The area of the region enclosed by the curve f(x)=\max sin x, \cos x\,-\pi \leq x \leq \pi and the x -axis is

Area Under the Curves PYQ — JEE Main 2023 · 13 Apr · Shift 1 · Q23 with Solution

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Correct answer: B

We need the area enclosed by the curve

$y=f(x)=\max\{\sin x,\cos x\}, \qquad -\pi\le x\le \pi$

and the $x$ -axis.

So the required area is

$A=\int_{-\pi}^{\pi} |f(x)|\,dx$

because the graph may lie above or below the $x$ -axis.

First, determine where $\sin x=\cos x$ .

$\sin x=\cos x \implies \tan x=1 \implies x=\frac{\pi}{4}+n\pi$

In $[-\pi,\pi]$ , the relevant points are

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

These points split the interval into three parts.

Find $f(x)=\max\{\sin x,\cos x\}$ on each interval.

For $x\in[-\pi,-3\pi/4]$ , take a test point, say $x=-\pi$ : $\sin(-\pi)=0,\quad \cos(-\pi)=-1$ so $\sin x>\cos x$ . Hence $f(x)=\sin x.$
For $x\in[-3\pi/4,\pi/4]$ , take $x=0$ : $\sin 0=0,\quad \cos 0=1$ so $\cos x>\sin x$ . Hence $f(x)=\cos x.$
For $x\in[\pi/4,\pi]$ , take $x=\pi/2$ : $\sin\frac\pi2=1,\quad \cos\frac\pi2=0$ so $\sin x>\cos x$ . Hence $f(x)=\sin x.$

Thus,

f(x)= \begin{cases} \sin x, & -\pi\le x\le -\frac{3\pi}{4},\\[4pt] \cos x, & -\frac{3\pi}{4}\le x\le \frac{\pi}{4},\\[4pt] \sin x, & \frac{\pi}{4}\le x\le \pi. \end{cases}

Now check the sign of $f(x)$ .

On $[-\pi,-3\pi/4]$ , $f(x)=\sin x\le 0$ .
On $[-3\pi/4,-\pi/2]$ , $f(x)=\cos x\le 0$ .
On $[-\pi/2,\pi/4]$ , $f(x)=\cos x\ge 0$ .
On $[\pi/4,\pi]$ , $f(x)=\sin x\ge 0$ on $[\pi/4,\pi]$ except at $x=\pi$ where it is $0$ .

So area is

$A=\int_{-\pi}^{-3\pi/4}(-\sin x)\,dx+\int_{-3\pi/4}^{-\pi/2}(-\cos x)\,dx+\int_{-\pi/2}^{\pi/4}\cos x\,dx+\int_{\pi/4}^{\pi}\sin x\,dx.$

Evaluate each integral.

First,

=\cos\left(-\frac{3\pi}{4}\right)-\cos(-\pi) =-\frac{\sqrt2}{2}-(-1)=1-\frac{\sqrt2}{2}.$$ Second, $$\int_{-3\pi/4}^{-\pi/2}(-\cos x)\,dx=[-\sin x]_{-3\pi/4}^{-\pi/2} =-\sin\left(-\frac{\pi}{2}\right)+\sin\left(-\frac{3\pi}{4}\right) =1-\frac{\sqrt2}{2}.$$ Third, $$\int_{-\pi/2}^{\pi/4}\cos x\,dx=[\sin x]_{-\pi/2}^{\pi/4} =\sin\frac\pi4-\sin\left(-\frac\pi2\right) =\frac{\sqrt2}{2}+1.$$ Fourth, $$\int_{\pi/4}^{\pi}\sin x\,dx=[-\cos x]_{\pi/4}^{\pi} =-\cos\pi+\cos\frac\pi4=1+\frac{\sqrt2}{2}.$$ 6. Add them: $$A=\left(1-\frac{\sqrt2}{2}\right)+\left(1-\frac{\sqrt2}{2}\right)+\left(1+\frac{\sqrt2}{2}\right)+\left(1+\frac{\sqrt2}{2}\right)=4.$$ Hence the area is $$\boxed{4}.$$ 7. Compare with options: - A: $2\sqrt2(\sqrt2+1)=4+2\sqrt2\ne 4$ - B: $4$ ✓ - C: $2(\sqrt2+1)\ne 4$ - D: $4\sqrt2\ne 4$ Therefore, the correct option is $$\boxed{\text{B}}.$$

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View written solutionFree

Correct answer: B

We need the area enclosed by the curve

$y=f(x)=\max\{\sin x,\cos x\}, \qquad -\pi\le x\le \pi$

and the $x$ -axis.

So the required area is

$A=\int_{-\pi}^{\pi} |f(x)|\,dx$

because the graph may lie above or below the $x$ -axis.

First, determine where $\sin x=\cos x$ .

$\sin x=\cos x \implies \tan x=1 \implies x=\frac{\pi}{4}+n\pi$

In $[-\pi,\pi]$ , the relevant points are

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

These points split the interval into three parts.

Find $f(x)=\max\{\sin x,\cos x\}$ on each interval.

For $x\in[-\pi,-3\pi/4]$ , take a test point, say $x=-\pi$ : $\sin(-\pi)=0,\quad \cos(-\pi)=-1$ so $\sin x>\cos x$ . Hence $f(x)=\sin x.$
For $x\in[-3\pi/4,\pi/4]$ , take $x=0$ : $\sin 0=0,\quad \cos 0=1$ so $\cos x>\sin x$ . Hence $f(x)=\cos x.$
For $x\in[\pi/4,\pi]$ , take $x=\pi/2$ : $\sin\frac\pi2=1,\quad \cos\frac\pi2=0$ so $\sin x>\cos x$ . Hence $f(x)=\sin x.$

Thus,

f(x)= \begin{cases} \sin x, & -\pi\le x\le -\frac{3\pi}{4},\\[4pt] \cos x, & -\frac{3\pi}{4}\le x\le \frac{\pi}{4},\\[4pt] \sin x, & \frac{\pi}{4}\le x\le \pi. \end{cases}

Now check the sign of $f(x)$ .

On $[-\pi,-3\pi/4]$ , $f(x)=\sin x\le 0$ .
On $[-3\pi/4,-\pi/2]$ , $f(x)=\cos x\le 0$ .
On $[-\pi/2,\pi/4]$ , $f(x)=\cos x\ge 0$ .
On $[\pi/4,\pi]$ , $f(x)=\sin x\ge 0$ on $[\pi/4,\pi]$ except at $x=\pi$ where it is $0$ .

So area is

$A=\int_{-\pi}^{-3\pi/4}(-\sin x)\,dx+\int_{-3\pi/4}^{-\pi/2}(-\cos x)\,dx+\int_{-\pi/2}^{\pi/4}\cos x\,dx+\int_{\pi/4}^{\pi}\sin x\,dx.$

Evaluate each integral.

First,

=\cos\left(-\frac{3\pi}{4}\right)-\cos(-\pi) =-\frac{\sqrt2}{2}-(-1)=1-\frac{\sqrt2}{2}.$$ Second, $$\int_{-3\pi/4}^{-\pi/2}(-\cos x)\,dx=[-\sin x]_{-3\pi/4}^{-\pi/2} =-\sin\left(-\frac{\pi}{2}\right)+\sin\left(-\frac{3\pi}{4}\right) =1-\frac{\sqrt2}{2}.$$ Third, $$\int_{-\pi/2}^{\pi/4}\cos x\,dx=[\sin x]_{-\pi/2}^{\pi/4} =\sin\frac\pi4-\sin\left(-\frac\pi2\right) =\frac{\sqrt2}{2}+1.$$ Fourth, $$\int_{\pi/4}^{\pi}\sin x\,dx=[-\cos x]_{\pi/4}^{\pi} =-\cos\pi+\cos\frac\pi4=1+\frac{\sqrt2}{2}.$$ 6. Add them: $$A=\left(1-\frac{\sqrt2}{2}\right)+\left(1-\frac{\sqrt2}{2}\right)+\left(1+\frac{\sqrt2}{2}\right)+\left(1+\frac{\sqrt2}{2}\right)=4.$$ Hence the area is $$\boxed{4}.$$ 7. Compare with options: - A: $2\sqrt2(\sqrt2+1)=4+2\sqrt2\ne 4$ - B: $4$ ✓ - C: $2(\sqrt2+1)\ne 4$ - D: $4\sqrt2\ne 4$ Therefore, the correct option is $$\boxed{\text{B}}.$$

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