1 Focus on Exam 1

8/12/2019 1 Focus on Exam 1

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Oxford Fajar Sdn. Bhd. (008974-T) 2012

CHAPTER 1 FUNCTIONS

Focus on Exam 1

1 (a) g(x) = 16x2

Forg(x) to be defined, 16x20

(4 +x)(4x) 0

x

4 4

Hence, the domain ofg(x) is

{x: 4 x4,xR}.

(b)

x

y

O 4

4

4

16 x2y =

The graph of g(x) is

actually part of a circlewith the equation

y2=16 x2x2+y2=42.

(c) The range is {y: 0 y4,yR}.

2 (a) f: x x29

f (x) = x29

Forf (x) to be defined,x29 0

(x +3)(x 3) 0

x

3 3

Hence, the domain off (x) is

{x:x3 orx3,xR}.

(b)

xO

y

3 3

x2 9y=

(c) The range is {y:y0,yR}.

3 First, consider onlyg(x) =12x 2. The graph ofg(x) is as shown below.

x

y

4

2

2

O

1

2y = x 2

1

2y = x + 2

1

2x+2, x


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ACE AHEADMathematics (T) First Term2

1

2x 2, x


3/26


Fully Worked Solution 3

(ii) g(x) =x2 3

x

y

O

y = x233

The domain ofg(x) is {x:xR}.

The range ofg(x) is

{y:y3,yR}.

(b) gf exists because RfDg.

gf =g[f(x)]

=g x2

= x22 3=x 5

(c) Forfg to be defined, RgDf .

Rg D

f

x2 3 2

x2 5 0

x+ 5x 50Hence, the required set of values of xis

{x:x 5orx 5,xR}.

6 (a) (i) f (x) = 25 x2

x

5

5 5O

y

25 x2y=

The domain off (x) is

{x: 5 x 5,xR}.The range off (x) is

{y: 0 y5,yR}.

(ii) g (x) =x2 5

x

y

O

5

y = x25

{ {

The domain ofg(x) is {x:x R}.

The range ofg(x) is

{y:y 5,yR}.

(b) fgdoes not exist because RgDf .

(c) Forfgto be defined, RgDf .

Df Rg Df

5 x2 5 5

0 x2 10

Hence, the required set of values of xis

{x: 10x 10,xR}.

7 (a) (i) f (x) =1

x 2

x

y

O21

2

The domain off (x) is

{x:xR,x 2}.

The range off (x) is

{y:yR,y 0}.

(ii) g(x) =2

x+4

xO

y

1

2

4

The domain ofg(x) is{x:xR,x 4}. (1)

The range ofg(x) is

{y:yR,y 0}.

(b) fg=f [g(x)]

=f 2x+4=

1

2x+4 2= x+4

2 2(x+4)

} } }


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=x+4

6 2x,x3 (2)

Combining (1) and (2), the domain of

fgis {x:xR,x 4,x 3}.

8 (a) Let y=f 1(x)

f (y)= x2 + y1= x

y1= x 2

y 1 =(x 2)2

y 1 =x2 4x +4

y=x2 4x +5

f 1(x) =x2 4x+5

The domain off 1is the same as the

range off, i.e. {x:x2,xR}.

The range off 1is the same as the

domain off, i.e. {y:y1,yR}. (b) The graphs ofy=f (x) andy=f 1(x)

are as shown below.

x

y

O1 2 3 4 5

1

2

3

4

5

y =f1(x)

y =f(x)

y=

x

The graph of y= f1(x) is

the reflectionof the graph of

y= f(x) in the straight line y= x.

The point of intersection of the graphs of

y=f(x) andy=f 1(x) is the same as the

point of intersection of the curvey=f1(x)

=x2 4x+5 and the straight liney=x.

y=x2 4x+5 (1)

y=x (2)

x2 4x+5 =xx2 5x+5 =0

x=(5) (5)2 4(1)(5)

2(1)

x=5 5

2x=1.38 or 3.62

x=1.38 is not accepted

x=3.62

y=x =3.62

Hence, the required point ofintersection is (3.62, 3.62).

9 (a) The graph ofy=f (x) =x2 3xis as

shown below.

x

y

O 3

12(

14

,2 )1

y =f(x) = x2 3x

f 1does not exist because fis not a

one-to-one function.

(b) In order forf1to exist, the domain off

must be restricted to only

x :x112,xR.Let y=f

1

(x)f (y) =x

y2 3y=x

y2 3y x=0

y=(3) + (3)

2 4(1)(x)

2(1)

y=3 + 9 + 4x

2

f1(x)=3+ 9 + 4x

2

The domain off1

is the same as the

range off, i.e. x :x214,xR.10 Since (x 2) is a factor of

P(x) =qx3 rx2+x 2,

P(2) =0

q(2)3 r(2)2+2 2 =0

8q 4r=0

2q r=0 (1)

P(x) has a remainder of 12 when it is

divided by (x+1). P(1) =12

q(1)3 r(1)2 1 2 =0

q r=3 (2)

(1) (2):

2q r=0

q r=3

3q =3

q =1

From (1), 2(1) r=0

r=2P(x) =x3+2x2+x 2


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x2+1

x 2 x3+2x2+x 2x3+2x2

x 2

x 2

0

P(x) =(x 2)(x2+1)=(x 2)(1 +x)(1 x)

Hence, the zeroes ofP(x) are 2, 1and 1.

11 Since Q(x) is divisible byx2+x 6 =

(x2)(x+3), then it is also divisible by

(x 2) and (x+3).

Q(2) =0

m(2)3 5(2)2+k(2) +54 =0

8m +2k=34

4m +k=17 (1)Q(3) =0

m(3)3 5(3)2+k(3) +54 =0

27m 3k=9

9m +k=3 (2)

(2) (1): 5m=20 m=4

From (1), 4(4) +k=17 k=33

12 Since (x+2) is a factor ofP(x), then

P(2) =0

(2)3+4(2)2h(2) +k=0

2h+k=8

k=2h 8 (1)

WhenP(x) is divided by (x h), the

remainder is h3.

P(h) =h3

h3+4h2 h2+k=h3

3h2+k=0 (2)

Substituting (1) into (2),

3h2 2h 8 =0

(3h+4)(h 2) =0

h=43

or 2

When h=4

3, k=2 43 8 =

16

3

When h=2, k=2(2) 8 =12

13 When a polynomialP(x) of degree n2

is divided by 2x2+3x 2 =(2x 1)(x+2),

the remainder is an expression in the

form ax+b, where aand bareconstants. i.e.

P(x) =(2x 1)(x+2) Q(x) +(ax+b)

WhenP(x) is divided by (2x 1), the

remainder is3

2.

P12

=(0)12

+2Q(x) + 12

a+b=3

2a+2b=3 (1)

WhenP(x) is divided by (x+2), the

remainder is 1.

P(2) =[2 (2) 1](0) Q(x)

+(2a+b) =1

2a+b=1 (2)

Solving (1) and (2), a=1, b=1

Hence, the remainder whenP(x) is divided

by 2x2+3x 2 is ax+b=x+1.

14 6x4 25x3 12x2+25x+6 =0

Dividing throughout byx2,

6x2 25x 12 +25

x+

6

x2=0

6x2+ 1x2 25x+1

x 12 =0 (1)

Consideringz2=x 1x2

,

z

2

=x

2

2 +

1

x2

x2+1

x2=z2+2

From (1), 6(z2+2) 25z 12 =0

6z2+12 25z 12 =0

6z2 25z=0

z(6z 25) =0

z=0 orz=25

6 Whenz=0, x

1

x=0

x2 1 =0x=1

Whenz=25

6,x

1

x=

25

66x2 6 =25x

6x2 25x 6 =0

x=25 (25)2 4(6)(6)

2(6)

x=25 769

12x=0.228or 4.39


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15 x3+x 2

x2 4 x5 3x3 2x2 4x+8x5 4x3

x3 2x2 4x+8

x3 4x

2x2 +8

2x2 +80

The remainder is 0.

Since the remainder is 0, (x2 4) is a factor

ofP(x).

P(x) =(x2 4)(x3+x 2)

Let Q(x) =x3+x 2.

Ifx=1, Q(x) =13+1 2 =0

(x 1) is a factor of Q(x).

P(x) =(x

2

4)(x 1)(x

2

+x+2) WhenP(x) =0,

x2 4 = 0 or x 1 = 0 or x2+x+2 =0

x=2, x=1,

No real solutions because

b2 4ac=12 4(1)(2) =7 (< 0)

The zeroes ofP(x) are 2and 1.

16x2 1 =(x 1)(x+1)

P(1) =12n (m+2)(1)2+m+1

=1 m 2 +m+1 =0

Thus, (x 1) is a factor ofP(x).

P(1) =(1)2n (m+2)(1)2+m+1

=1 m 2 +m+1 =0

Thus, (x+1) is a factor ofP(x).

Since (x 1) and (x+1) are factors ofP(x),

then (x 1)(x+1) =x2 1 is a factor ofP(x).

When m=8, P(x) =x2n (8 +2)x2+8 +1

=x2n 10x2+9

Since (x 3) is a factor, then

P(3) =0

32n 10(3)2+9 =032n=81

32n=34

2n=4

n=2

Hence,P(x) =x4 10x2+9 =(x2 9)(x2 1)

=(x+3)(x 3)(x+1)(x 1)

17 Let2x+1

(x2+x)2=

2x+1

x2(x +1)2

Ax +Bx2+ Cx+1+ D(x+1)2

2x+1 Ax(x+1)2+B(x+1)2

+Cx2(x+1) +Dx2

Lettingx=1, 1 =D(1)2D=1

Lettingx=0, B=1

Lettingx=1, 3 =4A+4B+2C+D

3 =4A+4(1) +2C 1

4A+2C=0 (1)Lettingx=2, 5 =18A+9B+12C+4D

5 =18A+9(1) +12C+4(1)

18A+12C=0

3A+2C=0 (2)

(1) (2): A=0

From (2), 3(0) +2C=0 C=0

Hence,2x+1

(x2+x)2=

1

x2

1

(x+1)2

18 4x2

x+3x3 1

= 4x2

x+3(x 1)(x2 +x+1)

A

x 1+

Bx +C

x2 +x+1

4x2x+3 A(x2+x+1) +(Bx+C)(x 1)

Lettingx=1, 6 =3A A=2

Lettingx=0, 3 =A +C(1)

3 =2 C

C=1

Lettingx=1, 8 =A+(B+C)(2)

8 =2 +(B 1)(2)8 =2 +2B+2

2B=4

B=2

4x2x+3

x3 1=

2

x 1+

2x 1

x2+x+1

19P(x) =x3+x2+px+q

P(x) =3x2+2x+p

Since (x 1) is a factor ofP(x),

P(1) =0. 13+12+p(1) +q=0

p+q=2 (1)

Since (x 1) is a factor ofP(x),

P(1) =0.

3(13) +2(1) +p=0

p=5

From (1): 5 +q=2 q=3

P(x) =x3+x2 5x+3

Since (x 1) is a factor ofP(x) andP(x),

thus (x 1)

2

[orx

2

2x+1] is a factorofP(x).


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x+3

x2 2x +1x3+x2 5x+3x3 2x2+x

3x2 6x+3

3x2 6x +3

0

P(x) =(x 1)2(x+3)20 4x

P(x)=

20 4x

(x 1)2(x+3)

A

x 1+

B

(x 1)2+

C

x+3

20 4xA(x 1)(x+3) +B(x+3)

+C(x 1)2

Lettingx=3, 32 =16CC=2

Lettingx=1, 16 =4BB=4

Lettingx=0, 20 =3A+3B+C

20 =3A+3(4) +23A=6

A=2

20 4x

(x 1)2(x+3)=

2

x 1+

4

(x 1)2+

2

x+3

20 Since the remainders whenP(x) is divided

by (x+1) is 0,P(1) =0.

P(1) =0

(1)3+m(1)2+15(1) +k=0

1 +m 15 +k=0m+k=16 (1)

Since the remainders whenP(x) is divided

by (x+2) is ( 4),P(2) = 4.

P(2) = 4

(2)3+m(2)2+15(2) +k= 4

8 +4m 30 +k= 4

4m+k=34 (2)

(2) (1): 3m=18 m=6

From (1): 6 +k=16 k=10

P(x) =x3

+6x2

+15x+10Since the remainders whenP(x) is divided

by (x+1) is 0, (x+1) is a factor ofP(x).

x2+5x+10

x +1x3+6x2+15x+10x3+x2

5x2+15x

5x2+5x

10x+10

10x+10

0P(x) =(x+1)(x2+5x+10)

x+7

P(x)=

x+7

(x+1)(x2+5x +10)

A

x+1+

Bx +C

x2+5x+10

x+7 A(x2+5x+10) +(Bx+C)(x+1)

Lettingx=1, 6 =6AA=1Lettingx=0, 7 =10A+C

7 =10(1) +C

C=3

Lettingx=1, 8 =16A+2B+2C

8 =16(1) +2B+2(3)

2B=2

B=1

x+7

(x+1)(x2+5x +10)=

1

x+1+

x 3

x2+5x+10

= 1x+1 x+3x2+5x+10

21 12 x2 7x10

The first inequality is

12 x2 7x

x2 7x+12 0

(x 3)(x 4) 0

x4 0

x3 0

+

++

x

3+ +

4

Therefore, (x 3)(x 4) 0 ifx3 or

x 4.

The second inequality is

x2 7x10

x2 7x+10 0

(x 2)(x 5) 0

x5 0

x2 0

+

++

x

2+ +

5

Therefore, (x 2)(x 5) 0 if 2 x5.

3 4x

52

2 x5

x3 or x4

The required set of values ofxis{x:2x3 or 4x5}.


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22 16 x3 4x2+4x 16 0

When 16 x3 4x2+4x 16,

x3 4x2+4x0

x(x2 4x+4) 0

x(x 2)20

Since (x 2)20,

in order thatx(x 2)20, thenx0 (1) Whenx3 4x2+4x 16 0,

we letf (x) =x3 4x2+4x 16.

f (4) =43 4(4)2+4(4) 16 =0

Thus, (x 4) is a factor off (x).

x2+4

x 4x3 4x2+4x 16x3 4x2

4x 16

4x 16

0 x3 4x2+4x 16 0

(x 4)(x2+4) 0

Sincex2+4 > 0,

in order that (x 4)(x2+4) 0, then

x 4 0 x4 (2)

Combining (1) and (2),

the required set of values ofxis

{x: 0x 4}.

233x 5

xx 3

3x 5

xx+3 0

3x 5 x2+3x

x0

x2+6x 5

x0

x2 6x+5

x0

(x 1)(x 5)

x0

x1 0

x5 0

x 0

+

+

+

+

+

+

+

+

x

0 1 5

The required set of values ofxis

{x:x


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25 The graphs ofy=|x+2| andy=1

x+1is as

shown below.

x

y

O2

2

1

y = 1x + 1

y =x +2

y = x 2

A

y=x+2 (1)

y=1

x+1 (2)


x+2 = 1x+1 x2+3x+2 =1

x2+3x+1 =0

x=3 32 4(1)(1)

2(1)

x=3 5

2

Thex-coordinate of pointAis

x=3 + 5

2

.

Based on the graphs, the solution set ofx

for which |x+2| >1

x+1is

x3 + 52 .

This is the set of values ofxwhere the

graph ofy=|x+2| is above the graph of

y=1

x+1.

26

xO11

1

1 3

y = x 1

y

Ay = x 1 x +1y=

To determine thex-coordinate of pointA, solve

y=x 1 (1)

y= x+ 1 (2)


x 1 = x+ 1(x 1)2=x+1

x2 2x+1 =x+1

x2 3x=0x(x 3) =0

Thus, thex-coordinate of pointAis x=3.

The part of thex-axis where the graph of

y= x+ 1is above the graph ofy=|x| 1

is 1 x3.

Hence, the required set of values ofxis

{x: 1x3}.

27P(x) =2x3+hx2+kx+36

Since (x 3) is a factor, then

P(3) =0

2(3)3+h(3)2+k(3) +36 =0

9h+3k=90

3h+k=30 (1)

P(x) =(x+2)f (x) 30 means that the

remainder whenP(x) is divided by (x+2)

is 30.

P(2) =30

2(2)3+h(2)2+k(2) +36 =30

4h 2k=50

2h k=25 (2) (1) +(2): 5h=55 h=11

From (1): 3(11) +k=30

k=3

Therefore,P(x) =2x3 11x2+3x+36.

2x2 5x 12

x 3 2x3 11x2+3x+362x3 6x2

5x2+3x

5x2+15x

12x+3612x+36

0

Therefore,P(x) =(x 3)(2x2 5x 12)

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

3 4

x

3

2

The sets of values ofxsuch thatP(x) 0

is x: 32x 3 orx 4.


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28P(x) =2x3+px2+qx+6

Since (2x+1) is a factor ofP(x), then

P 12=0

2 123

+p 122

+q 12+6 =0

1

4+

1

4p

1

2q+6 =0

1 +p 2q+24 =0

p 2q=23 (1)

WhenP(x) is divided by (x+3), the

remainder is 15.

P(3) =15

2(3)3+p(3)2+q(3) +6 =15

9p 3q=33

3p q=11 (2)

p 2q=23 (1) 6p 2q=22 (2) 2

5p = 45

p =9

From (1): 9 2q=23 q=16

P(x) =2x3+9x2+16x+6

x2+4x +6

2x+1 2x3+9x2+16x+62x3+x2

8x2+16x8x2+4x

12x +6

12x+6

0

Let Q(x) =x2+4x+6

=x2+4x+422

422

+6

=(x+2)2+2 [> 0] [Shown]

P(x) =(2x+1)(x2+4x+6)

Sincex2+4x+6 is positive for all real

values ofx, thenP(x) < 0 only if

2x+1 < 0 x< 1

2.

Hence, the solution set is x:x


11/26



When u=5

2, When u=3,

x

1

2=5

2 x

1

2=3

x1=522

1x

=254

x=4

25

33 8x+6(8x) =5

8x+6

8x=5

Let 8x=u

u+6

u

=5

u2+6 =5u

u2 5u+6 =0

(u 2)(u 3) =0

u=2 or u=3

8x=2 8x=3

23x=21 xlg 8 =lg 3

3x=1 x =lg 3

lg 8

x=1

3 x =0.528

34 log2x log

x8 +2log2h+hlogx4 =0

log2x

log28

log2x

+h+hlog

24

log2x=0

ylog

223

y+h+h

log222

y =0

y3

y+h+h2y=0

y23 +hy+2h=0

y2+hy+2h 3 =0[Shown]

When h=1

4,y2

1

4y+2 14 3 =0

4y2 y 14 =0

(4y+7)(y 2) =0

y=7

4or 2

Giveny=log2x, thenx=2y.

Wheny=7

4

,x=2

7

4=0.297.

Wheny=2, x=22=4.

[Not possible because

x

1

2> 0 for all real

values ofx.]

35 2 logcx 3 log

xc=5

2 logcx 3

logcc

logcx=5

2 logcx 3 1log

cx=5

Let logcx=u

2u 31u=52u2 3 =5u

2u2 5u 3 =0

(2u+1)(u 3) =0

u=1

2 or u=3

logcx=

1

2 log

cx=3

x=c

1

2

=

1

c x =c3

36 xlog432 ylog

82 =4

xlog

232

log24

ylog

22

log28

=4

xlog

225

log222

ylog

22

log223

=4

5x2

y

3=4

15x 2y=24 (1)

log2x+

1

5log

2y5=2 log46

log2x +

1

5(5 log

2y) =2

log26

log24

log2x+log2y=log26

log2xy=log26

xy=6

y=6

x (2)


15x 26x=2415x2 12 =24x

5x2 8x 4 =0

(x 2)(5x+2) =0

x=2 or 2

5

x=2

5is not accepted

x=2

Whenx=2, y=62

=3


12/26



37 (a) The graph ofy= f (x) = ln (x+ 1) is as

shownbelow.

1 O

y

x

y= f(x) = ln (x + 1)

f1exists becausefis a one-to-oneand

an ontofunction.

(b) Lety=f1(x)f (y) =x

ln (y+1) =x

y+1 =ex

y=ex 1

f1(x) =ex1

The domain off1is the same as the

range off, i.e. {x:xR}.

The range off1is the same as the

domain off, i.e. {y:y1,yR}.

(c) gf1

=g[f1

(x)]=g(ex 1)

= ex 1 + 1

=e1

2x

The domain ofgf1is the same as the

domain off1, i.e. {x:xR}.

The range ofgf1is {y:y>0,yR}.

x

y

y =e

O

1

1

2x

38 (a) fg=f [g(x)]

=f lnx 12 =1 +2e

lnx 1

2

=1 +2

x 1

2 =x

Since it is known thatf f 1(x) =x, by

comparisonf 1(x) =g(x) = ln x 12 . (b) The domain off 1is the same as the

range off, i.e. {x:x1,xR}.

The range off

1

is the same as thedomain off, i.e. {y:yR}.

(c) The graphs ofy=f (x) andy=f 1(x)

are as shown below.

1

1

O

y

x

y= f(x) = 1+ 2ex

3

y=

x

3

y= f1(x) = lnx 1

2 )(

39 (a) The graph ofy=|sinx| is as shownbelow.

y

x

1

O2pp

2

p 3p

2

y = sin x y = sin x

In the non-modulus form,y=|sin x| is

f(x) =sin x,

sin x,

0 xp,

px2p.

The graph ofy=sinxfor 0 x2p

is as shown below.

y

x

1

1

O2pp

2

p 3p

2


13/26



Hence, the functionf (x) =|sinx| sinx

in the non-modulus form is:

f(x) ={sinx sinx, 0 x


14/26



=

cosP (sinP cos Q+cosPsin Q)

sinP(cosPcos Q sinPsin Q)cosPcos (P+Q)

=

cosP sinP cos Q+cos2Psin Q

sinPcosPcos Q +sin2Psin Q

cosPcos (P+Q)

=cos2P sin Q +sin2Psin Q

cosPcos (P+Q)

=sin Q (cos2P +sin2P)

cosPcos (P+Q)

=sin Q (l)

cosPcos (P+Q)

=sin Q

cosPcos (P+Q)

=RHS

tan (P+Q)tanP

sinQ

cosPcos (P+Q) [Proven]

43 LHS =cosec 2q cot 2q

=1

sin 2q

cos 2q

sin 2q

=1 cos 2q

sin 2q

=1 (1 2 sin2q)

sin 2q

=2 sin2q

2 sin q cos q

=sin q

cos q=tan q

=RHS

cosec 2qcot 2qtan q [Proven]tan 22.5 =cosec 2(22.5) cot 2(22.5)

=1

sin 45

1

tan 45=

11

2

1

1

= 2 1 [Shown]

44 LHS =sin 3q

sin q

cos 3q

cos q

=3 sin q 4 sin3q

sin q

4 cos3

q

3 cosq

cos q

=3 4 sin2q 4 cos2q+3

=6 4(sin2q +cos2q)=6 4(1)

=2

=RHS

sin 3q

sin q

cos 3q

cosq2 [Proven]

45 LHS =

cos5

2A+cos

3

2A

sin3

2A+sin

A

2

=

2 cos1

252A+

3

2Acos 12

5

2A

3

2A

2 sin1

232A+

1

2Acos 12

3

2A

1

2A

=

2 cos 2Acos1

2A

2 sinAcos1

2A

=cos 2A

sinA

=cos2A sin2A

sinA

=cos2A

sinA

sin2A

sinA

=cosA

sinAcosA sinA

=cotA cosA sinA

=RHS

cos5

2A+cos

3

2A

sin3

2A+sin

1

2A

cotA cosA sinA

[Proven]

46 LHS = cos 2A 2 cos 4A +cos 6Acos 2A +2 cos 4A +cos 6A

=cos 6A+cos 2A2cos 4A

cos 6A +cos 2A +2 cos 4A

=

2 cos 6A +2A2 cos 6A 2A

2 2 cos 4A

2 cos 6A +2A2 cos 6A 2A

2 +2 cos 4A

= 2 cos 4Acos 2A2cos 4A2 cos 4Acos 2A +2 cos 4A


15/26



=2 cos 4A(cos 2A 1)

2 cos 4A (cos 2A+1)

=cos 2A1

cos 2A +1

=1 2 sin2A1

2 cos2A 1 +1

=2sin2A2 cos2A

=sin2A

cos2A

= tan2A

=RHS

cos 2A 2 cos 4A +cos 6A

cos 2A +2 cos 4A +cos 6A tan2A

[Proven]

47 LHS =

cosx+cos 2x cos 3x cos 4x

sinx +sin 2x+sin 3x+sin 4x

=cos 2x+cosx (cos 4x+cos 3x)

sin 2x +sinx+(sin 4x+sin 3x)

=

2 cos2x +x2 cos2xx

2 2 cos4x +3x2 cos

4x 3x

2

2 sin2x +x2 cos 2xx

2

+2 sin

4x +3x

2 cos

4x 3x

2 =

2 cos3x2cosx

2 2 cos7x

2cosx

22 sin3x2cos

x

2+2 sin7x

2cosx

2

/ /

/ /

=cos3x2 cos

7x

2sin 3x2+sin

7x

2

=

2 sin

3x +7x

4 sin

3x 7x

4 2 sin3x +7x4 cos

3x 7x

4 =

sin (x)

cos (x)

=sinx

cosx

=tanx

=RHS

cosx +cos 2x cos 3x cos 4x

sinx+sin 2x+sin 3x+sin 4x

tanx

[Proven]

48 SinceA,Band Care angles of a triangle,

thenA+B+C=180.

(a) tanA+tanB+tan C

=sinA

cosA+

sinB

cosB+

sin C

cos C

=sinAcosBcos C+sinBcosAcos C+sin CcosAcosB

cosAcosBcos C

=

cos C(sinAcosB+sinBcosA)

+sin CcosAcosB

cosAcosBcos C

=cos C[sin (A+B)]+sin CcosAcosB

cosAcosBcos C

=

cos C[sin (180 C)]+sin CcosAcosB

cosAcosBcos C

=cos Csin C+sin CcosAcosB

cosAcosBcos C

=sin C(cos C+cosAcosB)

cosAcosBcos C

=

sin C{cos [180 (A +B)]+cosAcosB}

cosAcosBcos C

= sin C{ cos (A +B) +cosAcosB}cosAcosBcos C

=

sin C(sinA sinB cosAcosB+cosAcosB)

cosAcosBcos C

=sinA sinBsin C

cosAcosBcos C

=tanAtanBtan C [Shown]

(b) sin 2A+sin 2B+sin 2C

=sin 2A

+sin 2C

+sin 2B

=2 sin 2A +2C2 cos2A 2C

2 +sin 2B=2 sin (A+C) cos (A C) +sin 2B

=2 sin (180 B)cos (A C) +sin 2B

=2 sinBcos (A C) +sin 2B

=2 sinBcos (A C) +2 sinBcosB

=2 sinB[cos (A C) +cosB]

=2 sinB2 cosA C +B2

cos A CB2


16/26



=2 sinB2 cos A +B C2 cos A (B+C)2

=2 sinB2 cos 180 C C2 cos A (180 A)2

=2 sinB2 cos 180 2C2 cos 2A 1802

=4 sinB[cos (90 C)cos (A 90)]

=4 sinB[sin CsinA]=4 sinAsinBsin C [Shown]

49 (a) LHS =2 sin A+p

4cos A+p

4=sin 2 A+p4=sin 2A+p2=sin 2Acos

p

2+cos 2Asin

p

2

=(sin 2A)(0) +(cos 2A)(1)

=cos 2A

=RHS

2 sin A+p4cos A+p

4cos 2A[Proven]

(b) LHS =2 cos B+p4cos Bp

4=cos B+p4+B

p

4+

cos B+p4 Bp

4

=cos 2B+cos

p

2

=cos 2B+ 0

=cos 2B

=RHS

2 cos B+p4cos Bp

4cos 2B[Proven]

50 cos 3x=cos2x

4 cos3x 3 cosx=cos2x

4 cos

3

x cos

2

x 3 cosx=0cosx(4 cos2x cosx 3)=0

cosx(4 cosx+3)(cosx 1) =0

cosx=0, 3

4, 1

When cosx=0,

x=90, 270

When cosx= 34,

x=138.6, 221.4

When cosx=1,

x=0, 360

x=0, 90, 138.6, 221.4, 270, 360

51 sin 3q+sin2q=2

sin 3q=3 sin q 4 sin3q

(3 sin q 4 sin3

q)+sin2

q 2 =04 sin3q sin2q 3 sin q+2 =0

(sin q+1)(4 sin2q 5 sin q+2) =0

sin q+1 =0 or 4 sin2q 5 sin q+2 =0

When sin q+1 =0

sin q=1

q=270

For 4 sin2q 5 sin q+2 =0, there are no

real roots because b2 4ac=(5)2 4(4)(2)

=7 (


17/26



When cos 2x=1

2,

2x=60, 300

x=30, 150

x=0, 30, 60, 120, 150, 180

x=0,

1

6p,

1

3p,

2

3p,

5

6p, p

x =x180

prad.

53 (a) 3 sin1x+cos1x=p

3 sin1x+p2 sin1x=p

2 sin1x=pp

2

2 sin1x=p

2

sin1x=p

4

x=sinp

4

x=1

2

(b) cos1x+sin1x=sin132 x

p2 sin1x+sin1x=sin132 x

p

2=sin132 x

sinp

2=

3

2 x

1 =3

2 x

x=12

54 tanx=

2 tanx

2

1 tan2x

2

Letting t=tanx

2,

tanx=2t

1 t2

From the right-angled triangle above,

x

1 t2

2t

1 +t2

cosx=1 t2

1 +t2

cotx=1

tanx

=1 t2

2t

cotx=cosx

1 t2

2t=

1 t2

1 +t2

(1 t2)(1 +t2) =(1 t2)(2t)

(1 t2)(1 +t2) (1 t2)(2t) =0

(1 t2)(1 +t2 2t) =0

(1 t2)(t2 2t+1) =0

(1 t2)(t 1)2=0

(1 +t)(1 t)(1 t)2=0

(1 +t)(1 t)3=0 [Shown]

1 +t= 0 or (1 t)3=0

t=1 t=1

When t=1, When t=1,

tanx

2=1 tan

x

2=1

basic =45 basic=45

x

2=180 45

x

2=45

x

2

=135 x=90

x=270

x=90, 270

55

q

1 t2

2t

1 +t2

(a) LHS =cosec q cot q=

1 +t2

2t

1 t2

2t

=1 +t2 1 +t2

2t

=2t2

2t

=t

=tanq

2

=RHS

cosecq cot qtanq

2


18/26



(b) LHS =sec q tan q

=1 +t2

1 t2

2t

1 t2

=1 +t2 2t

1 t2

=t2 2t+1

1 t2

=(t 1)2

(1 +t)(1 t)

=(1 t)2

(1 +t)(1 t)

=1 t

1 +t

=

tanp

4 tan

q

2

1 +tanp

4tanq

2

=tan p4q

2=RHS

sec q tan qtan p4q

2cosec q cot q=sec q tan q

tanq

2=tan p4

q

2q

2=p

4 q

2or q

2=p+p4 q2

q =p

4 or q=

5

4p

56

q

1 t2

2t

1 +t2

1 +sin q+cos q

1 +cos q

=1 +cos q

1 +cos q+

sin q

1 +cos q

=1 +

2t

1 +t2

1 +1 t2

1 +t2

=1 + 2t1 +t2+1 t2

First quadrant.

Third quadrant.

=1 +2t

2

=1 +t

=1 +tanq

2

0

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