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Mathematics · 2023 · 2 marks
Continuity and DifferentiabilityDerivatives of Functions in Parametric Forms2 marksApplyvery short answer
CBSE 2023 · Region 1 · Set 3 · Q25
If x =√(a^tan^-1 t), y =√(a^cot^-1 t), then show that x dy/dx+ y =0.
If $\displaystyle \mathrm{x}=\sqrt{\mathrm{a}^{\tan ^{-1} \mathrm{t}}}, \mathrm{y}=\sqrt{\mathrm{a}^{\cot ^{-1} \mathrm{t}}}$, then show that $\displaystyle \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$.
Marking-scheme solution
$$\begin{aligned}
\frac{\mathrm{dx}}{\mathrm{dt}}= & \frac{\sqrt{\mathrm{a}^{\tan ^{-1} \mathrm{t}}}}{2} \frac{\log \mathrm{a}}{1+\mathrm{t}^{2}} \\
& \frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{\sqrt{\mathrm{a}^{\cot ^{-1} \mathrm{t}}}}{2} \frac{\log \mathrm{a}}{1+\mathrm{t}^{2}} \\
& \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\sqrt{\mathrm{a}^{\cot ^{-1} \mathrm{t}}}}{\sqrt{\mathrm{a}^{\tan ^{-1} \mathrm{t}}}}=-\frac{\mathrm{y}}{\mathrm{x}} \\
& \Rightarrow \mathrm{x} \frac{d \mathrm{y}}{d \mathrm{x}}+\mathrm{y}=0
\end{aligned}
$$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.