MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Parametric differentiation is used to find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$, then substitute into the given expression.Explanation:Given $x = \sin t$ and $y = \cos(pt)$.First, $\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = -p \sin(pt)$.So, $\frac{dy}{dx} = \frac{-p \sin(pt)}{\cos t}$.Differentiate again with respect to $x$:$\frac{d^2y}{dx^2} = \frac{1}{\cos t} \cdot \frac{d}{dt}\left(\frac{-p \sin(pt)}{\cos t}\right)$.Applying the quotient rule:$\frac{d^2y}{dx^2} = \frac{1}{\cos t} \left[ -p^2 \cos(pt)\sec t - p \sin(pt)\sec t \tan t \right]$.Simplify:$\frac{d^2y}{dx^2} = -p^2 \cos(pt)\sec^2 t - p \sin(pt)\sec^2 t \tan t$.Now, $1 - x^2 = 1 - \sin^2 t = \cos^2 t$ and $x = \sin t$.Substitute into the given expression:$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx}$$= \cos^2 t\left[-p^2 \cos(pt)\sec^2 t - p \sin(pt)\sec^2 t \tan t\right] - \sin t\left(\frac{-p\sin(pt)}{\cos t}\right)$.This simplifies to:$-p^2 \cos(pt) - p \sin(pt)\tan t + p \sin(pt)\tan t$.The last two terms cancel, giving:$-p^2 \cos(pt)$.Since $y = \cos(pt)$, the expression equals $-p^2 y$.Answer:$-p^2 y$, which is option B.