Relations & Functions
Evaluation of functions
Grade 12
Question:
<p>Let \(f\) be a function defined on the set of real numbers such that for \(x \geq 0\), \(f(x) = 3\sin x + 4\cos x\). Then \(f(x)\) at \(x = -\dfrac{11\pi}{6}\) is equal to</p>
<p>\(\dfrac{3}{2} + 2\sqrt{3}\)</p>
<p>\(-\dfrac{3}{2} + 2\sqrt{3}\)</p>
<p>\(\dfrac{3}{2} - 2\sqrt{3}\)</p>
<p>\(-\dfrac{3}{2} - 2\sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Use the periodicity and symmetry properties of trigonometric functions. Since f is defined for x ≥ 0, convert the negative argument to an equivalent positive value using the periodic nature of sin and cos (period 2π), then apply even/odd properties.
<p><strong>Step 1:</strong> The function f is defined for x ≥ 0 as f(x) = 3sin x + 4cos x. For x = -11π/6 < 0, we must use periodicity to convert to an equivalent positive angle.</p><p><strong>Step 2:</strong> Add 2π repeatedly to -11π/6 until we get a positive angle: -11π/6 + 2π = -11π/6 + 12π/6 = π/6. Since sin and cos have period 2π, we have sin(-11π/6) = sin(π/6) and cos(-11π/6) = cos(π/6).</p><p><strong>Step 3:</strong> Calculate f(π/6) = 3sin(π/6) + 4cos(π/6) = 3(1/2) + 4(√3/2) = 3/2 + 2√3.</p><p><strong>Step 4:</strong> Simplify: f(-11π/6) = 3/2 + 2√3 = (3 + 4√3)/2.</p><p>∴ Answer: B</p>
Correct Answer: B