<p>Let <i>f</i>(<i>x</i>) be a continuous function satisfying \(\begin{cases} f(x) = x^2 - 6x + 8 & \text{for } -1 \le x \le a-1 \\ f(x) = f(x+a) \end{cases}\) where '<i>a</i>' is a constant, then find the sum of all possible values of '<i>a</i>'.</p>
Step-by-Step Solution
Key Concept: For continuity at the boundary point x = a-1, the quadratic value must equal the periodic continuation value f(a-1) = f(a-1+a) = f(2a-1). This constraint, combined with the periodicity condition, determines which values of 'a' are valid.
<p><strong>Step 1:</strong> Identify the piecewise definition. For -1 ≤ x ≤ a-1, f(x) = x² - 6x + 8. For x > a-1, f is periodic with period a, so f(x) = f(x-a).</p><p><strong>Step 2:</strong> For continuity at x = a-1, we need the left limit to equal the right limit. The left limit is f(a-1) = (a-1)² - 6(a-1) + 8 = a² - 8a + 15. The right limit uses periodicity: f((a-1)⁺) = f((a-1) - a)⁺ = f(-1)⁺ = (-1)² - 6(-1) + 8 = 15.</p><p><strong>Step 3:</strong> Set continuity condition: a² - 8a + 15 = 15, which gives a² - 8a = 0, so a(a - 8) = 0.</p><p><strong>Step 4:</strong> This yields a = 0 or a = 8. However, a = 0 makes the domain [-1, -1] (degenerate), so only a = 8 is physically valid. Checking a = 8: the domain becomes [-1, 7], and f(7) = 49 - 42 + 8 = 15 = f(-1) ✓</p><p><strong>Step 5:</strong> If the problem allows a = 0 as a limiting case or asks for algebraic solutions, both roots sum to 8. Otherwise, only a = 8 is valid.</p><p>∴ <strong>Answer: 8</strong> (if only non-degenerate case) or <strong>Sum = 8</strong> (if both algebraic roots count)</p>
Correct Answer: 8