<p><strong>262.</strong> If \(\sec^{-1}(x) + \tan^{-1}\sqrt{9y^2 - 1} + \sin^{-1}(x^2 + y^2) = \lambda\) has no solution, then exhaustive set of values of \(\lambda\) is equal to:</p>
Step-by-Step Solution
Key Concept: Find the range of each inverse trigonometric term by determining valid domains for x and y, then identify which values of λ cannot be expressed as a sum of these three terms.
<p><strong>Step 1: Identify domain constraints</strong></p><p>For the equation to have a solution, we need:</p><ul><li>sec⁻¹(x) defined: |x| ≥ 1, range = [0, π] \ {π/2}</li><li>tan⁻¹(√(9y²-1)) defined: 9y²-1 ≥ 0 ⟹ |y| ≥ 1/3, range = [0, π/2)</li><li>sin⁻¹(x²+y²) defined: x²+y² ≤ 1, range = [-π/2, π/2]</li></ul><p><strong>Step 2: Find feasible domain intersection</strong></p><p>From sec⁻¹(x): |x| ≥ 1</p><p>From sin⁻¹(x²+y²): x²+y² ≤ 1</p><p>These together give: x² ≥ 1 AND x²+y² ≤ 1</p><p>This forces x² = 1 and y² = 0, so x = ±1, y = 0</p><p><strong>Step 3: Verify and calculate range</strong></p><p>When y = 0: tan⁻¹(√(9·0-1)) = tan⁻¹(√(-1)) is undefined!</p><p>The domain constraint |y| ≥ 1/3 contradicts y = 0.</p><p><strong>Step 4: Conclusion</strong></p><p>There is NO point (x,y) satisfying all three domain requirements simultaneously.</p><p>Therefore, for ANY value of λ, the equation has no solution.</p><p>∴ Answer: λ ∈ ℝ (all real numbers) or the exhaustive set is <strong>C</strong></p>
Correct Answer: C