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  1. Chapter 1 Class 12 Relation and Functions
  2. Serial order wise

Transcript

Misc 4 Show that function f: R โ†’ {x โˆˆ R: โˆ’1 < x < 1} defined by f(x) = x/(1 + |๐‘ฅ| ) , x โˆˆ R is one-one and onto function. f: R โ†’ {x โˆˆ R: โˆ’1 < x < 1} f(x) = x/(1 + |๐‘ฅ| ) We know that |๐‘ฅ| = {โ–ˆ( ๐‘ฅ , ๐‘ฅโ‰ฅ0 @โˆ’๐‘ฅ , ๐‘ฅ<0)โ”ค So, ๐‘“(๐‘ฅ)={โ–ˆ(๐‘ฅ/(1 + ๐‘ฅ), ๐‘ฅโ‰ฅ0@&๐‘ฅ/(1 โˆ’ ๐‘ฅ), ๐‘ฅ<0)โ”ค For x โ‰ฅ 0 f(x1) = ๐‘ฅ_1/(1 + ๐‘ฅ_1 ) f(x2) = ๐‘ฅ_2/(1 + ๐‘ฅ_2 ) Putting f(x1) = f(x2) ๐‘ฅ_1/(1 + ๐‘ฅ_1 ) = ๐‘ฅ_2/(1 + ๐‘ฅ_2 ) ๐‘ฅ_1 (1 + ๐‘ฅ_2)= ๐‘ฅ_2 (1 + ๐‘ฅ_1) ๐‘ฅ_1+๐‘ฅ_1 ๐‘ฅ_2= ๐‘ฅ_2 +๐‘ฅ_2 ๐‘ฅ_1 ๐‘ฅ_1= ๐‘ฅ_2 For x < 0 f(x1) = ๐‘ฅ_1/(1 โˆ’ ๐‘ฅ_1 ) f(x2) = ๐‘ฅ_2/(1 โˆ’ ๐‘ฅ_2 ) Putting f(x1) = f(x2) ๐‘ฅ_1/(1 โˆ’ ๐‘ฅ_1 ) = ๐‘ฅ_2/(1 โˆ’ ๐‘ฅ_2 ) ๐‘ฅ_1 (1 โˆ’ ๐‘ฅ_2)= ๐‘ฅ_2 (1 โˆ’ ๐‘ฅ_1) ๐‘ฅ_1โˆ’๐‘ฅ_1 ๐‘ฅ_2= ๐‘ฅ_2 โˆ’๐‘ฅ_2 ๐‘ฅ_1 ๐‘ฅ_1= ๐‘ฅ_2 Hence, if f(x1) = f(x2) , then x1 = x2 โˆด f is one-one Checking onto ๐‘“(๐‘ฅ)={โ–ˆ(๐‘ฅ/(1 + ๐‘ฅ), ๐‘ฅโ‰ฅ0@&๐‘ฅ/(1 โˆ’ ๐‘ฅ), ๐‘ฅ<0)โ”ค For x โ‰ฅ 0 f(x) = ๐‘ฅ/(1 + ๐‘ฅ) Let f(x) = y, "y = " ๐‘ฅ/(1 + ๐‘ฅ) y(1 + x) = x y + xy = x y = x โ€“ xy x โ€“ xy = y x(1 โ€“ y) = y x = ๐‘ฆ/(1 โˆ’ ๐‘ฆ) For x < 0 f(x) = ๐‘ฅ/(1 โˆ’ ๐‘ฅ) Let f(x) = y "y = " ๐‘ฅ/(1 โˆ’ ๐‘ฅ) y(1 โ€“ x) = x y โ€“ xy = x y = x + xy x + xy = y x(1 + y) = y x = ๐‘ฆ/(1 + ๐‘ฆ) Thus, x = ๐‘ฆ/(1 โˆ’ ๐‘ฆ) , for x โ‰ฅ 0 & x = ๐‘ฆ/(1 + ๐‘ฆ) , for x < 0 Here, y โˆˆ {x โˆˆ R: โˆ’1 < x < 1} i.e. Value of y is from โ€“1 to 1 , โ€“ 1 < y < 1 If y = 1, x = ๐‘ฆ/(1 โˆ’ ๐‘ฆ) will be not defined, If y = โ€“1, x = ๐‘ฆ/(1 + ๐‘ฆ) will be not defined, But here โ€“ 1 < y < 1 So, x is defined for all values of y. & x โˆˆ R โˆด f is onto Hence, f is one-one and onto.

About the Author

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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.