CBSE 2026 · Region 1 · Set 3 · Q35 · 5 marks
If $\displaystyle x=a(\sin t-t \cos t)$ and $\displaystyle y=b(\cos t+t \sin t)$, then find $\displaystyle \frac{d y}{d x}$ and $\displaystyle \frac{d^{2} y}{d x^{2}}$.
Marking-scheme solution
$\displaystyle \dfrac{dx}{dt} = a(\cos t + t\sin t - \cos t) = at\sin t,$
$\displaystyle \dfrac{dy}{dt} = b(-\sin t + t\cos t + \sin t) = bt\cos t$
$\displaystyle \dfrac{dy}{dx} = \dfrac{bt\cos t}{at\sin t}$
$\displaystyle = \dfrac{b}{a}\cot(t)$
$\displaystyle \dfrac{d^2y}{dx^2} = -\dfrac{b}{a}\operatorname{cosec}^2 t \times \dfrac{dt}{dx}$
$\displaystyle = -\dfrac{b}{a}\operatorname{cosec}^2 t \times \dfrac{1}{at\sin t}$ or $\displaystyle -\dfrac{b\operatorname{cosec}^3 t}{a^2 t}$
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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.