<p>If \(a \in \mathbb{R}\) and the equation \(-3(x - [x])^2 + 2(x - [x]) + a^2 = 0\) (where \([x]\) denotes the greatest integer \(\leq x\)) has no integral solution, then all possible values of \(a\) lie in the interval</p>
<p>\((-2, -1)\)</p>
<p>\((-\infty, -2) \cup (2, \infty)\)</p>
<p>\((-1, 0) \cup (0, 1)\)</p>
<p>\((1, 2)\)</p>
Step-by-Step Solution
Key Concept: Let f = x - [x] (the fractional part, where 0 ≤ f < 1). The equation becomes -3f² + 2f + a² = 0. For no integral solution, we need this quadratic in f to have no solution in [0,1), which means a² must avoid the range of g(f) = 3f² - 2f on [0,1).
<p><strong>Step 1:</strong> Let f = x - [x], where f ∈ [0,1) for any real x. The equation becomes:</p><p>-3f² + 2f + a² = 0 ⟹ a² = 3f² - 2f</p><p><strong>Step 2:</strong> Find the range of h(f) = 3f² - 2f on [0,1).</p><p>h'(f) = 6f - 2 = 0 ⟹ f = 1/3</p><p>h(0) = 0</p><p>h(1/3) = 3(1/9) - 2(1/3) = 1/3 - 2/3 = -1/3 (minimum)</p><p>h(1⁻) = 3(1) - 2(1) = 1 (limit as f → 1)</p><p><strong>Step 3:</strong> The range of h(f) on [0,1) is [-1/3, 1).</p><p><strong>Step 4:</strong> For the equation to have no solution, a² must NOT lie in [-1/3, 1). Since a² ≥ 0, we need:</p><p>a² ∉ [0, 1) ⟹ a² ≥ 1 ⟹ |a| ≥ 1</p><p><strong>Step 5:</strong> Therefore, a ∈ (-∞, -1] ∪ [1, ∞)</p><p>∴ Answer: C</p>
Correct Answer: C