Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question