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
$f(x)=\{\begin{array}{ll}x-[x],& \text{ if }(x)\text{ is odd }\\ 1+[x]-x& \text{ if }(x)\text{ is even }\end{array}$
Then the value of $\frac{{\pi }^2}{10}\underset{-10}{\overset{10}{\int }}f(x)\cos \pi xdx$ is :

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (A)

$f(x)$ is periodic function whose period is 2
$\frac{{\pi }^2}{10}\underset{-10}{\overset{10}{\int }}f(x)\cos \pi xdx=\frac{{\bar{\pi }}^2}{10}\times 10\underset{0}{\overset{2}{\int }}f(x)\cos \pi xdx$
$={\pi }^2\left(\underset{0}{\overset{1}{\int }}(1-x)\cos \pi xdx+\underset{1}{\overset{2}{\int }}(x-1)\cos \pi xdx\right)$
Using by parts
$={\pi }^2\times \frac{4}{{\pi }^2}=4$