Question

$f\left(x\right)=\frac{x}{lo{g}_{e}x},x\ne 1$, is decreasing in interval

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

Answer 1

The correct option is A $(0,e)$

f'(x) = $\frac{lo{g}_{e}x.1-x.\frac{1}{x}}{(lo{g}_{e}x{)}^2}$
=$\frac{(lo{g}_{e}x-1)}{(lo{g}_{e}x{)}^2}<0$
It is decreasing at (0, e) – {1}