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 Function Exercise 2.4 Question 17 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 17 Maths Textbook Solution.

Answers (1)

best_answer

Answer:

                f^{-1}\left ( x \right )=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )

Given:

                f\left ( x \right )=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}   is invertible

Hint:

                f^{-1}\left ( x \right )=y\Rightarrow f\left ( y \right )=x

Solution:

Let          f^{-1}\left ( x \right )=y                                                                                                                                        … (i)

                f\left ( y\right )=x

                \frac{10^{y}-10^{-y}}{10^{y}+10^{-y}}=x

                \frac{10^{-y}\left ( 10^{2y}-1 \right )}{10^{-y}\left ( 10^{2y}+1 \right )}=x

                10^{2y}-1=x10^{2y}+x

                10^{2y}\left ( 1-x \right )=1+x

                10^{2y}=\frac{1+x}{1-x}

                2y=\log\left ( \frac{1+x}{1-x} \right )

                y=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )=f^{-1}\left ( x \right )

So,          f^{-1}\left ( x \right )=\frac{1}{2}\log\left ( \frac{1+x}{1-x} \right )

Posted by

infoexpert27

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

NSUI stages protest outside CBSE headquarters, demands rollback of OSM
anam-khan

CBSE revaluation payment system faces malicious attack; IIT experts, PSU banks step in
anam-khan

From next year, CBSE Class 12 answer sheets on Digilocker: Education ministry

Latest Question