Question:
Let $f:(-1, \infty) \rightarrow \mathbf{R}$ be defined by $f(0)=1$ and $f(x)=\frac{1}{x} \log _{e}(1+x), x \neq 0 .$ Then the function $f$ :
Correct Option: , 4
Solution:
$f^{\prime}(x)=\frac{\frac{x}{1+x}-\ln (1+x)}{x^{2}}$
$=\frac{x-(1+x) \ln (1+x)}{(1+x) x^{2}}<0, \forall x \in(-1, \infty)-\{0\}$
[For $x \in(-1,0), f^{\prime}(x)<0$ and for $x \in(0, \infty), f^{\prime}(x)<0$ ]
So, $f(x)$ is increases in $(-1, \infty)$.
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