First slide

Functions (XII)

Question

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

Moderate

A

$f (2008) = f (2004)$
B

$f (2006) = f (2010)$
C

$f (2006) = f (2002)$
D

$f (2006) = f (2018)$

Solution

$f (x + 1) = \frac{f (x) - 5}{f (x) - 3} (1)$
$\begin{array}{l} or f (x) f (x + 1) - 3 f (x + 1) = f (x) - 5 \\ or f (x) = \frac{3 f (x + 1) - 5}{f (x + 1) - 1} \end{array}$
Replacing x by (x - 1), we get
$f (x - 1) = \frac{3 f (x) - 5}{f (x) - 1} (2)$
Using(1), $f (x + 2) = \frac{f (x + 1) - 5}{f (x + 1) - 3} = \frac{\frac{f (x) - 5}{f (x) - 3} - 5}{\frac{f (x) - 5}{f (x) - 3} - 3}$
$= \frac{2 f (x) - 5}{f (x) - 2} (3)$
Using (2), $f (x - 2) = \frac{3 f (x - 1) - 5}{f (x - 1) - 1} = \frac{3 (\frac{3 f (x) - 5}{f (x) - 1}) - 5}{\frac{3 f (x) - 5}{f (x) - 1} - 1}$
$= \frac{2 f (x) - 5}{f (x) - 2} (4)$
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.

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