Let f : (-1 , infinite) rightarrow R be defined by f(0) = 1 and


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 $f$ :

  1. decreases in $(-1, \infty)$

  2. decreases in $(-1,0)$ and increases in $(0, \infty)$

  3. increases in $(-1, \infty)$

  4. increases in $(-1,0)$ and decreases in $(0, \infty)$

Correct Option: 1


$f^{\prime}(x)=\frac{\frac{x}{1+x}-\ell n(1+x)}{x^{2}}$

$=\frac{x-(1+x) \ell n(1+x)}{x^{2}(1+x)}$

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 $\mathrm{x} \in(-1$,$\infty)$



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