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

Answers (1)


Given : Here given that


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}

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