Question

Let $g\left(x\right)$ be a continuous function satisfying $g(x+a)+g\left(x\right)=0,\forall x\in R,a>0.$ If $\underset{b}{\overset{2k}{\int }}g\left(t\right)dt$ is independent of $b$ and $b,k,c$ are in A.P. ,then the least positive value of $c$ is

• A. $\sqrt{a}$
• B. $a$
• C. $a^2$
• D. $2a$

Answer 1

The correct option is D $2a$

We have
$g(x+a)+g\left(x\right)=0\Rightarrow g(x+2a)+g(x+a)=0\Rightarrow g(x+2a)=g\left(x\right)$
$\Rightarrow g\left(x\right)$ is periodic with period $2a$
$\Rightarrow \underset{b}{\overset{2k}{\int }}g\left(t\right)dt=\underset{b}{\overset{b+c}{\int }}g\left(x\right)dx[\because b,k,c\text{are in}A.P.]$
This is independent of $b,$ then the least value $c$ has to be $2a.$