Question

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

• A. $f\left(x\right)$ is an even function
• B. $f\left(x\right)$ is an odd function
• C. If $f\left(2\right)=a$ then $f(-2)=a$
• D. If $f\left(4\right)=b$ then $f(-4)=-b$

Answer 1

Answer: (A), (C)

$f\left(x\right)$ is an even function
If $f\left(2\right)=a$ then $f(-2)=a$

Substitute $x=y=0$
$f\left(0\right)+f\left(0\right)=2f\left(0\right)f\left(0\right)$
$\Rightarrow 2\times f\left(0\right)=2\times {\left(f\left(0\right)\right)}^2$
$f\left(0\right)=1$
$f(x+y)+f(x-y)=2f\left(x\right)f\left(y\right)$
Substitute $x=0$
Thus, $f\left(y\right)+f(-y)=2f\left(0\right)f\left(y\right)$
$\Rightarrow f\left(y\right)=f(-y)$
Therefore, $f\left(x\right)$ is even function
Hence, options 'A' and 'C' is correct.