# Misc. 4 - Chapter 1 Class 12 Relation and Functions (Term 1)

Last updated at Jan. 28, 2020 by

Last updated at Jan. 28, 2020 by

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Misc 4 Show that function f: R โ {x โ R: โ1 < x < 1} defined by f(x) = x/(1 + |๐ฅ| ) , x โ R is one-one and onto function. f: R โ {x โ R: โ1 < x < 1} f(x) = x/(1 + |๐ฅ| ) We know that |๐ฅ| = {โ( ๐ฅ , ๐ฅโฅ0 @โ๐ฅ , ๐ฅ<0)โค So, ๐(๐ฅ)={โ(๐ฅ/(1 + ๐ฅ), ๐ฅโฅ0@&๐ฅ/(1 โ ๐ฅ), ๐ฅ<0)โค For x โฅ 0 f(x1) = ๐ฅ_1/(1 + ๐ฅ_1 ) f(x2) = ๐ฅ_2/(1 + ๐ฅ_2 ) Putting f(x1) = f(x2) ๐ฅ_1/(1 + ๐ฅ_1 ) = ๐ฅ_2/(1 + ๐ฅ_2 ) ๐ฅ_1 (1 + ๐ฅ_2)= ๐ฅ_2 (1 + ๐ฅ_1) ๐ฅ_1+๐ฅ_1 ๐ฅ_2= ๐ฅ_2 +๐ฅ_2 ๐ฅ_1 ๐ฅ_1= ๐ฅ_2 For x < 0 f(x1) = ๐ฅ_1/(1 โ ๐ฅ_1 ) f(x2) = ๐ฅ_2/(1 โ ๐ฅ_2 ) Putting f(x1) = f(x2) ๐ฅ_1/(1 โ ๐ฅ_1 ) = ๐ฅ_2/(1 โ ๐ฅ_2 ) ๐ฅ_1 (1 โ ๐ฅ_2)= ๐ฅ_2 (1 โ ๐ฅ_1) ๐ฅ_1โ๐ฅ_1 ๐ฅ_2= ๐ฅ_2 โ๐ฅ_2 ๐ฅ_1 ๐ฅ_1= ๐ฅ_2 Hence, if f(x1) = f(x2) , then x1 = x2 โด f is one-one Checking onto ๐(๐ฅ)={โ(๐ฅ/(1 + ๐ฅ), ๐ฅโฅ0@&๐ฅ/(1 โ ๐ฅ), ๐ฅ<0)โค For x โฅ 0 f(x) = ๐ฅ/(1 + ๐ฅ) Let f(x) = y, "y = " ๐ฅ/(1 + ๐ฅ) y(1 + x) = x y + xy = x y = x โ xy x โ xy = y x(1 โ y) = y x = ๐ฆ/(1 โ ๐ฆ) For x < 0 f(x) = ๐ฅ/(1 โ ๐ฅ) Let f(x) = y "y = " ๐ฅ/(1 โ ๐ฅ) y(1 โ x) = x y โ xy = x y = x + xy x + xy = y x(1 + y) = y x = ๐ฆ/(1 + ๐ฆ) Thus, x = ๐ฆ/(1 โ ๐ฆ) , for x โฅ 0 & x = ๐ฆ/(1 + ๐ฆ) , for x < 0 Here, y โ {x โ R: โ1 < x < 1} i.e. Value of y is from โ1 to 1 , โ 1 < y < 1 If y = 1, x = ๐ฆ/(1 โ ๐ฆ) will be not defined, If y = โ1, x = ๐ฆ/(1 + ๐ฆ) will be not defined, But here โ 1 < y < 1 So, x is defined for all values of y. & x โ R โด f is onto Hence, f is one-one and onto.

Miscellaneous

Misc 1
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Misc 2 Deleted for CBSE Board 2022 Exams

Misc 3 Important Deleted for CBSE Board 2022 Exams

Misc. 4 Important You are here

Misc 5

Misc 6 Deleted for CBSE Board 2022 Exams

Misc 7 Deleted for CBSE Board 2022 Exams

Misc. 8 Important

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Misc 10 Important

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Misc 11 (ii) Deleted for CBSE Board 2022 Exams

Misc 12 Deleted for CBSE Board 2022 Exams

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Misc 15

Misc 16 (MCQ) Important

Misc 17 (MCQ) Important

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Misc 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

Chapter 1 Class 12 Relation and Functions (Term 1)

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About the Author

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.