Let $f$ be a real function such that $f(x-y), f(x) f(y)$ and $f(x+y)$ are in A.P. for all $x, y \in R$. If $f(0) \neq 0$, then
Answer & explanation
Correct answer: option 3
It is given that $f(x-y), f(x) f(y)$ and $f(x+y)$ are in A.P. for all $x, y \in R$.
∴ $2 f(x) f(y)=f(x-y)+f(x+y)$ for all $x, y \in R$ ......(i)
Putting x = y = 0, we get
$2\{f(0)\}^2=2 f(0)$
$\Rightarrow f(0)\{f(0)-1\}=0 \Rightarrow f(0)=1$ [∵ f(0) ≠ 0]
Putting x = 0 and y = x in (i), we get
$2 f(0) f(x)=f(-x)+f(x)$ for all $x \in R$
$\Rightarrow 2 f(x)=f(-x)+f(x)$ for all $x \in R$
$\Rightarrow f(-x)=f(x)$ for all $x \in R$
$\Rightarrow f(x)$ is an even function
$\Rightarrow f^{\prime}(x)$ is an odd function
∴ $f(-1)=f(1)$ and $f^{\prime}(-3)=-f^{\prime}(3)$
$\Rightarrow f(1)=f(-1)$ and $f^{\prime}(3)+f^{\prime}(-3)=0$