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 $\frac{\pi^2}{10}\int_{-10}^{10}f(x)\cos\pi x\,dx$ is :

JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 25 July 2022, Shift 1, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

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]

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 :

Official answer

Option 1

NTA final key (2022 Session 2).