Question

The set of real values of x for which $f(x)=\frac{x}{logx}$ increasing, is

• A. $\{x:x<e\}$
• B. $\{1\}$
• C. $\{x:x\ge e\}$
• D. empty

Answer 1

Answer: (C)

$f(x)=\frac{x}{\log x}$
$f^{\prime }(x)=\frac{\log x\cdot 1-x\cdot \frac{1}{x}}{(\log x{)}^2}=\frac{(\log x-1)}{(\log x{)}^2}$
We know that, $f(x)$ is increasing (strictly) When $f^{\prime }(x)>0$
$\Rightarrow \frac{(\log x-1)}{(\log x{)}^2}>0$
$\Rightarrow (\log x-1)>0$
$\Rightarrow \log x>1$
$\Rightarrow {\log }_{e}x>{\log }_{e}e$
$\Rightarrow x>e$
Hence, $x:x\ge e$