MHTCET 23 Apr 2025 Shift 1 Paper

Concept:In an infinite nested radical with a repeating pattern, replace the repeating inner core by the original variable $y$ to form an equation.Explanation:The given expression is:$$y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + \cdots}}}}$$After the first $x$, the inner part $\sqrt{y + \sqrt{x + \sqrt{y + \cdots}}}$ repeats with the same infinite pattern, so it equals $\sqrt{2y}$.Thus,$$y = \sqrt{x + \sqrt{2y}}$$Squaring both sides:$$y^2 = x + \sqrt{2y}$$$$\Rightarrow y^2 - x = \sqrt{2y}$$Squaring again:$$(y^2 - x)^2 = 2y$$Differentiate both sides with respect to $x$:$$2(y^2 - x)\left(2y \frac{dy}{dx} - 1\right) = 2 \frac{dy}{dx}$$Cancel $2$ and let $\frac{dy}{dx} = y'$:$$(y^2 - x)(2y y' - 1) = y'$$$$2y(y^2 - x)y' - (y^2 - x) = y'$$$$y'\left[2y(y^2 - x) - 1\right] = y^2 - x$$Therefore,$$\frac{dy}{dx} = \frac{y^2 - x}{2y^3 - 2xy - 1}$$Answer:Option C: $\frac{y^{2}-x}{2 y^{3}-2 x y-1}$