Trigonometry & Inverse Trigonometry
Trigonometric Inequalities
Grade 11
Question:
<p>Let \(f(x) = \sin^2 x - \sin x + k\), \(x \in R\). 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: Substitute t = sin x where t ∈ [-1, 1] to convert the trigonometric problem into analyzing a quadratic g(t) = t² - t + k on a bounded interval. The range of f depends on the extrema of this quadratic within [-1, 1].
<p><strong>Step 1:</strong> Let t = sin x, where t ∈ [-1, 1]</p><p><strong>Step 2:</strong> Then f(x) = g(t) = t² - t + k is a quadratic in t</p><p><strong>Step 3:</strong> Find the vertex: t = -(-1)/(2·1) = 1/2, which lies in [-1, 1]</p><p><strong>Step 4:</strong> Evaluate at critical points:</p><ul><li>g(1/2) = 1/4 - 1/2 + k = k - 1/4 (minimum, as parabola opens upward)</li><li>g(-1) = 1 + 1 + k = k + 2</li><li>g(1) = 1 - 1 + k = k</li></ul><p><strong>Step 5:</strong> The maximum value is max{k + 2, k} = k + 2 (at t = -1)</p><p><strong>Step 6:</strong> The minimum value is k - 1/4 (at t = 1/2)</p><p>∴ Range of f: [k - 1/4, k + 2]</p>
Correct Answer: A