Question

$f\left(x\right)=\frac{x}{\log x}-\frac{\log 5}{5}$ is decreasing in

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

Answer 1

Answer: (D)

$(1,e)$

$f\left(x\right)=\frac{x}{\log x}-\frac{log5}{5}$

Differentiating $f\left(x\right)$ w.r.t $x$, we have

$f^{\prime }\left(x\right)=\frac{\log x-x.\frac{1}{x}}{{\left(\log x\right)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{\log x-1}{{\left(\log x\right)}^2}$

As we know that for a decreasing function,

$f^{\prime }\left(x\right)<0$

$\frac{\log x-1}{{\left(\log x\right)}^2}<0$

• $\log x-1<0$

$\log x<1$

$x<e.....\left(1\right)$

• $\log x>0$
  

$\Rightarrow x>1.....\left(2\right)[\because \log 1=0]$

But the domain of $f\left(x\right)$ is

$x>0.....\left(3\right)$

Now from ${eq}^n\left(1\right),\left(2\right)\&\left(3\right)$, we have

$f\left(x\right)$ is decreasing for all $x\in (1,e)$