Q113 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

If $y=\left(\tan^{-1} 2 x\right)^2+\left(\cot ^{-1} 2 x\right)^2$, then $\left(1+4 x^2\right)^2 y^{\prime \prime}-16=$

  • A
    $8 x y^{\prime}$
  • Correct answer
    B
    $-8 x\left(1+4 x^2\right) y^{\prime}$
  • C
    $8 x\left(1+4 x^2\right) y^{\prime}$
  • D
    $-8 x y^{\prime}$

Solution

$y=\left(\tan ^{-1} 2 x\right)^2+\left(\cot ^{-1} 2 x\right)^2$ $\begin{aligned} & \frac{d y}{d x}=\frac{4 \tan ^{-1} 2 x}{1+4 x^2}-\frac{4 \cot ^{-1} 2 x}{1+4 x^2} \\ & \Rightarrow y^{\prime}=\frac{4 \tan ^{-1} 2 x-4 \cot ^{-1} 2 x}{1+4 x^2} \\ & \Rightarrow\left(1+4 x^2\right) y^{\prime}=4 \tan ^{-1} 2 x-4 \cot ^{-1} 2 x \\ & \Rightarrow 8 x y^{\prime}+\left(1+4 x^2\right) y^{\prime \prime}=\frac{8}{\left(1+4 x^2\right)}+\frac{8}{\left(1+4 x^2\right)} \\ & \Rightarrow\left(1+4 x^2\right)^2 y^{\prime \prime}-16=-8 x\left(1+4 x^2\right) y^{\prime}\end{aligned}$

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

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