Mathematics - Functions Question with Solution | TestHub

3 does not belong to the range of $f$, which implies 2 also cannot belong to the range of $f$. This is because if $f(x)=2$ for some $x\in R$, then $f(x+p)=\frac{2-5}{2-3}=3$, which is not in the range of $f$. Hence, 2 and 3 are not in the range of $f$. If $f(x+2p)=f(x)$, this implies:

$$\begin{array}{rl}f(x)& =f(x+p+p)\\ & =\frac{f(x+p)-5}{f(x+p)-3}\\ & =\frac{\frac{f(x)-5}{f(x)-3}-5}{\frac{f(x)-5}{f(x)-3}-3}\\ & =\frac{-4f(x)+10}{-2f(x)+4}\\ & =\frac{2f(x)-5}{f(x)-2}\end{array}$$

so that $[f(x)-2{]}^2=-1$, which is absurd. Therefore, $2p$ is not a period. Again:

$$\begin{array}{rl}f(x+3p)& =\frac{2f(x+p)-5}{f(x+p)-2}\\ & =\frac{3f(x)-5}{f(x)-1}\ne f(x)\end{array}$$

Now $f(x+4p)=f(x+3p+p)$:

$$\begin{array}{rl}& =\frac{f(x+3p)-5}{f(x+3p)-3}\\ & =\frac{\frac{3f(x)-5}{f(x)-1}-5}{\frac{3f(x)-5}{f(x)-1}-3}\\ & =\frac{-2f(x)}{-2}\\ & =f(x)\end{array}$$

Therefore, $4p$ is a period.