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

Answers (1)

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

Hint : Domain f=(2,\infty ) and Range (f)=(0,\infty )

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

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

            Here we have to compute f\; \circ \; f

Solution :

We know that

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

          \begin{aligned} &=\{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) x \geq 6 \\ &=(6, \infty) \end{aligned}


\begin{aligned} (f \circ f)(x) &=f(f(x)) \\ &=f(\sqrt{x-2}) \\ &=\sqrt{\sqrt{x-2}-2} \\ \therefore f \circ f:(6, \infty) \rightarrow R \end{aligned}

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

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