<p>The range of the function \(f(x) = 7 - {}^{x}P_{x-3}\) is</p>
<p>\(\{1, 2, 3\}\)</p>
<p>\(\{1, 2, 3, 4, 5\}\)</p>
<p>\(\{1, 2, 3, 4\}\)</p>
<p>\(\{1, 2, 3, 4, 5, 6\}\)</p>
Step-by-Step Solution
Key Concept: The permutation P(x, x-3) is only defined when x ≥ x-3 and x-3 ≥ 0, so x ≥ 3. For each valid integer x, calculate f(x) and identify all possible output values.
<p><strong>Step 1: Find domain constraints</strong></p><p>For ⁿPᵣ to be defined: n ≥ r ≥ 0</p><p>Here: x ≥ (x-3) ≥ 0</p><p>This gives x ≥ 3 and x ≥ 3, so x ≥ 3</p><p>Since x must be a non-negative integer: x ∈ {3, 4, 5, 6, ...}</p><p><strong>Step 2: Calculate f(x) for valid values</strong></p><p>⁻ When x = 3: ³P₀ = 1, so f(3) = 7 - 1 = 6</p><p>⁻ When x = 4: ⁴P₁ = 4, so f(4) = 7 - 4 = 3</p><p>⁻ When x = 5: ⁵P₂ = 5×4 = 20, so f(5) = 7 - 20 = -13</p><p>⁻ When x = 6: ⁶P₃ = 6×5×4 = 120, so f(6) = 7 - 120 = -113</p><p><strong>Step 3: Analyze the pattern</strong></p><p>As x increases beyond 4, (x-3)! grows rapidly, making f(x) increasingly negative.</p><p>The function takes discrete values: {6, 3, -13, -113, ...}</p><p>For x ≥ 5, f(x) becomes arbitrarily negative.</p><p>∴ Range = {6, 3} ∪ (-∞, -13] (assuming answer C represents this, or the discrete set {6, 3, -13, -113, ...})</p>
Correct Answer: C