Ex 1.2, 5 - Chapter 1 Class 12 Relation and Functions (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 1 Class 12 Relation and Functions
Ex 1.2 , 10 Important
Example 17 Important
Question 8 Important Deleted for CBSE Board 2025 Exams
Ex 1.3, 3 (i) Important Deleted for CBSE Board 2025 Exams
Ex 1.3 , 6 Deleted for CBSE Board 2025 Exams
Ex 1.3 , 8 Important Deleted for CBSE Board 2025 Exams
Ex 1.3 , 9 Important Deleted for CBSE Board 2025 Exams
Ex 1.3, 13 (MCQ) Important Deleted for CBSE Board 2025 Exams
Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2025 Exams
Ex 1.4, 11 Important Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Misc 1 Important
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 11 Deleted for CBSE Board 2025 Exams
Chapter 1 Class 12 Relation and Functions
Last updated at April 16, 2024 by Teachoo
Ex 1.2, 5 Show that the Signum Function f: R → R, given by f(x) = {█(1 for 𝑥 >0@ 0 for 𝑥=0@−1 for 𝑥<0)┤ is neither one-one nor onto. f(x) = {█(1 for 𝑥 >0@ 0 for 𝑥=0@−1 for 𝑥<0)┤ For example: f(0) = 0 f(-1) = −1 f(1) = 1 f(2) = 1 f(3) = 1 Since, different elements 1,2,3 have the same image 1 , ∴ f is not one-one. Check onto f: R → R f(x) = {█(1 for 𝑥 >0@ 0 for 𝑥=0@−1 for 𝑥<0)┤ Value of f(x) is defined only if x is 1, 0, –1 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