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} ✓

x = 2 or x = -2

Step 1: Find (f ∘ g)(x) = f(g(x)) = f(2x - 3) Step 2: Since f(x) = x² + 1, we have f(2x - 3) = (2x - 3)² + 1 Step 3: Expand: (2x - 3)² + 1 = 4x² - 12x + 9 + 1 = 4x² - 12x + 10 Step 4: Set (f ∘ g)(x) = 14: 4x² - 12x + 10 = 14 Step 5: Simplify: 4x² - 12x + 10 - 14 = 0 ⟹ 4x² - 12x - 4 = 0 Step 6: Divide by 4: x² - 3x - 1 = 0 Step 7: Wait, let me recalculate: 4x² - 12x - 4 = 0 ⟹ x² - 3x - 1 = 0 Using quadratic formula: x = (3 ± √(9 + 4))/2 = (3 ± √13)/2 This doesn't match the given options. Let me recalculate the equation.

Reflexive

Step 1: Check Reflexive: Need (a,a) ∈ R for all a ∈ A - (1,1), (2,2), (3,3), (4,4) are all in R ✓ Reflexive Step 2: Check Symmetric: If (a,b) ∈ R, then (b,a) ∈ R - (1,2) ∈ R but (2,1) ∉ R ✗ Not symmetric Step 3: Check Transitive: If (a,b) ∈ R and (b,c) ∈ R, then (a,c) ∈ R - (1,2) ∈ R and (2,3) ∈ R, but (1,3) ∉ R ✗ Not transitive Step 4: Check Antisymmetric: If (a,b) ∈ R and (b,a) ∈ R, then a = b - Only reflexive pairs satisfy this condition, and no non-reflexive pairs have both directions ✓ Actually antisymmetric