Q104 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics

If $y=\frac{\log x}{x}$, then $\frac{d^2 y}{d x^2}$ at $x=1$

  • A
    $-e^{-3}$
  • Correct answer
    B
    $-3$
  • C
    $3$
  • D
    $e^3$

Solution

$y=\frac{\log x}{x}$ $\Rightarrow \quad x y=\log x$ $\Rightarrow \quad x \frac{d y}{d x}+y=\frac{1}{x}$...(i) $\Rightarrow \quad \frac{d y}{d x}=\left(\frac{1}{x}-y\right)\left(\frac{1}{x}\right)$ From Eq. (i), $x \frac{d^2 y}{d x^2}+\frac{d y}{d x}+\frac{d y}{d x}=-\frac{1}{x^2}$ $\begin{aligned} & \Rightarrow \quad x \frac{d^2 y}{d x^2}+2\left(\frac{1}{x}-y\right)\left(\frac{1}{x}\right)=-\frac{1}{x^2} \\ & \text { At } \quad x=1,1\left(\frac{d^2 y}{d x^2}\right)+2\left(\frac{1}{1}-\frac{\log 1}{1}\right)\left(\frac{1}{1}\right)=-\frac{1}{12} \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=-3\end{aligned}$

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

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