The maximum value of the function f(x)= x x is ....... . - Vidyadip

MCQ

The maximum value of the function $f(x)=\frac{\log x}{x}$ is $....... .$

A
$e$
✓
$\frac{1}{e}$
C
$e^2$
D
$\frac{1}{e^2}$

✓

Answer

Correct option: B.

$\frac{1}{e}$

$\frac{1}{e}$
Hint: $f(x)=\frac{\log x}{x}$
$\therefore f^{\prime}(x) =\frac{x \cdot \frac{1}{x}-\log x}{x^2}=\frac{1-\log x}{x^2}$
and $ f^{\prime \prime}(x) =\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^4}$
$ =\frac{-x-2 x(1-\log x)}{x^4}$
$ =\frac{-1-2(1-\log x)}{x^3}$
Put $f^{\prime}(x)=0$
$\therefore \frac{1-\log x}{x^2}=0$
$\therefore 1-\log x=0$
$\therefore \log x=1,$
$ \therefore x=e$
$\therefore f^{\prime \prime}(e)=\frac{-1-2(1-\log e)}{e^3}$
$=\frac{-1-2(0)}{e^3}=-\frac{1}{e^3}<0$
$\therefore f(x)$ is maximum at $x=e$
$\therefore$ Maximum $f(x)=f(e)$
$=\frac{\log e}{e}=\frac{1}{e}$

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