Ex 1.2 , 3 - Chapter 1 Class 12 Relation and Functions
Last updated at Dec. 16, 2024 by Teachoo
To prove one-one & onto (injective, surjective, bijective)
One One function
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Chapter 1 Class 12 Relation and Functions
Concept wise
- Relations - Definition
- Empty and Universal Relation
- To prove relation reflexive, transitive, symmetric and equivalent
- Finding number of relations
- Function - Definition
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To prove one-one & onto (injective, surjective, bijective)
- Composite functions
- Composite functions and one-one onto
- Finding Inverse
- Inverse of function: Proof questions
- Binary Operations - Definition
- Whether binary commutative/associative or not
- Binary operations: Identity element
- Binary operations: Inverse
Transcript
Ex 1.2, 3 (Introduction) Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. f(x) = [x] = greatest integer less than equal to x Example: [1] = 1 [1.01] = 1 [1.2] = 1 [1.9] = 1 [1.99] = 1 [2] = 2 Ex 1.2, 3 Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. f(x) = [x] where [x] denotes the greatest integer less than equal to x Check one-one f(x) = [x] Example f(1) = [1] = 1, f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1, f(1.99) = [1.99] = 1, Since, different elements 1, 1.2, 1.9, 1.99 have the same image 1 , ∴ f is not one-one. Check onto f(x) = [x] Let y = f(x) y = [x] i.e. y = Greatest integer less than or equal to x Hence, value of y will always come an integer. But y is a real number Hence f is not onto.
