CBSE 2025 · Region 5 · Set 1 · Q23 · 2 marks
Differentiate $\displaystyle \frac{\sin \mathrm{x}}{\sqrt{\cos \mathrm{x}}}$ with respect to x .If $\displaystyle y=5 \cos \mathrm{x}-3 \sin \mathrm{x}$, prove that $\displaystyle \frac{d^{2} y}{d \mathrm{x}^{2}}+y=0$.
Differentiate $\displaystyle \frac{\sin \mathrm{x}}{\sqrt{\cos \mathrm{x}}}$ with respect to x .
If $\displaystyle y=5 \cos \mathrm{x}-3 \sin \mathrm{x}$, prove that $\displaystyle \frac{d^{2} y}{d \mathrm{x}^{2}}+y=0$.
Marking-scheme solution
Let $\displaystyle y=\frac{\sin \mathrm{x}}{\sqrt{\cos \mathrm{x}}}$
\[\begin{aligned}
\frac{d y}{d \mathrm{x}} & =\frac{\sqrt{\cos \mathrm{x}} \cdot \cos \mathrm{x}-\sin \mathrm{x} \cdot\left(\dfrac{-\sin \mathrm{x}}{2 \sqrt{\cos \mathrm{x}}}\right)}{\cos \mathrm{x}} \\
\Rightarrow \frac{d y}{d \mathrm{x}} & =\frac{2 \cos ^{2} \mathrm{x}+\sin ^{2} \mathrm{x}}{2(\cos \mathrm{x})^{3 / 2}} \text { or } \frac{1+\cos ^{2} \mathrm{x}}{2(\cos \mathrm{x})^{3 / 2}}
\end{aligned}
\]
$\displaystyle y=5 \cos \mathrm{x}-3 \sin \mathrm{x}$, then $\displaystyle \frac{d y}{d \mathrm{x}}=-5 \cdot \sin \mathrm{x}-3 \cdot \cos \mathrm{x}$
\[\begin{aligned}
\Rightarrow \frac{d^{2} y}{d \mathrm{x}^{2}}= & -5 \cdot \cos \mathrm{x}+3 \cdot \sin \mathrm{x}=-y \\
& \Rightarrow \frac{d^{2} y}{d \mathrm{x}^{2}}+y=0
\end{aligned}
\]
Continuity and DifferentiabilitySecond Order DerivativeApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.