# Ex 1.3 , 6 - Chapter 1 Class 12 Relation and Functions (Term 1)

Last updated at Jan. 28, 2020 by Teachoo

Last updated at Jan. 28, 2020 by Teachoo

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Ex 1.3, 6 Show that f: [โ1, 1] โ R, given by f(x) = ๐ฅ/(๐ฅ + 2) is one-one. Find the inverse of the function f: [โ1, 1] โ Range f. (Hint: For y โ Range f, y = f(x) = ๐ฅ/(๐ฅ + 2) , for some x in [โ1, 1], i.e., x = 2๐ฆ/(1 โ ๐ฆ) ) f(x) = x/(x+2) Check one-one f(x1) = ๐ฅ1/(๐ฅ1 + 2) f(x2) = ๐ฅ2/(๐ฅ2 + 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) ๐ฅ1/(๐ฅ1 + 2) = ๐ฅ2/(๐ฅ2 + 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 โด f is one-one Checking onto f(x) = ๐ฅ/(๐ฅ + 2) Putting f(x) = y y = ๐ฅ/(๐ฅ + 2) y(x + 2) = x yx + 2y = x yx โ x = โ2y x(y โ 1) = โ2y x = (โ2๐ฆ )/(๐ฆ โ1) x = (โ2๐ฆ )/(โ1(โ๐ฆ + 1) ) x = (2๐ฆ )/((1 โ ๐ฆ) ) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((2๐ฆ )/((1 โ ๐ฆ) )) = ((2๐ฆ )/((1 โ ๐ฆ) ))/((2๐ฆ )/((1 โ ๐ฆ) ) + 2) = ((2๐ฆ )/((1 โ ๐ฆ) ))/((2๐ฆ + 2(1 โ ๐ฆ) )/((1 โ ๐ฆ) )) = 2๐ฆ/(2๐ฆ + 2 โ 2๐ฆ) = y Thus, for every y โ Range f, there exists x โ [โ1, 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^(โ1) (๐ฆ) = (2๐ฆ )/((1 โ ๐ฆ) ) , y โ 1 Note: Here, y โ Range f is important Inverse is not defined for y โ R Because denominator in (2๐ฆ )/((1 โ ๐ฆ) ) will be 0 if y = 1

Ex 1.3

Ex 1.3, 1
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Chapter 1 Class 12 Relation and Functions (Term 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.