MHTCET 10 May 2024 Shift 1 Paper

Concept:The absolute value function changes its sign depending on which of $\sin x$ and $\cos x$ is greater. The point of intersection in $0$ to $\frac{\pi}{2}$ is $x = \frac{\pi}{4}$, so the integral must be split there.Explanation:For $0 \le x \le \frac{\pi}{4}$, we have $\sin x \le \cos x$. Therefore, $|\sin x - \cos x| = \cos x - \sin x$.For $\frac{\pi}{4} \le x \le \frac{\pi}{2}$, we have $\sin x \ge \cos x$. Therefore, $|\sin x - \cos x| = \sin x - \cos x$.Split the given integral into these two intervals:$$I = \int_{0}^{\frac{\pi}{4}}(\cos x - \sin x)\,dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x - \cos x)\,dx$$Integrate each term:$$I = \bigl[\sin x + \cos x\bigr]_{0}^{\frac{\pi}{4}} + \bigl[-\cos x - \sin x\bigr]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$Evaluate the limits:$$I = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 0 - 1\right) - \left(0 + 1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$$Simplify the expression:$$I = \left(\frac{2}{\sqrt{2}} - 1\right) - \left(1 - \frac{2}{\sqrt{2}}\right) = \frac{4}{\sqrt{2}} - 2$$Rationalise $\frac{4}{\sqrt{2}} = 2\sqrt{2}$, giving:$$I = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$$Answer:The value of the integral is $2(\sqrt{2} - 1)$, which matches option C.