<p>The equation \(4\cos 2x = 7 - 4x + 4x^2\) has no solution for \(\theta\) if \(x \in \mathbb{R}\).</p><p><em>State whether the statement is true or false.</em></p>
Step-by-Step Solution
Key Concept: Recognize that 4cos(2x) ∈ [-4, 4] while the RHS 4x² - 4x + 7 = 4(x - 1/2)² + 6 ≥ 6 for all real x. Since the ranges don't overlap, no solution exists.
<p><strong>Step 1:</strong> Analyze the LHS: Since -1 ≤ cos(2x) ≤ 1, we have <strong>-4 ≤ 4cos(2x) ≤ 4</strong>.</p><p><strong>Step 2:</strong> Analyze the RHS: 4x² - 4x + 7 = 4(x² - x) + 7 = 4(x - 1/2)² - 1 + 7 = 4(x - 1/2)² + 6.</p><p><strong>Step 3:</strong> Since (x - 1/2)² ≥ 0 for all x ∈ ℝ, we have <strong>4x² - 4x + 7 ≥ 6</strong> for all real x, with minimum value 6 at x = 1/2.</p><p><strong>Step 4:</strong> For a solution to exist, we need overlap between [-4, 4] and [6, ∞). Since 4 < 6, <strong>there is no overlap</strong>.</p><p><strong>Step 5:</strong> Therefore, the equation 4cos(2x) = 7 - 4x + 4x² has no solution for any x ∈ ℝ.</p><p>∴ <strong>The statement is TRUE</strong></p>
Correct Answer: A