Trigonometry & Inverse Trigonometry
Inverse Sine Function
Grade 12
Question:
<p>Let <span>0 < a < 2π</span>. If <span>\sin^{-1}(\sin a) < x^2 - 2x</span> for all <span>x \in \mathbb{R}</span>, then <span>a \in</span></p>
<p>(a) <span>(0, π + 1)</span></p>
<p>(b) <span>[π + 1, \frac{3π}{2})</span></p>
<p>(c) <span>[\frac{3π}{2}, 2π - 1]</span></p>
<p>(d) <span>(2π - 1, 2π)</span></p>
Step-by-Step Solution
Key Concept: Recognize that the range of sin⁻¹ is [-π/2, π/2], so sin⁻¹(sin a) depends on which interval a falls into. The minimum value of x² - 2x is -1.
<p><strong>Step 1:</strong> We need <span>\sin^{-1}(\sin a) < \min(x^2 - 2x) = -1</span></p><p><strong>Step 2:</strong> For <span>a \in (0, π]</span>: <span>\sin^{-1}(\sin a) = a</span>, so <span>a < -1</span> (impossible since <span>a > 0</span>)</p><p><strong>Step 3:</strong> For <span>a \in (π, \frac{3π}{2}]</span>: <span>\sin^{-1}(\sin a) = π - a</span>, so <span>π - a < -1</span>, giving <span>a > π + 1</span>. Thus <span>a \in [π + 1, \frac{3π}{2}]</span></p><p><strong>Step 4:</strong> For <span>a \in (\frac{3π}{2}, 2π)</span>: <span>\sin^{-1}(\sin a) = a - 2π</span>, so <span>a - 2π < -1</span>, giving <span>a < 2π - 1</span>. Thus <span>a \in [\frac{3π}{2}, 2π - 1]</span></p><p>∴ Answer is (b) and (c)</p>
Correct Answer: b,c