Question

Let $f(x$) be an odd continuous function which is periodic with period $2$. If $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$, then?

• A. $g\left(x\right)$ is an odd function
• B. $g\left(n\right)=0\forall n\in N$
• C. $g\left(2n\right)=0\forall n\in N$
• D. $g\left(x\right)$ is a non periodic function

Answer 1

Answer: (C)

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

It is given that $f\left(x\right)$ is an odd function. Therefore,
$g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ is an even function.
$\Rightarrow g(x+2)={\int }_0^{x+2}f\left(t\right)dt\Rightarrow g(x+2)={\int }_0^{2}f\left(t\right)dt+{\int }_2^{x+2}f\left(t\right)dt$
$\Rightarrow g(x+2)=g\left(2\right)+{\int }_0^{x}f\left(t\right)dt$ [$\because$ f(x) is periodic with period 2]
$\Rightarrow g(x+2)=g\left(2\right)+g\left(x\right)$ for all x ... (i)
Now,$g(x+2)=g\left(2\right)+g\left(x\right)$ $\Rightarrow g\left(1\right)=g\left(2\right)+g(-1)$ [Replacing x by - i]
$\Rightarrow g\left(2\right)=g\left(1\right)-g(-1)$ $\Rightarrow g\left(2\right)={\int }_0^{1}f\left(t\right)dt-{\int }_0^{-1}f\left(t\right)dt$
$\Rightarrow g\left(2\right)={\int }_0^{1}f\left(t\right)dt+{\int }_{-1}^{0}f\left(t\right)dt$
$\Rightarrow g\left(2\right)={\int }_{-1}^{1}f\left(t\right)dt$
$\Rightarrow g\left(2\right)=0$ [$\because f$ is an odd function]
Substituting $g\left(2\right)=0in\left(i\right)$, we get
$g(x+2)=g\left(x\right)$ for all x $\Rightarrow g\left(x\right)$ is periodic with period 2 units
Also, $g(x+2)=g\left(x\right)$ for all $x$
$\Rightarrow g\left(2\right)=g\left(4\right)=g\left(6\right)=.....=g\left(2n\right)$
$\Rightarrow g\left(2n\right)=0$ for all n $\in$ N