Question:
Let $f:(-1, \infty) \rightarrow R$ 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: 1
Solution:
$f^{\prime}(x)=\frac{\frac{x}{1+x}-\ell n(1+x)}{x^{2}}$
$=\frac{x-(1+x) \ell n(1+x)}{x^{2}(1+x)}$
Suppose $h(x)=x-(1+x) \ell n(1+x)$
$\Rightarrow h^{\prime}(x)=1-\ell n(1+x)-1=-\ell n(1+x)$
$h^{\prime}(x)>0, \forall x \in(-1,0)$
$h^{\prime}(x)<0, \forall x \in(0, \infty)$
$h(0)=0 \Rightarrow h^{\prime}(x)<0 \forall x \in(-1, \infty)$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})<0 \forall \mathrm{x} \in(-1, \infty)$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is a decreasing function for all $\mathrm{x} \in(-1$,$\infty)$