Relations and Functions — Practice Quiz

3

Step 1: We need to substitute x = 2 into the function f(x) = 2x² - 3x + 1. Step 2: f(2) = 2(2)² - 3(2) + 1. Step 3: Calculate each term: 2(4) = 8, 3(2) = 6. Step 4: f(2) = 8 - 6 + 1 = 3. Final Step: f(2) = 3. A common mistake is to compute (2x)² = 4x² instead of 2(x²). Always apply the exponent before multiplication.

81

Step 1: The formula for the total number of functions from set A to set B is [n(B)]^n(A). Step 2: Here n(A) = 4 and n(B) = 3. Step 3: Total functions = 3^4. Step 4: 3^4 = 3 × 3 × 3 × 3 = 81. Final Step: Answer is 81. A common error is computing n(A)^n(B) = 4^3 = 64 (swapping base and exponent). Remember: co-domain size is the base, domain size is the exponent.

One-one and onto (Bijection)

Step 1: Check if f is one-one: Assume f(x₁) = f(x₂). Then 3x₁ + 5 = 3x₂ + 5, which gives x₁ = x₂. So f is one-one (injective). Step 2: Check if f is onto: For any y ∈ R, we need to find x such that f(x) = y. Solving: 3x + 5 = y → x = (y-5)/3, which is a real number for every y ∈ R. Step 3: Since every y in the co-domain has a pre-image in R, f is onto (surjective). Step 4: Since f is both one-one and onto, f is a bijection. Final Step: Any linear function f(x) = mx + c with m ≠ 0 and domain = co-domain = R is always a bijection.

2

Step 1: f∘g(x) means f(g(x)). We first apply g, then apply f to the result. Step 2: Find g(2) first: g(2) = 2(2) - 3 = 4 - 3 = 1. Step 3: Now find f(g(2)) = f(1): f(1) = (1)² + 1 = 1 + 1 = 2. Step 4: Therefore f∘g(2) = 2. Final Step: A common mistake is to compute g∘f(2) instead of f∘g(2). In f∘g, we apply g FIRST, then f. Always read composition from right to left.