MHTCET 14 May 2023 Shift 2 Solved Paper

To solve this problem, let's first observe the nested square root structure within the equation $y = \sqrt{(x - \sin x) + \sqrt{(x - \sin x) + \sqrt{(x - \sin x) + \dots}}}$ Since the structure of nested square roots repeats indefinitely, we can rewrite this as $y = \sqrt{(x - \sin x) + y}$Now let's square both sides:$y^2 = (x - \sin x) + y$Isolate $y$ on one side:$y^2 - y = x - \sin x$$y^2 - y - (x - \sin x) = 0$ This equation gives us a relationship between $x$ and $y$ that we can differentiate with respect to $x$. We will use implicit differentiation, differentiating both sides of the equation with respect to $x$ :$\frac{d}{dx}(y^2 - y - (x - \sin x)) = \frac{d}{dx}(0)$When differentiating the left side, keep in mind that $y$ is a function of $x$ ( $y = f(x)$ ). Applying the chain rule and using the fact that $\frac{d}{dx}(\sin x) = \cos x$, we get:$2y\frac{dy}{dx} - \frac{dy}{dx} - (1 - \cos x) = 0$ Reorganize the terms:$2y\frac{dy}{dx} - \frac{dy}{dx} = 1 - \cos x$$\frac{dy}{dx}(2y - 1) = 1 - \cos x$Now, solving for $\frac{dy}{dx}$ :$\frac{dy}{dx} = \frac{1 - \cos x}{2y - 1}$This matches Option A:$\frac{1 - \cos x}{2y - 1}$