JEE Mains · Maths · STD 12 - 7.2 definite integral
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 \(f(x)=\left\{\begin{array}{cl}x-[x], & \text { if }(x) \text { is odd } \\ 1+[x]-x & \text { if }(x) \text { is even }\end{array}\right.\) Then the value of \(\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x d x\) is.
- A \(4\)
- B \(2\)
- C \(1\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(f ( x )\) is periodic function whose period is 2 \(\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x d x=\frac{\pi^{2}}{10} \times 10 \int_{0}^{2} f(x) \cos \pi x d x\) \(=\pi^{2}\left(\int_{0}^{1}(1-x) \cos \pi x d x+\int_{1}^{2}(x-1) \cos \pi x d x\right)\) Using by parts…
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