Q69 / 180 · KCET 2020 · Mathematics
The maximum value of $\frac{\log _{e} x}{x}$, if $x>0$ is
- A$e$
- B1
- Correct answerC$\frac{1}{e}$
- D$-\frac{1}{e}$
Solution
Let $y=\frac{\log _{e} x}{x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log _{e} x \cdot 1}{x^{2}}=\frac{1-\log _{e} x}{x^{2}}$
For maximum, put $\frac{d y}{d x}=0$
$\Rightarrow \quad 1-\log _{e} x=0$
$\Rightarrow \quad \log _{e} x=1$
$\Rightarrow \quad x=e$
Now, $\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(0-\frac{1}{x}\right)-\left(1-\log _{e} x\right)(2 x)}{x^{4}}$
$=\frac{-x-2 x\left(1-\log _{e} x\right)}{x^{4}}$
$\therefore \quad\left(\frac{d^{2} y}{d x^{2}}\right)_{\text {at } x=e} < 0$
$\therefore \quad y(e)=\frac{\log _{e} e}{e}=\frac{1}{e}$
Hence, the correct answer is option: 'C'.