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  1. Chapter 1 Class 12 Relation and Functions
  2. Concept wise

Transcript

Ex 1.3, 6 Show that f: [โˆ’1, 1] โ†’ R, given by f(x) = ๐‘ฅ/(๐‘ฅ + 2) is one-one. Find the inverse of the function f: [โˆ’1, 1] โ†’ Range f. (Hint: For y โˆˆ Range f, y = f(x) = ๐‘ฅ/(๐‘ฅ + 2) , for some x in [โˆ’1, 1], i.e., x = 2๐‘ฆ/(1 โˆ’ ๐‘ฆ) ) f(x) = x/(x+2) Check one-one f(x1) = ๐‘ฅ1/(๐‘ฅ1 + 2) f(x2) = ๐‘ฅ2/(๐‘ฅ2 + 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) ๐‘ฅ1/(๐‘ฅ1 + 2) = ๐‘ฅ2/(๐‘ฅ2 + 2) x1(x2 + 2) = x2(x1 + 2) x1x2 + 2x1 = x2x1 + 2x2 x1x2 โ€“ x2x1 + 2x1 = 2x2 0 + 2x1 = 2x2 2x1 = 2x2 x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 โˆด f is one-one Checking onto f(x) = ๐‘ฅ/(๐‘ฅ + 2) Putting f(x) = y y = ๐‘ฅ/(๐‘ฅ + 2) y(x + 2) = x yx + 2y = x yx โ€“ x = โ€“2y x(y โ€“ 1) = โ€“2y x = (โˆ’2๐‘ฆ )/(๐‘ฆ โˆ’1) x = (โˆ’2๐‘ฆ )/(โˆ’1(โˆ’๐‘ฆ + 1) ) x = (2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) ) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) )) = ((2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) ))/((2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) ) + 2) = ((2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) ))/((2๐‘ฆ + 2(1 โˆ’ ๐‘ฆ) )/((1 โˆ’ ๐‘ฆ) )) = 2๐‘ฆ/(2๐‘ฆ + 2 โˆ’ 2๐‘ฆ) = y Thus, for every y โˆˆ Range f, there exists x โˆˆ [โˆ’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 = ๐‘“^(โˆ’1) (๐‘ฆ) = (2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) ) , y โ‰  1 Note: Here, y โˆˆ Range f is important Inverse is not defined for y โˆˆ R Because denominator in (2๐‘ฆ )/((1 โˆ’ ๐‘ฆ) ) 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 9 years. He provides courses for Maths and Science at Teachoo.