CBSE 2024 · Region 2 · Set 1 · Q22 · 2 marks
If $\displaystyle \mathrm{f}(\mathrm{x})=|\tan 2 \mathrm{x}|$, then find the value of $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})$ at $\displaystyle \mathrm{x}=\frac{\pi}{3}$.If $\displaystyle \mathrm{y}=\operatorname{cosec}\left(\cot ^{-1} \mathrm{x}\right)$, then prove that $\displaystyle \sqrt{1+\mathrm{x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x}=0$.
If $\displaystyle \mathrm{f}(\mathrm{x})=|\tan 2 \mathrm{x}|$, then find the value of $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})$ at $\displaystyle \mathrm{x}=\frac{\pi}{3}$.
If $\displaystyle \mathrm{y}=\operatorname{cosec}\left(\cot ^{-1} \mathrm{x}\right)$, then prove that $\displaystyle \sqrt{1+\mathrm{x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x}=0$.
Marking-scheme solution
$$\begin{aligned}
& \mathrm{f}(\mathrm{x})=-\tan 2 \mathrm{x}, \frac{\pi}{4}<\mathrm{x}<\frac{\pi}{2} \\
& \mathrm{f}^{\prime}(\mathrm{x})=-2 \sec ^{2} 2 \mathrm{x}, \frac{\pi}{4}<\mathrm{x}<\frac{\pi}{2} \\
& \mathrm{f}^{\prime}\left(\frac{\pi}{3}\right)=-2(-2)^{2}=-8
\end{aligned}
\begin{aligned}
& \mathrm{y}=\sqrt{1+\cot ^{2}\left(\cot ^{-1} \mathrm{x}\right)}=\sqrt{1+\mathrm{x}^{2}} \\
& \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}} \\
& \Rightarrow \sqrt{1+\mathrm{x}^{2}} \frac{d \mathrm{y}}{d \mathrm{x}}-\mathrm{x}=0
\end{aligned}
$$
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.