JEE Mathematics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: The region is bounded by $y = |\cos x - \sin x|$ and $y = \sin x$ for $0 \leq x \leq \frac{\pi}{2}$.

Find: The area of the given region and the correct option.

From the solution, the intersection point of $\cos x - \sin x = \sin x$ is obtained from

$$\tan x = \frac{1}{2}$$

Let $\psi = \tan^{-1}\left(\frac{1}{2}\right)$. Then

$$\sin \psi = \frac{1}{\sqrt{5}}, \qquad \cos \psi = \frac{2}{\sqrt{5}}$$

The area is written as

$$\text{Area} = \int_{\psi}^{\pi/2} \left(\sin x - |\cos x - \sin x|\right) \, dx$$

Now split at $x = \frac{\pi}{4}$, where the expression inside the modulus changes sign:

$$\text{Area} = \int_{\psi}^{\pi/4} \left(\sin x - (\cos x - \sin x)\right) \, dx + \int_{\pi/4}^{\pi/2} \left(\sin x - (\sin x - \cos x)\right) \, dx$$

So,

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

Evaluating,

$$= \left[-2\cos x - \sin x\right]_{\psi}^{\pi/4} + \left[\sin x\right]_{\pi/4}^{\pi/2}$$

Substituting the limits,

$$= \left(-2 \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) + 2\cos \psi + \sin \psi + \left(1 - \frac{1}{\sqrt{2}}\right)$$

Using $\cos \psi = \frac{2}{\sqrt{5}}$ and $\sin \psi = \frac{1}{\sqrt{5}}$,

$$= -\frac{3}{\sqrt{2}} + \frac{4}{\sqrt{5}} + \frac{1}{\sqrt{5}} + 1 - \frac{1}{\sqrt{2}}$$ $$= 1 + \sqrt{5} - 2\sqrt{2}$$

Therefore, the area is $\sqrt{5} - 2\sqrt{2} + 1$. The correct option is D.

The solution marks option A, but its own working concludes $\sqrt{5} - 2\sqrt{2} + 1$, which matches option D.

Common mistakes

• Using the interval $0$ to $\frac{\pi}{2}$ directly without first finding where $\sin x = |\cos x - \sin x|$ is wrong, because the bounded region starts from the intersection point $x = \psi$. First solve $\cos x - \sin x = \sin x$.

• Not splitting the modulus at $x = \frac{\pi}{4}$ is incorrect, because $|\cos x - \sin x|$ changes form there. Use $\cos x - \sin x$ on $\left[\psi, \frac{\pi}{4}\right]$ and $\sin x - \cos x$ on $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$.

• Taking the lower curve as $\cos x - \sin x$ throughout can give negative area on part of the interval. The absolute value must be handled carefully so that the integrand remains upper curve minus lower curve.