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

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

           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}

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