Q111 / 160 · AP EAMCET 2009 · Mathematics

The maximum value of $\frac{\log x}{x}, 0 < x < \infty$ is

  • A
    $\infty$
  • B
    $e$
  • C
    1
  • Correct answer
    D
    $e^{-1}$

Solution

Let $\quad y=\frac{\log x}{x}$ $\Rightarrow \quad \frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log x}{x^2}=\frac{1-\log x}{x^2}$ Put $\quad \frac{d y}{d x}=0 \Rightarrow \log x=1$ $\Rightarrow \quad x=e$ Now, $\frac{d^2 y}{d x^2}=\frac{x^2\left(-\frac{1}{x}\right)-(1-\log x) 2 x}{\left(x^2\right)^2}$ $=-\frac{(3-2 \log x)}{x^3}$ $\Rightarrow\left(\frac{d^2 y}{d x^2}\right)_{(x=e)}=\frac{-(3-2)}{e^3}=-\frac{1}{e^3} < 0$, maxima Hence, maximum value at $x=e$ is $\frac{1}{e}$.

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

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