Question

A real valued function $f\left(x\right)$ satisfies the functional equation $f(x-y)=f\left(x\right)f\left(y\right)-f(3-x)f(3+y)$ where $f\left(0\right)=1,f(6-x)$ is equal to

• A. $f\left(x\right)$
• B. $f\left(3\right)$
• C. $f\left(3\right)+f(3-x)$
• D. $-f\left(x\right)$

Answer 1

The correct option is D

$-f\left(x\right)$

Explanation for the correct option:

Find the value of $f(6-x)$:

Given that, $f(x-y)=f\left(x\right)f\left(y\right)-f(3-x)f(3+y)\forall x,y\in R$

By Putting $y=0$, we get

$f\left(x\right)=f\left(x\right)-f(3-x)f\left(3\right)$

$\Rightarrow$$f(3-x)f\left(3\right)=0$

$\Rightarrow$ $f\left(3\right)=0$ $\left[\because f(a-x)\ne 0\right]$
Now, $f(6-x)=f\left[3-(x-3)\right]$
$$\begin{gathered} =f\left(3\right)f(3-x)-f(3-3)f(3+x-3) \\ =0-f\left(x\right) \end{gathered}$$

$\therefore f(6-x)=\mathbf{-}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

Hence, Option ‘D’ is Correct.