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`f`is an odd function, It is also known that `f(x)`is continuous for all values of `x`and is periodic with period 2. If `g(x)=int_0^xf(t)dt ,`then`g(x)i sod d`(b) `g(n)=0,n in N``g(2n)=0,n in N`(d) `g(x)`is non-periodicA. `g(x)` is oddB. `2(n)=0, n epsilonN`C. `g(2n)=0,n epsilonN`D. `g(x)` is non-periodic |
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Answer» Correct Answer - C `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`. |
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