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

Answer : $(f \circ f \circ f)(38)=0$

Hint : $\text { Domain }(f)=(2, \infty) \text { and Range }(f)=(0, \infty)$

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

Now, we will compute $(f \circ f \circ f)(38)$

Solution :

Clearly the range $(f)$ is not a subset of domain of $(f)$

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

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

$\text{Domain of}((f \circ f) \circ f)=\{x: x \in \text {Domain of } f \text { and } f(x) \in \text {Domain}(f \circ f)\}$

$\therefore \text { Domain of }((f \circ f) \circ f)=\{x: x \in[2, \infty) \text { and } \sqrt{x-2} \in[\not \subset, \infty)\}$

\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\} \\ &=[38, \infty) \end{aligned}

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

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

Again we will compute $(f \circ f \circ f)(x)$

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

Now, $(f \circ f \circ f)(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$

\begin{aligned} &=\sqrt{\sqrt{\sqrt{36}-2}-2} \\ &=\sqrt{\sqrt{6-2}-2} \\ &=\sqrt{2-2} \\ &=0 \end{aligned}

Hence, $(f \circ f \circ f)(38)=0$