`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

Question

`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

`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-periodic
A. `g(x)` is odd
B. `2(n)=0, n epsilonN`
C. `g(2n)=0,n epsilonN`
D. `g(x)` is non-periodic

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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`.