Let `f: R->R` be a function defined by `f(x+1)=(f(x)-5)/(f(x)-3)AAx in R`. Then which of the following statement(s) is/are true?

Options

A. `f(2008)=f(2004)`
B. `f(2006)=f(2010)`
C. `f(2006)=f(2002)`
D. `f(2006)=f(2018)`

Answer

A, B, C, D

Explanation

`f(x+1)=(f(x)-5)/(f(x)-3)" (1)" `
`or f(x)f(x+1)-3f(x+1)=f(x)-5`
`or f(x)=(3f(x+1)-5)/(f(x+1)-1)`
Replacing x by `(x - 1)`, we get
`f(x-1)=(3f(x)-5)/(f(x)-1) " (2)" `
Using (1), `f(x+2)=(f(x+1)-5)/(f(x+1)-3)=((f(x)-5)/(f(x)-3)-5)/((f(x)-5)/(f(x)-3)-3)`
`=(2f(x)-5)/(f(x)-2) " (3)" `
Using (2), `f(x-2)=(3f(x-1)-5)/(f(x-1)-3)=(3((3f(x)-5)/(f(x)-1))-5)/((3f(x)-5)/(f(x)-1)-1)`
`=(2f(x)-5)/(f(x)-2) " (4)" `
Using (3) and (4) , we have `f(x+2) =f(x-2).`
Therefore, `f(x+4)=f(x),` i.e., `f(x)` is periodic with period 4.