Question

Let $f$ be a function satisfies the relation $f(x+y)+f(x-y)=2f\left(x\right)f\left(y\right)\forall x,y\in R.$ If $f\left(5\right)=10$ and $f\left(0\right)\ne 0,$ then $f(-5)=$

Answer 1

$f(x+y)+f(x-y)=2f\left(x\right)f\left(y\right)$
Put $x=y=0,$ we get
$f\left(0\right)+f\left(0\right)=2\left(f\right(0){)}^2$
$\Rightarrow f\left(0\right)=1[\because f(0)\ne 0]$

Now put $x=0,y=5$ we get
$f\left(5\right)+f(-5)=2f\left(0\right)f\left(5\right)$
$\Rightarrow f(-5)=f\left(5\right)[\because f(0)=1]$
$\therefore f\left(5\right)=f(-5)=10$

Alternate Solution:
$f(x+y)+f(x-y)=2f\left(x\right)f\left(y\right)$
Put $x=y=0,$ we get
$f\left(0\right)+f\left(0\right)=2\left(f\right(0){)}^2$
$\Rightarrow f\left(0\right)=1[\because f(0)\ne 0]$

Now put $x=0,$ we get
$f\left(y\right)+f(-y)=2f\left(0\right)f\left(y\right)$
$\Rightarrow f(-y)=f\left(y\right)[\because f(0)=1]$
$\therefore f\left(5\right)=f(-5)=10$