Trigonometry & Inverse Trigonometry
Trigonometric Inequalities
Grade 11
Question:
<p>Let \(f(x) = \sin^2 x - \sin x + k\), \(x \in \mathbb{R}\). Then:</p>
<p>\(f(x) \geq 0\) if \(k \geq \dfrac{1}{4}\)</p>
<p>\(f(x) \geq 0\) if \(k \leq \dfrac{1}{4}\)</p>
<p>\(f(x) \leq 0\) if \(k \geq -2\)</p>
<p>\(f(x) \leq 0\) if \(k \geq -2\)</p>
Step-by-Step Solution
Key Concept: Recognize f(x) as a quadratic in sin(x). Since sin(x) ∈ [-1,1], substitute t = sin(x) to find the range of f(t) = t² - t + k on [-1,1].
<p><strong>Step 1:</strong> Let t = sin(x), so t ∈ [-1, 1]. Then f(x) = g(t) = t² - t + k.</p><p><strong>Step 2:</strong> Find the vertex: g'(t) = 2t - 1 = 0 ⟹ t = 1/2 ∈ [-1,1].</p><p><strong>Step 3:</strong> Evaluate at critical point and endpoints:</p><ul><li>g(1/2) = 1/4 - 1/2 + k = k - 1/4 (minimum)</li><li>g(-1) = 1 + 1 + k = k + 2</li><li>g(1) = 1 - 1 + k = k</li></ul><p><strong>Step 4:</strong> Range of f is [k - 1/4, k + 2] (assuming the question asks for range conditions or specific properties of k).</p><p>∴ Answer: A</p>
Correct Answer: A