<p>Let \(f(x) = (1 + b^2)x^2 + 2bx + 1\) and let \(m_b()\) be the minimum value of \(f(x)\). As b varies, the range of \(m_b()\) is</p>
<p>(a) \([0, 1]\)</p>
<p>(b) \(\left[0, 1\right)\)</p>
<p>(c) \(\left[\frac{1}{2}, 1\right)\)</p>
<p>(d) \((0, 1]\)</p>
Step-by-Step Solution
Key Concept: For a quadratic function with positive leading coefficient, find the minimum value using the vertex formula, then treat this minimum as a function of the parameter b to find its range.
<p><strong>Step 1:</strong> Identify the form of f(x). We have f(x) = (1 + b²)x² + 2bx + 1, where the coefficient of x² is (1 + b²) > 0 for all real b. So the parabola opens upward and has a minimum.</p><p><strong>Step 2:</strong> Find the x-coordinate of the vertex using x = -2b/[2(1 + b²)] = -b/(1 + b²).</p><p><strong>Step 3:</strong> Calculate the minimum value m_b using the vertex formula: m_b = f(x_vertex) = (1 + b²)x² + 2bx + 1, or use m_b = c - b'²/(4a) where a = 1 + b², b' = 2b, c = 1.</p><p>m_b = 1 - (2b)²/[4(1 + b²)] = 1 - 4b²/[4(1 + b²)] = 1 - b²/(1 + b²) = (1 + b² - b²)/(1 + b²) = 1/(1 + b²)</p><p><strong>Step 4:</strong> Find the range of m_b = 1/(1 + b²) as b varies over all real numbers.</p><p><strong>Step 5:</strong> Analyze the behavior: Since b² ≥ 0 for all real b, we have 1 + b² ≥ 1, which means 1/(1 + b²) ≤ 1.</p><p><strong>Step 6:</strong> Check boundary values:</p><p>• When b = 0: m_b = 1/(1 + 0) = 1</p><p>• As |b| → ∞: m_b = 1/(1 + b²) → 0⁺ (approaches 0 from above, but never equals 0)</p><p><strong>Step 7:</strong> Since 1/(1 + b²) is continuous and strictly decreasing as |b| increases, it takes all values in (0, 1], achieving the maximum value 1 when b = 0, and approaching 0 asymptotically without reaching it.</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b