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Created by Abdul Muiz., S.Pd.,
Matematika Dikrit STKIP PGRI Bangkalan
KUMPULAN SOAL FORMULA DISKRIT BAGIAN PERTAMAOleh : ABDUL MUIZ., S.Pd., M.Pd.,
SOAL 01.
a. f ( x ) = ex
b. f ( x ) = e−xc. f ( x ) = −ex
d. f ( x ) = −e−x
SOAL 02.
a. f ( x ) = e2 x
b. f ( x ) = e−2 x
c. f ( x ) = −e2 x
d. f ( x ) = −e−2 x
SOAL 03.a. f ( x ) = 3ex c. f ( x ) = −3 ex
b. f ( x ) = 3e−x d. f ( x ) = −3 e−x
SOAL 04.a. f ( x ) = 3e2 x c. f ( x ) = −3 e2 x
b. f ( x ) = 3e−2 x d. f ( x ) = −3 e−2 x
SOAL 05.
a. f ( x ) = e25x c. f ( x ) = −e
25x
b. f ( x ) = e−
25x d. f ( x ) = −e
−25x
SOAL 06.
a. f ( x ) = 34ex c. f ( x ) = − 3
4ex
b. f ( x ) = 34e−x d. f ( x ) = − 3
4e−x
Page | 2
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
SOAL 07.
a. f ( x ) = 3e25x c. f ( x ) = −3 e
25x
b. f ( x ) = 3 e−
25x d. f ( x ) = −3 e
−25x
SOAL 08.
a. f ( x ) = 34e2 x c. f ( x ) = − 3
4e2 x
b. f ( x ) = 34e−2 x d. f ( x ) = − 3
4e−2 x
SOAL 09.
a. f ( x ) = 34e
25x
c. f ( x ) = − 34e
25x
b. f ( x ) = 34e−
25x
d. f ( x ) = − 34e−
25x
3 | Page
Matematika Dikrit STKIP PGRI Bangkalan
KUNCI JAWABAN SOAL FORMULA DISKRIT BAGIAN PERTAMAOleh : ABDUL MUIZ., S.Pd., M.Pd.,
SOAL 01.
a. f ( x ) = ex
Jawaban:
f ( x ) = ex P( x ) = ( 1 ) [ ∑n =0
∞ xn
n ! ]= ( 1 ) [1 + x + x
2
2+ x
3
6+⋯]
f ( x ) = ex P( x ) = 1 + x + x2
2+ x3
6+ ⋯
b. f ( x ) = e−x
Jawaban:
f ( x ) = e−x P( x ) = ( 1 ) [ ∑n =0
∞ (−x )n
n ! ]= ( 1 ) [1 − x + x
2
2− x3
6± ⋯]
f ( x ) = e−x P( x ) = 1 − x + x2
2− x3
6± ⋯
c. f ( x ) = −ex
Jawaban:
f ( x ) = −ex P( x ) = ( −1 ) [ ∑n =0
∞ xn
n ! ]= ( −1 ) [1 + x + x
2
2+ x
3
6+⋯]
Page | 4
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
f ( x ) = −ex P( x ) = −1 − x − x2
2− x3
6− ⋯
d. f ( x ) = −e−x
Jawaban:
f ( x ) = −e−x P( x ) = ( −1 ) [ ∑n =0
∞ (−x )n
n ! ]= ( −1 ) [1 − x + x2
2− x3
6± ⋯]
f ( x ) = −e−x P( x ) = −1 + x − x2
2+ x3
6∓ ⋯
SOAL 02.
a. f ( x ) = e2 x
Jawaban:
f ( x ) = e2 x P( x ) = ( 1 ) [ ∑n =0
∞ (2 x )n
n ! ]= ( 1 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
f ( x ) = e2 x P( x ) =
1 + 2x + 4 x2
2+ 8x3
6+ ⋯
b. f ( x ) = e−2 x
Jawaban:
f ( x ) = e−2 x P( x ) = ( 1 ) [ ∑n = 0
∞ (−2 x )n
n ! ]= ( 1 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
5 | Page
Matematika Dikrit STKIP PGRI Bangkalan
f ( x ) = e−2 x P( x ) =
1 − 2x + 4 x2
2− 8 x3
6± ⋯
c. f ( x ) = −e2 x
Jawaban:
f ( x ) = −e2 x P( x ) = ( −1 ) [ ∑n =0
∞ (2 x )n
n ! ]= ( −1 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
f ( x ) = −e2 x P( x ) =
−1 − 2 x − 4 x2
2− 8 x3
6− ⋯
d. f ( x ) = −e−2 x
Jawaban:
f ( x ) = −e−2 x P( x ) = ( −1 ) [ ∑n =0
∞ (−2 x )n
n ! ]= ( −1 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
f ( x ) = −e−2 x P( x ) =
−1 + 2x − 4 x2
2+ 8 x3
6∓ ⋯
SOAL 03.
a. f ( x ) = 3ex
Jawaban:
f ( x ) = 3ex P( x ) = ( 3 ) [ ∑n =0
∞ xn
n ! ]Page | 6
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
= ( 3 ) [1 + x + x2
2+ x
3
6+⋯]
f ( x ) = 3ex P( x ) = ( 3 ) [1 + x + x2
2+ x3
6+ ⋯]
b. f ( x ) = 3e−x
Jawaban:
f ( x ) = 3e−x P( x ) = ( 3 ) [ ∑n =0
∞ (−x )n
n ! ]= ( 3 ) [1 − x + x2
2− x3
6± ⋯]
f ( x ) = 3e−x P( x ) = ( 3 ) [1 − x + x2
2− x3
6± ⋯]
c. f ( x ) = −3 ex
Jawaban:
f ( x ) = −3 ex P( x ) = ( −3 ) [ ∑n =0
∞ xn
n ! ]= ( −3 ) [1 + x + x
2
2+ x
3
6+⋯]
f ( x ) = −3 ex P( x ) = ( −3 ) [1 + x + x2
2+ x3
6+ ⋯]
d. f ( x ) = −3 e−x
Jawaban:
7 | Page
Matematika Dikrit STKIP PGRI Bangkalan
f ( x ) = −3 e−x P( x ) = ( −3 ) [ ∑n = 0
∞ (−x )n
n ! ]= ( −3 ) [1 − x + x2
2− x3
6± ⋯]
f ( x ) = −3 e−x P( x ) = ( −3 ) [1 − x + x2
2− x3
6± ⋯]
SOAL 04.
a. f ( x ) = 3e2 x
Jawaban:
f ( x ) = 3e2 x P( x ) = ( 3 ) [ ∑n = 0
∞ (2 x )n
n ! ]= ( 3 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
f ( x ) = 3e2 x P( x ) = ( 3 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
b. f ( x ) = 3e−2 x
Jawaban:
f ( x ) = 3e−2 x P( x ) = ( 3 ) [ ∑n =0
∞ (−2 x )n
n ! ]= ( 3 ) [1 − 2 x + 4 x2
2− 8 x3
6± ⋯]
f ( x ) = 3e−2 x P( x ) = ( 3 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
c. f ( x ) = −3 e2 x
Page | 8
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
Jawaban
f ( x ) = −e2 x P( x ) = ( −3 ) [ ∑n = 0
∞ (2 x )n
n ! ]= ( −3 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
f ( x ) = −e2 x P( x ) = ( −3 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
d. f ( x ) = −3 e−2 x
Jawaban:
f ( x ) = −3 e−2 x P( x ) = ( −3 ) [ ∑n =0
∞ (−2 x )n
n ! ]= ( −3 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
f ( x ) = −3 e−2 x P( x ) = ( −3 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
SOAL 05.
a. f ( x ) = e25x
Jawaban:
f ( x ) = e25x
P( x ) = ( 1 ) [ ∑n = 0
∞ ( 25 x )n
n ! ]= ( 1 ) [1 + 2x
5+ 4 x2
50+ 8 x3
750+ ⋯]
f ( x ) = e25x
P( x ) = 1 + 2 x
5+ 4 x2
50+ 8 x3
750+ ⋯
9 | Page
Matematika Dikrit STKIP PGRI Bangkalan
b. f ( x ) = e−
25x
Jawaban:
f ( x ) = e−
25x
P( x ) = ( 1 ) [ ∑n = 0
∞ (−25 x )nn ! ]
= ( 1 ) [1 − 2 x5
+ 4 x2
50− 8 x3
750± ⋯]
f ( x ) = e−
25x
P( x ) = 1 − 2x
5+ 4 x2
50− 8x3
750± ⋯
c. f ( x ) = −e25x
Jawaban:
f ( x ) = −e25x
P( x ) = ( −1 ) [ ∑n =0
∞ ( 25 x )n
n ! ]= ( −1 ) [1 + 2x
5+ 4 x2
50+ 8 x3
750+ ⋯]
f ( x ) = −e25x
P( x ) = −1 − 2x
5− 4 x2
50− 8 x3
750− ⋯
d. f ( x ) = −e−
25x
Jawaban
f ( x ) = −e−
25x
P( x ) = ( −1 ) [ ∑n = 0
∞ (−25 x )nn ! ]
= ( −1 ) [1 − 2 x5
+ 4 x2
50− 8 x3
750± ⋯]
f ( x ) = −e−
25x
P( x ) = −1 + 2 x
5− 4 x2
50+ 8 x3
750∓ ⋯
Page | 10
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
SOAL 06.
a.f ( x ) = 3
4ex
Jawaban:
f ( x ) = 34ex
P( x ) = (34 ) [ ∑n =0
∞ xn
n ! ]= (
34 )
[1 + x + x2
2+ x
3
6+⋯]
f ( x ) = 3
4ex
P( x ) = (34 ) [1 + x + x2
2+ x3
6+ ⋯]
b.f ( x ) = 3
4e−x
Jawaban:
f ( x ) = 34e−x
P( x ) = (34 ) [ ∑n =0
∞ (−x )n
n ! ]= (
34 )
[1 − x + x2
2− x3
6± ⋯]
f ( x ) = 3
4e−x
P( x ) = (34 ) [1 − x + x
2
2− x3
6± ⋯]
c.f ( x ) = − 3
4ex
Jawaban:
f ( x ) = − 34ex
P( x ) = (−34 ) [ ∑n =0
∞ xn
n ! ]= (−
34 )
[1 + x + x2
2+ x
3
6+⋯]
11 | Page
Matematika Dikrit STKIP PGRI Bangkalan
f ( x ) = − 3
4ex
P( x ) = (−34 ) [−1 − x − x2
2− x3
6− ⋯]
d.f ( x ) = − 3
4e−x
Jawaban:
f ( x ) = − 34e−x
P( x ) = (−34 ) [ ∑n = 0
∞ (−x )n
n ! ]= (−
34 )
[1 − x + x2
2− x3
6± ⋯]
f ( x ) = − 3
4e−x
P( x ) = (−34 ) [−1 + x − x2
2+ x3
6∓ ⋯]
SOAL 07.
a. f ( x ) = 3e25x
Jawaban:
f ( x ) = 3e25x
P( x ) = ( 3 ) [ ∑n =0
∞ ( 25 x )nn ! ]
= ( 3 ) [1 + 2x5
+ 4 x2
50+ 8 x3
750+ ⋯]
f ( x ) = 3e25x
P( x ) = ( 3 ) [1 + 2x5
+ 4 x2
50+ 8 x3
750+ ⋯]
b. f ( x ) = 3e−
25x
Jawaban:
f ( x ) = 3e−
25x
P( x ) = ( 3 ) [ ∑n =0
∞ (−25 x )nn ! ]
Page | 12
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
= ( 3 ) [1 − 2 x5
+ 4 x2
50− 8 x3
750± ⋯]
f ( x ) = 3 e−
25x
P( x ) = ( 3 ) [1 − 2 x5
+ 4 x2
50− 8 x3
750± ⋯]
c. f ( x ) = −3 e25x
Jawaban:
f ( x ) = −3 e25x
P( x ) = ( −3 ) [ ∑n =0
∞ ( 25 x )nn ! ]
= ( −3 ) [1 + 2x5
+ 4 x2
50+ 8 x3
750+ ⋯]
f ( x ) = −3 e25x
P( x ) = ( −3 ) [1 + 2x5
+ 4 x2
50+ 8 x3
750+ ⋯]
d. f ( x ) = −3 e−
25x
Jawaban
f ( x ) = −3 e−
25x
P( x ) = ( −3 ) [ ∑n = 0
∞ (−25 x )n
n ! ]= ( −3 ) [1 − 2 x
5+ 4 x2
50− 8 x3
750± ⋯]
f ( x ) = −3 e−
25x
P( x ) = ( −3 ) [1 − 2 x5
+ 4 x2
50− 8 x3
750± ⋯]
SOAL 08.
a.f ( x ) = 3
4e2 x
Jawaban:
13 | Page
Matematika Dikrit STKIP PGRI Bangkalan
f ( x ) = 34e2 x
P( x ) = (34 ) [ ∑n = 0
∞ (2 x )n
n ! ]= (
34 )
[1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
f ( x ) = 3
4e2 x
P( x ) = (34 ) [1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
b.f ( x ) = 3
4e−2 x
Jawaban:
f ( x ) = 34e−2 x
P( x ) = (34 ) [ ∑n =0
∞ (−2 x )n
n ! ]= (
34 ) [1 − 2 x + 4 x2
2− 8 x3
6± ⋯]
f ( x ) = 3
4e−2 x
P( x ) = (34 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
c.f ( x ) = − 3
4e2 x
Jawaban
f ( x ) = − 34e2 x
P( x ) = (−34 ) [ ∑n = 0
∞ (2 x )n
n ! ]= (−
34 )
[1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
f ( x ) = − 3
4e2 x
P( x ) = (−34 )
[1 + 2 x + 4 x2
2+ 8 x3
6+ ⋯]
d.f ( x ) = − 3
4e−2 x
Jawaban:
Page | 14
Dosen Pembina – Abdul Muiz., S.Pd., M.Pd.,
f ( x ) = − 34e−2 x
P( x ) = (−34 ) [ ∑n =0
∞ (−2 x )n
n ! ]= (−
34 )
[1 − 2x + 4 x2
2− 8 x3
6± ⋯]
f ( x ) = − 3
4e−2 x
P( x ) = (−34 ) [1 − 2x + 4 x2
2− 8 x3
6± ⋯]
SOAL 09.
a.f ( x ) = 3
4e
25x
Jawaban:
f ( x ) = 34e
25x
P( x ) = (34 ) [ ∑n = 0
∞ ( 25 x )nn ! ]
= (34 )
[1 + 2x5
+ 4 x2
50+ 8 x3
750+ ⋯]
f ( x ) = 3
4e
25x
P( x ) = (34 ) [1 + 2x
5+ 4 x2
50+ 8 x3
750+ ⋯]
b.f ( x ) = 3
4e−
25x
Jawaban:
f ( x ) = 34e−
25x
P( x ) = (34 ) [ ∑n =0
∞ (−25 x )nn ! ]
= (34 ) [1 − 2 x
5+ 4 x2
50− 8 x3
750± ⋯]
f ( x ) = 3
4e−
25x
P( x ) = (34 ) [1 − 2 x
5+ 4 x2
50− 8 x3
750± ⋯]
15 | Page
Matematika Dikrit STKIP PGRI Bangkalan
c.f ( x ) = − 3
4e
25x
Jawaban:
f ( x ) = − 34e
25x
P( x ) = (−34 ) [ ∑n =0
∞ ( 25 x )nn ! ]
= (−34 )
[1 + 2x5
+ 4 x2
50+ 8 x3
750+ ⋯]
f ( x ) = − 3
4e
25x
P( x ) = (−34 ) [1 + 2x
5+ 4 x2
50+ 8 x3
750+ ⋯]
d.f ( x ) = − 3
4e−
25x
Jawaban
f ( x ) = − 34e−
25x
P( x ) = (−34 ) [ ∑n = 0
∞ (−25 x )n
n ! ]= (−
34 ) [1 − 2 x
5+ 4 x2
50− 8 x3
750± ⋯]
f ( x ) = − 3
4e−
25x
P( x ) = (−34 ) [1 − 2 x
5+ 4 x2
50− 8 x3
750± ⋯]
Page | 16
