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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 2 · Set 3 · Q28
(a) If y =( log x )^2, prove that x^2 y^′ ′+ xy^′=2.
(a)
If $\displaystyle \mathrm{y}=(\log \mathrm{x})^{2}$, prove that $\displaystyle \mathrm{x}^{2} \mathrm{y}^{\prime \prime}+\mathrm{xy}^{\prime}=2$.
OR
(b)
If $\displaystyle \mathrm{y}=\sin \left(\tan ^{-1} \mathrm{e}^{\mathrm{x}}\right)$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \mathrm{x}=0$.
Marking-scheme solution
Differentiating both sides w.r.t. x , we get
$$\begin{aligned}
& \mathrm{y}^{\prime}=\frac{2 \log \mathrm{x}}{\mathrm{x}} \\
& \Rightarrow \mathrm{x} \mathrm{y}^{\prime}=2 \log \mathrm{x} \\
& \Rightarrow \mathrm{x} \mathrm{y}^{\prime \prime}+\mathrm{y}^{\prime}=\frac{2}{\mathrm{x}} \\
& \Rightarrow \mathrm{x}^{2} \mathrm{y}^{\prime \prime}+\mathrm{xy}^{\prime}=2
\end{aligned}
$$
$$\begin{aligned}
& \frac{d \mathrm{y}}{d \mathrm{x}}=\cos \left(\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)\right) \times \frac{\mathrm{e}^{\mathrm{x}}}{1+\mathrm{e}^{2 \mathrm{x}}} \\
& \left(\frac{d \mathrm{y}}{d \mathrm{x}}\right)_{\mathrm{x}=0}=\cos \frac{\pi}{4} \times \frac{1}{2}=\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
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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.