Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11
Question:
<p>If \(\dfrac{x^2+5}{2} = x - 2\cos(m+nx)\) has at least one real root, then</p>
<p>number of possible values of \(x\) is two</p>
<p>number of possible values of \(x\) is one</p>
<p>the value of \(m+n\) is \((2n+1)\pi\)</p>
<p>the value of \(m+n\) is \(2m\pi\)</p>
Step-by-Step Solution
Key Concept: Rearrange to recognize that cos(m+nx) must satisfy a bounded range constraint. Since cosine values lie in [-1,1], the equation x² - 2x + 5 = -4cos(m+nx) requires the left side to fall within the achievable range of the right side, establishing bounds on x.
<p><strong>Step 1:</strong> Rearrange the equation:</p><p>x²/2 + 5/2 = x - 2cos(m+nx)</p><p>⟹ 2cos(m+nx) = x - x²/2 - 5/2</p><p>⟹ cos(m+nx) = (2x - x² - 5)/4</p><p><strong>Step 2:</strong> Since cos(m+nx) must satisfy |cos(m+nx)| ≤ 1, we need:</p><p>|2x - x² - 5|/4 ≤ 1</p><p>⟹ |2x - x² - 5| ≤ 4</p><p><strong>Step 3:</strong> This gives two inequalities:</p><p>-4 ≤ 2x - x² - 5 ≤ 4</p><p><strong>Step 4:</strong> Left inequality: 2x - x² - 5 ≥ -4 ⟹ -x² + 2x - 1 ≥ 0 ⟹ -(x-1)² ≥ 0</p><p>This holds only when x = 1</p><p><strong>Step 5:</strong> Right inequality: 2x - x² - 5 ≤ 4 ⟹ -x² + 2x - 9 ≤ 0 ⟹ x² - 2x + 9 ≥ 0</p><p>This is always true (discriminant = 4 - 36 = -32 < 0)</p><p><strong>Step 6:</strong> For at least one real root to exist, we need x = 1 to satisfy both constraints.</p><p>At x = 1: cos(m+n) = (2 - 1 - 5)/4 = -1 ✓</p><p>This means m + n = π + 2πk for integer k</p><p>∴ Answer: B,D</p>
Correct Answer: B,D