JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f(x)=\begin{cases}x-[x], & \text{if } [x] \text{ is odd}\\ 1+[x]-x, & \text{if } [x] \text{ is even}\end{cases}$. Then the value of $\frac{\pi^2}{10}\int_{-10}^{10}f(x)\cos\pi x\,dx$ is :
Definite Integration
Solution
Concept. $f$ is a triangle wave with period $2$ and $\cos\pi x$ also has period $2$; find the integral over one period and multiply by $10$. Formulas used. (1) With $t=x-2m$: $f=1-t$ on $[0,1)$ and $f=t-1$ on $[1,2)$; $\cos\pi x=\cos\pi t$(2) $\int_0^1u\cos\pi u\,du=-\dfrac2{\pi^2}$ (by parts)Step 1 - The graph. By (1):Step 2 - First half. By parts, $$\int_0^1(1-t)\cos\pi t\,dt=\Big[\frac{(1-t)\sin\pi t}{\pi}\Big]_0^1+\frac1\pi\int_0^1\sin\pi t\,dt=\frac2{\pi^2}.$$ Step 3 - Second half. With $u=t-1$, $\cos\pi t=-\cos\pi u$. By (2), $$\int_1^2(t-1)\cos\pi t\,dt=-\int_0^1u\cos\pi u\,du=\frac2{\pi^2}.$$ Step 4 - Total. One period gives $\frac4{\pi^2}$; $[-10,10]$ has $10$ periods: $\frac{40}{\pi^2}$. So $$\frac{\pi^2}{10}\cdot\frac{40}{\pi^2}=4.$$ Answer: (A)