Sequence And Series Question 41
Question: If $ y=-( x^{3}+\frac{x^{6}}{2}+\frac{x^{9}}{3}+….. ) $ , then $ x= $
[MNR 1975]
Options:
A) $ \frac{1+e^{y}}{3} $
B) $ \frac{1-e^{y}}{3} $
C) $ {{(1-e^{y})}^{1/3}} $
D) $ {{(1-e^{y})}^{3}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=-{ x^{3}+\frac{{{(x^{3})}^{2}}}{2}+\frac{{{(x^{3})}^{3}}}{3}+…… }=\log (1-x^{3}) $
$ \Rightarrow e^{y}=1-x^{3}\Rightarrow x={{(1-e^{y})}^{1/3}} $ .
BETA
