jika 0 < x < 1 dan deret : 1 + x'log 2 + (x'log)² y ..... konvergen maka a. 0 < x < 1/4 b. 0 < x < 1/2 c. 1/4 < x < 1/2 d. 1/2 < x < 1 e. 1/4 < x < 1
Matematika
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Pertanyaan
jika 0 < x < 1 dan deret :
1 + x'log 2 + (x'log)² y ..... konvergen maka
a. 0 < x < 1/4
b. 0 < x < 1/2
c. 1/4 < x < 1/2
d. 1/2 < x < 1
e. 1/4 < x < 1
1 + x'log 2 + (x'log)² y ..... konvergen maka
a. 0 < x < 1/4
b. 0 < x < 1/2
c. 1/4 < x < 1/2
d. 1/2 < x < 1
e. 1/4 < x < 1
1 Jawaban
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1. Jawaban Anonyme
Jawab
r = x'log 2
konvergen
-1 < r < 1
- 1 > x'log 2 < 1
x'log 2 > -1 → 2 > x⁻¹ → x > 1/2
x'log 2 < 1 → 2 < x