CBSE 2024 · Region 2 · Set 2 · Q25 · 2 marks
Check the differentiability of $\displaystyle \mathrm{f}(\mathrm{x})=|\cos \mathrm{x}|$ at $\displaystyle \mathrm{x}=\frac{\pi}{2}$.If $\displaystyle y=A \sin 2 \mathrm{x}+B \cos 2 \mathrm{x}$ and $\displaystyle \frac{d^{2} y}{d \mathrm{x}^{2}}-k y=0$, find the value of $\displaystyle k$.
Check the differentiability of $\displaystyle \mathrm{f}(\mathrm{x})=|\cos \mathrm{x}|$ at $\displaystyle \mathrm{x}=\frac{\pi}{2}$.
If $\displaystyle y=A \sin 2 \mathrm{x}+B \cos 2 \mathrm{x}$ and $\displaystyle \frac{d^{2} y}{d \mathrm{x}^{2}}-k y=0$, find the value of $\displaystyle k$.
Marking-scheme solution
$$\begin{aligned}
& \mathrm{f}(\mathrm{x})=|\cos \mathrm{x}|=\left\{\begin{array}{cc}
\cos \mathrm{x} & 0 \leq \mathrm{x} \leq \frac{\pi}{2} \\
-\cos \mathrm{x} & \frac{\pi}{2} \leq \mathrm{x} \leq \pi
\end{array}\right. \\
& \text { LHD at } \frac{\pi}{2}=\lim _{h \rightarrow 0} \frac{\mathrm{f}\left(\dfrac{\pi}{2}-h\right)-\mathrm{f}\left(\dfrac{\pi}{2}\right)}{-h}=\lim _{h \rightarrow 0} \frac{\cos \left(\dfrac{\pi}{2}-h\right)-0}{-h}=\lim _{h \rightarrow 0} \frac{\sin h}{-h}=-1
\end{aligned}
\begin{aligned}
& \frac{d y}{d \mathrm{x}}=2 A \cos 2 \mathrm{x}-2 B \sin 2 \mathrm{x} \\
& \Rightarrow \frac{d^{2} y}{d \mathrm{x}^{2}}=-4 A \sin 2 \mathrm{x}-4 B \cos 2 \mathrm{x}=-4 y \\
& \Rightarrow \frac{d^{2} y}{d \mathrm{x}^{2}}+4 y=0 \\
& \Rightarrow k=-4
\end{aligned}
$$
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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.