We have, $\quad f(x)=\frac{x-1}{x+1}$

(i) $f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{(1-x) / x}{(1+x) / x}=\frac{1-x}{1+x}=\frac{-(x-1)}{x+1}=-f(x)$

(ii) $f\left(-\frac{1}{x}\right)=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}=\frac{(-1-x) / x}{(-1+x) / x} \Rightarrow f\left(-\frac{1}{x}\right)=\frac{-(x+1)}{x-1}$

$\begin{array}{ll}\text { Now, } & \frac{-1}{f(x)}=\frac{-1}{\frac{x-1}{x+1}}=\frac{-(x+1)}{x-1} \\ \therefore & f\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}\end{array}$

We have, $f(x)=\sqrt{x}$ and $g(x)=x$ be two function defined in the domain $R^{+} \cup\{0\}$.

(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$

(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$

(ii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}$

(iv) $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$

We have, $\quad f(x)=\frac{1}{\sqrt{x-5}}$

$f(x)$ is defined, if $x-5>0 \Rightarrow x>5$

$\therefore \quad$ Domain of $f=(5, \infty)$

Let $$f(x)=y$$

$$\therefore \quad y=\frac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\frac{1}{y}$$

$$\begin{array}{ll} \Rightarrow & x-5=\frac{1}{y^2} \\ \therefore & x=\frac{1}{y^2}+5 \\ \because & x \in(5, \infty) \Rightarrow y \in R^{+} \end{array}$$

Hence, range of $f=R^{+}$

We have, $\quad f(x)=y=\frac{a x-b}{c x-a}$

$\begin{aligned} \therefore \quad f(y) & =\frac{a y-b}{c y-a}=\frac{a\left(\frac{a x-b}{c x-a}\right)-b}{c\left(\frac{a x-b}{c x-a}\right)-a} \\ & =\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)}=\frac{a^2 x-a b-b c x+a b}{a c x-b c-a c x+a^2} \\ & =\frac{a^2 x-b c x}{a^2-b c}=\frac{x\left(a^2-b c\right)}{\left(a^2-b c\right)}=x \\ \therefore \quad f(y) & =x\quad \text{Hence proved.}\end{aligned}$