Q109 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

If $y=\frac{1}{x}+\cos 2 x$, then $\frac{d^2 y}{d x^2}$ is equal to

  • A
    $\frac{2}{x^3}+4 y-\frac{4}{x}$
  • B
    $4 y-\frac{4}{x}-\frac{2}{x^3}$
  • Correct answer
    C
    $\frac{2}{x^3}+\frac{4}{x}-4 y$
  • D
    $4 y-\frac{4}{x}-\frac{1}{x^3}$

Solution

$\begin{aligned} & y=\frac{1}{x}+\cos 2 x \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-1}{x^2}-2 \sin 2 x \Rightarrow \frac{d^2 y}{d x^2}=\frac{2}{x^3}-4 \cos 2 x \\ & =\frac{2}{x^3}-4\left(y-\frac{1}{x}\right)=\frac{2}{x^3}+\frac{4}{x}-4 y \\ & \end{aligned}$

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

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