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, ∞)
• D. None of these

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}