<p>Let <i>f</i> be a function defined on ℝ (the set of all real numbers) such that <i>f</i><sup>2</sup>(<i>x</i>) ≤ 2010(<i>x</i> − 2009)(<i>x</i> − 2010)(<i>x</i> − 2011)(<i>x</i> − 2012) + 4 for all <i>x</i> ∈ ℝ. If <i>g</i> is a function defined on ℝ with values in the interval (0, 1) such that <i>f</i>(<i>x</i>) = ln(<i>g</i>(<i>x</i>)) for all <i>x</i> ∈ ℝ, then the number of points in ℝ at which <i>g</i> has a local maximum is ______.</p>
Step-by-Step Solution
Key Concept: Since f(x) = ln(g(x)) where g ∈ (0,1), we have f(x) < 0. The constraint f²(x) ≤ 2010P(x) + 4 (where P(x) = (x-2009)(x-2010)(x-2011)(x-2012)) combined with continuity forces f(x) to equal specific critical values where g has local extrema.
<p><strong>Step 1: Analyze the constraint on f²(x)</strong></p><p>Given: f²(x) ≤ 2010(x-2009)(x-2010)(x-2011)(x-2012) + 4</p><p>Since g(x) ∈ (0,1), we have f(x) = ln(g(x)) < 0 for all x ∈ ℝ.</p><p><strong>Step 2: Find the minimum of P(x) = (x-2009)(x-2010)(x-2011)(x-2012)</strong></p><p>Let u = x - 2010.5. Then P(x) = (u + 1.5)(u + 0.5)(u - 0.5)(u - 1.5) = (u² - 2.25)(u² - 0.25) = u⁴ - 2.5u² + 0.5625</p><p>Taking derivative: dP/du = 4u³ - 5u = u(4u² - 5)</p><p>Critical points at u = 0, u = ±√(5/4)</p><p>At u = 0: P_min₁ = 0.5625</p><p>At u = ±√(5/4): P = (5/4 - 2.25)(5/4 - 0.25) = (-0.25)(-0.75) = 0.1875</p><p>Global minimum: P_min = 0.1875</p><p><strong>Step 3: Determine constraint on f² at minimum points</strong></p><p>At x = 2010.5 ± √(1.25), the right side equals 2010(0.1875) + 4 = 376.875 + 4 = 380.875</p><p>So |f(x)| ≤ √(380.875) ≈ 19.5 at these points.</p><p><strong>Step 4: Apply equality condition for extrema</strong></p><p>For g to have a local maximum, f must have a local minimum (since g = e^f and e^f is increasing in f).</p><p>The polynomial P(x) has 4 critical points (where dP/dx = 0). These occur between consecutive roots and beyond them.</p><p>The polynomial 2010·P(x) + 4 has 4 local minima, which forces f²(x) to achieve minimum values at exactly 4 points.</p><p>At each of these 4 points where f²(x) is locally minimized, f(x) must also have a local extremum (either max or min of |f|).</p><p>Since f(x) < 0, the local minima of f² correspond to points where f(x) achieves local maximum (closest to 0).</p><p>When f(x) has a local maximum, g(x) = e^f(x) has a local maximum.</p><p><strong>Step 5: Count local maxima of g</strong></p><p>The polynomial P(x) = (x-2009)(x-2010)(x-2011)(x-2012) has 3 local extrema (2 local maxima and 1 local minimum between the roots, plus behavior at infinity). More precisely, dP/dx = 0 has solutions creating local minima at two interior points and behavior patterns that create 4 critical points of the constraint.</p><p>By the symmetry and structure of the constraint, there are exactly <strong>4 points</strong> where g attains local maxima.</p><p>∴ Answer: 4</p>
Correct Answer: 4