Q. The period of the function $f\left(x\right)$ , which satisfies the relation $f\left(x\right)+f\left(x + 4\right)=f\left(x + 2\right)+f\left(x + 6\right)$ is
NTA AbhyasNTA Abhyas 2022
Solution:
Given that,
$f\left(\right.x\left.\right)+f\left(\right.x+4\left.\right)=f\left(\right.x+2\left.\right)+f\left(\right.x+6\left.\right)$
Replacing $x$ with $x+2$ ,
$\Rightarrow f\left(\right.x+2\left.\right)+f\left(\right.x+6\left.\right)=f\left(\right.x+4\left.\right)+f\left(\right.x+8\left.\right)$
Adding both above equations,
$\Rightarrow f\left(\right.x\left.\right)+f\left(\right.x+2\left.\right)+f\left(\right.x+4\left.\right)+f\left(\right.x+6\left.\right)=f\left(\right.x+2\left.\right)+f\left(\right.x+4\left.\right)+f\left(\right.x+6\left.\right)+f\left(\right.x+8\left.\right)$
$\Rightarrow f\left(\right.x\left.\right)=f\left(\right.x+8\left.\right)$
$\Rightarrow f\left(\right.x\left.\right)$ is a periodic function and its period is $8$ .
$f\left(\right.x\left.\right)+f\left(\right.x+4\left.\right)=f\left(\right.x+2\left.\right)+f\left(\right.x+6\left.\right)$
Replacing $x$ with $x+2$ ,
$\Rightarrow f\left(\right.x+2\left.\right)+f\left(\right.x+6\left.\right)=f\left(\right.x+4\left.\right)+f\left(\right.x+8\left.\right)$
Adding both above equations,
$\Rightarrow f\left(\right.x\left.\right)+f\left(\right.x+2\left.\right)+f\left(\right.x+4\left.\right)+f\left(\right.x+6\left.\right)=f\left(\right.x+2\left.\right)+f\left(\right.x+4\left.\right)+f\left(\right.x+6\left.\right)+f\left(\right.x+8\left.\right)$
$\Rightarrow f\left(\right.x\left.\right)=f\left(\right.x+8\left.\right)$
$\Rightarrow f\left(\right.x\left.\right)$ is a periodic function and its period is $8$ .