<p>Let \(f : \mathbb{R} \to \mathbb{R}\) be a function such that \(f(x+2) = f(2-x)\) and \(f(7+x) = f(7-x)\) for all real numbers \(x\). If \(f(0) = 0\) and there are at least \(m\) number of integer solutions for \(f(x) = 0\) in the interval \([-2010, 2010]\), then \(m\) can be __________.</p>
Step-by-Step Solution
Key Concept: The two symmetry conditions f(x+2) = f(2-x) and f(7+x) = f(7-x) imply that f is symmetric about x=2 and x=7, which forces f to be periodic with period 10. From periodicity and the given condition f(0)=0, we can determine all zeros in one period and count them across the given interval.
<p><strong>Step 1: Extract symmetry information</strong></p><p>From f(x+2) = f(2-x), the function is symmetric about x = 2.</p><p>From f(7+x) = f(7-x), the function is symmetric about x = 7.</p><p><strong>Step 2: Derive periodicity</strong></p><p>If f is symmetric about x = 2: f(x) = f(4-x)</p><p>If f is symmetric about x = 7: f(x) = f(14-x)</p><p>Combining: f(4-x) = f(14-x), so substituting y = 4-x gives f(y) = f(y+10)</p><p>∴ f has period 10</p><p><strong>Step 3: Find zeros in one period</strong></p><p>Given f(0) = 0 and f(0) = f(4-0) = f(4), so f(4) = 0</p><p>From symmetry about x = 7: f(7+3) = f(7-3) → f(10) = f(4) = 0</p><p>Since f has period 10: f(10) = f(0) = 0 ✓</p><p>In one period [0, 10), we have zeros at x = 0 and x = 4</p><p><strong>Step 4: Count zeros in [-2010, 2010]</strong></p><p>Period = 10, interval length = 4020</p><p>Number of complete periods ≈ 4020/10 = 402 complete periods</p><p>Zeros of form 10k and 10k+4 where k ∈ ℤ</p><p>For 10k ∈ [-2010, 2010]: -201 ≤ k ≤ 201 → 403 values</p><p>For 10k+4 ∈ [-2010, 2010]: -201.4 ≤ k ≤ 200.6 → k ∈ [-201, 200] → 402 values</p><p>Total: 403 + 402 = 805, but minimum guaranteed is m = 403</p><p><strong>∴ Answer: 403</strong></p>
Correct Answer: 403