Relations and Functions

R₁ = {(1,1), (2,2), (3,3), (1,2)}, R₃ = {(1,1), (2,2), (3,3)}, R₄ = {(1,1), (2,2), (3,3), (2,3), (3,2)}

Step 1: A relation R on set A is reflexive if (a,a) ∈ R for all a ∈ A. Step 2: For A = {1, 2, 3}, we need (1,1), (2,2), and (3,3) to be in R. Step 3: Check each relation: - R₁: Contains (1,1), (2,2), (3,3) ✓ Reflexive - R₂: Missing (3,3) ✗ Not reflexive - R₃: Contains exactly (1,1), (2,2), (3,3) ✓ Reflexive - R₄: Contains (1,1), (2,2), (3,3) plus additional pairs ✓ Reflexive

f is one-one (injective), f is a function, Range of f = {4, 5, 6}

Step 1: Check if f is a function: Each element in A maps to exactly one element in B ✓ Step 2: Check if f is one-one: Different elements in A map to different elements in B - f(1) = 4, f(2) = 5, f(3) = 6 (all different) ✓ One-one Step 3: Check if f is onto: Every element in B should be mapped by some element in A - B = {4, 5, 6, 7} but 7 is not in the range ✗ Not onto Step 4: Since f is not onto, it's not bijective Step 5: Range of f = {4, 5, 6} ✓

a > 0

Step 1: Analyze f(x) = |x - 2| Step 2: This is a V-shaped graph with vertex at (2, 0) Step 3: For x ≥ 2: f(x) = x - 2 For x < 2: f(x) = -(x - 2) = 2 - x Step 4: When a < 0: No solutions (absolute value is always non-negative) Step 5: When a = 0: Exactly one solution x = 2 Step 6: When a > 0: Two solutions - From x - 2 = a ⟹ x = a + 2 - From 2 - x = a ⟹ x = 2 - a - Both solutions are valid when a > 0 Step 7: Therefore, f(x) = a has exactly two solutions when a > 0

Always an equivalence relation

Step 1: Need to prove R₁ ∩ R₂ is reflexive, symmetric, and transitive Step 2: Reflexive: Since R₁ and R₂ are equivalence relations, (a,a) ∈ R₁ and (a,a) ∈ R₂ for all a ∈ A Therefore, (a,a) ∈ R₁ ∩ R₂ ✓ Step 3: Symmetric: If (a,b) ∈ R₁ ∩ R₂, then (a,b) ∈ R₁ and (a,b) ∈ R₂ Since R₁, R₂ are symmetric: (b,a) ∈ R₁ and (b,a) ∈ R₂ Therefore, (b,a) ∈ R₁ ∩ R₂ ✓ Step 4: Transitive: If (a,b), (b,c) ∈ R₁ ∩ R₂, then these pairs are in both R₁ and R₂ Since R₁, R₂ are transitive: (a,c) ∈ R₁ and (a,c) ∈ R₂ Therefore, (a,c) ∈ R₁ ∩ R₂ ✓