JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The area of the region $A = \{(x, y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\}$ is:

Choose the correct answer:

Explanation

Solution:

Condition: $|\cos x - \sin x| \leq \sin x$

If $\cos x \geq \sin x$ (i.e., $0 \leq x \leq \frac{\pi}{4}$ ):

$\cos x - \sin x \leq \sin x \Rightarrow \cos x \leq 2\sin x \Rightarrow \tan x \geq \frac{1}{2}$

Intersection point: $x = \alpha$ where $\tan \alpha = \frac{1}{2}$ ( $\sin \alpha = \frac{1}{\sqrt{5}}, \cos \alpha = \frac{2}{\sqrt{5}}$ ).
If $\sin x > \cos x$ (i.e., $\frac{\pi}{4} < x \leq \frac{\pi}{2}$ ):

$\sin x - \cos x \leq \sin x \Rightarrow -\cos x \leq 0 \Rightarrow \cos x \geq 0$ (Always true in this range).

Area Calculation:

\text{Area} = \int_{\alpha}^{\pi/4} (\sin x - (\cos x - \sin x)) dx + \int_{\pi/4}^{\pi/2} (\sin x - (\sin x - \cos x)) dx

\text{Area} = \int_{\alpha}^{\pi/4} (2\sin x - \cos x) dx + \int_{\pi/4}^{\pi/2} \cos x dx

Performing integration:

= [-2\cos x - \sin x]_{\alpha}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}

= (-2\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - (-2\cos \alpha - \sin \alpha) + (1 - \frac{1}{\sqrt{2}})

= -\frac{3}{\sqrt{2}} + (2 \cdot \frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}) + 1 - \frac{1}{\sqrt{2}}

= \frac{5}{\sqrt{5}} - \frac{4}{\sqrt{2}} + 1 = \sqrt{5} - 2\sqrt{2} + 1

Correct Option: (4)