Question

For any real number $x$, let $\left[x\right]$ denotes the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f\left(x\right)=\{\begin{array}{c}x-\left[x\right],\text{if}\left[x\right]\text{is odd}\\ 1+\left[x\right]-x,\text{if}\left[x\right]\text{is even}\end{array}$
Then, the value of $\frac{{\pi }^2}{10}\underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx$ is

• A. $0$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

We have, $f\left(x\right)=\{\begin{array}{c}x-\left[x\right],\text{if}\left[x\right]\text{is odd}\\ 1+\left[x\right]-x,\text{if}\left[x\right]\text{is even}\end{array}$ $f\left(x\right)$ and $\cos \pi x$ are both periodic with period $2$ and both are even.
$\therefore \underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx=2\underset{0}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx$ $=10\underset{0}{\overset{2}{\int }}f\left(x\right)\cos \pi xdx$
Now,
$\underset{0}{\overset{1}{\int }}f\left(x\right)\cos \pi xdx=\underset{0}{\overset{1}{\int }}(1-x)\cos \pi xdx=-\underset{0}{\overset{1}{\int }}u\cos \pi udu$
$\underset{1}{\overset{2}{\int }}f\left(x\right)\cos \pi xdx=\underset{1}{\overset{2}{\int }}(x-1)\cos \pi xdx=-\underset{0}{\overset{1}{\int }}u\cos \pi udu\therefore \underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx=-20\underset{0}{\overset{1}{\int }}u\cos \pi udu=\frac{40}{{\pi }^2}\Rightarrow \frac{{\pi }^2}{10}\underset{-10}{\overset{10}{\int }}f\left(x\right)\cos \pi xdx=4$