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

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

Answers (1)

Answer : (f \circ f \circ f)(38)=0

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

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

Now, we will compute (f \circ f \circ f)(38)

Solution :

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

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

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

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

      \therefore \text { Domain of }((f \circ f) \circ f)=\{x: x \in[2, \infty) \text { and } \sqrt{x-2} \in[\not \subset, \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\} \\ &=[38, \infty) \end{aligned}

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

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

Again we will compute (f \circ f \circ f)(x)

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

Now, (f \circ f \circ f)(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}

                                         \begin{aligned} &=\sqrt{\sqrt{\sqrt{36}-2}-2} \\ &=\sqrt{\sqrt{6-2}-2} \\ &=\sqrt{2-2} \\ &=0 \end{aligned}

Hence, (f \circ f \circ f)(38)=0

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