CBSE 2024 · Region 2 · Set 1 · Q26 · 3 marks
If $\displaystyle \mathrm{x}=\mathrm{e}^{\cos 3 \mathrm{t}}$ and $\displaystyle \mathrm{y}=\mathrm{e}^{\sin 3 \mathrm{t}}$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y} \log \mathrm{x}}{\mathrm{x} \log \mathrm{y}}$.Show that : \[\frac{\mathrm{d}}{\mathrm{dx}}(|\mathrm{x}|)=\frac{\mathrm{x}}{|\mathrm{x}|}, \mathrm{x} \neq 0 \]
If $\displaystyle \mathrm{x}=\mathrm{e}^{\cos 3 \mathrm{t}}$ and $\displaystyle \mathrm{y}=\mathrm{e}^{\sin 3 \mathrm{t}}$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y} \log \mathrm{x}}{\mathrm{x} \log \mathrm{y}}$.
Show that : \[\frac{\mathrm{d}}{\mathrm{dx}}(|\mathrm{x}|)=\frac{\mathrm{x}}{|\mathrm{x}|}, \mathrm{x} \neq 0 \]
Marking-scheme solution
$$\begin{aligned}
& \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \mathrm{t}}=\mathrm{e}^{\cos 3 \mathrm{t}} \times(-\sin 3 \mathrm{t}) \times 3 \\
& \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{t}}=\mathrm{e}^{\sin 3 \mathrm{t}} \times(\cos 3 \mathrm{t}) \times 3
\end{aligned}
\begin{aligned}
& \mathrm{x}=\mathrm{e}^{\cos 3 \mathrm{t}} \Rightarrow \cos 3 \mathrm{t}=\log \mathrm{x} \\
& \mathrm{y}=\mathrm{e}^{\sin 3 \mathrm{t}} \Rightarrow \sin 3 \mathrm{t}=\log \mathrm{y} \\
& \therefore \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}=\frac{-\mathrm{y} \log \mathrm{x}}{\mathrm{x} \log \mathrm{y}}
\end{aligned}
\begin{aligned}
& \frac{\mathrm{d}(|\mathrm{x}|)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}\left(\sqrt{\mathrm{x}^{2}}\right)}{\mathrm{d} \mathrm{x}}, \mathrm{x} \neq 0 \\
& =\frac{1}{2}\left(\mathrm{x}^{2}\right)^{-\frac{1}{2}} \times \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}} \\
& =\frac{1}{2 \sqrt{\mathrm{x}^{2}}} 2 \mathrm{x}=\frac{\mathrm{x}}{|\mathrm{x}|}
\end{aligned}
$$
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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.