Question

The function of $x-\frac{\log (1+x)}{x}(x>0)$ is increasing in

• A. $(1,\infty )$
• B. $(0,\infty )$
• C. $(2,2e)$
• D. $(1/e,2e)$

Answer 1

The correct option is A $(1,\infty )$

$Letf\left(x\right)=x-\frac{\log \left(1+x\right)}{x},\left(x>0\right)f^{\prime }\left(x\right)=1-\frac{\left\{\left(\frac{1}{1+x}\right).x-1.\log \left(1+x\right)\right\}}{x^2}>0=1.x^2\frac{x}{1+x}+\log \left(1+x\right)>0=x^2\left(1+x\right)-x+\log \left(1+x\right).\left(1+x\right)>0=x^2\left(1+x\right)+\left(1+x\right)\log \left(1+x\right)>x$

$Take\left(1,\infty \right)inthisinternal{f}^{\prime }\left(x\right)>0So,\left(1,\infty \right)iscorrect.$

Hence, this is the answer.