Question

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

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

Answer 1

Answer: (C)

$\\{x:x \geq e\\}$

Answer 2

$f(x)= \frac{x}{\log x}$
$f'(x)=\frac{\log x \cdot 1-x \cdot \frac{1}{x}}{(\log x)^{2}}=\frac{(\log x-1)}{(\log x)^{2}}$
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, $x: x \geq e$