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Last updated at May 29, 2018 by Teachoo

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Ex 1.3, 2 (Introduction) Let f, g and h be functions from R to R. Show that (f + g)oh = foh + goh (f.g)oh = (foh).(goh) Let f(x) = x, g(x) = sin x , h(x) = log x We will show (f + g) oh = foh + goh Let f(x) = x, g(x) = sin x , h(x) = log x To prove: We will show (f .g) oh = foh . goh (f .g) oh = foh . goh Ex 1.3, 2 Let f, g and h be functions from R to R. Show that (f + g)oh = foh + goh & (f.g) oh = (foh).(goh) Proving (f + g) oh = foh + goh L.H.S (f + g)oh = (f + g) (h(x)) = f (h(x)) + g (h(x)) = foh + goh = R.H.S Hence, (f + g) oh = foh + goh Proving (f .g) oh = foh . goh L.H.S (f . g)oh = (f . g) (h(x)) = f (h(x)) . g (h(x)) = foh . goh = R.H.S Hence, (f . g) oh = (foh) . (goh)

Composite functions

Composition of functions

Example 15 Not in Syllabus - CBSE Exams 2021

Ex 1.3, 1 Not in Syllabus - CBSE Exams 2021

Example 16 Not in Syllabus - CBSE Exams 2021

Ex 1.3, 3 Important Not in Syllabus - CBSE Exams 2021

Ex 1.3 , 13 Important Not in Syllabus - CBSE Exams 2021

Misc 3 Important Not in Syllabus - CBSE Exams 2021

Example 17 Not in Syllabus - CBSE Exams 2021

Example 26 Not in Syllabus - CBSE Exams 2021

Misc 18

Ex 1.3, 2 Not in Syllabus - CBSE Exams 2021 You are here

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
- 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

About the Author

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.