JEE Main 2022 — Definite Integration Question with Solution

Case 1 :

Let $$0\le x<1$$

then $$\left[x\right]=0$$, which is even

$$\therefore$$ $$f(x)=1+\left[x\right]-x$$

$$=1+0-x$$

$$=1-x$$

Case 2 :

Let $$1\le x<2$$

then $$\left[x\right]=1$$, which is odd

$$\therefore$$ $$f(x)=x-\left[x\right]$$

$$=x-1$$

Case 3 :

Let $$2\le x<3$$

then $$\left[x\right]=2$$, which is even

$$\therefore$$ $$f(x)=1+\left[x\right]-x$$

$$=1+2-x$$

$$=3-x$$

Case 4 :

Let $$3\le x<4$$

then $$\left[x\right]=3$$, which is odd

$$\therefore$$ $$f(x)=x-\left[x\right]$$

$$=x-3$$

$$\therefore$$ f(x) = \left\{ {\matrix{ {1 - x} & ; & {0 \le x < 1} \cr {x - 1} & ; & {1 \le x < 2} \cr {3 - x} & ; & {2 \le x < 3} \cr {x - 3} & ; & {3 \le x < 4} \cr } } \right.

$$\therefore$$ $$f(x)$$ is periodic and period of $$f(x)=2$$

And period of $$\cos \pi x=\frac{2\pi }{\pi }=2$$

$$\therefore$$ Period of $$f(x)\cos \pi x=2$$

Now,

$$I=\frac{{\pi }^2}{10}{\int }_{-10}^{10}f(x)\cos \pi xdx$$

$$=\frac{{\pi }^2}{10}{\int }_{-10}^{-10+10\times 2}f(x)\cos \pi xdx$$

$$=\frac{{\pi }^2}{10}{\int }_0^{10\times 2}f(x)\cos \pi xdx$$

$$=\frac{{\pi }^2}{10}\times 10{\int }_0^{2}f(x)\cos \pi xdx$$

$$={\pi }^2{\int }_0^{2}f(x)\cos \pi xdx$$

$$\therefore$$ $$I={\pi }^2\left[{\int }_0^{1}f(x)\cos \pi xdx+{\int }_1^{2}f(x)\cos \pi xdx\right]$$

$$={\pi }^2\left[{\int }_0^1(1-x)\cos \pi xdx+{\int }_1^2(x-1)\cos \pi xdx\right]$$

$$={\pi }^2\left[{\int }_0^1\cos \pi xdx-{\int }_0^{1}x\cos \pi xdx+{\int }_1^{2}x\cos \pi xdx-{\int }_1^2\cos \pi xdx\right]$$

$$={\pi }^2\left[\frac{1}{\pi }{\left[\sin \pi x\right]}_0^1-{\int }_0^{1}x\cos \pi xdx+{\int }_1^{2}x\cos \pi xdx-\frac{1}{\pi }{\left[\sin \pi x\right]}_1^2\right]$$

$$={\pi }^2\left[0-{\int }_0^{1}x\cos \pi xdx+{\int }_1^{2}x\cos \pi xdx-0\right]$$

$$={\pi }^2\left[-{\left[x\frac{\sin \pi x}{\pi }+\frac{1}{{\pi }^2}\cos \pi x\right]}_0^1+{\left[x\frac{\sin \pi x}{\pi }+\frac{1}{{\pi }^2}\cos \pi x\right]}_1^2\right]$$

$$\left[\mathrm{As}\int x\cos \pi xdx=x.\int \cos \pi x-\int \left(1.\frac{\sin \pi x}{\pi }\right)dx=x.\frac{\sin \pi x}{\pi }+\frac{1}{{\pi }^2}\cos \pi x+c\right]$$

$$={\pi }^2\left[-\left[\left(1.\frac{\sin \pi }{\pi }+\frac{1}{{\pi }^2}.\cos \pi \right)-\left(0+\frac{1}{{\pi }^2}.\cos 0\right)\right]+\left[\left(2.\frac{\sin 2\pi }{\pi }+\frac{1}{{\pi }^2}\cos 2\pi \right)-\left(1.\frac{\sin \pi }{\pi }+\frac{1}{{\pi }^2}\cos \pi \right)\right]\right]$$

$$={\pi }^2\left[-\left\{\left(-\frac{1}{{\pi }^2}\right)-\left(\frac{1}{{\pi }^2}\right)\right\}+(\left(+\frac{1}{{\pi }^2}\right)-\left(-\frac{1}{{\pi }^2}\right)\right]$$

$$={\pi }^2\left[-\left(-\frac{2}{{\pi }^2}\right)+\frac{2}{{\pi }^2}\right]$$

$$={\pi }^2\left[\frac{2}{{\pi }^2}+\frac{2}{{\pi }^2}\right]$$

$$={\pi }^2\times \frac{4}{{\pi }^2}$$

$$=4$$