MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Implicit differentiation of the given radical equation, followed by rationalization to match the required form of $\frac{dy}{dx}$.Explanation:Let $u=\sqrt{y+x}$ and $v=\sqrt{y-x}$.Given $u+v=c$.Differentiate both sides with respect to $x$:$$\frac{1}{2\sqrt{y+x}}\left(\frac{dy}{dx}+1\right)+\frac{1}{2\sqrt{y-x}}\left(\frac{dy}{dx}-1\right)=0$$Multiply by 2 and solve for $\frac{dy}{dx}$:$$\frac{dy}{dx}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}$$Rationalize the numerator:$$\frac{dy}{dx}=\frac{(y+x)-(y-x)}{\left(\sqrt{y+x}+\sqrt{y-x}\right)^2}$$Since $\left(\sqrt{y+x}+\sqrt{y-x}\right)^2=2y+2\sqrt{y^2-x^2}$, we get:$$\frac{dy}{dx}=\frac{2x}{2y+2\sqrt{y^2-x^2}}=\frac{x}{y+\sqrt{y^2-x^2}}$$Rationalize the denominator:$$\frac{dy}{dx}=\frac{y-\sqrt{y^2-x^2}}{x}$$Split the fraction:$$\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}$$This matches $\frac{dy}{dx}=f(x)-\sqrt{[f(x)]^2-1}$.Answer:$f(x)=\frac{y}{x}$, so the correct option is A. $\frac{y}{x}$.