CBSE 2024 · Region 1 · Set 1 · Q21 · 2 marks
Check whether the function $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}|\mathrm{x}|$ is differentiable at $\displaystyle \mathrm{x}=0$ or not.If $\displaystyle \mathrm{y}=\sqrt{\tan \sqrt{\mathrm{x}}}$, prove that $\displaystyle \sqrt{\mathrm{x}} \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{1+\mathrm{y}^{4}}{4 \mathrm{y}}$.
Check whether the function $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}|\mathrm{x}|$ is differentiable at $\displaystyle \mathrm{x}=0$ or not.
If $\displaystyle \mathrm{y}=\sqrt{\tan \sqrt{\mathrm{x}}}$, prove that $\displaystyle \sqrt{\mathrm{x}} \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{1+\mathrm{y}^{4}}{4 \mathrm{y}}$.
Marking-scheme solution
$$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}
\mathrm{x}^{3}, \mathrm{x} \geq 0 \\
-\mathrm{x}^{3}, \mathrm{x} \leq 0
\end{array}\right.
$\displaystyle \because R H D=L H D=0, \quad$ So $\displaystyle \mathrm{f}(\mathrm{x})$ is differentiable at $\displaystyle \mathrm{x}=0$
\begin{aligned}
& \mathrm{y}=\sqrt{\tan \sqrt{\mathrm{x}}} \\
& \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\sec ^{2} \sqrt{\mathrm{x}}}{2 \sqrt{\tan \sqrt{\mathrm{x}}}} \times \frac{1}{2 \sqrt{\mathrm{x}}} \\
& \begin{aligned}
\sqrt{\mathrm{x}} \frac{d \mathrm{y}}{d \mathrm{x}} & =\frac{\sec ^{2} \sqrt{\mathrm{x}}}{4 \sqrt{\tan \sqrt{\mathrm{x}}}} \\
& =\frac{1+(\tan \sqrt{\mathrm{x}})^{2}}{4 \sqrt{\tan \sqrt{\mathrm{x}}}}=\frac{1+\mathrm{y}^{4}}{4 \mathrm{y}}
\end{aligned}
\end{aligned}
$$
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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.