A) \[x\in (-1,\,1)\]
B) Any positive (+) real x
C) \[x\in (-1,\,1]\]
D) Any positive (+) real \[x(x\ne 1)\]
Correct Answer: C
Solution :
\[{{\log }_{e}}(1+x)=\sum\limits_{i=1}^{\infty }{\frac{{{(-1)}^{i+1}}{{x}^{i}}}{i}}\]is defined for \[x\in (-1,1]\] Because \[{{\log }_{e}}(1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+....\infty \] is defined for \[(-1<x\le 1)\].
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