Question

Which of the following is (are) correct about the function $f\left(x\right)=\frac{x}{\log x}$

• A. f(x) is increasing on $\left(e,\infty \right)$
• B. f(x) is decreasing on $\left(0,\infty \right)$
• C. f(x) is increasing on $\left(0,1\right)\cup (1,e)$
• D. f(x) is decreasing on $\left(0,1\right)\cup (1,e)$

Answer 1

Answer: (A), (D)

f(x) is decreasing on $\left(0,1\right)\cup (1,e)$

$\mathrm{Given}\mathrm{the}\mathrm{function}f\left(x\right)=\frac{x}{\log x}\Rightarrow f\text{'}\left(x\right)=\frac{\log x-1}{{\left(\log x\right)}^2}\mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{an}\mathrm{increasing}\mathrm{function},\mathrm{we}\mathrm{must}\mathrm{have}f\text{'}\left(x\right)>0\Rightarrow \frac{\log x-1}{{\left(\log x\right)}^2}>0\Rightarrow \log x-1>0[\because {\left(\log x\right)}^2>0\mathrm{for}x>0]\Rightarrow \log x>1\Rightarrow x>e\Rightarrow x\in \left(e,\infty \right)\mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}a\mathrm{decreasing}\mathrm{function},\mathrm{we}\mathrm{must}\mathrm{have}f\text{'}\left(x\right)<0\Rightarrow \frac{\log x-1}{{\left(\log x\right)}^2}<0\Rightarrow \log x-1<0[\because {\left(\log x\right)}^2>0\mathrm{for}x>0]\Rightarrow \log x<1\Rightarrow x<e\Rightarrow x\in \left(0,e\right)-\left\{1\right\}[\because f(x)\mathrm{is}\mathrm{defined}\mathrm{for}x>0\mathrm{and}x\ne 1]\mathrm{So},f\left(x\right)\mathrm{is}\mathrm{decreasing}\mathrm{on}(0,1)\cup (1,e).$