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

Given : Here given that

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

Now we have to compute $f^{2}$ and

To prove : We have to prove that $f \circ f \neq f^{2}$

Solution :

First, we will compute $f^{2}$

\begin{aligned} \mathrm{f}^{2}(\mathrm{x}) &=\mathrm{f}(\mathrm{x}) \times \mathrm{f}(\mathrm{x}) \\ &=\sqrt{x-2} \times \sqrt{x-2} \\ &=\mathrm{x}-2 \end{aligned}

Clearly, Domain $(f)=[2, \infty) \text { and Range }(f)=[0, \infty]$

We observe that the range $(f)$ is not subset of domain of f

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

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

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

and again for $f^{2}$

\begin{aligned} &f^{2}(x)=[f(x)]^{2} \\ &=[\sqrt{x-2}]^{2} \\ &=x-2 \end{aligned}

Here we see that $f \circ f \neq f^{2}$

Hence, $f \circ f \neq f^{2}$