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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 2 · Q32
If x=3 sin t - sin 3 t, y =3 cos t - cos 3 t find dy/(d x) and prove that (d^2 y)/(d x^2)=(- cosec^3 2 t · cosec t)/3
If $\displaystyle x=3 \sin \mathrm{t}-\sin 3 \mathrm{t}, \mathrm{y}=3 \cos \mathrm{t}-\cos 3 \mathrm{t}$ find $\displaystyle \frac{\mathrm{dy}}{\mathrm{d} x}$ and prove that $\displaystyle \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}=\frac{-\operatorname{cosec}^{3} 2 \mathrm{t} \cdot \operatorname{cosec} \mathrm{t}}{3}$
Marking-scheme solution
$\displaystyle \dfrac{dx}{dt} = 3\cos t - 3\cos 3t,$
$\displaystyle \dfrac{dy}{dt} = -3\sin t + 3\sin 3t$
$\displaystyle \dfrac{dy}{dx} = \dfrac{3(\sin 3t - \sin t)}{3(\cos t - \cos 3t)}$
$\displaystyle = \dfrac{2\cos 2t \sin t}{2\sin 2t \sin t}$
$\displaystyle = \cot 2t$
$\displaystyle \dfrac{d^2y}{dx^2} = -2\operatorname{cosec}^2 2t \times \dfrac{dt}{dx}$
$\displaystyle = -\dfrac{2\operatorname{cosec}^2 2t}{6\sin 2t \sin t}$
$\displaystyle = -\dfrac{\operatorname{cosec}^3 2t \operatorname{cosec} t}{3}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.