1. Diketahui fungsi f dan g dirumuskan oleh f ( x ) = 3x2 – 4x + 6 dan g ( x ) = 2x – 1. Jika nilai

( f o g )( x ) = 101, maka nilai x yang memenuhi adalah …

a. 3 dan –2

b. –3 dan 2

c. dan 2

d. –3 dan –2

e. – dan –2

2. Diketahui ( f o g )( x ) = 42x + 1. Jika g ( x ) = 2x – 1, maka f ( x ) = …

a. 4x + 2

b. 42x + 3

c. 24x + 1 +

d. 22x + 1 +

e. 22x + 1 + 1

3. Jika f ( x ) = dan ( f o g )( x ) = 2 , maka fungsi g adalah g ( x ) = …

a. 2x – 1

b. 2x – 3

c. 4x – 5

d. 4x + 3

e. 5x – 4

4. Ditentukan g ( f ( x ) ) = f ( g ( x ) ). Jika f ( x ) = 2x + p dan g ( x ) = 3x + 120, maka nilai

p = …

a. 30

b. 60

c. 90

d. 120

e. 150

5. Fungsi f : R R didefinisikan sebagai f ( x ) = –

, x .

Invers dari fungsi f adalah f–1 ( x ) = …

a. –

, x –

b. –

, x

c. –

, x

d. –

– , x

e. , x –

6. Diketahui f ( x – 1 ) = –

, x – dan f–1 ( x ) adalah invers dari f ( x ).

Rumus f–1 ( 2x – 1 ) = …

a. – –

, x –

b. –

– , x

c. –

, x –

d. –

, x

e. –

, x 2

7. Diketahui fungsi f ( x ) = 6x – 3, g ( x ) = 5x + 4, dan ( f o g )( a ) = 81. Nilai a = …

a. – 2

b. – 1

c. 1

d. 2

e. 3

8. Diketahui fungsi f ( x ) = 2x + 1 dan ( f o g )( x + 1 ) = –2x2 – 4x – 1. Nilai g (– 2 ) = …

a. – 5

b. – 4

c. – 1

d. 1

e. 5

9. Diketahui f ( x ) = –

, x – . Jika f–1 ( x ) adalah invers fungsi f, maka f–1 ( x – 2 ) = …

a. –

– , x

b. – –

– , x

c. –

, x

d. , x

e. –

, x

PEMBAHASAN:

1. Jawab: A

f ( x ) = 3x2 – 4x + 6 ; g ( x ) = 2x – 1 ; ( f o g )( x ) = 101

( f o g )( x ) f ( g ( x ) ) = 101

3 ( 2x – 1 )2 – 4 ( 2x – 1 ) + 6 = 101

12x2 – 12x + 3 – 8x + 10 = 101

12x2 – 20x – 88 = 0

3x2 – 5x – 22 = 0

( x + 2 ) ( 3x – 11 ) = 0

x = –2 x = = 3

2. Jawab: A

( f o g )( x ) = 42x + 1 ; g ( x ) = 2x – 1

Misal f ( x ) = Aax + b

( f o g )( x ) f ( g ( x ) ) = 42x + 1

Aa( 2x – 1 ) + b = 42x + 1

A2ax – a + b = 42x + 1

A = 4 ; 2ax =2x x = 1 ; – a + b = 1 b = 2

Sehinnga f ( x ) = 4x + 2

3. Jawab: D

f ( x ) = ; ( f o g )( x ) = 2

Misal g ( x ) = ax + b

( f o g )( x ) f ( g ( x ) ) = 2

= 2

=

ax + b + 1 = 4x + 4

ax = 4x a = 4 ; b + 1 = 4 b = 3

Sehingga g ( x ) = 4x + 3

4. Jawab: B

g ( f ( x ) ) = f ( g ( x ) ) ; f ( x ) = 2x + p ; g ( x ) = 3x + 120

g ( f ( x ) ) = f ( g ( x ) ) 3( 2x + p ) + 120 = 2( 3x + 120 ) + p

6x+ 3p + 120 = 6x + 240 + p

2p = 120

p = 60

5. Jawab: C

f ( x ) = –

, x a = 2 ; b = –1 ; c = 3 ; d = 4

f–1 ( x ) = –

–

– –

–

–

–

–

6. Jawab: B

f ( x – 1 ) = –

, x – ; f–1 ( x ) adalah invers dari f ( x ). Rumus f–1 ( 2x – 1 ) = …

Mencari f ( x ):

Misal x – 1 = y, sehingga x = y + 1

f ( y ) = –

= f ( x ) =

Mencari f–1 ( x ):

f ( x ) = = a = 1 ; b = 2 ; c = 2 ; d = 1

f–1 ( x ) = –

– =

–

–

Mencari f–1 ( 2x – 1 ):

f–1 ( 2x – 1 ) = – –

– – =

–

–

5

7. Jawab: D

f ( x ) = 6x – 3 ; g ( x ) = 5x + 4 ; ( f o g )( a ) = 81

( f o g )( a ) f ( g ( a ) ) = 81

6( 5a + 4 ) – 3 = 81

30a + 24 = 84

30a = 60

a = 2

8. Jawab: B

f ( x ) = 2x + 1 ; ( f o g )( x + 1 ) = –2x2 – 4x – 1

Mencari ( f o g )( x ):

Misal x + 1 = y x = y – 1

( f o g )( y ) = –2( y – 1 )2 – 4( y – 1 ) – 1

= –2( y2 – 2y + 1 ) – 4y + 4 – 1

= –2y2 + 4y – 2 – 4y + 3

= –2y2 + 1

( f o g )( x ) = –2x2 + 1

Mencari g ( x ):

Misal g ( x ) = ax2 + bx + c

( f o g )( x ) = f ( g ( x ) ) 2 ( ax2 + bx + c ) + 1 = –2x2 + 1

2ax2 + 2bx + 2c + 1 = –2x2 + 1

2ax2 = –2x2 sehingga a = –1

2bx = 0 sehingga b = 0

2c + 1 = 1 sehingga c = 0

Rumus g ( x ) = –x2

Sehinnga g ( –2 ) = –( –2 )2 = –4

9. Jawab: A

f ( x ) = –

, x – a = –3 ; b = 2 ; c = 4 ; d = 1

f–1 ( x ) = –

– =

– f–1 ( x – 2 ) =

– –

– =

–

–