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

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

Answers (1)

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

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

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

We have to compute f \circ f \circ f

Solution :

Clearly the range f is not subset of domain of f.

\begin{aligned} &\therefore \text { Domain }(f \circ 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-2 \geq 6 \\ &=(6, \infty) \end{aligned}

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

\begin{aligned} &\therefore \text { Domain of }(f \circ f \circ f)=\{x: x \in(2, \infty) \text { and } \sqrt{x-2} \in(6, \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\} \end{aligned} \end{aligned}

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

First we compute (f \circ f )(x)

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

Now,

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

                              =\sqrt{\sqrt{\sqrt{x-2}-2}-2}

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

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