Let f be an odd continous function which is periodic with priod 2 if g(x) = `int_0^x f(t)` dt then

Options

A. `g(x)` is odd
B. `2(n)=0, n epsilonN`
C. `g(2n)=0,n epsilonN`
D. `g(x)` is non-periodic

Answer

C

Explanation

`g(x)=int_(0)^(x)f(t)dt`
`g(-x)=int_(0)^(-x)f(t)dt=-int_(0)^(x)f(-t)dt=int_(0)^(x)f(t)dt` as `f(-t)=-f(t)`
or `g(-x)=g(x)`
Thus `g(x)` is even
Also `g(x+2)=int_(0)^(x+2)f(t)dt`

`=int_(0)^(2)f(t)dt+int_(2)^(2+x)f(t)dt`
`g(2)+int_(0)^(x)f(t+2)dt`
`=g(2)+int_(0)^(x)f(t)dt`
`=g(2)+g(x)`
Now `g(2) =int_(0)^(2)f(t)dt int_(0)^(1)f(t)dt+int_(1)^(2)f(t)dt`
`=int_(0)^(1)f(t)dt+int_(-1)^(0)f(t+2)dt`
`=int_(0)^(1)f(t)dt+int_(-1)^(0)f(t)dt`
`=int_(-1)^(1)f(t)dt=0` as `f(t)` is odd
`g(2)=0implies g(x+2)=g(x)`.
i.e. `g(x)` is periodic with period 2.
`:. g(4)0` or `g(6)=0, g(2n)=0, n epsilon N`.