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 $f$

• A. decreases in $(-1, \infty)$
• B. decreases in (-1,0) and increases in $(0, \infty)$
• C. increases in $(-1, \infty)$
• D. increases in (-1,0) and decreases in $(0, \infty)$

Answer 1

The Correct Option is A

$f'(x)=\frac{\frac{x}{1+x}-\ell n(1+x)}{x^{2}}$ $=\frac{x-(1+x) \ell n(1+x)}{x^{2}(1+x)}$ Suppose $h(x)=x-(1+x) \ell n(1+x)$ $\Rightarrow h'(x)=1-\ell n(1+x)-1=-\ln (1+x)$ $h ^{\prime}( x )>0, \forall x \in(-1,0)$ $h'( x )<0, \forall x \in(0, \infty)$ $h (0)=0 \Rightarrow h'( x )<0 \forall x \in(-1, \infty)$ $\Rightarrow f'( x )<0 \forall x \in(-1, \infty)$ $\Rightarrow f ( x )$ is a decreasing function for all $x \in(-1, \infty)$