Relations and Functions — Practice Quiz

R = {(1,a),(2,b),(3,c)}

Step 1: A function requires every element in A to have exactly one image in B. Step 2: Option A: element 1 has two images (a and b) — NOT a function. Step 3: Option B: 1→a, 2→b, 3→c — each element has exactly one image — this IS a function. Step 4: Option C: element 3 has no image — NOT a function. Option D: element 2 has two images — NOT a function. Common mistake: Accepting option C thinking partial mapping is okay.

3

Step 1: Calculate f(2) = 2(2)² - 3(2) + 1 = 2(4) - 6 + 1 = 8 - 6 + 1 = 3. Step 2: Calculate f(1) = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0. Step 3: f(2) - f(1) = 3 - 0 = 3. Step 4: Verify by substituting again carefully. Common mistake: Calculating f(2) as 2×2² = 16 without correctly applying order of operations.

All non-negative real numbers

Step 1: The modulus function is defined as f(x) = x when x ≥ 0 and f(x) = -x when x < 0. Step 2: When x ≥ 0, f(x) = x ≥ 0 (non-negative values). Step 3: When x < 0, f(x) = -x > 0 (positive values, since negative of negative is positive). Step 4: So f(x) is always ≥ 0. The range is [0, ∞), i.e., all non-negative real numbers. Common mistake: Thinking the range is all real numbers because the domain is all real numbers.

64

Step 1: The total number of relations from A to B = number of subsets of A × B. Step 2: n(A × B) = n(A) × n(B) = 3 × 2 = 6. Step 3: Number of subsets of A × B = 2^(n(A×B)) = 2^6 = 64. Step 4: This is because each subset of A × B is a possible relation. Common mistake: Computing 2^(n(A) × n(B)) correctly as 2^6 = 64 but writing 32 by computing 2^5 by error.