IIT-JEE 2010 · Paper 1 · Mathematics Q52

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MathematicsInteger+3Question 52 of 84

For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [−10, 10] by
$f(x)=\begin{cases}x-[x]&\text{if }[x]\text{ is odd,}\\1+[x]-x&\text{if }[x]\text{ is even.}\end{cases}$
Then the value of $\dfrac{\pi^2}{10}\displaystyle\int_{-10}^{10}f(x)\cos\pi x\,dx$ is

Your answer

Answer and worked solution

Official answer: 4

Shape of $f$ . On $[2k,2k+1)$ ( $[x]$ even) $f=1-\{x\}$ , and on $[2k+1,2k+2)$ ( $[x]$ odd) $f=\{x\}$ . So $f(x)=|x-(2k+1)|$ on $[2k,2k+2]$ : a triangle wave of period 2.

One period. With $u=x-1$ :
$\int_0^2|x-1|\cos\pi x\,dx=-\int_{-1}^1|u|\cos\pi u\,du=-2\int_0^1u\cos\pi u\,du=-2\left(-\frac2{\pi^2}\right)=\frac4{\pi^2}.$

Whole interval. $[-10,10]$ is 10 periods: $\frac{40}{\pi^2}$ . Then $\frac{\pi^2}{10}\cdot\frac{40}{\pi^2}=4$ .

Answer: 4

Topics:Properties of definite integrals Graphs of functions

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IIT-JEE 2010 · Paper 1, Mathematics question 52