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)`={:{(x-[x]","" if"[x]" is odd "),(1+[x]-x","" if"[x]" is even"):}`
Then the value of `(pi^(2))/(10)int_(-10)^(10) f(x) cos pi x dx`, is

Options

A. 3
B. 4
C. 6
D. 9

Answer

B

Explanation

Clearly, f(x) is a periodic function with period2. Also, `cos pi` x is periodic with period 2.
`:. I=overset(10)underset(-10)int f(x) cos pi xdx`
`rArr I=2overset(10)underset(0)int f(x) cos pi xdx`
` rArr I=2xx5overset(2)underset(0)int f(x)cos pix dx`
`rArr 10{overset(1)underset(0)int (1-x)cospixxdx+overset(2)underset(1)int (x-1)cos pi xdx}`
`rArr I=10[[(1)/(pi)(1-x)sinpix]_(0)^(1)-[(cos pix)/(pi^(2))]_(0)^(1)+[(1)/(pi)(x-1)sin pi x]_(1)^(2)+[(cos pix)/(pi^(2))]_(1)^(2)]`
`rArr I=10[-(1)/(pi^(2))(-1-1)+(1)/(pi^(2))(1+1)]=(40)/(pi^(2))`
Hence,`(pi^(2))/(10)overset(10)underset(-10)int f(x)dx=(pi^(2))/(10)xx(40)/(pi^(2))=4`