NCERT Solutions Class 12 Maths Chapter 1 – Relations and Functions
Complete NCERT Solutions for Class 12 Maths Chapter 1 – Relations and Functions. This chapter covers types of relations, types of functions, composition of functions, and invertible functions.
Chapter Overview
| Topic | Weightage in Board |
|---|---|
| Types of Relations | 4-6 marks |
| Types of Functions | 4-6 marks |
| Composition of Functions | 4 marks |
| Invertible Functions | 4 marks |
| Total Chapter Weightage | 8-10 marks |
Key Concepts
1. Types of Relations
A relation R on set A is:
- Reflexive: (a, a) ∈ R for all a ∈ A
- Symmetric: If (a, b) ∈ R, then (b, a) ∈ R
- Transitive: If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R
- Equivalence: Reflexive + Symmetric + Transitive
2. Types of Functions
- One-One (Injective): f(a) = f(b) ⟹ a = b
- Onto (Surjective): Range of f = Codomain
- Bijective: Both one-one and onto
3. Composition of Functions
If f: A → B and g: B → C, then gof: A → C where (gof)(x) = g(f(x))
4. Invertible Functions
A function f is invertible if it is bijective. If f: A → B is bijective, then f⁻¹: B → A exists such that f⁻¹(f(x)) = x
Exercise 1.1 Solutions
Question 1
Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3, …, 13, 14} defined as R = {(x, y) : 3x – y = 0}
R = {(1,3), (2,6), (3,9), (4,12)}
Reflexive: (1,1) ∉ R. So R is not reflexive.
Symmetric: (1,3) ∈ R but (3,1) ∉ R. So R is not symmetric.
Transitive: (1,3) ∈ R, (3,9) ∈ R but (1,9) ∉ R. So R is not transitive.
Question 2
Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive nor symmetric nor transitive.
Not Reflexive: For a = 1/2, we need 1/2 ≤ (1/2)² = 1/4, which is false. So (1/2, 1/2) ∉ R.
Not Symmetric: (1, 4) ∈ R since 1 ≤ 16, but (4, 1) ∉ R since 4 ≤ 1 is false.
Not Transitive: (3, 2) ∈ R (3 ≤ 4) and (2, 1.5) ∈ R (2 ≤ 2.25), but (3, 1.5) ∉ R since 3 ≤ 2.25 is false.
Exercise 1.2 Solutions
Question 1
Show that the function f : R → R defined by f(x) = 1/x is one-one and onto.
Note: f is defined on R* (non-zero reals)
One-One: Let f(x₁) = f(x₂)
⟹ 1/x₁ = 1/x₂
⟹ x₁ = x₂
Hence f is one-one.
Onto: For any y ∈ R*, let x = 1/y
Then f(x) = f(1/y) = 1/(1/y) = y
Hence f is onto.
Therefore, f is bijective.
Important Formulas
| Concept | Formula/Definition |
|---|---|
| Composition | (gof)(x) = g(f(x)) |
| Inverse | f⁻¹(y) = x where f(x) = y |
| Identity Function | I(x) = x for all x |
| Constant Function | f(x) = c for all x |
Important Tips for Board Exam
- Always check all three properties (R, S, T) separately for relations
- For proving one-one: Start with f(x₁) = f(x₂) and prove x₁ = x₂
- For proving onto: Take arbitrary element in codomain and find pre-image
- Draw diagrams for better understanding
- Practice previous year questions from this chapter
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