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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 5 · Set 3 · Q27
If y =( tan^-1 x)^2, show that (x^2+1)^2 (d^2 y)/(d x^2)+2 x(x^2+1) dy/dx=2.
If $\displaystyle \mathrm{y}=\left(\tan ^{-1} x\right)^{2}$, show that $\displaystyle \left(x^{2}+1\right)^{2} \frac{\mathrm{~d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2 x\left(x^{2}+1\right) \frac{\mathrm{dy}}{\mathrm{dx}}=2$.
Marking-scheme solution
$$\begin{aligned}
\mathrm{y}=\left(\tan ^{-1} x\right)^{2} & \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\frac{2 \tan ^{-1} x}{1+x^{2}} \\
& \Rightarrow\left(1+x^{2}\right) \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=2 \tan ^{-1} x, \text { differentiating with respect to ' } x \\
& \Rightarrow\left(1+x^{2}\right)^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2 x\left(x^{2}+1\right) \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=2
\end{aligned}
$$
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.