Saved Bookmarks
| 1. |
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, thenA. `phi(x)` is also an even functionB. `phi(x)` is an odd functionC. if `f(a-x)=-f(x)`, then `phi(x)` is an even functionD. if `f(a-x)=-f(x)` then `phi(x)` is an odd function |
|
Answer» Correct Answer - D 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. |
|