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

Then the value of $\mathrm{\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x d x}$ is :
Option: 1
$4$

Option: 2
$2$

Option: 3
$1$

Option: 4
$0$

Answers (1)

best_answer

$\mathrm{f(x)}$ is periodic function where period is $\mathrm{2}$

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

$\mathrm{=\pi^{2}\left(\int_{0}^{1}(1-x) \cos \pi x d x+\int_{1}^{2}(x-1) \cos \pi x d x\right) }\\$

$\text { using by parts } \\$

$\mathrm{=\pi^{2} \times \frac{4}{\pi^{2}}=4}$

Hence correct option is 1

Posted by

shivangi.bhatnagar

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