<p>Let \(f_k(x) = \dfrac{1}{k}(\sin^k x + \cos^k x)\) where \(x \in \mathbb{R}\) and \(k \geq 1\). Then \(f_4(x) - f_6(x)\) equals:</p>
Step-by-Step Solution
Key Concept: Express sin⁴x + cos⁴x and sin⁶x + cos⁶x using the substitution u = sin²x, then leverage the identity sin²x + cos²x = 1 to create a common framework for comparison.
<p><strong>Step 1:</strong> Let u = sin²x and v = cos²x, so u + v = 1 and uv = sin²x·cos²x.</p><p><strong>Step 2:</strong> For f₄(x): sin⁴x + cos⁴x = (sin²x)² + (cos²x)² = u² + v² = (u+v)² - 2uv = 1 - 2sin²x·cos²x</p><p>Therefore: f₄(x) = (1 - 2sin²x·cos²x)/4</p><p><strong>Step 3:</strong> For f₆(x): sin⁶x + cos⁶x = (sin²x)³ + (cos²x)³ = u³ + v³ = (u+v)³ - 3uv(u+v) = 1 - 3sin²x·cos²x</p><p>Therefore: f₆(x) = (1 - 3sin²x·cos²x)/6</p><p><strong>Step 4:</strong> f₄(x) - f₆(x) = (1 - 2sin²x·cos²x)/4 - (1 - 3sin²x·cos²x)/6</p><p>= 1/4 - 1/6 - sin²x·cos²x(1/2 - 1/2) = 1/12 - sin²x·cos²x·0</p><p>= 1/12</p><p>∴ Answer: B</p>
Correct Answer: B