JEE Mathematics 2023 Question with Solution

Given: $f(x) = \max\{\sin x, \cos x\}$ on $-\pi \leq x \leq \pi$.

Find: The area enclosed by the curve and the x-axis.

The solution analyzes where $\sin x = \cos x$, so we first find the switching points:

$$x = \frac{\pi}{4}, \; -\frac{3\pi}{4}$$

Thus, as extracted from the solution, the curve is taken as:

• $f(x) = \sin x$ for $-\frac{3\pi}{4} \leq x \leq \frac{\pi}{4}$
• $f(x) = \cos x$ for $\frac{\pi}{4} \leq x \leq \frac{3\pi}{4}$

So the area is written as

$$A_1 = \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \sin x \, dx$$

and

$$A_2 = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos x \, dx$$

Now evaluate:

$$A_1 = -\cos x \Big|_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}$$

the solution simplifies this to $\sqrt{2}$.

Also,

$$A_2 = \sin x \Big|_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$$

the solution simplifies this to $\sqrt{2}$.

Hence,

$$\text{Total Area} = A_1 + A_2 = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$$

However, the same the solution finally states that the total area is $4$ and concludes that the correct option is C. Since the source solution concludes with option (3), the extracted answer is C.

Therefore, the correct option is C.

Given: $f(x) = \max\{\sin x, \cos x\}$.

Find: The area enclosed by the curve and the x-axis.

The provided solution contains an internal inconsistency:

1. It computes

$$A_1 = \sqrt{2}$$

2. It computes

$$A_2 = \sqrt{2}$$

3. Then it adds them to get

$$A_1 + A_2 = 2\sqrt{2}$$

4. But immediately after that, it states the total area is $4$ and marks option (3) as correct.

the final stated conclusion on the solution is taken as the authoritative answer. Therefore, despite the intermediate mismatch, the answer is recorded as C.

Therefore, the correct option is C.