Question

If $f(x-y),f\left(x\right).f\left(y\right)$ and $f(x+y)$ are in AP for all $x,y$ and $f\left(0\right)\ne 0$, then

• A. $f\left(2\right)=f(-2)$
• B. $f\left(3\right)+f(-3)=0$
• C. $f^{{}^{\prime }}\left(2\right)+f^{{}^{\prime }}(-2)=0$
• D. $f^{{}^{\prime }}\left(3\right)=f^{{}^{\prime }}(-3)$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(2\right)=f(-2)$
C $f^{{}^{\prime }}\left(2\right)+f^{{}^{\prime }}(-2)=0$

Given, $f(x-y),f\left(x\right),f\left(y\right)$ and $f(x+y)$ are in AP
$\Rightarrow f(x-y)+f(x+y)=2f\left(x\right)f\left(y\right)$ ....(1)
Substitute $x=0,y=0$ in (1)
$\Rightarrow 2f\left(0\right)=2f\left(0\right)f\left(0\right)$
$\Rightarrow f\left(0\right)\left(f\right(0)-1)=0$
$\Rightarrow f\left(0\right)=1(\because f(0)\ne 0$ given)
Substitute $x=0,y=x$ in (1), we get
$f(-x)+f\left(x\right)=2f\left(0\right)f\left(x\right)$
$\Rightarrow f(-x)=f\left(x\right)$ ....(2)
$\Rightarrow f(-2)=f\left(2\right),f(-3)=f\left(3\right)$
Differentiating (2) w.r.t x
$f^{\prime }(-x)(-1)=f^{\prime }\left(x\right)$
$f^{\prime }(-x)+f^{\prime }\left(x\right)=0$
$f^{\prime }(-2)+f^{\prime }\left(2\right)=0$ and $f^{\prime }(-3)+f^{\prime }\left(3\right)=0$