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 :

(1) $4$ Official answer
(2) $2$
(3) $1$
(4) $0$

Official answer

Option 1

NTA final key (2022 Session 2).

Chapter: Integrals
Topic: Definite Integral Properties
Idea tested: Periodic Function Property

Same idea in other shifts

Asked 6× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.