Question

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

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

Answer 1

The correct option is C

$f\text{'}\left(2\right)+f\text{'}\left(-2\right)=0$

Explanation for the correct option:

Finding the relation:

Since, $f\left(x-y\right),f\left(x\right)f\left(y\right)$and $f\left(x+y\right)$ are in AP

$f\left(x-y\right)+f\left(x+y\right)=2f\left(x\right)f\left(y\right)...\left(1\right)$

Substitute $x=0,y=0$ in $\left(1\right)$

$\begin{array}{rcl}2f\left(0\right)& =& 2f\left(0\right)f\left(0\right)\\ & \Rightarrow & f\left(0\right)\left(f\left(0\right)-1\right)=0\\ & \Rightarrow & f\left(0\right)=1or0\\ & \Rightarrow & f\left(0\right)=1\mathbf{[}\mathbf{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{\ne }\mathbf{0}\mathbf{]}\end{array}$

Now, substitute $x=0,y=x$ in $\left(1\right)$

$$\begin{gathered} f\left(-x\right)+f\left(x\right)=2f\left(0\right)f\left(x\right) \\ \Rightarrow f\left(-x\right)=f\left(x\right)...\left(2\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{f}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{]} \end{gathered}$$

By differentiating $\left(2\right)$ with respect to $x$.

$\begin{array}{rcl}f\text{'}\left(-x\right)\left(-1\right)& =& f\text{'}\left(x\right)\\ & \Rightarrow & f\text{'}\left(x\right)+f\text{'}\left(-x\right)=0\end{array}$

Substitute $x=2$, we get

$f\text{'}\left(2\right)+f\text{'}\left(-2\right)=0$

Hence, the correct option is C.