For a real number x let [x] denote the largest integer les than or eqaul to x and {x} =x -[x]. Let n be a positive integer. Then`int_(0)^(1) cos(2pi[x]{x})dx` is equal to

Options

A. 0
B. 1
C. n
D. 2n-1

Answer

B

Explanation

`underset(0)overset(1)intcos(2pi[x]{x})dx`
`=underset(0)overset(n)intcos(o)dx+underset(1)overset(2)intcos(2pi(x-1))dx+underset(2)overset(3)intcos(4pi(x-2))dx+....+underset(n-1)overset(n)intcos(2pi(n-1)(x-(n-1)))dx`
`=(1-0)+underset(1)overset(2)intcos2pixdx+underset(2)overset(3)intcso4pixdt+....+underset(n-1)overset(n)intcos(2pi(n-1)x)dx`
`=1+(sin2pix)/(2pi):|_(1)^(2)++(sin4pix)/(4pi):|_(2)^(3)+....+(sin2pi(n-1))/(2pi(n-1)):|_(n-1)^(n)`
`1+0=1`