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Answers for Extension 1 HSC Course for Maths in Focus
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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
base angles of isosceles
corresponding sides in congruent s
`
`
+ ++ +
/∆ ∆
∆
∆
= ==
=
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
base angles of isosceles
`
+ +
+ ++ ++ +
<
∆
=====
)(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
base angles of isosceles
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,
(opposite sides of gram)
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
==
=
564 Maths In Focus Mathematics Extension 1 HSC Course
24.
`
`
(corresponding sides in congruent s)∆
°( )—
AD BCD C
DE ECADE BCE
AE BE
E
90
by SAS,
(opposite sides of gram)
(given)
is the midpoint given
+ +
/
<
∆ ∆
== ==
=
25. (a)
`
(
AD BCAB DC
DBADB BCD
is commonby SSS,
(opposite sides of gram)
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
+ +++++
+
+
+ ∆
= == +=== − += −
566 Maths In Focus Mathematics Extension 1 HSC Course
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
(corresponding sides in congruent s)
+ ++ ++ +
+
/∆ ∆
∆
=== =
=
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
corresponding sides in congruent s
`
`
`
+ + +
/∆ ∆
∆
= =
=
=
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
(corresponding sides in congruent s)
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
(corresponding sides in congruent s)
`
`
/∆ ∆
∆=
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
(corresponding sides in congruent s)
`
`
`
`
`
+ +
/∆ ∆
∆
=
= =
= = =
568 Maths In Focus Mathematics Extension 1 HSC Course
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
s in alternate segment)
sum of
in semicircle)
`
`
`
++++
+ ++
++ +
+
+
+
+
∆
=== −== += − +== =
45. ( , )
(
ACE DECEDC ACEEDC DECEDC s
DE BA
is isoscele
alternate s
s in alternate segment)
(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
s in alternate segment)
(alternate s,
`
+ ++ +
+
+
<
<
∆
==
48. °
°° °
°.
(
( )
EFC EDF
FEC EDFEFC FEC
AC BC
90
9090 90180
These are supplementary cointerior s
the diagram is impossible
s in alternate segment)
similarly
`
`
`
+ +
+ ++ +
+
+
<
= =
= =+ = +
=
49. (a) (
(
EDF ECDBAE ECD
EDF BAE
s in alternate segment)
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
s in alternate segment)
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)
+ ++ ++ +
+
+
+
∆===
570 Maths In Focus Mathematics Extension 1 HSC Course
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
(corresponding sides in congruent s)
`
`
+ +
/∆ ∆
∆∆
= ==
=
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
s in alternate segment)
+ ++ ++ + +
+ ++
+
+
<=== += +=
(b) °
°
(
a
CBA CAEOAECAE OAE CAOCBA CAO
90
90
s in alternate segment)
(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
572 Maths In Focus Mathematics Extension 1 HSC Course
(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
574 Maths In Focus Mathematics Extension 1 HSC Course
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)
576 Maths In Focus Mathematics Extension 1 HSC Course
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.
578 Maths In Focus Mathematics Extension 1 HSC Course
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
580 Maths In Focus Mathematics Extension 1 HSC Course
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
= − + −= − + +
− += − +
582 Maths In Focus Mathematics Extension 1 HSC Course
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)
584 Maths In Focus Mathematics Extension 1 HSC Course
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
Challenge exercise 2
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
586 Maths In Focus Mathematics Extension 1 HSC Course
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+−
++
588 Maths In Focus Mathematics Extension 1 HSC Course
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+− + +
Challenge exercise 3
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
590 Maths In Focus Mathematics Extension 1 HSC Course
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
592 Maths In Focus Mathematics Extension 1 HSC Course
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
594 Maths In Focus Mathematics Extension 1 HSC Course
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
Challenge exercise 4
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=
596 Maths In Focus Mathematics Extension 1 HSC Course
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
598 Maths In Focus Mathematics Extension 1 HSC Course
(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
600 Maths In Focus Mathematics Extension 1 HSC Course
(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
602 Maths In Focus Mathematics Extension 1 HSC Course
(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.
604 Maths In Focus Mathematics Extension 1 HSC Course
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
Ans_PART_2.indd 605Ans_PART_2.indd 605 6/30/09 12:09:23 PM6/30/09 12:09:23 PM
606 Maths In Focus Mathematics Extension 1 HSC Course
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=
Challenge exercise 5
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
= +
=
= + −= −
608 Maths In Focus Mathematics Extension 1 HSC Course
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
610 Maths In Focus Mathematics Extension 1 HSC Course
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= − +
612 Maths In Focus Mathematics Extension 1 HSC Course
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=
614 Maths In Focus Mathematics Extension 1 HSC Course
(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
616 Maths In Focus Mathematics Extension 1 HSC Course
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
Challenge exercise 6
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
618 Maths In Focus Mathematics Extension 1 HSC Course
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.
Practice assessment task set 2
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!= -
620 Maths In Focus Mathematics Extension 1 HSC Course
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)
622 Maths In Focus Mathematics Extension 1 HSC Course
(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
−− −
624 Maths In Focus Mathematics Extension 1 HSC Course
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
Challenge exercise 7
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 +−
626 Maths In Focus Mathematics Extension 1 HSC Course
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!=
628 Maths In Focus Mathematics Extension 1 HSC Course
(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
8. $78 700
9. (a) Get Rich $949.61, Capital Bank $491.27 (b) $33 427.80 more through Capital Bank
10. $43 778.80 11. $61 292.20
12. NSW Bank $175.49 a month ($5791.25 altogether) Sydney Bank $154.39 a month ($5557.88 altogether) Sydney Bank is better
630 Maths In Focus Mathematics Extension 1 HSC Course
13. (a) $249.60 (b) $13 485.12
14. (a) $13 251.13 (b) $374.07 (c) $20 199.77
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
So the statement is true for .n 1=
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 .
4. Step 1: Prove true for n 1=
´
21
21
1
2 121
221
1
LHS
RHS
LHS RHS
1 1
0
1
=
=
=
= −
=
==
−
d n
So the statement is true for n 1= .
Ans_PART_2.indd 630Ans_PART_2.indd 630 6/18/09 12:39:49 PM6/18/09 12:39:49 PM
631ANSWERS
Step 2: Assume true for n k=
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
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
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 the statement is true for .n 1=
Step 2: Assume true for n k=
So … kk k
2 5 8 3 12
3 2
+ + + + − = +] g
Step 3: Prove true for n k 1= +
… – –( ) ( )
… –( ) ( )
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
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
6. ( ) . . .
2 4 8 2. . .
2 2 2 2 2r
r
nn
n1
1 2 3= + + + +
= + + + +=
/
Step 1: Prove true for n 1=
222 2 12 12
LHS
RHS
LHS RHS
1
1
=== −===
]
]
g
g
So the statement is true for .n 1=
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
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
632 Maths In Focus Mathematics Extension 1 HSC Course
7. ( ) ( ) ( ) ( ) . . . ( )
5 10 15 . . . 5
´ ´ ´ ´r n
n
5 5 1 5 2 5 3 5r
n
1= + + + +
= + + + +=
/
Step 1: Prove true for n 1=
( ) ( )
´
´
5 15
25
1 1 1
25
2
5
LHS
RHS
LHS RHS
==
= +
=
==
So the statement is true for n 1= .
Step 2: Assume true for n k=
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
So the statement is true for n 1= .
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 the statement is true for n 1= .
Step 2: Assume true for n k=
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
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
633ANSWERS
10. ( )
. . .
. . .3 3 3 3 3
9 27 81 3
r
r
nn
n2
2 3 4= + + + +
= + + + +=
/
Step 1: Prove true for n 2=
( )
( )
( )
39
2
9 3 1
2
9 3 1
2
9 2
9
LHS
RHS
LHS RHS
2
2 1
==
=−
=−
=
==
−
So the statement is true for .n 2=
Step 2: Assume true for n k=
So ( )
. . .9 27 81 32
9 3 1k
k 1
+ + + + =−−
Step 3: Prove true for n 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
11. Step 1: Prove true for n 1=
( )
( )
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 the statement is true for .n 1=
Step 2: Assume true for n k=
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
@
@
@
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
12. Step 1: Prove true for n 1=
( )´
11
61
1 1 1 2 1 1
61
2 3
1
LHS
RHS
LHS RHS
2==
= + +
=
==
]
] ]
g
g g
So the statement is true for .n 1=
Step 2: Assume true for n k=
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
634 Maths In Focus Mathematics Extension 1 HSC Course
( )
( )( ) ( )
( )( ) ( )
( )( ) ( )
( ) ( ) ( )
. . . 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
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
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
/
Step 1: Prove true for n 1=
2 2
( )
( )
´
´
2 1 1111 2 1 12 11
LHS
RHS
LHS RHS
3
3
= −=== −= −==
So the statement is true for .n 1=
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
14. Step 1: Prove true for n 1=
7 1 7 16 which is divisible by 6
1 − = −=
So the statement is true for .n 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
Step 3: Prove true for n k 1= +
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
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
15. Step 1: Prove true for n 1=
– 3 1 9 18 which is divisible by 8
´2 1 = −=
So the statement is true for .n 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
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
16. Step 1: Prove true for n 1=
–5 1 5 14 which is divisible by 4
1 − ==
So the statement is true for .n 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
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
17. Step 1: Prove true for n 2= 2 2 2 2 4
8 which is divisible by 4+ =
=] ]g g
So the statement is true for .n 2=
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
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
20. Step 1: Prove true for n 1=
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=== −= −
So the statement is true for .n 1=
636 Maths In Focus Mathematics Extension 1 HSC Course
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
22. Step 1: Prove true for n 2=
≥
4163 2 713
LHS
RHS
LHS RHS
2=== +=
] g
So the statement is true for .n 2=
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
23. Step 1: Prove true for n 3= 5 3
125 31224 2064 2084
LHS
RHS
LHS > RHS
3
3
= −= −== += +=
So the statement is true for .n 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
24. Step 1: Prove true for n 1=
≥
332 13
LHS
RHS
LHS RHS
1
1
=== +=
So the statement is true for .n 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
25. Step 1: Prove true for n 1=
≥
553 25
LHS
RHS
LHS RHS
1
1 1
=== +=
So the statement is true for .n 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
/
Step 1: Prove true for n 1=
( )
( )
5 315
2
15 3 1
2
15 2
15
LHS
RHS
LHS RHS
1
1
==
=−
=
==
] g
So the statement is true for n 1= .
Step 2: Assume true for n k=
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
= + + + + +
=−
+
=−
+
=− +
=− +
=− +
=− +
=−
=−
=
+
+
+
+
+
+
+ +
+
+
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
(b) Step 1: Prove true for n 1=
( )
´
´
3 1 14
2
1 3 1 5
28
4
LHS
RHS
LHS RHS
= +=
=+
=
==
So the statement is true for .n 1=
638 Maths In Focus Mathematics Extension 1 HSC Course
Step 2: Assume true for n k=
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
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
(c) Step 1: Prove true for n 2= ´ ´2 2 2 1 4 1
4 4which is divisible by− =
=] g
So the statement is true for .n 2=
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>
Challenge exercise 8
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
17. Step 1: Prove true for n 1=
( )
ara
r
a r
a1
1
LHS
RHS
LHS RHS
1 1
1
==
=−−
==
−
So the statement is true for .n 1=
Step 2: Assume true for n k=
So ( )
. . .a ar ar arr
a r
1
1– k
k2 1+ + + + =
−−
Step 3: Prove true for n k 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
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
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 the statement is true for .n 1=
Step 2: Assume true for n k=
So …x x xxx
111–k
k2 1+ + + + =
−−
Step 3: Prove true for n k 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 .
Practice assessment task set 3
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
20. Step 1: Prove true for n 0= 3 2
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
640 Maths In Focus Mathematics Extension 1 HSC Course
. . .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)
39. Step 1: Prove true for n 5=
2 1315 5 227
LHS
RHS
5= −== +=
] g
LHS RHS> So true for n 5=
Step 2: Assume true for n k=
. .k
k2 1 5 2
2 5 3i e>
>
k
k
− ++
Step 3: Prove true for n k 1= +
. .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=
Step 2: Assume true for n k=
…1 3 3 32
3 1kk
21
+ + + + = −+
Step 3: Prove true for n k 1= +
…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
642 Maths In Focus Mathematics Extension 1 HSC Course
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
644 Maths In Focus Mathematics Extension 1 HSC Course
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=
Challenge exercise 9
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
646 Maths In Focus Mathematics Extension 1 HSC Course
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
648 Maths In Focus Mathematics Extension 1 HSC Course
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−
650 Maths In Focus Mathematics Extension 1 HSC Course
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/
Challenge exercise 10
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
652 Maths In Focus Mathematics Extension 1 HSC Course
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
654 Maths In Focus Mathematics Extension 1 HSC Course
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
Challenge exercise 11
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
Practice assessment task set 4
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
` =
==
656 Maths In Focus Mathematics Extension 1 HSC Course
(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
658 Maths In Focus Mathematics Extension 1 HSC Course
(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
− =
=
=
=
=
660 Maths In Focus Mathematics Extension 1 HSC Course
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)
662 Maths In Focus Mathematics Extension 1 HSC Course
(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)
664 Maths In Focus Mathematics Extension 1 HSC Course
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< − −
666 Maths In Focus Mathematics Extension 1 HSC Course
(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 @
668 Maths In Focus Mathematics Extension 1 HSC Course
. . .
( ) ( )
[ ( )( ) ]
( ) [ ( ) ( )]
( )( )
( )( )( )
( )( ) [ ( ) ]
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
670 Maths In Focus Mathematics Extension 1 HSC Course
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
672 Maths In Focus Mathematics Extension 1 HSC Course
(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