Ex 1.2, 5 - Chapter 1 Class 12 Relation and Functions (Important Question)
Last updated at Jan. 28, 2020 by Teachoo
Last updated at Jan. 28, 2020 by Teachoo
Transcript
Ex 1.2, 5 Show that the Signum Function f: R โ R, given by f(x) = {โ(1 for ๐ฅ >0@ 0 for ๐ฅ=0@โ1 for ๐ฅ<0)โค is neither one-one nor onto. f(x) = {โ(1 for ๐ฅ >0@ 0 for ๐ฅ=0@โ1 for ๐ฅ<0)โค For example: f(0) = 0 f(-1) = โ1 f(1) = 1 f(2) = 1 f(3) = 1 Since, different elements 1,2,3 have the same image 1 , โด f is not one-one. Check onto f: R โ R f(x) = {โ(1 for ๐ฅ >0@ 0 for ๐ฅ=0@โ1 for ๐ฅ<0)โค Value of f(x) is defined only if x is 1, 0, โ1 For other real numbers(eg: y = 2, y = 100) there is no corresponding element x Hence f is not onto Thus, f is neither one-one nor onto
Chapter 1 Class 12 Relation and Functions
Ex 1.2 , 10 Important
Example 23 Important Deleted for CBSE Board 2022 Exams
Example 25 Important Deleted for CBSE Board 2022 Exams
Ex 1.3, 3 (i) Important Deleted for CBSE Board 2022 Exams
Ex 1.3 , 6 Deleted for CBSE Board 2022 Exams
Ex 1.3 , 8 Important Deleted for CBSE Board 2022 Exams
Ex 1.3 , 9 Important Deleted for CBSE Board 2022 Exams
Ex 1.3, 13 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 1.4, 11 Important Deleted for CBSE Board 2022 Exams
Misc 3 Important Deleted for CBSE Board 2022 Exams
Misc. 4 Important
Misc 14 Important Deleted for CBSE Board 2022 Exams
Misc 18 Deleted for CBSE Board 2022 Exams
Chapter 1 Class 12 Relation and Functions
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