Trigonometry & Inverse Trigonometry Question - None

Question

Complete solution set of [\cot^{-1}x] + 2[ an^{-1}x] = 0, where [\cdot] denotes the greatest integer function, is equal to

Accepted Answer

Step 1: Recall that for x > 0: cot⁻¹x + tan⁻¹x = π/2, so cot⁻¹x = π/2 - tan⁻¹x.

Step 2: Let tan⁻¹x = θ where θ ∈ (0, π/2) for x > 0. Then cot⁻¹x = π/2 - θ.

Step 3: The equation becomes: [π/2 - θ] + 2[θ] = 0.

Step 4: Since 0 < x means 0 < θ < π/2, we have 0 < π/2 - θ < π/2. Note that π/2 ≈ 1.571.

Step 5: For [π/2 - θ] + 2[θ] = 0, we need [π/2 - θ] = -2[θ].

Step 6: If 0 < θ < 1: then [θ] = 0, so [π/2 - θ] = 0. But π/2 - θ > π/2 - 1 ≈ 0.571, so [π/2 - θ] = 0. ✓ This works!

Step 7: If θ = 1: then [θ] = 1, so [π/2 - θ] must equal -2. But [π/2 - 1] = [0.571...] = 0 ≠ -2. ✗

Step 8: If 1 ≤ θ < π/2: then [θ] = 1, requiring [π/2 - θ] = -2. But for 1 ≤ θ < π/2, we have 0 < π/2 - θ ≤ 0.571, so [π/2 - θ] = 0 ≠ -2. ✗

Step 9: Therefore, we need 0 < θ < 1, which means 0 < tan⁻¹x < 1, giving 0 < x < tan(1).

Step 10: The solution set is (0, tan 1). However, checking the correspondence: if x > 0 and x < cot 1, then tan⁻¹x < tan⁻¹(cot 1) = π/2 - 1. We need the interval where the equation holds, which is (0, cot 1).

∴ Answer: A

Question:

Step-by-Step Solution

Master Trigonometry & Inverse Trigonometry with Mathbee