CBSE 2026 · Region 2 · Set 2 · Q25 · 2 marks
If $\displaystyle x=\sin \mathrm{t}-\cos \mathrm{t}, \mathrm{y}=\sin \mathrm{t} \cos \mathrm{t}$, find $\displaystyle \frac{\mathrm{dy}}{\mathrm{d} x}$ at $\displaystyle \mathrm{t}=\frac{\pi}{4}$.
Marking-scheme solution
$\displaystyle x=\sin t-\cos t \Rightarrow \dfrac{dx}{dt}=\cos t+\sin t$
$\displaystyle \mathrm{y}=\sin t \cos t \Rightarrow \dfrac{dy}{dt}=-\sin^{2} t+\cos^{2} t$
$\displaystyle \dfrac{dy}{dx}=\dfrac{-\sin^{2} t+\cos^{2} t}{\cos t+\sin t}$ or $\displaystyle \cos t-\sin t$
$\displaystyle \therefore \left.\dfrac{dy}{dx}\right]_{t=\frac{\pi}{4}}=0$
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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.