#### Please solve RD Sharma Class 12 Chapter Functions Exercise 2.3 Question 11 Sub question (ii)  maths textbook solution.

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

Hint : Domain $(f)=(2, \infty)$ and range of $(f)=(0, \infty)$

Given : Here f be a real function given by $f(x)=\sqrt{x-2}$

We have to compute $f \circ f \circ f$

Solution :

Clearly the range f is not subset of domain of f.

\begin{aligned} &\therefore \text { Domain }(f \circ f)=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(2, \infty)\} \\ &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \geq 2\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 4\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 6 \\ &=(6, \infty) \end{aligned}

Clearly the range of $f=(0, \infty) \not \subset \text { Domain of }(f \circ f)$

\begin{aligned} &\therefore \text { Domain of }(f \circ f \circ f)=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(6, \infty)\\ &\begin{aligned} &=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \geq 6\} \\ &=\{x: x \in(2, \infty) \text { and } x-2 \geq 36\} \\ &=\{x: x \in(2, \infty) \text { and } x \geq 38\} \end{aligned} \end{aligned}

Now, we will compute $(f \circ f \circ f)(x)$

First we compute $(f \circ f )(x)$

\begin{aligned} (f \circ f)(x) &=f(f(x)) \\ &=f(\sqrt{x-2}) \\ (f \circ f(x)) &=\sqrt{\sqrt{x-2}-2} \end{aligned}

Now,

\begin{aligned} (f \circ f \circ f)(x) &=(f \circ f)(f(x)) \\ &=(f \circ f)(\sqrt{x-2}) \end{aligned}

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

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