Let f(x) be an odd continous function which is periodic with period 2. If ` g(x)=int_(0)^(x) f(t) dt` , then

Options

A. g(x) is an odd function
B. `g(n) =0 "for all" n in N`
C. `g(2n)=0 "for all" n in N`
D. g(x) is non periodic

Answer

C

Explanation

It is given that f(x) is odd function. Therefore, `g(x)=underset(0)overset(x)intf(t)dt` dt is an even function.
`rArr g(x+2)=underset(0)overset(x+2)intf(t)dt`
`rArr g(x+2)=underset(0)overset(2)intf(t)dtunderset(0)overset(+2)intf(t)dt`
`rArr g(x+2)=g(2)+underset(0)overset(x)intf(t) [ " " :. "f is periodic with period" 2]`
`rAr g(x+2)=g(2)+g(x) "for all" " " ...(i)`
Now,
`g(x+2)=g(2)+g(x)`
`rArr g(1)=g(2)+g(-1) " "` [Replacing x by-1]
`rArr g(2)=g(1)-g(-1)`
`rArr g(2)=underset(0)overset(1)intf(t)dt-underset(0)overset(-1)intf(t)dt`
`rArr g(2)=underset(0)overset(1)intf(t)dt+underset(-1)overset(0)intf(t)dt`
`rArr g(2)=underset(-1)overset(1)intf(t)dt`
`rArr g(2)=0 " " [ :. "f is an odd function"]`
Putting g g(2) =0 in (i), we get ltrbgt `g(x+2)=g(x) "for all" x`
`rArr g(x)` is periodic with period 2 units
So,
`g(x+2)=g(x)` for all x `
`g(2)=g(4)=g(6)=....g(2n)
g (2n) 0 for all `n in N`