Nama :........................ N P M .........................: Kelas .........................: Created by Abdul
Nama :..................................................................N P M :..................................................................Kelas :..................................................................
Created by Abdul Muiz., S.Pd.,
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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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
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Matematika Dikrit STKIP PGRI Bangkalan
b.f ( x )= 1
( 12 ) + x d.
f ( x )= −1
( 12 ) − x
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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 ±⋯ )
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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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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)
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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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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 ±⋯ )
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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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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 + ⋯)
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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 ± ⋯ )
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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
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:
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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
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
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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
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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 + ⋯)
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