Question

State whether true or false.

If $f\left(x\right)$ be continuous function in $[0,2\pi ]$ and $f\left(0\right)=f\left(2\pi \right)$, then there exists a point $cϵ(0,\pi )$ such that $f\left(c\right)=f(c+\pi )$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $g\left(x\right)=f\left(x\right)-f(x+\pi )$ ...(i)

at $x=\pi ,g\left(\pi \right)=f\left(\pi \right)-f\left(2\pi \right)$ ....(ii)

at $x=0,g\left(0\right)=f\left(0\right)-f\left(\pi \right)$ ...(iii)

Adding Eqs. (ii) and (iii), we have

$g\left(0\right)+g\left(\pi \right)=f\left(0\right)-f\left(2\pi \right)=0$

$\Rightarrow g\left(0\right)=-g\left(\pi \right)$

$\Rightarrow g\left(0\right)$ and $g\left(\pi \right)$ are of opposite sign.

$\Rightarrow$ There is a point c between 0 and $\pi$ such that

$g\left(c\right)=0$ ....(iv)

From Eq. (i) putting $x=c$, we have

$g\left(c\right)=f\left(c\right)-f(\pi +c)$ ...(v)

From Eqs. (iv) and (v); we have

$f\left(c\right)-f(\pi +c)=0$

Hence, $f\left(c\right)=f(\pi +c)$