Question

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 $\frac{\pi^2}{10} \int\limits_{-10}^{10} f(x) \cos \pi x d x$ is :

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (A)

$f(x)$ is periodic function whose period is 2
$ \frac{\pi^2}{10} \int\limits_{-10}^{10} f(x) \cos \pi x d x=\frac{\bar{\pi}^2}{10} \times 10 \int\limits_0^2 f(x) \cos \pi x d x $
$ =\pi^2\left(\int\limits_0^1(1-x) \cos \pi x d x+\int\limits_1^2(x-1) \cos \pi x d x\right)$
Using by parts
$=\pi^2 \times \frac{4}{\pi^2}=4$