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 (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}

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question