Q102 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics

If $y=\tan ^{-1} \frac{x}{1+2 x^2}+\tan ^{-1} \frac{x}{1+6 x^2}+\tan ^{-1} \frac{x}{1+12 x^2}$, then $\left(\frac{d y}{d x}\right)_{x=\frac{1}{2}}=$

  • A
    $1$
  • B
    $-1$
  • Correct answer
    C
    $0$
  • D
    $\frac{1}{2}$

Solution

$\begin{aligned} & y=\tan ^{-1} \frac{x}{1+2 x^2}+\tan ^{-1} \frac{x}{1+6 x^2}+\tan ^{-1} \frac{x}{1+12 x^2} \\ & \frac{d y}{d x}=\frac{1-2 x^2}{\left(1+2 x^2\right)^2+x^2}+\frac{1-6 x^2}{\left(1+6 x^2\right)^2+x^2}+\frac{1-12 x^2}{\left(1+12 x^2\right)^2+x^2} \\ & \left(\frac{d y}{d x}\right)_{x=\frac{1}{2}}=\frac{1-\frac{1}{2}}{\left(1+\frac{1}{2}\right)^2+\frac{1}{4}}+\frac{1-\frac{3}{2}}{\left(1+\frac{3}{2}\right)^2+\frac{1}{4}}+\frac{1-3}{(1+3)^2+\frac{1}{4}} \\ & \left(\frac{d y}{d x}\right)_{x=\frac{1}{2}}=\frac{1}{5}-\frac{1}{13}-\frac{8}{65}=0\end{aligned}$

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

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