Ex 1.3, 6 - Show f(x) = x/x+2 is one-one. Find inverse of f.

Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions - Part 2
Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions - Part 3
Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions - Part 4
Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions - Part 5
Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions - Part 6

  1. Chapter 1 Class 12 Relation and Functions (Term 1)
  2. Concept wise

Transcript

Ex 1.3, 6 Show that f: [βˆ’1, 1] β†’ R, given by f(x) = π‘₯/(π‘₯ + 2) is one-one. Find the inverse of the function f: [βˆ’1, 1] β†’ Range f. (Hint: For y ∈ Range f, y = f(x) = π‘₯/(π‘₯ + 2) , for some x in [βˆ’1, 1], i.e., x = 2𝑦/(1 βˆ’ 𝑦) ) f(x) = x/(x+2) Check one-one f(x1) = π‘₯1/(π‘₯1 + 2) f(x2) = π‘₯2/(π‘₯2 + 2) Rough 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(x1) = f(x2) π‘₯1/(π‘₯1 + 2) = π‘₯2/(π‘₯2 + 2) x1(x2 + 2) = x2(x1 + 2) x1x2 + 2x1 = x2x1 + 2x2 x1x2 – x2x1 + 2x1 = 2x2 0 + 2x1 = 2x2 2x1 = 2x2 x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking onto f(x) = π‘₯/(π‘₯ + 2) Putting f(x) = y y = π‘₯/(π‘₯ + 2) y(x + 2) = x yx + 2y = x yx – x = –2y x(y – 1) = –2y x = (βˆ’2𝑦 )/(𝑦 βˆ’1) x = (βˆ’2𝑦 )/(βˆ’1(βˆ’π‘¦ + 1) ) x = (2𝑦 )/((1 βˆ’ 𝑦) ) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((2𝑦 )/((1 βˆ’ 𝑦) )) = ((2𝑦 )/((1 βˆ’ 𝑦) ))/((2𝑦 )/((1 βˆ’ 𝑦) ) + 2) = ((2𝑦 )/((1 βˆ’ 𝑦) ))/((2𝑦 + 2(1 βˆ’ 𝑦) )/((1 βˆ’ 𝑦) )) = 2𝑦/(2𝑦 + 2 βˆ’ 2𝑦) = y Thus, for every y ∈ Range f, there exists x ∈ [βˆ’1, 1] 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) = (2𝑦 )/((1 βˆ’ 𝑦) ) , y β‰  1 Note: Here, y ∈ Range f is important Inverse is not defined for y ∈ R Because denominator in (2𝑦 )/((1 βˆ’ 𝑦) ) will be 0 if y = 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.