<p>Find the number of integral values of parameter '<i>a</i>' so that the inequality \((2a - a^2) \leq x^2 - 3x + 2 \leq 3 - a^2\) holds for any real <i>x</i> in the interval \([0, 2]\).</p>
Step-by-Step Solution
Key Concept: For the compound inequality to hold for all x ∈ [0,2], we need the minimum value of f(x) = x² - 3x + 2 on [0,2] to be ≥ (2a - a²), and the maximum value to be ≤ (3 - a²). This requires finding the range of f(x) on the given interval first.
<p><strong>Step 1:</strong> Find the range of f(x) = x² - 3x + 2 on [0, 2].</p><p>f(x) = x² - 3x + 2 has vertex at x = 3/2 ∈ [0,2].</p><p>• f(0) = 0 - 0 + 2 = 2</p><p>• f(3/2) = 9/4 - 9/2 + 2 = 9/4 - 18/4 + 8/4 = -1/4</p><p>• f(2) = 4 - 6 + 2 = 0</p><p>Therefore, min f(x) = -1/4 and max f(x) = 2 on [0,2].</p><p><strong>Step 2:</strong> Apply the left inequality constraint: 2a - a² ≤ -1/4 (minimum of f(x)).</p><p>-a² + 2a + 1/4 ≤ 0</p><p>-a² + 2a + 1/4 ≤ 0</p><p>Multiply by -4: 4a² - 8a - 1 ≥ 0</p><p>Using quadratic formula: a = (8 ± √(64 + 16))/8 = (8 ± √80)/8 = (8 ± 4√5)/8 = (2 ± √5)/2</p><p>Since the parabola opens upward: a ≤ (2 - √5)/2 or a ≥ (2 + √5)/2</p><p>(2 - √5)/2 ≈ (2 - 2.236)/2 ≈ -0.118 and (2 + √5)/2 ≈ 2.118</p><p><strong>Step 3:</strong> Apply the right inequality constraint: 2 ≤ 3 - a² (maximum of f(x)).</p><p>a² ≤ 1</p><p>-1 ≤ a ≤ 1</p><p><strong>Step 4:</strong> Find intersection of both constraints.</p><p>From Step 2: a ≤ -0.118 or a ≥ 2.118</p><p>From Step 3: -1 ≤ a ≤ 1</p><p>The intersection is: -1 ≤ a ≤ (2 - √5)/2 ≈ -1 ≤ a ≤ -0.118</p><p><strong>Step 5:</strong> Count integral values in [-1, -0.118].</p><p>The only integer in this range is a = -1.</p><p>Verification: a = -1 gives 2(-1) - (-1)² = -3 and 3 - (-1)² = 2.</p><p>Check: -3 ≤ f(x) ≤ 2 for all x ∈ [0,2]? Since -1/4 ≤ f(x) ≤ 2, this holds. ✓</p><p><strong>∴ Answer: 1</strong></p>
Correct Answer: 1