Let f: A → B be a function defined as f(x) = (2x + 3)/(x - 3) , where

A = R − {3} and B = R − {2}. Is the function f one–one and onto? Is f invertible? If yes, then find its inverse

Let f(x) = (2x + 3)/(x - 3). If function f one-one and onto? Is f

Question 27 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 27 - CBSE Class 12 Sample Paper for 2020 Boards - Part 3 Question 27 - CBSE Class 12 Sample Paper for 2020 Boards - Part 4 Question 27 - CBSE Class 12 Sample Paper for 2020 Boards - Part 5 Question 27 - CBSE Class 12 Sample Paper for 2020 Boards - Part 6

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 27 Let 𝑓: A β†’ B be a function defined as 𝑓(π‘₯)=(2π‘₯ + 3)/(π‘₯ βˆ’ 3) , where A = R βˆ’ {3} and B = R βˆ’ {2}. Is the function f one–one and onto? Is f invertible? If yes, then find its inverse 𝑓(π‘₯)=(2π‘₯ + 3)/(π‘₯ βˆ’ 3) Checking one-one Let π‘₯_1 , π‘₯_2 ∈ A 𝑓(π‘₯_1 )=(2π‘₯_1+ 3)/(π‘₯_1βˆ’ 3) 𝑓(π‘₯_2 )=(2π‘₯_2+ 3)/(π‘₯_2 βˆ’ 3) One-one Steps 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Putting 𝑓(π‘₯_1 ) = 𝑓(π‘₯_2 ) (2π‘₯_1+ 3)/(π‘₯_1βˆ’ 3) = (2π‘₯_2+ 3)/(π‘₯_2 βˆ’ 3) (2π‘₯_1+ 3) (π‘₯_2 βˆ’ 3) = (2π‘₯_2+ 3) (π‘₯_1 βˆ’ 3) 2π‘₯_1 "(" π‘₯_2 βˆ’ 3")"+ 3 (π‘₯_2 βˆ’ 3) = 2π‘₯_2 "(" π‘₯_1 βˆ’ 3")"+ 3(π‘₯_1 βˆ’ 3) 2π‘₯_1 π‘₯_2 βˆ’6π‘₯_1+ 3π‘₯_2βˆ’9 = 2π‘₯_2 π‘₯_1 βˆ’6π‘₯_2+ 3π‘₯_1 βˆ’9 βˆ’6π‘₯_1+ 3π‘₯_2 = βˆ’6π‘₯_2+ 3π‘₯_1 3π‘₯_2+6π‘₯_2 = 3π‘₯_1+6π‘₯_1 9π‘₯_2= 9π‘₯_1 π‘₯_2= π‘₯_1 If 𝑓(π‘₯_1 ) = 𝑓(π‘₯_2 ) , then π‘₯_1= π‘₯_2 ∴ f is one-one Checking onto 𝑓(π‘₯)=(2π‘₯ + 3)/(π‘₯ βˆ’ 3) Let y ∈ B where y = (2π‘₯ + 3)/(π‘₯ βˆ’ 3) y(x – 3) = 2x + 3 xy – 3y = 2x + 3 xy – 2x = 3y + 3 x(y – 2) = 3(y + 1) x = (3(𝑦 + 1))/((𝑦 βˆ’ 2)) For y = 2 , x is not defined But it is given that y ∈ R – {2} Hence , x = (3(𝑦 + 1))/((𝑦 βˆ’ 2)) ∈ R – {3} Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((3(𝑦 + 1))/((𝑦 βˆ’ 2))) = (2(3(𝑦 + 1)/((𝑦 βˆ’ 2) ))+ 3)/((3(𝑦 + 1)/((𝑦 βˆ’ 2) ))βˆ’ 3) = (((6(𝑦 + 1)+3(π‘¦βˆ’2))/((𝑦 βˆ’ 2) )))/((3(𝑦 + 1)/((𝑦 βˆ’ 2) ))βˆ’ 3) Also, y = f(x) Putting y in f(x) Hence, f is onto = (2(3(𝑦 + 1)/((𝑦 βˆ’ 2) ))+ 3)/((3(𝑦 + 1)/((𝑦 βˆ’ 2) ))βˆ’ 3) = (((6(𝑦 + 1) + 3(𝑦 βˆ’ 2))/((𝑦 βˆ’ 2) )))/(((3(𝑦 + 1) βˆ’ 3(𝑦 βˆ’ 2))/((𝑦 βˆ’ 2) )) ) = (6(𝑦 + 1) + 3(𝑦 βˆ’ 2))/(3(𝑦 + 1) βˆ’ 3(𝑦 βˆ’ 2)) = (6𝑦 + 6 + 3𝑦 βˆ’ 6)/(3𝑦 + 3 βˆ’ 3𝑦 + 6) = 9𝑦/9 = y Thus, for every y ∈ B, there exists x ∈ A such that f(x) = y Hence, f is onto Since f(x) is one-one and onto, So, f(x) is invertible And Inverse of x = 𝑓^(βˆ’1) (𝑦) = (3(𝑦 + 1))/((𝑦 βˆ’ 2))

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.