Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Let \(\cos^{-1}(4x^3 - 3x) = a + b\cos^{-1}x\).</p><p>If \(x \in \left(\frac{1}{2}, 1\right]\), then \(\lim_{y \to a} b\cos y\) is:</p>
<p>(a) \(-\frac{1}{3}\)</p>
<p>(b) \(-3\)</p>
<p>(c) \(\frac{1}{3}\)</p>
<p>(d) \(3\)</p>
Step-by-Step Solution
Key Concept: Identify the values of \(a\) and \(b\) from the given relation, then evaluate the limit using continuity of the cosine function.
<p><strong>Solution:</strong> For \(x \in \left(\frac{1}{2}, 1\right]\), the identity becomes \(\cos^{-1}(4x^3 - 3x) = 3\cos^{-1}x\), giving \(a = 0\) and \(b = 3\).</p><p>Therefore, \(\lim_{y \to a} b\cos y = \lim_{y \to 0} 3\cos y = 3 \cdot \cos 0 = 3\).</p>
Correct Answer: D