For any real number x , let \([\mathrm{x}]\) denote the largest integer less than equal to x . Let f be a real valued function defined on the interval \([-10,10]\) by \(f(x)=\left\{\begin{array}{cc}x-[x], & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.\) Then the value of \(\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x d x\) is :

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Definite Integrals

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) = {x - [x], 1 + [x] - x if [x] is odd if [x] is even

Then the value of

\frac{π ^{2}}{10} \int_{- 10}^{10} f (x) cos π x d x

is :

Answer: (A)

The function

f (x)

is periodic with period 2. Using the property of periodic functions:

\frac{π ^{2}}{10} \int_{- 10}^{10} f (x) cos π x d x = \frac{π ^{2}}{10} \times 10 \int_{0}^{2} f (x) cos π x d x

= π^{2} (\int_{0}^{1} (1 - x) cos π x d x + \int_{1}^{2} (x - 1) cos π x d x)

Using integration by parts on each integral:

For

\int_{0}^{1} (1 - x) cos π x d x

: Let

u = 1 - x

d v = cos π x d x

After evaluation, each integral contributes

\frac{2}{π ^{2}}

, giving:

π^{2} \times \frac{4}{π ^{2}} = 4

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option