essential specialist34 solutions ch1

Essential Specialist Mathematics Complete Worked Solutions 1

Chapter 1 A toolbox

Solutions to Exercise 1A 1

a

i. 720 = (720 p180

)c = 4c

ii. 540 = (540 p180

)c = 3c

iii. 450 = (450 p180

)c =

-5p2

c

iv. 15 = (15 p180

)c = p12

c

v. 10 = (10 p180

)c =

-p18

c

vi. 315 = (315 p180

)c =

-7p4

c

b

i. 5p4

c = ( 5p

4 180

p) = 225

ii.

-2p3

c = (

-2p3

180p

) = 120

iii. 7p12

c = ( 7p

12 180

p) = 105

iv.

-11p6

c = (

-11p6

180p

) = 330

v. 13p9

c = (13p

9 180

p) = 260

vi.

-11p12

c = (

-11p12

180p

) = 165

2

a

i. 7 = (7 p180

)c 0.12c

ii. 100 = (100 p180

)c 1.75c

iii. 25 = (25 p180

)c 0.44c

iv. 51 = (51 p180

)c 0.89c

v. 206 = (206 p180

)c 3.60c

vi. 410 = (410 p180

)c 7.16c

b

i. 1.7c = (1.7 180p

) 97.40

ii. 0.87c = (0.87 180p

) 49.85

iii. 2.8c = (2.8 180p

) 160.43

iv. 0.1c = (0.1 180p

) 5.73

v. 3c = (3 180p

) 171.89

vi. 8.9c = (8.9 180p

) 509.93

3

a sin (2p3

) = sin ( 2p3

)

= sin (p3

)

= 3

2

b cos (3p4

) = cos ( 3p4

)

= cos (p4

)

=

- 22

c cos (

-p3

) = cos (p3

)

= 12


d cos (5p4

) = cos ( + p4

)

= cos (p4

)

=

- 22

e cos (9p4

) = cos (2 + p4

)

= cos (p4

)

= 2

2

f sin (11p

3) = sin (4

p3

)

= sin (p3

)

=

- 32

g cos (31p6

) = cos (5 + p6

)

= cos (p6

)

=

- 32

h cos (29p

6) = cos (5

p6

)

= cos (p6

)

=

- 32

i sin (

-23p6

) = sin( 23p6

)

= sin (4 p6

)

= sin (p6

)

= 12

4

a sin (135) = sin (180 45)

= sin (45)

= 2

2

b cos (300) = cos (300)

= cos (360 60)

= cos (60)

= 12

c sin (480) = sin (540 60)

= sin (180 60)

= sin (60)

= 3

2

d cos (240) = cos (180 + 60)

= cos (60)

=

-12

e sin (225) = sin (225)

= sin (180 + 45)

= sin (45)

= 2

2

f sin (420) = sin (360 + 60)

= sin (60)

= 3

2

5

a

sin2 (x) + cos2 (x) = 1

0.25 + cos2 (x) = 1

cos2 (x) = 34

cos (x) = 34

cos (x) =

- 32

as 90 < x < 180

b

tan (x) = sin(x)

cos (x)

2

3

2

1

= 3

2

2

1

= 3

1

= 3

3


6

a

sin2 (x) + cos2 (x) = 1

sin2 (x) + 0.49 = 1

sin2 (x) = 51

100

sin (x) = 51100

sin (x) = 51

10 as 180 < x < 270

b

tan (x) = sin(x)

cos (x)

10

7

10

51

7

10

10

51

7

51

7

a

sin2 (x) + cos2 (x) = 1

0.25 + cos2 (x) = 1

cos2 (x) = 34

cos (x) = 34

cos (x) = 3

2 as < x

3p2

b

tan (x) = sin(x)

cos (x)

2

3

2

1

3

2

2

1

3

3

8

a

sin2 (x) + cos2 (x) = 1

0.09 + cos2 (x) = 1

cos2 (x) = 91

100

cos (x) = 91100

cos (x) = 91

10 as

3p2

< x 2

b

tan (x) = sin(x)

cos (x)

10

91

10

3

91

10

10

3

91

913

9

a

sin x =

- 32

x = 4p3

, 5p3

as x [0, 2]

b

sin (2x) = 3

2, x [0, 2]

2x [0, 4]

2x = p3

, 2p3

, 2 + p3

, 2 + 2p3

as 2x [0, 4]

x = p6

, p3

, 7p6

, 4p3

as x [0, 2]

c

2 cos 2x = 1

cos 2x = 12

, x [0, 2]

2x [0, 4]

2x = 2p3

, 4p3

, 2 + 2p3

, 2 + 4p3

as 2x [0, 4]

x = p3

, 2p3

, 4p3

, 5p3

as x [0, 2]


d

sin (x + p3

) = 12

, x [0, 2]

x + p3

[p3

, 7p3

]

x + p3

= 7p6

, 11p

6

as x + p3 [

p3

, 7p3

]

x = 5p6

, 3p2

as x [0, 2]

e

2 cos (2 (x + p3

)) = 1

cos (2 (x + p3

)) = 12

, x [0, 2]

x + p3 [

p3

, 7p3

]

2 (x + p3

) [2p3

, 14p

3]

2 (x + p3

) = 2p3

, 4p3

, 2 + 2p3

,

2 + 4p3

, 4 + 2p3

as 2 (x + p3

) [2p3

, 14p

3]

x + p3

= p3

, 2p3

, 4p3

, 5p3

, 7p3

x = 0, p3

, , 4p3

, 2 as x [0, 2]

CAS: Type

solve(2cos 2 + = -1, ) | 0 23x x x

For part e we have,

f

2 sin (2x + p3

) = 3

sin (2x + p3

) =

- 32

, x [0, 2]

2x [0, 4]

2x + p3

[p3

, 13p

3]

2x + p3

= 4p3

, 5p3

, 2 +4p3

, 2 + 5p3

as 2x + p3 [

p3

, 13p

3]

2x = , 4p3

, 3, 10p

3

x = p2

, 2p3

, 3p2

, 5p3

as x [0, 2]

10

a

tan 5

4

= tan +

4

= tan

4

= 1

b

tan

2

3

= tan +

3

= tan

3

= 3

c

tan

29

6

= tan

5 +

6

= tan

+

6

= tan

6

= 3

3

d

tan (240 ) = tan (180 + 60 )

= tan (60 )

= 3


11

h2 = 1 + 16

h = 17

a

sin x = 1

17 =

17

17

b

cos x = 4

17 =

4 17

17

c

Since x 3

2

3

2 x

2 x

tan ( x) = 1

4 as

2 x

d

Since x 3

2

3

2 x 2

tan ( x) = 1

4 as

3

2 x 2

12

h2

= 3 + 4

h = 7

a

sin x = 3

7 =

21

7

b

cos x = 2

7 =

2 7

7

c

Since

2 x

x

2 x

3

2

tan ( x) = 3

2 as x

3

2

d

Since

2 x

2 x 0

tan (x ) = 3

2 as

2 x 0

-1

-4

x

hx

3

-2

h


13

a

tan x = 3

x =

3 , 2

3

x = 2

3 ,

5

3 as x [0 , 2 ]

b

tan

3x

6

= 3

3

as x [0, 2] 3x [0, 6]

3x

6

6 ,

35

6

3x

6 =

6 ,

7

6 ,

6+ 2 ,

7

6+ 2 ,

6+ 4 ,

7

6+ 4

3x

6 =

6 ,

7

6 ,

13

6 ,

19

6 ,

25

6 ,

31

6

3x =

3 ,

4

3 ,

7

3 ,

10

3 ,

13

3 ,

16

3

x =

9 ,

4

9 ,

7

9 ,

10

9 ,

13

9 ,

16

9

CAS: Type

3solve(tan 3 = , ) | 0 26 3x x x

Use the right arrow key to view all solutions.

c

2tan x

2

+ 2 = 0

tan x

2

= 1

and x

2 [0, ]

x

2 =

3

4 as

x

2 [0 , ]

x = 3

2

d

3tan

2+ 2x

= 3

tan

2+ 2x

= 1

as x [0, 2]

2+ 2x

2 ,

9

2

2+ 2x =

3

4 ,

7

4 ,

3

4+ 2 ,

7

4+ 2

2+ 2x =

3

4 ,

7

4 ,

11

4 ,

15

4

2x =

4 ,

5

4 ,

9

4 ,

13

4

x =

8 ,

5

8 ,

9

8 ,

13

8

14

a

f(x) = sin 2x , x [0, 2]

The transformation from the graph of

g(x ) = sin x is a dilation from the y axis of

factor 12

.


b

f (x) = cos (x + p3

), x [p3

, ]

The transformation from the graph of

g(x ) = cos x is a translation of p3

to the

left.

f (

-p3

) = cos 0 = 1

f (0) = cos p3

= 12

f () = cos 4p3

= cos p3

= 12

c

f (x) = cos (2(x + p3

)), x [0, ]

The transformations from the graph of

g(x ) = cos x are a dilation from the y axis

of factor 12

and a translation of p3

to the left.

f(0) = cos 2

3

= 1

2

f() = cos 8

3

= 1

2

d

f (x) = 2 sin (3x) + 1, x [0, ]


g(x) = sin x are a dilation from the y axis of

factor 13

, a dilation from the x axis of factor 2

and a translation of 1 in the positive direction

of the y axis.

To find x axis intercepts for f (x), solve

f (x) = 0

i.e. 2 sin (3x) + 1 = 0, x [0, ]

sin (3x) = 12

, 3x [0, 3]

3x = 7p6

, 11p6

x = 7p18

, 11p18

f (0) = 1, f () = 2 sin (3) + 1

= 1


e

f (x) = 2 sin (x p4

) + 3 , x [0, 2]


g(x) = sin x are a dilation from the x axis of

factor 2, a translation of p4

to the right and a

translation of 3 in the positive direction of

the y axis.

f(0) = 2sin

4

+ 3

= 2 sin

4

+ 3

= 3 2

f(2) = 2sin 7

4

+ 3

= 3 2 To find x axis intercepts for f (x), solve

f (x) = 0

i.e. 2 sin (x p4

) + 3 = 0, x [0, 2]

sin (x p4

) =

- 32

,

x p4

[p4

, 7p4

]

x p4

= 4p3

, 5p3

x = 19p12

, 23p12

15

a

f(x) = tan (2x)

Period: 2

n

Asymptotes:

x =(2k + 1)

2n

x = (2k + 1)

4

x =

4 ,

3

4 as x [0 , ]

x-intercepts:

as x [0, ]

2x [0, 2] tan (2x) = 0 2x = 0, , 2

x = 0 ,

2,

y-intercept:

f(0) = tan (0) = 0

b

f(x) = tan

x

3

Period:

n

Asymptotes:

x =(2k + 1)

2n +

3

x = (2k + 1)

2 +

3

x =

2+

3 as x [0 , ]


x = 5

6 x-intercepts:

as x [0, ]

x

3

3 ,

2

3

tan

x

3

= 0

x

3 = 0

x =

3 y-intercept:

f(0) = tan

3

= tan

3

= 3

Endpoint:

f() = tan 2

3

= 3

c

f(x) = 2tan

2x +

3

= 2tan

2

x +

6

Period: 2

n

Asymptotes:

x =(2k + 1)

2n

6

x = (2k + 1)

4

6

x =

4

6 ,

3

4

6 as x [0 , ]

x =

12 ,

7

12

x-intercepts:

as x [0, ]

2x +

3

3 ,

7

3

tan

2x +

3

= 0

2x +

3 = , 2

x =

3 ,

5

6 y-intercept:

f(0) = 2tan

3

= 2 3

Endpoint:

f() = 2tan 7

3

= 2 3

d

f(x) = 2 tan

2x +

3

2

= 2 tan

2

x +

6

2

Period: 2

n

Asymptotes:

x =(2k + 1)

2n

6

x = (2k + 1)

4

6

x =

4

6 ,

3

4

6 as x [0 , ]

x =

12 ,

7

12


x-intercepts:

as x [0, ]

2x +

3

3 ,

7

3

tan

2x +

3

= 1

2x +

3 =

5

4 ,

9

4

x = 11

24 ,

23

24 y-intercept:

f(0) = 2tan

3

2 = 2 3 2

Endpoint:

f() = 2tan 7

3

2 = 2 3 2


Solutions to Exercise 1B 1

a

AC2 = 82 + 52 = 89

AC = 89

tan x = 85

,

cos x = 589

= 5 89

89,

sin x = 889

= 8 89

89

b

AC2 + 52 = 72

AC2 = 24

AC = 2 6

tan x = 52 6

= 5 612

,

cos x = 2 6

7,

sin x = 57

c

AC2 + 72 = 92

AC2 = 32

AC = 4 2

tan x = 4 27

,

cos x = 79

, sin x = 4 29

2

a

sin 30 = a12

a = 12 sin 30

= 12 12

= 6

b

sin 45 = 6a

a = 6sin 45

= 61

2

= 6 2

c

sin 30 = 5x

x = 5sin 30

= 512

= 10

cos 30 = xa

= 10a

a = 10cos 30

= 103

2

= 203

= 20 3

3

5

8

x

A B

C

5

7x

A B

C

9

7

xA B

C

x5

a

30

30


3

a

a2 = 12 + 52

= 1 + 25

= 26

a = 26

b

a2 = 12 + 22

= 5

a = 5

b2 = 12 + a2

= 1 + 5

= 6

b = 6

c2 = b2 + 1

= 6 + 1

= 7

c = 7

c

Since the triangle has two sides of length 3,

it is isosceles and hence a = 1.

h2

+ 12

= 32

h

2= 8

h = 8 = 2 2 d

sin 30 = 1a

12

= 1a

a = 2

b2 + 12 = a2

b2 = 22 1 = 3

b = 3

tan 45 = cb

= c

3

1 = c3

c = 3

d2 = b2 + c2

= ( 3 )2 + ( 3 )2

= 6

d = 6

4

a

sin 45 = 1a

1

2 = 1

a

a = 2

45 + z = 60

z = 15

sin 30 = 12

1x + w

= 12

x + w = 2

cos 30 = 3

2

1 + m2

= 3

2

1 + m = 3 , so 3 1m

cos 30 = 3

2

w

3 - 1 =

32

w =

3( 3 - 1)

2=

3 - 32

Now x + w = 2

x = 2 w

= 2 (

3 - 32

)

=

4 - (3 - 3)

2

x = 1 + 3

2

1

x + w30

45

1

m

45

3

za

3 1

w 30

y


sin 30 = 12

y

3 - 1 =

12

y =

3 - 12

b

sin (15) =

3 - 12

2

=

3 - 12

1

2

=

3 - 1

2 2 2

2

=

6 - 24

cos (15) = 1 + 3

2 2

= 1 + 3

2

22

= 2 + 6

4

tan (15) =

3 - 12

1 + 3

2

=

3 - 12

2

3 + 1

=

3 - 1

3 + 1

3 - 1

3 - 1

=

( 3 - 1)2

2

=

4 - 2 32

= 2 3

CAS:

Change to Degree/Deg mode

c

sin (75) = 1 + 3

2 2

= 1 + 3

2

22

= 2 + 6

4

cos (75) =

3 - 12

2

=

3 - 1

2 2 2

2

=

6 - 24

tan (75) = 1 + 3

2

3 - 12

= 13

2

2

31

= 13

13

13

31

=

2

312

= 2

324

= 2 + 3

1 + 32 75

152

23 1


Solutions to Exercise 1C 1

A + B + C = 180

B = (180 (73 + 55)) = 52

a

Applying the sine rule: 10

sin 55 = a

sin 73

BC = a = 10 sin 73

sin 55 11.67

BC is 11.67 cm, correct to two decimal places.

b

10sin 55

= bsin 52

AC = b = 10 sin 52

sin 55 9.62

AC is 9.62 cm, correct to two decimal places.

2

a

Applying the cosine rule:

AC2 = AB2 + BC2 2(AB)(BC) cos B

= 6.52 + 82 2(6.5)(8) cos 58

= 51.13839

AC = 7.15111


b

Applying the sine rule: 65

sin C = AC

sin 58

C = sin1 (65 sin 58

AC)

= (50.42874)

or C = 180 sin1 (65 sin 58

AC)

= (129.57125)

Therefore BCA = 50.43, correct to two

decimal places. (A triangle with two angles

of 58 and 129.57 cannot be formed.)

3

Since AD | | BC,

ABC = 180 BAC

= (180 67)

= 113


AC2 = AB2 + BC2 2(AB)(BC) cos ABC

= 92 + 112 2(9)(11) cos 113

= 279.36476

AC = 16.71420

The length of the longer diagonal is 16.71

cm, correct to two decimal places.

4

The two possible triangles are:

5573A

B

C

c = 10 cm a

b

5865 cm

A

B

C

8 cm

9 cm

11 cm

67

A

B C

D


a

Applying the sine rule: sin 34

56 =

sin B

85

B = sin1 (85 sin34

56)

= (58.07867)

or B = 180 sin1 (85 sin34

56)

= (121.92132)

ABC is either 58.08 or 121.92, correct to

two decimal places.

b

If ABC = 58.08, then

BAC = (180 (58.08 + 34)) = 87.92


BC2 = AB2 + AC2 2(AB)(AC) cos BAC

= 5.62 + 8.52 2(5.6)(8.5) cos 87.92

= 100.15472

BC = 10.00773


If ABC = 121.92, then

BAC = (180 (121.92 + 34)) = 24.08


BC2 = AB2 + AC2 2(AB)(AC) cos BAC

= 5.62 + 8.52 2(5.6)(8.5) cos 24.08

= 16.69462

BC = 4.08590


5

a


AC2 = AB2 + BC2 2(AB)(BC) cos A

= 102 + 4.72 2(10)(4.7) cos 35

= 45.08970

AC = 6.71488


b

Applying the sine rule: sin C

10 =

sin 35

AC

ACB = C = sin1 (10 sin35

AC)

= (58.66995)

or C = 180 sin1 (10 sin35

AC)

= (121.33004)

If C = 58.67 then

A = (180 (58.67 + 35)) = 86.33

But | AB | > | BC |

C > A

C = 121.33

ACB is 121.33, correct to two decimal

places.

6

AB

sin 60 = AC

sin 45

AB = 12

32

1

2

= 6 6

AB is 6 6 cm.

7

QR2 = PQ2 + PR2 2(PQ)(PR) cos P

= 22 + 32 2(2)(3) cos 60

= 4 + 9 12 12

= 7

QR = 7

QR is 7 cm.

35A

B

C

10 cm

47 cm

45

A

B

C12 cm

60

2 cm

P

Q

R3 cm

60


8

Applying the sine rule:

20

40sin

18

sin

C

...)34573.35(

10

40sin9sin

1C

Hence, 65.10435.3540180A

Applying the sine rule:

40sin

20

65.104sin

BC

40sin

65.104sin20BC

...102322.30BC

BC is 30.10 cm

9

The ambiguous case applies in this instance as

the smaller known side is opposite the known

angle.

Applying the cosine rule: 2 2 2

2( )( ) cos 30AB AC BC AC BC

2 364 100 2(10)( )

2BC BC

210 3 36 0BC BC

10 3 300 144

2BC

=

10 3 156

2

=

10 3 2 39

2

= 5 3 39

BC is 5 3 39 cm

10

a

cos B = 10

2 5

2 12

2

2 (5)(12 )

=

69

120

=

23

40

B = cos1

23

40

= 54.90

ABC = 54.90

b

cos A = 12

2 10

2 5

2

2 (10 )(5)

=

19

100

A = cos1

19

100

= 100.95

BAC = 100.95

A

B

C

18 cm

20 cm

40

10 cm

8 cm

30

C

A

B

8 cmB'

A

B

C

12 cm

10 cm

5 cm


Solutions to Exercise 1D 1

z + 68 = 150 (vertically opposite)

z = 82

a = 82 (alternate)

y = 180 150 = 30 (supplementary)

x = 30 (vertically opposite)

2

a

RTW = (180 105) = 75

(opposite angle of a cyclic quadrilateral)

b

TSW = 62

(angle subtended by the arc TW)

c

RTS = 37 (angle subtended by the arc RS)

STW = (75 37) = 38

SRW = 38 (angle subtended by the arc SW)

TRS = (62 + 38) = 100

d

RST = (105 62) = 43

RWT = 43 (angle subtended by the arc RT)

3

c = 50 (angle between tangent and chord)

a = 40 (angle between tangent and chord)

b = 180 (50 + 40) (angles in a triangle)

= 90

4

a

ABX = BXA = XAB = 60

(angles of an equilateral triangle, ABX)

DAX = XBC = (90 60) = 30

ADX = AXD = BXC = BCX

= (

180 - 302

)

= 75

(angles of a triangle, isosceles triangles ADX,

BCX)

d = 180 (80 + 60) = 40

(angles in a triangle)

DXC = (360 (75 + 60 + 75)) = 150 (angles at a point)


b

XDC = (90 75) = 15

5

a = 69 (alternate)

b = 47 (alternate)

c = 180 105 = 75 (supplementary)

d = 180 (105 + 47) = 28

(angles of a triangle, WOZ)

e = 180 (69 + 75) = 36

(vertically opposite angles of WOX)

6

a + (180 b + c) + (180 x) = 180

a + 180 b + c + 180 x = 180

a b + c + 180 = x

7

x = 80 (angle subtended by arc at centre)

y = 180 40 = 140

(opposite angle of a cyclic quadrilateral)

8

a = 180 (70 + 50) = 60

(angles in a triangle)

b = 180 (50 + 50) = 80

(angle between tangent and chord, angles in a

triangle)

c = 180 (60 + 60) = 60

(angle between tangent and chord,

angles in a triangle)

180

180 80 60

40

d b c

x

b

a

a

c (180 b)

(180 c (180 b))

= (b c)

(180 (b c))

= (180 b + c)

(180 x)


9

Triangle XAB is isosceles

(tangents from a common point)

XAB = XBA = 70

x = 70 (alternate segment theorem)

y = 110

(opposite angles in a cyclic quadrilateral)

10

20 + y = x + 50 and y = 2x

(angle subtended by arc at centre)

20 + 2x = x + 50

x = 30

y = 60

A

B

XO y

x40


Solutions to Exercise 1E 1

6,1311

ttt

nn

t2 = 3t

1 1 t

3 = 3t

2 1

= 3 6 1 = 3 17 1

= 17 = 50

TI: Open a Lists & Spreadsheet application.

Press Menu3:Data1:Generate Sequence

and input as shown below

You will now have the sequence of numbers

listed in column A like shown.

Scroll down to cell A8 to find the value of .8

t

120298 t

CP: Open the Sequence application and

input the following: an + 1 = 3an 1

a 0 = 6 Tap 8 and change the Table End value to 10.

Now tap # to generate the sequence. Read

the value of t8 from the table (this occurs

when n is 7)

2

5,6211

yyy

nn

y2 = 2y1 + 6 y3 = 2y2 + 6

= 2 5 + 6 = 2 16 + 6

= 16 = 38

TI: In a new Lists & Spreadsheet, enter the

values from 1 to 10 into column A. Give

column A the name n. Give column B the

name term and generate the following

sequence into column B (as per question 1)

Formula: 6)1(2 nu

Initial Terms: 5

n0: 1

nMax: 10

nstep: 1

Ceiling Value (upper limit): 6000

Scroll down to cell B10 to find the value of

10y .

562610

y


Open a Data & Statistics page. Add the

variable n along the horizontal axis and add

the variable term along the vertical axis.

or sketching by hand we have:

3

t = 1 t6 = t5 + t4 = 8

t2 = 1 t7 = t6 + t5 = 13

t3 = t2 + t1 = 2 t8 = t7 + t6 = 21

t4 = t3 + t2 = 3 t9 = t8 + t7 = 34

t5 = t4 + t3 = 5 t10 = t9 + t8 = 55

The first ten terms are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55

4

a = 3, d = 4, n = 10

Sn = n2

[2a + (n 1)d]

S10 = 102

[2 3 + (10 1) 4]

= 5[6 + 9 4]

= 5 42

= 210

5

S = ra

1 , a = 1, r =

-13

=

11 - (-1

3)

=

3

4

1

= 34

6

a x

x + 5 =

x - 4x

x2 = (x + 5)(x 4)

= x2 + x 20

x = 20

b

r = xx + 5

= 2025

= 45

c

S Sn =

a1 - r

a(1 - rn )

1 - r

=

arn

1 - r,

a = x + 5 = 25, r = 45

, n = 10

S S10 =

25 (45)10

1 - 45

= 5 25 45

)10

= 7

10

5

4

7

Sn =

a(rn - 1)

r - 1,

a = 6, r = 3, n = 8

S8 =

6((-3)8 - 1)

-3 - 1

=

-32

((3)8 1)

= 9840

6000

5000

4000

3000

2000

1000

0 1 2 3 4 5 6 7 8 9 10

yn

n

(1, 5) (2, 16) (3, 38) (4, 82) (5, 170) (6, 346)(7, 698)

(8, 1402)

(9, 2810)

(10, 5626)


8

S =

a1 - r

, a = a, r = a

2 a =

1

2

=

a

1 - 12

=

a

2 - 1

2

=

2 a

2 - 1

2 + 1

2 + 1

= a(2 + 2)

1

= a (2 + 2 )

9

a

Sn =

a(rn - 1)

r - 1, n = 10, a = 1, r = x

2

S10 =

1(( x2

)10 - 1)

x2

- 1

=

2x - 2

(( x2

)10 1)

When x = 1.5, S10 =

10

4

314

b

i.

S =

a1 - r

, a = 1, r = x2

=

1

1 - x2

, x 2

=

22 - x

Now 1 < r < 1

1 < x2

< 1

2 < x < 2

The infinite sum exists for 2 < x < 2

ii.

Let S =

22 - x

, x 2

Given S = 2S10,

22 - x

=

4x - 2

(( x2

)10 1)

( x2

)10 1 = 12

( x2

)10 = 12

x2

= ( 12

)1

10

x = 2 2

-110

= 109

2

10

a

S =

a1 - r

, a = 1, r = sin

=

11 - sinq

b

11 - sinq

= 2

2(1 sin ) = 1

1 sin = 12

sin = 12

= p6

, 5p6

, p6

2, 5p6

2, p6

4,

= p6

+ 2k, 5p6

+ 2k, k Z


Solutions to Exercise 1F 1

a

(x 2)2 + (y 3)2 = 1

b

(x + 3)2 + (y 4)2 = 25

c

x2 + (y + 5)2 = 25

d

(x 3)2 + y2 = 2

2

a

x2 + y2 + 4x 6y + 12 = 0

Completing the square in x and y gives:

(x2 + 4x + 4) + (y2 6y + 9) + 12 = 13

(x + 2)2 + (y 3)2 = 1

A circle with centre (2, 3) and radius 1 is

described.

b

x2 + y2 2x 4y + 1 = 0

Completing the square in x and y gives:

(x2 2x + 1) + (y2 4y + 4) + 1 = 5

(x 1)2 + (y 2)2 = 4


described.

c

x2 + y2 3x = 0

(x2 3x + 9

4) + y2 =

9

4

(x 3

2)2 + y2 =

9

4

A circle with centre (3

2, 0) and radius

3

2 is

described.

d

x2 + y2 + 4x 10y + 25 = 0

(x2 + 4x + 4) + (y2 10y + 25) + 25 =

29

(x + 2)2 + (y 5)2 = 4


described.

3

a

2x2 + 2y2 + x + y = 0

2[x2 + y2 + 12

x + 12

y] = 0

(x2 + 12

x + 116

) + (y2 + 12

y + 116

) = 18

(x + 14

)2 + (y + 14

)2 = 18

centre ( 14

, 14

), radius 1

2 2 =

2

4

When x = 0, 116

+ (y + 14

)2 = 18

(y + 14

)2 = 116

y + 14

= 14

y = 0, 12

Similarly when y = 0, x = 0, 12

b

x2 + y2 + 3x 4y = 6

(x2 + 3x + 9

4) + (y2 4y + 4) =

49

4

(x + 3

2)2 + (y 2)2 =

49

4

centre (3

2, 2), radius

7

2

When x = 0, 9

4 + (y 2)2 =

49

4

(y 2)2 = 10

y 2 = 10

y = 2 10

When y = 0, (x + 3

2)2 + 4 =

49

4

(x + 3

2)2 =

33

4

x + 3

2 =

33

2

x = 12

(3 33 )


c

x2 + y2 + 8x 10y + 16 = 0

(x2 + 8x + 16) + (y2 10y + 25) + 16 =

41

(x + 4)2 + (y 5)2 = 25

centre (4, 5), radius 5

When x = 0, 16 + (y 5)2 = 25

(y 5)2 = 9

y 5 = 3

y = 2, 8

When y = 0, (x + 4)2 + 25 = 25

(x + 4)2 = 0

x + 4 = 0

x = 4

d

x2 + y2 8x 10y + 16 = 0

(x2 8x + 16) + (y2 10y + 25) + 16 = 41

(x 4)2 + (y 5)2 = 25


When x = 0, 16 + (y 5)2 = 25

(y 5)2 = 9

y 5 = 3

y = 2, 8

When y = 0, (x 4)2 + 25 = 25

(x 4)2 = 0

x 4 = 0

x = 4

e

2x2 + 2y2 8x + 5y + 10 = 0

2(x2 + y2 4x + 5

2y + 5) = 0

(x2 4x + 4) + (y2 + 5

2y +

25

16) + 5 =

89

16

(x 2)2 + (y + 5

4)2 =

9

16

centre (2,

5

4), radius

3

4

f

3x2 + 3y2 + 6x 9y = 100

3(x2 + 2x + 1) + 3(y2 3y + 9

4) =

439

4

3(x + 1)2 + 3(y 3

2)2 =

439

4

(x + 1)2 + (y 3

2)2 =

439

12

centre (1, 3

2), radius 1

2439

3 =

1317

6

When x = 0, 1 + (y 3

2)2 =

439

12

(y 3

2)2 =

427

12

y = 3

2

1281

6

When y = 0, (x + 1)2 + 9

4 =

439

12

(x + 1)2 = 41212

x = 1 309

3


4

a

x2 + y2 16

b

x2 + y2 9

c

(x 2)2 + (y 2)2 < 4

d

(x 3)2 + (y + 2)2 > 16

For (x 3)2 + (y + 2)2 = 16

When x = 0, 9 + (y + 2)2 = 16

(y + 2)2 = 7

y + 2 = 7

y = 2 7

When y = 0, (x 3)2 + 4 = 16

(x 3)2 = 12

x 3 = 2 3

x = 3 2 3

e

x2 + y2 16 and x 2

f

x2 + y2 9 and y 1

5

Length of diameter = (8 2)2

+ (4 2)2

= 36 + 4

= 40

= 2 10

r = 10

The centre of the circle lies at the midpoint of

the diameter and has coordinates

(8 2

2, 4 2

2) i.e. (5, 3)



6


(x 2)2 + (y + 3)2 = 9

7 (x h)2 + (y k)2 = r2 At (3, 1), (3 h)2 + (1 k)2 = r2 9 6h + h2 + 1 2k + k2 = r2

10 6h + h2 2k + k2 = r2 1

At (8, 2), (8 h)2 + (2 k)2 = r2 64 16h + h2 + 4 4k + k2 = r2

68 16h + h2 4k + k2 = r2 2

At (2, 6), (2 h)2 + (6 k)2 = r2 4 4h + h2 + 36 12k + k2 = r2

40 4h + h2 12k + k2 = r2 3

1 2 58 + 10h + 2k = 0

k = 29 5h 4

3 1 30 + 2h 10k = 0

15 + h 5k = 0 5

Substituting 4 in 5 yields

15 + h 5(29 5h) = 0

15 + h 145 + 25h = 0

26h = 130

h = 5

Substituting h = 5 in 4 yields

k = 29 5 5

= 29 25

= 4

Substituting h = 5, k = 4 in 1 yields

10 6 5 + 52 2 4 + 42 = r2

r2 = 10 30 + 25 8 + 16

= 13

(x 5)2 + (y 4)2 = 13 is the circle

passing through (3, 1), (8, 2) and (2, 6)

8

053676604422

yxyx

x2 + y2 15x 19y + 134 = 0

(x 15

2)2 + (y

19

2)2 =

25

2

centre (15

2,

19

2), radius

5 2

2

x2 + y2 10x 14y + 49 = 0 1

(x2 10x + 25) + (y2 14y + 49) + 49 = 74

(x 5)2 + (y 7)2 = 25 centre (5, 7), radius 5

To find points of intersection, let

x2 + y2 15x 19y + 134 = x2 + y2 10x 14y + 49

5x + 5y = 85

x + y = 17

y = 17 x 2

Substituting 2 in 1 yields

x2 + (17 x)2 10x 14(17 x) + 49 = 0 x2 + 289 34x + x2 10x 238 + 14x + 49 = 0

2x2 30x + 100 = 0 x2 15x + 50 = 0 (x 5)(x 10) = 0 x = 5 or x = 10 When x = 5, y = 17 5 = 12 When x = 10, y = 17 10 = 7

The points of intersection of the two circles

are (5, 12) and (10, 7)

TI: Type

solve(4x2+4y

260x76y+536=0 and x

2+y

2

10x 14y+49=0,x)

CP: Type

solve({4x2+4y

260x76y+536=0, x

2+y

210x

14y+49=0},{x,y})

9

a

Substituting y = x into x2 + y2 = 25

yields

x2 + x2 = 25

2x2 = 25

x2 = 25

2


x = 52

= 5 22

Hence 5 2

2y x

The points of intersection are ( 5 22

, 5 22

)

and (

5 2

2,

5 2

2)

b

Substituting y = 2x into x2 + y2 = 25

yields

x2 + 4x2 = 25

5x2 = 25

x2 = 5

Hence 2

2 5

y x

The points of intersection are ( 5 , 2 5 )

and ( 5 , 2 5 )

TI: Type

solve(x2+y

2=25 and y=x,x)

CP: Type

solve({x2+y

2=25, y=x},{x,y})


Solutions to Exercise 1G 1

a

x2

9 +

y2

16 = 1

ellipse, centre (0, 0)

b

25x2 + 16y2 = 400

x2

16 +

y2

25 = 1


c

(x - 4)2

9 +

(y - 1)2

16 = 1


When x = 0, 169

+

(y - 1)2

16 = 1

(y - 1)2

16 =

79

no y axis intercepts

When y = 0,

(x - 4)2

9 + 1

16 = 1

(x - 4)2

9 =

1516

(x 4)2 = 9 15

16

x = 4 3 15

4

d

x2 +

(y - 2)2

9 = 1


When x = 0,

(y - 2)2

9 = 1

(y 2)2 = 9

y = 2 3

= 1, 5

When y = 0, x2 + 49

= 1

x2 = 59

x = 5

3

e 9x2 + 25y2 54x 100y = 44

9(x2 6x + 9) + 25(y2 4y + 4) = 225

9(x 3)2 + 25(y 2)2 = 225

(x - 3)2

25 +

(y - 2)2

9 = 1


When x = 0, 9

25 +

(y - 2)2

9 = 1

(y - 2)2

9 =

1625

(y 2)2 = 9 1625

y = 2 125

=

-25

, 225

(x + 2)2 = 365

x

y

0

4

3

4

3

x

y

0

5

4

5

4

3 15

44

x

y

0

(4, 1)

3 154

4 +

x

y

0

5

(0, 2)

15

3

53


When y = 0,

(x - 3)2

25 + 4

9 = 1

(x - 3)2

25 =

59

(x 3)2 = 25 59

x = 3 5 5

3

f

9x2 + 25y2 = 225

x2

25 +

y2

9 = 1


g 5x2 + 9y2 + 20x 18y 16 = 0

5(x2 + 4x +4) + 9(y2 2y +1) 16 29 = 0

5(x + 2)2 + 9(y 1)2 = 45

(x + 2)2

9 +

(y - 1)2

5 = 1


When x = 0, 49

+

(y - 1)2

5 = 1

(y - 1)2

5 =

59

(y 1)2 = 259

y = 1 53

=

-23

, 83

When y = 0, (x + 2)2

9 + 1

5 = 1

(x + 2)2

9 = 4

5

x = 2 6 5

5

h

16x2 + 25y2 32x + 100y 284 = 0

16(x22x+1) + 25(y2+4y+4) 284 116 = 0

16(x 1)2 + 25(y + 2)2 = 400

(x - 1)2

25 +

(y + 2)2

16 = 1


When x = 0, 125

+ (y + 2)2

16 = 1

(y + 2)2

16 = 24

25

(y + 2)2 = 16 2425

= 38425

y = 2 8 6

5

When y = 0,

(x - 1)2

25 + 1

4 = 1

(x - 1)2

25 =

34

(x 1)2 = 754

x = 1 5 3

2

5 53 3

x

y

0

(3, 2)

25

2 25

5 53 +3

x

y

0

3

5

3

5

x

y

0

(2, 1)

23

83

6 52 +

56 52

5

5 3 1 +

25 3

1 2

8 62 +

5

8 62 5

x

y

0

(1, 2)


i

(x - 2)2

4 +

(y - 3)2

9 = 1


j

2(x 2)2 + 4(y 1)2 = 16

(x - 2)2

8 +

(y - 1)2

4 = 1


When x = 0,

14

1

2

12

y

2

1

4

12

y

212 y

21 y

When y = 0,

14

1

8

22

x

4

3

8

22

x

64

242

2 x

62 x

2

a

x2

16

y2

9 = 1

y2

9 = x

2

16 1

y2 = 9x2

16 9

= 9x2

16(1

16

x2)

As x , 16

x2 0

y2 9x2

16

y 3x4

Equations of asymptotes: y = 3x4

When y = 0, x2 = 16 x = 4 centre (0, 0)

b

y2

16

x2

9 = 1

This is the reflection of x2

16

y2

9 = 1 in the

line

y = x

Asymptotes are x = 34

y

y = 43

x

The y axis intercepts are (0, 4) and (0, 4)

x

y

0

(2, 3)

4

6

2

3

2 + 6

1 + 2

x

y

0

(2, 1)

1 2

2 6


c

x2 y2 = 4

x2

4

y2

4 = 1

y2

4 = x

2

4 1

y2 = 4x2

4 4

= x2 (1 4

x2)

As x , 4

x2 0

y2 x2

y x

Equations of asymptotes: y = x

When y = 0, x2 = 4

x = 2

centre (0, 0)

d

2x2 y2 = 4

x2

2

y2

4 = 1

y2

4 =

x2

2 1

y2 = 2x2 4

= 2x2 (1 2

x2)

As x , 2

x2 0

y2 2x2

y 2 x

Equations of asymptotes: y =

2 x

When y = 0, 2x2 = 4

x2 = 2

x = 2

centre (0, 0)

e

x2 4y2 4x 8y 16 = 0

(x2 4x + 4) 4(y2 + 2y + 1) 16 = 0

(x 2)2 4(y + 1)2 = 16

(x - 2)2

16

(y + 1)2

4 = 1

(y + 1)2

4 =

(x - 2)2

16 1

(y + 1)2 =

(x - 2)2

4 4

=

(x - 2)2

4

2

2

161

x

As x ,

16

(x - 2)2

0

(y + 1)2

(x - 2)2

4

y + 1

x - 22

y 1

x - 22

Equations of asymptotes: y = 1

x - 22

i.e. y =

x - 42

and y =

- x2

=

12

x 2 = 12

x

When y = 1,

(x - 2)2

16 = 1

(x 2)2 = 16

x 2 = 4

x = 2, 6

centre (2, 1)

When y = 0,

(x - 2)2

16 1

4 = 1

(x - 2)2

16 =

54

(x 2)2 = 20

x = 2 2 5


f

9x2 25y2 90x + 150y = 225

9(x2 10x + 25) 25(y2 6y + 9) = 225

9(x 5)2 25(y 3)2 = 225

(x - 5)2

25

(y - 3)2

9 = 1

(y - 3)2

9 =

(x - 5)2

25 1

(y 3)2 =

9(x - 5)2

25 9

=

9(x - 5)2

25

2

5

251

x

As x ,

25

(x - 5)2

0

(y 3)2

9(x - 5)2

25

y 3

3(x - 5)

5

y 3

3(x - 5)

5

Equations of asymptotes:

y = 3 +

3(x - 5)

5 and y = 3

3(x - 5)

5

=

15 + 3x - 155

=

15 - 3x + 155

= 35

x =

30 - 3x5

= 6 35

x

When y = 3,

(x - 5)2

25 = 1

(x 5)2 = 25

x 5 = 5

x = 0, 10

centre (5, 3)

When y = 0,

(x - 5)2

25 1 = 1

(x - 5)2

25 = 2

(x 5)2 = 50

x = 5 5 2

g

(x - 2)2

4

(y - 3)2

9 = 1

(y - 3)2

9 =

(x - 2)2

4 1

(y 3)2 =

9(x - 2)2

4 9

=

9(x - 2)2

4

2

2

41

x

As x ,

4

(x - 2)2

0

(y 3)2

9(x - 2)2

4

y 3

3(x - 2)

2

y 3

3(x - 2)

2


y = 3 +

3(x - 2)

2 and y = 3

3(x - 2)

2

=

6 + 3x - 62

=

6 - 3x + 62

= 32

x =

12 - 3x2

= 6 32

x

When y = 3,

(x - 2)2

4 = 1

(x 2)2 = 4

x = 2 2 = 0, 4

centre (2, 3)

When y = 0,

(x - 2)2

4 1 = 1

(x - 2)2

4 = 2

(x 2)2 = 8

x = 2 2 2


h

4x2 8x y2 + 2y = 0

4(x2 2x + 1) (y2 2y + 1) = 3

4(x 1)2 (y 1)2 = 3

4(x - 1)2

3

(y - 1)2

3 = 1

(y - 1)2

3 =

4(x - 1)2

3 1

(y 1)2 = 4(x 1)2 3

= 4(x 1)2

2

14

31

x

As x ,

3

4(x - 1)2

0

(y 1)2 4(x 1)2

y 1 2(x 1)

y 1 2(x 1)


y = 1 + 2(x 1) and y = 1 2(x 1)

= 1 + 2x 2 = 1 2x + 2

= 2x 1 = 3 2x

When y = 1,

4(x - 1)2

3 = 1

4(x 1)2 = 3

(x 1)2 = 34

x = 1 3

2

centre (1, 1)

When y = 0,

4(x - 1)2

3 1

3 = 1

4(x - 1)2

3 = 4

3

(x 1)2 = 1

x = 1 1 = 0, 2

When x = 0, 43

(y - 1)2

3 = 1

(y - 1)2

3 = 1

3

(y 1)2 = 1

y = 1 1 = 0, 2

i

9x2 16y2 18x + 32y 151 = 0

9(x22x+1) 16(y22y+1) 151 + 7 = 0

9(x 1)2 16(y 1)2 = 144

(x - 1)2

16

(y - 1)2

9 = 1

(y - 1)2

9 =

(x - 1)2

16 1

(y 1)2 =

9(x - 1)2

16 9

=

9(x - 1)2

16

2

1

161

x

As x ,

16

(x - 1)2

0

(y 1)2

9(x - 1)2

16

y 1 4

)1 (3 x

y 1 4

)1 (3 x


y = 1 + 4

)1 (3 x and y = 1

4

)1 (3 x

=

4 + 3x - 34

=

4 - 3x + 34

= 1 + 3x

4 =

7 - 3x4

= 34

x + 14

= 74

34

x


When y = 1,

(x - 1)2

16 = 1

(x 1)2 = 16

x = 1 4 = 3, 5

centre (1, 1)

When y = 0,

(x - 1)2

16 1

9 = 1

(x - 1)2

16 =

109

(x 1)2 = 160

9

x = 1 4 10

3

j

25x2 16y2 = 400

x2

16

y2

25 = 1

y2

25 =

x2

16 1

y2 = 25x2

16 25

= 25x2

16

2

161

x

As x , 16x2

0

y2 25x2

16

y 54

x

Equations of asymptotes: y = 54

x

When y = 0, 25x2 = 400

x2 = 16

x = 4

centre (0, 0)

3

a

Substituting y = 12

x into x2 y2 = 1 gives

x2 ( 12

x)2 = 1

x2 14

x2 = 1

34

x2 = 1

x2 = 43

x = 2 3

3

Now y = 12

x

y = 3

3 when x =

2 33

and y =

- 33

when x =

-2 33

The points of intersection are (2 3

3,

33

) and

(

-2 33

,

- 33

)


b

Substituting y = 12

x into x2

4 + y2 = 1 gives

x2

4 + ( 1

2x)2 = 1

x2

4 +

x2

4 = 1

x2

2 = 1

x2 = 2

x = 2

When x = 2 , y = 2

2

When x = 2 , y =

- 22

The points of intersection are ( 2 , 2

2) and

( 2 ,

- 22

)

4

Substituting y = x + 5 into x2 + y2

4 = 1

gives

x2 + (x + 5)2

4 = 1

4x2 + x2 + 10x + 25 = 4

5x2 + 10x + 21 = 0

5(x2 + 2x + 1) + 16 = 0

5(x + 1)2 = 16

(x + 1)2 =

-165

But (x + 1)2 0 there is no intersection

point.

5

Since x2

9 +

y2

4 = 1 is a reflection of the

ellipse x2

4 +

y2

9 = 1 in either of the lines

y = x, the points of intersection of the two

ellipses occur when y = x.

Substituting y = x into x2

9 +

y2

4 = 1

gives

x2

9 +

x2

4 = 1

4x2 + 9x2 = 36

13x2 = 36

x2 = 3613

x = 613

= 6 13

13

Hence the points of intersection are (

-6 1313

,

-6 1313

), (6 13

13,

6 1313

),

(

-6 1313

, 6 13

13) and (

6 1313

,

-6 1313

).

These four points are all equidistant from the

origin and hence form the vertices of a square.

6

5x = 4y y = 54

x

Substituting y = 54

x into x2

16 +

y2

25 = 1

gives

x2

16 +

(54 x )

2

25 = 1

x2

16 +

25x2

16 25 = 1

x2

16 +

x2

16 = 1

x2

8 = 1

x2 = 8

x = 2 2

When x = 2 2 ,

y = 54

2 2 = 5 2

2

The points of intersection are ( 2 2 ,

-5 22

)

and (2 2 , 5 2

2)

7

x2 + y2 = 9

A circle with centre (0, 0) and radius 3

x2 y2 = 9 x2

9

y2

9 = 1

A hyperbola with centre (0, 0) and asymptotes at

y = x

When y = 0, x2 = 9

x = 3


8

a

x2 y2 1

b

x2 y2 4

x2

4

y2

4 1

c

y2 x2

4 1

x2

4 y2 1

d

x2

9 +

y2

4 < 1

e

x2 y2 1 and x2 + y2 4

f

(x - 3)2

16 +

y2

9 1

For

(x - 3)2

16 +

y2

9 = 1

When x = 0, 9

16 +

y2

9 = 1

y2

9 =

716

y2 = 6316

y = 3 7

4

3 7

4

x

y

0 3

(3, 3)

1 7

3 7

4

(3, 3)


g

x2 y2 4 and x2

9 + y2 1

144

22

yx

and x2

9 + y2 1

h

x2 y2 > 1 and x2 + y2 4

i

(x - 2)2

9 + y2 4

(x - 2)2

36 +

y2

4 1

For

(x - 2)2

36 +

y2

4 = 1

When x = 0, 436

+ y2

4 = 1

y2

4 =

89

y2 = 329

y = 4 23

j

x2

4 + y2 1 and y x

4 2

3

x

y

0 24 8

4 2

3

x

y

0

1

2 2

1

y = x

2

2


Solutions to Exercise 1H 1 x = 2cos 3t and y = 2sin 3t ran (x) = [2 , 2]

= dom(cartesian equation) ran (y) = [2 , 2]

= ran(cartesian equation)

x

2 = cos 3t and

y

2 = sin 3t

Squaring both sides of each equation gives

x2

4 = cos

2 3 t and

y2

4 = sin

2 3 t

Adding these two equations together gives

x2

4 +

y2

4 = 1

x2 + y

2 = 4 , dom = [2 , 2]

ran = [2 , 2]

2

a

x = sec t and y = tan t, t

2 ,

3

2


x2 = sec

2 t 1 and y

2 = tan

2 t 2

1 2

x2 y

2 = sec

2 t tan

2 t

x2 y

2 = 1

For the function x = sec t , t

2 ,

3

2

the range is ( , 1] . Hence the domain of the cartesian equation is ( , 1]

which corresponds to the left branch of the

hyperbola.

Equations of asymptotes: xy

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

y = x

y = x

b x = 3cos 2t and y = 4sin 2t ran (x) = [3 , 3]

= dom(cartesian equation) ran (y) = [4 , 4]


x

3 = cos 2t and

y

4 = sin 2t


x2

9 = cos

2 2 t and

y2

16 = sin

2 2 t


x

2

9 +

y2

16 = 1

x

2

9 +

y2

16 = 1 , dom = [3 , 3]

ran = [4 , 4]

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5


c

x = 3 3cos t and y = 2 + 2sin t ran (x) = [3 + 3, 3 + 3]

= [0 , 6]

= dom(cartesian equation) ran (y) = [2 + 2, 2 + 2]

= [0 , 4]


x 3 = 3cos t and y 2 = 2sin t Squaring both sides of each equation gives

(x 3)2 = 9cos

2 t and (y 2)

2 = 4sin

2 t

(x 3)

2

9 = cos

2 t and

(y 2)2

4 = sin

2 t


(x 3)2

9 +

(y 2)2

4 = 1

(x 3)

2

9 +

(y 2)2

4 = 1 , dom = [0

6]

ran = [0, 4]

x1 2 3 4 5 6 7 8

y

1

2

3

4

5

6

A CAS calculator has the capability to sketch

parametric equations.

In order to sketch a graph for part c:

Note: ensure your handheld unit is set to

radian/Rad mode.

TI: Open a Graphs page. Press

Menu3:Graph

Entry/Edit3:Parametric

Now type the following information:

ttx cos33)(1 tty sin22)(1

13.020 tstept and press ENTER

Set the window to:

Xmin = -0.5

Xmax = 8

Ymin = -0.5

Ymax = 6

CP: In the Graph & Table application tap n

and select xt =. Input the equations into the

corresponding positions followed by EXE.

Tap $ to see the graph.

d

x = 3sin t and y = 4cos t , t

2 ,

2

ran (x) = [3 , 3] for t

2 ,

2

= dom(cartesian equation)

ran (y) = [0 , 4] for t

2 ,

2


x

3 = sin t and

y

4 = cos t


x2

9 = sin

2 t and

y2

16 = cos

2 t


x2

9 +

y2

16 = 1

x

2

9 +

y2

16 = 1 , dom = [3 , 3]

ran = [0, 4]


x-4 -3 -2 -1 1 2 3 4

y

1

2

3

4

5

e

x = sec t and y = tan t, t

2 ,

2


x2 = sec

2 t 1 and y

2 = tan

2 t 2

1 2

x2 y

2 = sec

2 t tan

2 t

x2 y

2 = 1

For the function x = sec t , t

2 ,

2

the range is [1, ) . Hence the domain of the cartesian equation is [1, ) which corresponds to the right branch of the

hyperbola.

Equations of asymptotes: xy

x-5 -4 -3 -2 -1 1 2 3 4 5 6 7

y

-5

-4

-3

-2

-1

1

2

3

4

5y = x

y = x

f

x = 1 sec (2t) and y = 1 + tan (2t) ,

where t

4 ,

3

4

x 1 = sec (2t) and y 1 = tan (2t) Squaring both sides of each equation gives

(x 1)2 = sec

2(2t) 1

and (y 1)2 = tan

2(2t) 2

1 2

(x 1)2 (y 1)

2 = 1

For the function x = 1 sec (2t) where

t

4 ,

3

4

the range is [2 , ) .

Hence the domain of the cartesian equation is [2, ) which corresponds to the right branch of the hyperbola.

Equations of asymptotes: )1(1 xy

y = x and y = 2 x

x-3 -2 -1 1 2 3 4 5 6

y

-5

-4

-3

-2

-1

1

2

3

4

5

y = 2 x

y = x

( 1 , 1 )

3

a

x2 + y

2 = 16

x2 + y

2 = 4

2

In general for x2

+ y2

= a2 the most basic

parametric equations have the form x = a cos t and y = a sin t

Hence the parametric equations are x = 4cos t and y = 4sin t


b

x2

9

y2

4 = 1

x

2

32

y2

22 = 1

In general for x

2

a2

y2

b2 = 1 the most basic

parametric equations have the form x = a sec t and y = a tan t

Hence the parametric equations are

x = 3sec t and y = 2tan t

c

(x 1)2 + (y + 2)

2 = 9

(x 1)2 + (y + 2)

2 = 3

2

centre (1, 2) radius is 3

In general for (x h)2 + (y k)

2 = a

2 the

parametric equations have the form

x = h + a cos t and y = k + a sin t

Hence the parametric equations are

x = 1 + 3cos t and y = 3sin t 2

d

(x 1)2

9 +

(y + 3)2

4 = 9

(x 1)

2

92

+ (y + 3)

2

62

= 1

In general for (x h)

2

a2

+ (y k)

2

b2

= 1

the parametric equations have the form

x = h + a cos t and y = k + b sin t

Hence the parametric equations are x = 1 + 9cos t and y = 6sin t 3

4

centre (1, 3) and radius 2

(x 1)2 + (y 3)

2 = 2

2

As the parametric equations are in the form

x = a + b cos (2t) and y = c + d sin (2t) x = 1 + 2cos (2t) and

y = 3 + 2sin (2t)

a = 1, b = 2, c = 3 and d = 2

5

Ellipse: x intercepts at (4 , 0) and (4, 0) y intercepts at (0, 3) and (0, 3) Hence a possible cartesian equation for this

ellipse is x

2

16 +

y2

9 = 1

Thus a possible pair of parametric equations

for the above ellipse is

x = 4cos t and y = 3sin t

6

x = 2cos (2t) and y = 2sin (2t) a

For a dilation of factor 3 from the x-axis the point (x, y) is mapped onto (x, 3y) i.e. (x, y) (x, 3y) Thus to find the equation of the image curve

under the dilation (x, y) (x, 3y),

replace y with y

3 .

y

3 = 2sin (2t)

y = 6sin (2t)

Hence one possible pair of parametric

equations for the image curve is x = 2cos (2t) and y = 6sin (2t)

b x = 2cos (2t) and y = 6sin (2t)

x

2 = cos (2t) and

y

6 = sin (2t)


x2

4 = cos

2(2 t) and

y2

36 = sin

2(2 t)


x2

4 +

y2

36 = 1 as cos

2(kt) + sin

2(kt) = 1


Hence the cartesian equation is

x2

4 +

y2

36 = 1

7

x = 3 2cos t

2

and y = 4 + 3sin t

2

a

For a translation of 3 units in the negative

direction of the x axis and a translation of 2 units in the negative direction of the y

axis:

(x, y) (x 3, y 2) Let (x ', y') be the coordinates of the image of (x, y) so x ' = x 3, y ' = y 2 Rearranging gives x = x ' + 3 and y = y ' + 2

So x = 3 2cos t

2

becomes

x ' + 3 = 3 2cos t

2

x ' = 2 cos t

2

and y = 4 + 3sin t

2

becomes

y ' + 2 = 4 + 3sin t

2

y ' = 2 + 3sin t

2

Thus the parametric equations of the image

curve are

x = 2 cos t

2

and y = 2 + 3sin t

2

b

x = 2 cos t

2

and y = 2 + 3sin t

2

x

2 = cos

t

2

and y 2

3 = sin

t

2


x2

4 = cos

2

t

2

and (y 2)

2

9 = sin

2

t

2


x2

4 +

(y 2)2

9 = 1

Hence the cartesian equation is

x2

4 +

(y 2)2

9 = 1

8

x = 2 + 3sin (2t) and y = 4 + 2cos (2t) a

ran (x) = [2 , 5] for t

0 , 1

4


ran (y) = [4 , 6] for t

0 , 1

4


and the cartesian equation is

(x 2)2

9 +

(y 4)2

4 = 1

(x 2)

2

9 +

(y 4)2

4 = 1 , dom = [2 , 5]

ran = [4, 6]

b

ran (x) = [2 , 5] for t

0 , 1

2


ran (y) = [2 , 6] for t

0 , 1

2


(x 2)

2

9 +

(y 4)2

4 = 1 , dom = [2 , 5]

ran = [2, 6]


c

ran (x) = [1 , 5] for t

0 , 3

2


ran (y) = [2 , 6] for t

0 , 3

2


When x = 0 , 4

9 +

(y 4)2

4 = 1

(y 4)

2

4 =

5

9

(y 4)2 =

20

9

y 4 = 2 5

3

y = 4 2 5

3


Chapter review: multiple-choice solutions 1 t3 = 4 and t8 = 128

For a geometric sequence tn = arn 1

Using t3 : 4 = ar2 1

Using t8 : 128 = ar7 2

2 1 gives

r5 = 32

r = 2

Hence tn = a(2)n 1

As t3 = 4 then 4 = a(22)

a = 1 Thus the first term of the sequence is 1

Answer is B

2

The first term of the arithmetic sequence is

not known thus the following sequence

should be used. tn = tn 1 + d

If 5, x and y are in arithmetic sequence then

x = 5 + d 1

y = x + d 2

Rearranging 1 for d gives:

d = x 5 3

Substituting 3 into 2 gives

y = x + (x 5) y = 2x 5

Answer is D

3

2cos x 2 = 0

cos x = 2

2

45 for [0 , 90 ]x x

Answer is C

4 As the range of the given graph is [1 , 1] ,

response D is incorrect.

Clearly, the given graph has period . Thus

response B and E are also incorrect.

The graph also has a y-intercept of 1.

Response C clearly does not pass through

the point (0, 1) while response A does

Hence the given graph is y = sin 2

x

4

A quick sketch of response A on your CAS

calculator will alleviate all doubt.

Answer is A

5

sin 2

3 cos

4 tan

6

sin

3

2

2

3

3

sin

3

6

6

3

2

6

6

18

12

3 2

12

2

4

Answer is C


6

180 (100 35 ) 45BAX

) triangle,a of angles( BAX

XDC = 45 BC) arcby subtended angle(

Answer is C

7

t2 = 24 and t4 = 54

A geometric sequence is given by tn = arn 1

Using t2 : 24 = ar 1

Using t4 : 54 = ar3 2

2 1 gives

r2 =

9

4

r = 3

2 as r > 0

Hence tn = a 3

2

n 1

As t2 = 24 then 24 = a 3

2

1

a = 16 Thus the geometric sequence is given by

tn = 16 3

2

n 1

The sum of the first 5 terms of this sequence

is S5 where

S n = a(r

n 1)

r 1

S 5 =

16

3

2

5

1

3

2 1

= 211

2 2

= 211

Hence the sum of the first 5 terms is 211

Answer is B

8

Using the cosine rule,

c2 = 30

2 + 21

2 2(30 )(21 )cos C

= 1341 1260 51

53

= 1341

64260

53

=

6813

53

c = 6813

53 as c > 0

c = 11.33786 . . . Thus c = 11 rounded to the nearest whole number.

Answer is C

9

x2 8x + y

2 2y = 8

(x2 8x + 16 ) + (y

2 2y + 1) = 25

(x 4)2 + (y 1)

2 = 25

centre (4, 1)

Answer is D

10

From the graph:

The centre occurs at (2, 0) Responses A, C and E are incorrect The vertices occur at (7 , 0) and (11 , 0)

Generally the vertices of a hyperbola occur

at ( a + h, k)

For response B: a = 3, h = 2 and k = 0 So the vertices are ( 3 + 2, 0) i.e. (1 , 0) and (5, 0) Response B is incorrect

For response D: a = 9, h = 2 and k = 0 So the vertices are ( 9 + 2, 0) i.e. (7 , 0) and (11 , 0) Thus response D is correct.

Answer is D


Chapter review: short answer questions 1 fn = 5fn 1 , f0 = 1

This defines a geometric sequence with first term 5 and common ratio 5.

Therefore fn = 1 5n 1

= 5n 1

2 AP = 10 cm. Form triangle PAO which is right-angled at A. Triangle AOP is congruent to triangle BOP as AO = BO (radii of a circle) and AP = BP (tangents from a common point). Also PO is a common side.

Therefore PO bisects angle APB. Then APOP = cos and OP =

10

cos

3

The centre of the ellipse is (2, 3). The minor axis has length 4 and the major axis

length 8. Hence using the general equation(x h)2

a2 +

(y k)2

b2 = 1

gives (x + 2)2

a2 +

(y 3)2

b2 = 1

(0, 3) is on the ellipse. Hence 4

a2 = 1 and a

2 = 4.

Also (2, 7) is on the ellipse. Hence 16

b2 = 1 and b

2 = 16

Hence the equation is (x 2

4 +

(y 3)2

16 = 1

x

(2, 7)

(0, 3)

0

y

O

A

B

P


4

The triangle is right-angled and so the hypotenuse has length 49 + 64 = 113 .

Therefore sin (= 7

113

5 x

9 = sin (30)

Therefore x = 9 sin (30) = 9

2

6 a

X is the midpoint of AB and OX is perpendicular to AB. OA 2 = AX 2 + OX 2

Hence OA = 25 + 9 = 34 cm

b Let angle AOX have magnitude

Then tan = 53 and = tan

1 53

Angle AOB = 2

= 2 tan1 53

7 a cos (315) = cos (360 45)

= cos (45) = 2

2

b If tan (x) = 34 and 180 < x < 270,

use 1 + tan2 = sec2

sec2 (x) = 1 +

9

16

sec (x)=

25

16

and sec (x) = 5

4

Hence cos (x) = 4

5 as 180 < x < 270.

c sin A = sin 330.

7 8

O

A B X


One possible answer is A = 210. The entire set of solutions is 330 + 360n where n is an integer, and 210 + 360n where n is an integer.

8

a Triangle ABD is isosceles with BD = AB (given). Therefore angle BDA = x By the alternate segment theorem angle BCD = x

b Triangle ABD is similar to triangle CDA.

and ADCA =

BDAD

That is, y

a + b = ay

Hence y2 = a (a + b)

and y = a (a + b)

9

The triangle ABC is right-angled at B, and AB = BC = 1 cm.

Pythagoras' theorem gives that AC = 2 . Triangle ABC is isosceles, and X is the midpoint of AC.

Using Pythagoras' theorem again gives BX = 1 12 =

22

Let angle BXP have magnitude .

Then tan = 3 2

2

= 6

2

= 3 2

Therefore = tan1 3 2 )

B

O x A

D

C B

A

B C

P

A C X


10 a 2 cos (2x + ) 1 = 0 implies 2 cos (2x) = 1

and thereforecos (2x) = 12

2x = , 4

3 , 2

3 ,

2

3 ,

4

3 , ...

x = 2

3 ,

3 ,

3 or

2

3

b

c From the graph, 2 cos (2x + ) < 1 for

23

3

3

2

3

11

a Triangle ABC is a right-angled triangle at C as AC 2 + CB 2 = AB 2. b In triangle DAC, AC = DC = 9 cm.

The triangle is isosceles with a right angle at C. Therefore DAC = 45

For angle DBC, tan (DBC) = 9

12

= 34

and hence DBC = tan1 34

2

3

(0, 3)

x

y

3

4

2

0

2

4

3

2

3

(,3) (, 3)

(

2 , 1) (

2 , 1)

A

C B

D


12 a The cosine rule gives AB

2 = 24

2 + 33

2 2 24 33 cos (60) = 873

AB = 3 97 km

The distance apart after three hours is 3 97 nautical miles.

b The speeds are 8 nautical miles per hour and 11 nautical miles per hour. Therefore the distances travelled are 40 nautical miles and 55 nautical miles respectively. The new triangle formed is similar to the triangle of part a, with a scale

factor of 53 . The distance apart is 5 97 nautical miles after 5 hours.

13

Using the sine rule givesx

sin (30) = 18

sin (45)

Therefore x = 18

sin (45) sin (30)

= 18 2 12 = 9 2

14 a

b The triangle ABC is right-angled at C.

AC

480 = cos (45)

Therefore AC = 240 2

c The triangle is isosceles and so the total distance flown = 480 2 km.

A B

C

480 km

45

45

P A

B

60

33

24

18 cm x cm

30 45


15 For x2 (y 2)2

9 = 15

Rearrange to give (y 2)2

9 = x2 15

and hence y 2 = 3 x2 15

and hence y 2 = 3x

1

15

x2

12

It now can be observed that the asymptotes will have equations

y = 3x + 2 or y = 3x + 2 and y = 3x + 2

16 For x = 3 cos (2t) + 4 and y = sin (2t) 6,

first rearrange each of the equations.

cos (2t) = x 4

3 and sin (2t) = y + 6

Square each of these equations and add

cos2 (2t) + sin

2 (2t) =

(x 4)2

9 + (y + 6)2

Therefore the cartesian equation is(x 4)

2

9 + (y + 6)2 = 1

17 a

The quadrilateral is cyclic and therefore 3x = 180 which implies x = 60.

b

Firstly d = 60 by the alternate segment theorem. The angle at X subtended by

the diameter is 90. Angle OXD is 60 as triangle DOX is isosceles (radii of a

circle). Therefore a = 90 60 = 30.

Triangle BOX is also isosceles. Therefore b = a = 30.

Angle c = 120 (angle sum of a triangle).

P

R

a b

c

d

60

O

X

D

B

x

2x


18 For x = 2 cos (t) and y = 2 sin (t) + 2,

first rearrange; cos (t) = x

2 and sin (t) =

y 2

2

Squaring and adding gives

cos2 (t) + sin

2 (t) =

x2

4 +

(y 2)2

4

Hence the cartesian equation is x2 + (y 4)

2 = 4

19 a

b 2 cos

x

4 = 0

implies cos

x

4 = 0

x

4 =

2 or 32 or

x = 34 or

74

c 2 cos x 0 is equivalent to cos x 0

From the graph for x [0, 2], cos x 0 for

0

2

3

2 2

20 a sin = 12

=

6 or 56

b cos = 3

2

=

6 or 11

6

c tan = 1

=

4 or

5

4

x

y

4

4

2

0

2

4

3

34

54

7

4


21 For x = a + b cos (2t) and y = c + d sin (2t)

rearranging gives

x a

b = cos (2t) and

y c

d = sin (2t)

Squaring and adding gives

(x a)2

b2 +

(y c)2

d2 = 1

The centre of the circle is (1, 2) and the radius is 3.

Hence a = 1, c = 2 and b = d = 3

22

a ADB = 180 (90 + 40)

= 50 (angle sum of triangle)

as BA is a tangent to the circle at A, and perpendicular to AD.

b AEC = 50, as ADB and AEC are subtended by the same arc at the circle.

c In DAC, right-angled at C,DAC = (90 50) = 40

23 x2 + 8x + y2 12y + 3 = 0 Completing the square gives x2 + 8x + 16 + y2 12y + 36 + 3 = 52 (x 4)2 + (y 6)2 = 49 The centre of the circle is the point with coordinates (4, 6) and the radius is 7.

24 x2

81 + y2

9 = 1

When x = 0, y2 = 9 and y = 3 or 3 When y = 0, x2 = 81 and x = 9 or 9

25 a i Use tn = a + (n 1)d 17p + 17 = 3p + 5 + 2(n 1)

14p + 12 = 2(n 1)

Therefore n = 7p + 7

O

A

C D

B E

40


ii The sum of the sequence, Sn = 7p + 7

2 (3p + 5 + 17p + 17)

= 7(p + 1)(10p + 11)

= 7(10p2

+ 21p + 11)

=70p2 + 147p + 77

b sum = 7(p + 1)(10p + 11)

If p is even, p + 1 is odd and 10p + 1 is odd. Therefore the sum is not divisible

by 14.

If p is odd, p + 1 is even and hence the sum is divisible by 14.

26 a The nth

term is 3n 1

b 30 31 32 3n 1 = 30 + 1 + 2 ++ (n 1)

= 31 + 2 + 3 + + (n 1)

1 + 2 + 3 + + 19 = 19(19 1)

2

= 190

Therefore the product of the first 20 terms is 3190

.


Chapter review: extended-response questions 1

a AB 2

= 25 + 49 70 cos (115)

Therefore AB = 10.2 km, correct to two decimal places.

b Then using the sine rule, 7

sin =

AB

sin (115)

Therefore sin = 7 sin (115)

AB

which gives = 38.56...

and the bearing of B from A is given by 10 + 38.56...

The bearing is 049.

c

i The magnitude of angle BAP = (80 ( + 10)) = (31.43...)

Using the cosine in triangle APB gives

BP 2 = AB

2 + 4

2 8 AB cos (31.43...)

Therefore BP = 7.079...

The total distance travelled by the second hiker

= 4 + 7.079...

= 11.08 km, correct to two decimal places.

ii Use the cosine rule to find the size of angle APB.

cos P = AB

2 AP

2 PB

2

2 AP PB

and so the magnitude of angle APB is 131.42

The bearing is therefore given by 131.42 100

The bearing is 031.

A

B

10

75

115

A

B

10

75

115

P


d

In this diagram, AC = CB and the bearing of C from A is 80.

Triangle ACB is isosceles,

therefore cos (= AX

AC

and AC = AX

cos (

AX = 1

2 AB

From the above, = 31.43 and AX = 5.088

Therefore AC = 5.963...

The total distance travelled = 11.93 km, correct to two decimal places.

2

a i The centre of the ellipse is (0, 3) and so the minor axis has endpoints

( 2 , 3) and ( 2 , 3). The domain is [ 2 , 2 ]

ii The major axis has endpoints (0, 3 + 5 and (0, 3 5 .

The range is [3 5 , 3 + 5 ]

iii The centre is (0, 3)

x

y

2 1 0 1 2

6

4

2

A

B

10

75

115

C

X


b The centre of the ellipse has coordinates

3 + 1

2 1 + 5

2= (1, 2)

The major axis (parallel to y axis) has length 6 and the minor axis (parallel to

x axis) length 4.

Hence the equation of the ellipse is (x 1)

2

4 +

(y 2)2

9 = 1.

So a = 2, b = 3, h = 1, k = 2.

c The line y = x 2 intersects the ellipse (x 1)

2

4 +

(y 2)2

9 = 1 at the point

(1, 1) and another point.

Substituting, 9(x 1)2 + 4( x 4)

2 = 36

Expanding and simplifying gives

9(x2 2x + 1) + 4(x

2 8x + 16) = 36

and 13x2 50x + 37 = 0

(x 1) is a factor.

Therefore (x 1)(13x 37) = 0

The line intersects the ellipse at (1, 1) and

37

13

11

13.

P has coordinates

37

13

11

13.

d The line perpendicular to the line with equation y = x 2, and which passes

through

37

13

11

13, has equation

y 11

13 =

x

37

13

Rearranging gives y = x + 48

13

The coordinates of Q are

0

48

13

e There is a right angle at P and hence AQ is a diameter.

The coordinates of A, P and Q are (1, 1),

37

13

11

13 and

0

48

13 respectively.

The centre of AQ is

1

2

35

26

The diameter =

61

13

2

+ 1

= 3890

13

The equation of the circle is

x

1

2

2

+

y

35

26

2

= 3890

676


3 a x2 + y

2 2ax 2ay + a

2 = 0

Completing the square gives

x2 2ax + a

2 + y

2 2ay + a

2 + a

2 = 2a

2

(x a)2 + (y a)

2 = a

2

The centre is at (a, a) and the radius is a.

Therefore the circle touches both axes at (0, a) and (a, 0).

essential specialist34 solutions ch1

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