MHTCET 25 Apr 2025 Shift 2 Paper

Concept:This problem is based on parametric differentiation and the elimination of the parameter.We use $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$, then differentiate again with respect to $x$.Explanation:Given $x=\sin t$ and $y=\sin(pt)$.Differentiate both with respect to $t$:$\frac{dx}{dt}=\cos t$ and $\frac{dy}{dt}=p\cos(pt)$.Therefore,$\frac{dy}{dx}=\frac{p\cos(pt)}{\cos t}$.Since $\cos t=\sqrt{1-x^{2}}$ and $\cos(pt)=\sqrt{1-y^{2}}$, we get:$\frac{dy}{dx}=\frac{p\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$.Now, differentiate $\frac{dy}{dx}$ with respect to $x$ using the quotient rule.Let $u=\sqrt{1-y^{2}}$ and $v=\sqrt{1-x^{2}}$.Then $y'=\frac{p u}{v}$.Using $\frac{d}{dx}\sqrt{1-y^{2}}=\frac{-y y'}{\sqrt{1-y^{2}}}$ and $\frac{d}{dx}\sqrt{1-x^{2}}=\frac{-x}{\sqrt{1-x^{2}}}$:$\frac{d^{2}y}{dx^{2}}=p\frac{\frac{-y y'}{\sqrt{1-y^{2}}}\sqrt{1-x^{2}}-\sqrt{1-y^{2}}\left(\frac{-x}{\sqrt{1-x^{2}}}\right)}{1-x^{2}}$.Multiplying both sides by $(1-x^{2})$:$(1-x^{2})\frac{d^{2}y}{dx^{2}}=-\frac{py y'\sqrt{1-x^{2}}}{\sqrt{1-y^{2}}}+\frac{px\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$.Substitute $\frac{dy}{dx}=\frac{p\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$:First term: $-\frac{p y \cdot p\sqrt{1-y^{2}}\sqrt{1-x^{2}}}{\sqrt{1-y^{2}}\sqrt{1-x^{2}}}=-p^{2}y$.Second term: $\frac{px\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}=x\frac{dy}{dx}$.So,$(1-x^{2})\frac{d^{2}y}{dx^{2}}=-p^{2}y+x\frac{dy}{dx}$.Rearranging:$(1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+p^{2}y=0$.Answer:The value of the given expression is $0$.Hence, the correct option is A. 0.