<p>Sum of positive integral values of <span>\(a\)</span> for which <span>\(a^2 - 6\sin x - 5a \leq 0\)</span> for all <span>\(x \in \mathbb{R}\)</span> is</p>
Step-by-Step Solution
Key Concept: For an inequality involving a trigonometric function to hold for all x ∈ ℝ, it must hold at the extreme values of that function. Since sin x ranges from -1 to 1, we need the inequality to hold for the maximum value of the trigonometric expression.
<p><strong>Step 1:</strong> We need a² - 6sin x - 5a ≤ 0 to hold for all x ∈ ℝ.</p><p><strong>Step 2:</strong> Rearrange: a² - 5a ≤ 6sin x for all x ∈ ℝ.</p><p><strong>Step 3:</strong> Since this must hold for all x, it must hold for the minimum value of 6sin x. The minimum value of sin x is -1, so the minimum value of 6sin x is -6.</p><p><strong>Step 4:</strong> Therefore: a² - 5a ≤ -6 (this is the most restrictive condition)</p><p><strong>Step 5:</strong> Rearrange: a² - 5a + 6 ≤ 0</p><p><strong>Step 6:</strong> Factor: (a - 2)(a - 3) ≤ 0</p><p><strong>Step 7:</strong> This inequality holds when 2 ≤ a ≤ 3.</p><p><strong>Step 8:</strong> The positive integral values in the range [2, 3] are: a = 2 and a = 3</p><p><strong>Step 9:</strong> Verify: For a = 2: 4 - 5(2) + 6 = 0 ≤ 0 ✓; For a = 3: 9 - 5(3) + 6 = 0 ≤ 0 ✓</p><p><strong>Step 10:</strong> Sum of positive integral values = 2 + 3 = 5</p><p><strong>∴ Answer:</strong> R</p>
Correct Answer: R