Question

Let $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$ and $f\left(x\right)$ is a continuous function. If $f(x+4)=f(x+6)$ and $f\left(x\right)+f(-x)=0,$ then which of the following is/are correct?

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

Answer 1

Answer: (B), (D)

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

Given:
$f(x+4)=f(x+6)\cdots \left(i\right)f(-x)=-f\left(x\right)\cdots \left(ii\right)$
$\therefore f\left(x\right)$ is an odd function with period $2.$
$g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
Putting $t=-z\Rightarrow dt=-dz$
$\Rightarrow g\left(x\right)=-\underset{0}{\overset{-x}{\int }}f(-z)dz\Rightarrow g(-x)=-\underset{0}{\overset{x}{\int }}f(-z)dz\Rightarrow g(-x)=\underset{0}{\overset{x}{\int }}f\left(z\right)dz=\underset{0}{\overset{x}{\int }}f\left(t\right)dt(\because f(-x)=-f(x\left)\right)\therefore g\left(x\right)=g(-x)$
Thus $g\left(x\right)$ is an even function.

Also,
$g(x+2)=\underset{0}{\overset{x+2}{\int }}f\left(t\right)dt\Rightarrow g(x+2)=\underset{0}{\overset{2}{\int }}f\left(t\right)dt+\underset{2}{\overset{2+x}{\int }}f\left(t\right)dt\Rightarrow g(x+2)=g\left(2\right)+g\left(x\right)\cdots \left(iii\right)(\because f(x\left)\text{has period}2\right)$

Now,
$g\left(2\right)=\underset{0}{\overset{2}{\int }}f\left(t\right)dt\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{1}{\overset{2}{\int }}f\left(t\right)dt$
Replacing $t$ with $t+2$ in second integral, we get
$\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{-1}{\overset{0}{\int }}f(t+2)dt\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{-1}{\overset{0}{\int }}f\left(t\right)dt\Rightarrow g\left(2\right)=\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=0$
As $f\left(t\right)$ is odd function.
$\Rightarrow g\left(2\right)=0$
From equation $\left(iii\right)$, we get
$g(x+2)=g\left(x\right)$
Therefore, $g\left(x\right)$ is even function with period $2.$
and $g\left(2n\right)=\underset{0}{\overset{2n}{\int }}f\left(t\right)dt=n\cdot g\left(2\right)$
[$\because f\left(x\right)$ has period $2$]
Hence, $g\left(2n\right)=n\cdot 0=0$