<p>For all \(x \in \mathbb{R}\), a function \(f\) satisfies \(f(2+x) = f(2-x) = f[7-(5+x)] = f[7+(5+x)] = f(12+x)\). It is also given that \(f(0) = 0\). Find the number of integers \(n\) in \([-2010, 2010]\) for which \(f(n) = 0\).</p>
Step-by-Step Solution
Key Concept: Extract the period and symmetries from the given functional equations, then use f(0) = 0 with periodicity to find all zeros in the given interval. The function has period 10 and satisfies f(x) = 0 at regular intervals.
<p><strong>Step 1: Analyze the symmetry conditions</strong></p><p>From f(2+x) = f(2-x): The function is symmetric about x = 2.</p><p>From f(2+x) = f(12+x): We have f(2+x) = f(12+x), which gives f(y) = f(y+10) for y = 2+x. Therefore, <strong>f has period 10</strong>.</p><p><strong>Step 2: Verify the period using other conditions</strong></p><p>From f(7-(5+x)) = f(7+(5+x)): We get f(2-x) = f(12+x), which confirms periodicity of 10.</p><p>All four conditions are consistent with: <strong>f has period 10 and is symmetric about x = 2</strong>.</p><p><strong>Step 3: Use symmetry to find zeros</strong></p><p>Given f(0) = 0 and f is symmetric about x = 2:</p><p>f(0) = f(2-2) = f(2+2) = f(4) = 0</p><p>So f(0) = f(4) = 0 in one period [0, 10).</p><p><strong>Step 4: Apply periodicity</strong></p><p>Since f(n) = 0 at n ≡ 0 (mod 10) and n ≡ 4 (mod 10), the zeros occur at:</p><p>n ∈ {..., -20, -16, -10, -6, 0, 4, 10, 14, 20, 24, ...}</p><p><strong>Step 5: Count zeros in [-2010, 2010]</strong></p><p>For n ≡ 0 (mod 10): n = 10k where -2010 ≤ 10k ≤ 2010</p><p>This gives -201 ≤ k ≤ 201, so k ∈ {-201, -200, ..., 0, ..., 200, 201}</p><p>Number of values: 201 - (-201) + 1 = <strong>403</strong></p><p>For n ≡ 4 (mod 10): n = 10k + 4 where -2010 ≤ 10k + 4 ≤ 2010</p><p>This gives -201.4 ≤ k ≤ 200.6, so k ∈ {-201, -200, ..., 199, 200}</p><p>Number of values: 200 - (-201) + 1 = <strong>402</strong></p><p>Total zeros: 403 + 402 = <strong>805</strong></p><p><strong>∴ Answer: 805</strong></p>
Correct Answer: 805