Sequence And Series Question 139
Question: If $ y=x-\frac{x^{2}}{2,!}+\frac{x^{3}}{3!}-\frac{x^{4}}{4,!}+……, $ then $ x= $
Options:
A) $ {\log_{e}}(1-y) $
B) $ \frac{1}{{\log_{e}}(1-y)} $
C) $ {\log_{e}}\frac{1}{1-y} $
D) $ {\log_{e}}(1+y) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=x-\frac{x^{2}}{2\ !}+\frac{x^{3}}{3\ !}-\frac{x^{4}}{4\ !}+…….=1-{e^{-x}} $
$ \Rightarrow {e^{-x}}=1-y\Rightarrow -x={\log_{e}}(1-y)\Rightarrow x={\log_{e}}\frac{1}{1-y} $ .
BETA
