Let f be a real function given by

$f(x)=\sqrt{x-2}$

For domain,

$x-2 \geq 0$

$\Rightarrow x \geq 2$

Domain of $f=[2, \infty)$

Since $f$ is a square-root function, range of $f=(0, \infty)$

So, $f:[2, \infty) \rightarrow(0, \infty)$

(i) fof

Range of $f$ is not a subset of the domain of $f$.

$\Rightarrow$ Domain $(f o f)=\{x: x \in$ domain of $f$ and $f(x) \in$ domain of $f\}$

$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \in[2, \infty)\}$

$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \geq 2\}$

$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $x-2 \geq 4\}$

$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $x \geq 6\}$

$\Rightarrow$ Domain $($ fof $)=\{x: x \geq 6\}$

$\Rightarrow$ Domain $($ fof $)=[6, \infty)$

$f o f:[6, \infty) \rightarrow R$

$(f o f)(x)=f(f(x))$

$=f(\sqrt{x-2})$

$=\sqrt{\sqrt{x-2}-2}$

(ii) fofof $=(f o f)$ of

We have, $f:[2, \infty) \rightarrow(0, \infty)$ and fof : $[6, \infty) \rightarrow R$

$\Rightarrow$ Range of $f$ is not a subset of the domain of $f o f$.

Then, domain $((f o f) o f)=\{x: x \in$ domain of $f$ and $f(x) \in$ domain of $f o f\}$

$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \in[6, \infty)\}$

$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \geq 6\}$

$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $x-2 \geq 36\}$

$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $x \geq 38\}$

$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \geq 38\}$

$\Rightarrow$ Domain $(($ fof $) o f)=[38, \infty)$

fof $:[38, \infty) \rightarrow R$

So, $((f$ of $)$ of $)(x)=($ fof $)(f(x))$

$=(f$ of $)(\sqrt{x-2})$

$=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$

(iii) We have, (fofof) $(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$

So, $($ fofof $)(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$

$=\sqrt{\sqrt{\sqrt{36}-2}-2}$

$=\sqrt{\sqrt{6-2}-2}$

$=\sqrt{2-2}$

$=0$

(iv) We have, $f o f=\sqrt{\sqrt{x-2}-2}$ $\Rightarrow f^{2}(x)=f(x) \times f(x)=\sqrt{x-2} \times \sqrt{x-2}=x-2$

So, fof $\neq f^{2}$