CBSE 2026 · Region 2 · Set 1 · Q22 · 2 marks
If $\displaystyle x=\mathrm{a}^{2} \sin ^{3} \mathrm{t}, \mathrm{y}=\mathrm{b} \cos ^{3} \mathrm{t}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{d} x}$ at $\displaystyle \mathrm{t}=\frac{\pi}{4}$.
Marking-scheme solution
$\displaystyle \dfrac{d x}{d t}=3 a \sin^{2} t \cos t, \quad \dfrac{d y}{d t}=-3 b \cos^{2} t \sin t$
$\displaystyle \dfrac{d y}{d x}=-\dfrac{b}{a} \cot t$
$\displaystyle \left.\dfrac{d y}{d x}\right]_{t=\frac{\pi}{4}}=-\dfrac{b}{a}$
Continuity and DifferentiabilityDerivatives of Functions in Parametric FormsApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.