Trigonometry & Inverse Trigonometry
Inverse Trigonometric Inequalities
Grade 12
Question:
<p>All \( x \) satisfying the inequality \( (\cot^{-1}x)^2 - 7(\cot^{-1}x) + 10 > 0 \), lie in the interval:</p>
<p>\( (-\infty, \cot 5) \cup (\cot 4, \cot 2) \)</p>
<p>\( (\cot 2, \infty) \)</p>
<p>\( (-\infty, \cot 5) \cup (\cot 2, \infty) \)</p>
<p>\( (\cot 5, \cot 4) \)</p>
Step-by-Step Solution
Key Concept: Substitute y = cot⁻¹(x) to convert the inequality into a quadratic form, then solve the quadratic inequality and map back to x using the properties of inverse cotangent function (range [0,π], decreasing).
<p><strong>Step 1:</strong> Let y = cot⁻¹(x). The inequality becomes: y² - 7y + 10 > 0</p><p><strong>Step 2:</strong> Factor the quadratic: (y - 2)(y - 5) > 0</p><p>This gives: y < 2 or y > 5</p><p><strong>Step 3:</strong> Since the range of cot⁻¹ is (0, π) and π ≈ 3.14, we have y > 5 is impossible (5 > π). Only y < 2 is valid.</p><p><strong>Step 4:</strong> Solve cot⁻¹(x) < 2: Since cot⁻¹ is a decreasing function, cot⁻¹(x) < 2 ⟹ x > cot(2)</p><p><strong>Step 5:</strong> As y → 0⁺, cot⁻¹(x) → 0⁺ means x → +∞. As y → π⁻, cot⁻¹(x) → π⁻ means x → -∞. For cot⁻¹(x) ∈ (0, 2), we get x ∈ (cot(2), +∞)</p><p>∴ Answer: C</p>
Correct Answer: C