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Page 1: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Nama :..................................................................N P M :..................................................................Kelas :..................................................................

Created by Abdul Muiz., S.Pd.,

Page 2: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Matematika Dikrit STKIP PGRI Bangkalan

KUMPULAN SOAL FORMULA DISKRIT BAGIAN KEDUAOleh : ABDUL MUIZ., S.Pd., M.Pd.,

SOAL 01.

a.f ( x )= 1

1 − x

b.f ( x )= 1

1 + x

c.f ( x )= −1

1 + x

d.f ( x )= −1

1 − x

SOAL 02.

a.f ( x )= 1

1 − 2 x

b.f ( x )= 1

1 + 2x

c.f ( x )= −1

1 + 2 x

d.f ( x )= −1

1 − 2 x

SOAL 03.

a.f ( x )= 1

2 − x

b.f ( x )= 1

2 + x

c.f ( x )= −1

2 + x

d.f ( x )= −1

2 − x

SOAL 04.

a.f ( x )= 2

1 − x

b.f ( x )= 2

1 + x

c.f ( x )= −2

1 + x

d.f ( x )= −2

1 − x

SOAL 05.

a.f ( x )= 1

2 − 3 x c.f ( x )= −1

2 + 3x

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Page 3: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

b.f ( x )= 1

2 + 3 x d.f ( x )= −1

2 − 3 xSOAL 06.

a.f ( x )= 2

3 − x

b.f ( x )= 2

3 + x

c.f ( x )= −2

3 + x

d.f ( x )= −2

3 − x

SOAL 07.

a.f ( x )= 2

1 − 3 x

b.f ( x )= 2

1 + 3x

c.f ( x )= −2

1 + 3x

d.f ( x )= −2

1 − 3 x

SOAL 08.

a.f ( x )= 2

3 − 4 x

b.f ( x )= 2

3 + 4 x

c.f ( x )= −2

3 + 4 x

d.f ( x )= −2

3 − 4 x

SOAL 09.

a.f ( x )= 1

1 − ( 12 x )

b.f ( x )= 1

1 + ( 12 x)

c.f ( x )= −1

1 + ( 12 x)

d.f ( x )= −1

1 − ( 12 x )

SOAL 10.

a.f ( x )= 1

( 12 ) − x c.

f ( x )= −1

( 12 ) + x

3 | Page

Page 4: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Matematika Dikrit STKIP PGRI Bangkalan

b.f ( x )= 1

( 12 ) + x d.

f ( x )= −1

( 12 ) − x

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Page 5: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

KUNCI JAWABAN SOAL FORMULA DISKRIT BAGIAN KEDUAOleh : ABDUL MUIZ., S.Pd., M.Pd.,

SOAL 01.

a.f ( x )= 1

1 − x

Jawaban:

f ( x )= 11 − x P( x ) = ( 1 ) (∑n = 0

( x )n)= ( 1 ) (1 + x + x2 + x3 + ⋯)

f ( x )= 1

1 − x P( x ) = 1 + x + x2 + x3 + ⋯

b.f ( x )= 1

1 + x

Jawaban:

f ( x )= 11 + x P( x ) = ( 1 ) (∑n = 0

(−x )n)=( 1 ) ( 1 − x + x2 − x3 ±⋯ )

f ( x )= 1

1 + x P( x )= 1 − x + x2 − x3 ± ⋯

c.f ( x )= −1

1 + x

Jawaban:

f ( x )= −11 + x P( x ) = ( − 1 ) (∑n = 0

(−x )n)=( − 1 ) ( 1 − x + x2 − x3 ±⋯ )

5 | Page

Page 6: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Matematika Dikrit STKIP PGRI Bangkalan

f ( x )= 1

1 − x P( x ) = −1 + x − x2 + x3 ∓ ⋯

d.f ( x )= −1

1 − x

Jawaban:

f ( x )= −11 − x P( x ) = ( − 1 ) (∑n = 0

( x )n)= ( − 1 ) (1 + x + x2 + x3 + ⋯)

f ( x )= −1

1 − x P( x ) = −1 − x − x2 − x3 − ⋯

SOAL 02.

a.f ( x )= 1

1 − 2x

Jawaban:

f ( x )= 11 − 2x P( x ) = ( 1 ) (∑n = 0

(2x )n)= ( 1 ) (1 + 2x + 4 x2 + 8 x3 +⋯)

f ( x )= 1

1 − 2x P( x ) = 1 + 2 x + 4 x2 + 8 x3 + ⋯

b.f ( x )= 1

1 + 2 xJawaban:

f ( x )= 11 + 2 x P( x ) = ( 1 ) (∑n = 0

(− 2 x )n)= ( 1 ) (1 − 2x + 4 x2 − 8 x3 ±⋯)

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Page 7: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

f ( x )= 1

1 + 2 x P( x ) = 1 − 2x + 4 x2 − 8 x3 ±⋯

c.f ( x )= −1

1 + 2 x

Jawaban:

f ( x )= −11 + 2 x P( x ) = ( − 1 ) (∑n = 0

(− 2 x )n)= ( − 1 ) (1 − 2x + 4 x2 − 8 x3 ±⋯)

f ( x )= −1

1 + 2 x P( x ) = − 1 + 2x − 4 x2 + 8 x3 ∓⋯

d.f ( x )= −1

1 − 2x

Jawaban:

f ( x )= −11 − 2x P( x ) = ( − 1 ) (∑n = 0

(2x )n)= ( − 1 ) (1 + 2x + 4 x2 + 8 x3 +⋯)

f ( x )= −1

1 − 2x P( x ) = − 1 − 2 x − 4 x2 − 8 x3 − ⋯

SOAL 03.

a.f ( x )= 1

2 − x

Jawaban:

f ( x )= 12 − x P( x ) = (

12 )(∑n = 0

( 12x )n)

7 | Page

Page 8: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Matematika Dikrit STKIP PGRI Bangkalan

= (12 ) (1 + 1

2x + 1

4x2 + 1

8x3 +⋯)

f ( x )= 1

2 − x P( x ) = (12 ) (1 + 1

2x + 1

4x2 + 1

8x3 +⋯)

b.f ( x )= 1

2 + x

Jawaban:

f ( x )= 12 + x P( x ) = (

12 )(∑n = 0

(− 12x )n)

= (12 ) (1 − 1

2x + 1

4x2 − 1

8x3 ± ⋯)

f ( x )= 1

2 + x P( x ) = (12 ) (1 − 1

2x + 1

4x2 − 1

8x3 ± ⋯)

c.f ( x )= −1

2 + x

Jawaban:

f ( x )= −12 + x P( x ) = (−

12 )(∑n = 0

(− 12x )n)

= (−12 ) (1 − 1

2x + 1

4x2 − 1

8x3 ± ⋯)

f ( x )= −1

2 + x P( x ) = (−12 ) (1 − 1

2x + 1

4x2 − 1

8x3 ± ⋯)

d.f ( x )= −1

2 − x

Jawaban:

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Page 9: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

f ( x )= −12 − x P( x ) = (−

12 )(∑n = 0

( 12x )n)

= (−12 ) (1 + 1

2x + 1

4x2 + 1

8x3 +⋯)

f ( x )= −1

2 − x P( x ) = (−12 )

(1 + 1

2x + 1

4x2 + 1

8x3 +⋯)

SOAL 04.

a.f ( x )= 2

1 − x

Jawaban:

f ( x )= 21 − x P( x ) = ( 2 )(∑n = 0

( x )n)= ( 2 ) ( 1 + x + x2 + x3 + ⋯ )

f ( x )= 2

1 − x P( x ) = ( 2 ) ( 1 + x + x2 + x3 + ⋯ )

b.f ( x )= 2

1 + x

Jawaban:

f ( x )= 21 + x P( x ) = ( 2 )(∑n = 0

(− x )n)= ( 2 ) ( 1 − x + x2 − x3 ±⋯ )

f ( x )= 2

1 + x P( x ) = ( 2 ) ( 1 − x + x2 − x3 ±⋯ )

9 | Page

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Matematika Dikrit STKIP PGRI Bangkalan

c.f ( x )= −2

1 + x

Jawaban:

f ( x )= −21 + x P( x ) = ( − 2 )(∑n = 0

(− x )n)= ( − 2 ) ( 1 − x + x2 − x3 ±⋯ )

f ( x )= −2

1 + x P( x ) = ( − 2 ) ( 1 − x + x2 − x3 ±⋯ )

d.f ( x )= −2

1 − x

Jawaban:

f ( x )= −21 − x P( x ) = ( − 2 )(∑n = 0

( x )n)= ( − 2 ) ( 1 + x + x2 + x3 + ⋯ )

f ( x )= −2

1 − x P( x ) = ( − 2 ) ( 1 + x + x2 + x3 + ⋯ )

SOAL 05.

a.f ( x )= 1

2 − 3 x

Jawaban:

f ( x )= 12 − 3 x P( x ) = (

12 )(∑n = 0

( 32x )n)

= (12 ) (1 + 3

2x + 9

4x2 + 27

8x3 + ⋯)

f ( x )= 1

2 − 3 x P( x ) = (12 ) (1 + 3

2x + 9

4x2 + 27

8x3 + ⋯)

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Page 11: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

b.f ( x )= 1

2 + 3x

Jawaban:

f ( x )= 12 + 3x P( x ) = (

12 )(∑n = 0

(−32x )n)

= (12 ) (1 − 3

2x + 9

4x2 − 27

8x3 ± ⋯)

f ( x )= 1

2 + 3x P( x ) = (12 ) (1 − 3

2x + 9

4x2 − 27

8x3 ± ⋯)

c.f ( x )= −1

2 + 3x

Jawaban:

f ( x )= −12 + 3x P( x ) = (−

12 )(∑n = 0

(−32x )n)

= (−12 ) (1 − 3

2x + 9

4x2 − 27

8x3 ± ⋯)

f ( x )= −1

2 + 3x P( x ) = (−12 ) (1 − 3

2x + 9

4x2 − 27

8x3 ± ⋯)

d.f ( x )= −1

2 − 3 x

Jawaban:

f ( x )= −12 − 3 x P( x ) = (−

12 )(∑n = 0

( 32x )n)

= (−12 ) (1 + 3

2x + 9

4x2 + 27

8x3 + ⋯)

f ( x )= −1

2 − 3 x P( x ) = (−12 )

(1 + 3

2x + 9

4x2 + 27

8x3 + ⋯)

11 | Page

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Matematika Dikrit STKIP PGRI Bangkalan

SOAL 06.

a.f ( x )= 2

3 − x

Jawaban:

f ( x )= 23 − x P( x ) = (

23 )(∑n = 0

( 13x )n)

= (23 ) ( 1 + 1

3x + 1

9x2 + 1

27x3 +⋯ )

f ( x )= 2

3 − x P( x ) = (23 ) ( 1 + 1

3x + 1

9x2 + 1

27x3 +⋯ )

b.f ( x )= 2

3 + x

Jawaban:

f ( x ) = 23 + x P( x ) = (

23 )(∑n = 0

(− 13x)n)

= (23 ) ( 1 − 1

3x + 1

9x2 − 1

27x3 ± ⋯ )

f ( x ) = 2

3 + x P( x ) = (23 ) ( 1 − 1

3x + 1

9x2 − 1

27x3 ± ⋯ )

c.f ( x )= −2

3 + x

Jawaban:

f ( x ) = −23 + x P( x ) = (−

23 )(∑n = 0

(− 13x)n)

= (−23 ) ( 1 − 1

3x + 1

9x2 − 1

27x3 ± ⋯ )

Page | 12

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Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

f ( x ) = −2

3 + x P( x ) = (−23 ) ( 1 − 1

3x + 1

9x2 − 1

27x3 ± ⋯ )

d.f ( x )= −2

3 − x

Jawaban:

f ( x )= −23 − x P( x ) = (−

23 )(∑n = 0

( 13x )n)

= (−23 )( 1 + 1

3x + 1

9x2 + 1

27x3 +⋯ )

f ( x )= −2

3 − x P( x ) = (−23 )( 1 + 1

3x + 1

9x2 + 1

27x3 +⋯ )

SOAL 07.

a.f ( x )= 2

1 − 3 x

Jawaban:

f ( x )= 21 − 3 x P( x ) = ( 2 )(∑n = 0

(3 x )n )= ( 2 ) ( 1 + 3x + 9 x2 + 27 x3 + ⋯ )

f ( x )= 2

1 − 3 x P( x ) = ( 2 ) ( 1 + 3x + 9 x2 + 27 x3 + ⋯ )

b.f ( x )= 2

1 − 3 x

Jawaban:

f ( x )= 21 − 3 x P( x ) = ( 2 )(∑n = 0

(−3 x )n)13 | Page

Page 14: Web viewMatematika Dikrit STKIP PGRI Bangkalan. Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Dosen Pembina – Abdul Muiz., S.Pd., M.Pd., Matematika Dikrit STKIP PGRI Bangkalan

Matematika Dikrit STKIP PGRI Bangkalan

= ( 2 ) ( 1 − 3 x + 9 x2 − 27 x3 ±⋯ )

f ( x )= 2

1 − 3 x P( x ) = ( 2 ) ( 1 − 3 x + 9 x2 − 27 x3 ±⋯ )

c.f ( x )= −2

1 − 3 x

Jawaban:

f ( x )= −21 − 3 x P( x ) = ( − 2 )(∑n = 0

(−3 x )n)= ( − 2 ) ( 1 − 3 x + 9 x2 − 27 x3 ±⋯ )

f ( x )= −2

1 − 3 x P( x ) = ( − 2 ) ( 1 − 3 x + 9 x2 − 27 x3 ±⋯ )

d.f ( x )= −2

1 − 3 x

Jawaban:

f ( x )= −21 − 3 x P( x ) = ( − 2 )(∑n = 0

(3 x )n )= ( − 2 ) ( 1 + 3x + 9 x2 + 27 x3 + ⋯ )

f ( x )= −2

1 − 3 x P( x ) = ( − 2 ) ( 1 + 3x + 9 x2 + 27 x3 + ⋯ )

SOAL 08.

a.f ( x )= 2

3 − 4 x

Jawaban:

Page | 14

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Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

f ( x )= 23 − 4 x P( x ) = (

23 )(∑n = 0

( 43x )n)

= (23 )

( 1 + 43x + 16

9x2 + 64

27x3 +⋯ )

f ( x )= 2

3 − 4 x P( x ) = (23 )

( 1 + 43x + 16

9x2 + 64

27x3 +⋯ )

b.f ( x )= 2

3 + 4 x

Jawaban:

f ( x )= 23 + 4 x P( x ) = (

23 )(∑n = 0

(− 43x )n)

= (23 )

( 1 − 43x + 16

9x2 − 64

27x3 ± ⋯ )

f ( x )= 2

3 + 4 x P( x ) = (23 )

( 1 − 43x + 16

9x2 − 64

27x3 ± ⋯ )

c.f ( x )= −2

3 + 4 x

Jawaban:

f ( x )= −23 + 4 x P( x ) = (−

23 )(∑n = 0

(− 43x )n)

= (−23 )

( 1 − 43x + 16

9x2 − 64

27x3 ± ⋯ )

f ( x )= −2

3 + 4 x P( x ) = (−23 )

( 1 − 43x + 16

9x2 − 64

27x3 ± ⋯ )

d.f ( x )= −2

3 − 4 x

Jawaban:15 | Page

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Matematika Dikrit STKIP PGRI Bangkalan

f ( x )= −23 − 4 x P( x ) = (−

23 )(∑n = 0

( 43x )n)

= (−23 )

( 1 + 43x + 16

9x2 + 64

27x3 +⋯ )

f ( x )= −2

3 − 4 x P( x ) = (−23 )

( 1 + 43x + 16

9x2 + 64

27x3 +⋯ )

SOAL 09.

a.f ( x )= 1

1 − 12 x

Jawaban:

f ( x )= 11 − 1

2 x P( x ) = ( 1 ) (∑n = 0

( 12 x)

n)= ( 1 ) (1 + 1

2x + 1

4x2 + 1

8x3 + ⋯)

f ( x )= 1

1 − 12 x P( x ) =

1 + 12x + 1

4x2 + 1

8x3 + ⋯

b.f ( x )= 1

1 + 12 x

Jawaban:

f ( x )= 11 + 1

2 x P( x ) = ( 1 ) (∑n = 0

(− 12 x )n)

= ( 1 ) (1 − 12x + 1

4x2 − 1

8x3 ± ⋯)

f ( x )= 1

1 + 12 x P( x ) =

1 − 12x + 1

4x2 − 1

8x3 ± ⋯

c.f ( x )= −1

1 + 12 x

Page | 16

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Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,

Jawaban:

f ( x )= −11 + 1

2 x P( x ) = ( − 1 ) (∑n = 0

(− 12 x )n)

= ( − 1 ) (1 − 12x + 1

4x2 − 1

8x3 ± ⋯)

f ( x )= −1

1 + 12 x P( x ) =

− 1 + 12x − 1

4x2 + 1

8x3 ∓ ⋯

d.f ( x )= − 1

1 − 12 x

Jawaban:

f ( x )= − 11 − 1

2 x P( x ) = ( − 1 ) (∑n = 0

( 12 x)

n)= ( − 1 ) (1 + 1

2x + 1

4x2 + 1

8x3 + ⋯)

f ( x )= − 1

1 − 12 x P( x ) =

− 1 − 12x − 1

4x2 − 1

8x3 −⋯

SOAL 10.

a.f ( x )= 1

12 − x

Jawaban: f ( x )= 1

12 − x P( x ) = ( 2 )(∑n = 0

( x )n)= ( 2 ) (1 + x + x2 + x3 + ⋯)

f ( x )= 1

12 − x P( x ) = ( 2 ) (1 + x + x2 + x3 + ⋯)

b.f ( x )= 1

12 + x

17 | Page

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Matematika Dikrit STKIP PGRI Bangkalan

Jawaban:f ( x )= 1

12 + x P( x ) = ( 2 )(∑n = 0

(− x )n)= ( 2 ) (1 − x + x2 − x3 + ⋯)

f ( x )= 1

12 + x P( x ) = ( 2 ) (1 − x + x2 − x3 + ⋯)

c.f ( x )= −1

12 + x

Jawaban:f ( x )= −1

12 + x P( x ) = ( − 2 )(∑n = 0

(− x )n)= ( − 2 ) (1 − x + x2 − x3 + ⋯)

f ( x )= −1

12 + x P( x ) = ( − 2 ) (1 − x + x2 − x3 + ⋯)

d.f ( x )= − 1

12 − x

Jawaban: f ( x )= − 1

12 − x P( x ) = ( − 2 )(∑n = 0

( x )n)= ( − 2 ) (1 + x + x2 + x3 + ⋯)

f ( x )= − 1

12 − x P( x ) = ( − 2 ) (1 + x + x2 + x3 + ⋯)

Page | 18