(A) f(x) ∈ <span class="math-inline">[0, 1]</span> ∀ <span class="math-inline">x ∈ ℝ</span>
(B) f(x) = 0 has no solution
(C) f(x) ∈ <span class="math-inline">[\frac{1}{4}, 1]</span> ∀ <span class="math-inline">x ∈ ℝ</span>
(D) f(x) is an injective function
Step-by-Step Solution
Key Concept: Use algebraic identities to simplify sin⁶(x) + cos⁶(x) into a function of sin²(x)cos²(x), then find its range by analyzing the constraint that sin²(x) + cos²(x) = 1.
<p><strong>Step 1: Simplify using sum of cubes formula</strong></p><p>Let a = sin²(x) and b = cos²(x). Note that a + b = 1 and a, b ∈ [0,1].</p><p>f(x) = sin⁶(x) + cos⁶(x) = a³ + b³</p><p><strong>Step 2: Apply sum of cubes factorization</strong></p><p>a³ + b³ = (a + b)(a² - ab + b²) = 1·(a² - ab + b²)</p><p>= a² + b² - ab</p><p><strong>Step 3: Express in terms of single variable</strong></p><p>Since a + b = 1, we have b = 1 - a.</p><p>a² + b² = a² + (1-a)² = a² + 1 - 2a + a² = 2a² - 2a + 1</p><p>ab = a(1-a) = a - a²</p><p>Therefore: f(x) = 2a² - 2a + 1 - (a - a²) = 3a² - 3a + 1</p><p><strong>Step 4: Find the range</strong></p><p>Let g(a) = 3a² - 3a + 1 where a ∈ [0,1]</p><p>g'(a) = 6a - 3 = 0 ⟹ a = 1/2</p><p>g(0) = 1</p><p>g(1/2) = 3(1/4) - 3(1/2) + 1 = 3/4 - 3/2 + 1 = 1/4</p><p>g(1) = 3 - 3 + 1 = 1</p><p><strong>Step 5: Determine the range</strong></p><p>The minimum value is g(1/2) = 1/4 (achieved when sin²(x) = cos²(x) = 1/2, i.e., x = π/4 + nπ/2)</p><p>The maximum value is 1 (achieved when x = 0, π/2, π, ...)</p><p>Since g(a) is continuous on [0,1], the range is [1/4, 1].</p><p><strong>Verification of options:</strong></p><p>(A) False: f(x) ≥ 1/4, not ≥ 0</p><p>(B) False: f(x) = 1 > 0 always, but f(x) never equals 0</p><p>(C) True: f(x) ∈ [1/4, 1] ✓</p><p>(D) False: f(x) is periodic with period π/2, not injective</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C