MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Use implicit differentiation twice on $x^{3}+y^{3}=6$ to find $\frac{d^{2}y}{dx^{2}}$ and $\frac{d^{2}x}{dy^{2}}$, then multiply them and compare with $\frac{m}{(xy)^{n}}$ to get $m$ and $n$.Explanation:Implicit differentiation treats $y$ as a function of $x$, so every $y$-term is multiplied by $\frac{dy}{dx}$.Differentiate $x^{3}+y^{3}=6$ with respect to $x$:$$3x^{2}+3y^{2}\frac{dy}{dx}=0$$$$\Rightarrow \frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$$Differentiate again with respect to $x$ using the quotient rule and chain rule:$$\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left(-x^{2}y^{-2}\right)$$$$=-2xy^{-2}+2x^{2}y^{-3}\frac{dy}{dx}$$Substitute $\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$:$$\frac{d^{2}y}{dx^{2}}=-\frac{2x}{y^{2}}-\frac{2x^{4}}{y^{5}}$$Factor using $x^{3}+y^{3}=6$:$$\frac{d^{2}y}{dx^{2}}=-\frac{2x(x^{3}+y^{3})}{y^{5}}=-\frac{12x}{y^{5}}$$Since the equation is symmetric in $x$ and $y$, differentiating with respect to $y$ gives:$$\frac{d^{2}x}{dy^{2}}=-\frac{12y}{x^{5}}$$Multiply the two second-order derivatives:$$\frac{d^{2}y}{dx^{2}}\cdot\frac{d^{2}x}{dy^{2}}=\left(-\frac{12x}{y^{5}}\right)\left(-\frac{12y}{x^{5}}\right)$$$$=\frac{144xy}{x^{5}y^{5}}=\frac{144}{x^{4}y^{4}}=\frac{144}{(xy)^{4}}$$Compare with $\frac{m}{(xy)^{n}}$: we get $m=144$ and $n=4$.Therefore,$$\frac{m}{n}=\frac{144}{4}=36$$Answer:$\frac{m}{n}=36$. Hence, the correct option is A. 36.