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

CBSE issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee
anam-khan

CBSE Supplementary Exam 2025: Registrations open for Class 10, 12 private students today
anam-khan

CBSE opens application to access 10th answer scripts; verification, revaluation for Class 12 starts on May 31

Latest Question