MHTCET 11 Apr 2026 Shift 1 Paper

Concept:The area of the shaded region equals the sum of definite integrals of the given curves over their respective intervals.Explanation:From the figure, the shaded region is split into two parts.The first part lies under $y = \sin x$ from $x = 2\pi$ to $x = \frac{9\pi}{4}$.The second part lies under $y = \cos x$ from $x = \frac{9\pi}{4}$ to $x = \frac{5\pi}{2}$.Therefore, the total area is:$$A = \int_{2\pi}^{\frac{9\pi}{4}} \sin x \, dx + \int_{\frac{9\pi}{4}}^{\frac{5\pi}{2}} \cos x \, dx$$Evaluating the first integral:$$\int_{2\pi}^{\frac{9\pi}{4}} \sin x \, dx = [-\cos x]_{2\pi}^{\frac{9\pi}{4}}$$$$= -\cos\frac{9\pi}{4} + \cos 2\pi$$Since $\cos\frac{9\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos 2\pi = 1$:$$= 1 - \frac{1}{\sqrt{2}}$$Evaluating the second integral:$$\int_{\frac{9\pi}{4}}^{\frac{5\pi}{2}} \cos x \, dx = [\sin x]_{\frac{9\pi}{4}}^{\frac{5\pi}{2}}$$$$= \sin\frac{5\pi}{2} - \sin\frac{9\pi}{4}$$Since $\sin\frac{5\pi}{2} = 1$ and $\sin\frac{9\pi}{4} = \frac{1}{\sqrt{2}}$:$$= 1 - \frac{1}{\sqrt{2}}$$Adding both areas:$$A = \left(1 - \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$$Answer:$2 - \sqrt{2}$ sq.units, which is Option A.