Question

If $f:R\to R$ is a function satisfying the equation $f(3x+1)+f(3x+10)=10,\forall x\in R$, then the period of $f(x)$ is

Answer 1

Answer: 18

$\because f(3x+1)+f(3x+10)=10$ ...(i)
Replaceing $x\to x+3$
$f(3(x+3)+1)+f(3(x+3)+10)=10$
$f(3x+10)+f(3x+19)=10$... (ii)
eq. (ii) - (i)
$f(3x+19)-f(3x+1)=0$
$f(3x+1)=f(3x+19)$
Now, replace $x\to \frac{x}{3}-\frac{1}{3}$
$f\left(3\left(\frac{x}{3}-\frac{1}{3}\right)+1\right)=f\left(3\left(\frac{x}{3}-\frac{1}{3}\right)+19\right)$
$f(x)=f(x+18)\forall x\in R$
Hence, period of $f(x)$ is $18$ .