For any real number x , let [x] denote the largest integer less...

For any real number $x$ , let \left[x\right] $denote the largest integer less than or equal to$ x $. Let$ f be a real-valued function defined on the interval \left[- 10, 10\right] $by $ f\left(x\right) = \left\{ $\begin{cases}x - \left[x\right], & \text{if}\left[x\right]\text{is odd} \\ 1 + \left[x\right] - x, & \text{if}\left[x\right]\text{is even}\end{cases}$ \right)

Then, the value of $\dfrac{{\pi}^{2}}{10}{∫}_{- 10}^{10}f\left(x\right)cos \pi xdx$ is

For any real number x , let [x] denote the largest integer less than or equal to x . Let f be a real-valued function

Options

A $4$
B $2$
C $1$
D $0$

Correct answer

A. $4$

Step-by-step solution

$f\left(x\right) = \left\{$ $\begin{array}{ll}x - 1, \\ 1 - x,\end{array}$ $$\begin{array}{ll}1 \le x < 2 \\ 0 \le x < 1\end{array}$ \right)
$f\left(x\right)$ is periodic with period $2$ , and is even.
$∴ I = {∫}_{- 10}^{10}f\left(x\right)cos \pi xdx$
$= 2{∫}_{0}^{10}f\left(x\right)cos \pi xdx$
$= 2 \times 5{∫}_{0}^{2}f\left(x\right)cos \pi xdx$
$= 10\left[{∫}_{0}^{1}\left(1 - x\right)cos \pi xdx + {∫}_{1}^{2}\left(x - 1\right)cos \pi xdx\right] = 10\left({I}_{1} + {I}_{2}\right)$
${I}_{2} = {∫}_{1}^{2}\left(x - 1\right)cos \pi xdx$
Put \left(x - 1\right) = t $$ {I}_{2} = - {∫}_{0}^{1}tcos \pi tdt $$ {I}_{1} = {∫}_{0}^{1}\left(1 - x\right)cos \pi xdx = - {∫}_{0}^{1}xcos\left(\pix\right)dx $$ ∴ I = 10\left[- 2{∫}_{0}^{1}xcos \pi xdx\right] = - 20{\left[x\dfrac{sin \pix}{\pi} + \dfrac{cos \pix}{{\pi}^{2}}\right]}_{0}^{1} $$ = - 20\left[- \dfrac{1}{{\pi}^{2}} - \dfrac{1}{{\pi}^{2}}\right] = \dfrac{40}{{\pi}^{2}} $$ ∴ \dfrac{{\pi}^{2}}{10}I = 4$

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