Sequence And Series Question 36
Question: $ {\log_{e}}\sqrt{\frac{1+x}{1-x}}= $
Options:
A) $ {\log_{e}}\frac{1}{2} $
B) $ 2,[ x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+…..\infty ] $
C) $ 2,[ x^{2}+\frac{x^{4}}{4}+\frac{x^{6}}{6}+…..\infty ] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {\log_{e}}\sqrt{\frac{1+x}{1-x}}=\frac{1}{2}{\log_{e}}\frac{1+x}{1-x} $ $ =\frac{1}{2}.2{ x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+…… }=x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+…… $
BETA
