Local maximum value of the function x x is - Vidyadip

MCQ

Local maximum value of the function ${{\log x} \over x}$ is

A
$e$
B
$1$
✓
${1 \over e}$
D
$2e$

✓

Answer

Correct option: C.

${1 \over e}$

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

For maximum or minimum value of $f(x),\,\,f'(x) = 0$

==> $f'(x) = \frac{{1 - {{\log }_e}x}}{{{x^2}}} = 0$ or $\frac{{1 - {{\log }_e}x}}{{{x^2}}} = 0$

$\therefore {\log _e}x = 1$ or $x = e$, which lie in $(0,\infty )$.

For $x = e,\,\,\frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{{e^3}}}$, which is $ - ve$.

Hence $ y $ is maximum at $x = e$ and its maximum value $ = \frac{{\log e}}{e} = \frac{1}{e}$.

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