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

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  1. Class 12
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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))

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.