Q111 / 180 · KCET 2008 · Mathematics

The maximum value of $\frac{\log x}{x}$ in $(2, \infty)$ is

  • A
    1
  • B
    $\frac{2}{\mathrm{e}}$
  • Correct answer
    C
    $\frac{1}{\mathrm{e}}$
  • D
    None of the above

Solution

$$ \text { Let } y=\frac{\log x}{x} $$ On differentiating w.r.t. x, we get $$ \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log x \cdot 1}{x^{2}}=\frac{1-\log x}{x^{2}} $$ For maxima, put $\frac{d y}{d x}=0$ $\begin{array}{lc}\Rightarrow & \frac{1-\log x}{x^{2}}=0 \\ \Rightarrow & \log x=1 \\ \Rightarrow & x=e\end{array}$ Now, $\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) 2 x}{\left(x^{2}\right)^{2}}$ At $\mathrm{x}=\mathrm{e}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx^{2 }} \leq 0, \text { maxima }}$ $\therefore$ The maximum value at $\mathrm{x}=\mathrm{e}$ is $$ \mathrm{y}=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}} $$

Hence, the correct answer is option: 'C'.

Relevant Questions