FE

opposite sides in gram

( )similarly

<===

`

( )

( )

AD BCBC FEAD FE

ABCD

BCEF

Alsoand

is a gram

is a gram

<

<

<

<

<

Since one pair of sides is both parallel and equal, AFED is a parallelogram.

16.

`

`

( )

( )

DEC DCEDEC ECBDCE ECB

CE BCD

AD BCAlso,

bisects


alternate angles,

+ ++ ++ +

+

<

∆===

17.

`

`

(corresponding sides in congruent s)∆

.

AB CDBAC DCA

ACABC ADC

AD BC

ABCD

is commonby SAS,

Since two pairs of opposite sides are equal,is a parallelogram

(given)

(given)+ +

/∆ ∆

==

=

18.

`

`

(a)( )diagonals bisect each other in gram<

(corresponding sides incongruent s)∆

°

AE EC

AEB CEBEB

ABE CBEAB BC

90is commonby SAS,

(given)+ +

/∆ ∆

=

= =

=

(b) (corresponding angles in congruent s)∆

ABE CBE+ +=

19.

Let ABCD be a rectangle

`

BCD/ ∆`

°

AD BCD C

DCADC

AC DB

90is common

by SAS,

(opposite sides in gram)

(corresponding sides in congruent s)

+ +<

∆

∆

== =

=

20.

Let ABCD be a rectangle with °D 90+ =

`

)

( )

D C AD BC

B C AB DC

( and cointerior angles,

and cointerior angles,

+ +

+ +

<

<

( )A B AD BCand cointerior angles,+ + <

° °

°° °

°° °

°

C

B

A

180 90

90180 90

90180 90

90all angles are right angles

` +

+

+

= −

== −

== −

=

21.

`

`

( )

(

AD CDAD BCAB CDAB AD BC CD

Also

all sides of the rhombus are equal

given

(opposite sides of gram)

similarly)

<

==== = =

22.

`

`

(

)

BE ADAD BEAD BCBE BC

BCD BECADC BECADC BCD

AD BE

ConstructThenButThen

Also,


given)

(base angles of isosceles )

(corresponding angles,

+ ++ ++ +

<

<

<

∆

======

23.

`

`

(corresponding angles in congruent s)∆

AD ABDC BC

ACADC CADC ABC

is commonby SSS,

(given)

(given)

+ +/∆ ∆ΑΒ

==

=


24.

`

`

(corresponding sides in congruent s)∆

°( )—

AD BCD C

DE ECADE BCE

AE BE

E

90

by SAS,


(given)

is the midpoint given

+ +

/

<

∆ ∆

== ==

=

25. (a)

`

(

AD BCAB DC

DBADB BCD

is commonby SSS,


similarly)

/

<

∆ ∆

==

(b) (corresponding angles in congruent s)∆

ABE CBE+ +=

`

(c)( )

AB BCABE CBE

BEABE CBE

is commonby SAS,

(adjacent sides in rhombus)

found+ +

/∆ ∆

==

`

(d)

But

(corresponding angles in congruent s)∆

°°

(

AEB BEC

AEB BECAEB BEC

AEC18090

is a straight angle)

+ +

+ ++ +

=

+ == =

Exercises 1.2

1. (a) 14 452 mm2 (b) 67 200 mm3 2. 90 m3

3. V x x x2 3 23 2= + −

4. ππ

π

π

V r hr h

hr

hr

250250

250

2

2

2

==

=

=

5. πV 5 3= r 6. Ab2

3 2

= 7. V xx x x

26 12 8

3

3 2

= += + + +] g

8. πS h24 2=

9. V x xx x x3 2

4 12 9

2

3 2

= −= − +

] g

10. 262 cm 3 11. V h h3 22= +

12. V h h2 52= +

13. ( )V x x x31

6 5 34 153 2= − − −

14. (a) V x x x18 12 23 2= − + (b) S x x54 30 42= − +

15. l h r2 2= +

16. Vx y

π4

2

= 17. hrπ

4002

= 18. h rr

ππ750 2

=−

19. l rr

ππ850 2

=−

20. y x810 000 2= −

Exercises 1.3

1. Show m m 4AB CD= = and

m m74

AD BC= = −

2. Show ;´ ´m m47

74

1AC BC = − = −

∴ right-angled triangle with °C 90+ =

3. (a) ,AB AC BC73 6= = = (b) 8 units (c) 24 units 2

4. Show m m51

XY YZ= =

5. (a) Show ,AB AD 26= =

BC CD 80= = (b) Show ´m m 1AC BD = −

( , ), ,E CEAE

1 2 72 6 218 3 2

(c) = − = == =

6. r 1=

7. (a) x y2 3 13 0− + = (b) Substitute (7, 9) into the equation (c) Isosceles

8.

`

°AOB COD

OBOD

OAOC

OBOD

OAOC

90

24

2

714

2

+ += =

= =

= =

=

Since 2 pairs of sides are in proportion and their included angles are equal, .OCD∆;OAB<∆

9. (a) OB is common OA BC

AB OCOAB OCB

5

20by SSS` /∆ ∆

= == =

(b) m m m m131

2Show andOA BC AB OC= = = = −

10. °ABC 90+ = and AB BC 2= = So ABC is isosceles CAB ACB` + += But °CAB ACB 90+ ++ = (angle sum of triangle) CAB ACB 45` + += = Similarly, other angles are 45°.

11. PR QS 145= = Since diagonals are equal, PQRS is a rectangle.

12. (a) ( , ), ,X Y2 2 1 0= − = −^ h (b) m m 2XY BC= = − So XY BC< (c) ,XY BC5 20 2 5= = = So BC XY2=

565ANSWERS

13. ´ ´m m 1 1 1AC BD = − = − So AC and BD are perpendicular.

,AC BDa a2 2

Midpoint midpoint= = −d n

So AC and BD bisect each other. So AC and BD are perpendicular bisectors.

14. (a) X Y 1Distance from distance from unit= =

(b) ,Z41

0= d n

(c) 141

units 2

15. Midpoint AB : ,W 2 121= −d n

Midpoint BC : ,X 2 3= − −^ h

Midpoint CD : ,Y 421

21= −d n

Midpoint AD : ,Z21

2= −d n

m m83

WX ZY= =

So WX ZY<

m m57

XY WZ= = −

So XY WZ< WXYZ is a parallelogram.

Exercises 1.4

1. (a) AOC BOC+ += (equal s+ on equal arcs) But °AOC BOC 180+ ++ = ( AOB+ straight line) `

°AOC BOCOC AB

90=

+ += =

`

( )

( )

( )

( )

° °°°° °°

÷OAC OCA

OCA

OCBACB

180 90 2454545 4590

(b) base s of isosceles

sum of

similarly

+ +

+

++

+

+

∆

∆

== −=== +=

2. (a) 40°

(b) πr9

2units

3.

`

`

( )

( )

( )

(

OC OACOD AOB

OD OBOCD OAB

CD ABby SAS,

equal radii

given

equal radii

corresponding sides in congruent s)

+ +

/∆ ∆

∆

===

=

4.

`

`

( )

( )

( )

( )

AB CDOA PCOB PD

AOB CPDAOB CPD

by SSS,

given

equal radii

similarly

corresponding s in congruent s

+ ++

/∆ ∆

∆

===

=

5.

`

`

)s in semicircle°

(

( )

( )

ABC ADCAC

BC DCABC ADCBAC DAC

s

90is common

by RHS,given

corresponding s in congruent

+ +

+ +

+

+

/∆ ∆

∆

= =

=

=

6.

`

`

( )

( )°°°°

AOC y

AOC xy xy x

x y

2

360 22 360 2

180180

Reflex

Butat centre is double the at the circumference

of revolution

+

++ +

+

=

= −= −= −

+ =

7.

`

(a) ( )

( , )

BCA BACBCA CADBAC CAD

AC BAD

BC AD

bisects

base s of isosceles

alternate s

`

+ ++ ++ +

+

+

+ <

∆===

`

But`

(b)

`

°°

°

( )

( )

( )

ACDCAD ADCCAD BACBAC ADC

BAC ADC

9090

90and are complementary

in a semicircle

sum of

found

++ ++ ++ +

+ +

+

+ ∆== −== −

8.

`

`

`

`

`

But

( )

( )

( )

( )

( )

( )

( )

( )

°

°

°°

AB BCAOB COB

AO COOAC OCA

ADO AOB OAC

COB OCACDO

ADO CDOADO CDO

BO AC

ADC

180

180

18090

given

equal chords subtend equal s

equal radii

base s of isosceles

sum of

straight

=

+ +

+ +

+ + +

+ ++

+ ++ +

+

+

+

+ +

∆

∆

==

==

= − +

= − +=

+ == =

9.

`

`

`

( )

( )

( )

( )

( )

( )

XOZ XYZ

OX OYYXO XYO

OY OZYZO ZYOXYZ XYO ZYO

YXO YZOXOZ YXO YZO

2

2

at centre is double at circumference

equal radii

base s of isosceles

equal radii

base s of isosceles

+ +

+ +

+ ++ + +

+ ++ + +

+ +

+

+

∆

∆

=

===== += += +

10.

`

Let

( )( )

( )°°

BAE y DAE xBAD x yBDC yDBC xBCD x y

BADBCD180

180

andThen

( s in same segment)

similarly

sum in

+ +++++

+

+

+ ∆

= == +=== − += −


11.

`

° °

° ( ° ° )

( )

( )

RQP xMQPMQR xQMR x

xQMR RQP

MQR

9090180 90 90

(a) Letin semicircle

sum of

`

++++

+ +

+

+ ∆

=== −= − + −

==

`

`

`

( )

( )°

RQP NMRNMR QMRNRM QRM

MRMNR MQR

MN MQ

90

(b)

is commonby AAS,

s in same segment

given


+ ++ ++ +

+

/∆ ∆

∆

=== =

=

12.

`

`

( )

( )

DAC CDECDE BACDAC BAC

AE DABbisects

given

s in same segment

+ ++ ++ +

+

+

===

13. ° (

( )

( )

FEH FGHFH

EF FGEFH GFH

EH GH

EFGH

90is common

by RHS,

is a kite

s in semicircle)

given


`

`

`

+ + +

/∆ ∆

∆

= =

=

=

14. °°

°

(

( )

( )

ADBCDB ADB

BC

ADC

CDB

9018090

is a diameter

in semicircle)

straight

angle in semicircle`

++ +

+

+ +

+

== −=

15.

`

`

(

( )

( )

BDC BACAO CO

BAC ACOBDC ACO

s in same segment)

equal radii

base s of isosceles

+ +

+ ++ +

+

+ ∆

====

16. (

( )

OAC ECBC DCAB AC BC

EC DCDE

perpendicular from bisects chord)

similarly

=== −= −=

17.

`

`

(

( )

OF OE

OEFOFE OEF

is isoscelesequal chords are same distance from centre)

base s of isosceles+ + +

∆∆

=

=

18. °

( )

( )

( )

OXE OYEOE

AB CDOX OY

90is common

given

given

equal chords are same distance from centre

`

+ += =

==

` OXE OYEby RHS, /∆ ∆

`

`

( )

( )

O

XE YE

AB CD

AX AB

CD

DY

21

21


perpendicular from bisects chord

similarly

∆=

=

=

=

=

Also,

` AE AX XEDY YEDE

CE DC DEAB AEBE

= += +== −= −=

19.

( )OBD 3 pairs of s equal+;∆

`

`

( )

(

( )

( )

( )

°° BC)

ABCODBABC ODB

OB OCOBD ACBBAC BODABC

OD

9090

(a) in semicircle

bisects chord

equal radii

base s of isosceles

sum of s

`

`

+++ +

+ ++ +

+

+

+

<∆

∆∆

======

`

.[ ]ABC ODC

AB OD

(b)These are equal corresponding angles

from (a)+ +

<

=

20.

`

`

( )

( )

CE EGAD EO FB

EFED

OBAO

AO OBED EFCD CE ED

EG EFFG

OE CGbisects

given

(equal intercepts)

(equal radii)

< <

=

=

=== −= −=

21. ( )

( )

( )

OA OBPA PB

OAPBOP AB

(a)

is a kite

equal radii

similarly

diagonals of kite

`

` =

==

(b) OA 18cm=

22. (a) Since CE is the perpendicular bisector of AB , it must pass through the centre of the circle. DE is the perpendicular bisector of AB similarly, so it passes through the centre. CF and GD are diameters (b) .CE 28 125 cm=

23. AB DC AD BC ABCDand so is a parallelogram.< <

`

`

°

°° °

°°

( )

( )

)

A C

A BC BC B

C BC B

ABCD

AD BC

AB DC

180

180180 180

18090

But

is a rectangle

opposite s of cyclic quad.

cointerior s,

(cointerior s,

`

`

+ +

+ ++ ++ +

+ ++ +

+

+

+

<

<

= −

= −− = −

=+ =

= =

567ANSWERS

24.

°

°° °

°

( )

)

B D

AC

180180 9090

90is a diameter

opposite s in cyclic quad.

( in semicircle is`

+ + +

+

= −= −=

25.

( )But

is a diameter

(given)

(given)

sum of

( of cyclic quadrilateral)

( in semicircle is 90 )

°

°AC °

DAC BACBCA DCAADC ABCADC ABC

ADC ABC

180

90opposite s

`

`

`

+ ++ ++ + ++ +

+ ++

+

∆

==== −

= =

26.

` ( )

equal radii

(base of isosceles )

(opposite of cyclic quad.)

°

°°

OB OCOBC OCB yABC ABO OBC

y zODC ABC

x y zx y z

180

180180

s

s

`

`

`

+ ++ + +

+ +

+

+

∆== == += += −

= − ++ + =

^ h

27.

( )

(base s of isosceles )

( sum of )

( in cyclic quad.)

°°° °

θθ

θθ

BAC BCAABCADC ABC

180 2180180 180 22

opposite s

`

+ +++ +

+

+

+

∆∆

= == −= −= − −=

28.

CED;∆

is common(ext. equals int. opposite )

( sum of )

DDEC ABDDCE DABADB

`

`

++ ++ +

+ +

+

<∆∆

==

29.

These are equal corresponding angles.

(base of isosceles )

( equals int. opposite )

DEC DCEDEC ABDDCE ABD

AB EC

s

ext.

`

`

+ ++ ++ +

+

+ +

<

∆===

30. LetThen

is a cyclic quadrilateral.

( )

Also, is a cyclic quadrilateral.

( )

( of cyclic quad.)

(opposite of cyclic quad.)

°

°

° °

°

FAB xBCD x

BCDEBED BCD

xx

ABEFBEF x

AF CD180

180

180 180

180

cointerior s,

opposite s

s

`

`

++

+ +

+

+

+

+

<

== −

= −

= − −=

= −

is a straight angleand are collinear

°°

FED BED BEFx x

FEDF E D

180180

,`

`

+ + +

+

= += + −=

31.

is common

by ,

( radius)

equal radii

°BAO BCO

OBOA OC

OAB OCB

90(a)

RHS

tangent

`

=

+ +

/∆ ∆

= =

= ^ h

(b) (corresponding in congruent s)

AOB COBs

+ ++ ∆

=

32. Let

( )

(tangent radius)

( in semicircle)

( sum of )

°°°

° ° °

ACD xACODCE xCDECED x

xACD CED

909090180 90 90

`

`

`

=

+++++

+ +

+

+ ∆

=== −== − + −

==

33. ( )°°

OAB OCBOAB OCB

90180

tangent radius

`

=+ ++ +

= == −

`

( )

°

° °°

OAB OCB AOC ABC

AOC ABCAOC ABC

360

180 360180

sum of quad.

+ + + +

+ ++ +

+

+ + + =

+ + =+ =

Opposite angles are supplementary. OABC is a cyclic quadrilateral.

34. BD and AB are both tangents to the larger circle. AB BD` = BD and CB are both tangents to the smaller circle. BD CB

AB CB`

`

==

35. ( )

( )

( )

AE BECE DE

CEAE

DEBE

AB CD

tangents from external point are equal

similarly

equal intercepts

`

` <

==

=

36. ,

( )°AD BD

ADO BDOOD

D AB

90is common

( bisects given)

tangent radius=+ +== =

, AOD BODAO BO

by SAS


`

`

/∆ ∆

∆=

37. (a) ( )

AE BEDE CEAC AE CE

BE DEBD

(tangents from external point are equal)

similarly

=== += +=

( )

ECAE

EDBE

AB DC

(b)

equal intercepts` <

=

38. ( )

( , )

( )

( )

BC DCABD CBDABD ADBCBD BDC

BD ABC

ABD

DBC

tangents from external point are equal

bisects given

base s of isosceles

base s of isosceles

+ ++ ++ +

+

+

+

∆∆

====

ADB BDCBDADB CDB

AB BC AD DC

AB BC AD DCABCD

Also, is commonby AAS,

and

is a rhombus


`

`

`

`

`

+ +

/∆ ∆

∆

=

= =

= = =


39.

(OCD 3 pairs of s equal)+;∆

°( )

( )

( , )

( )

OXA OYC

AB CD

OAX OCY

OBX ODYOAB

AB CD

90(a)

and

tangent radius

corresponding s equal

corresponding s

similarly

`

`

`

=

+ +

+ +

+ +

+

+

<

<

<

∆

= =

=

=

(b) .CD 19 2 cm=

40.

.

( )

( )

°°°°

OBAPBA

OBA PBAOBPOBP

O P B

9090180180i.e.

is a straight angleSo , and are collinear

tangent radius

similarly

`

`

=++

+ +++

==

+ ==

41.

`

`

`

( , )

( )

ECB xBCA xECB BAC

BAC xECA x x

xADC xADC BAC

BC ECA

222

LetThenAlso, and are s inalternate segments

bisects given

s in alternate segment

+++ + +

++

++ +

+

+

==

== +===

42. CE is a diameter of the smaller circle (line through centres passes through point of contact).

CDE∆;

(

(

°CDE ABCBCA DCEBAC CEDABC

90 in semicircle)

(vertically opposite s)

sum of )

`

`

`

+ ++ ++ +

+

+

+

<∆∆

= ===

43. (

(

EAB BCABOA BCA

BOA EAB

2

2

s in alternate segment)

at centre double at circumference)

`

+ ++ +

+ +

+

+ +

==

=

44.

,

(

( )

(

°°

° °

BCE xBAC xACF xACB

ACF FCBx FCB

x FCBFCB BCE x

BC FCE

ACF9090

90 90

LetThen

Sincebisects


sum of

in semicircle)

`

`

`

++++

+ ++

++ +

+

+

+

+

∆

=== −== += − +== =

45. ( , )

(

ACE DECEDC ACEEDC DECEDC s

DE BA

is isoscele

alternate s


(base s equal)

`

`

+ ++ ++ +

+

+

+

<

∆

===

46. ( )( )

´´´

CB BD BEBD BD DEBD BD BD

BD

CB BDCB AB

AB BD

B

45

5

5

But(tangents from external point are equal)

2

2

`

`

== += +===

=

47.

BEC;∆

(

)

EBC EDBDBE BECBDE

DB EC


(alternate s,

`

+ ++ +

+

+

<

<

∆

==

48. °

°° °

°.

(

( )

EFC EDF

FEC EDFEFC FEC

AC BC

90

9090 90180

These are supplementary cointerior s

the diagram is impossible


similarly

`

`

`

+ +

+ ++ +

+

+

<

= =

= =+ = +

=

49. (a) (

(

EDF ECDBAE ECD

EDF BAE


ext. equal to int. opposite in cyclic quad.)

`

+ ++ +

+ +

+

+ +

==

=

(b) Ð D is common

;

ECD BADADB EDC

[from (a)]

`

+ +

<∆ ∆=

(c) [ ]EDF BAE from (a)+ += These are equal alternate angles. AB FG` <

50. (a) ( , )AY BYAZ AY XB BY

Y

andmidpoint given

(tangents from same external point are equal)

== =

`

( )

BX AZCZ CXAC AZ CZ

BX CXBC

Also, similarly

=== += +=

(b) ( ,

AZ XBABC AC BCis isosceles

[from (a)]

found)∆=

=

( )

ZAY XBY

AY YBAZY BXYby SAS

base s of isosceles

(given)

`

`

+ ++

/∆ ∆

∆=

=

)(corresponding sides in congruent s∆

( s in alternate segment)+

( )

AYZ YXZ

AZY ZYXZY XY

XYZ AZY

(c)

by AAS,

similarly

`

+ +

+ +

/∆ ∆

=

==

569ANSWERS

51. (a) OA is diameter of the smaller circle (line of centres passes through point of contact A ). °

°.

(

( )

OCABDC

OCA BDCOC BD

9090

and are equal corresponding angles

in semicircle)

similarly

`

`

++

+ +

+

<

==

( , ,AE OA BO OAgiven and equal radii)= =

`

( )( )

´´´

EF AE BEAE BO OA AEAE AE

AE

EF AE

3

3

3

(b) 2

2

== + +=

==

(c) .OC 3 5 cm=

52.

, ,

(

( )

( )

)

( )

°°

°°

CEG x DEF yBAD x ABC y

DCE CEG xBCD DCE

xBCD BADCDE DEF yADC CDE

yADC ABC

ABCDA B C D

DC FG

BCE

DC FG

ADE

180180

180180

Let andThen and

and are supplementary

and are supplementarySince opposite angles are supplementary,

is a cyclic quadrilateral.and are concyclic points


alternate s,

straight

(alternate s,

straight

`

`

`

+ ++ +

+ ++ +

+ ++ ++ +

+ +

+

+

+ +

+

+ +

<

<

= == =

= == −= −

= == −= −

Test yourself 1

1. (a) AB AC= (given) So BD EC= (midpoints)

)DBC ECB (base s in isoceles+ + + T= BC is common ( )BEC BDC SAS` /∆ ∆ (b) s)TBE DC (corresponding sides in` /=

2. (a) °, °x y94 86= = (b) ° °

°A D 94 86

180+ ++ = +

=

These are supplementary cointerior s.+ AB DC` <

3.

12

y12

x

c a b

x y

x y

x y

cx y

x y

2 2

4 4

4

4

2

2 2 2

2 2

2 2

2 2

2 2

2 2

= +

= +

= +

=+

=+

=+

d en o

4. 2 2

2 2

2 2

( ) ( )

( ) ( )

( ) ( )

AB

BC

AC

7 4 5 15

7 1 5 310

4 1 1 35

= − + − − −== − + − −== − + − −=

Since ,≠AB AC BC= ABC∆ is isosceles.

5. h ππ50 2

=−r

r

6. °° °°

( )

( )

ADEEAD

ADE

BC AD4590 4545

(a)

So is isosceles.

corresponding s

sumof

++

+

+

<

∆

∆== −=

(b) AE DE yCD yAB yAB CE

( , )

(

CD DE

(isoceles )

given

oppositesides in gram)

(given)<

= === <

∆=

So ABCE is a trapezium.

( )

( )´ ´

´ ´

A h a b

y y y

y y

y

21

21

2

21

3

2

3 2

= +

= +

=

=

7.

`

`

( )

( )

( )

°

°

COB x y

OC OBOCB OBC

OCBx y

x y

2

2

180 2

90

( at centre twice at circumference)

(equal radii)

(base s in isosceles )

+

+ +

+

+ +

+ ∆

= +

==

=− +

= − +

8. (a)

`

BACB

GFCG

GFCG

DECD

BACB

DECD

(equal ratio of intercepts)

(similarly)

=

=

=

(b) 20.4 cm

9.

,

m

m

m

m

BC AD CD ABABCD

2 54 4

0

5 64 3

7

1 63 3

0

2 14 3

7

So is a parallelogram.

BC

CD

AD

AB

< <

=− −

− − − =

=−

− − =

=− −

− =

=− − −− − =

10.

`

.

DEB DBEDEB CBDDBE CBD

BD EBCSo bisects

(base s of isosceles )

( s in alternate segment)

+ ++ ++ +

+

+

+

∆===


11.

` °

DC BC

DBC

ABCD

12 5144 2516913

90So is a rectangle.

2 2 2 2

2

2

(Pythagoras)+

+ = += +====

12.

WXY∆;`

..

..

.

39°Y PPQR

3 49 18

1 95 13

2 7

(two pairs of sides in proportion, with included s equal)

(given)+ +

+

<∆

= =

= =

13.

`

`

OA ODOB OC

AOB CODABO CDO

AB CD

(equal radii)

(similarly)

(given)

(SAS)

(corresponding sides in s)

+ +/

/

∆ ∆∆

===

=

14.

`

—

( )

BAD

BAC xDAC xDCA xBAC DCA

A

DAC

LetThen (given C bisects )

base s of isosceles

++++ +

+

+ ∆

====

These are equal alternate angles.

°

°°

°

( )

( )

AB ECEDA xDEA DAE

DAEx

xEAC x x

EAC ACB

AED

2

2180 2

909090

base s of isosceles

sum of isosceles

)DAC(exterior of`

`

++ +

+

+

+ +

+

+

<

∆

∆

==

= −

= −= − +==

+ ∆

These are equal alternate angles. AE CB` < So ABCE is a parallelogram.

15. ,1 0−^ h

16.

( )CDE AAA∆;

A DB E

ACB DCEABC

(a) ( s in same segment)

(similarly)

(vertically oppositeangles)

`

+ ++ +

+ +

+

<∆

===

(b) . .x y2 9 7 7cm, cm= =

17. (a) x y4 3 1 0− − = (b) 2.4 units (c) 12 units 2

18.

`

AB ADBC DC

ACABC ADC

(a)

is common(SSS)

(given)

(given)

/∆ ∆

==

AB ADBAE DAE

AEABE ADE

(b)

is common(corresponding s in s)

(SAS)

(given)

`

+ + +

/

/

∆ ∆

∆==

(c)`

`

°°

.

BE DEAC BDBEA DEA

BEA DEABEA DEA

AC BD

18090

bisects

But

So is perpendicular to

(straight )

(corresponding sides in s)

(corresponding s in s)+ ++ ++ +

+

=

=+ =

= =

+

/

/

∆

∆

19. : , :

´

AB m BC m

m m

AB BC

21

2

21

2 1

(a)

So and are perpendicular.

1 2

1 2

= = −

= − = −

(b) ( , )3 2−

(c) ( , )321

(d) 5 units

20. (a) x xhx xhx xh

xx

h

500 4 6500 4 6250 2 3

3250 2

2

2

2

2

= +− =− =− =

(b) V x h

xx

x

xx

x x

2

23

250 2

23

250 2

3500 4

2

22

2

3

=

= −

= −

= −

e

e

o

o

Challenge exercise 1

1. BD is common °

( )—

ADB CDBAD DC

ABD CBDAB BC

ABC

BD AC

90

by SAS,

So is isosceles.

(given)

bisects given


`

`

+ +

/∆ ∆

∆∆

= ==

=

2.

`

`

`

AD AB

ABAD

AE AC

ACAE

ABAD

ACAE

A

21

21

21

21

(a)

is common+

=

=

=

=

=

Since two pairs of sides are in proportion and their included angles are equal, ;

`

ABC∆ADEADE ABC

DE BCThese are equal corresponding angles`

+ +<

<

∆=

571ANSWERS

`

,ABC; Δ

`

ADE

ABAD

BCDE

BCDE

DE BC

21

21

21

(b) Since <Δ

= =

=

=

3.

Let ABCD be a rhombus with AD DC= To prove: ADE CDE+ +=

Proof `

`

`

`

( )

AD DCADC

DAE DCEAE EC

ADE CDEADE CDE

is isosceles

by SAS,

(given)

diagonals bisect each other

(corresponding angles in congruent s)

+ +

+ +/

Δ

Δ Δ

Δ

=

==

=

(Note: We can prove other pairs of angles equal similarly.)

4. ´1189 841mm mm

5. S x x240002= +

6.

°

°

( )

( )

´n

n

nn

n

2 180

180 360

180360

Each angle =−

=−

= −

°

c m

7. Assume XY is not a tangent to the circle. Draw a circle through, A, B and C so that XY cuts the circle at C and D .

`

`

`

`

°

( a )

BAC xCDB xBDY xBAC BCYBDY BCY x

BC BDXY

CDY

180LetThen

These are equal corresponding angles.

is a tangent to the circle

(opposite s in cyclic quad.)

str ight

(given)

(impossible!)

++++ ++ +

+

+ +

<

== −=== =

8. 70 cm

9. (a) ( , )

(

OCA CABBAD BCAOCB OCA BCA

CAB BADCAD

CO BAalternate s


+ ++ ++ + +

+ ++

+

+

<=== += +=

(b) °

°

(

a

CBA CAEOAECAE OAE CAOCBA CAO

90

90


(tangent r dius)

`

=

+ +++ + ++ +

+=== += +

10.

EAF ACDADC ABCABC AEFADC AEF

(alternate angles, )

(opposite angles in gram)

(corresponding angles, )

AB DC

BC EF

`

+ ++ ++ ++ +

<

<

<

====

Since 2 pairs of angles equal, third is equal by angle sum of Δ ADC;ΔAEF` <Δ

11.

(a) Midpoint of : , ( , )BDa a b b

a2 2

0+ + − =d n

Midpoint of : , ( , )ACa

a2

0 22

0 00

+ + =d n

diagonals BD and AC bisect each other at E (a, 0) BD is vertical and AC is horizontal the diagonals are perpendicular

Ans_PART_1.indd 571Ans_PART_1.indd 571 6/30/09 12:06:16 PM6/30/09 12:06:16 PM


(b) 2

2

2

( ) ( )

( ) ( )

( ) ( )

( ) ( )

AB a ba b

BC a a ba b

CD a a ba b

AD a ba b

0 0

2 0

2 0

0 0

2

2 2

2

2 2

2 2

2 2

2

2 2

= − + −= += − + −= += − + += += − + − −= +

all sides are equal

(c)

`

`

(from (a))(from (a))

is common

So bisects(

BC CDBE EDCESSS CBE

BCE DCEAC BCD

bycorresponding s in congruent+ +

++

/∆ ∆∆

==

= s)

CDE

12. °

DE BEAEB AED 90

(digonals bisect in gram)

(given)+ +<=

= =

AE is common by SAS, ADE ABE/∆ ∆ AB AD (corresponding sides in congruent` ∆= s)

13. P RPACP

RBCR

11

( and are midpoints)= =

PR AB (equal ratios on lines)` < < Similarly PQ CB QR ACand< < PQ CB

AC RQ

QPR PRCCPR PRQ

(alternate s, )

(alternate s, )

+ ++ +

+

+

<

<

==

PR is common by AAS, PQR CPR/∆ ∆

14.

`

`

`

°°°

( ))

( )

OBD xADBOAD xOBCDCB xOBD DCB

ABD

ABC

909090

(a) Let( in semicircle)

sum of(tangent radius

sum of

=

++++++ +

+

+

+

∆

∆

=== −===

(b) °

° ( ° )

( )

( )

( )

OA ODOAD ODA x

AOD xx

DCB

AOD

90

180 2 9022

equal radii

base s of isosceles

sum of

+ +

+

+

+

+

∆∆

== = −

= − −==

15.

`

( )

)

ACD DGCACE EFCDCE ACE ACD

EFC DGCFCG DGC EFC

FCG EFC DGCDCE

FGC

( s in alternate segment)similarly

(external of

+ ++ ++ + +

+ ++ + +

+ + ++

+

+ ∆

=== −= −

+ == −=

16. Draw in OD . °

( )

( )

O

D

ODA ODC

OC OACD AD

90Then(line bisecting chord from is to it)

equal radii

given, midpoint

=

+ += =

==

OCD OADAOD COD

COA AOD CODAOD

COA CBA

AOD CBA

22

by RHS,

But

(corresponding s in congruent s)

( at centre twice at circumference)

`

`

`

+ +

+ + ++

+ +

+ +

+

+ +

/∆ ∆

∆=

= +==

=

17. (a) ( ) r2 3 units+ (b) ( )r2 2 3 units+ (c) ( )r2 5 3 units+

18. FE EB EB EDand= = (tangents from external point are equal) circle can be drawn through F , B and D with centre E FBD`+ is the angle in a semicircle °FBD 90`+ =

19. R r3 + units

20. 188 mm

21. P SDADP

DCDS

D21

(a)

is common

( , are midpoints)

+

= =

Since 2 pairs of sides are in proportion and the included angles are equal, .DAC;∆DPS<∆

(b) PADP

SCDS

11= =

(PS AC equal rations on lines)` < <

Similarly, QA

BQ

RCBR=

QR AC` <

PS QR` <

(c) ( ,PDAP

QB

AQP Q

11

are midpoints)= =

PQ DB (equal ratios on lines)` < <

Similarly SR DB< PQ SR` < Since , .PS QR PQ SR< < PQRS is a parallelogram.

573ANSWERS

Chapter 2: Geometrical applications

of calculus

Problem

. , .0 25 1 125− −^ h

Exercises 2.1

1. (a)

(b)

(c)

(d)

2. x41< 3. x 0<

4. (a) .x 1 5< (b) .x 1 5> (c) .x 1 5=

5. ( )x 2 0<= −fl for all x

6. y x3 0>2=l for all ≠x 0

7. ( , )0 0 8. ,x 3 2= − 9. (a) ( , )1 4− (b) ( , )0 9

(c) ( , )1 1 and ( , )2 0 (d) ( , ),0 1 ( , )1 0 and ( , )1 0−

10. ( , )2 0 11. x1 1< <− 12. ,x x5 3< >− −

13. (a) ,x 2 5= (b) x2 5< < (c) ,x x2 5< >

14. p 12= −

15. ,a b121

6= = −

16. (a) dx

dyx x3 6 272= − +

(b) The quadratic function has a 0>b ac4 288 0<2 − = −

x x x3 6 27 0So for all>2 − +The function is monotonic increasing for all x.

17.

x2

y

18.

x4

y

19.

1

y

x

20.

5-2

y

x


21.

x

y

3

2

22.

x

y

-1-2

23. ( , )2 0 and ,32

38113

c m

24. ; ,x

xx

x

x

2 11

2 1

3 232

92 3

++ + =

++ −

−f p

25. .a 1 75= −

26. x2

10! 27.

x3

04!

−

Exercises 2.2

1. ( , ); y y0 1 0 0on LHS, on RHS< >− l l

2. ( , )0 0 minimum 3. ( , )0 2 infl exion

4. ( , );2 11− show ( )f x 0>l on LHS and ( )f x 0<l on RHS.

5. ( , )1 2− − minimum 6. ( , )4 0 minimum

7. ( , )0 5 maximum, ( , )4 27− minimum

8. ( ) , ( )f f x0 0 0>=l l on LHS and RHS

9. ( , )0 5 maximum, ( , )2 1 minimum

10. ( , )0 3− maximum, ( , )1 4− minimum, ( , )1 4− − minimum

11. ( , )1 0 minimum, ( , )1 4− maximum

12. m 6121= −

13. x 3= − minimum 14. x 0= minimum, x 1= − maximum

15. x 1= infl exion, x 2= minimum

16. (a) dxdP

x2

502

= −

(b) (5, 20) minimum, ,5 20− −^ h maximum

17. ,121

d n minimum

18. (2.06, 54.94) maximum, . , .20 6 54 94− −^ h minimum

19. (4.37, 54.92) minimum, . , .4 37 54 92− −^ h maximum

20. (a) dxdA

xx

x

x

x

36003600

3600

3600 2

2

2

2

2

2

= − −−

=−

−

(b) (42.4, 1800) maximum, . ,42 4 1800− −^ h minimum

Exercises 2.3

1. ; ;x x x x x x7 10 4 1 42 40 126 4 3 5 3 2− + − − + ;x x x x x210 120 24 840 240 244 2 3− + − +

2. ( )f x x72 7=m 3. ( ) , ( )f x x x f x x x10 3 40 64 2 3= − = −l m

4. ( ) , ( )f f1 11 2 168= − =l m

5. ; ;x x x x x x7 12 16 42 60 486 5 3 5 4 2− + − + x x x210 240 964 3− +

6. ,dx

dyx

dx

d y4 3 4

2

2

= − =

7. ( ) , ( )f f1 16 2 40− = − =l m 8. ,x x4 205 6− − −

9. ( )g 4321= −m 10.

dtd h

262

2

= when t 1=

11. x187= 12. x

31>

13. ( ) ; ( )x x20 4 3 320 4 34 3− −

14. ( ) ;f xx2 2

1= −−

l

( )( )

f xx4 2

13

= −−

m

15. ( )

; ( )( )

xx

xx3 1

163 1

962 3

= −−

=−

ffl m^ h

16. dtd v

t24 162

2

= + 17. b32= 18. ( )f 2

4 2

38

3 2= =m

19. ( )f 1 196=m 20. .b 2 7= −

Exercises 2.4

1. x31> − 2. x 3< 3. y 8 0<= −m 4. y 2 0>=m

5. x 231< 6. ( , )1 9 7. ( , )1 17− and ( , )1 41− −

8. ( , ); y y0 2 0 0on LHS, on RHS< >− m m 9. x2 1< <−

10. (a) No—minimum at (0, 0) (b) Yes—infl exion at (0, 0) (c) Yes—infl exion at (0, 0) (d) Yes—infl exion at (0, 0) (e) No—minimum at (0, 0)

575ANSWERS

11. y

x

12.

x1

y

13. None: (2, 31) is not an infl exion since concavity does not change.

14. ( )f xx12

4=m

x 0>4 for all x 0!

So x12

0>4

for all x 0!

So the function is concave upward for all .x 0!

15. (a) (0, 7), (1, 0) and ,1 14−^ h (b) (0, 7)

16. (a) dx

d yx12 24

2

2

2= +

≥x 02 for all x So ≥x12 02 for all x ≥x12 24 242 + So ≠x12 24 02 + and there are no points of infl exion. (b) The curve is always concave upwards.

17. a 2= 18. p 4= 19. ,a b3 3= = −

20. (a) , , ,0 8 2 2−^ ^h h

(b) dx

dyx x6 15 215 4= − +

At

, :

, :

≠

≠

dx

dy

dx

dy

0 8 6 0 15 0 21

210

2 2 6 2 15 2 21

270

At 5 4

5 4

− = − +

=

= − +

= −

^ ] ]

^ ] ]

h g g

h g g

So these points are not horizontal points of infl exion.

Exercises 2.5

1. (a) ,dx

dy

dx

d y0 0> >

2

2

(b) ,dx

dy

dx

d y0 0< <

2

2

(c) ,dx

dy

dx

d y0 0> <

2

2

(d) ,dx

dy

dx

d y0 0< >

2

2

(e) ,dx

dy

dx

d y0 0> >

2

2

2. (a) ,dtdP

dtd P

0 0> <2

2

(b) The rate is decreasing.

3.

4. (a)

(b)

(c)

(d)


5.

6.

7. ,dtdM

dtd M

0 0< >2

2

8. (a) The number of fi sh is decreasing. (b) The population rate is increasing. (c)

9. The level of education is increasing, but the rate is slowing down.

10. The population is decreasing, and the population rate is decreasing.

Exercises 2.6

1. (1, 0) minimum 2. (0,1) minimum

3. ( , ), y2 5 6 0>− =m 4. ( . , . ), y0 5 0 25 0 so maximum<m

5. ( , ); ( ) ( , ),f x0 5 0 0 5at− = −m ( ) , ( )f x f x0 0on LHS RHS< >m m

6. Yes—infl exion at (0, 3)

7. ,2 78− −^ h minimum, ,3 77− −^ h maximum

8. (0, 1) maximum, ,1 4− −^ h minimum, ,2 31−^ h minimum

9. (0, 1) maximum, (0.5, 0) minimum, . ,0 5 0−^ h minimum

10. (a) (4, 176) maximum, (5, 175) minimum (b) (4.5, 175.5)

11. (3.67, 0.38) maximum

12. ,0 1−^ h minimum, ,2 15−^ h maximum, ,4 1− −^ h minimum

13. (a) a32= − (b) maximum, as y 0<m

14. m 521= − 15. ,a b3 9= = −

Exercises 2.7

1.

2.

3.

4.

577ANSWERS

5.

6.

7.

8. (a) ,0 7−^ h minimum, ,4 25−^ h maximum (b) ,2 9−^ h (c)

9.

10.

11.

12.


13.

14. ( )

≠dx

dy

x12

02

=+−

15.

Exercises 2.8

1.

2.

3.

(0, 0)

(-3, -3)

-1

y

x12

4.

5.

6.

579ANSWERS

7.

8.

9.

10.

11.

x1

1

2 13 2 3 41

2

3

4

5

6

2

3

4

5

4

( 4.24, 0.1)

(0.24, 0.66)

y

12. y

x

⎛⎝

⎞⎠

1, - 12

⎛⎝

⎞⎠

-1,12

13. y

x

−2 2

(−2.16, −30.24)

(2.16, 30.24)

14. f(t)

t1

(2.41, 4.83)

(−0.41, −0.83)−1


15.

0, 214

x

h1

1

−2 2 3 4−1

−3

−4

6

2

3

4

5

−1

−

−2

−3−4

−5

⎛⎝

⎛⎝

Exercises 2.9

1. Maximum value is 4.

2. Maximum value is 9, minimum value is −7.

3. Maximum value is 25.

4. Maximum value is 86, minimum value is −39.

5. Maximum value is −2.

6. Maximum value is 5, minimum value is .1631−

7. Absolute maximum 29, relative maximum −3, absolute minimum −35, relative minimum −35, −8

8. Minimum −25, maximum 29

9. Maximum 3, minimum 1

581ANSWERS

10. Maximum ∞, minimum −∞

Problem

The disc has radius cm.7

30 (This result uses Stewart’s

theorem—check this by research.)

Exercises 2.10

1.

´

A xyxy

x y

P x y

x x

x x

5050

2 2

2 250

2100

`

==

=

= +

= +

= +

2.

( )

x yy xy xA xy

x xx x

2 2 1202 120 2

60

6060 2

+ == −= −== −= −

3. xy

y xS x y

x x

2020

20

=

=

= +

= +

4. 400

2 2

2 2

2

ππ

ππ π

π ππ

π

V r hr h

rh

S r rh

r rr

r r

400

400

800

2

2

2

2

22

2

==

=

= +

= +

= +

e o

5. x yy x

3030

(a)`

+ == −

(b) The perimeter of one square is x , so its side is

.x41

The other square has side .y41

2( )

( )

A x y

x y

x y

x x

x x x

x x

x x

x x

41

41

16 16

16

16

30

16900 60

162 60 900

16

2 30 450

830 450

2 2

2 2

2 2

2

2 2

2

2

2

= +

= +

=+

=+ −

= + − +

= − +

=− +

= − +

d dn n

6. (a) x y

y x

y x

28078 40078 400

78 400

2 2 2

2 2

2

+ === −= −

(b) A xy

x x78 400 2

== −

7.

( ) ( )( )( )

V x x xx x x xx x x

x x x

10 2 7 270 20 14 470 34 4

70 34 4

2

2

2 3

= − −= − − += − += − +

8. Profit per person Cost Expenses( ) ( )

For people, ( )

x xx xx

x P x xx x

900 100 200 400900 100 200 400700 500

700 500700 500 2

= −= − − += − − −= −= −= −

9.

After t hours, Joel has travelled 75 t km. He is t700 75− km from the town. After t hours, Nick has travelled 80 t km. He is t680 80− km from the town.

2( ) ( )d t tt t

t tt t

700 75 680 80490 000 105 000 5625 462 400108 800 6400952 400 213 800 12 025

2

2

2

2

= − + −= − + +

− += − +


10. The river is 500 m, or 0.5 km, wide Distance AB :

..

Speedtime

distance

Timespeed

distance

.

d xx

tx

0 50 25

50 25

2 2

2

2

`

= += +

=

=

=+

Distance BC :

Timespeed

distanced x

tx

7

47

= −

=

= −

So total time taken is:

.

tx x

50 25

472

=+

+ −

Exercises 2.11

1. 2 s, 16 m 2. 7.5 km

3.

( )

( ) from ( )

x yy xy xA xy

x xx x

2 2 602 60 2

30 1

30 130 2

+ == −= −== −= −

Max. area 225 m2

4.

`

(a)

( )

from ( )

A xy

y xP x y

x x

x x

40004000

1

2 2

2 24000

1

28000

= =

=

= +

= +

= +

c m

(b) 63.2 m by 63.2 m(c) $12 332.89

5. 4 m by 4 m 6. 14 and 14 7. -2.5 and 2.5

8. . ,x 1 25 m= .y 1 25 m=

9. (a) ( ) ( )( )

V x x xx x x

x x x

30 2 80 22400 220 4

2400 220 4

2

2 3

= − −= − += − +

(b) cmx 632=

(c) 7407.4 cm 3

10.

( )

π π

ππ

π

π

ππ

V r h

hr

rS r r h

r rr

r r

5454

54

2

254

2108

2

2

2

2

2

= =

=

=

= +

= +

= +

d n

Radius is 3 m.

11. (a) 2πS r r17 2002= + (b) 2323.7 m 2

12. 72 cm 2

13. (a)

( ) ( )

xy

y xA y x

xy y x

x x x x

x x

x x

400400

10 1010 10 100

40010

40010 100

4004000

10 100

500 104000

`

=

=

= − −= − − +

= − − +

= − − +

= − −

c cm m

(b) 100 cm 2

14. 20 cm by 20 cm by 20 cm 15. 1.12 m 2

16. (a) 7.5 m by 7.5 m (b) 2.4 m

17. 301 cm 2 18. cm16061 2 19. 1.68 cm, 1.32 cm

20. ( ) ( )

km

d t tt t

t tt t

200 80 120 6040 000 32 000 6400

14 400 14 400 360010 000 46 400 54 40024

2 2 2

2

2

2

= − + −= − +

+ − += − +

21. (a) d x x x xx x x x

x x

2 5 42 5 4

2 6 5

2 2

2 2

2

= − + − −= − + − += − +

] ]g g (b) unit21

22. (a) Perimeter ( ) whereπ

π

π

π

π

´

x y r ry

x yy

x yy

yy

x

y yx

y yx

π

221

22

1200 221

22

22

12002

2

6002 4

4

2400 2

= + + =

= + +

= + +

− − =

− − =

− −=

e o

583ANSWERS

(b)

2

π

ππ

π π

π π

π

A xy r

y yy

y

y y y y

y y y y

y y y

21

4

2400 2

21

2

4

2400 2

8

8

4800 4 2

8

8

4800 4

2

2 2 2

2 2 2

2 2

= +

=− −

+

=− −

+

=− −

+

=− −

e eo o

(c) m, mx y168 336= =

23. (a) Equation AB : y mx b

ab

x b

= +

= +

Substitute ,1 2−^ h

( )ab

b

ab

b

a b abb ab a

aa

b

2 1

21

1

12

= − +

= − +

= − += − += −

−=

]

]

g

g

(b) ,a b2 4= =

24. 26 m

25. (a)

So

std

t sd

s1500

=

=

=

Cost of trip taking t hours:

´

C s t

s s

s s

s s

9000

90001500

15009000 1500

15009000

2

2

= +

= +

= +

= +

]

]

c

g

g

m

(b) 95 km/h (c) $2846

Exercises 2.12

1. (a) x x C32 − + (b) x

x x C3

43

2+ + +

(c) x

x C6

64− + (d)

xx x C

3

32− + +

(e) x C6 +

2. (a) ( )f x xx

C22

32

= − + (b) ( )f xx

x x C5

75

3= − + +

(c) ( )f xx

x C2

22

= − + (d) ( )f xx

x x C3

33

2= − − +

(e) 2

( )f xx

C3

2= +

3

3. (a) y x x C95= − + (b) yx

x C3

23

1= − + +−

−

(c) yx x

C20 3

4 3

= − + (d) y x C2= − +

(e) yx x

x C4 3

4 2

= − + +

4. (a) x

C3

2 3

+ (b) x

C2

2

− +−

(c) x

C71

7− + (d)

2 3x x C2 6+ +1 1

(e) x

x C6

26

1− + +−

−

5. yx

x x4

5414

3= − + − 6. ( )f x x x2 7 112= − +

7. ( )f 1 8= 8. y x x2 3 192= − + 9. x 1631=

10. y x x4 8 72= − + 11. y x x x2 3 23 2= + + −

12. ( )f x x x x 53 2= − − + 13. ( ) .f 2 20 5=

14. yx x

x3 2

12 24213 2

= + − + 15. yx

x3

415 14

313

= − −

16. yx

x x3

2 3 4323

2= − + − 17. f x x x x x2 4 24 3 2= − + + −] g

18. y x x3 8 82= + + 19. f 2 77− =] g 20. y 0=

Test yourself 2

1. ( , )3 11− − maximum, ( , )1 15− − minimum

2. x 161> 3. y x x x2 6 5 333 2= + − −

4. (a) -8 (b) 26 (c) -90 5. 50 m

6. (0, 0) minimum 7. x 1> −

8. (a) (0, 1) maximum, ( , )4 511− − minimum, ( , )2 79− minimum

(b)


9. ,21

1−c m 10. ( )f xx

x x2

56 49 59

32= + − +

11. (a) ππ

ππ π

π ππ

π

V r hr h

rh

S r rh

r rr

r r

375375

2 2

2 2375

2750

2

2

2

2

22

2

==

=

= +

= +

= +

d n

(b) 3.9 cm

12. (a) (0, 0) and ( , )1 1− (b) (0, 0) minimum, ( , )1 1− point of infl exion (c)

13. (a)

( )

S x xhx xh

x xh

xx

h

x

xh

xx

h

2 4250 2 4

250 2 4

4250 2

4

2 125

2125

2

2

2

2

2

2

= += +

− =− =

−=

− =

h

( )

V x

xx

x

x x

x x

2125

2

125

2125

2

22

2

3

=

= −

=−

= −

d n

(b) . cm . cm . cm´ ´6 45 6 45 6 45

14. x 3< 15. y x x3 33 2= + + 16. 150 products

17. For decreasing curve, dx

dy0<

dx

dyx3 2= −

( )≠x x0 0 0since for all< >2 monotonic decreasing function

18. (a)

(b)

(c)

19. (a) x yy x

y x

A xy

x x

5 2525

25

21

21

25

2 2 2

2 2

2

2

+ = == −= −

=

= −

(b) 6.25 m 2

20. (a) ( , )4 171− minimum, ( , )6 329− maximum (b) ( , )1 79− (c)

21. x 1<

22. f x x x x2 3 31 683 2= − − +] g

23. (0, 1) and ,3 74−^ h

24. 179

585ANSWERS

25. y

2x


1.

5

( );

( )

( )x

x x

x

x x x4 1

20 120 1

4 1

8 60 420 9 15

2 4

2

2

3 2

+− −

+− − − +

2.

3. ,x x21

4< >−

4. 16 m 2

5. 27; -20.25

6. ( . ) ( . )f f0 6 0 6 0= =l m and concavity changes

7. Show sum of areas is least when .r s 12 5= =

8. 2565

9. (a) ≠dx

dy

x2 1

10=

−

(b) Domain: ;≥x 1 range: ≥y 0 (c)

10. . cm,r 3 17= . cmh 6 34=

11. yx

x x3

15 13

2= − − −

12. 110 km/h

13. may be other solution.)y x x2 3 (There2= + +

14. (a) ,021

c m maximum

(b) Domain: all real numbers x ; range: ≤y021<

(c) 0; 0 (d)

15. (a) (3, 2) minimum; ( , )3 2− − maximum

(b) yx

x3

92

= +

(c) As ,

As ,

x

x

y

yx

0

3

" "

" "!

3

3

(d)

16. (a) ( )f xx x x4 3

22

34 3 2

= − −

(b) ( , ), , , ,0 0 3 1141

1127− − −c cm m

(c)

17. m m m´ ´4 4 4


18. ( )f 3 2261= − 19. (a) -2 (b) -1

20. y 0=l at (0, 0); (a) y 0>m on LHS and RHS (b) y 0<m on LHS, y 0>m on RHS

21. cm2131 3 22. (a) (0, 1) (b) , , , , . . .k 2 4 6 8=

23.

24. minimum −1; maximum 51−

25. 87 kmh −1

Chapter 3: Integration

Exercises 3.1

1. 2.5 2. 10 3. 2.4 4. 0.225 5. (a) 28 (b) 22

6. 0.39 7. 0.41 8. 1.08 9. 0.75 10. 0.65

11. 0.94 12. 0.92 13. 75.1 14. 16.5 15. 650.2

Exercises 3.2

1. 48.7 2. 30.7 3. 1.1 4. 0.41 5. (a) 3.4475 (b) 3.4477 6. 2.75 7. 0.693 8. 1.93 9. 72 10. 5.25 11. 0.558 12. 0.347 13. 3.63 14. 7.87 15. 175.8

Exercises 3.3

1. 8 2. 10 3. 125 4. -1 5. 10 6. 54 7. 331

8. 16 9. 50 10. 5232

11. 32

12. 2141

13. 0

14. 432

15. 141

16. 431

17. 0 18. 231

19. 0

20. 692

21. 10141

22. 1243− 23. 22

32

24. 231

25. 0.0126

Exercises 3.4

1. x

C3

3

+ 2. x

C2

6

+ 3. x

C5

2 5

+ 4. m

m C2

2

+ +

5. t

t C3

73

− + 6. h

h C8

58

+ + 7. y

y C2

32

− +

8. x x C42 + + 9. b b

C3 2

3 2

+ + 10. a a

a C4 2

4 2

− − +

11. x

x x C3

53

2+ + + 12. x x x x C44 3 2− + − +

13. xx x

C5 2

65 4

+ + + 14. x x

x C8 7

39

8 7

− − +

15. x x x

x C2 3 2

24 3 2

+ − − + 16. x x

x C6 4

46 4

+ + +

17. x x

x C3

42

58

3 2

− − + 18. x x x

C5

32 2

5 4 2

− + +

19. x x

x C2

33

54

4 3

+ − + 20. xx

x C2

232

1− − − +−−

−

21. x

C71

7− + 22.

3xC

4

3+

4

23. x

x x C4

43 2− + +

24. x xx

C23

423

− + + 25. x x

x C3 2

310

3 2

+ − + 26. x C3− +

27. x

C21

2− + 28. x x

x xC

423

47

2 4− − + − +

29. y y

y C3 6

53 6

+ + +−

30. t t

t t C4 3

2 44 3

2− − + +

31. x

C3

2 3

+ 32. t

C21

4− + 33.

xC

43 43

+

34. x

C5

2 5

+ 35. x

x C3

2 3

+ +

Exercises 3.5

1. (a) (i) x x x C3 12 163 2− + + (ii) 3( )x

C9

3 4−+

(b) ( )x

C5

1 5++ (c)

( )xC

50

5 1 10−+

(d) ( )y

C24

3 2 8−+ (e)

( )xC

15

4 3 5++

(f) ( )x

C91

7 8 13++ (g)

( )xC

7

1 7

−−

+

(h) ( )x

C3

2 5 3−+ (i)

( )xC

9

2 3 1 3

−+

+−

(j) ( )x C3 7 1− + +− (k) ( )x

C16 4 5

12

−−

+

(l) ( )x

C16

3 4 33 4++ (m) 2( )x C2 2− − +

1

(n) 5( )t

C5

3++2 (o)

( )xC

35

5 2 7++2

2. (a) 288.2 (b) 141− (c)

81− (d) 60

32

(e) 61

(f) 71

(g) 432

(h) 81− (i) 1

51

(j) 53

587ANSWERS

Exercises 3.6

1. units131 2 2. 36 units 2 3. 4.5 units 2 4. units10

32 2

5. units61 2 6. 14.3 units 2 7. 4 units 2 8. 0.4 units 2

9. 8 units 2 10. 24.25 units 2 11. 2 units 2 12. units931 2

13. units1132 2 14. units

61 2 15. units

32 2 16. units

31 2

17. units531 2 18. 18 units 2 19. . unitsπ 3 14 2=

20. unitsa2

42

Exercises 3.7

1. units2131 2 2. 20 units 2 3. 4

32

units2

4. 1.5 units 2 5. 141

units2 6. 231

units2

7. 1032

units2 8. 61

units2 9. 397

units2

10. 2 units 2 11. 1141

units2 12. 60 units 2

13. 4.5 units 2 14. 131

units2 15. 1.9 units 2

Exercises 3.8

1. 131

units2 2. 131

units2 3. 61

units2

4. 1032

units2 5. 2065

units2 6. 8 units 2

7. 32

units2 8. 16632

units2 9. 0.42 units 2

10. 32

units2 11. 121

units2 12. 31

units2

13. 36 units2 14. 232

units2 15. ( )π 2 units2−

Problem

π15

206units3

Exercises 3.9

1. π

5243

units3 2. π

3485

units3

3. π

15376

units3 4. π7

units3 5. 39π2

units3 6. 758π

3units3

7. 2π3

units3 8. 992π

5units3 9.

5π3

units3 10. 9π2

units3

11. 27π2

units3 12. 64π3

units3 13. 16 385π

7units3

14. 25π2

units3 15. 65π2

units3 16. 1023π

5units3

17. 5π3

units3 18. 13π units3 19. 344π27

units3

20. 3π5

units3 21. 2π5

units3 22. 72π5

units3

23. `

( )

π

π

π

π

π

π

y r xy r x

V y dx

r x dx

r xx

rr

rr

r r

r

3

3 3

32

32

34

units

a

b

r

r

r

r

2 2

2 2 2

2

2 2

23

33

33

3 3

33

= −= −

=

= −

= −

= − − − −−

= − −

=

−

−

]

d e

d

g

n o

n

<

>

F

H

#

#

Exercises 3.10

1. (a) ( )x

C24

3 4 8−+ (b)

3( )xC

3

2 3++

(c) 6( )x

C12

92 −+ (d)

5( )x xC

10

4 12 + ++

(e) ( )x x

C2 3 1

12 2

−+ −

+ (f) x

C2

3 2 12 −+

(g) 3( )x

C9 1

13

−+

+

2.

n

n

n

1

1

+

+

n( ) ( )

( )

( )

( )

u ax bdu a dx

ax b dx a ax b a dx

a u du

a nu

C

a n

ax bC

a n

ax bC

1

1

11

11

1

n

n 1

= +=

+ = +

=

=+

+

=+

++

=+

++

+d

e

n

o

# #

#

3. x C42 − +

4. ( ) ( )x x

C10

2 7

6

7 2 75 3+−

++


5. x

x C3

2 28 2

3+− + +

] g

6. ( ) ( )x x

C3

10 5

5

2 53 5− −+

−+

7. ( )x x

C6

2 5

25 2 53−

+−

+

8. ( )( )

xx

C2 43

40 45

3

+ −+

+

9. ( )

yx2 1

165

2= −

−−

10. ( )

yx

12

3 1 112 2

=− +

11. ( )( )

f xx

15

21

3 5

=−

+

12. (a) Domain: all real x 12−

(b) y 432=

Exercises 3.11

1. (a) 143

(b) 7 2− (c) 0 (d) 81

(e) 25615

(f) 24.51 (g) 331−

2. (a) 63.05 (b) 4 (c) 41− (d)

7219−

(e) 2 22

24 3 2 4+ − = −

3. 10.2 units 2 4. 83

units2 5. 65 536 14

units2

6. π

15211

units3 7. π

340

units3 8. 232

9. 2

65

units 2

10. 64 units 2

Test yourself 3

1. (a) 0.535 (b) 0.5

2. (a) x

x C2

3 2

+ + (b) x

x C2

5 2

− + (c) x

C3

2 3

+

(d) ( )x

C16

2 5 8++

3. 14.83 4. (a) 2 (b) 0 (c) 251

5. (a)

(b)

6. 3 units 2 7. 1.1 units 2

8. 232

units2 9. π9 units3 10. 421

11. 43

units2

12. 12( )x

C84

7 3++ 13. 3 units 2 14. (a)

π15

206units3

(b) π2

units3 15. (a) ±x y 3= − (b) 332

units2

(c) π

25

units3

16. 36 17. 10.55 18. 8531

units2 19. π

53

units3

20. (a) ( )x

C2

2 1 5−+ (b)

xC

8

6

+

(c) 3( )

( )x

x C5

6 12 1

5+− + +


1. (a) 121

units2 (b) π

352

units3

2. (a) Show ( ) ( )f x f x− = − (b) 0 (c) 12 units 2

3. 27.2 units 3 4. 9 units 2 5. (a) 8( )x x36 13 4 −

(b) 9( )x

C36

14 −+

6. (a) ( )x

x3 4

222 2−

− (b)

81

7. 7.35 units 2 8. π

32

units3

9. ( ) ∞f 001= =

10. (a)

(b) 3.08 units 2

11. 6

17 17units2 12.

215π6

units3

13. (a) ( )

x

x

x

x

2 3

3 6

2 3

3 2

++ =

+

+ (b)

x xC

32 3+

+

589ANSWERS

14. (a) 632

(b) 632

15. 125

units2 16. (a) 3 units2 (b) π

1558

units3

17. (a)

(b) 1.69 units 2

18. 0

19. (a) a3

8units

22

(b) πa2 units3 3

Practice assessment task set 1

1.

Let ABCD be a parallelogram with diagonal AC . ACD BAC+ += (alternate , AB DCs+ < )

DAC BCA+ += (alternate , AD BCs+ < )

AC is common by AAS ACD ACB/Δ Δ AB DC` = and AD BC= )(corresponding sides in congruent

opposite sides are equal`

Δs

2. x21< 3. x x x C3 2− + + 4. 24 5. 8 m

6. AC FD= (opposite sides of < gram equal)

( )BC FEAB AC BC

FD FEED

given

`

== −= −=

Also AB ED< (since ACDF is < gram) since AB ED= and ,AB ED< ABDE is a parallelogram

7. x

x C3

29

2+ +

8.

9. (a) π

7198

units3 (b) π

596

units3

10. 131

units2 11. ( ) , ( )f f3 20 2 16= − = −l m 12. 68

13. AB ACBD CD

(given)

(given)

==

AD is common. ABD ACD

ADB ADCby SSS`

` + +/Δ Δ=

(corresponding s+ in congruent sΔ )

But °ADB ADC 180+ ++ = ( BDC+ is a straight + )

°ADB ADC 90` + += = AD BC` =

14. (a) 78.7 units 3 (b) 1.57 units 3 15. x97> −

16. ( ) ;1 18=( ) , f1 2= −( ) ,f 1 3= fl m curve is decreasing and concave upwards at ( , )1 3

17. P x yy xy x

8 4 44 4 8

1 2

= + == −= −

A x y

x xx x xx x

3

3 1 23 1 4 47 4 1

2 2

2 2

2 2

2

= += + −= + − += − +

] g

Rectangle ,´72

76

m m square with sides 73

m

18. ( )f 1 0− = 19. 12 20. 0.837

Ans_PART_1.indd 589Ans_PART_1.indd 589 6/24/09 9:43:55 AM6/24/09 9:43:55 AM


21. AB

BC

AC

AB BC

AC

24576321024401600576 10241600

2 2

2 2

2 2

2 2

2

======

+ = +==

ABC` ∆ is right angled at B+ (Pythagoras’ theorem)

22. 8( )x

C24

3 5++ 23. 5

31− 24. (1, 1)

25. (a) 1.11 (b) 1.17

26. ,0 3^ h maximum, ( , )1 2 minimum, ( , )1 2− minimum

27. 2158

28. (a) .1 58 units2 (b) π

25

units3

29. 1232

30. 1032

units2

31. B+ is common °BDC ACB 90+ += = (given) CBD;∆ ( )ABC AAA` <∆

32. Show ( ) ( ) ( )f x f x f x0 0and >= =l m m on both LHS and RHS of ( , )0 0

33. 232

m3 34. ( )f 2 16= − 35. 9 units2

36. ( )x

C3

2 53 3−+

37. .119 3 m2 38. ( )f x x x x4 3 203 2= − − +

39. ( , )0 1− maximum

40. °OCA OCBOA OB

90 (given)

(equal radii)

+ += ==

OC is common OAC OBCby RHS` /∆ ∆ AC BC` = (corresponding sides in congruent ∆ s)

OC bisects AB

41. 2021

42. 5 3( ) ( )

yx x

5

2 4

3

8 4=

−+

−

43.

44.

Let ABCD be a rhombus with AC x= and .BD y= °AEB 90+ =

(diagonals perpendicular in rhombus)

DE BE y21= =

(diagonals bisect each other)

ACB∆ has area ´ ´x y xy21

21

41=

ADC∆ has area ´ ´x y xy21

21

41=

ABCD has area xy xy xy41

41

21+ =

45. ( )

( )

ACAB

ADAG

ADAG

AEAF

ACAB

AEAF

BG CD

GF DE

equal ratios of intercepts,

equal ratios of intercepts,

`

<

<

=

=

=

46. ( )f 2 132=

47. (a) nx

C1

n 1

++

+

(b) Since ( )dxd

C 0= , the primitive function

could include C.

48. AB2

3

49. units124987 2

591ANSWERS

50.

(opposite angles in cyclic quadrilateral)

(angle sum of triangle)( )

ADB xBCD x

A x

ABD x x

xADB

2180 2

180 180 2

LetThen (given)

++

+

+

+

=== −

= − − +

==

So ADB is an isosceles triangle.

51. ( )

3

4 6 6 1− 52. (c), (d) 53. (a), (b) 54. (c)

55. (d) 56. (b) 57. (a) 58. (b) 59. (d)

60. (b)

Chapter 4: Exponential and

logarithmic functions

Exercises 4.1

1. (a) 4.48 (b) 0.14 (c) 2.70 (d) 0.05 (e) -0.14

2. (a)

(b)

(c)

3. (a) e9 x (b) ex− (c) e x2x + (d) x x e6 6 5 x2 − + −

(e) e e3 1x x + 2] g (f) e7 x x 6( )e 5+ (g) e e4 2 3x x −] g

(h) e x 1x +] g (i) ( )

x

e x 1x

2

− (j) xe x 2x +] g

(k) ( ) ( )x e e e x2 1 2 2 3x x x+ + = + (l) ( )

( )

x

e x

7 3

7 10x

2−−

(m) ( )

ee xe

e

x5 5 5 1x

x x

x2

− =−

4. ( ) e1 6= −( ) ;e f1 6= −fl m 5. e 6. ee15

5− = −−

7. 19.81 8. ex y 0+ = 9. x e y e3 03 6+ − − =

10. , mine11− −c m

11. ;dx

dye

dx

d ye y7 7x x

2

2

= = =

12.

`

`

;dx

dye

dx

d ye

y ey e

dx

d yy

2 2

2 11 2

1

x x

x

x

2

2

2

2

= =

= +− =

= −

Exercises 4.2

1. (a) e7 x7 (b) e x− − (c) e6 x6 2− (d) xe2 x 1+2

(e) ( )x e53 x x2 5 73+ + + (f) e5 x5 (g) e2 x2− − (h) e10 x10

(i) e2 1x2 + (j) x e2 2 x1+ − − (k) ( ) ( )e x e5 1 4 x x4 4 4+ +

(l) ( )e x2 1x2 + (m) ( )

x

e x3 2x

3

3 − (n) ( )x e x5 3x2 5 +

(o) ( )

( )

x

e x

2 5

4 2x

2

2 1

+++

2. ( ) ( )e e e28 1 7 1x x x2 2 5 2+ +

3. ( ) e0 9=( ) ;e f1 3=f 2−m 4. 5

5. x y 1 0+ − = 6. e31

3−

7. y ex e2= − 8. ( ) e1 18 4 2− = − −f m


9. 2−( , ) ; ( , )min maxe0 0 1−

10.

( )

dx

dye e

dx

d ye e

e ey

4 4

16 16

1616

x x

x x

x x

4 4

2

2

4 4

4 4

= −

= +

= +=

−

−

−

11.

( ) ( )

y e

dx

dye

dx

d ye

dx

d y

dx

dyy

e e ee e e

3

6

12

3 2

12 3 6 2 312 18 60

LHS

RHS

x

x

x

x x x

x x x

2

2

2

2

2

2

2

2 2 2

2 2 2

=

=

=

= − +

= − += − +==

dx

d y

dx

dyy3 2 0

2

2

` − + =

12. y ae

dx

dybae

dx

d yb ae

b y

bx

bx

bx2

2

2

2

=

=

=

=

13. n 15= −

14.

15.

Exercises 4.3

1. (a) e C21 x2 + (b) e C

41 x4 + (c) e Cx− +− (d) e C

51 x5 +

(e) e C21 x2− +− (f) e C

41 x4 1 ++ (g) e C

53 x5− +

(h) e C21 t2 + (i) e x C

71

2x7 − + (j) ex

C2

x 32

+ +−

2. (a) ( )e51

15 − (b) ee

11

122

− = −− (c) ( )e e32

17 9 −

(d) ( )e e1921

14 2− − (e) e21

1214 + (f) e e 1

212 − −

(g) e e21

1216 3+ −−

3. (a) 0.32 (b) 268.29 (c) 37 855.68 (d) 346.85 (e) 755.19

4. ( )e e e e 1 units4 2 2 2 2− = − 5. ( )e e41

units3 2− −

6. 2.86 units 2 7. 29.5 units 2 8. ( )π

e2

1 units6 3−

9. 4.8 units 3 10. 7.4 11. (a) ( )x x e2 x+ (b) x e Cx2 +

12. e C31 x 13 ++ 13. ( )e

21

14 − 14. πe units3

15. ( )e21

5 units4 2−

Exercises 4.4

1. (a) 4 (b) 2 (c) 3 (d) 1 (e) 2 (f ) 1 (g) 0 (h) 7

2. (a) 9 (b) 3 (c) −1 (d) 12 (e) 8 (f ) 4 (g) 14 (h) 14 (i) 1 (j) 2

3. (a) −1 (b) 21

(c) 21

(d) −2 (e) 41

(f ) 31− (g)

21−

(h) 31

(i) 121

(j) 121−

4. (a) 3.08 (b) 2.94 (c) 3.22 (d) 4.94 (e) 10.40 (f ) 7.04 (g) 0.59 (h) 3.51 (i) 0.43 (j) 2.21

593ANSWERS

5. (a) log y x3 = (b) log z x5 = (c) log y 2x = (d) log a b2 = (e) log d 3b = (f ) log y x8 = (g) log y x6 = (h) log y xe = (i) log y xa = (j) log Q xe =

6. (a) 3 5x = (b) a 7x = (c) a3b = (d) x y9 = (e) a by = (f ) 2 6y = (g) x3y = (h) 10 9y = (i) e 4y = (j) x7y =

7. (a) x 1 000 000= (b) x 243= (c) x 7= (d) x 2= (e) x 1= − (f ) x 3= (g) .x 44 7= (h) x 10 000= (i) x 8= (j) x 64=

8. y 5= 9. 44.7 10. 2.44 11. 0 12. 1

13. (a) 1 (b) (i) 3 (ii) 2 (iii) 5 (iv) 21

(v) -1 (vi) 2

(vii) 3 (viii) 5 (ix) 7 (x) 1 (xi) e

14. Domain: ;x 0> range: all real y

15. Curves are symmetrical about the line .y x=

16. x ey=

Exercises 4.5

1. (a) log y4a (b) log 20a (c) log 4a (d) logb5a

(e) zlog yx3 (f ) log y9k

3 (g) logyx

a 2

5

(h) log zxy

a

(i) log ab c104 3 (j) log

r

p q3 2

3

2. (a) 1.19 (b) -0.47 (c) 1.55 (d) 1.66 (e) 1.08 (f ) 1.36 (g) 2.02 (h) 1.83 (i) 2.36 (j) 2.19

3. (a) 2 (b) 6 (c) 2 (d) 3 (e) 1 (f ) 3 (g) 7

(h) 21

(i) -2 (j) 4

4. (a) x y+ (b) x y− (c) 3 x (d) 2 y (e) 2 x (f ) x y2+ (g) x 1+ (h) y1 − (i) x2 1+ (j) y3 1−

5. (a) p q+ (b) 3 q (c) q p− (d) 2 p (e) p q5+ (f ) p q2 − (g) p 1+ (h) q1 2− (i) q3 + (j) p q1− −

6. (a) 1.3 (b) 12.8 (c) 16.2 (d) 9.1 (e) 6.7 (f ) 23.8 (g) -3.7 (h) 3 (i) 22.2 (j) 23

7. (a) x 4= (b) y 28= (c) x 48= (d) x 3= (e) k 6=

Exercises 4.6

1. (a) 1.58 (b) 1.80 (c) 2.41 (d) 3.58 (e) 2.85 (f ) 2.66 (g) 1.40 (h) 4.55 (i) 4.59 (j) 7.29

2. (a) .x 1 6= (b) .x 1 5= (c) .x 1 4= (d) .x 3 9= (e) .x 2 2= (f ) .x 2 3= (g) .x 6 2= (h) .x 2 8= (i) .x 2 9= (j) .x 2 4=

3. (a) .x 2 58= (b) .y 1 68= (c) .x 2 73= (d) .m 1 78= (e) .k 2 82= (f ) .t 1 26= (g) .x 1 15= (h) .p 5 83= (i) .x 3 17= (j) .n 2 58=

4. (a) .x 0 9= (b) .n 0 9= (c) .x 6 6= (d) .n 1 2= (e) .x 0 2= − (f ) .n 2 2= (g) .x 2 2= (h) .k 0 9= (i) .x 3 6= (j) .y 0 6=

5. (a) .x 5 30= (b) .t 0 536= (c) .t 3 62= (d) .x 3 81= (e) .n 3 40= (f ) .t 0 536= (g) .t 24 6= (h) .k 67 2= (i) .t 54 9= (j) .k 43 3= −

Exercises 4.7

1. (a) x11+ (b) x

1− (c) x3 13+

(d) x

x4

22 −

(e) x x

x5 3 9

15 33

2

+ −+

(f) x

xx

x x5 1

52

5 110 2 52

++ =

++ +

(g) x x6 51+ + (h)

x8 98−

(i) ( ) ( )x x

x2 3 16 5

+ −+

(j) ( ) ( )x x x x4 1

42 7

24 1 2 7

30+

−−

=+ −

−

(k) 4( )logx x5

1 e+ (l) ( )lnx x x91

1 8− −c m

(m) ( )logx x4

e3 (n) 5( )logx x x x6 2

1e

2+ +c m

(o) log x1 e+ (p) log

x

x1 e

2

−

(q) logxx

x2 1

2 e

+ + (r) ( )logx

xx x

13 1e

32

++ +

(s) logx x

1

e

(t) ( )

log

x x

x x x

2

2 e

2−

− −

(u) ( )

( )

log

log

x x

e x x2 1

e

xe

2

2 −

(v) loge x x1x

e+c m

(w) log

x

x10 e


2. ( )1 = −21

fl 3. logx 10

1

e

4. logx y2 2 2 2 0e− − + =

5. y x 2= − 6. 52− 7. logx y5 5 25 0e+ − − =

8. logx y5 19 19 19 15 0e− + − = 9. , log21

21

21

41

e −d n

10. ,e e1

c m maximum

11. (a)

(b)

(c)

12. ( ) logx2 5 3

2

e+ 13. (a) ln3 3x (b) ln10 10x

(c) ´ln3 2 2 x3 4−

14. ln x y4 4 4 0$ − + = 15. log logx y3 3 1 9 3 0e e$ + − − =

Exercises 4.8

1. (a) ( )log x C2 5e + + (b) ( )log x C2 1e2 + +

(c) ( )ln x C25 − + (d) log logx C x C21

21

2ore e+ +

(e) ln x C2 + (f) log x C35

e + (g) ( )log x x C3e2 − +

(h) ( )ln x C21

22 + + (i) ( )log x C23

7e2 + +

(j) ( )log x x C21

2 5e2 + − +

2. (a) ( )ln x C4 1− + (b) ( )log x C3e + +

(c) ( )ln x C61

2 73 − + (d) ( )log x C121

2 5e6 + +

(e) ( )log x x C21

6 2e2 + + +

3. (a) 0.5 (b) 0.7 (c) 1.6 (d) 3. 1 (e) 0.5

4. .log log log3 2 1 5 unitse e e2− = 5. log 2 unitse

2

6. ( . )log0 5 2 unitse2+ 7. 0.61 units 2

8. 3π log unitse3 9. 2 9π log unitse

3

10. 47.2 units 2 11. ( )π

e e2

1 units2 4 3−

12. (a)

( ) ( )

( )

( )( )

( )

x x

x x

x

x x

x

xx x

xx

31

32

3 3

1 3

3 3

2 3

93 2 6

93 3

RHS

LHS

2

2

=+

+−

=+ −

−

++ −

+

=−

− + +

=−+

=

xx

x x93 3

31

32

2`

−+ =

++

−

(b) ( ) ( )log logx x C3 2 3e e+ + − +

13. (a) x

xx

x

xx

11

5

11

15

16

RHS

LHS

= −−

=−− −

−

=−−

=

xx

x16

11

5`

−− = −

−

(b) ( )logx x C5 1e− − +

14. log

C2 3

3

e

x2 1

+−

15. .1 86 units2

Test yourself 4

1. (a) 6.39 (b) 1.98 (c) 3.26 (d) 1.40 (e) 0.792 (f) 3.91 (g) 5.72 (h) 72.4 (i) 6 (j) 2

2. (a) e5 x5 (b) e2 x1− − (c) x1

(d) x4 54+

(e) ( )e x 1x +

(f) ln

xx1

2

− (g) 9( )e e10 1x x +

3. (a) e C41 x4 + (b) ( )ln x C

21

92 − + (c) e Cx− +−

(d) ( )ln x C4+ +

595ANSWERS

4. x y3 3 0− + = 5. e

e12

2

−+

6. ( )e e21

1 units4 6 2−

7. ( )π

e e6

1 units6 6 3−

8. (a) 0.92 (b) 1.08 (c) 0.2 (d) 1.36 (e) 0.64

9. ( )e e 1 units2 2−

10. (a) 2.16 units 2 (b) x ey= (c) .2 16 units2

11. (a) .x 1 9= (b) .x 1 9= (c) x 3= (d) x 36= (e) .t 18 2=

12. (a) ( )e23

12 −

(b) ln31

10

(c) ln861

3 2+

13. e x y e3 04 4− − =

14. 0.9

15. (a) ( )e e 1 units2−

(b) ( )π

e e2

1 units2 2 3−

16. (a) ylog xa5 3

(b) log3x

2pk

17. lnx y2 2 4 0+ − − =

18. (0, 0) point of infl exion, ( , )e3 27 3− − − minimum

19. 5.36 units 2

20. (a) 0.65 (b) 1.3


1. ( )

( ) ( ) log

e x

e x x e x1

2 1

x

x xe

2 2

2 2

+

+ − + 2. e C

21 x2 + 3. 2 e

4. (a) 2.8 (b) 1.8 (c) 2.6

5. ( )loge x e x9 41x x

e4 4 8+ +c m 6. .0 42 units2

7. x2 32−−

8. 12 units3 9. log5 5xe

10. e C31 x 13 +−

11. ( ) ( );log log logdxd

x x x x1 2 18 3e e e2 = + 12.

logC

33

e

x

+

13. (a) (1, 0) (b) ; logx y x y1 0 10 1 0e $− − = − − =

(c) log

110

1units

e

−f p 14. 0.645 units 2

15. ( )log log

e

e x xe x1x

xe

xe

2

+ −

log log

e

x x x1x

e e=+ −

16. e C61 x3 12 ++

17.

( )

y e e

dx

dye e

dx

d ye e

e ey

x x

x x

x x

x x

2

2

= +

= −

= − −

= +=

−

−

−

−

18. ( )e ee

e31

316 2

2

8

− = −−

19.

( ) ( )

y e

dx

dye

dx

d ye

dx

d y

dx

dyy

e e ee e e

3 2

15

75

4 5 10

75 4 15 5 3 2 1075 60 15 10 100

LHS

RHS

x

x

x

x x x

x x x

5

5

2

2

5

2

2

5 5 5

5 5 5

= −

=

=

= − − −

= − − − −= − − + −==

20. ( )f x e x3 6x2= −

21.

22. y e21 x2=


Chapter 5: Trigonometric functions

Exercises 5.1

1. (a) 36° (b) 120° (c) 225° (d) 210° (e) 540° (f) 140° (g) 240° (h) 420° (i) 20° (j) 50°

2. (a) 43π

(b) π6

(c) 5π6

(d) 4π3

(e) 5π3

(f) 7π20

(g) π

12

(h) 5π2

(i) 5π4

(j) 2π3

3. (a) 0.98 (b) 1.19 (c) 1.78 (d) 1.54 (e) 0.88

4. (a) 0.32 (b) 0.61 (c) 1.78 (d) 1.54 (e) 0.88

5. (a) 62° 27’ (b) 44° 0’ (c) 66° 28’ (d) 56° 43’ (e) 18° 20’ (f) 183° 21’ (g) 154° 42’ (h) 246° 57’ (i) 320° 51’ (j) 6° 18’

6. (a) 0.34 (b) 0.07 (c) 0.06 (d) 0.83 (e) −1.14 (f) 0.33 (g) −1.50 (h) 0.06 (i) −0.73 (j) 0.16

Exercises 5.2

1.

π3

π4

π6

sin 23

2

121

cos 21

2

1

23

tan 3 1 3

1

cosec 3

2 2 2

sec 2 23

2

cot 3

1 1 3

2. (a) 31

(b) 21

(c) 8

3 3 (d)

34 3

(e) 0 (f) 2

2 3 1+

(g) 2 3− (h) 2

2 2+ (i)

23 2 2−

(j) 2

2 3+

3. (a) 1

41

(b) 4

6 2− (c)

23

(d) 1 (e) 441

4. (a)

46 2+

(b) 3 2−

5. π π π π

π π π π

´ ´

´ ´

cos cos sin sin

sin cos cos sin

3 4 3 4

21

2

123

2

1

2 2

1

2 2

3

4 6 4 6

2

123

2

121

2 2

3

2 2

1

LHS

RHS

LHS RHS

+

= +

= +

= +

= +

=

= +

=

So π π π π π π

cos cos sin sin sin cos3 4 3 4 4 6

+ = +π π

cos sin4 6

6. (a) π π π

ππ

43

44

4

4

= −

= −

(b) 2 nd (c) 2

1−

7. (a) π π π

ππ

65

66

6

6

= −

= −

(b) 2 nd (c) 21

8. (a) π π π

ππ

47

48

4

24

= −

= −

(b) 4 th (c) 1−

9. (a) π π π

ππ

34

33

3

3

= +

= +

(b) 3 rd (c) 21−

10. (a) π π π

ππ

35

36

3

23

= −

= −

(b) 4 th (c) 23

−

11. (a) 1− (b) 23

(c) 3− (d) 2

1− (e) 3

1

12. (a) (i) π π π

ππ

613

612

6

26

= +

= +

(ii) 1 st (iii) 23

(b) (i) 2

1 (ii) 3 (iii)

2

1− (iv) 3

1 (v)

23

−

13. (a) ,π π3 3

5 (b) ,

π π4

54

7 (c) ,

π π4 4

5 (d) ,

π π3 3

4 (e) ,

π π6

56

7

14. (a) sin θ (b) tan x− (c) − cos α (d) sin x (e) cot θ

597ANSWERS

15. (a) π π π π π6 4 12

2 3125

+ =+

= (b) ( )

2 2

3 14

2 3 1−=

−

16. (a) θ θsin cos

23+

(b) ( )θ θ θ θcos sin cos sin

2 2

2−=

−

(c) tantan

xx

11

+−

(d) cos siny y

2

3− (e)

θ θcos sin2

3+

17. (a) θsin2 (b) αsin (c) θπ

tan3

+d n

(d) θ θ θcos sin cos 22 2− = (e) θtan

18. ;cos sinx x54

53= − =

19. (a) , , ,π π

πx 03

23

42= (b) , , ,

π π π πx

4 43

45

47

=

(c) π

x3

= (d) π

x2

= (e) , , , ,π ππ π

x 0 22 2

3=

20. (a) ±ππ

x n23

= (b) ππ

x n4

= − (c) n( )ππ

x n 13

= + −

(d) ±ππ

x n6

= (e) , 2π πx n nπ

12

n= + −^ h

(f) ,π, ππ

ππ

x2 2

n n= −( 1) ( 1)n n n+ − −

(g) 2 ±ππ

x n6

=

Exercises 5.3

1. (a) π4 cm (b) π m (c) π

325

cm (d) π2

cm (e) π4

7mm

2. (a) 0.65 m (b) 3.92 cm (c) 6.91 mm (d) 2.39 cm (e) 3.03 m

3. 1.8 m 4. 7.5 m 5. π

212

6. 25 mm 7. 1.83

8. 1397

mm 9. 25.3 mm 10. π

SA36

175cm ,2=

π

V648

125 35cm3=

Exercises 5.4

1. (a) π8 cm2 (b) π2

3m2 (c)

π3

125cm2 (d)

π4

3cm2

(e) π

849

mm2

2. (a) .0 48 m2 (b) .6 29 cm2 (c) .24 88 mm2 (d) .7 05 cm2 (e) .3 18 m2

3. .16 6 m2 4. θ 494= 5. 6 m 6. (a)

π6

7 cm (b)

π12

49cm2

7. π8

6845mm2 8. 75 cm 2 9. 11.97 cm 2

10. , 3π

rθ15

cm= =

Exercises 5.5

1. (a) π8 cm2 (b) π

46 9 3

m2−

(c) π3

125 75cm2−

(d) ( )π

4

3 3cm2

− (e)

( )π8

49 2 2mm2

−

2. (a) .0 01 m2 (b) 1.45 cm 2 (c) 3.65 mm 2 (d) 0.19 cm 2 (e) 0.99 m 2

3. 0.22 cm 2 4. (a) π7

3 cm (b)

π149

cm2 (c) 0.07 cm 2

5. 134.4 cm 6. (a) 2.6 cm (b) π6

5 cm (c) 0.29 cm 2

7. (a) 10.5 mm (b) 4.3 mm 2

8. (a) π

425

cm2 (b) 0.5 cm 2

9. (a) °77 22l (b) 70.3 cm 2 (c) 26.96 cm 2 (d) 425.43 cm 2

10. 9.4 cm 2

11. (a) π

911

cm (b) π π

2218

12118

396 121cm2− =

−

(c) ( )π π

229

119

11 18cm+ =

+

12. (a) 5 cm 2 (b) 0.3% (c) 15.6 cm

13. (a) π10 cm (b) π24 cm2

14. (a) ( )π π

89

209

4 18 5cm+ =

+ (b) 3:7

15. (a) π

2225

cm3 (b) ( )π π

2105

1802

15 7 24cm2+ =

+

Exercises 5.6

1. (a) 0.045 (b) 0.003 (c) 0.999 (d) 0.065 (e) 0.005

2. 41

3. 31

4. (a)

( )

π π π

´ ´

sin sin cos cos sin

cos sin

x x x

x x

x

x

3 3 3

23

21

23

121

21

3

Z

+ = +

= +

+

= +

d n


(b)

( )

( )

π π π

´ ´

´

cos cos cos sin sin

cos sin

x x x

x x

x

x

x

4 4 4

2

1

2

1

2

11

2

1

2

11

2

2

22

1

Z

− = +

= +

+

= +

= +

d n

(c) π

π

π

´

tantan tan

tan tan

tantan

xx

x

xx

xx

4 14

4

1 11

11

+ =−

+

=−

+

−+

Z

d n

5. 1 343 622 km 6. 7367 m

Exercises 5.7

1. (a) y

xπ2

3π2

π 2π

-1

1

(b) y

xπ2

3π2

π 2π

-2

2

(c) y

xπ2

3π2

π 2π

1

-1

2

(d) y

xπ2

3π2

π 2π

3

2

1

(e) y

xπ2

π 2π

3

-3

3π2

(f) y

xπ 2π

4

-4

3π2

π2

599ANSWERS

(g)

xπ 2π

y

3

4

2

1

3π2

π2

(h)

x

y

5

2πππ2

3π2

(i)

3

x

y

π 2π3π2

π2

(j)

xπ 2π

y

1

3π2

π2

2. (a)

x

y

1

-1

2ππ 3π2

π4

7π4

5π4

3π4

π2

(b) y

xπ 2π3π

45π4

3π2

7π4

π2

π4

(c) y

1

–1

x2ππ

32π3

4π3 3

5ππ

(d)

3

–3

x

y

π4

3π4

π2

5π4

3π2

7π 2ππ4

(e) y

-6

x

6

2π3

5π3

2π4π3

ππ3

(f) y

x2ππ


(g) y

x5π3

4π3

2π3

π3

2ππ

(h) y

-3

3

x2ππ 3π

2

π2

(i) y

−2

2

x2ππ 3π

2π2

(j) y

-4

4

x2ππ 3π

2π2

3. (a) y

-1

1

xπ-π

- - - 3π4

3π4

π4

π2

π4

π2

(b) y

-7

7

xπ-π - - 3π

4π2

π4

π4

π2

3π4

-

(c)

xπ-π

y

- 3π4

π2

π4

π2

π4-3π

4-

(d) y

5

x

-5

-π - - - π3π4

π4

π4

π2

3π4

π2

(e) y

-2

xπ-π --

π2

π2

-3π4

2

π4

π4

3π4

601ANSWERS

4.

πx

y

8

-8

3π 4π2π

5. (a)

x

y

1

-1

π 2π3π2

π2

(b)

x

y

3π2

π2

π 2π

(c)

x

y

1

-1

π 2ππ2

3π2

(d)

x

y

3

-3

π 2π3π2

π2

(e)

x

y

2

-2

π 2π3π2

π2

(f)

x

y

4

-4

π 2π5π4

3π4

π2

π4

3π2

7π4

(g)

x

y

1

-1

π 2π5π4

3π2

π4

π2

7π4

3π4

(h)

x

y

1

π 2π7π4

3π4

π4

π2

5π4

3π2

(i)

3π2

π2

x

y

3

2

1

-1π 2π


(j)

x

y

2

1

5

4

3

-1π 2π3π

2π2

6. (a)

x

y

1

-1

-2 -1 1 2

(b)

x

y

3

-3

-2 -1 1 2

7. (a)

y = sin x

y = sin 2x

x

y

1

-1

π 3π2

π2

2π

(b)

x

y

1

2

-1

-2

ππ2

3π2

2π

8. (a)

x

y

1

2

3

4

5

−1

−2

−3

−4

−5

π

y = 2 cos x

y = 3 sin x

π2

3π2

2π

(b)

y = 2 cos x + 3 sin x

x

y

1

−1

−2

−3

−4

−5

2

3

4

5

π 2π3π2

π2

9.

x

y

-1

-2

1

2 y = cos 2x - cos x

π 2π3π2

π2

603ANSWERS

10. (a)

y = cos x + sin x

-1

-2

1

2

y

x2πππ

23π2

(b)

y = sin 2x – sin x

–1

–2

1

2

y

xπ 3π

2π2

2π

(c)

x

–1

–2

1

2

–3

y = sin x + 2 cos 2x

2ππ

y

3π2

π2

(d)

y=3 cos x − cos 2x

x

−1

−2

1

2

−3

−4

3

y

2πππ2

3π2

(e)

y = sin x - sinx2

3π2

2πx

y

-1

-2

1

2

-3

-4

3

π2

π

Exercises 5.8

1. (a)

x

y

1

-1y = sin x

6

y = x2

541 2π2

2π3π2

π3

There are 2 points of intersection, so there are 2 solutions to the equation. (b)

x

y

1

-π2

π2

-1

y = sin x

y = x

2

1-1-2 32 π-π -3

There are 3 points of intersection, so there are 3 solutions to the equation.


2. x 0= 3. .x 1 5= 4. , .x 0 4 5= 5. ,x 0 1=

6. . ,x 0 8 4=

π2

3π2

π 2π

1

-1

y

x

y = cos x

y = sin x

7. (a) Period 12 months, amplitude 1.5 (b) 5.30 p.m.

8. (a) 1300 (b) (i) 1600 (ii) 1010 (c) Amplitude 300, period 10 years

9. (a)

0

2

4

6

8

10

12

14

16

18

January February March April May June July August

(b) It may be periodic - hard to tell from this data. Period would be about 10 months. (c) Amplitude is 1.5

10. (a)

(b) Period 24 hours, amplitude 1.25 (c) 2.5 m

Exercises 5.9

1. (a) 4 cos 4 x (b) sin x3 3− (c) sec x5 52 (d) ( )sec x3 3 12 + (e) ( )sin x− (f) 3 cos x

(g) ( )sin x20 5 3− − (h) ( )sinx x6 2 3−

(i) ( )secx x14 52 2 + (j) cos sinx x3 3 8 8−

(k) ( )πsec x x22 + + (l) sec tanx x x2 +

(m) sin sec tan cosx x x x3 2 3 2 3 22 +

(n) cos sin cos sin

xx x x

xx x x

42 2

22 2

− = −

(o) ( )

sin

sin cos

x

x x x

5

3 5 5 3 4 52

− +

(p) ( ) ( )sec tanx x x9 2 7 7 2 72 8+ +

(q) ( )sin cos sin cosx x x x2 45 5 5r 2−

(s) ( ) ( )sin cos loge x x x2 21

1txe+ − −

(u) ( ) ( )cose e x1x x+ + (v) sincos

cotxx

x=

(w) ( )sin coscos sin

e x e xe x x2 2 3 2

3 2 2 2

x x

x

3 3

3

− += −

(x)

( )tan

tan sec

tan

tan secx

e x e x

x

e x x7

2 7 7 7

7

2 7 7 7

x x

x

2

2 2 2

2

2 2

−

−=

2. ( )

cos sin sinsin cos sin

x x xx x x

44

2 3 5

3 2 2

−= −

3. 12 4. πx y6 3 12 6 3 0− + − =

5. cossin

tanxx

x− = − 6.

3 3

29

2 3− = −

7. sec xetanx2 8. πx y8 2 48 72 2 2 0+ − − =

9.

( )

cos

sin

cos

y x

dx

dyx

dx

d y

x

2 5

10 5

25 2 5

2

2

=

= −

= −

cos x

y

50 5

25

= −

= −

10. ( )( )( )

( )

sincos

sin

f x xf x xf x x

f x

22

2

= −= −== −

l

m

11. [ ( )]log tan

tansec

tantan

tantan

tantan cot

dxd

x

xx

xx

xx

xx x

1

1

LHS

RHS

e

2

2

2

=

=

= +

= +

= +=

( )log tan tan cotdxd

x x xe` = +7 A

12. ,π π3

33

−d n maximum,

,π π3

53

35

− −d n minimum

0

0.5

1

1.5

2

2.5

3

3.5

4

6.20

am

11.55 am

6.15

pm

11.48 pm

6.20

am

11.55 am

6.15

pm

11.48 pm

6.20

am

11.55 am

6.15

pm

11.48 pm

605ANSWERS

13. (a) °π

sec x180

2 (b) °π

sinx60−

(c) °π

cosx900

14.

( )

sin cos

cos sin

sin cos

sin cos

y x x

dx

dyx x

dx

d yx x

x xy

2 3 5 3

6 3 15 3

18 3 45 3

9 2 3 5 39

2

2

= −

= +

= − +

= − −= −

15. ,a b7 24= − = −

Exercises 5.10

1. (a) sin x C+ (b) cos x C− + (c) tan x C+

(d) °π cos x C45− + (e) cos x C

31

3− + (f) cos x C71

7 +

(g) tan x C51

5 + (h) ( )sin x C1+ +

(i) ( )cos x C21

2 3− − + (j) ( )sin x C21

2 1− +

(k) ( )πcos x C− + (l) ( )πsin x C+ +

(m) tan x C72

7 + (n) cosx

C82

− +

(o) tanx

C93

+ (p) ( )cos x C3− − +

2. (a) 1 (b) 33

13

2 3− = (c)

2

22= (d)

31−

(e) π1

(f) 21

(g) 43

(h) 51−

3. 4 units 2 4. 3 2

162

= units 2 5. 0.86 units 2

6. 0.51 units 3 7. π4

units 3 8. e Ctan x + 9. 241

10. π π

33 3

3 3− =

− units 2 11. 2 2 units 2

12. (a)

2

( )

( )

π

π

π

ππ

ππ

cos

sin

sin sin

V y dx

x dx

x

20

1 0units

πa

b2

0

2

0

3

=

=

=

= −

= −=

π

; E

#

#

(b) .3 1 units 3

13. (a) ( )

( )

cos cos sincos cos

coscos cos

cos cos

x x xx x

xx x

x x

21

2 12 1 2

21

2 1

2 2

2 2

2

2

2

= −= − −= −

+ =

+ =

(b) sin x x C41

221+ +

14. (a) ( )cos x21

1 2− (b) ( )π

π8

2− units 3

15. siny x2 3= −

Exercises 5.11

1. (a) sin x x C41

221− + (b) sinx x

xC

21

41

23

3

− + +

(c) sin xx

C43

227− + (d) sin sinx x x C

21

2− − +

(e) sin tanx

x x C27

47

271

7+ + +

(f) sin sinx xx

C41

225+ + +

(g) cos sinx x x C21

221

41

2− − + +

(h) tan sinx x x C21

41

2+ − +

(i) sin sinx x x C31

321

41

2+ + + (j) sinx x C21

81

4+ +

2. (a) π4

(b) π2

3 (c)

( )π π8

2 3+ (d)

π3

(e) π π12 8

324

2 3 3− =

−

3. ( )π

π8

2− units 3 4. π

82−

units 2

5. ( )π

π4

3 8+ units 3 6. (a) ( )sinx x C21

61

6− +

(b) ( )sinx x C21 + + (c) °( )π sinx x C

21 90

2− +; E

(d) ( )sinxa

ax C21

21

2+ + (e) sinxx

C23

45

54− +d n

7. (a) sin sin sinx x x3 3 4 3= −

(b) ( )41

38

23 3

241

16 9 3− = −e o

8.

`

( ) ( ) ( )

[ ( ) ( )]

sin sinsin cos cos sin

sin cos cos sinsin cos

sin sin

sin cos

x x x xx x x x

x x x xx x

x x x x

x x

7 3 7 37 3 7 3

7 3 7 32 7 3

21

7 3 7 3

7 3

a + + −= +

+ −=

+ + −

=

(b) 101

9. (a) cos x C41

2− + (b) ( )sin sinx x x C21

21

2+ − +

(c) ( )θ θsin C81

41

4− + (d) cosx x C21

2− +

(e) θ θsin sin C41

31

3 3+ +c m

10. (a) ( )21

2 2 3 1− − units 2 (b) ( )π4

2 3− units 3



Test yourself 5

1. (a) π6

5cm (b)

π12

25cm2 (c) 0.295 cm 2

2. (a) 3 (b) 23

(c) 23

(d) 2

1−

3. (a) ,π π

x4

34

7= (b) ,

π πx

6 65

=

4. (a)

(b)

5. (a) sin x− (b) cos x2 (c) sec x2 (d) cos sinx x x+

(e) sec tan

xx x x

2

2 − (f) sin x3 3− (g) sec x5 52

6. (a) cos x C21

2− + (b) sin x C3 + (c) tan x C51

5 +

(d) cosx x C− +

7. (a) 2

1 (b)

32 3

(c) π

124 3 3−

(d) π

12

8. π

x y3 2 14

30+ − − =

9. cos

sin

x t

dtdx

t

dtd x

2

2 2

2

2

=

= −

cos t

x

4 2

4

= −

= −

10. 2

1 units 2 11.

π3

units 3 12. (a) 5 (b) 2

13. 3 3− 14. (a) π7

8cm2 (b) 0.12 cm 2

15. (a)

(b) .x 0 6=

16. (a) 2

3 2− units 2 (b)

( )π

π24

6 3 3− + units 3

17. 2 units 2 18. πx y4 8 8 0+ − − = 19. cosy x3 2= −

20. (a)

(b) . , . ,x 0 9 2 3 3=


1. 0.27 2. 21

13

16

3 3− =

−e o 3. r 64= units, θ

π512

=

4. (a) ,2 3Period amplitude= =

(b)

5. (a) siny x3= −

(b)

( )sin sinsin sin

dx

d yy

x xx x

9

9 3 9 39 3 9 30

LHS

RHS

2

2

= +

= + −= −==

607ANSWERS

6.

7. °π

sec x180

2 8. 2

9. (a)

´

´

tansec

cossincos

cos sincos

cos sinsec cosec

xx

xxx

x xx

x xx x

1

1

1 1

RHS

LHS

2

2

2

=

=

=

=

==

sec cosectansec

x xxx2

` =

(b) log log321

3e e=

10. ( )cos sinx x x e2 2 2 sinx x2+ 11. (a) ,π4

4d n and ,π3

24

d n

(b) 4Maximum = (c) 1Amplitude =

12. cos x C31 3− +

13. ( )π π

83 2

33

4 2 3 3cm cm2 2− =

−e o

14. °π cosx C180− + 15.

21−

16. 0.204 units 3 17. sin coscos sin

x xx x

+−

18. ( )π π

29

21

4

9 2cm2− =

−d n

19.

20. , , , , , , ,π π π π8

085

089

08

130d d d dn n n n

21. (a) 2

121

22 1

− =−

units 2 (b) ( )π

π24

6 3 3+ − units 3

22. n n( ) , ( )±ππ

ππ

x n n12

14

= + − − 23. cos

cosx

x C3

3

− +

24. (a)

sin sin cosx x x2 =

( )sin sinsin cos cos sin

sin cos

x x xx x x x

x x

2

2

21

= += +=

(b) π64

2−

25. ( )( )( )

( )( )

cossincoscos

f x xf x xf x x

xf x

2 36 318 39 2 39

== −= −= −= −

l

m

Chapter 6: Applications of calculus to

the physical world

Exercises 6.1

1. (a) R t20 8= − (b) R t t15 42= + (c) R x16 4= −

(d) R t t15 4 24 3= − + (e) R et= (f) θsinR 15 5= −

(g) πRr

2100

3= − (h) R

x

x

42=

− (i) R

r800

4002

= −

(j) πR r4 2=

2. (a) h t t C2 42 3= − + (b) A x x C2 4= + +

(c) πV r C34 3= + (d) cosd t C7= − +

(e) s e t C4 3t2= − +

3. 20 4. 1 5. 6 e 12 6. 13 7. 900 8. e2 53 +

9. y x x x 63 2= − + + 10. ;RdtdM

t R1 4 19= = − = −

[i.e. melting at the rate of 19 g per minute (g min -1 )]

11. .11 079 25 cm− per second (cms -1 ) 12. 21 000 L

13. 165 cm 2 per second (cm 2 s -1 ) 14. 0.25−

15. 41 cm 2 per minute (cm 2 min -1 ) 16. 31 cm 3

17. 108 731 people per year 18. (a) 27 g (b) .2 7− (i.e. decaying at a rate of 2.7 g per year)

19. y e

dx

dye

y

4

4

x

x

4

4

=

=

=

20.

( )

( )( )

S e

dtdS

e

eS

2 3

2 2

2 2 3 32 3

t

t

t

2

2

2

= +

=

= + −= −


Exercises 6.2

1. (a) 8 x 3 (b) x54 2 (c) x6 3− + (d) 10 e 2 x (e) sin x−

2. (a) 297 (b) 4 e 4 (c) 6084− (d) ln3 4 3+ (e) 20−

3. 2

63 2= 4.

176

5. 95− 6. 426 7. 289 8. 44

9. 141

10. 6 11. 8100 mm 3 s -1 12. 0.287 mm 3 s -1

13. 205.84 cm 2 s -1 14. 159.79 cm 3 s -1

15. 40 units per second 16. 34 560− , i.e. decreasing at the rate of 34 560 mm 3 s -1

17. -614, i.e. decreasing by 614 radios per week

18. 2411.5 cms -2 19. -11.12, i.e. decreasing by 11.12 mm 3 s -1

20. -2.5, i.e. decreasing by 2.5 ms -1

21. -2765, i.e. decreasing by 2765 rabbits per day

22. 1.02 cms -1 23. 2.14 houses per year 24. 0.92 mh -1

25. -0.01, i.e. decreasing at the rate of 0.01 cms -1

26. (a) °

´

sinA x

x

x

21

60

2 23

43

2

2

2

=

=

=

(b) 32

cms -1

27. (a) 663.5 mm 2 s -1 (b) 29 194.2 mm 3 s -1

28. 0.66 mm 2 s -1 29. 0.57 mm 3 s -1

30. .30 48− , i.e. decreasing at the rate of 30.48 cm 2 s -1

31. .0 52− ms -1 , i.e. moving down at the rate of 0.52 ms -1

32. 458 kmh -1 33. 2.6 ms -1

34. Rate of volume decrease is proportional to surface area.

( )

( )

π

π

π

π

ππ

´

´

dtdV

kS

k r

V r

drdV

r

dVdr

r

dtdr

dVdr

dtdV

rk r

k

4

34

4

41

41

4

i.e.

2

3

2

2

22

`

= −

= −

=

=

=

=

= −

= −

` radius will decrease at a constant rate of k

35. 2.55 m per minute

36. (a) α

α

α

tan

tan

tan

DD

D

5

55

=

=

=

(b) α αsind

dD 52

= −

(c) -39 (decreasing by 39 radians/hour) (d) No, the angle will reach zero after a few seconds.

Exercises 6.3

1. (a) 80 (b) 146 (c) 92 days (d)

2. (a) 99 061 (b) 7 hours

3. `

`

( ) ,

,

.

.

.ln

t MM et M

ee

kk

M e

0 1001005 95

95 1000 950 95 50 01

100

a When

When

So .

kt

k

k

t

5

5

0 01

= === ==== −==

−

−

−

−

(b) 90.25 kg (c) 67.6 years

4. (a) 35.6 L (b) 26.7 minutes

5. (a) P 50000 = (b) .k 0 157= (c) 12 800 units (d) 8.8 years

6. 2.3 million m 2 7. (a) P e50 000 . t0 069= (b) 70 599 (c) 4871 people per year (d) 2040

8. (a) . °65 61 C (b) 1 hour 44 minutes

9. (a) 92 kg (b) Reducing at the rate of 5.6 kg per hour (c) 18 hours

10. (a) ; .M k200 0 002530 = = (b) 192.5 g (c) Reducing by 0.49 g per year (d) 273.8 years

11. (a) B e15 000 . t0 073= (b) 36 008 (c) 79.6 hours

12. 11.4 years 13. (a) 19% (b) 3200 years

609ANSWERS

14. (a) ( ) ( )( )

( )

( )

P t P t e

dt

dP tkP t e

kP t

kt

kt

0

0

=

= −

= −

−

−

(b) 23% (c) 2% decline per year (d) 8.5 years

15. 12.6 minutes 16. 12.8 years

17. (a) 76.8 mg/dL (b) 9 hours 18. 15.8 s 19. 8.5 years

20. (a) Q Ae

dt

dQkAe

kQ

kt

kt

=

=

=

(b)

´

ln

lnln

dt

dQkQ

dQdt

kQ

tkQ

dQ

k QdQ

kQ C

kt Q C

kt C Q

e Qe e Q

Ae Q

1

1

1 1

1

So

kt C

kt C

kt

1

1

1

1

=

=

=

=

= +

= +− =

===

−

−

#

#

Exercises 6.4

1. (a)

( )( )

dtdx

Ae

Aex

2

2 100 1002 100

t

t

2

2

=

= + −= −

(b) .A 0 198=

(c) .t 2 76=

2. (a) .

. ( )

. ( )

dtdN

Ae

AeN

0 14

0 14 45 450 14 45

.

.

t

t

0 14

0 14

=

= + −= −

(b) .A 27 96= (c) .N 101 3= (d) .t 7 05=

3. (a)

( )( )

dtdv

kAe

k Aek v

5000 50005000

kt

kt

=

= + −= −

(b) , .A k82 000 0 0414= = (c) 1 615 609.47 kL (d) 3 days, 23 h

4. (a)

( )( )

dtdN

kAe

k P Ae Pk N P

kt

kt

=

= + −= −

(b) .t 6 27=

5.

`

`

`

( ),

,

.

ln ln

ln

ln

T Aet

AeA

T et

ee

e

e

k ek

k

kT e

180 80

80 1862

18 6215 68

68 18 6250 62

6250

6250

1515

156250

0 014318 62

aWhen

When

.

kt

kt

k

k

k

k

t

0

15

15

15

15

0 0143

Τ

Τ

= += == +== += == +=

=

=

= −= −

−=

== +

−

−

−

−

−

−

−

(b) After 114.5 minutes (c) ∞, T0 18As t e . t0 0143" " "`−

6. (a) 25 464 (b) After 68.8 weeks

7. (a) c5.2 C− (b) After 8 minutes

8. (a)

( )( )

dtdv

kAe

k P Ae Pk v P

kt

kt

= −

= − + −= − −

−

−

(b) , .A k500 0 008 58= − = (c) 78.85 ms -1 (d) 500 ms -1

9. (a) c7.2 C (b) 68 s or 1 m 8 s

10. (a) 3738 (b) After 17.1 years

11. 5.1

12. (a) 8.4 years (b) 7.5 years

13. 125.7 years

14. 16 minutes

15. 50.1%

16. (a) 2800 (b) 17.7 years


17. ( )

( )

( )( )

T e

dtdT

k e

k ek T

28 172

172

28 172 2828

a

So

kt

kt

kt

= +

= −

= − + −= − −

−

−

−

(b) 28° C (c) 40 minutes

18. (a) 17.6% (b) 156 years

19. (a) 32.5% (b) 80.8 years

20. ( )

( )

( )

( )

( )( )

´

ln

lnln

dtdN

k N P

dNdt

k N P

tk N P

dN

k N PdN

kN P C

kt N P C

kt C N P

e N Pe e N P

Ae N PP Ae N

1

1

1 1

1

So

kt C

kt C

kt

kt

1

1

1

1

= −

=−

=−

=−

= − +

= − +− = −

= −= −= −

+ =

−

−

#

#

Exercises 6.5

1. (a) v

t

a

t

(b) v

t

a

t

(c)

t t

v a

(d)

(e)

2. (a) , ,t t t2 4 6 (b) 0 to , ,t t t1 3 5 (c) 0 to t1 (d) t5

3. (a)

(b)

4. (a) , , ,O t t t2 4 6 (b) , ,t t t1 53 (c) t5 (d) (i) At rest, accelerating to the left . (ii) Moving to the left with zero acceleration.

5. (a) , , ,…π π π4 4

34

5 (b) , , , ,…

ππ

π0

2 23

6. (a) At the origin, with positive velocity and positive constant acceleration (moving to the right and speeding up). (b) To the right of the origin, at rest with negative constant acceleration. (c) To the left of the origin, with negative velocity and positive acceleration (moving to the left and slowing down). (d) To the right of the origin, with negative velocity and acceleration (moving to the left and speeding up) . (e) To the left of the origin, at rest with positive acceleration.

611ANSWERS

Exercises 6.6

1. (a) 18 cms -1 (b) 12 cms -2 (c) When , ;t x0 0= = after 3 s (d) After 5 s

2. (a) 8 ms 1− − (b) ;a 4= constant acceleration of 4 ms -2 (c) 13 m (d) after 2 s (e) 5 m−

(f)

3. (a) 4 m (b) 40 ms -1 (c) 39 m (d) 84 m

(e)

4. (a) 2 cm (b) After 1 s (c) -4 cm (d) 6 cm (e) -7 cms -1

5. (a) 2 ms -1 (b) e4 ms2 2− (c) ( )a e e v4 2 2 2t t2 2= = =

(d)

6. (a) sinv t2 2= − (b) cosa t4 2= − (c) 1 cm

(d) , , , , . . .π

ππ

02 2

3s (e) ±1 cm (f) , , . . .

π π π4 4

34

5s

(g) cosa t x4 2 4= − = −

7. (a) ;v t t a t3 12 2 6 122= + − = + (b) 266 m (c) 133 ms -1 (d) 42 ms -1

8. (a) 4 3( ) , ( )x t x t20 4 3 320 4 3• •= − = −• (b) , , ,x x x1 20 320cm cms cms1 2= = =− −o p

(c) The particle is on the RHS of the origin, travelling to the right and accelerating.

9. (a) v t5 10= − (b) 95 ms 1− − (c) a 10 g= − =

10. 32( )

,( )

vt

at3 1

173 1

102=+

=+

−

11. (a) At the origin (b) 61

cms 1− (c) 361

cms 2− −

(d) The particle is moving to the right but decelerating (e) ( )e 1 s3 −

12. (a) 3 ms -1 (b) When , ,t 0 1 3s s s= (c) 10 ms -2

13. (a)

(b) ,cos sinx t x t6 2 12 2= = −o p (c) 6 3− cms -2 (d) ( )sin sinx t t x12 2 4 3 2 4= − = − = −p

14. (a) 7 m (b) 16 m (c) After 7 s

(d)

(e) 10 m

15. (a) 18.75 m (b) -15 ms -1 (c) 5 s

16. (a) At the origin: x 0=

( )

,

t t tt t t

t t t

2 3 42 02 3 42 00 2 3 42 0

3 2

2

2

− + =− + =

= − + =

Since ,t 0= the particle is initially at the origin .

( ) ( ) ( )

t tb ac2 3 42 0

4 3 4 2 42327

0<

2

2 2

− + =− = − −

= −

So the quadratic equation has no real roots . So the particle is never again at the origin .

(b) dtdx

t t6 6 422= − +


At rest:

( ) ( )( )

dtdx

t tt tb ac

0

6 6 42 07 0

4 1 4 1 727

0<

2

2

2 2

=

− + =− + =− = − −

= −

So the quadratic equation has no real roots. So the particle is never at rest .

17. (a) 0 cm (at the origin) (b) , , , . . .π π π4 4

34

5s (c) ±12 cm

18. (a) 8 e 16 cms -1 (b) 0 s (initially) (c) 1 cm

19. (a) 7 s (b) 2

7 or

27 2

s (c) 49 cm

20. (a) ( )sinx t 2=

sin cos

sindtdx

t t

t

2

2

=

=

(b) , , , , . . .π

ππ

02 2

3s (c) 1 ms -2

Exercises 6.7

1. 12 cm 2. 28 m 3. -42.5 cm

4. (a) 570 cms -2 (b) 135 cm (c) After 0.5 s

5. ( 1)e cm3 + 6. 163 m 7. (a) 95 cms -1 (b) 175 cm

8. .h t t4 9 4 22= − + + 9. 262 m 10. ( )e 3 m5 −

11. -744 cm 12. ( )π2 3 cm− 13. 1.77 m 14. 893 m

15. (a) ( )3 3 m+ (b) 4 3 ms 2− −

16. (a) 154

ms -1 (b) ( )lnxn

n3

22 3 m= − +

(c)

( )

( )

( )

≥

vt

t

t

tt

tt

t

32

32

3 3

2 3 6

3 32 6 6

3 92

0

= −+

=+

+ −

=+

+ −

=+

When :t v0 0= = When :t v0 0> > Also t2 0> and t3 9 0>+ when t 0>

≤

t t

tt

v

2 3 9

3 92

1

0 1

So <

<

<`

+

+

17. (a) 5 e 45 ms −1 (b) e 30 m (c) x ex

2525

t5==

p (d) 50 ms -2

18. (a) 20 cms -2 (b) 9

828

cm

19. 3

12 8 2− ms -1

20. (a) 1− cms -1 (b) π

82

cm−

(c) , , , . . .π π π4 4

34

5s

Exercises 6.8

1. (a) x t8 2= + (b) xt4 12= −−

(c) xt

714= +

(d) xt24 1

23

= −−

(e) x t24 83= +

2. x t10 1= + 3. 0.93 m 4. xt

29 61

3= −

5. (a) ( )e21

1 s3 − (b) 1.2 m 6. x t4 3= − −

7. (a) x e5 t4= (b) e5 cm12 (c) e20 cms12 1− (d) At the origin, .x e0 5 0t4`= = This has no solution. ≠x 0` (particle is never at the origin)

(e)

( )

x ex ex e

ex

5208016 516

t

t

t

t

4

4

4

4

=====

o

p

8. (a) After 0.05 s

(b) cost x t101

2 0>`= − only when cos x2 0<

When ,≤ ≤ ≤ ≤π π

x x04

0 22

.≥cos x2 0 But t cannot be negative

displacement is never between 0 and π4

.

9.

`

`

`

`

x

( )

( ),

( )

( )

ln

lnln

ln

dtdx

x

dxdt

x

tx

dx

x Cx

CC

CC

xe x

e x

dtdx

edtd

e

x

5

51

51

50 6

0 6 51

0

55

5

a

When

and

velocity acceleration for all

t

t

t t2

2

= −

=−

=−

= − += == − += += +== −= −

+ =

= =

=

t

t

#

(b) 60 m

10.

ln

dtdx

x

dxdt

x

tx

dx

t x C

5

51

5

51

`

`

=

=

=

= +

#

613ANSWERS

Now ln x is real for x 0> displacement is always positive

,v xxxv

50

5 00

Since >>>`

=

So velocity is always positive.

11. (a) 3 s (b) 2 s

12. 321ln

m

Exercises 6.9

1. v x x2 10 92= + +

2. v x x x2 4 2 684 2= − − + −

3. 8 ms -1 4. (a) 2.8 ms -1

( ) ,

.

ππ

ππ

π π

sin

sin

sin

vx

x

x

012

24 0

122

4

2 124

1 05

b For

i.e.

= + =

= −

=

= −

This has no solution, so ≠v 0 (particle is never at rest) .

5. 7.9 cms -1 6. 1.55 cms -1 7. ( )v n a x2 2 2 2= −

8. 0.37 m

9. (a) v x x2 4 162= − + + (b) 3 2 ms 1− (c) Between -2 m and 4 m

10. ( )

vx

5

3 792 5

=− +

11. (a) 4 kms -1

( ) , v

x

0400

0

400 0

b At rest

i.e.

=

=

=

This is impossible . the rocket never comes to rest

12. ( )

vx

k x

3200

6400=

−

Exercises 6.10

1. (a) cosx t2=

(b) sinv t2= −

(c) cosa t2= −

2. (a) sinx t5=

(b) cosv t5=


(c) sina t5= −

3. (a) cosx t4 2=

(b) , , , , , . . . ; ±π

ππ

πt x02 2

32 4= =

(c) sinv t8 2= −

(d) v 0= (e) cosa t16 2= −

(f) a 0=

4. (a) cossincos

x tx tx t

x

22 24 24

== −= −= −

o

p

(b) , π1amplitude period= =

(c) cosx t2=

5. (a)

( )

cossin

coscos

x tx tx t

tx

2 36 318 39 2 39

== −= −= −= −

o

p

(b) ± 2 (c) v 0= (d) ;±v a6 0= =

6. (a)

( )

cossin

coscos

x tx tx t

tx

7 535 5175 525 7 525

== −= −= −= −

o

p

(b) , , , , . . . ; ±π π π

t x05 5

25

37= = (c)

π5

2Period =

7. (a)

( )

sincos

sinsin

x tx tx t

tx

3 412 4

48 416 3 416

=== −= −= −

o

p

(b) , , , . . .π π π

t8 8

38

5= (c) ;± ±x x3 48= =p

8. (a) x x36= −p (b) 12 ms -1

(c) , , , , . . . ; ±π π π

t x06 3 2

12 ms 1= = −o

(d) ±v x144 36 2= −

9. (a) ;π

cosx t24

= +b l

;

,

π πsin cosx t x t

x

24

24

SHM`

= − + = − +

= −

o pb bl l

(b) , , , . . .π π π

t4 4

54

9= (c) π2 (d) ±x 2=

10. (a)

( )

cos sinsin coscos sincos sin

x t tx t tx t t

t tx

5 3 2 315 3 6 345 3 18 39 5 3 2 39

= += − += − −= − += −

o

p

(b) 16.2 ms -1

11. (a) ( )( )( )

[ ( )]

πππ

π

cossincoscos

x tx tx t

tx

4 312 336 39 4 39

= += − += − += − += −

o

p

(b) 2 3 cm (c) ,π

43

2Amplitude period= =

615ANSWERS

12. (a) 0 m, 4 m (b) 2 m (c) 4 ms -1 (d) x x2= −p

13. (a) .±1 5 m (b) 5 ms -1

14. (a) ±

vx

381 2

=−

(b) ±x 3 5 cm=

15. . ; .v a6 5 5 5cms cm–1 2= − = − −

16. Period ,π3

2 amplitude ;

34 3

π

π

cos

sin

x t

x t

34 3

36

34 3

33

or= −

= +

c

c

m

m

17. (a) Between x 1= − and x 9= (b) Yes—centre of motion is x 4=

:

( )

( )

X x

dtd x

x

x

X n

4

8 2

2 4

2 2

Let

2

2

= −

= −

= − −= − =

18. (a) a x1600= − (b) π20

Period = (c) 30 cms -1

19. (a) ( )

sin cossin cos

an nt bn ntn a nt b ntn x

2 2

2

2

= − −= − += −

xp

(b) Amplitude ;a b2 2+ πn

2period = (c) n a b2 2+

20. (a) π

65 3

cms 1− −

(b) π9

5cms

22−

21. (a) sin cosx t t2 3 3= − ( ) ( )

( ) ( )

( )

cos sincos sin

sin cossin cos

sin cos

x t tt t

x t tt t

t tx

2 3 3 3 36 3 3 36 3 3 3 3 3

18 3 9 39 2 3 39

= − −= += − += − += − −= −

o

p

(b) Amplitude 5 , period π3

2

(c) 45 3 5=

22. (a) Equilibrium ,x 1= endpoints ,x x0 2= =

(b) Period π6

2

23. (a) Centre ,x 3= endpoints ,x x0 6= = (b) cosx t3 3 3= −

( )

( )

( )( )( )

sinsin

coscos

coscos

x tt

x tt

tt

x

3 3 39 39 3 327 3

9 3 39 3 3 3 39 3

= − −==== − −= − − −= − −

o

p

24. (a) v x x4 32 2= − −

( )

v xx

dxd

v x

x xxx

21

22 2

3

21

2

222

So

22

2

= − −

= −

= −= − += − −

p

c m

This is in the form ( )x n x x20= − −p so SHM.

(b) Centre ,x 2= endpoints ,x x1 3= = (c) ,a n1 1= =

25.

1

( )( ) ( )

( )

( )

( )( )( )( )( )

coscos

cos sinsin cos

sin cossin

coscoscos sincos coscos cos

cos

x tt

x t tt t

t tt

x tt

t tt tt t

tx

22

2 2 2 24 2 22 2 2 22 42 4 48 48 2 28 2 1 28 2 1 28 2 2 18 2 1

2

2

2 2

2 2

2 2

2

=== −= −= −= −= −= −= − −= − − −= − − += − −= − −

o

p

5 ?

This is in the form ( )x n x x20= − −p so SHM.

Exercises 6.11

1. (a) (i) xt

215 2

= (ii) y tt

52

15 22= − +

(b) 2

3 2s

2. (a) 6 3 s (b) 540 m 3. (a) 2.3 s (b) 5.8 m

4. (a) 2.59 m (b) 7.3 m 5. 2.8 s

6. (a) 4 minutes (b) 102 km

7. (a) , ,cos cosx x u a x ut a0= = =p o , ,sin

sin

y g y gt u a

ygt

ut a2

2

= − = − +

= − +

p o

(b) 15 m

8. ( )θ θtan tanyx

x2565

12

2= − + +

9. ,β βcos sinx vt ygt

vt h2

2

= = − + +

10. °20 34l 11. 1.8 ms -1 12. 28.86 ms -1

13. ° , °2 45 87 15l l 14. 9.3 ms -1

15. (a) 6 s (b) 91.7 m 16. ° °63 6 50 52orl l 17. 2 m

18. 8 m 19. 0.12 m 20. (a) 8 ms -1 (b) °53 8l (c) 3.2 m

21. (a) 4.8 s (b) 100.9 m


22. (a) Second stone is fi rst by 1.46 s (b) 1st stone:

( )

x t10

10 2 320 3 m

===

2nd stone:

( ( ))x t10 3

10 10 3 220 3 m

===

So both land at the same place .

23. ° °63 26 39 27l l

24. 2.7 s

25. (a) 29.5 m

(b) . , .x y11 5 8 2= = or (11.5, 8.2); No they will not collide—they reach this point at different times.

Test yourself 6

1. -2 ms -1

2. (a) 0 m, 0 ms -1 , 8 ms -2 (b) 0, 0.8 s (c) 0.38 m

3. (a)

( )( )

T Ae

dtdT

kAe

k Aek T

25

25 2525

kt

kt

kt

= +

= −

= − + −= − −

−

−

−

(b) , .A k295 0 042= = (c) c.108 4 (d) 97 min or 1 h 37 min

4. -76 m, -66 ms -2 5. 39.6 years

6. (a) 6 cms 1− (b) 145 855.5 cms -2 (c)

( )

x ex ex e

ex

26189 29

t

t

t

t

3

3

3

3

=====

o

p

7. 1 m

8.

( )

sincos

sinsin

x tx tx t

tx

2 36 3

18 39 2 39

=== −= −= −

o

p

9. (a) 2, 6 s (b) (i) 16 cm (ii) 15 cms 1− (iii) 18 cms 2− − (c) Particle is 16 cm to the right of the origin, travelling at 16 cms 1− to the right. Acceleration is 18 cms 2− − (to the left), so the particle is slowing down.

10. (a) (i) 18 ms 2− (ii) 15 ms 1− (iii) -28 m

(b) Particle is 28 m to the left of the origin, travelling at 15 ms 1− to the right, with 18 ms 2− acceleration (to the right), so the particle is speeding up.

11. (a) , ,t t t1 3 5 (b) ,t t2 4 (c) andt tafter3 5

(d) (i)

(ii)

12. (a) 48.2% (b) 1052.6 years

13. (a) , , , , . . .π π π π6 6

56

136

17s (b) , , , , . . .

π π π π3 3

53

73

11s

(c) 2

1ms 2− −

14. (a) 15 m (b) 20 m (c) 4 s

15. (a) 16 941 (b) 1168 birds/year (c) 18.3 years

16. 0.0193 mms -1

17. (a) (i)

(ii)

(b) ,t t1 3

18. (a) Period π7

2 (b) 7 ms 1−

(c) , , , , . . .π π π π5 13 17

21 21 21 21seconds

19. ° , °80 58 16 38l l 20. 55 033 m

617ANSWERS


1. (a) ( )

coscos

x tt

32

338

3

2 4 3= − + =

−

(b) x x938= − −p c m

(c) Yes. ,π

32

32

Amplitude period= =

2. (a) 1 m, 0 ms -1 (b) . ´3 26 107 ms -2 (c) Show ( )t 1 03 6+ = has no solution for ≥t 0

3. (a) 0.75 cm

(b) xe

21

cm10

= −

4. , , ,x xV

xVt

02 2

= = =p o

, ,y g y gtV

ygt Vt

2 2 2

2

= − = − + = − +p o

: ,ygt Vt

tgt V

t

tg

V

02 2

0

2 20

0

2

2

Range when

or

2

`

= − + =

− + =

=

=

f p

,

( )

´tg

Vx

V

g

V

gV

2

2

2 2

2

1

When

2

= =

=

y

gtV

tg

V

0

20

2

Height is maximum when

i.e.

=

− + =

=

o

,

[ ]

tg

Vy

g

g

V V

g

V

gV

gV

gV

x

2 2 2 2 2

4 2

41

41

When

from (1)

2

2

= = − +

= − +

=

=

2 2

f f

e

p p

o

range is 4 times maximum height

5. 0.25 cms -1

6. (a) π π

π π

π π

π π

sin cos

cos sin

sin cos

sin cos

x t t

x t t

x t t

t t

x

2 42

3 42

8 42

12 42

32 42

48 42

16 2 42

3 42

16

= + + +

= + − +

= − + − +

= − + + +

= −

o

p

c c

c c

c c

c c

m m

m m

m m

m m= G

(b) 3.6 m (c) 14.4 ms -1

7. (a) 36 208 (b) 67 795 bacteria per hour (c) 32 000

8. ° , °88 35 10 52l l

9. (a) 7.85 mm 2 per hour (b) 31 416 mm 2

10. (a) cosx t4= (b) a x16Yes, = − (c) ±2 3 cms -1

11.

`

( )

( )

,

( )

cos

cos

sin

sin

sin

cos

sin

v t

x t dt

t Ct x

CC

x t

adtd

t

tx

5 5

5 5

50 0

0 0

5

5 5

25 525

a

When

=

== += == +==

=

= −= −

#

(b) 25 ms 2− (c) .7 5ms–2− (d) ±v x5 1 2= −

12. (a) 0.0024 cms -1 (b) 0.75 cm 3 s -1

13. (a) 19.9 years (b) 16%

14. (a) 16.1 ms -1 (b) ,

´ ´

vx x

03 2 225 0

2 4 3 2250

If

<

2

2∆

=+ + =

= −

no solutions ≠v 0`

15. Carla jumps further by 0.8 m

16. ( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

NbN k bN e

kN

dtdN

bN k bN e

kN k k bN e

bN k bN e

k N k bN e

bN k bN e

k N bN k bN e bN

bN k bN e

k N bN k bN e

bN k bN e

bk N

bN k bN e

k N

bbN k bN e

kN

kN bN

kt

kt

kt

kt

kt

kt

kt

kt

kt

kt

kt

kt

0 0

0

0 02

0 0

0 02

20 0

0 02

20 0 0 0

0 02

20 0 0

0 02

202

0 0

20

0 0

02

2

=+ −

=+ −

− − −

=+ −

−

=+ −

+ − −

=+ −

+ −

−+ −

=+ −

−+ −

= −

−

−

−

−

−

−

−

−

−

−

−

−f p

7

7

7

7

7

7

7

7

A

A

A

A

A

A

A

A


17. (a) θtanx

y=o

o

(b) ( ) ( )V x y2 2 2= +o o (by Pythagoras’ theorem)

( ) ( )V x y2 2` = +o o

18. (a) For small ,θ θ θsin Z

θ

π θsindtd

2

22

2

= −

π θZ −

(b) v v22 212+= π θ− (c) Yes. 2Period =

19. (a) e3 cms 2− (b) log 7 se

20. SHM is a continuous oscillation. There is no displacement

when .π π

tantn

x a2 2

as= =

( )( ) ( )

coscos sin

tan sec

x an ntx an nt n nt

an nt nt2

2

Also, 2

3

2 2

== − −=

−

−

o

p

This is not in the form ,x n x2= −p so the particle is not in SHM.


1. 3.2 years 2. 1.099 3. 27 m 4. 2−

5. , .x e x 3 42y= = 6. x e3 2 x2 2+ 7. ±x21=

8. (a) 47.5 g (b) 3.5 g/year (c) After 9.3 years

9. (a) 0 cms 1− (b) sina t x18 3 9= − = −

10. (a) 7750 L (b) 28 minutes

11. .622 1 units3 12. x4 312

+

13. ( )log x x C31

3 3 2e2 + − +

14. (a) 100 L (b) 40 L (c) 16 L− per minute, i.e. leaking at the rate of 16 L per minute (d) 12.2 minutes

15. logx x x C2 4 e3 2− + + 16. log3 2e

17. (a) 7.8 cm (b) .0 06 cms 2− −

18. e x C41 x4 + + 19.

ex1 2

x2

−

20. π2

21. (a) .k 0 101= (b) 2801 (c) 20 days

(d) (i) 11 people per day (ii) 283 people per day

22. (a) 1.77 (b) logx 3

1

e

23. ( )πe

e2

1 units2

4 3−

24. (a) ,v a6 0ms ms1 2= =− − (b) 3 m

(c) , , , . . .π π π4 4

34

5seconds

(d) ( )

sinsin

a tt

x

12 24 3 24

= −= −= −

25. .4 67 units2 26. 27 m− 27. .x 0 28=

28. (a)

( )

cossincos

cos

x tx tx t

tx

2 510 550 525 2 525

== −= −= −= −

o

p

(b) ; , , , . . .±π π π

t1010 10

32

cms 1 =− (c) ±2 cm

29. x y 2 0− + = 30. ,e2

121− −c m minimum

31. (a) .1 60 cm2 (b) .0 17 cm2

32. 15 months

33. (a) 1.2 s (b) .5

36 312 5 mZ (c) 1.8 m

34. ±ππ

x n24

=

35. 16 cm s ,3 1− − i.e. decreasing by 16 cm s3 1−

36. (a) π6

5cm (b)

π12

25cm2

37. e Cx 32 +−

38.

39. 1 40. cot x 41. ( )sece e5 1x x5 2 5 + 42. 3

619ANSWERS

43. (a)

(b) 2 units2

44. 0.348

45. (a) ( )sin cose x xx + (b) x xtan sec3 2 2

(c) π

sin x6 32

− −c m

46. (a) 546 ms 1− (b) ( )

a ee

x

204 54

t

t

2

2

===

(c) 20 ms -2

47. .29 5 4 ms 1Z −

48. (a) ( )πsin sinβ β+ = − (b) θcos 10

(c) 2π

θ θcos sin− =c m

49. ln ln ln8 338− = 50. ´6 12m m

51. (a) 5500 (b) 9.5 years

52. ° , °85 25 15 54l l 53. π

23

units3

54.

55. π

x y6 12

30− − − =

56. π π

33 3

3 3units2− =

−

57. ( ) ( )sin cose e e3 x x x2

58. (a) sincos

sin

x tx tx t

x

22 2

4 24

=== −= −

o

p

(b) , , , , . . .π

ππ

t 02 2

3s=

(c) ±v 2 cms 1= −

59. ( )e5 units2−

60. (a) 2

1 (b)

23

−

61. (a) 21 π cm 2 (b) ( )π21 9 cm2+

62. ( )sec log

x

x 1e2 +

63. lnx x x C3 25− − +

64. .8 32 units3 65. π πcose x C51 1x5 + +

66. sinx x C21

41

2+ + 67. π2

units32

68. .x 1 8Z

69. ( )e 1 units2 2− 70. 1−

71. ( ) .f 1 0 519=

72. (a) π π

sin cosx x dx x C2 2

12

2 2− = − − +c cm m#

(b) 3π

x2

=

73. (c) 74. (d) 75. (a)

76. (d) 77. (b) 78. (c)

79. (a), (c), (d) 80. (d)

Chapter 7: Inverse functions

Exercises 7.1

1. yx3

= 2. y x= − 3. ( )f x x51 =− 4. y x3=

5. yx7

= 6. ( )f x x 11 = −− 7. y x 5= +

8. ( )f x x 31 = −− 9. 3y x y xor3= =1

10. ( ) logf x x12=− 11. y 4x= 12. 5y x y xor= =

15

13. ( )f x x 91 = +− 14. ( )f x x51 = −− 15. yx3

= −

16. y x!= 17. y x7= 18. y ex= 19. y x9=

20. y x8!= -


Exercises 7.2

1. yx5

= 2. yx

23= +

3. y x 53= −

4. ( )f x x 11 7= +− 5. y x 23= + 6. y x2=

7. y x3

5= − 8. y x2 1= − 9. ( )f x x 21 2= −−

10. y x 73= + 11. yx9

2= 12. y

x5

1= −3

13. yx

32= +

5 14. ( )( )

f xx

4

51

2

=−

− 15. yx

54273

= −

16. lny x= 17. ln

yx

2= 18. y ex= 19. y e 1x= −

20. ( )( )ln

f xx

3

11 =

−− 21. ±y x= 22. ±y

x2

= 4

23. ±y x 5= − 24. ±y x 3= +6 25. ±y x 16 4= + −

26. ±y x4 2= − + 27. ±y x 2 1= − +

28. ±y x 26 5= + − 29. ±y x 12 3= + +

30. ±y x 47 6= + −

Exercises 7.3

1. Yes 2. No 3. No 4. Yes

5. Yes 6. No 7. Yes 8. No

9. No 10. No 11. Yes 12. No

Exercises 7.4

1. (a) : ;f y x1 3=− domain: all real x , range: all real y

(b) : ;f yx

321 = +− domain: all real x , range: all real y

(c) : ;lnf y x1 =− domain: ,x 0> range: all real y

(d) ( ) ;f x x21 =− domain: all real ,≠x 0 range: all real ≠y 0

(e) : ;f y x1

11 = −− domain: all real ,≠x 0 range: all real ≠y 1−

2. (a) ;yx2

= domain: ,≥x 0 range: ≥y 0

(b) ;y x 2= − domain: ,≥x 2 range: ≥y 0

(c) ;y x 3= + domain: ,≥x 0 range: ≥y 3

(d) ;y x 1 1= + + domain: ,≥x 1− range: ≥y 1

(e) ;y x6= domain: ,≥x 0 range: ≥y 0

(f) ;y x1= − − domain: ,≤x 1 range: ≤y 0

(g) ;y x 14= + domain: ,≥x 1− range: ≥y 0

(h) ;yx

1= − domain: ,x 0> range: y 0<

3. (a) ≥x 3− (b) ;y x 9 3= + − domain: ,≥x 9− range: ≥y 3− (c) ≤x 3− (d) ;y x 9 3= − + − domain: ,≥x 9− range: ≤y 3−

4. (a) y x= − (b) yx

31= − +

(c) ( )f x x 21 4= − +

(d) y x3= − (e) y x

24= −

5. (a) (i) 1y x 1= + + (ii) 1y x 1= − + +

(b) (i) y x 24= + (ii) y x 24= − +

(c) (i) yx

21

4= − (ii) y

x2

14= − −

(d) (i) 3y x 8= + + (ii) 3y x 8= − + +(e) (i) y x 7 2= + − (ii) y x 7 2= − + −

Exercises 7.5

1. (a) [ ( )] ( )( )

[ ( )] ( )( )

f f x f xx

xf f x f x

xx

77 7

77 7

1 1

1

= += + −== += − +=

− −

−

(b) [ ( )] ( )

[ ( )]

f f x f xx

x

f f x fx

x

x

3

33

3

33

1 1

1

=

=

=

=

=

=

− −

− d

d

n

n

(c) [ ( )]

[ ( )] ( )

( )

f f x f x

xx

f f x f x

xx

1 1

2

1 2

2

======

− −

−

^

^

h

h

(d) [ ( )] ( )( )

[ ( )] ( )

lnln

ln

f f x f ee

x ex

f f x f xex

ln

x

x

x

1 1

1

=======

− −

−

(e) [ ( )] ( )( )

[ ( )]

f f x f xx

x

x

f f x fx

x

x

3 1

3

3 1 1

33

31

33

11

1 1

1

= +

=+ −

=

=

= −

= − +

=

− −

− d

d

n

n

2. (a) Domain: all real ,≠x 1 range: all real ≠y 0

(b) ( )f x x2

11 = +−

(c) Domain: all real ,≠x 0 range: all real ≠y 1

3. (a)

621ANSWERS

(b) y ex= (c) — : ,lny x x 0domain >= range: all real y ; —y e domain:x= all real x , range: y 0>

4. (a) (i) x5 4 (ii) 5x y=1

(iii) dydx

x51 4= − (iv) ´ ´

dx

dy

dydx

x x551

14 4= =−

(b) (i) x1

2− (ii) x y

1=

(iii) dydx

x2= − (iv) ( )´dx

dy

dydx

xx

11

22= − − =

(c) (i) x 1

1+

(ii) x e 1y= −

(iii) dydx

x 1= + (iv) ( )´dx

dy

dydx

xx

11

1 1=+

+ =

(d) (i) e x− − (ii) lnx y= −

(iii) dydx

ee

1x

x= − = −− (iv) ( )´dx

dy

dydx

e e 1x x= − − =−

(e) (i) x2 3

1

− (ii) x y 32= +

(iii) dydx

x2 3= −

(iv) ´ ´dx

dy

dydx

xx

2 3

12 3 1=

−− =

5. (a) (i) x y3 1= + (ii) yx

31= −

(iii) 31

(iv) 3 (v) ´ ´dx

dy

dydx

31

3 1= =

(b) (i) x y3 5= (ii) 5

yx3

=1

d n (iii) 5x

151

3

− 4

d n

(iv) 5x

153

4

d n (v) 5 5

´ ´dx

dy

dydx x x

151

315

31= =

− 4 4

d dn n

(c) (i) xy 3

2=+

(ii) y x2

3= − (iii) x2

2−

(iv) x2

2

− (v) ´ ´dx

dy

dydx

xx22

12

2

= − − =

(d) (i) x y 73= − (ii) y x 73= + (iii) ( )x3 7

123 +

(iv) ( )x3 7 23 +

(v) ( )

( )´ ´dx

dy

dydx

xx

3 7

13 7 1

23

23=+

+ =

(e) (i) x e y5 1= + (ii) ln

yx5

1= − (iii)

x51

(iv) 5 x (v) ´ ´dx

dy

dydx

xx

51

5 1= =

Exercises 7.6

1. (a) π2

(b) 0 (c) 0 (d) π6

(e) π4

−

(f) π2

− (g) π2

(h) π4

(i) π6

(j) π3

− (k) π4

(l) π6

5 (m)

π6

−

2. (a) 0 (b) 1− (c) π2

(d) 2

1 (e)

2

1

(f) 3 (g) 0 (h) 3− (i) 23

(j) 1−

3. (a) 0.41 (b) 1.04 (c) 0.97 (d) .0 64− (e) .1 31−

4. (a) 0.67 (b) .0 14− (c) 1.64 (d) 0.97 (e) .0 90−

5. (a)

(b)

(c)

(d)

(e)


(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)

(n)

6. (a) ≤ ≤x1 1− (b) ≤ ≤x1 1− (c) ≤ ≤π π

x4 4

−

7. (a) π (b) 0 (c) 0

(d) π2

(e) π2

(f) π2

8. (a) 54

(b) 53

(c) 125

(d) 58

3 (e)

π4

(f) π4

−

9. (a) θ θsin cos2 (b) 2524

10. (a) odd (b) odd (c) even (d) neither (e) odd (f) odd (g) odd

11. (a) ( )

( )

πtan

tan

14

1

1

1

− = −

= −

−

−

(b) ( )

( )

πsin

sin

12

1

1

1

− = −

= −

−

−

(c) ( ) .( )

tantan

3 1 2493

1

1

Z− −= −

−

−

623ANSWERS

(d)

π

π

ππ

π

cos

cos

21

32

21

3

32

LHS

RHS

1

1

= −

=

= −

= −

=

−

−

c m

(e) π

sin

sin

2

14

2

1

1

1

− = −

= −

−

−

e

e

o

o

12. (a) ( )θ ππ

n 13

n= + − (b) 2θ πn= (c) θ ππ

n3

= +

(d) ( )θ ππ

n 14

n= + − (e) θ ππ

n4

= − (f) ±θ ππ

n26

=

(g) ±θ ππ

n22

= (h) n( )θ ππ

n 16

= + −

(i) n( )θ ππ

n 14

= − − (j) θ ππ

n6

= −

13. (a) n( )θ ππ

n 12

= − − (b) ,θπ π2 2

3= −

14. (a) ±θ ππ

n23

= (b) ,± ±θπ π3

53

7=

15 (a) θ ππ

n3

= − (b) , ,θπ π π3

23

53

8=

16. (a) .±πx n2 1 53= (b) n( ) .πx n 1 0 39= + − (c) .πx n 0 92= + (d) n( ) .πx n 1 1 04= − − (e) .πx n 0 75= − (f) .±πx n2 1 80=

17. (a) and (b)

y = cos-1 x

x

y

-

1-1

y = sin-1 x

y = sin-1 x + cos-1 x

π2

π2

π

18. (a)

θπ

θ

π

sin cos73

73

2

2

1 1+ = + −

=

− −

(b) .sin

sin

95

0 589

95

1

1

− = −

= −

−

−

d

d

n

n

(c)

.

.

π

cos

cos

52

1 9823

52

1 9823

LHS

RHS

1

1

= −

=

= −

=

−

−

d

d

n

n

So πcos cos52

521 1− = −− −d dn n

(d) .tan

tan

107

0 61

107

1

1

− = −

= −

−

−

d

d

n

n

Exercises 7.7

1. (a) x1

12

−−

(b) x1

22− (c)

x11

2+

(d) x1 9

32

−−

(e) x1 4

82− (f)

x

x

1

24−

(g) 2( )x x x1 2 1

22 2 1

12+ −

=− +

(h) x1 64

402

−−

(i) x9

12

−−

(j) x4

22+

(k) x36

32− (l)

( )x x2 1

3

−−

(m) x49

12

−−

(n) 2( )x x x1 3 2

15

9 12 3

152− +

=− − −

(o) cosx

xx

1 2

1

−− + − (p) ( )tan

xx

15

12

1 4

++−

2. (a) 1− (b) 1 (c) 2( )logx x1

1

e−

(d) e

e1 x

x

2+ (e)

sinx x1

12 1− −

(f) 2( ) ( )tanx x1

12 1

−+ −

(g) 2( )cosx x1 1 1

12 1− + +

−−6 @

(h) x1

12

−+

(i) x x x

2 12

1

1

4

12 2

− +

=− −

d n

(j) x

e

1

cos x

2

1

−− −


3. (a) (i) 1− (ii) 1

(b) (i) 153

(ii) 85−

(c) (i) π

6 3

2

(ii) π

6 32

−

(d) (i) 31− (ii) 3

(e) (i) 51

(ii) 5−

4. y x2= 5. πx y40 100 25 8 0+ − − =

6. (a) ( )sin cosdxd

x xx x1

1

1

1

0

1 1

2 2+ =

−+

−−

=

− −

(b) π

sin cosx x2

1 1+ =− −

π

dxd

20` =d n

7. ( ) ( )

,

,

sin sin

sin

sin

sin

dxd

x xx

xx

dx

dy

x

xx

xx

x

xx y

1

0

10

100 0 00

a

For

satisfies this equationWhen

1

2

1

2

1

1

2

1

=−

+

=

−+ =

=−

−

== ==

− −

−

−

−

( , )0 0 is a stationary point` (b) Domain: ,≤ ≤x1 1− range: ≤ ≤

πy0

2

(c)

8. (a) 3( )x

x

9 2−−

(b) ( ) ( )tan

tan

x x

x x

1

1 22 1 2

1

+− −

−

−

6 @

9. 1− for ,π π

x2 2

< <− 1 for ,≤ ≤ππ π

πx x2 2

< <− −

10. (a) 0

(b)

11. (a) e

e

1

2x

x

4

2

−−

(b) ( ) tanx x1

12 1+ −

(c) ( )lnx x1

12+ 5 ?

(d) x1

12

−−

(e) x

e1

tan x

2

1

+

−

12. (a) θ

θ

sin

sin

h

h6

61`

=

= − d n

(b) °0 0 33l m per second

Exercises 7.8

1. (a) sin x C1 +− (b) cos x C2 1 +− or sin x C2 1− +−

(c) tan x C1 +− (d) tanx

C31

31 +−

(e) sinx

C2

1 +− (f) tanx

C25

21 +−

(g) sin x C23 1 +− (h) sin

xC

51

41 +−

(i) tanx

C3

1

31 +− (j) sin

xC

51 +−

2. (a) tanx

C61

61 +− (b) sin x C

21

21 +−

(c) sint

C3

1 +− (d) tan x C31

31 +−

(e) sinx

C51

251 +− d n (f) tan

xC

41

341 +− d n

(g) sint

C521 +− d n (h) ( )tan x C

55

51 +−

(i) tanx

C63

23

1 +− e o (j) cosx

C152

531 +− d n

3. (a) π2

(b) π4

(c) π3

(d) π12

(e) π6

(f) π3

(g) π (h) π9

(i) π

283

(j) π

3 5

4. 1.1 units 2

5. (a) π6

units2 (b) ( )π6

7 3 12 units3−

6. (a) sinx x1

12 1− −

(b) 0.3

7. 0.1

8. 2

1units2

9. sin x C31 1 3 +−

10. π

sinyx3 14

51= −− d n

625ANSWERS

11. (a)

( ) ( )

( )( )

x x

x xx x

x xx

11

41

1 44 1

1 42 5

RHS

LHS

2 2

2 2

2 2

2 2

2

=+

++

=+ ++ + +

=+ +

+

=

(b) 0.48

12. π12

units2

3

13. (a) cos x1− (b) π6

123

units2+ −e o

14. π2

1 units2−d n

15. tanv x2 1= −

Test yourself 7

1. (a) π4

(b) π3

(c) π4

3

(d) 0 (e) π6

−

2. (a) tan x C3 1 +− (b) sinx

C4

1 +−

3. (a) x1

12− (b)

x1 93

2+ (c)

x1 25

102

−−

4. π

125

units2

5. ( )f xx

231 = −−

6. 2 ±θπ

πn4

=

7. ( )( )ln

f xx

3

11 =

−−

8. (a) 0

(b) π

sin cosx x2

1 1+ =− − and π

dxd

20=d n

9. tanx

xx

1 21

++ −

10. (a) Domain: all real ;≥x 1 range: all real ≥y 0 (b) ( )f x x 11 2= +− Domain: all real ≥x 0

11.

12.

ππ

π

cos

cos

23

6

23

LHS

RHS

1

1

= −

= −

= −

=

−

−

e

e

o

o

13. (a) Domain: all real ;≠x 2− range: all real ≠y 0

(b) ( )f x x1

21 = −−

(c) Domain: all real ;≠x 0 range: all real ≠y 2−

14. (a) π6

units2 (b) 1.73 units 3

15. (a) π2

(b) 34

16. (a) Domain: ;≤ ≤x1 1− range: ≤ ≤π πy−

(b)

17. πx y4 3 18 6 3 3 0− − + =

18. (a) x 2> (b) ( )f x x 4 21 = + +−

19. (a) 2 (b) 10.5

20. (a) π3

(b) π

147


1. (a) ,y x1 2= − − domain: ,≤ ≤x0 1 range:

≤ ≤y1 0−

(b) ;y x1

1= − + domain: , ,≤x x0 1> −

range: ,≤ ≠y y0 1−

2. ; : ≠sin

x x

x x xx

1

10no

2 2

2 1

−

− − −

3. ( )tan x C21 1 2 +−


4. (a)

θπ

θ

π

tan tan54

45

2

2

LHS

RHS

1 1= +

= + −

=

=

− −

(b) 0 (c)

θπ

θ

π

tan tan

tan tan

x xx

x

1

11

2

2

LHS

RHS

1 1

1 1

= +

= +

= + −

=

=

− −

− −

5. .1 4 ms 1−

6. π π

63

21

63 3

units2− =−

7. Let θsin x1 =−

Then θsinx1

1 =−

By Pythagoras’ theorem,

θ

θ

cos

cos

sin cos

BC xx

xx

x x

1

11

11

1

2

2

2

1 2

1 1 2

`

`

= −

=−

= −= −= −

−

− −

8. ( )sin x C61

31 2 +−

9.

10. (a) 0.9 units 3 (b) π units 3

Chapter 8: Series

Exercises 8.1

1. 14, 17, 20 2. 23, 28, 33 3. 44, 55, 66

4. 85, 80, 75 5. , ,1 1 3− − 6. 87, 83, 79

7. , ,2 221

3 8. 3.1, 3.7, 4.3 9. 16, 32, 64

10. 108, 324, 972 11. 16, 32− , 64 12. 48, 96− , 192

13. , ,161

321

641

14. , ,13516

40532

121564

15. 36, 49, 64 16. 125, 216, 343 17. 35, 48, 63

18. 38, 51, 66 19. 126, 217, 344 20. 21, 34, 55

Exercises 8.2

1. (a) , ,T T T3 11 191 2 3= = = (b) , ,T T T5 7 91 2 3= = = (c) , ,u u u5 11 171 2 3= = = (d) , ,T T T3 2 71 2 3= = − = − (e) , ,t t T19 18 171 2 3= = = (f) , ,u u u3 9 271 2 3= = = (g) , ,Q Q Q9 11 151 2 3= = = (h) , ,t t t2 12 581 2 3= = = (i) , ,T T T8 31 701 2 3= = = (j) , ,T T T2 10 301 2 3= = =

2. (a) , ,T T T1 4 71 2 3= = = (b) , ,t t t4 16 641 2 3= = = (c) , ,T T T2 6 121 2 3= = =

3. (a) 349 (b) 105 (c) 248 (d) -110 (e) -342

4. (a) 1029 (b) -59 039 (c) 1014 (d) 53 (e) 1002

5. (a), (b), (d) 6. (b), (d) 7. 16 th term 8. Yes

9. 7 th term 10. 23 rd term 11. (a) 1728 (b) 25 th term

12. (a) -572 (b) 17 th term

13. n 33= 14. n 9= 15. , , , . . .n 88 89 90=

16. , ,n 41 42 43= 17. n 501= 18. n 151=

19. (a) n 14= (b) -4 20. -6

627ANSWERS

Exercises 8.3

1. (a) 128 (b) 54 (c) 70 (d) 175 (e) 220

(f) 6047

(g) 40 (h) 21 (i) 126 (j) 1024

2. (a) 65 (b) 99 (c) 76 (d) 200 (e) 11 (f) 39 (g) 97 (h) 66 (i) 75 (j) 45

3. (a) n2 1n 1

6

−=

/ (b) n7n 1

10

=/ (c) n

n

3

1

5

=/ (d) k6 4

k

n

1−

=/

(e) kk

n2

3=/ (f) ( )n

n 1

50

−=

/ (g) .3 2k

k

n

0=/ (h)

21

nn 0

9

=/

(i) ( )a k d1k

n

1+ −

=/ (j) ark

k

n1

1

−

=/

Exercises 8.4

1. (a) y 13= (b) x 4= − (c) x 72= (d) b 11= (e) x 7=

(f ) x 4221= (g) k 4

21= (h) x 1= (i) t 2= − (j) t 3=

2. (a) 46 (b) 78 (c) 94 (d) -6 (e) 67

3. (a) 590 (b) -850 (c) 414 (d) 1610 (e) -397

4. (a) -110 (b) 12.4 (c) -8.3 (d) 37 (e) 1554

5. T n2 1n = +

6. (a) T n8 1n = + (b) T n2 98n = + (c) T n3 3n = +

(d) T n6 74n = + (e) T n4 25n = − (f) T n20 5n = −

(g) Tn

86

n = + (h) T n2 28n = − − (i) .T n1 2 2n = +

(j) Tn4

3 1n = −

7. 28 th term 8. 54 th term 9. 30 th term

10. 15 th term 11. Yes 12. No

13. Yes 14. n 13= 15. , , . . .n 30 31 32=

16. -2 17. 103 18. 785

19. (a) d 8= (b) 87

20. d 9= 21. ,a d12 7= =

22. 173 23. a 5=

24. 280 25. 1133

26. (a) log loglog log

log

log loglog log

log

T T x x

x x

x

T T x x

x x

x

2

3 2

2 1 52

5

5 5

5

3 2 53

52

5 5

5

− = −= −=

− = −= −=

Since T T T T2 1 3 2− = − it is an arithmetic series with .logd x5= (b) log logx x80 or5 5

80 (c) 8.6

27. (a)

´

´ ´

T T

T T

12 3

4 3 32 3 3

327 12

9 3 4 33 3 2 3

3

2 1

3 2

− = −

= −= −=

− = −

= −= −=

Since T T T T2 1 3 2− = − it is an arithmetic series

with .d 3=

(b) 50 3

28. 26 29. 122 b 30. 38 th term

Exercises 8.5

1. (a) 375 (b) 555 (c) 480

2. (a) 2640 (b) 4365 (c) 240

3. (a) 2050 (b) 2575−

4. (a) 4850− (b) 4225

5. (a) 28 875 (b) 3276 (c) 1419− (d) 6426 (e) 6604 (f) 598 (g) 2700− (h) 11 704 (i) 290− (j) 1284

6. (a) 700 (b) 285− (c) 1170 18 terms− ] g (d) 6525 (e) 2286−

7. 21 8. 8 9. 11 10. ,a d14 4= =

11. ,a d3 5= − = 12. 2025 13. 3420

14. 8 and 13 terms 15. 1010

16. (a) ( ) ( )x x2 4 1+ − +

( ) ( )x xx3 7 2 4

3= + − += +

(b) ( )x25 51 149+

17. 1290 18. 16

19. S S T

S S Tn n n

n n n

1

1`

= +− =

−

−

20. 4234

Exercises 8.6

1. (a) No (b) Yes, r43= − (c) Yes, r

72=

(d) No (e) No (f ) No (g) Yes, .r 0 3=

(h) Yes, r53= − (i) No (j) Yes, r 8= −

2. (a) x 196= (b) y 48= − (c) ±a 12=

(d) y32= (e) x 2= (f ) ±p 10=

(g) ±y 21= (h) ±m 6= (i) x 4 3 5!=


(j) 3 7±k 1= (k) ±t61= (l) ±t

32=

3. (a) T 5nn 1= − (b) .T 1 02n

n 1= − (c) T 9 –n

n 1=

(d) .T 2 5nn 1= − (e) .6 3Tn

n 1= − (f) .T 8 2

2n

n

n

1

2

==

−

+

(g) ·T41

4

4

n

n 2

=

= −

n 1− (h) T 1000 10

10n

n

n

1

2

= −= − −

−

+

]

]

g

g

(i) T 3 3

3n

n

n

1= − −= −

−]

]

g

g

(j) T31

52

n

n 1

=−

d n

4. (a) 1944 (b) 9216 (c) -8192

(d) 3125 (e) 72964

5. (a) 256 (b) 26 244 (c) 1.369

(d) -768 (e) 1024

3

6. (a) 234 375 (b) 268.8 (c) -81 920

(d) 156 250

2187 (e) 27

7. (a) ´3 219 (b) 7 19 (c) 1.04 20

(d) 41

21

2119

21=d n (e)

43 20

d n

8. 11 49 9. 6 th term

10. 5 th term 11. No

12. 7 th term 13. 11 th term

14. 9 th term 15. n 5= 16. r 3=

17. (a) r 6= − (b) −18

18. , ±a r1

210

= = 19. n 7= 20. 20872

Exercises 8.7

1. (a) 2 097 150 (b) 7 324 218

2. (a) 720 600 (b) 26 240

3. (a) 131 068 (b) 65 53632 769

4. (a) 7812 (b) 356455

(c) 8403 (d) 273 (e) 255

5. (a) 255 (b) 729364

(c) 97 656.2

(d) 1128127

(e) 87 376

6. (a) 1792 (b) 3577

7. 148.58 8. 133.33

9. n 9= 10. 10 terms

11. a 9= 12. 10 terms

13. (a) $33 502.39 (b) $178 550.21

14. (a) 1−( )2 5 k

k

n

1−

=/ (b)

( ) ( )S

3

5 1

3

1 5n

n n

= −− −

=− −

15. 2146

Puzzles

1. Choice 1 gives $465.00. Choice 2 gives $10 737 418.23!

2. 382 apples

Exercises 8.8

1.(a) Yes LS 1321= (b) No (c) Yes LS 12

54= (d) No

(e) Yes LS 3= (f) Yes LS 3225= (g) No

(h) Yes LS 1225= − (i) No (j) Yes LS 1

73=

2. (a) 80 (b) 42632

(c) 6632

(d) 12 (e) 107

(f) 54

(g) 1072− (h)

209

(i) 48 (j) 3916−

3. (a) 127

(b) 274

(c) 12500

1 (d)

641

(e) 40963645

4. (a) 141

(b) 52

(c) 481

(d) 221

(e) 3 (f) 5 (g) 52

(h) 531− (i) 1

54

(j) 56

5. a 4= 6. r52= 7. a 5

53= 8. r

87= 9. r

41= −

10. r32= − 11. 3, ,a r a r

32

631

and= = = =

12. 192, , 153a r41

53

LS= = − =

13. 1, , 3, 1, ,a r a r32

32

43

LS LS= = = = − = − = −

14. 150, , 375a r53

LS= = =

15. , , 1a r52

32

51

LS= = = 16. , ,a r a r352

253

and= = = =

17. x3221= 18. (a) k1 11 1− (b)

52− (c) k

43=

19. (a) p21

21

1 1− (b) 75

(c) p141=

20.

Sr

ar

a r

r

a a r

ra a ar

rar

1 1

1

1

1

1

1

LS n− =−

−−

−

=−

− −

=−

− +

=−

n

n

n

n

^

^

h

h

Exercises 8.9

1. (a) 210 (b) 13th (c) 57

2. (a) 39 (b) 29th (c) 32

3. (a) n3 3+

629ANSWERS

(b) [ ( ) ]

[ ( ) ]

( )

( )

( )

´ ´

Sn

a n d

n n

n n

n n

n n

22 1

21

2 6 1 3

21

12 3 3

21

3 9

23

3

n = + −

= + −

= + −

= +

= +

4. (a) (i) $23 200 (ii) $26 912 (iii) $31 217.92 (b) $102 345.29 (c) 6.2 years

5. (a) (i) 93% (ii) 86.49% (iii) 80.44% (b) 33.67% (c) 19 weeks

6. (a) 0.01 m (b) 91.5 m

7. (a) 49 (b) 4 mm

8. (a) 3 k m (b) ( )k k3 3 m+ (c) 9

9. (a) 96.04% (b) 34 (c) 68.6

10. (a) 77.4% (b) 13.5 (c) 31.4

11. (a) 94

(b) 97

(c) 192

(d) 9925

(e) 2119

(f) 307

(g) 19043

(h) 14507

(i) 990131

(j) 2999361

12. 0.625 m 13. 15 m 14. 20 cm 15. 3 m

16. (a) 74.7 cm (b) 75 m

17. (a) 4.84 m (b) After 3 years

18. 300 cm 19. 3.5 m 20. 32 m

21. (a) 1, 8, 64, … (b) 16 777 216 people (c) 19 173 961 people

Exercises 8.10

1. (a) $740.12 (b) $14 753.64 (c) $17 271.40 (d) $9385.69 (e) $5298.19

2. (a) $2007.34 (b) $2015.87 (c) $2020.28

3. (a) $4930.86 (b) $4941.03

4. $408.24 5. $971.40

6. $1733.99 7. $3097.06

8. $22 800.81 9. $691.41

10. $1776.58 11. $14 549.76 12. $1 301 694.62

13. (a) $4113.51 (b) $555.32 (c) $9872.43 (d) $238.17 (e) $10 530.59

14. $4543.28 15. 4 years 16. 8 years

17. (a) x 7= (b) x 5= (c) x 8=

(d) .x 6 5= (e) .x 8 5=

18. $7.68 19. Kate $224.37

20. Account A $844.94

Exercises 8.11

1. $27 882.27 2. $83 712.95

3. $50 402.00 4. $163 907.81

5. $40 728.17 6. $29 439.16

7. $67 596.72 8. $62 873.34

9. $164 155.56 (28 years) 10. $106 379.70

11. $3383.22 12. $65 903.97

13. $2846.82 14. $13 601.02

15. $6181.13 16. $4646.71 17. $20 405.74

18. (a) $26 361.59 (b) $46 551.94

19. $45 599.17

20. (a) $7335.93 (b) $1467.18

21. $500 for 30 years 22. Yes, $259.80 over

23. No, shortfall of $2013.75

24. (a) $14 281.87 (b) $9571.96 (c) No, they will only have $23 853.83 .

25. $813.16

Exercises 8.12

1. $1047.62 2. $394.46 3. $139.15

4. (a) $966.45 (b) $1265.79

5. $2519.59

6. (a) $592.00 (b) $39 319.89

7. (a) $77.81 (b) $2645.42

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12. NSW Bank $175.49 a month ($5791.25 altogether) Sydney Bank $154.39 a month ($5557.88 altogether) Sydney Bank is better


13. (a) $249.60 (b) $13 485.12

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15. (a) $1835.68 (b) $9178.41

Exercises 8.13

1. Step 1: Prove true for n 1= –

–

( )

( )´

n

nn

7 47 1 43

27 1

21

7 1 1

3

LHS

RHS

LHS RHS

===

= −

= −

==

] g

So the statement is true for .n 1=

Step 2: Assume true for n k=

So … – ( )kk

k3 10 17 7 42

7 1+ + + + = −] g

Step 3: Prove true for n k 1= + Prove … – [ ]k k3 10 17 7 4 7 1 4+ + + + + + −] ^g h

( [ ] )k

k2

17 1 1= + + −

i.e. … –

( )

k kk

k

3 10 17 7 4 7 3

21

7 6

+ + + + + +

= + +

] ]g g

… –

( )

( )( )

( ) ( )

( )( )

( )( )

k kk

k k

kk

k

k k k

k k k

k k

k k

kk

3 10 17 7 4 7 3

27 1 7 3

27 1

2

2 7 3

2

7 1

2

2 7 3

27 14 6

27 13 6

2

1 7 6

2

17 6

LHS

RHS

2

2

= + + + + + +

= − + +

= − ++

=−

++

= − + +

= + +

=+ +

=+

+

=

] ]

]

g g

g

So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on . The statement is true for all integers .≥n 1

2. Step 1: Prove true for n 1=

( )´

n

n n

8 38 1 35

4 11 4 1 15

LHS

RHS

LHS RHS

= −= −== += +==

]

]

g

g


Step 2: Assume true for n k= So … –k k k5 11 19 8 3 4 1+ + + + = +] ]g g

Step 3: Prove true for n k 1= + Prove … k k5 11 19 8 3 8 1 3+ + + + − + + −] ]g g5 ?

k k1 4 1 1= + + +] ]g g5 ?

i.e. … –k k5 11 19 8 3 8 5+ + + + + +] ]g g

k k1 4 5= + +] ]g g

… k kk k k

k k kk kk k

5 11 19 8 3 8 54 1 8 5

4 8 54 9 5

1 4 5

LHS

RHS

2

2

= + + + + − + += + + += + + += + += + +=

] ]

] ]

] ]

g g

g g

g g

So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on. The statement is true for all integers .≥n 1

3. Step 1: Prove true for .n 0= ·

( )

5 255 2 15 2 15

LHS

RHS

LHS RHS

0

0 1

=== −= −==

+

] g

So the statement is true for n 0= .

Step 2: Assume true for n k= So … . ( )5 10 20 5 2 5 2 1k k 1+ + + + = −+

Step 3: Prove true for n k 1= + Prove … · · ( )5 10 20 5 2 5 2 5 2 1k k k1 1 1+ + + + + = −+ + + i.e. … · · ( )5 10 20 5 2 5 2 5 2 1k k k1 2+ + + + + = −+ + … · ·

·· ·

·· ··

5 10 20 5 2 5 25 2 1 5 25 2 5 5 210 2 55 2 2 55 2 55 2 1

LHS

RHS

k k

k k

k k

k

k

k

k

1

1 1

1 1

1

1

2

2

= + + + + += − += − += −= −= −= −=

+

+ +

+ +

+

+

+

+

]

]

g

g

So the statement is true for n k 1= + . The statement is true for n 0= so it must be true for n 1= . It is true for n 1= so it must be true for n 2= and so on . The statement is true for all integers ≥n 0 .


´

21

21

1

2 121

221

1

LHS

RHS

LHS RHS

1 1

0

1

=

=

=

= −

=

==

−

d n



631ANSWERS


So …121

41

21

2 121

k k1+ + + + = −

−d n

Step 3: Prove true for n k 1= +

Prove …121

41

21

21

2 12

1k k k1 1 1 1

+ + + + + = −− + − +

d n

i.e. …121

41

21

21

2 12

1k k k1 1

+ + + + + = −− +

d n

…

·

´

121

41

21

21

2 121

21

2 221

21

221

221

22

22

2

2 12

1

LHS

RHS

k k

k k

k k

k

k

k

k

1

1

1

= + + + + +

= − +

= − +

= −

= −

= −

= −

=

−

+

+

d

e

n

o


5. ( ) ( ) ( ) ( ) . . .

( ). . .

´ ´ ´r

nn

3 1 3 1 1 3 2 1 3 3 1

3 12 5 8 3 1

r

n

1− = − + − + − +

+ −= + + + + −

=

] g

/

Step 1: Prove true for n 1=

2( )

n

n n

3 13 1 12

23

2

3 1 1

2

LHS

RHS

LHS RHS

2

= −= −=

= +

=+

==

] g



So … kk k

2 5 8 3 12

3 2

+ + + + − = +] g


… – –( ) ( )

… –( ) ( )

k kk k

k kk k k

k k k

k k

2 5 8 3 1 3 1 1

2

3 1 1

2 5 8 3 1 3 2

2

3 2 1 1

23 6 3 1

23 7 4

Prove

i.e.

2

2

2

2

+ + + + + +

=+ + +

+ + + + + +

=+ + + +

= + + + +

= + +

] ]

] ]

g g

g g

5 ?

… –

( )

k kk k

k

k k k

k k k

k k k

k k

2 5 8 3 1 3 2

23

3 2

23

2

2 3 2

23

26 4

23 6 4

23 7 4

LHS

RHS

2

2

2

2

2

= + + + + + +

= + + +

= + ++

= + + +

= + + +

= + +

=

] ]

]

g g

g


6. ( ) . . .

2 4 8 2. . .

2 2 2 2 2r

r

nn

n1

1 2 3= + + + +

= + + + +=

/


222 2 12 12

LHS

RHS

LHS RHS

1

1

=== −===

]

]

g

g


Step 2: Assume true for n k= So …2 4 8 2 2 2 1k k+ + + + = −] g

Step 3: Prove true for n k 1= + Prove … –2 4 8 2 2 2 2 1 k k k1 1+ + + + + =+ +] g …

– –

· ––

2 4 8 2 22 2 1 22 2 22 2 22 2 1

LHS

RHS

k k

k k

k k

k

k

1

1

1 1

1

1

= + + + + += += +===

+

+

+ +

+

+

]

]

g

g



7. ( ) ( ) ( ) ( ) . . . ( )

5 10 15 . . . 5

´ ´ ´ ´r n

n

5 5 1 5 2 5 3 5r

n

1= + + + +

= + + + +=

/


( ) ( )

´

´

5 15

25

1 1 1

25

2

5

LHS

RHS

LHS RHS

==

= +

=

==



So … ( )k k k5 10 15 525

1+ + + + = +

Step 3: Prove true for n k 1= + Prove …

( )( )

k k

k k

5 10 15 5 5 1

25

1 1 1

+ + + + + +

= + + +

] g

i.e. … ( )( )k k k k5 10 15 5 5 125

1 2+ + + + + + = + +] g

…

( )

( )( )

( ) ( )

[ ( ) ( )]

( )

( )

( )( )

´

k k

k k k

k kk

k k k

k k k

k k k

k k

k k

5 10 15 5 5 1

25

1 5 1

25

12

5 1 2

25

125

2 1

25

1 2 1

25

2 2

25

3 2

25

1 2

LHS

RHS

2

2

$

= + + + + + +

= + + +

= + ++

= + + +

= + + +

= + + +

= + +

= + +

=

]

]

g

g

So the statement is true for n k 1= + . The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on . The statement is true for all integers ≥n 1 .

8. Step 1: Prove true for n 1= 2 1

21 1 12

LHS

RHS

LHS RHS

= −= −= − += −=

]

]

g

g


Step 2: Assume true for .n k= So . . . k k k2 4 6 2 1− − − − − = − +] g

Step 3: Prove true for n k 1= + Prove … k k k k2 4 6 2 2 1 1 1 1− − − − − − + = − + + +] ] ]g g g i.e. … k k k k2 4 6 2 2 1 1 2− − − − − − + = − + +] ] ]g g g … ( )

( ) ( )k k

k k kk k kk kk kk k

2 4 6 2 2 11 2 1

2 23 23 2

1 2

LHS

RHS

2

2

2

= − − − − − − += − + − += − − − −= − − −= − + += − + +=

]

] ]

g

g g

So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on . The statement is true for all integers ≥n 1 .

9. Step 1: Prove true for .n 1=

2( ) ( )

5 1 49

2

5 1 13 1

218

9

LHS

RHS

LHS RHS

= +=

=+

=

==

] g



So … kk k

9 14 19 5 42

5 132

+ + + + + = +] g

Step 3: Prove true for n k 1= + Prove

2

…( ) ( )

k kk k

9 14 19 5 4 5 1 4

2

5 1 13 1+ + + + + + + +

=+ + +

] ]g g5 ?

…( )

k kk k k

k k k

k k

9 14 19 5 4 5 9

2

5 2 1 13 13

25 10 5 13 13

25 23 18

i.e.2

2

2

+ + + + + + +

=+ + + +

= + + + +

= + +

] ]g g

…

( )

k kk k

k

k k k

k k k

k k k

k k

9 14 19 5 4 5 9

25 13

5 9

25 13

2

2 5 9

25 13

210 18

25 13 10 18

25 23 18

LHS

RHS

2

2

2

2

2

= + + + + + + +

= + + +

= + ++

= + + +

= + + +

= + +

=

] ]

]

g g

g


633ANSWERS

10. ( )

. . .

. . .3 3 3 3 3

9 27 81 3

r

r

nn

n2

2 3 4= + + + +

= + + + +=

/


( )

( )

( )

39

2

9 3 1

2

9 3 1

2

9 2

9

LHS

RHS

LHS RHS

2

2 1

==

=−

=−

=

==

−



So ( )

. . .9 27 81 32

9 3 1k

k 1

+ + + + =−−


( )

( )

. . .9 27 81 3 32

9 3 1

2

9 3 1

Prove k kk

k

11 1

+ + + + + =−

=−

++ −

…( )

( ) ( )

· ·

· ·

·

·

·

( )

9 27 81 3 3

2

9 3 13

2

9 3 1

2

2 3

29 3 9 2 3

23 3 9 2 3

23 9 2 3

23 3 9

29 3 9

2

9 3 1

LHS

RHS

k k

kk

k k

k k

k k

k k

k

k

k

1

11

1 1

1 1

2 1 1

1 1

1

= + + + + +

=−

+

=−

+

= − +

= − +

= − +

= −

= −

=−

=

+

−+

− +

− +

− +

+ +

+

So the statement is true for .n k 1= +

The statement is true for n 2= so it must be true for n 3= . It is true for n 3= so it must be true for n 4= and so on. The statement is true for all integers .≥n 2


( )

( )

4 24 24

3

4 2 1

34 2 1

3

4 3

4

LHS

RHS

LHS RHS

– 1 1

0

1

= − −= − −= −

=− −

=− −

=−

= −=

]

]

g

g

6

5

@

?



So k

– – …( )

4 8 16 4 23

4 2 1– k 1+ − − − =

− −] g

6 @

Step 3: Prove true for n k 1= + Prove – – …

( )4 8 16 4 2 4 2

3

4 2 1

– – k k

k

1 1 1

1

+ − − − − −

=− −

+

+

] ]g g

6 @

i.e. – – … – ( )

4 8 16 4 2 4 2

3

4 2 1

– k k

k

1

1

+ − − − −

=− −+

] ]g g

6 @

k

k

k

k 1+

k 1+

…( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

( )

( )

´

´

4 8 16 4 2 4 2

3

4 2 14 2

3

4 2 1

3

3 4 2

3

4 2 4

3

12 2

3

4 2 4 12 2

3

8 2 4

3

4 2 2 4

3

4 2 4

3

4 2 1

LHS

RHS

– k k

kk

k k

k

k

k

1= − + − − − − − −

=− −

− −

=− −

−−

=− −

−−

=− − − −

=− − −

=− − −

=− −

=− −

=

] ]

]

g g

g6

6

6

@

@

@



( )´

11

61

1 1 1 2 1 1

61

2 3

1

LHS

RHS

LHS RHS

2==

= + +

=

==

]

] ]

g

g g



So . . . k k k k1 2 361

1 2 12 2 2 2+ + + + = + +] ]g g

Step 3: Prove true for n k 1= + Prove ( ). . . k k1 2 3 12 2 2 2 2+ + + + + +

( )k k k61

1 1 1 2 1 1= + + + + +] ]g g5 ?

. ( ). . . k k

k k k

1 2 3 1

61

1 2 2 3

i e. 2 2 2 2 2+ + + + + +

= + + +] ] ]g g g


( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

( ) ( ) ( )

. . . k k

k k k k

k k k k

k k k k

k k k k

1 2 3 1

61

1 2 1 1

61

1 2 166

1

61

1 2 1 6 1

61

1 2 1 6 1

LHS 2 2 2 2 2

2

2

2

= + + + + + +

= + + + +

= + + + +

= + + + +

= + + + +

6

7

@

A" ,

[ ]

[ ]

k k k k

k k k

k k k

61

1 2 6 6

61

1 2 7 6

61

1 2 2 3

RHS

2

2

= + + + +

= + + +

= + + +

=

] ]

] ]

] ] ]

g g

g g

g g g


13. 3 3 3 3

3

3

( ) ( ) ( ) ( )

. . . ( )

1 3 5 . . .

´ ´ ´

´

r

n

n

2 1 2 1 1 2 2 1 2 3 1

2 1

2 1

r

n

1

3 3 3

− = − + − + −

+ + −= + + + + −

=

] g

/


2 2

( )

( )

´

´

2 1 1111 2 1 12 11

LHS

RHS

LHS RHS

3

3

= −=== −= −==


Step 2: Assume true for n k= So 31 3 5 . . . k k k2 1 2 13 3 3 2 2+ + + + − = −] ]g g

Step 3: Prove true for n k 1= + Prove 3 331 3 5 . . . (2 1) (2 1)k k

k k

1

1 2 1 1

3 3

2 2

+ + + + − + + −= + + −] ^g h

5

5

?

?

i.e. 3 3 3

2

31 3 5 . . . (2 1) (2 1)( ) ( )

k kk k k1 2 2 1 1

3

2

+ + + + − + += + + + −5 ?

( ) ( )( )( )

1 3 5 . . . (2 1) (2 1)( ) ( )

k k k kk k k kk k k k k k k kk k k k

k kk k k

2 1 2 4 2 12 1 2 4 1

2 4 4 8 2 2 4 12 8 11 6 1

2 1 2 1LHS

2 2

2 2

4 3 2 3 2 2

4 3 2

3 3 3 3 3

2 2 3

= + + + + −= + + + += + + + + + + + += + + + += + + + + − + += − + +

k k k k kk k k k

2 8 12 6 12 8 11 6 1RHS

4 2 3 2

4 3 2

= − + + + += + + + +=

So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for

.n 2=

It is true for n 2= so it must be true for n 3= and so on. The statement is true for all integers .≥n 1


7 1 7 16 which is divisible by 6

1 − = −=


Step 2: Assume true for n k= So –7 1k is divisible by 6. i.e. p7 1 6k − = where p is an integer


Prove 7 1k 1 −+ is divisible by 6 i.e. – q7 1 6k 1 =+ where q is an integer

7 7 42

´p

ppppp

7 1 67 7 1 7 67 7 42

6 67 1 42 6

6 7 16 where is an integerq q

k

k

k

k

k

1

1

1

`

− =− =− =

− + = +− = +

= +=

+

+

+

^

^

h

h



– 3 1 9 18 which is divisible by 8

´2 1 = −=


Step 2: Assume true for n k= So 3 1k2 − is divisible by 8. i.e. p3 1 8k2 − = where p is an integer

Step 3: Prove true for n k 1= + Prove 3 1( )k2 1 −+ is divisible by 8 i.e. q3 1 8 k2 2 − =+ where q is an integer

´

pp

pppp

q

3 1 83 3 1 3 8

3 9 723 9 8 72 8

3 1 72 88 9 18 where is an integerq

k

k

k

k

k

2

2 2 2

2 2

2 1

2 2

`

− =− =− =

− + = +− = +

= +=

+

+

+

]

^

g

h



–5 1 5 14 which is divisible by 4

1 − ==


Step 2: Assume true for n k= So 5 1k − is divisible by 4. i.e. p5 1 4k − = where p is an integer

Factorise by taking out a common factor of k 1.+

635ANSWERS

Step 3: Prove true for n k 1= + Prove 5 1k 1 −+ is divisible by 4 i.e. q5 1 4k 1 − =+ where q is an integer

´p

ppppp

5 1 45 5 1 5 45 5 20

5 5 4 20 45 1 20 4

4 5 14 where is an integerq q

k

k

k

k

k

1

1

1

`

− =− =− =

− + = +− = +

= +=

+

+

+

^

^

h

h


17. Step 1: Prove true for n 2= 2 2 2 2 4

8 which is divisible by 4+ =

=] ]g g


Step 2: Assume true for n k= where k is even So k k 2+] g is divisible by 4.

i.e. k k p2 42 + = where p is an integer

Step 3: Prove true for n k 2= + (the next even integer) Prove k k2 2 2+ + +] ]g g is divisible by 4 i.e. k k q2 4 4+ + =] ]g g where q is an integer k k k k k

k k kp k

p k

2 4 4 2 82 4 8

4 4 84 24 where is an integerq q

2

2

+ + = + + += + + += + += + +=

] ]

^

g g

h

So the statement is true for .n k 2= + The statement is true for n 2= so it must be true for 4n = . It is true for n 4= so it must be true for n 6= and so on. The statement is true for all even positive integers.

18. Step 1: Prove true for n 1= 1 3 4+ = which is divisible by 4 So the statement is true for .n 1=

Step 2: Assume true for n k= where k is odd So k k 2+ +] g is divisible by 4. i.e. k k p2 4+ + = where p is an integer k p2 2 4+ =

Step 3: Prove true for n k 2= + (the next odd integer) Prove k2 2 2+ +] g is divisible by 4 i.e. k q2 2 2 4+ + =] g where q is an integer k q

k qk k

pp

q

2 4 2 42 6 42 6 2 2 4

4 44 14 where is an integerq

+ + =+ =+ = + +

= += +=

^ h

So the statement is true for .n k 2= + The statement is true for n 1= so it must be true for n 3= .

It is true for n 3= so it must be true for n 5= and so on. The statement is true for all odd positive integers.

19. Step 1: Prove true for n 1= 5 3 5 3

8 which is divisible by 2

1 1+ = +=

So the statement is true for n 1=

Step 2: Assume true for n k= So 5 3k k+ is divisible by 2. i.e. p5 3 2k k+ = where p is an integer

p5 2 3k k= −

Step 3: Prove true for n k 1= + Prove 5 3k k1 1++ + is divisible by 2 i.e. q5 3 2k k1 1+ =+ + where q is an integer · ·

·· ··

pppp

5 3 5 5 3 35 2 3 3 310 5 3 3 310 2 32 5 32 where is an integerq q

k k k k

k k

k k

k

k

1 1+ = += − += − += −= −=

+ +

^

^

h

h



7 3 7 310 which is divisible by 10

1 1+ = +=

So the statement is true for n 1= Step 2: Assume true for n k= where k is odd So 7 3k k+ is divisible by 10. i.e. p7 3 10k k+ = where p is an integer p7 10 3k k= −

Step 3: Prove true for n k 2= + (next odd integer) Prove 7 3k k2 2++ + is divisible by 10 i.e. q7 3 10k k2 2+ =+ + where q is an integer · ·

· ··

· ···

ppp

p

7 3 7 7 3 349 7 9 349 10 3 9 3490 49 3 9 3490 40 310 49 4 310 where is an integerq q

k k k k

k k

k k

k k

k

k

2 2 2 2+ = += += − += − += −= −=

+ +

^

^

h

h

So the statement is true for .n k 2= + The statement is true for n 1= so it must be true for n 3= . It is true for n 3= so it must be true for n 5= and so on. The statement is true for all odd integers .≥n 1

21. Step 1: Prove true for n 1= 1

11 5

4

LHS

RHS

LHS > RHS

2=== −= −



Step 2: Assume true for n k= So k k 5>2 −

Step 3: Prove true for n k 1= + Prove k k

k k kk k k

1 1 52 1 4

2 5

>>>

2

2

2

+ + −+ + −

+ −

] g

We know that k k 5>2 − So k k k2 5>2 + − since k k2 0 0for> > So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on. The statement is true for all integers .≥n 1


≥

4163 2 713

LHS

RHS

LHS RHS

2=== +=

] g


Step 2: Assume true for n k= So ≥ k4 3 7k +

Step 3: Prove true for n k 1= + Prove ≥ k4 3 1 7k 1 + ++ ] g i.e. ≥ k4 3 3 7k 1 + ++ ≥ k4 3 10k 1 ++

≥≥≥

kk

k

44 44 3 712 283 10

LHS k

k

1==

++

+

+

]

]

g

g

since 12 3≥k k and ≥28 10 So the statement is true for .n k 1= + The statement is true for n 2= so it must be true for n 3= . It is true for n 3= so it must be true for n 4= and so on. The statement is true for all integers .≥n 2


125 31224 2064 2084

LHS

RHS

LHS > RHS

3

3

= −= −== += +=


Step 2: Assume true for n k= So 5 3 4 20>k k− + i.e. 5 4 23>k k +

Step 3: Prove true for n k 1= + Prove 5 3 4 20>k k1 1− ++ + 5 3 5 5 3

5 4 23 35 4 115 35 4 1124 4 204 20

>>>>>

k k

k

k

k

k

k

1

1

− = −+ −+ −++

+

+

+

]

]

]

]

]

g

g

g

g

g

since 5 4 4 4>k k] ]g g and 112 20> The statement is true for n 3= so it must be true for n 4= . It is true for n 4= so it must be true for n 5= and so on. The statement is true for all integers .≥n 3


≥

332 13

LHS

RHS

LHS RHS

1

1

=== +=


Step 2: Assume true for n k= So ≥ k3 2k k +

Step 3: Prove true for n k 1= + Prove ≥ k3 2 1k k1 1 + ++ +

≥≥≥≥

kk

kk

33 33 23 2 32 2 12 1

LHS k

k

k

k

k

k

1

1

==

+++ +

+ +

+

+

]

]

]

]

g

g

g

g

since 3 2≥2 2k k] ]g g and ≥k k3 1+ for 1≥k So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on. The statement is true for all integers .≥n 1


≥

553 25

LHS

RHS

LHS RHS

1

1 1

=== +=


Step 2: Assume true for n k= So ≥5 3 2k k k+

Step 3: Prove true for n k 1= + Prove 5 3 2≥k k k1 1 1++ + +

≥≥≥≥

55 55 3 25 3 5 23 3 2 23 2

LHS k

k

k k

k k

k k

k k

1

1 1

==

++++

+

+ +

]

]

] ]

] ]

g

g

g g

g g

since ≥5 3 3 3k k] ]g g and ≥5 2 2 2k k] ]g g So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on. The statement is true for all integers .≥n 1

Test yourself 8

1. (a) T n4 5n = + (b) T n14 7n = −

(c) T 2 3nn 1

:= − (d) T 20041

n

n 1

=−

d n

(e) T 2nn= −] g

637ANSWERS

2. (a) 2 (b) 1185 (c) 1183 (d) T S S15 15 14= −

S S T15 14 15= +

(e) n 16=

3. (a) 11 125 (b) 114013

(c) 3 985 785 (d) 34 750

(e) 21

4. (a) Each slat rises 3 mm so the bottom one rises up ´30 3 mm or 90 mm. (b) 87 mm (c) 90, 87, 84, . . . which is an arithmetic sequence with a 90= , d 3= − (d) 42 mm (e) 1395 mm

5. $3400.01

6. (a) (i) (b) (ii) (c) (i) (d) (iii) (e) (i) (f) (ii) (g) (ii) (h) (i) (i) (i) (j) (i)

7. n 108=

8. (a) $24 050 (b) $220 250

9. ,a d33 13= − =

10. (a) 59 (b) 80 (c) 18 th term

11. (a) x 25= (b) ±x 15=

12. (a) 94

(b) 1813

(c) 13319

13. x 3=

14. (a) 136 (b) 44 (c) 6

15. 12121

16. $8066.42

17. (a) T n4 1n = + (b) .T 1 07nn 1= −

18. (a) x1 1< <− (b) 221

(c) x31=

19. d 5=

20. (a) 39 words/min (b) 15 weeks

21. (a) $59 000 (b) $15 988.89

22. 4.8 m 23. ,x172

2= −

24. (a) $2385.04 (b) $2392.03

25. 1300

26. (a) 735 (b) 4315

27. (a) $1432.86 (b) $343 886.91

28. n 20=

29. n 11=

30. (a) r 1 2( ) ( ) ( ) ( ) ( ). . .

. . .

5 3 5 3 5 3 5 3 5 3

3r

nn

n1

3= + + + +=

15 45 135 5= + + + + ] g

/


( )

( )

5 315

2

15 3 1

2

15 2

15

LHS

RHS

LHS RHS

1

1

==

=−

=

==

] g



So …( )

15 45 135 5 32

15 3 1k

k

+ + + + =−

] g

Step 3: Prove true for n k 1= + Prove … ( ) ( )15 45 135 5 3 5 3k k 1+ + + + + +

( )

2

15 3 1k 1

=−+

(3) (3)3

(3)

3 (3)

3 (3)

3 3

3 3

3 3

3

3

…( )

( )

( )

( )

´

´ ´

´ ´ ´

´ ´

´

15 45 135 5 5

2

15 15

2

15 1

2

2 5

2

15 1 10

215 15 10

25 3 15 10

25 15 10

215 15

2

15 1

LHS

RHS

k k

kk

k k

k k

k k

k k

k k

k

k

1

1

1

1

1

1

1 1

1

1

= + + + + +

=−

+

=−

+

=− +

=− +

=− +

=− +

=−

=−

=

+

+

+

+

+

+

+ +

+

+


(b) Step 1: Prove true for n 1=

( )

´

´

3 1 14

2

1 3 1 5

28

4

LHS

RHS

LHS RHS

= +=

=+

=

==




So …( )

kk k

4 7 10 3 12

3 5+ + + + + =

+] g

Step 3: Prove true for n k 1= + Prove … k k4 7 10 3 1 3 1 1+ + + + + + + +] ]g g5 ?

( ) ( [ ] )k k

2

1 3 1 5=

+ + +

…( )( )

k kk k

k k k

k k

4 7 10 3 1 3 42

1 3 8

23 8 3 8

23 11 8

i.e.

2

2

+ + + + + + + =+ +

= + + +

= + +

] ]g g

…( )

( ) ( )

( ) ( )

´

k kk k

k

k k k

k k k

k k k

k k

4 7 10 3 1 3 4

2

3 53 4

2

3 5

2

2 3 4

2

3 5 2 3 4

23 5 6 8

23 11 8

LHS

RHS

2

2

= + + + + + + +

=+

+ +

=+

++

=+ + +

= + + +

= + +

=

] ]

]

g g

g


(c) Step 1: Prove true for n 2= ´ ´2 2 2 1 4 1

4 4which is divisible by− =

=] g


Step 2: Assume true for n k= So ( )k k2 1− is divisible by 4. i.e. –k k p2 2 42 = where p is an integer

Step 3: Prove true for n k 1= + Prove –k k2 1 1 1+ +] ]g g5 ? is divisible by 4

i.e. –k k q2 1 1 1 4+ + =] ]g g5 ? where q is an integer

k k qk k k k

k k kp k

p kq

2 1 42 1 2 2

2 2 44 444

2

2

+ =+ = +

= − += += +=

]

]

^

g

g

h

So the statement is true for .n k 1= + The statement is true for n 2= so it must be true for n 3= . It is true for n 3= so it must be true for n 4= and so on. The statement is true for all integers .n 1>


1. (a) 8.1 (b) 19 th term

2. (a) π4

(b) π4

9 (c)

π4

33

3. (a) 2 097 170 (b) -698 775

4. (a) $40 (b) $2880

5. 6 th term 6. 17 823

7. 5 terms 8. , ,n 1 2 3=

9. 56− 10. $1799.79

11. x83= 12. $8522.53 13. k 20=

14. (a) $10 100 (b) $11 268.25 (c) $4212.41 (d) $2637.23

15. (a) cosec 2 x (b) ≤ ≤cos x1 1− So ≤ ≤cos x0 12 | | ≤cos x 12 So the limiting sum exists.

16. $240 652.62


( )

ara

r

a r

a1

1

LHS

RHS

LHS RHS

1 1

1

==

=−−

==

−



So ( )

. . .a ar ar arr

a r

1

1– k

k2 1+ + + + =

−−


Prove ( )

. . .a ar ar ar arr

a r

1

1– – k k

k2 1 1 1

1

+ + + + + =−

−+

+

i.e. ( )

. . .a ar ar ar arr

a r

1

1– k

k2 1

1

+ + + + + =−

−+k

. . .( )

( ) ( )

( ) ( )

( )

a ar ar ar ar

r

a rar

r

a r

r

ar r

r

a r ar r

rar a ar ar

ra ar

r

a r

1

1

1

1

1

1

1

1 1

1

1

1

1

LHS

RHS

– k

k

k k

k k

k

k

2 1

1

1

1

= + + + + +

=−−

+

=−−

+−

−

=−

− + −

=−

− + −

=−

− +

=−

−

=

+

+

+

k

k

k k

k


639ANSWERS

18. …

…

x x x x x

x x x1 –

r n

r

n

n

1 1 1 2 1 3 1 1

12 1

= + + + +

= + + + +

− − − − −

=/

To prove …x x xxx

111n

n2 1+ + + + =

−−−

Step 1: Prove true for n 1= x

xx

1

11

1

LHS

RHS

LHS RHS

– 1 1

1

==

=−−

==



So …x x xxx

111–k

k2 1+ + + + =

−−


Prove …x x x xx

x1

11– k k

k2 1 1 1

1

+ + + + + =−

−+ −+

] g

i.e. …x x x xx

x1

11–k k

k2 1

1

+ + + + + =−

− +

…

( )

x x x x

xx

x

xx

x

x x

xx

xx x

xx x x

xx

1

11

11

1

1

11

1

11

11

LHS

RHS

– k k

kk

k k

k k k

k k k

k

2 1

1

1

1

= + + + + +

=−− +

=−− +

−−

=−− +

−−

=−

− + −

=−

−

=

+

+

+

So the statement is true for .n k 1= + The statement is true for n 1= so it must be true for n 2= . It is true for n 2= so it must be true for n 3= and so on. The statement is true for all integers .≥n 1 .


1.

2. π6

3. cosx x1

12 1

−− −

4. 625324

5. ( )e

ee e

e1 1

32 23

x

x

x x

x

3 2

3

6 3

3

+ +=

+ +

6. 0.73

7. (a)

(b) 21

unit2

8. 94

9. 43

9

10. (a) x x1

1 12−

+ (b) ( )tanx

x1

42

1 3

+−

(c) x1 9

62−

−

11. π18

units2

3 12. $2929.08

13. 0.22 14. …± ±104 52 26+ +

15. tan x C21

21 +−

16. .π

367

0 61Z

17. ( )

( )

cos

cos

cos

cosx

xx

x

x x

x x x

1

1

1

1 2

1

2

2 1 2

2 1

+−

=−

− +

−

−

−

−

18. πx y9 12 2 3 3 0− + − = 19. 44 th term


33 2 13 2 13

LHS

RHS

0

0 1 −

==== −=

+

]

]

]

g

g

g

So true for n 0=

Step 2: Assume true for n k= . . .3 6 12 3 2 3 2 1k k 1 −+ + + + = +] ]g g

Step 3: Prove true for n k 1= + Prove . . .3 6 12 3 2 3 2 3 2 1 k k k1 1 1+ + + + + = −+ + +] ] ]g g g i.e. . . .3 6 12 3 2 3 2 3 2 1k k k1 2+ + + + + = −+ +] ] ]g g g


. . .3 6 12 3 2 3 23 2 1 3 23 2 3 3 26 2 33 2 2 33 2 33 2 1

LHS

RHS

k k

k k

k k

k

k

k

k

1

1 1

1 1

1

1

2

2

−−−

−−

−

= + + + + += += +=====

+

+ +

+ +

+

+

+

+

^ ^

^ ^

^ ^

^

] ^

^

^

h h

h h

h h

h

g h

h

h

So it is true for .n k 1= + Since it is true for n 0= , then it is true for .n 1= If true for n 1= , then it is true for n 2= and so on. So it is true for all ≥ .n 0

21. $945

22. (a) 724

(b) π4

− (c) π3

23. a

65619841

24. 2.4 m

25. ( )

( )

log

log log

x x

x x x

1 11

2 21

e

e e

2

2

+ +

=+ +

7

7

A

A

26. tanx

C32

231 +− d n

27. (a) 3 000 000 (b) 3 000 336 (c) 146 insects per day

28. (2, 0), infl exion

29. (a)

( ) ( ) ( )d t tt t t t

t t

230 65 125 8052 900 29 900 4225 15 625 20 000 640010 625 49 900 68 525

Pythagoras2 2 2

2 2

2

= − + −= − + + − += − +

(b) 2.3 h (c) 109.7 km

30. sinx

C2

1 +− 31. 199; 5050

32. (a) , , ,T T T T4 11 18 811 2 3 12= = = =

(b) 1410 (c) 29 th term

33. 23

−

34. ( ) ;f x x 11 3= +− domain: all real x , range: all real y

35. (a) 2

1− (b) π3

− (c) π2

36. .2 98 units3

37. sin cosx x x x C11 2+ − + +−

38. (a)

(b)

(c)

(d)

641ANSWERS

(e)


2 1315 5 227

LHS

RHS

5= −== +=

] g

LHS RHS> So true for n 5=


. .k

k2 1 5 2

2 5 3i e>

>

k

k

− ++


. .k

kk

2 1 5 1 22 1 5 72 5 8

Provei e

>>

k

k

k

1

1

1

− + +− +− +

+

+

+

] g

k

kk

2 2 22 5 310 65 8

>>>

k k1 =+

++

+ ^

]

h

g

since k k10 6 5 8>+ + for k 0> So it is true for n k 1= + . Since it is true for n 5= , then it is true for n 6= . If true for n 6= , then it is true for n 7= and so on. So it is true for all .n 4>

40. – . . .15 4 7− + +

41. y x1

1= + or ;y xx1= +

domain: all real ,≠x 0 range: all

real ≠y 1

42. $2851.52

43.

44. (a) sinv t12 4= − (b) cosa t48 4= − (c) 3 cm

(d) , , , . . .π π

t 04 2

s= (e) ±3 cm (f) , , , . . .π π π

t8 8

385

s=

(g) ( )cos

cosa t

tx

48 416 3 416

= −= −= −

45. 458

46. $180.76

47. . .

,.

ACAB BC

AC AB BCABC B

16 2569 6 12 8256

Sinceis right angled at

2 2

2 2 2 2

2 2 2

+∆

= =+ = +

== +

48. x65=

49. …

…

3 3 3 3 3

1 3 3 3

r n

r

n

n

0 1 2

02

= + + + +

= + + + +=

/

Step 1: Prove true for n 0= 3

1

23 1

23 1

22

1

LHS

RHS

LHS RHS

0

0 1

==

= −

= −

=

==

+

So true for n 0=


…1 3 3 32

3 1kk

21

+ + + + = −+


…1 3 3 3 32

3 1 k kk

2 11 1

+ + + + + = −++ +

. . . . .1 3 3 3 32

3 1i e k k

k2 1

2

+ + + + + = −++

…

( )

( )

( )

1 3 3 3 3

23 1

3

23 1

2

2 3

2

3 1 2 3

2

3 3 1

23 1

LHS

RHS

k k

kk

k k

k k

k

k

2 1

11

1 1

1 1

1

2

= + + + + +

= − +

= − +

=− +

=−

= −

=

+

++

+ +

+ +

+

+

So it is true for .n k 1= +

Since it is true for n 0= , then it is true for n 1= . If true for n 1= , then it is true for n 2= and so on. So it is true for all .≥n 0


50.

Let ABCD be a rhombus with .AB BC=

AB DC= and AD BC=

(opposite sides in a parallelogram)

AB BC DC ADall sides are equal

`

`

= = =

51. (a) .k 0 025Z (b) after 42.4 years (c) 20.6 years

52. (a) , ,π

πx 03

22= (b) , ,

π π πx

6 65

23

=

(c) ,π π

x3 3

4=

53. 450 cm2

54. (a) 12 ms 1− (b) e48 ms4 2− (c)

( )

x e

x e

x ee

x

3 2

12

484 12

4

t

t

t

t

4

4

4

4

= +

=

==

=

..

.

.

(d)

55. n 4=

56. (a) π3

(b) 23

57. 101

58. (a) Square . .´46 3 46 3m m, rectangle . .´30 9 92 7m m (b) $8626.38

59.

60. (a) 1.3 s (b) 6 m (c) 2.15 m

61. (a) . . .. . .

. . .

log log loglog log loglog log log

3 9 273 3 33 2 3 3 3

2 3

+ + += + + += + + +

Arithmetic series, since log log log log

log2 3 3 3 3 2 3

3− = −

=

(b) 210 log 3

62.

63. (a) 12.6 mL (b) 30 minutes

64. (a) , .A k24 000 0 038Z= (b) 37.4 years

65. 9 cms 1− −

66. (a)

(b) 2

1units2

67. (a) ( )θ ππ

n 16

n= + − (b) ±θ ππ

n6

=

68. x y3 0+ = 69. $277.33

70. (a) x 4= (b) Amplitude 1

71. (b) 72. (c), (d)

73. (a)

74. (b), (d)

75. (d) 76. (a)

77. (c)

78. (c)

79. (b), (c), (d)

80. (d)

643ANSWERS

Chapter 9 : Polynomials 2

Exercises 9.1

1. (a) ( ) , ( )f f3 0 4 0< > (b) ( ) , ( )f f3 0 4 0> < (c) ( 3) 0, ( 2) 0P P< >− − (d) ( ) , ( )P P0 0 1 0> < (e) ( ) , ( )f f2 0 3 0< >

2. (a) ( ) , ( )f f0 0 1 0< > root lies between 0 and 1 (b) ( . )f 0 5 0> root lies between 0 and 0.5

3. ( ) , ( )f f0 0 1 0> > and minimum turning point at (0.8, 0.9) no root lies between 0 and 1

4. (a) ( ) , ( )P P1 0 2 0< > (b) ( . ) ,P 1 5 0> so the root lies between 1 and 1.5; ( . )P 1 25 0=

1.25 is a zero of ( )P x

5. (a) ( ) , ( )f f1 0 2 0< > (b) ( . ) ,f 1 5 0> so the root lies between 1 and 1.5

6. ( ) , ( ) ,f f5 0 4 0> <− − so the root lies between 5− and ;4− ( . ) ,f 4 5 0<− so the root lies between 5− and . ;4 5− ( )f 5 1− = and ( . ) . ,f 4 5 1 25− = − so x 5= − is the better approximation

7. x 1Z

8. .x 3 25Z

9. x 4Z −

10. (a)

(b) ( )f x has 1 zero between 1− and 0 (c) 0

11. (a) For – –sinf x e x 3x=] g

..

ff

1 1 1 02 3 5 0

<>

= −=

]

]

g

g

So the root lies between x 1= and .x 2= (b) .x 1 375=

12. (a) –xx

x

12 012

12

3

3

3

===

(b) P 2 4 0<= −] g P 3 15 0>=] g So the root lies between x 2= and .x 3= (c) .x 2 52=

13. (a) .f 2 0 51 0>=] g .f 3 0 46 0<= −] g So the root lies between x 2= and .x 3= (b) .x 2 5=

14. (a) . .f 0 1 0 026 0>=] g . .f 0 2 0 058 0<= −] g So the root lies between .x 0 1= and . .x 0 2= (b) . .f 0 15 0 014 0<= −] g So the root lies between .x 0 1= and . .x 0 15= . .f 0 125 0 0063 0>=] g So the root lies between .x 0 125= and . .x 0 15=

15. P 1 1 0>=] g P 2 4 0<= −] g So the root lies between x 1= and .x 2= . .P 1 5 0 55 0<= −] g So the root lies between x 1= and . .x 1 5= . .P 1 25 0 43 0>=] g So the root lies between .x 1 25= and . .x 1 5= . .P 1 375 0 006 0<= −] g So the root lies between .x 1 25= and . .x 1 375=

Exercises 9.2

1. 2.39

2. .1 4−

3. 0.772

4. (a) ,f f0 0 1 0> <] ]g g (b) 0.448

5. (a) ,f f0 0 1 0< >] ]g g (b) 0.379

6. (a) ,f f2 0 1 0< >− −] ]g g (b) x 1Z − (c) .x 1 18Z −

7. (a) ( ) , ( )f f0 0 1 0< > (b) .x 0 75Z (c) .x 0 8Z (d) .x 0 78Z

8. ( ) xx

x

9 09

9

a 3

3

3`

− ===

(b) Between x 2= and x 3= (c) Using .x 2 5= as 1st approximation, .x 2 15Z


9. (a) 1.74 (b) 3.33 (c) 2.76 (d) 1.91

10. . ;x 2 31Z no since . .f 2 31 9 2Z] g 11. .x 1 34Z

12. .x 1 99Z 13. .x 0 42Z 14. (a) x 2= and x 3= (b) .x 2 75Z

15. .x 0 74Z

Test yourself 9

1. .x 1 39= 2. .x 1 87= 3. (a) ( ) , ( )f f2 1 3 49= − = (b) .x 2 05=

4. (a) ( ) , ( )f f2 9 3 5− = − = − (b) .x 2 75= −

5. (a)

(b) , . , .x 1 0 4 49= − 6. .x 1 375= 7. .x 1 34=

8. (a) 1.61x = (b) .x 1 58= 9. (a) ,f f0 1 1 1= = −] ]g g

(b) ’( )f 1 0= (c) x31= (d) .x 0 25=

10. (a) .x 2 25= (b) .x 2 3=


1. (a)

(b) . , .x 0 17 3 21Z − − (c) .x 3 21Z − (d) x 3Z −

2. –( )

( )

( )

P x x aP x x

x bP b

P b

bb

b a

bb

bb a

bb b a

bb a

3

3

33

3

33

32

3

2

1

2

3

2

3

2

3

2

3 3

2

3

==

= −

= −−

= −−

=− +

=+

l

l

] g

3. (a) .f 6 0 09 0<= −] g . .f 6 5 0 06 0>=] g So the root lies between x 6= and . .x 6 5= (b) . .f 6 25 0 01 0<= −] g So the root lies between .x 6 25= and 6.5.x = . .f 6 375 0 027 0>=] g So the root lies between .x 6 25= and . .x 6 375= (c) .x 6 253=

Chapter 10: The binomial theorem

Exercises 10.1

1. (a) 5040 (b) 40 320 (c) 6 (d) 120 (e) 1 (f) 3 628 798 (g) 72 (h) 3600 (i) 72 576 (j) 1680

2. (a) 15 (b) 45 (c) 84 (d) 165 (e) 1 (f) 1 (g) 4 (h) 792 (i) 2 (j) 45

3. (a) ( )! !

!

! !!

( )! !!

! !!

95 9 5 5

9

4 59

94 9 4 4

9

5 49

=−

=

=−

=

c

c

m

m

So 95

94

=c cm m

(b) ( )! !

!

! !!

( )! !!

! !!

C

C

7 2 27

5 27

7 5 57

2 57

72

75

=−

=

=−

=

So C C27

5=7

(c) ( )! !

!

! !!

( )! !!

! !!

1212 5 5

12

7 512

1212 7 7

12

5 712

=−

=

=−

=

5

7

c

c

m

m

So 12 12

=5 7

c cm m

(d) ( )! !

!

! !!

( )! !!

! !!

C

C

11 3 311

8 311

11 8 811

3 811

113

118

=−

=

=−

=

So C C113

118=

645ANSWERS

(e) ( )! !

!

! !!

( )! !!

! !!

C

C

10 1 110

9 110

10 9 910

1 910

101

109

=−

=

=−

=

So C C101

109=

(f) ( )! !

!

! !!

( )! !!

( )! !!

! !!

! !!

! !!

!!

! !!

!!

! !

( !)

( !)

( !)

! ( !)

( !)

! !

( !)

´´

´´

97 9 7 7

9

2 79

86

87 8 6 6

88 7 7

8

2 68

1 78

2 68

78

2 6 78 7

7 28 2

2 7

7 8

2 7

2 8

2 7

7 8

2 7

2 8

=−

=

+ =−

+−

= +

= +

= +

= +

= +

c

c c

m

m m

(since ! )2 2=

! !

( !) ( !)

! !

( !)

! !!

2 7

7 8 2 8

2 7

9 8

2 79

97

=+

=

=

= c m

(g) ( )! !

!

! !!

( )! !!

( )! !!

! !!

! !!

( ! !)

( !)

( ! !)

( !)

! !

( !)

! !

( !)

! !

( !)

! !!

1111 6 6

11

5 611

10 1010 5 5

1010 6 6

10

5 510

4 610

6 5 5

6 10

5 4 6

5 10

5 6

6 10

5 6

5 10

5 6

11 10

5 611

11

=−

=

+ =−

+−

= +

= +

= +

=

=

=

6

5 6

6

c

c c

c

m

m m

m

(h) ( )! !

!

! !!

( )! !!

( )! !!

! !!

! !!

! !

( !)

! !

( !)

! !

( !) ( !)

! !

( !)

! !!

C

C C

C

7 5 57

2 57

6 4 46

6 5 56

2 46

1 56

2 5

5 6

2 5

2 6

2 5

5 6 2 6

2 5

7 6

2 57

75

64

65

75

=−

=

+ =−

+−

= +

= +

=+

=

=

=

(i) ( )! !

!

! !!

( )! !!

( )! !!

! !!

! !!

( ! !)

( !)

( ! !)

( !)

! !

( !)

! !

( !)

! !

( !)

! !!

1010 6 6

10

4 610

95

96 9 5 5

99 6 6

9

4 59

3 69

6 4 5

6 9

4 3 6

4 9

4 6

6 9

4 6

4 9

4 6

10 9

4 610

10

=−

=

+ =−

+−

= +

= +

= +

=

=

=

6

6

c

c c

c

m

m m

m

(j) ( )! !

!

! !!

( )! !!

( )! !!

! !!

! !!

( ! !)

( !)

( ! !)

( !)

! !

( !)

! !

( !)

! !

( !)

! !!

73 7 3 3

7

4 37

62

63 6 2 2

66 3 3

6

4 26

3 36

3 4 2

3 6

4 3 3

4 6

4 3

3 6

4 3

4 6

4 3

7 6

4 37

73

=−

=

+ =−

+−

= +

= +

= +

=

=

=

c

c c

c

m

m m

m

4. ( )! !

!

[ ( )]!( )!!

( )!( )!!

!( )!

nk n k k

n

n k n n k n kn

n n k n kn

k n kn

nk

=−

− =− − −

=− + −

=−

=

n

c

c

c

m

m

m


5. x 5= 6. y 9= 7. a 3= 8. n 11=

9. k 6= 10. (a) ( ) !k k2 + (b) ( ) !r r− (c) ( ) !n n2+ (d) ( ) ( )!k k k1 12 + + − (e) ( ) ( )!p p p1 12 + − − (f) ( ) ( )!t t t2 1 12 + − −

11. (a) k1

(b) ( )k k 1+ (c) –n 1 (d) m 1

1+

(e) ( ) ( )k k k1 2

1+ +

12. (a) ! !5 32

(b) ! !4 59

(c) ! !2 46

(d) ! !5 472

(e) ! !7 390

13. (a) 1 (b) 1 (c) 241

(d) 73

(e) 87

(f) 43

(g) 21

14. k

n k 1− +

15. ( )! !

!nk n k k

n=−

c m

[ ( )]!( )!

( )!

( )! !

( )!

( )! ( )!

( )!

( )! !

( )!

( )! ( )!

( )!

( ) ( )! !

( ) ( )!

( )! !

( )!

( )! !

( ) ( )!

( ) ! !

( ) ( )!

( )! !

( )!

( )! !!

nk

nn k k

n

n k k

n

n k k

n

n k k

n

k n k k

k n

n k n k k

n k n

n k k

k n

n k k

n k n

n k k

k n k n

n k k

n n

n k kn

nk

11

11 1 1

1

1

1

1

1

1

1

1

1

1

1

1 1

1

1

−− +

−=

− − − −−

+− −

−

=− −

−+

− −−

=− −

−+

− − −− −

=−

−+

−− −

=−

+ − −

=−

−

=−

=

kc c

c

m m

m

Exercises 10.2

1. (a) (i) ( )x x x x1 1 3 33 2 3+ = + + + (ii) 8 (iii) 4

(iv) ( )x C x1 kk

k

3 3

0

3

+ ==

/

(b) (i) ( )x x x x x1 1 4 6 44 2 3 4+ = + + + + (ii) 16 (iii) 5

(iv) ( )x C x1 kk

k

4 4

0

4

+ ==

/

(c) (i) ( )x1 7+x x x x x x x1 7 21 35 35 21 72 3 4 5 6 7= + + + + + + +

(ii) 128 (iii) 8 (iv) ( )x xk

17

k

k

7

0

7

+ ==c m/

(d) (i) ( )x1 6+ 1 6 15 20 15 6x x x x x x2 3 4 5 6= + + + + + +

(ii) 64 (iii) 7 (iv) 6( )x C x1 kk

k

6

0

6

+ ==

/

(e) (i) ( )x x x x x x1 1 5 10 10 55 2 3 4 5+ = + + + + +

(ii) 32 (iii) 6 (iv) ( )x xk

15

k

k

5

0

5

+ ==c m/

2. (a) (i) 2 25 (ii) 26 (b) (i) 2 34 (ii) 35 (c) (i) 2 17 (ii) 18 (d) (i) 2 63 (ii) 64 (e) (i) 2 40 (ii) 41

3. (a) (i) ( )x1 4+ (ii) x x x x1 4 6 42 3 4+ + + + (b) (i) ( )x1 7+ (ii) 1 7 21 35 35 21 7x x x x x x x2 3 4 5 6 7+ + + + + + +

(c) (i) ( )x1 3+ (ii) x x x1 3 3 2 3+ + + (d) (i) ( )x1 6+ (ii) x x x x x x1 6 15 20 15 62 3 4 5 6+ + + + + +

(e) (i) ( )x1 8+ (ii) x x x x x1 8 28 56 70 562 3 4 5+ + + + + x x x28 86 7 8+ + +

4. (a) 35 (b) 126 (c) 15 (d) 70 (e) 5

5. (a) 84 (b) 10 (c) 165 (d) 120 (e) 4

6. (a) 3 (b) 15 (c) 45 (d) 21 (e) 91

7. (a) 252 (b) 792 (c) 56 (d) 3003 (e) 21

8. (a) 5 (b) 35 (c) 70 (d) 126 (e) 330

9. (a) 28 x 2 (b) 3 x 2 (c) 6 x 2 (d) 21 x 2 (e) 15 x 2

10. (a) C x–n

kk1

11+ − (b) C x–

nk

k21

1− (c) C x––n

kk1

11−

(d) C x––n

kk2 1

11+ (e) C x–

–nk

k3 11

1−

Exercises 10.3

1. (a) 6 (5 )x12

k k

k

12

0

12−

= kc m/ (b) ( )a b

18k k

k

18

0

18

−−

= kc m/

(c) (3 ) 2y24

k k

k

24

0

24−

= kc m/ (d) ( 2 )x y

16k k

k

16

0

16

−−

= kc m/

(e) 6 (5 )x12

k k

k

12

0

12−

= kc m/

2. (a) x ax x4+ +a a x a4 64 3 2 2 3 4+ + (b) a a x a x a x a x ax x6 15 20 15 66 5 4 2 3 3 2 4 5 6+ + + + + + (c) a a x a x a x ax x5 10 10 55 4 3 2 2 3 4 5+ + + + + (d) a a a8 12 6 13 2+ + + (e) – – –x x x x x x x14 84 280 560 672 448 1287 6 5 4 3 2+ + + − (f) x x x x256 768 864 432 818 6 4+ + + + (g) – x x x x729 2916 4860 4320 21602 3 4+ − +

x x576 645 6− + (h) –b ab b300 125+–a a64 2403 2 2 3 (i) 32 240 720 1080 810 243m m m m m2 3 4 5+ + + + + (j) x x x x x1 16 112 448 1120 17922 3 4 5− + − + −

x x x1792 1024 256 86 7+ − +

3. (a) (i) ( )a x 3+ (ii) a a x ax x3 33 2 2 3+ + + (b) (i) ( )x3 2 5− (ii) – – –x x x x x243 810 1080 720 240 325 4 3 2+ + (c) (i) ( )x y2 6+ (ii) y240y x+160y x+ +60y x+12x x6 5 4 2 3 3 2 4+ xy y192 64+5 6 (d) (i) ( )a b2 5 4+ (ii) bb a600+a a16 1604 3 2 2+ ab b1000 6253 4+ + (e) (i) ( )x y 7+ (ii) y +y x21+y x35+y x35+y x21+x x77 6 5 2 4 3 3 4 2 5+ xy y7 +6 7 (f) (i) ( )p q4 3 3− (ii) –q pq q108 27+–p p64 1443 2 2 3 (g) (i) ( )n3 2 5+ (ii) n n n243 810 1080 7202 3+ + + n n240 324 5+ + (h) (i) ( )a b2 6− (ii) b −b a60+b a160−b a240+a a64 1926 5 4 2 3 3 2 4− ab b12 +5 6 (i) (i) ( )ab c3 4 4− (ii) c 768 256b abc c− +864c a+b432b a−81a4 4 3 3 2 2 2 3 4 (j) (i) ( )x2 7− (ii) 128 448 672 560 280 84x x x x x2 3 4 5− + − + − + x x14 6 7−

647ANSWERS

4. (a) ( )a 7 10+ (b) ( )y6 9− (c) ( )a b3 4 7+ (d) ( )x 32 8+ (e) ( )p q5 11− (f) ( )x4 9 n− (g) ( )a b3 n2 + (h) ( )x3 2 n 1− + (i) ( )a b6 n2 1+ − (j) ( )x y8 7 n2 2+

5. (a) 41 29 2+ (b) 208 120 3− (c) 2 9970 − (d) 124 32 15+ (e) 89 3 109 2−

(f) xx x x

81 542

272

316

2 3 4

+ + + +

(g) x x x x x x5 10

10 5 15 33 5

+ + + + + (h) x x x

123

43

8

2 3

− + −

(i) x x xx x x

12 60 160240 192 646 4 2

2 4 6+ + + + + +

(j) ba a ab b

6 4 827

3 2 2 3

− + −

6. ,a b45 29= = 7. ,a b161 72= = −

8. ,a b29 656 20 880= = − 9. ,x y76 5808= − =

10. ,p q452 165 888= = 11. ,a b11 9= =

12. (a) 0.9703 (b) 0.9606 (c) 0.9510 (d) 0.9415 (e) 0.9321

13. (a) 0.9604 (b) 0.9039 (c) 0.9412 (d) 0.9224 (e) 0.8858

14. (a) 1.0303 (b) 1.0510 (c) 1.0406 (d) 1.0721 (e) 1.0615

15. (a) 1215 (b) 40 (c) 65 625 (d) 314 928 (e) 103 680

16. (a) 1120 (b) 280 (c) 8960 (d) 326 592 (e) 1215

17. (a) 15

610

5c m (b) 4

1055 5−

5c m (c) 3

1429 5

5c m

(d) 595

34 5− c m (e) 520

715 5−5

c m

18. (a) k

x8

5k k8 −c m (b) ( )a12

2 3k k12 −

kc m (c) ( ) ( )

ka b

65 k k6 −−c m

(d) ( ) ( )x y15

4 k k15 −−

kc m (e) ( ) ( )a b

213 2k k21 −−

kc m

19. (a) k

x1

2 k k10 1

−− −

9c m (b) ( ) ( )

kx y

15 2k k6 1

−− −

5c m

(c) ( )k

d1

3 2k k9 1

− −− −8

c m (d) ( )k

m n1

6k k14 1

− −− −13

c m

(e) ( ) ( )k

a b1

3 2k k21 1

− −− −20

c m

20. (a) k

x2

3 k k8 2

−− −

6c m (b)

ka

22k k11 2

−− −

9c m

(c) ( )k

a2

5 3k k8 2

−− −

6c m (d) ( )

kx

23 4k k20 2

− −− −18

c m

(e) ( ) ( )k

x y2

3 2k k2 10 2

− −− −8

c m

21. sin sin sin sin sinx x x x x1 5 10 10 42 3 4 5+ + + + + 22. 34

Exercises 10.4

1. x

y22 6803

8

2. x448 12− 3. 32

21879

4. (a) x x x x x x

243 405270 90 15

53 7 11 15+ + + + + (b) 405

5. 729112

6. (a) 1280− (b) 13 608

(c) 720 (d) 2268− (e) 12

7. (a) 30343

(b) 2716

2 (c) 279 936 (d) 3 784 704

(e) 1021

8. 10 777 536 9. 765421

10. (a) 483− (b) 2

163

11. (a) 65 536 (b) 216 (c) 1120 (d) 489 888 (e) 210

12. (a) 175 000 (b) 8 660 520 13. 945− 14. 112 640

15. n 19= 16. , ,a b a b2 1 2 1or= = − = − =

17. , ,a b a b2 3 2 3or= = − = − = 18. (a) k

k9 − (b) 70

19. (a) k

k7 − (b) 240

20. (a) 78 732 (b) 15 360 (c) 5760 (d) 1792 (e) 1 082 565 (f) 240 (g) 241 920 (h) 10 450 944 (i) C 2 515

114 11 (j) C 4 712

84 8

21. (a) 672 (b) ±80 (c) 78 732 (d) 1792 (e) 11 547 360−

Exercises 10.5

1. (a) 2 040 714

(b) 53

52

20+ =c cm m

(c) 20

83

21

82

22

81

120+ + =c c c c c cm m m m m m

(d) 40

73

41

72

42

71

43

70

165+ + + =c c c c c c c cm m m m m m m m

2. From ( ) ,x1 6+ the coeffi cient of x2 is .62c m

From 3 3( ) ( ) ,x x1 1+ + the coeffi cient of x2 is

.30

32

31

31

32

30

+ +c c c c c cm m m m m m

62

30

32

31

31

32

30

30

32

31

31

= + +

2= +

c c c c c c c

c c c c

m m m m m m m

m m m m

3. Coeffi cient of x4 from ( )x1 10+ is .104

c m

Coeffi cient of x4 from ( )( )x x1 1 9+ + is .94

93

+c cm m

10 9

493

` = +4

c c cm m m

4. Coeffi cient of xk from ( )x1 n 1+ + is .n

k1+

c m

Coeffi cient of xk from ( )( )x x1 1 n+ + is

.nk k 1

+ −n

c cm m

n

knk

nk

11

`+

= + −c c cm m m


5. (a) 5( ) ( )x x2 1+ + (b) 30 6. 92 7. n 9=

8. (a) n( )x C x1 nk

k

k

n

0+ =

=/

Substitute x 6=

( ) C

C

1 6 6

7 6

n nk

k

k

n

n nk

k

k

n0

0

+ =

=

=

=

/

/

(b) ( )x C x1 n nk

k

k

n

0+ =

=/

Substitute x 2=

( ) C

C

1 2 2

3 2

n nk

k

k

n

n nk

k

k

n0

0

+ =

=

=

=

/

/

(c) ( )x C x1 n nk

k

k

n

0+ =

=/

Substitute x 3=

n( )

( )

C

C

C

C

1 3 3

4 3

2 3

2 3

nk

k

k

n

n nk

k

k

n

n nk

k

k

n

n nk

k

k

n

0

0

2

0

2

0

+ =

=

=

=

=

=

=

=

/

/

/

/

9. ( )

. . .

x C x

n nx

nx

nx

nn

x

1

0 1 2 3

n nk

k

k

n

n

0

2 3

+ =

= + + + + +

=

c c c c cm m m m m

/

Differentiating:

( ) . . .n xn n

xn

xnn

nx11 2

23

3n n1 2 1+ = + + + +− −c c c cm m m m

Substitute x 1=

( ) ( ) ( )

. . . ( )

. . .

nn n n

nn

n

n n nn

nn

n knk

1 11 2

2 13

3 1

1

12

23

3

2

n

n

n

k

n

1 2

1

1

1

+ = + +

+ +

= + + + +

=

−

−

−

=

c c c

c

c c c c

c

m m m

m

m m m m

m/

10. (a) x a+( )

. . .

a xn

an

an

x

na x

nn

x

0 1 2

3

n n n n

n n

1 2 2

3 3

+ = +

+ + +

− −

−

c c c

c c

m m m

m m

Substitute x 0=

n( ) ( ) ( )

( ) . . . ( )

1

an

an

an

a

na

nn

an

a

n

00 1

02

0

30 0

0

0

n n n

n n

n n

1 2 2

3 3

+ = + +

+ + +

=

=

− −

−

c c c

c c

c

c

m m m

m m

m

m

(b) ( )

. . .

a xn

an

a xn

a x

na x

nn

x

0 1 2

3

n n n n

n n

1 2 2

3 3

+ = + +

+ + +

− −

−

c c c

c c

m m m

m m

Substitute x 1=

n( ) ( ) ( )

( ) . . . ( )

. . .

an

an

an

a

na

nn

na

na

na

na

nn

nk

a

10 1

12

1

31 1

0 1 2

3

n n n

n n

n n n

n

k

nn k

1 2 2

3 3

1 2

3

0

+ = + +

+ + +

= + +

+ + +

=

− −

−

− −

−

=

−

c c c

c c

c c c

c c

c

m m m

m m

m m m

m m

m/

(c) x

+

x a+

. . .x x

n

+ +

( )a xn

an

an

na

nn

0 1 2

3

n n n

n n

1 2 2

3 3

+ = + − −

−

c c c

c c

m m m

m m

Substitute x 1= −

a

n( ) ( ) ( )

( ) . . . ( )

( )

an

an

an

a

na

nn

nk

10 1

12

1

31 1

1

n n n

n n

k

k

nn k

1 2 2

3 3

0

− = + − + −

+ − + + −

= −

− −

−

=

−

c c c

c c

c

m m m

m m

m/

(d) n( )

. . .

a xn

an

a xn

a x

na x

nn

x

0 1 2

3

n n n

n n

1 2 2

3 3

+ = + +

+ + +

− −

−

c c c

c c

m m m

m m

Differentiating:

( )

. . .

. . .

n a xn

an

a x

na x

nn

nx

na

na x

na x n

nn

x

knk

a x

1 22

33

12

2

33

n n n

n n

n n

n n

k

nn k k

1 1 2

3 2 1

1 2

3 2 1

1

1

+ = +

+ + +

= +

+ + +

=

− − −

− −

− −

− −

=

− −

c c

c c

c c

c c

c

m m

m m

m m

m m

m/

(e) From (d):

n 1−( )

. . .

n an

an

a x

na x n

nn

x

11

22

33

n n

n n

1 2

3 2 1

+ = +

+ + +

− −

− −

c c

c c

m m

m m

Substitute x 0=

n 1−( ) ( ) ( )

. . . ( )

n an

an

an

a

nnn

nan

a

nn

01

22

0 33

0

0

1

1

n n n

n

n n

1 2 3 2

1

1 1

+ = + +

+ +

=

=

− − −

−

− −

c c c

c

c

c

m m m

m

m

m

11. n( ) . . .xn n

xn

xn

xnn

x10 1 2 3

n2 3+ = + + + + +c c c c cm m m m m

Substitute x 2= −

n( ) ( ) ( ) ( ) . . .

( )

( ) . . . ( )

n n n n

nnn n n n n

n

1 20 1

22

23

2

2

10

21

42

83

2

n

n n

2 3− = + − + − + − +

+ −

− = − + − + + −

c c c c

c

c c c c c

m m m m

m

m m m m m

649ANSWERS

12. (a) Let .x 1=

Then n( ) . . . .

. . .

n n n nn

n n n nn

1 10 1

12

1 1

20 1 2

n

n

2+ = + + + +

= + + + +

c c c c

c c c c

m m m m

m m m m

(b) Let x 1= −

Then n

n

n

( ) ( ) ( )

. . . ( )

0 . . . ( 1)

n n n

nn

n n n nn

1 10 1

12

1

1

0 1 2

2− = + − + −

+ + −

= − + − + −

c c c

c

c c c c

m m m

m

m m m m

13. (a) Let

( )

xn

n

n

1

1 12

1

22

42

n

k

nk

n

k

n

n

k

n

2

0

2

2

0

2

0

2

=

+ =

=

=

=

=

= k

k

kc

c

c

m

m

m

/

/

/

(b) n2( )

. . .

xn n

xn

x

nn

x

120

21

22

22

n

2

2

+ = + +

+ +

c c c

c

m m m

m

Differentiating:

` .

( ) . . .

Let

( ) . . .

( )

n xn n

x nnn

x

x

nn n

nnn

n kn

k n n

n

2 12

22

222

1

2 1 12

22

1 222

1

2 22

2 2

4

n n

n n

n

k

n

k

nn

n

2 1 2 1

2 1 2 1

2 1

1

2

1

22

+ = + + +

=

+ = + + +

=

=

=

− −

− −

−

=

=

1

k

2

1 2

k

c c c

c c c

c

a

m m m

m m m

m

k

/

/

14.

`

( )

( )

( )

x dxnr

x dx

n

xC

nr r

xC

n

x nr r

xC

1

1

1

1

1

1

1

n

r

nr

n

r

n r

n

r

n r

01

10

1

2

1

0

1

+ =

++

+ =+

+

++

=+

+

=+

=

+

+

=

+

c

c

c

m

m

m

/

/

/

##

:

( )

( )

( )

x

nnr r

C

nC

n

x nr r

xn

nr r

xn

x

n

n

x

0

11

10

11

1

1

1 11

1 1

1

11

1

1 1

Let

r

n r

n

r

n r

r

n r n

n

0

1

1

0

1

0

1 1

1

`

`

`

=

+=

++

+=

++

=+

++

+=

++

−+

=+

+ −

=

+

+

=

+

=

+ +

+

c

c

c

m

m

m

/

/

/

r n1 1+ +

r

Let( ) ( )

( )

x

nr r n

nnr r n

1

1

1

1

1 1 1

11

1

1

11

r

n

r

n

0

0`

= −

+−

=+

− −

=+

−

+−

=+

=

=

c

c

m

m

/

/

15.

. . .( )

nx

n x n x

nn n

xn

x0 1 2 2 3

1 1

1 1n n

2 3

1 1

+ +

+ ++

=+

+ −+ +

c c c

c

m m m

m

(from question 14)

Let

. . .

( )

. . .

xn n n

nn n n

n n nn

nn

n

1

01

1 21

2 31

11

1

1 1 1

0 21

1 31

2 11

12 1

n n

n

2 3

1 1

1

=

+ + +

++

=+

+ −

+ + + ++

=+

−

+ +

+

c c c

c

c c c c

m m m

m

m m m m

Test yourself 10

1. r

x x12 3

r

r

0

1212

=

−r

c cm m/

2. Differentiating: LHS ( )n x1 n 1= + −

RHS . . .n n

xn

xnn

nx1 2

23

3 n2 1= + + + + −c c c cm m m m

Let x 1= : LHS ( )n n1 1 2n n1 1= + =− −

RHS . . .n n n n

nn r

nr1 2

23

3r

n

1= + + + + =

=c c c c cm m m m m/

3. 6048−

4. Coeffi cient of x3 in ( )x183

8+ = c m

Coeffi cient of x3 in 5( ) ( )x x1 13+ +

33

50

32

51

31

52

30

53

= + + +c c c c c c c cm m m m m m m m

5. (a) Let x 0= (b) Let x 1= − 6. 924 7. 152

8. Coeffi cient of xk in ( )xnk

1 n+ = c m

Coeffi cient of xk in ( )( )x x1 1 n 1+ + −

n

knk

1 11

− −−= +c cm m

9. x x y x y x yxy y

32 240 720 1080810 243

5 4 3 2 2 3

4 5

+ + ++ +

10. 43

50

42

51

41

52

40

53

84+ + + =c c c c c c c cm m m m m m m m

11. 960 740 352 12. 15 360 13. ,a b3 4= = −

14. (a) kk10 1= − +

(b) 295 245

15. n 8= 16. 105 17. 489 888

18. (a) p p p p p243 810 1080 720 240 322 3 4 5+ + + + + (b) 2889 12 592−

(c) x xx x x

1254 108 814

2 5 8− + − +

(d) a a b ab b8 12 63 2 2 3− + − (e) 184 592 130 728 2−


19. ,a b220 284= = −

20. (a) ( )a xn

a x0

n n k k

k

n

0+ = −

=c m/

Substitute x 2=

( )an

a20

2n n k k

k

n

0+ = −

=c m/

(b) ( )

. . .

a xn

an

a xn

a x

na x

nn

x

0 1 2

3

n n n n

n n

1 2 2

3 3

+ = + +

+ + +

− −

−

c c c

c c

m m m

m m

Differentiating:

( )

. . .

. . .

Substitute

( ) ( ) ( )

. . . ( )

. . .

n a xn

an

a xn

a x

nnx

na

na x

na x

nnn

x

x

n an

an

an

a

nnn

na

na

na

nnn

knk

a

1 22

33

4

12

23

3

1

11

22

1 33

1

1

12

23

3

n n n n

n

n n n

n

n n n n

n

n n n

k

nn k

1 1 2 3 2

1

1 2 3 2

1

1 1 2 3 2

1

1 2 3

1

+ = + +

+ +

= + +

+ +

=

+ = + +

+ +

= + +

+ +

=

− − − −

−

− − −

−

− − − −

−

− − −

=

−

c c c

c

c c c

c

c c c

c

c c c

c

c

m m m

m

m m m

m

m m m

m

m m m

m

m/

(c) ( )

. . .

a xn

an

a xn

a x

na x

nn

x

0 1 2

3

n n n n

n n

1 2 2

3 3

+ = + +

+ + +

− −

−

c c c

c c

m m m

m m

Integrating both sides:

n 1+

LHS ( )

( )

RHS

. . .

. . .

So( )

. . .

( )

. . .

Substitute :( )

. . .

a x dx

n

a xC

na

na x

na x

nn

x dx

na x

na

x

na

x nn n

xC

n

a xC

na x

na

x

na

x nn n

xC

n

a xC

na x

na

x

na

x nn n

x

x

n

aC

na

na

na

nn n

1

0 1

2

0 1 2

2 3 1

1 0 1 2

2 3 1

1 0 1 2

2 3 10

1

0

00

1 20

2 30

10

n

n n

n n

n n

nn

nn n

nn

nn n

nn

nn n

nn

1

1

2 2

12

23 1

2

1

11

2

23 1

2

1

31

2

23 1

1

31

2

23 1

= +

=+

++

= +

+ + +

= +

+ + ++

+

++

+ = +

+ + ++

+

++

+ = +

+ + ++

=

++

+ = +

+ + ++

−

−

−

−+

+−

−+

+−

−+

+−

−+

c c

c c

c c

c c

c c

c c

c c

c c

c c

c c

m m

m m

m m

m m

m m

m m

m m

m m

m m

m m

#

#

So( )

. . .

( )

. . .

na

C

Cna

n

a x

na n

a xn

ax

na

x nn n

x

n

a x a na x

n a x

n a x nn n

x

nk n

a x

10

1

1 1 0 1 2

2 3 1

1 0 1 2

2 3 1

1

n

n

n nn n

nn

n nn

n

n n

n k k

k

n

1

3

3

1

1 11

2

23 1

1 1 1 2

2 3 1

1

0

++ =

= −+

++

−+

= +

+ + ++

++ −

= +

+ + ++

=+

+

+

+ +−

−+

+ + −

− +

− +

=

c c

c c

c c

c c

c

m m

m m

m m

m m

m/


1. Coeffi cient of x4 from 20( )x1 + is .204

c m

Coeffi cient of x4 from 10 10( ) ( )x x1 1+ + is

100

104

101

103

102

102

103

101

104

100

204

100

104

101

103

102

102

103

101

104

100

100

104

101

103

102

2

+ +

+ +

+

` = + +

+ +

2= +

c c c c c c

c c c c

c c c c c c c

c c c c

c c c c c

m m m m m m

m m m m

m m m m m m m

m m m m

m m m m m= G

2. (a)

2

LHS ( )

( )

( )

( )( )

( )( )RHS

x x x

x x x

x x xx x xx x

xx

1 11

1 11

11

12 1

11

n nn

n

n

n

n

n

n

2

2

2

2

= + +

= + +

= + +

= + + += + += += +=

c

c

c

m

m

m

<

<

6

F

F

@

(b) Coeffi cient of xn in ( )x1 n2+ is .n

n2c m

Coeffi cient of xn in ( )x x x1 11n n

n

+ +c m

comes from the terms independent of x in.

.( )

i.e. . . .

. . .

x xn n n n n n n

nnn

n n n n nn

1 11

0 0 1 1 2 22

0 1 2

nn

2 2 2 2

`

+ +

+ + + +

= + + + +n

c

c c c c c c c c

c c c c c

m

m m m m m m m m

m m m m m

3. Coeffi cient of xn in ( ) ( )x x1 1n n+ − is

( 1) ( 1) . . . ( 1) .n n

nn n

nnn

n0 1 1 0

n n 1 0− + − − + + −−c c c c c cm m m m m m

0i.e. ( 1) ( 1) . . . ( 1)nn n

n1 0

n n2

12 2

− + − − + + −−n

c c cm m m

since nk

nn k

= −c cm m

651ANSWERS

i.e. ( )

( ) ( )

nk

x x

1

1

k

k

n

n k

k

n0

2

2 2

0

−

− = −

=

=

c m/

/

For the coeffi cient of xn ( )i.e.

x xk n

kn

2

2

k n2− ==

=

If n is odd, k is not an integer. there is no coeffi cient of xn if n is odd

( )nk

1 0k

k

n

0

2

` − ==

c m/

4. (a) Ck is the coeffi cient of xk

i.e. k

153 2k k15 −c m

Ck 1+ is the coeffi cient of xk 1+

i.e.k15

13 2( )k k15 1 1

+− + +c m

( )! !!

( )]!( )!!

( )]!( )! !

! ( )! !

( )

( )

´

´

´ ´

´ ´

C

C k

k k

k k

k k

k k

k

k

153 2

13 2

1515

3

15 1 115

2

15 1 1 15 3

15 2 15

3 1

2 15

( )

k

k

k k

k k

1

15

15 1 1

=+

=

−

− + +

=− + +

−

=+−

+

−

− + +

k

15

c

c

m

m

6

6

(b) C155

3 2510 5= c m

5. n

n

2

2 1+

( )

( )

Let :

x dxn

x dx

n

x nkx

C

x

nn

kC

nC

12

2 1

1 21

0

2 11 2

10

2 11

k

k

n

k

k

n

k

k

n

0

2

1

0

2

1

0

2

+ =

++

=+

+

=

+=

++

+=

=

+

=

+

=

k

k

k

c

c

c

m

m

m

/

/

/

##

( )

( )

( )

Let :( )

n

x nkx

nn

kx

n

x

n

n

x

x

nk n

n

2 1

1 21 2 1

1

21 2 1

1

2 11

2 1

1 1

2

21

22 1

1 2 1

2 13 1

n k

k

n

k

n k n

n

k

n k n

n

2 1 1

0

2

0

2 1 2 1

2 1

0

2 1 2 1

2 1

`

`

++

=+

++

+=

++

−+

=+

+ −

=

+=

++ −

=+

−

+ +

=

=

+ +

+

=

+ +

+

k

k

k

c

c

c

m

m

m

/

/

/

6. ( )

( )

x dxnr

x dx

n

x nr r

x

n nnr r

nr r

nnr r

1

1

1

1

13

11

12

10

13 1

12

n

r

nr

n

r

n r

n n

r

n r

r

n r

n

r

n r

000

1

0

2

0

1

0

2

1 1

0

1

0

1

1

0

1

+ =

++

=+

+−

+=

+−

+

+− =

+

=

+

=

+

+ +

=

+

=

+

+

=

+

22c

c

c c

c

m

m

m m

m

= =G G

/

/

/ /

/

##

7. ( )1 . . .

( 1)( 2) . . . ( 1)for ≤ ≤

´ 2 ´ 3 ´ ´S n C

k

n n n n kk n1n

k= =− − − +

Step 1: Check that S(2) is true by inspection of .( )x1 2+ ( )

, ,..

x x xC C x C x

C C C

1 1 2

112

1 22 1

1

2 2

20

21

22

2

20

21

22

+ = + += + +

= = = =

∴ S(n) is true for n 2= .

Step 2: Assume ( )S n 1− is true.

( ):. . .

( ) ( )( ) . . . [( ) ]

. . .

( ) ( )( ) . . . ( )

If :

If :

( definition)

´ 2 ´ 3 ´ ´

´ 2 ´ 3 ´ ´

S n Ck

n n n n k

k

n n n n k

k C

k Cn

11

1 2 3 1 1

1

1 2 3

0 1

11

1by

nk

n

n

1

10

11

− =− − − − − +

=− − − −

= =

= = −

−

−

−

Step 3: Prove S(n) is true.C C C

n

n

11

1

1

n n n1

10

11= +

= + −

=

− −

∴ S(n) is true for k 1=

… ( )

( )( )…( )

. . .

( ) ( ) . . . ( )

. . . ( )

( )( ) . . . ( )

. . . ( )

( )( ) . . . ( )

. . . ( )

( )( ) . . . ( )

´ 2 ´ 3 ´ ´

´ 2 ´ 3 ´ ´

´ 2 ´ 3 ´ ´

´ 2 ´ 3 ´ ´

´ 2 ´ 3 ´ ´ ´

C C C

k

n n n k

k

n n n k

k

n n n k

kn k

k

n n n k

kn

k k

n n n n k

1 1

1 2 1

1

1 2

1 1

1 2 11

1 1

1 2 1

1 1

1 2 1

nk

nk

nk

11

1= +

=−

− − − +

+− − −

=−

− − − ++ −

=−

− − − +

=−

− − − +

−−

−

=

;

G

E

∴ S(n) is proved for all ≥k 2, and has been proved for k 1= above.

Chapter 11 : Probability

Exercises 11.1

1. 301

2. 521

3. 61

4. 401

5. 20 000

1

6. (a) 74

(b) 73

7. 373

8. 121

9. (a) 2011

(b) 43


10. (a) 61

(b) 21

(c) 31

11. (a) 621

(b) 313

(c) 21

(d) 12499

12. (a) 158

(b) 157

(c) 151

13. 501

14. ,21

1 15. 4423

16. (a) 317

(b) 317

(c) 3112

17. 1751

18. 8 19. 4325

20. 34 21. 31

22. (a) 61

(b) 31

(c) 65

23. (a) False: outcomes are not equally likely. Each horse and rider has different skills . (b) False: outcomes are not equally likely. Each golfer has different skills . (c) False: outcomes are not dependent on the one before. Each time the coin is tossed, the probability is the same . (d) False: outcomes are not dependent on the one before. Each birth has the same probability of producing a girl or boy. (e) False: outcomes are not equally likely. Each car and driver has different skills.

Exercises 11.2

1. 115

2. 92

3. 99.8% 4. 0.73 5. 38%

6. 98.5% 7. 2322

8. 185

9. 0.21 10. 91.9%

11. 87

12. 4946

13. (a) 152

(b) 1513

14. 117

15. 1615

Exercises 11.3

1. (a) 103

(b) 53

(c) 2011

(d) 107

2. (a) 51

(b) 21

(c) 53

(d) 53

(e) 5019

3. (a) 265

(b) 269

(c) 1312

4. (a) 10029

(b) 2013

(c) 259

5. (a) 4527

(b) 94

(c) 32

6. (a) 143

(b) 2813

(c) 289

7. (a) 8021

(b) 8017

(c) 4021

8. (a) 101

(b) 2011

(c) 207

9. (a) 257

(b) 152

(c) 7544

10. (a) 103

(b) 52

(c) 103

Exercises 11.4

1. 361

2. 41

3. 81

4. 41

5. 12125

6. (a) 0.0441 (b) 0.6241 7. 80.4% 8. 32.9%

9. (a) 499

(b) 9115

10. 2075

3 11.

9919

12. 16 170

1

13. (a) 35 93729 791

(b) 35 937

8 (c)

35 93735 929

14. (a) 2400

1 (b)

5760 0001

(c) 5760 0005755 201

15. (a) 7776

1 (b)

77763125

(c) 77764651

16. (a) 25 000 000

9 (b)

25 000 00024 970 009

(c) 25 000 000

29 991

17. (a) 41

(b) 100

9 (c)

1009

18. (a) 221

(b) 111

(c) 227

(d) 2215

19. (a) 61.41% (b) 0.34% (c) 99.66%

20. (a) 21

n (b)

21

n (c) 1

21

22 1

n n

n

− = −

Exercises 11.5

1. (a) 41

(b) 41

(c) 21

2. (a) 81

(b) 83

(c) 87

3. (a) 900

1 (b)

9001

(c) 450

1 4. (a)

251

(b) 252

5. (a) 16925

(b) 16980

6. (a) 27.5% (b) 23.9% (c) 72.5%

7. (a) 0.42 (b) 0.09 (c) 0.49 8. (a) 1000189

(b) 1000441

(c) 1000657

9. (a) 0.325 (b) 0.0034 (c) 0.997

10. (a) 12160

(b) 116

11. (a) 274

(b) 61

12. (a) 251

(b) 825

1 (c)

82564

(d) 165152

(e) 16513

13. (a) 1 249750

19 (b)

124 975498

(c) 1 249 7501 239771

14. (a) 7516

(b) 7538

15. (a) 20251936

(b) 2025

88

16. (a) 2011

(b) 203

17. (a) 1296

1 (b)

324125

(c) 1296671

18. (a) 1 000 000

84 681 (b)

1 000 000912 673

(c) 1 000 000

27

653ANSWERS

19. (a) 17.6% (b) 11% (c) 21.2%

20. (a) 30251488

(b) 1211

21. (a) 191

(b) 956

(c) 19021

(d) 3817

22. (a) 42522

(b) 425368

(c) 4257

23. (a) 6517

(b) 715133

(c) 2145496

24. (a) 0.23 (b) 0.42 (c) 0.995 25. (a) 33% (b) 94%

26. (a) 216

1 (b)

725

(c) 21691

27. (a) 101

(b) 103

(c) 52

28. (a) 8125

(b) 8140

(c) 8156

29. (a) 1331343

(b) 1331336

(c) 1331988

30. (a) 8000

1 (b)

80006859

(c) 80001141

Exercises 11.6

1. 10 000 2. 2 600 000 3. 387 420 489

4. 10 000 5. 10 6. 248 832 7. 523 566

1

8. 1 757 600 9. 1320 10. 1 000 000

1

11. 1000 12. 90 000 000 13. (a) 7776 (b) 7776

1

14. 280

1 15. 84

Exercises 11.7

1. (a) 360 (b) 5040 (c) 9 (d) 60 (e) 20 160

2. (a) 90 (b) 151 200 (c) 30 240 (d) 720 (e) 1 814 400

3. (a) 120 (b) 60 (c) 40 4. (a) 3024 (b) 1680 (c) 672

5. (a) 64 (b) 16 6. 990 7. 24

8. 60 9. (a) 5040 (b) 720 10. (a) 24 (b) 6

11. (a) 40 320 (b) 1152 (c) 10 080 (d) 2880 (e) 283

12. (a) 5040 (b) (i) 72

(ii) 354

13. (a) 3 628 500 (b) 362 880

14. (a) 120 (b) 53

(c) 36

15. (a) 362 880 (b) 40 320 (c) 20 160 16. 192

17. (a) 120 (b) 24 (c) 20 160 (d) 2520 (e) 1260 (f) 20 (g) 907 200 (h) 604 800 (i) 277 200 (j) 9 979 200

18. (a) 720 (b) 240 (c) 144 (d) 72 (e) 600 (f) 480

19. (a) 362 880 (b) 40 320 20. 19 958 400

Exercises 11.8

1. 190 2. 56 3. 15

4. (a) 1365 (b) 364 (c) 78 5. (a) 735 (b) 4915

6. (a) 2300 (b) 10077

7. 38760

1 8. 84 672 315

9. (a) 8 145 060 (b) 2 036 265

1 (c)

407 2535

10. 2 760 681

10 11. (a) 84 (b) 45 12. 100 100

13. (a) 126 (b) 40 14. (a) 86 493 225

1 (b) 3 837 240

15. (a) 4368 (b) 1764 (c) 212

16. (a) 350 658 000 (b) 75112

17. (a) 10 (b) 6

18. (a) 560 (b) 14

19. (a) 350 (b) 140 (c) 3512

20. (a) 5 852 925 (b) 66 (c) 452

Exercises 11.9

1. (a) 512105

(b) 12815

(c) 1024

45 (d)

5125

(e) 1024

11

2. (a) 128

7 (b)

1287

(c) 12829

3. (a) 419 904109 375

(b) 209 952

875 (c)

419 904175

(d) 1679 616

41

(e) 1679 6161 015 625

4. (a) 21625

(b) 324

5 (c)

324125

(d) 432425

(e) 432

7

5. (a) 0.254 (b) 0.01 (c) 0.0467 (d) 0.198 (e) 0.552

6. (a) 625144

(b) 3125243

(c) 62548

(d) 3125992

(e) 31252133

7. (a) 90.1% (b) 0.4% (c) 99.6%

8. (a) 11.4% (b) 16.4% (c) 1.2%

9. (a) 5764 801

69 984 (b)

823 54346 656

(c) 823 54357 591

(d) 823 543147 456

(e) 5764 801

65 536 10. C

113

11820

7

7 13

d dn n

11. (a) (i) 2512

(ii) 2516

(b) (i) 0.25 (ii) 0.167

12. (a) 0.23 (b) 0.07 (c) 0.03

13. ´C C61

65

113

118820

3

3 17

2

2 6

d d d dn n n n


14. ´C C43

41

65

61812

3

3 9

5

5 3

d d d dn n n n 15. 0.000 003 4

16. 7 people 17. (a) 8 sixes (b) 50C61

65

8

8 42

d dn n

18. (a) 25 times (b) 0.0918 19. (a) (i) 7235

(ii) 7231

(iii) 7237

(b) 144101

(c) 0.113 20. (a) 11253

(b) 0.127

Test yourself 11

l . (a) 80.4% (b) 1.4% (c) 99.97%

2. (a) 1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1

6 5 4 3 2 1 0

(b) (i) 61

(ii) 61

(iii) 21

3. (a) (i) 401

(ii) 4039

(b) 79639

4. (a) 154

(b) 101

5. False: the events are independent and there is the

same chance next time 41

d n

6. (a) 21

(b) 10029

(c) 51

(d) 2511

(e) 2516

7. (a) 52

(b) 157

(c) 152

8. (a) 7235

(b) 6635

9. 561

10. (a) 0.009% (b) 12.9% 11. (a) 60 (b) 75

12. (a) 131

(b) 133

(c) 265

13. (a) 3125972

(b) 15 625

640

14. (a) 6325

12 (b) 4320 15. (a)

125

(b) 31

16. (a) 39 916 800 (b) 32 659 200 (c) 112

17. (a) 409

(b) (i) 103

(ii) 16027

(iii) 254

18. (a) 200

1 (b)

20081

(c) 10011

19. (a) 151

(b) 54

20. 60 480 21. (a) 01

5 (b)

7450147

(c) 1

3725 (d)

33

725577

22. (a) 36180

(b) 17140

23. (a) 92

(b) 31

24. (a) (i) 24364

(ii) 729728

(b) 0.000 092

25. (a) 5021

(b) 253

(c) 5023


1. (a) 71

(b) 74

2. (a) 6720 (b) 1440 (c) 71

3. 34 650 4. (a) 0.04 (b) 0.75 (c) 0.25

5. (a) 54 145

1 (b)

1 26433

73

6. (a) 121

k− (b) C

21k

k k3− (c) C21k

k9

7. (a) 0.18% (b) 10.7% (c) . .C 0 73 0 27kk k20 20 −] ]g g

8. (a) 39 916 800 (b) 967 680 (c) 119

9. (a) 24 (b) 12

10. (a) 5 852 925 (b) 113 400 (c) 101

11. (a) !n 1−] g (b) n 1

2−

(c) Probability ( )!

! ( )!

n

n

1

3 3=

−−

( ) ( ) ( )( ) . . .

( ) ( ) . . .

( ) ( )

´ ´ ´

n n n n

n n

n n

1 2 3 4

3 2 1 3 4

1 26

=− − − −

− −

=− −

(d) ( )!

! ( )! !n

k n k

n n n n kk

1 1 2 3 1f−−

=− − − − +] ] ] ]g g g g

12. (a) 134

(b) 5225

(c) 134

13. No—any combination of numbers is equally likely to win.

14. (a) 0 (b) 101

(c) 103

15. (a) 7776

1 (b)

12961

(c) 3888125

16. (a) 103

(b) 14512

17. (a) 144

1 (b)

1445

(c) 144

7 (d)

1443


1. 2016− 2. 3360 3. 20 160

4. x448 10− 5. 3888125

6. 45 697 600

7. 80 8. 660 660 9. (a) ( ) , ( )f f2 0 3 0< > (b) .x 2 25Z

655ANSWERS

10. 31

11. (a) 13.3% (b) 14.1%

12. (a) 24011296

(b) 2401864

(c) 24011105

13. (a) 361

(b) 61

(c) 3611

(d) 365

(e) 125

14. (a) 507

(b) 2011

15. (a) 56 9531253764 768

(b) 170 859 375170 859 374

16. 1 400 000

17. x x

y

x

y

x

y

x

y1 18 135 540 12156 5 4

2

3

3

2

4

− + − +

xy

y1458

7295

6− +

18. 110

1 19. a a b ab b6 12 83 2 2 3− + −

20. (a) ( ) , ( )f f0 0 1 0> < (b) .x 0 76Z

21. ADC 90+ = BDC 90` + = ( ADB+ straight + )

BC` is a diameter ( + in semicircle)

AC is a diameter (similarly)

22. 2

5103− 23. (a) . ´6 5 1011 (b) 101 606 400

24. (a) 335

(b) 6635

25. (a) 121

(b) 41

26. x80

4− 27.

390 625108 864

28. a a b a b ab b81 216 216 96 164 3 2 2 3 4− + − + 29. 10 000

30. 40000

1 31. 76 473 32. πx y6 3 3 03− + − =

33.

2

2

x x

x

11

2

1 210

310

0

Substitute

k

k

k

k

k

k

10

0

10

10

0

10

10

0

10

+ =

=

+ =

=

=

=

=

k

k

k] c

] c

c

g m

g m

m

/

/

/

34.

( )

sin

cos

sin

sin

y x

dx

dyx

dx

d yx

xy

7

7 7

7 7 7

49 749

2

2

=

=

= −

= −= −

35. ( )f xx

211 3= −−

36. (a) 365

(b) 125

(c) 61

(d) 367

(e) 21

37. ff

0 3 01 1 0

<>

= −=

]

]

g

g

So the root lies between 0 and 1.

38. π8

3 39. (a) °θ 76 52= l (b) 0.92 cm 2

40. $18 399.86 41. (b) 42. (a) 43. (d) 44. (b)

45. (d) 46. (d) 47. (c) 48. (a)

49. (a), (c) 50. (d)

Sample examination papers

Mathematics—Paper 1

1. (a) 0.75 (b) ( ) ( )x x3 2 3− −

(c) ( )

( )

x x

x xx x

xx

16

2 16

31

6 5

3 2 1 303 2 2 30

2 3028

− − =

− − =− + =

+ ==

d dn n

(d)

.

π

π

π

π

r

r

r

r

r

1231

3636

36

3 39

2

2

2

=

=

=

=

=

(e) 3300 x

xxx

xx

3 74

3 73 7

1010 4

(f)

or

<<<>>< <`

+

− ++ −

−−

] g

2. (a) (ii) Since , : :AB BD AB AD 1 2= = ( )

( )

ABC ADEACB AEDA

BC DE

is common

corresponding s,

similarly

+ ++ ++

+ <==

ADE∆;ABC

ADAB

AEAC

AE ACAC CE AC

CE AC

21

22

`

`

`

`

<∆

= =

=+ =

=

(iii) ADAB

DEBC

21= =

.

..´

DEDE

3 421

2 3 46 8 cm

` =

==


(b) (ii) )

° °

( )

(

NOMNMO

10532

straight angle

angle sum of

++ T

==

° °

°°

sin sin

sin sin

sinsin

Aa

Bb

MO

MO

43 325

325 43

=

=

=

.6 4 mZ

(iii)

m

° °.

°. °

sin sin

sin sin

sinsin

Aa

Bb

MP

MP

75 536 4

536 4 75

8Z

=

=

=

3. (a) (i) 2

2

x x x x

dx

dyx x

xx

5 1 5 1

21

152

115

3 3

2 2

+ + = + +

= + = +

1

1−

(ii)

´

ln lnx x x x

dx

dyx x

x x

xx

31

3

31

3 1

3 1

1

2

2

2

+ = +

= −

= −

= −

−

−

(iii) ´dx

dyx

x

5 2 3 2

10 2 3

4

4

= +

= +

]

]

g

g

(b) (i) x

e Cx

e C2 1

12

x x2 2

−−

+ = + +− −

(ii) π

θ θ π π

ππ

cos cos cos 0 0

1 12

− + = − + − − +

= − − + += +

0

] ]

]

g g

g

; E

(c) (i) ( )

´2 1

5

2 1

2 1

2 1

5 2 5

15 2 5

5 2 5

2 2− +

+=

−

+

=+

= +

( )

,

´

a b

5 2 5 5 5 2

5 25 2

5 505 50

ii

`

+ = += += +

= =

4. (a) (i)

( ) ( , )( ) ( )

( )

A x y3 2 3 4 17 03 3 4 2 17 0

9 8 17 00 0

ii Substitute into

true

+ − =+ − =

+ − ==

A` lies on the line

( , )( ) ( )

( )

B x y1 5 3 4 17 03 1 4 5 17 0

3 20 17 00 0

Substitute into

true

− + − =− + − =

− + − ==

B` lies on the line Since both A and B lie on the line ,x y3 4 17 0+ − = this is the equation of AB

(iii) | |

| ( ) ( ) |

| |

.

da b

ax by c

3 4

3 0 4 0 17

9 16

17

25

17

517

3 4 units

2 2

1 1

2 2

=+

+ +

=+

+ −

=+

−

=

=

=

(iv) Length AB : 2 2

2

( ) ( )

( ) ( )

d x x y y

3 1 2 5

16 9

255

2 1 2 1

2

= − + −

= − − + −= +==

6 @

:

.

.

´ ´

OAB A bh21

21

5 3 4

8 5

Area

units2

∆ =

=

=

(b) °

.

.

coscos

c a b CBC

BC

ab26 4 2 6 4 8749 49

49 497 cm

2 2 2

2 2 2

Z

Z

= + −= + −

=

] ]g g

5. (a) (i) % . %´24 600

800100 3 25=

(ii) Arithmetic series with ,a d24 600 800= = and n 12=

–

–

T a n d

T

1

24 600 12 1 80033 400

n

12

= += +=

]

]

g

g

So Kate earns $33 400 in the 12 th year.

(iii) [ ( ) ]

[ ( ) ]´

Sn

a n d2

2 1

212

2 24 600 12 1 800

348 000

n = + −

= + −

=

So Kate earns $348 000 over the 12 years.

657ANSWERS

(b) (i) yy

y x x xx xx

3 9 23 6 96 6

3 2

2

= − − += − −= −

l

m

For stationary points, 0=yl

( )( ),

x xx x

x

3 6 9 03 3 1 0

3 1

i.e. 2

`

− − =− + =

= −

,

,

x y

x y

3 3 3 3 9 3 225

1 1 3 1 9 1 27

When

When

3 2

3 2

= = − − += −

= − = − − − − − +=

] ]

] ] ]

g g

g g g

So ( , )3 25− and ( , )1 7− are stationary points. ( , ), ( )

( , ), ( )( )

( )

y

y

3 25 6 3 60

1 7 6 1 60

At

Atminimum point

maximum point

>

<

− = −

− = − −

m

m

( , )1 7` − is a maximum, ( , )3 25− minimum stationary point ( )

xxx

06 6 0

6 61

iii.e.

=− =

==

yFor inflexions, m

When ,x 1= 3 ( ) ( )y 1 3 1 9 1 29

2= − − += −

( , )1 9` − is a point of infl exion

(iii)

6. (a)

(i) ( , ) ( ) ( )( )

´ ´ ´ ´

´ ´

P P PP

2 1

72

72

75

72

75

72

75

72

72

34360

W N WWN WNWNWW

= ++

= +

+

=

(ii) ( ) ( )

´ ´

P P1

175

75

75

343218

at least one W NNN= −

= −

=

(b) °, ° °°, °

x 60 180 6060 120

(first, second quadrants)= −=

(c) (i) ( )v t dt

t t C

6 4

3 42

= += + +

#

`

`

,

,

t v

CC

v t tt

v

0 0

0 3 0 4 0

3 45

3 5 4 595

When

When

cms

2

1

= == + +== +== += −

2

2

] ]

] ]

g g

g g

(ii) ( )x t t dt

t t C

3 4

2

2

2

= += + +3

#

,

,

t x

CCt t

t

0 0

0 2 0

22

2 2 216

When

When

cm

3 2

2

2

`

`

`

= == + +== +== +=

x

3

3

x

0 ] ]

] ]

g g

g g

7. (a)

c cc

( )

1

( )

(DC AB

DX AD

ADXDAX DXA

ADX

DX DC

2

180 60 260

21

ii

is isosceles

is equilateral

opposite sides of gram)

given`

`

'

`

+ +

<

∆

= =

= =

= = −=

= −c m

c c

c

( , )ADC XCB AD BCcointerior s,+ + + <

[ ,XC CB 1 similar to part (ii)]= =

( ) 180 60

120

XCB

CXB

iii

is isosceles

+

∆

= −

=

c cc

c c c

c

(180 120 ) 230

180 (60 30 )

90(

¸CXB CBX

AXB

AXB

DXC

is right angled

straight angle)

`

`

+ +

++

∆

= = −=

= − +

=

( ) (AXa b

BXBX

BX

BX

ADX1

2 11

3

iv equilateral)2 2 2

2 2 2

2

2

`

∆== += += +==

c

43


(b) Sketch y x2= and x y 3+ = as unbroken lines.

2

( , ) ≥≥

y x1 00 1

Substitute into(false)

2

( , ) ££

x y1 0 31 0 3

Substitute into(true)

++

(c) (i)

(ii) The curve is increasing so .dtdP

0>

The curve is concave downwards so .dtd P

0<2

2

8. (a) (i) x 1 2 3 4 5

y 0 0.301 0.477 0.602 0.699

( ) ( )

.

log log log log1 2 2 3

1 73Z

= + + +

( ) ( ) [ ( ) ( )]

( ) ( ) ( )

( ) ( ) ( )

( ) [ ( ) ( )]

( ) [ ( ) ( )]

( ) ( )

log

log log log log

f x dx b a f a f b

x dx f f

f f

f f

f f

21

21

2 1 1 2

21

3 2 2 3

21

4 3 3 4

21

5 4 4 5

21

21

21

3 421

4 5

(ii)a

101

5

= − +

= − +

+ − +

+ − +

+ − +

+ + + +

b

6

6

@

@

#

#

(b) ( ) ,

,

.

ln lnln

ln

t PP e

P

P et P

e

e

ee

k ek

k

k

0 2020

206 100

100 20

20100

55

66

65

0 268

i When

When( )

kt

k

k

k

k

00

0

6

6

6

6

`

Z

= ===== ==

=

====

=

When( )

,P etP e

e

20102020292

ii

mice

.

. ( )

.

t0 268

0 268 10

2 68

=====

(iii) ,

.

.

.

ln lnln

ln

Pe

e

ee

t et

t

t

500500 20

20500

2525

0 2680 268

0 26825

12

When

(i.e. after 12 weeks)

.

.

.

.

t

t

t

t

0 268

0 268

0 268

0 268

==

=

====

=

=

9. (a) ( )( )π

π

π

Sr r h

r hr

hr

r

1602 160

2160

2160

i`

=+ =

+ =

= −

ππ

π ππ

r r

V r h

r r r

r r

80

80

80

2

2

3

= −

=

= −

= −

c m

rZ

.

VV

±

±

π

ππ

π

π

r

rr

r

r

80 30

80 3 080 3

380

380

(ii)For max./min. volume,i.e

2

2

2

2

= −=

− ==

=

=

.2 91

l

l

. , ( . )

.

VV

ππ

rr

r

62 91 6 2 91

02 91

When

cm(maximum)<

`

= −= = −

=

m

m

659ANSWERS

(iii) . , ( . ) ..

πr V2 91 80 2 91 2 914206

Whencm

3

3

= = −=

] g

(b) 1996 to 2025 inclusive is 30 years. ( . ) ( . ) ( . ) . . .

( . )500(1.12 1.12 . . . 1.12 )500(1.12 1.12 . . . 1.12 )

A 500 1 12 500 1 12 500 1 12500 1 12

30 29 28

1

30 29 1

1 2 30

= + + ++

= + + += + + +

. . . . . .. , .( )

.

. ( . )

.( . )

$ .

a r

Sr

a r

S

A

1 12 1 12 1 121 12 1 12

1

1

1 12 1

1 12 1 12 1

270 29500 270 29

135146 30

is a geometric series

n

n

1 2 30

30

30

`

Z

+ + += =

=−

−

=−

−

==

10.

( , )( )lnln

log

y y ee

x

44

44 4

(a) (i) Substitute into

point of intersection isby definition of

x

x

`

`

= ==

=

(ii)

ln 4

ln4

ln

ln

A OABC e dx

e4 4

4 4 4 1

Area of rectangleln

x

x

0

4

0

2

= −

= −

= − +( )

( )

ln

ln

e e4 4

4 4 3 units

= − −

= −

06 @

#

(b) (i) For real, equal roots, 0∆ = i.e.

( )

( ) ( ) ( )( )

±

±

±

±

±

b ac

k kk k k

k k

ka

b b ac

4 0

1 4 1 02 1 4 0

6 1 0

24

2 1

6 6 4 1 1

26 32

26 4 2

3 2 2

2

2

2

2

2

2

− =− − =− + − =

− + =

=− −

=− − − −

=

=

=

] ] ]g g g

,( )

( )( )

,

kx k x k x x

ab ac

a x xx

51 4 5

04

4 4 1 54

00 0 4 5 0

(ii) When

Since andfor all

>

<> < >

2 2

2

2

2

`

∆

∆∆

=+ − + = + +

= −= −= −

+ +

(c) (i)

( )

y x px qy q x px

y q p x px p

y q p x p

222

2

2

2 2 2

2 2

= + +− = +

− + = + +− − = +^ h

This is in the form ( ) ( ),x h a y k42− = − where a is the focal length and ( , )h k is the vertex. h p= − and k q p2= −

( , )p q pvertex is 2` − −

(ii) a

a

4 1

41

`

=

=

Count up 41

units for the focus

focus is ,p q p412− − +c m

(iii) For P : x m= since it is vertically below ,m m q3 2 +^ h

,

x mm ym

y

P mm

8

8

8

When

So

2

2

2

==

=

= d n

≥

m qm

mq

mq m q

38

823

823

0 0

Distance

since and >

22

2

22

= + −

= +

= +

( ) m qq m

55

iv`

+ == −

Dm

q

mm

dmdD m

823

823

5

846

1

2

2

= +

= + −

= −

For stationary points dmdD

0=

m

m

m

m

846

1 0

846

1

46 8

468

234

− =

=

=

=

=


So there is a stationary point at .m234=

To determine its nature

dmd D

846

0>2

2

=

So concave upwards. minimum turning point

m234

When =

2

Dm

m8

235

8

23234

5234

42321

2

= + −

= + −

=

d n

So minimum distance is 42321

units.

Mathematics—Paper 2

1. (a) (i) xx3 5

8− =

=

(ii) xx3 5

2− = −

= −

(b)

±

xx

xx

5 49 0

93

2

2

2

− = −− =

==

(c)

( )

π

ππ

π

sin

sin

sin

65

6

6

21

2nd quadrant

= −

=

=

c m

(d) ( ) ( )

( )( )( )( )( )

a a aa aa a a

a a

2 4 22 42 2 2

2 2

2

2

2

− − −= − −= − + −= + −] ]g g

(e) ´ ´ ´´ ´ ´

2 4 6 25 6

2 2 6 5 6

4 6 5 6

6

−= −= −= −

(f) ( )

. ..

´

´

log log

log loglog log

50 5 2

5 22 5 22 1 3 0 433 03

a a

a a

a a

2

2

== += += +=

(g) ,

,

,

Px x y y

2 2

23 0

24 2

121

1

1 2 1 2=+ +

= − + + −

= −

f

d

c

p

n

m

2.

(a) Substitute ,A 5 121−c m into x y3 4 9 0+ − =

´ ´3 5 4 121

9

0

LHS

RHS

= + − −

==

A lies on the line

Substitute ,B 1 121

c m into x y3 4 9 0+ − =

´ ´3 1 4 121

9

0

LHS

RHS

= + −

==

` B lies on the line AB has equation x y3 4 9 0+ − = ( ) x y

y x

y x

m

3 4 9 04 3 9

43

49

43

b

1`

+ − == − +

= − +

= −

l is perpendicular to AB , so m m 11 2 = −

m

m

43

1

34

2

2`

− = −

=

Equation of l :

( )

( )

( )

y y m x x

y x

y xxx y

134

4

3 3 4 44 16

0 4 3 13

1 1− = −

− − = − −

+ = += += − +

(c)

( ) :´2 3

( )( )

( ) : ( )( )

( ) ( ):

´

x yx y

x yx y

xxx

4 3 13 0 13 4 9 0 2

1 4 16 12 52 0 39 12 27 0 4

3 4 25 25 025 25

1

− + =+ − =

− + =+ − =

+ + == −= −

661ANSWERS

Substitute x 1= − in (1): ´ y

yy

y

4 1 3 13 09 3 0

9 33

− − + =− =

==

So point of intersection is .( , )1 3−

(d) ( ) ( ):

( )

( )

:

( ) ( )

.

´ ´

d x x y y

AB

d

CP

d

A bh

5 1 121

121

4 3

16 9

255

1 4 3 1

3 4

9 16

255

21

21

5 5

12 5 units

2 12

2 12

22

2 2

2 2

2 2

2

= − + −

= − + − −

= + −= +==

= − − − + − −= += +==

=

=

=

c m

( ) ,

,

,

( , )

ACx x y y

D x y

xx x

x

xx

yy y

y

y

2 2

24 5

2

1 121

21

141

2

21

21

1 10

2

141

2

121

221

121

e Midpoint

where

1 2 1 2

1 2

1 2

=+ +

= − +− + −

= −

=

=+

= +

= +=

=+

− =+

− = +

AC BD

y4

Midpoint midpoint=

− =

f

f

c

p

p

m

(f) ( , )D 0 4So = −

3. (a) (i)

( )cos sincos sin

dx

dyu v v u

x x xx x x

1$

= +

= + −= −

l l

(ii) dx

dye5 x5=

(iii) dx

dy

xx

2 142

=−

(b) (i) ( )

( )´

xC

xC

3 5

3 2

15

3 2

5

5

−+

=−

+

(ii) ´ cos

cos

x C

x C

321

2

23

2

− +

= − +

(c)

3

[ ] [ ]

e e

e e

e e e ee ee e

11

1 12

x x

x x

0

3

3 3 0 0

3 3

3 3

−−

= += + − += + − −= + −

−

−

− −

−

−

0

<

6

F

@

(d) ( )dx

dyx dx

x x C

18 6

9 62

= −

= − +

#

( , ),dx

dy

CC

C

2 1 0

0 9 2 6 224

24

At

2

− =

= − += +

− =

] ]g g

( )

dx

dyx x

y x x dx

x x x C

9 6 24

9 6 24

3 3 24

2

2

3 2

` = − −

= − −= − − +

#

Substitute ( , )2 1− :

CC

Cy x x x

1 3 2 3 2 24 236

353 3 24 35

3 2

3 2`

− = − − += − +== − − +

] ] ]g g g

4. (a)

(i) ( ) ´P52

83

203

WW =

=

(ii) ( ) ( ) ´ ´P P52

85

53

83

4019

WL LW+ = +

=

(iii) ( ) ( )

´

P P1 1

153

85

85

at least W LL= −

= −

=

(b) (i)


(ii)

.

( )

( )( )

( )

.

´ ´

A bh

A x dx

xx

21

21

5 5

12 5

2

22

23

2 32

22 2

12 5

units

or

units

2

2

2

3

2 2

2

2

=

=

=

= +

= +

= + −−

+ −

=

−

−

3

<

< =

F

F G

#

or

`

( )

( )

( )

( ) ( )

π

π

π

π

π

π

π

π

π

´

´

y x

y x

V y dx

x dx

x

V r h

2

2

2

1 3

2

3

3 2

3

2 2

3125

0

3125

31

31

5 5

3125

iii

units

units

a

2 2

2

2

2

3

2

3

3 3

2

2

3

3

= += +

=

= +

=+

=+

−− +

= −

=

=

=

=

−

−

b

3

]

]

g

g

=

=

<

G

G

F

#

#

(c) (i)

(ii) x1 2< <−

5. (a) y x x x

dx

dyx x

dx

d yx

2 9 12 7

6 18 12

12 18

3 2

2

2

2

= − + −

= − +

= −

,

( ) ( ),

dx

dy

x xx x

x xx

0

6 18 12 03 2 0

2 1 01 2

(i) For stationary points

2

2

=

− + =− + =

− − ==

, ( ), ( )

x yx y

1 2 1 9 1 12 1 7 22 2 2 9 2 12 2 7 3

WhenWhen

3 2

3 2

= = − + − = −= = − + − = −

] ]

] ]

g g

g g

So ( , )1 2− and ( , )2 3− are stationary points.

At ( , ) ( )dx

d y1 2 12 1 18 6

2

2

− = − = −

( , )1 2− is a maximum turning point

At ( , ) ( )dx

d y2 3 12 2 18 6

2

2

− = − =

` ( , )2 3− is a minimum turning point

.

dx

d y

xxx

0

12 18 012 18

1 5

(ii) For points of inflexion2

2

=

− ===

. ,( . ) ( . ) ( . ).

xy

1 52 1 5 9 1 5 12 1 5 7

2 5

When3 2

== − + −= −

Check concavity:

x 1.25 1.5 1.75

dx

d y2

2

3− 0 3

Concavity changes, so . )2 5( . ,1 5 − is a point of infl exion. ,

3,

2 9 12 7

x

y

x

y

3

2 3 9 3 12 3 7178

3 3 32

(iii) When

When

3 2

3 2

= −= − − + − −= −== − + −=

] ] ]

] ] ]

g g g

g g g

(b) (i)

(ii)

°´ ´

coscos

c a b ab C

c

2850 1200 2 850 1200 1203182 500

31825001784

2

2 2 2

2

= + −= + −===

So the plane is 1784 km from the airport.

663ANSWERS

(c) ( ) ( )θ θ θ θθ θ

θ θ

cosec cot cosec cotcosec cot

cot cot11

2 2

2 2

+ −= −= + −=

6. (a)

`

At ( , ) ( )

dx

dyx

dx

dy

m

2

2 4 2 2

4

(i)

1

=

− = −

= −

Normal is perpendicular to tangent

( )

m m

m

m

y y m x x

y x

y xx y

14 1

41

441

2

4 16 20 4 18 1

1 2

2

2

1 1

` = −− = −

=

− = −

− = − −

− = += − +

_

]

i

g

(ii) y x2= (2) Substitute (2) in (1):

( ) ( ),

,

x xx x

x xx x

x x

x

0 4 184 18 0

2 4 9 02 0 4 9 0

2 4 9

241

2

2

= − +− − =

+ − =+ = − =

= − =

=

`

( ):

,

x

y

Q

241

2

241

5161

241

5161

Substitute in

2

=

=

=

=

d

c

n

m

(iii)

:PQ x yx y

xy

4 18 018 4

4 418

− + =+ =

+ =

4

4

( ) ( ) ( )

.

xx dx

x x x

4 418

8 418

3

8

241

4

18 241

3

241

8

2

4

18 2

3

2

12 8

Area

units

2

2

2 3

2

2

2 3

2 3

2

= + −

= + −

= + −

−−

+−

−−

=

−

−

1

1

2

c

c c c

m

m m m

R

T

SSSS

<

=

V

X

WWWW

F

G

#

(b) (i) The particle is at the origin when ,x 0= i.e. at ,t t1 3 and t5

(ii) At rest, dtdx

0= (at the stationary points,

i.e. t2 and t4 )

,

,

.

ln ln

ln

ln

T T e

t TT

T et T

e

e

e

k ek

k

kT e

0 9797

975 84

84 97

9784

9784

55

59784

0 02997

(c)When

When

So .

´

kt

kt

k

k

k

t

0

0

5

5

5

0 029

`

== ==== ==

=

=

= −= −

−=

==

−

−

−

−

−

−

(i) tT e

159763

When. ´0 029 15

===

−

So the temperature is 63°C after 15 minutes. (ii)

.

.

..

ln ln

ln

ln

Te

e

e

t et

t

t

2020 97

9720

9720

0 0290 029

0 0299720

54 9

When.

.

.

t

t

t

0 029

0 029

0 029

==

=

=

= −= −

−=

=

−

−

−

So the temperature is 20°C after 54.9 minutes.

7. (a) (i)


4

( ) ( ) ( )

ππ

π π π

tan tan tan

tan tan tan

f x dxh

y y y y y

x dx

34 2

316

04

416 16

32

8

a0 4 1 3 2

0

2

Z

Z

+ + + +

+

+ + +

π

.0 35 unitsZ

b

c

c

m

m

7

;

A

G

#

#

(ii) cossin

tandx

dyxx

x

= −

= −

(iii)

) 44 (

[ ( )]

.

π

tan ln cos

ln cos ln cos

x dx x

40

0 35

00

= −

= − − −

=

ππ

c m

:

=

D

G

#

(b) (i)

° °

°°

.

sin sin

sin sin

sinsin

Aa

Bb

AD

AD

23 11040

11040 23

16 6 m

=

=

=

=

(ii) °.

. °.

sin

sin

BD

BDBD

4716 6

16 6 4712 2

=

==

So the height is 12.2 m

( )sum of quadrilateral+

°° °°° °

°° ( ° ° °)

°

( )

( )

( )

CBE

CBEDCB

ABC

DAB

DAB DCB ABC ADC

5050 50100130 5080360 100 80 80

100

(c)

andABCD is a parallelogram

base s of isosceles

exterior of

opposite s equal

`

`

++

+

+

+ + + +

+

+

+

∆∆

== +== −== − + +

== =

8. (a) (i) & (ii) ,π

π322

Amplitude period= = =

(iii) 4 points of intersection, so 4 roots (b)

°, ° °°, °

( )

sinsin

sin

xx

x

x

2 1 02 1

21

30 180 3030 150

1st, 2nd quadrants

− ==

=

= −=

(c) (i) ´log log

log loglog logq p

12 2 3

2 32 2 32

x x

x x

x x

2

2

== += += +

] g

(ii) log log logx x

q

2 21

x x x= += +

(d) (i) . . .1 3 5+ + + is an arithmetic series with ,a d1 2= =

( )( ) ´

nT a n d

T

121

1 12 1 223

When

n

12

== + −= + −=

So there are 23 oranges in the 12th row. (ii) Total number of oranges is 289, so S 289n =

[ ( ) ]

[ ( ) ]

[ ]

´

´

Sn

a n d

nn

n nn nn

n

nn

22 1

2892

2 1 1 2

578 2 2 22

2289

28917

n

2

2

= + −

= + −

= + −=====

So there are 17 rows of oranges altogether.

9. (a)

(b) The statement would only be true if there were equal numbers of each colour. It is probably false. ( )

( )( ),

( )≠

ln ln

ln

x xx x

x xx x

x

x

2 32 3

2 3 03 1 0

3 1

13

1

(c)

Butso the solution is

does not exist

x

2

2

2

`

= += +

− − =− + =

= −−

=−

(d) (i)

Whendx

dye

x k

dx

dye

x

k

=

=

=

665ANSWERS

So gradient m ek=

(ii) ,( )

( )

( )

x k y ey y m x x

y e e x ke x ke

y e x ke ee x k 1

When k

k k

k k

k k k

k

1 1

= =− = −− = −

= −= − += − +

(iii) Substitute ( , )2 0 into the equation ( )

( )e ke k

kk

0 2 13

3 03

k

k

= − += −

− ==

`

θ

°

°°

°°

°

.

ππ

π

π

π

´

´ ´

A r

180

1180

53180

53

18053

21

21

718053

22 7

(e) radians

cm

2

2

2

=

=

=

=

=

=

=

10. (a) (i)

(ii)`

( )

( )

( ) ( )

π

π

π

π

π

π

´

y xy x

V y dx

x dx

x

22

2

1 5

2

5

1 2

5

2 2

5243

0

5243

units

a

b

2

2 4

2

4

2

1

5

2

1

5 5

3

= += +

=

= +

=+

=+

−− +

= −

=

−

−

]

]

g

g

=

=

<

G

G

F

#

#

(b) (i) std

=

t sd

s3000

So =

=

Cost of trip taking t hours:

´

C s t

s s

s s

s s

7500

75003000

30007500 3000

30007500

2

2

= +

= +

= +

= +

]

]

c

g

g

m

( )

C s ss s

dsdC

s

s

30007500

3000 7500

3000 1 7500

3000 17500

(ii)

1

2

2

= +

= +

= −

= −

−

−

c

]

d

m

g

n

For minimum cost, dsdC

0=

.( )

s

s

ss

s

3000 17500

0

17500

0

17500

7500

750086 6 km/h

speed is positive

2

2

2

2

− =

− =

=

===

d n

Check:

( )

.

.

dsd C

s

ss

dsd C

3000 15000

300015000

86 6

300086 6

15000

0

When

>

2

23

3

2

2

3

=

=

=

=

−

d

d

n

n

Concave upwards So minimum when .s 86 6=

..

$ .

C 3000 86 686 67500

519 6155196 15

(iii)

cents

= +

==

d n

Extension 1—Paper 1

1. (a) £x 5

31

+ 2x 5+( )

( ) [( ) ]( )( )

( ( ) )

£££

x xx xx xx x

3 5 50 5 3 50 5 5 30 5 2

multiplying both sides by2

2

+ ++ − ++ + −+ +

£ ]

] ]

g

g g

For ( ) ( ) , , .≥ ≥£x x x x5 2 0 5 2+ + − − But .≠x 5− Solution is , .≥x x5 2< − −


(b)

´ ´

xk l

kx lx

3 23 3 2 2

1

2 1=+

+

=+

+ −

=

´ ´

yk l

ky ly

3 23 3 2 8

5

2 1=+

+

=++

=

So point is ( , ) .1 5

(c)

π

π

sin sin sinx2

1 0

20

2

1

0

21 1= −

= −

=

− − −< F

(d) −

2 2[ ] ( )

( )

dxd

x x x

x

x

121

1 2

1

2 2

2 3

− = − − −

=−

−1 3

] ]g g

(e)

5

( ) ( )

( )

´

u x

dxdu

x

du x dx

x x dx x x dx

u du

uC

uC

xC

5

3

3

531

5 3

31

31

6

18

18

5

Let 3

2

2

2 3 5 3 2

5

6

6

3 6

= −

=

=

− = −

=

= +

= +

=−

+

# #

#

2. (a) (i) ( )

( )

EBD CDEBEO EBD CDE

CDE

s

2

base of isosceles

exterior of

+ ++ + +

+

+

+

∆∆

== +=

( )

( )

π

π

ABE

BAO BEO

2

2

(ii) in semicircle

sum of`

+

+

+

+ ∆

=

= −

(iii)

´´

DE BEDA DE AE

DF DE DA

DF

5

5 712

5 1260

60

2 15 cm

2

= == += +======

(b)

.0 72Z −

( )( . )

x0 5−

a

a

.

( . ) .

( . )

.( . )

( ( . ) )

x

f ee xe

a af

e

e

0 5

0 5 0 52

2 0 5

0 52 0 5

0 5

Let.

.

.

.

x

0 5 2

0 5

1

0 5

0 5 2

= −− = − −

= −= − −

= −

= − −− −− −

−

−

−

−

f

f

f ll

l

]

]

]

g

g

g

(c) (i)

π

π

sinf2

12

2

1

24

2

1=

=

=

−e

d

o

n

(ii) £ £siny x x1 1has domain1= −−

2

π π

π π

π π

£ £

£ £

£ £

£ £

sin

sin

sin

sin

y x x

y x y

x

y x

y

2 1 1

2 2

2 2

has domain

has range

i.e.

has range

i.e.

1

1

1

1 1

= −

= −

−

= −

−

−

−

−

− −2π π£ £sin x

(iii)

`

( )

π

π

dx

dy

x

x

dx

dy

y y m x x

y x

x

x y

1

2

2

1

121

2

21

2

2 2

22 2

2

1

2 2 2

0 2 2 22

When

2

1 1

=−

=

=−

=

=− = −

− = −

= −

= − − +

e o

3. (a) (i) cosc a b ab C22 2 2= + − c

cc

cc

2( )( ) 5510 10 2 (10) 55

0 20 5520 55

20 55 ( 0)

( )

≠

coscos

coscos

cos

AB BC AC BC ACBC BCBC BCBC BC

BC BC

ABCfrom2 2 2

2 2 2

2

`

∆= + −= + −= −= −=

.11 5 cmZ

] g

(ii)

c

.0 8665Z −150 3

cos C

BCD+

=

=`

( . ) ( ).

aba b c

2

2 11 5 511 5 5 16

2 2 2

2 2 2

+ −

= + −

l

667ANSWERS

`

( )

( )( )

, ,±

π π

cos coscos sincos cos

coscoscos cos

cos

x xx xx x

xxx x

xx

2

12 1

0 11 1

10 2

(b) 2

2 2

2 2

2

2

== −= − −= −= −= + −==

(c) (i) !11 39 916 800= (ii) To alternate, a girl must be fi rst in line as there is one more girl. The number of positions available for the girls is 6! The boys then have 5! positions possible. So the number of arrangements is 6!5!. i.e. 86 400 (iii) Let the 2 girls be 1 position (with 2! arrangements between them possible). Then there are ! !´2 10 arrangements. i.e. 7 257 600

4. (a) (i)

( )1

( ) ( )

θθθ

θθ

coscoscos

sinsin

x V

x tyy ty t t

1515

1010 155 15 2 not a full proof2

`

==== −= − += − +

o

p

o

(ii) From (1):

15 θcos

tx=

Put in (2):

( ) ( )

θ θθ

θθ

θ θ

θ θ

cos cossin

costan

sec tan

tan tan

yx x

xx

xx

xx

515

1515

2255

45

451 3

22

2

2

2

22

= − +

= − +

= − +

= − + +

d dn n

(iii) When ,x y5 2= = Put in (3):

c c

( )

( )

( ) ( )( )

,

±

±

θ θ

θ θ

θ θ

θ

θ

tan tan

tan tan

tan tan

tana

b b ac

2455

1 5

095

95

5 2

5 5 45 18

24

2 5

45 45 4 5 23

83 15 28 33

22

2

2

2

2

= − + +

= − − + −

= − − + −

=− −

=− −

= l l

θ θtan tan5 45 23 02 − + =

(b) 10( . ) ( . )1810

0 01 0 99 8c m

5. (a) (i)

( , ),

( )

( )

ya

x

dx

dy

ax

aq aqdx

dy

a

aq

qy y m x x

y aq q x aqqx aq

y qx aq

4

2

22

2

22

At

2

2

1 12

2

2

=

=

=

=− = −

− = −= −= −

(ii)

( )

( ) ( )

x x

y y

x x

y y

x aq

y aq

ap aq

ap aq

a p q

a p q p q

p q

2 2 2

2

2

1

1

2 1

2 1

2 2 2

−−

= −−

−

−=

−

−

=−

+ −

=+

( ) ( ) ( )( )

( )

( )

y aq p q x aqy aq p q x apq aq

y p q x apq

y p q x apq

2 22 2 2 2

2 2 0

21

0

2

2 2

− = + −− = + − −

− + + =

− + + =

Focal chord passes through ( , )a0

`

( )a p q apq

a apqapq apq

21

0 0

0

1

i.e. − + + =

+ == −= −

(iii) .

..

( . ) ( . ). .

°

θ

θ

tan

l q

PQp q

m m

m m

0 2

2 23 0 2

1 4

1

1 0 2 1 40 2 1 4

65 46

has gradient

has gradient

1 2

1 2

= −+

= − =

=+

−

=+ −− −

= l

(b) . .

( ) ( )( )

n 11 1

61

1 2 3 1

LetThen LHS

RHS

2

== =

= =

true for n 1= Assume true for .n k=

i.e. . . . ( ) ( )k k k k1 261

1 2 12 2 2+ + + = + +

Prove true for .n k 1= +

i.e. 2. . . ( )

( ) ( ) ( )

k k

k k k

1 2 1

61

1 2 2 1 1

2 2 2+ + + + +

= + + + +6 @


. . .

( ) ( )

[ ( )( ) ]

( ) [ ( ) ( )]

( )( )

( )( )( )

( )( ) [ ( ) ]

k k

k k k k

k k k k

k k k k

k k k

k k k

k k k

1 2 1

61

1 2 1 1

61

1 2 1 6 1

61

1 2 1 6 1

61

1 2 7 6

61

1 2 2 3

61

1 2 2 1 1

LHS

RHS

2 2 2 2

2

2

2

= + + + + +

= + + + +

= + + + +

= + + + +

= + + +

= + + +

= + + + +

=

]

]

]

g

g

g

6. (a) dxd

v x

vx

C

xC

vx

C

21

900

21

2900

450

900

(i) 2 3

22

1

2 1

22

= −

=−

− +

= +

= +

−

−

c m

When ,x v10 3= =

( )

C

CC

vx

vx

v x

V

310900

9 90

900

900

30

3 0since >

22

22

2

`

= +

= +=

=

=

=

=

(ii)

`

dtdx

x

dxdt x

tx

dx

xC

30

30

30

60

2

=

=

=

= +

#

,

,

t x

C

C

tx

x

t

s

0 10

06010

132

601

32

100

60100

132

165

When

When

2

2

2

`

= =

= +

− =

= −

=

= −

=

(b) (i)

( )( )

T C Ae

dtdT

kAe

k Ae C Ck T C

kt

kt

kt

= +

= −

= − + −= − −

−

−

−

c

( , ,

,

41.2

ln ln

ln

ln

t T CAe

AT Aet T

ee

e

e

k e

k

T et

T e

0 90 2590 2575

25 751 70

70 25 7545 75

7545

7545

17545

25 75325 75

ii) When and

When

When

C

.

. ´

kt

k

k

k

k

t

0

0 51

0 51 3

`

`

= = == +== += == +=

=

=

= −

−=

= +== +=

−

−

−

−

−

−

−

.k 0 51Z

7. (a) (i) xnknk

1

1 1 1

2

Let

n

k

nk

n

k

n

0

0

=

+ =

=

=

=

] c

c

g m

m

/

/

(ii)

( )

dxd

xdxd n

kx

n xnk

kx

x

nnk

k

nnk

k

1

1

1

1 1 1

2

Let

n

k

nk

n

k

nk

n

k

nk

n

k

n

0

1

1

1

1

1

1

1

1

+ =

+ =

=

+ =

=

=

−

=

−

−

=

−

−

=

] c

] c

] c

c

g m

g m

g m

m

= G/

/

/

/

(b) (i) Maximum speed is the amplitude of velocity. Maximum acceleration is the amplitude of acceleration . ( )

( )

( )

ε

ε

ε

cos

sin

cos

x a bt

x ab bt

x ab t2

SHM:

2

`

= +

= − +

= − +...

Max. speed is 4 ab

ab

44

So =

=

(1)

Max. acceleration is 8

( )

ab

bb

bb

84

8

4 82

So 2

2

=

=

==

669ANSWERS

Substitute in (1)

2 (2 )εcos

a

x t

24

2So

=

== +

π π

π

a

b

22

22

Amplitude

Period

=

=

=

(ii) x 0= when π

t6

=

( )επ

ε

πε

πε

cos

cos

cos

cos

x t2 2

0 2 26

23

3

= +

= +

= +

= +

d

d

d

n

n

n

= G

π

επ

επ π

π π

π

πcosx t

3 2

2 3

63

62

6

2 26

`

+ =

= −

= −

=

= +d n

(iii) επ

π

π

cos

cos

cos

cos

x ab bt

t

t

t

x

2 2 26

8 26

4 2 26

4

2

2

= − +

= − +

= − +

= − +

= −

..]

] d

d

d

g

g n

n

n= G

(iv)

`

( )The particle is at rest at the endpoints of the motion.,±

±

xdxd

v

v x dx

xC

v x C

x v

C

C

C

v xx

x

21

21

4

24

42 0

0 4 216

16

4 164 44 4

When

So

2

2

2

2 2

2 2

2

2

2

1

1

1

1

−

=

= −

= − +

= − += =

= − += − +== − += − +=

. 2.d

]

]

]

n

g

g

g

#

Extension 1—Paper 2

1. (a) ( ) ( ) ( )x x x x3 8 3 2 2 43 2+ = + − +

(b) (i) tan x4 ( )

π

π

tan tan4 3 0

43

0

34

1

0

31 1= −

= −

=

− − −

d n

: D

,

,

u x

dxdu

x

du x dxx u

x u

1

2

22 1 2

50 1 0

1

(ii) Let

When

When

2

2

2

= +

=

== = +

== = +

=

2

2

x

xdx

x

xdx

u

du

u du

u

u

1 21

1

2

21

21

21

21

21

2

5 1

5 1

20 20

1

1

1

5

1

5

+=

+

=

=

=

=

= −= −

− 1

1

2 2

5

5

R

T

SSSS

>

V

X

WWWW

H

# #

#

#

(c) 11 letters mean 11! arrangements. There are 2 C’s, E’s, T’s and I’s, each with 2! possible arrangements.

No. of arrangements ! ! ! !

!´ ´ ´2 2 2 2

11

2 494 800

=

=

(d)

c

:

:

( )

( )

81 52

θ

θ

tan

x y m

x y m

m m

m m

3 5 031

2 4 0 2

1

131

2

31

2

7

1

2

1 2

1 2

− + = =

+ − = = −

=+

−

=+ −

− −

== l

2. (a)

BDZ

. .

.

. .

.

´´

AB BD BCBD

CD BC BD

7 2 9 35 6

9 3 5 63 7 cm

2

2

==

= −= −=

(b) ( )T x x8

32

kk

k

12 8= −+

−

kc cm m


For the term with ,x4

`

( ) ( )

( )

k kk

kk

T x x

xx

x

2 8 416 3 4

12 34

84

32

70 8116

90 720

52 8 4

4

84

4

− + − =− =

==

= −

=

=

−c ] c

d

m g m

n

the coeffi cient of x4 is 90 720

(c) To prove: …( )

5 5 54

5 5 1nn

2+ + +−

=

:

( )

n

n

15

4

5 5 1

51

LetLHS

RHS

true for

1

`

==

=−

==

Assume true for n k=

i.e. . . .( )

5 54

5 5 1k

k

+ + =−

Prove true for n k 1= +

. . .( )

5 5 54

5 5 1i.e. k k

k1

1

+ + + =−

++

. . .( )

( ) ´

5 5 5

4

5 5 15

4

5 5 1 4 5

LHS k k

kk

k k

1

1

1

= + + +

=−

+

=− +

+

+

+

( )

´

´4

5 5 4 5

45 5 5

4

5 5 1

RHS

k k

k

k

1 1

1

1

= − +

= −

=−

=

+ +

+

+

Since it is true for ,n k 1= + then it is true for all .≥n 1

3. (a)

:

( )

( )

ya

x

dx

dy

ax

Pdx

dy

a

ap

m p

m m

mp

y y m x x

y app

x ap

py ap x apx py ap ap

4

2

2

2

11

12

22 0

(i)

For normal,

At

2

1

1 2

2

1 1

2

3

3

`

=

=

=

== −

= −

− = −

− = − −

− = − ++ − − =

(ii) Solve y a= − with (1): ( )

( , )( ( ), )

x p a ap apx ap ap

M ap ap aap p a

2 033

3

3

3

3

2

+ − − − == += + −= + −

(b) (i) 2

dx

dy

xx x x x

x x

x

x

x

x

x

1

11

21

1 2

1

1

1

1

1

1

2 2

2

2 2

2 2

2

2

2

2

2

=−

+ − + − −

=−

+−

− −−

=−

−

− 1

] ]g g

(ii)

sin x x+2

( )

π

π

sin

x

xdx

x

xdx

x

1

121

1

2 2

21 1

21

21

21

141

1 0

21

6 43

241

2 3 3

2

2

0

2

1

2

2

0

2

1

1 20

1

1 2

−− =

−−

= −

= + −

−

= +

= +

−

−

−

1

( )sin 0 0− +

e

e

o

o

:

>

D

G

# #

4. (a) (i)

( )

α β γ ab

1

3

3

+ + = −

=− −

=

(ii) αβ βγ αγ ac

12

2

+ + =

= −

= −

(iii) αβγ ad

11

1

= −

= −

= −

(iv) α β γ αβγ

βγ αγ αβ1 1 1

12

2

+ + =+ +

=−−

=

α β γ α β γ αβ βγ αγ2 2 2(v) 2 2 2+ + = + + + + +2^ h

( )

α β γ α β γ αβ βγ αγ23 2 213

2 2

2

2` + + = + + − + +

= − −=

2 ^ ^h h

(b)

π

π

π

ππ

π

´

´

dtdV

V r

drdV

r

dtdV

drdV

dtdr

rdtdr

r dtdr

S r

drdS

r

10

34

4

10 4

410

4

8

3

2

2

2

2

=

=

=

=

=

=

=

=

671ANSWERS

,

.

ππ

´

´

dtdS

drdS

dtdr

rr

r

rdtdS

8410

20

88

20

2 5

When

2

=

=

=

= =

=

So surface area is expanding at ..2 5 cm s2 1−

(c) :1 4Use ratio −

,

´ ´

´ ´

xk l

kx lx

yk l

ky ly

P

1 41 7 4 2

5

1 41 1 4 5

631

5 631

2 1

2 1

`

=+

+

=− +

− + −

= −

=+

+

=− +

− +

=

= −c m

5. (a) (i) π

π

π

sin

cos

cos

x t

x t

t

x

6 33

18 33

9 2 33

9

= − +

= − +

= − +

= −

.

.

.

d

d

d

n

n

n= G

(ii) Amplitude ,2= period π

3

2=

(iii)

, , , . . .

, , , . . .

, , , . . .

π

π

π π π π

π π π

π π π

cos

cos

x

t

t

t

t

t

0

2 33

0

33

0

33 2 2

325

36 6

76

13

18 187

1813

At the origin,

s

=

+ =

+ =

+ =

=

=

c

c

m

m

(b) (i)

x x x2 2 3− +x x x x x x

x x xx x xx x x

x x xx x x

1 2 5 42 2 2

2 52 2 2

3 33 3 3

3 2

2 5 3 2

5 4 3

4 3 2

4 3 2

3 2

3 2

+ − − + + −+ −− + +− − +

+ ++ −

x4 4−

g

(ii) ( ) ( ) ( ) ( )P x x x x x x x1 2 2 3 4 42 3 2= + − − + + −

(iii) P 1 2 1 1 5 1 1 43

5 3 2= − + + −=

] ] ]g g g

remainder is 3 when dividing by x 1−

(iv) .x 0 5=

( )

( . )

a

0 5

( . ) . . . ..

( ). . . . .

( )

.( . )

...

P

P x x x x

P

a af

f a

f

f

0 5 2 0 5 0 5 5 0 5 0 5 42 3125

10 3 10 10 5 10 0 5 3 0 5 10 0 5 1 5 875

0 50 5

0 55 8752 3125

5 3 2

4 2

4 2

1

= − + + −= −= − + += − + + =

= −

= −

= − −

.0 89Z

l

l

l

l

] ] ] ]

] ] ] ]

g g g g

g g g g

(c)

( )

θπ

θπ

θπ

θ θ θ

θ θ θ

sin

sin cos cos sin

sin cos cos

sin cos cos

26

26

26

223

2 121

321

LHS

RHS

2

2

= −

= −

= − −

= − +

=

d n

6. (a) (i) tk is the ( )k 1 th+ coeffi cient of the binomial since k 0= gives the fi rst term

T t12

2 5k kk k

112` = =+

−

kc m

(ii)

! !( ) ( )!

!

( ) ! !

( )

( )

¸

´

T tk

t

t

k

k k

k k

k

k

12 5

12 5

122 5

12 1 112 2 5

2 5 12

12

2 1

5 12

( )

( )

k kk k

k

k k k k k

k k

k k

2 112 1 1

1 12 1 1 12

11 1

12

= = +

= +

=− + +

−

=+−

+ +− + +

+ − + + −

− +

−

k

12

12

c

c c

m

m m

6 @

`

( )

( )t

tt t

k

k

k kk

k

k

t

1

2 1

5 121

60 5 2 258 7

872

812

2 5

(iii) when> >

>

>>

>

k

k

k k

1

1

84 8

+−

− +

=

=

+

+

8c m

(b) (i) !7 5040= (The 1st person can be seated anywhere.) (ii) For placing one person anywhere and then placing the required person opposite, there are 2! ways these can be arranged. The other 6 people can be arranged in 6! ways. ! !´2 6 1440` = arrangements


(iii) If one of these people is placed anywhere, the other can sit in a choice of 6 other seats (not opposite). The other 6 people can be arranged 6! ways.

!

. !P

76 6

76

=

=

7. (a) (i)

(ii) π

cos21

31 =− d n

(iii) cosx y=

sinyπ

3

π

π π π

π

π

´

cos

sin sin

ydy

21

3

6 2 3

61

23

66 3 3

Area Rectangle

units2

= +

= +

= + −

= + −

=+ −

π

π

π

3

2

2

< F

#

(iv) πV y dxa

b2= #

f+( )f f0 4+

( ) ( ) ( )

( )

.

π

π

π

cos

cos cos cos

f xb a

f a fa b

f b

V x dx

64

2

6

21

0

2

021

21

120 4

41

21

2 75 units

a

b

1 2

0

2

1

1 2 12

12

3

`

Z

Z

Z

−+

++

=

− +

= + +

−

− − −

d

cf c

] d d

n

mp m

g n n

R

T

SSSS

=

=

V

X

WWWW

G

G

#

#

(b) (i)

( )

π

π

π

D R H

RH

DH

R

D HR

D HR

D HR

V r h

D HH

H D H

221

44

44

44

24

2

44

44

16

4

4

2 22

22

22

2

22

2

2

2

2 2

2 2

= +

= +

− =

− =

−=

−=

=

=−

=−

2

2

2

2

] d

e

g n

o

(ii) π π

π π

VHD H

dHdV D H

4 16

4 163

2 3

2 2

= −

= −

For maxima or minima, dHdV

0=

π π

π π

π

´

D H

D H

DH

DH

DH

DH

dHd V H

H

4 163

0

4 163

34

34

3

2

3

3

32 3

166

0 0

i.e.

for< >

2 2

2

2

2

2

2

− =

=

=

=

=

=

= −

2

2

∴ maximum

π

π

π

π

HD

VD

DD

DD

D

D D D

D

32 3

16 32 3

43

2 3

243

43

4

243

312 4

93

When

units

2

2

22

2 2

33

=

= −

= −

= −

=

e eo o>

<

<

H

F

F

Answers [EXTENSION 1 MATHEMATICS HSC MATHS IN FOCUS]

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