JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q9

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MathematicsSingle correct+4 / -1Question 9 of 90

For any real number $x$ , let $[x]$ denote the largest integer less than 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 $\dfrac{\pi^2}{10}\displaystyle\int_{-10}^{10}f(x)\cos\pi x\,dx$ is :

Answer and worked solution

Official answer: A

Periodic. $f$ has period 2 and on $[0,1)$ equals $1-x$ , on $[1,2)$ equals $x-1$ : it is $|1-x|$ repeated, an even triangle wave.

One period. $\int_0^2|1-x|\cos\pi x\,dx=\frac4{\pi^2}$ ; ten periods give $\frac{40}{\pi^2}$ , and $\frac{\pi^2}{10}\cdot\frac{40}{\pi^2}=4$ .

(A)

Topics:Properties of definite integrals

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JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 9