Check sibling questions

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

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Transcript

Ex 1.3, 6 Show that f: [−1, 1] → R, given by f(x) = 𝑥/(𝑥 + 2) is one-one. Find the inverse of the function f: [−1, 1] → Range f. (Hint: For y ∈ Range f, y = f(x) = 𝑥/(𝑥 + 2) , for some x in [−1, 1], i.e., x = 2𝑦/(1 − 𝑦) ) f(x) = x/(x+2) Check one-one f(x1) = 𝑥1/(𝑥1 + 2) f(x2) = 𝑥2/(𝑥2 + 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) 𝑥1/(𝑥1 + 2) = 𝑥2/(𝑥2 + 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) = 𝑥/(𝑥 + 2) Putting f(x) = y y = 𝑥/(𝑥 + 2) y(x + 2) = x yx + 2y = x yx – x = –2y x(y – 1) = –2y x = (−2𝑦 )/(𝑦 −1) x = (−2𝑦 )/(−1(−𝑦 + 1) ) x = (2𝑦 )/((1 − 𝑦) ) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((2𝑦 )/((1 − 𝑦) )) = ((2𝑦 )/((1 − 𝑦) ))/((2𝑦 )/((1 − 𝑦) ) + 2) = ((2𝑦 )/((1 − 𝑦) ))/((2𝑦 + 2(1 − 𝑦) )/((1 − 𝑦) )) = 2𝑦/(2𝑦 + 2 − 2𝑦) = y Thus, for every y ∈ Range f, there exists x ∈ [−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^(−1) (𝑦) = (2𝑦 )/((1 − 𝑦) ) , y ≠ 1 Note: Here, y ∈ Range f is important Inverse is not defined for y ∈ R Because denominator in (2𝑦 )/((1 − 𝑦) ) will be 0 if y = 1

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.