<p>Let <i>f</i> : ℝ → ℝ is a function satisfying f(10 + x) = f(x) and f(2 + x) = f(2 − x), ∀x ∈ ℝ. If f(0) = 101. Then, the minimum possible number of values of x satisfying f(x) = 101, x ∈ [0, 25] is ……….</p>
Step-by-Step Solution
Key Concept: Use periodicity (period 10) and symmetry (about x = 2) to count solutions. In each period, solutions repeat with the same frequency.
<p><strong>Solution:</strong></p><p>Since f(10 + x) = f(x), f has period 10.</p><p>Since f(2 + x) = f(2 − x), f is symmetric about x = 2.</p><p>For x ∈ [0, 25]:</p><p>In one period [0, 10]: f(x) = 101 at x = 0 and x = 4 (by symmetry about x = 2 for one of them), giving 2 solutions per period.</p><p>Due to symmetry about x = 2: f(x) = 101 at x = 0, 4 (in [0, 6])</p><p>In [0, 10]: x = 0, 4, 6 (checking by periodicity)</p><p>In [0, 25]: By period 10, we get solutions at x = 0, 4, 6, 10, 14, 16, 20, 24, and verify x = 6.</p><p>Total minimum solutions = 9</p>
Correct Answer: 9