Find the intervals in which the function f(x) =x/(log x) is strictly…

The correct answer is Option (1) → Strictly increasing on $[e,∞)$ and strictly decreasing on $[0,1]∪[1,e].

Given $f(x) =\frac{x}{\log x}$

For $D_f, x > 0$ and $\log x ≠0⇒ x > 0$ and $x ≠ 1$

$⇒ D_f= (0,∞) - \{1\}$.

Differentiating (i) w.r.t. x, we get

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

Now $f'(x)>0$ iff $\frac{\log x-1}{(\log x)^2}⇒\log x-1>0$

$⇒\log x>1⇒x>e^1⇒x>e$

= f(x) is strictly increasing in $[e, ∞)$.

And $f'(x)<0$ iff $\frac{\log x-1}{(\log x)^2}⇒\log x-1<0$

$⇒ \log x <1⇒x<e^1⇒x<e$ but $x > 0$ and $x ≠ 1$

⇒ f(x) is strictly decreasing in $(0, e]- \{1\}$.