Question

Let $f:R\to R$ be a function defined by $f(x+1)=\frac{f\left(x\right)-5}{f\left(x\right)-3}\forall x\in R$. Then, which of the following statement is are true

• A. $f\left(2008\right)=f\left(2004\right)$
• B. $f\left(2006\right)=f\left(2010\right)$
• C. $f\left(2006\right)=f\left(2002\right)$
• D. $f\left(2006\right)=f\left(2018\right)$

Answer 1

The correct option is A $f\left(2008\right)=f\left(2004\right)$

$f(x+1)=\frac{f\left(x\right)-5}{f\left(x\right)-3}$
put $x\to x+2$
$f(x+2)=\frac{f(x+1)-5}{f(x+2)-3}$
$f(x+2)=\frac{\frac{f\left(x\right)-5}{f\left(x\right)-3}-5}{\frac{f\left(x\right)-5}{f\left(x\right)-3}-3}$
$f(x+2)=\frac{f\left(x\right)-5-5f\left(x\right)+15}{f\left(x\right)-5-3f\left(x\right)+9}$
$f(x+2)=\frac{-4f\left(x\right)+10}{-2f\left(x\right)+4}$
$f(x+2)=\frac{-2f\left(x\right)+5}{-f\left(x\right)+2}$
put $x\to x+2$
$f(x+4)=\frac{-2f(x+2)+5}{-f(x+2)+2}$
$f(x+4)=\frac{-2\left[\frac{-2f\left(x\right)+5}{-f\left(x\right)+2}\right]+5}{-\left(\frac{-2f\left(x\right)+5}{-f\left(x\right)+2}\right)+2}$
$f(x+4)=\frac{+4f\left(x\right)-10+5f\left(x\right)+10}{+2f\left(x\right)-5-2f\left(x\right)+4}$
$f(x+4)=\frac{-f\left(x\right)}{-1}=f\left(x\right)$
$f(x+4)=f\left(x\right)$