Question

$f$ is an odd function. It is also known that $f\left(x\right)$ is continuous for all values of $x$ and is periodic with period $2.$ If $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt,$ then

• A. $g\left(x\right)$ is even
• B. $g\left(n\right)=0,n\in N$
• C. $g\left(2n\right)=0,n\in N$
• D. $g\left(x\right)$ is non-periodic

Answer 1

Answer: (A), (C)

$g\left(2n\right)=0,n\in N$

$g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
$g(-x)=\underset{0}{\overset{-x}{\int }}f\left(t\right)dt=-\underset{0}{\overset{x}{\int }}f(-t)dt\Rightarrow g(-x)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt[\because f(-t)=-f(t\left)\right]$
$\Rightarrow g\left(x\right)=g(-x)$
Thus, $g\left(x\right)$ is even function.

Also,
$g(x+2)=\underset{0}{\overset{x+2}{\int }}f\left(t\right)dt\Rightarrow g(x+2)=\underset{0}{\overset{2}{\int }}f\left(t\right)dt+\underset{2}{\overset{x+2}{\int }}f\left(t\right)dt\Rightarrow g(x+2)=g\left(2\right)+\underset{0}{\overset{x}{\int }}f(t+2)dt\Rightarrow g(x+2)=g\left(2\right)+\underset{0}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow g(x+2)=g\left(2\right)+g\left(x\right)$
Now,
$g\left(2\right)=\underset{0}{\overset{2}{\int }}f\left(t\right)dt\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{1}{\overset{2}{\int }}f\left(t\right)dt\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{-1}{\overset{0}{\int }}f(t+2)dt\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{-1}{\overset{0}{\int }}f\left(t\right)dt\Rightarrow g\left(2\right)=\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=0$
As $f\left(t\right)$ is odd function.
So,
$g\left(2\right)=0\Rightarrow g(x+2)=g\left(x\right)$
i.e. $g\left(x\right)$ is periodic with period $2.$
$\therefore g\left(4\right)=0$ or $f\left(6\right)=0,$ $\Rightarrow g\left(2n\right)=0,n\in N.$