Then the function f :

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 $f$ :

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

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

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

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


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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