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`Iff(2-x)=f(2+x)a n df(4-x)=f(4+x)`for all `xa n df(x)`is a function for which `int_0^2f(x)dx=5,t h e nint_0^(50)f(x)dx`is equal to125 (b) `int_(-4)^(46)f(x)dx``int_1^(51)f(x)dx`(d) `int_2^(52)f(x)dx`A. `125`B. `int_(-4)^(46)f(x)dt`C. `int_(1)^(51)f(x)dx`D. `int_(2)^(52)f(x)dx` |
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Answer» Correct Answer - A::B::D `f(2-x)=f(2+x),f(4-x)=f(4+x)` or `f(4+x)=f(4-x)=f(2+2-x)=f(2-(2-x))=f(x)` Thus, the period of `f(x)` is 4. `int_(0)^(50)f(x)dx=int_(0)^(48)f(x)dx+int_(48)^(50)f(x)dx` `=12 int_(0)^(4)f(x)dx+int_(0)^(2)f(x)dx` [In second integral, replacing `x` by `x+48` and then using `f(x)=f(x+48)`] `=12(int_(0)^(2)f(x)dx+int_(0)^(2)f(4-x)dx)+5` `=12(int_(0)^(2)f(x)dx+int_(0)^(2)f(4+x)dx)+5` `=24int_(0)^(2)f(x)dx+5=125` `int_(-4)^(46)f(x)dx=int_(-4)^(-2)f(x)dx+int_(-2)^(-2+48)f(x)dx` `=int_(0)^(2)f(x+4)dx+12int_(0)^(4)f(x)dx` `=int_(0)^(2)f(x)dx+24int_(0)^(2)f(x)dx` `=125` also `int_(2)^(52)f(x)dx=int_(2)^(4)f(x)dx+int_(4)^(4+48)f(x)dx` `=int_(0)^(2)f(4-x)dx+12int_(0)^(4)f(x)dx` `=int_(0)^(2)f(4+x)dx+24int_(0)^(2)f(x)dx` `=int_(0)^(2)f(x)dx+24int_(0)^(2)f(x)dx` `=125` `int_(1)^(51)f(x)dx=int_(1)^(3)f(x)dx+int_(3)^(3+48)f(x)dx` `=int_(1)^(3)f(x)dx+12int_(0)^(4)f(x)dx` `=int_(0)^(2)f(x+1)dx+24int_(0)^(2)f(x)dx` `!=125` |
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