Question

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

• A. $f\left(4\right)=f(-4)$
• B. $f\left(2\right)+f(-2)=0$
• C. $f^{\prime }\left(4\right)+f^{\prime }(-4)=0$
• D. $f^{\prime }\left(2\right)=f^{\prime }(-2)$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(4\right)=f(-4)$
C $f^{\prime }\left(4\right)+f^{\prime }(-4)=0$

$f(x-y),f\left(x\right)f\left(y\right)$ and $f(x+y)$ are in A.P.
$f(x-y)+f(x+y)=2f\left(x\right)f\left(y\right)$
Putting $x=0=y$
$\Rightarrow f\left(0\right)+f\left(0\right)=2f\left(0\right)f\left(0\right)\Rightarrow f\left(0\right)=1[\because f(0)\ne 0]$
Putting $x=0,y=x$
$f\left(x\right)+f(-x)=2f\left(0\right)f\left(x\right)\Rightarrow f(-x)=f\left(x\right)\cdots \left(1\right)f\left(2\right)=f(-2);f\left(4\right)=f(-4)$
Differentiating the equation (1) w.r.t. $x$
$f^{\prime }\left(x\right)=-f^{\prime }(-x)\Rightarrow f^{\prime }\left(x\right)+f^{\prime }(-x)=0f^{\prime }\left(4\right)+f^{\prime }(-4)=0;f^{\prime }\left(2\right)+f^{\prime }(-2)=0$