Relations and Functions - I — Practice Quiz

11

Step 1: We are given f(x) = 2x + 3 and need to find f(4). Step 2: Substitute x = 4 into the function: f(4) = 2(4) + 3. Step 3: Calculate: 2 × 4 = 8, then 8 + 3 = 11. Step 4: So f(4) = 11. Final: A common mistake is computing 2 + 4 + 3 = 9 (adding instead of multiplying) or 2(4+3) = 14 (wrong order of operations). Always follow BODMAS.

R − {3}

Step 1: The domain is the set of all values of x for which the function is defined. Step 2: The function y = 1/(x − 3) is undefined when the denominator equals zero. Step 3: Set x − 3 = 0, which gives x = 3. Step 4: So x = 3 must be excluded from the domain. Final: Domain = R − {3} (all real numbers except 3). A common mistake is to write R − {−3}, which would be the answer if the function were 1/(x + 3).

{0, 1, 4, 9}

Step 1: The range is the set of all output values f(x) = x² for each x in the domain. Step 2: Calculate: f(−3) = 9, f(−2) = 4, f(−1) = 1, f(0) = 0. Step 3: f(1) = 1, f(2) = 4, f(3) = 9. Step 4: Collecting all unique values: {0, 1, 4, 9}. Final: Since x² gives the same value for x and −x (e.g., (−3)² = 3² = 9), we do not repeat values in the range. Option B is wrong because squaring never gives negative values.

x ≥ 4

Step 1: For a square root function to be defined (give real values), the expression inside must be greater than or equal to zero. Step 2: We need x − 4 ≥ 0. Step 3: Solving: x ≥ 4. Step 4: At x = 4, y = √0 = 0, which is valid. So x = 4 is included. Final: Domain = {x : x ≥ 4} or [4, ∞). Option A (x > 4) wrongly excludes x = 4 itself, which is a valid input.