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 even function, then

Options

A. `phi(x)` is also an even function
B. `phi(x)` is an odd function
C. `phi(x)` is an neither even nor odd function
D. None of these

Answer

D

Explanation

If `f(x)` is an even function, then
`phi(-x)=-int_(-a)^(x)f(t)dt`
`=-int_(-a)^(a)f(t)dt-int_(a)^(x)f(t)dt`
`=-2int_(0)^(a)f(t)dt-int_(a)^(x)f(t)dt` [as `f(x)` is an even function]
Now `int_(0)^(a)f(t)dt=int_(0)^(a)f(a-t)dt`
`-int_(0)^(a)-f(t)dt` [using `f(a-x)=-f(x))`]
or `int_(0)^(a)f(t)dt=0`
or `phi(-x)=-int_(a)^(x)f(t)dt=-phi(x)`
Thus `phi(x)` is an odd function.