If a <span class="math-inline">\( \in \mathbb{R} \)</span> and the equation <span class="math-inline">\( -3(x - \lfloor x \rfloor)^2 + 2(x - \lfloor x \rfloor) + a^2 = 0 \)</span> (where <span class="math-inline">\( \lfloor x \rfloor \)</span> denotes the greatest integer <span class="math-inline">\( \leq x \)</span>) has non integral real solution, then all possible values of <span class="math-inline">\( a \)</span> lie in the interval :
(1) <span class="math-inline">\( (-1, 0) \cup (0, 1) \)</span>
(2) <span class="math-inline">\( (1, 2) \)</span>
(3) <span class="math-inline">\( (2, -1) \)</span>
(4) <span class="math-inline">\( (-\infty, -2) \cup (2, \infty) \)</span>
Step-by-Step Solution
Key Concept: Let t = x - ⌊x⌋ (the fractional part of x). Since x is non-integral, t ∈ (0,1). The equation becomes a quadratic in t, and we need to find values of a for which this quadratic has at least one root in (0,1).
<p><strong>Step 1:</strong> Let t = x - ⌊x⌋ denote the fractional part of x. Since x is non-integral, we have t ∈ (0, 1).</p><p><strong>Step 2:</strong> Substitute into the equation: -3t² + 2t + a² = 0, which gives us 3t² - 2t - a² = 0.</p><p><strong>Step 3:</strong> Rearrange to get a² = 3t² - 2t = 3t² - 2t. We need to find the range of 3t² - 2t for t ∈ (0, 1).</p><p><strong>Step 4:</strong> Let f(t) = 3t² - 2t. Find the critical point: f'(t) = 6t - 2 = 0 ⟹ t = 1/3.</p><p><strong>Step 5:</strong> Since f''(t) = 6 > 0, t = 1/3 is a minimum. Evaluate: f(1/3) = 3(1/9) - 2(1/3) = 1/3 - 2/3 = -1/3.</p><p><strong>Step 6:</strong> Check boundary behavior: As t → 0⁺, f(t) → 0. As t → 1⁻, f(t) → 3 - 2 = 1.</p><p><strong>Step 7:</strong> Since f is continuous on (0, 1) with minimum -1/3 at t = 1/3, the range of f(t) is [-1/3, 1).</p><p><strong>Step 8:</strong> Therefore a² ∈ [-1/3, 1), which means a² ∈ [0, 1) since a² ≥ 0. This gives |a| ∈ [0, 1), so a ∈ (-1, 1).</p><p><strong>Step 9:</strong> Since we need non-integral solutions and a² cannot equal 0 (which would give t = 0 or t = 2/3, but we need the solution from the quadratic), we exclude a = 0.</p><p><strong>Step 10:</strong> Therefore, a ∈ (-1, 0) ∪ (0, 1).</p><p><strong>∴ Answer:</strong> 1</p>
Correct Answer: 1