Question

If the function $f$ 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)

If $f\left(2\right)=a$, then $f(-2)=a$

$f(x+y)+f(x-y)=2f\left(x\right)f\left(y\right)$ ....$\left[1\right]$
Putting $x=0$ in $\left[1\right]$, we get
$f\left(y\right)+f(-y)=2f\left(0\right)f\left(y\right)$ ....$\left[2\right]$
Putting $x=0,y=0$ in $\left[1\right]$, we get
$f\left(0\right)+f\left(0\right)=2f\left(0\right)f\left(0\right)\Rightarrow f\left(0\right)=1$
$\left(\text{As}f\right(0)\ne 0)$
$\therefore f(-y)=f\left(y\right)$ $\left(\text{From}\right[2\left]\right)$
Hence, the function is even.
Then, $f(-2)=f\left(2\right)=a$