Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b`
If `f(x)` is an odd function, then

Options

A. (a) `phi(x)` is also an odd function
B. (b) `phi(x)` is an even function
C. (c) `phi(x)` is neither an even nor an odd function
D. (d) for `phi(x)` to be an even function, it must satisfy `int_(0)^(a)f(x)dx=0`

Answer

B

Explanation

`f(x)` is an odd function. Thus, `f(x)=-f(-x)`
`phi(-x)=int_(0)^(-x)f(t)dt`
Put `t=-y`
`:. phi(-x)=int_(-a)^(x)f(-t)(-dt)`
`=int_(-a)^(x)f(t)dt=int_(-a)^(a)f(t)dt+int_(a)^(x)f(t)dt`
`=0+int_(a)^(x)f(t)dt=phi(x)`.