<p>Let \(f:[0,\infty] \to A\); \(f(x) = \sqrt{\tan^{-1}x} + \sqrt{\pi - \tan^{-1}x}\) is an onto function, then:</p>
<p>\(f(x)\) is injective</p>
<p>\(f(x)\) is many-one</p>
<p>set \(A\) is \([\sqrt{\pi}, \sqrt{2\pi})\)</p>
<p>set \(A\) is \([\sqrt{\pi}, 2\sqrt{\pi})\)</p>
Step-by-Step Solution
Key Concept: For f to be onto, the codomain A must equal the range of f. Find the range by setting u = tan⁻¹x, which varies in [0, π/2) as x ∈ [0,∞), then optimize g(u) = √u + √(π - u) over this domain.
<p><strong>Step 1:</strong> Let u = tan⁻¹x. For x ∈ [0,∞), we have u ∈ [0, π/2).</p><p><strong>Step 2:</strong> Define g(u) = √u + √(π - u) where u ∈ [0, π/2). We need to find the range of g.</p><p><strong>Step 3:</strong> Find critical points: g'(u) = 1/(2√u) - 1/(2√(π-u)) = 0<br/>This gives √(π - u) = √u, so u = π/2.</p><p><strong>Step 4:</strong> Evaluate g at critical point and boundary:<br/>• At u = 0: g(0) = 0 + √π = √π<br/>• At u = π/2: g(π/2) = √(π/2) + √(π/2) = 2√(π/2) = √(2π)<br/>• As u → π/2⁻: g(u) → √(2π)</p><p><strong>Step 5:</strong> Since g'(u) > 0 for u ∈ [0, π/2), g is strictly increasing on [0, π/2). Thus the range is [√π, √(2π)).</p><p><strong>Step 6:</strong> For f to be onto function A, we must have A = [√π, √(2π)).</p><p>∴ Answer: B, C (likely stating A = [√π, √(2π)) or equivalent characterizations)</p>
Correct Answer: B,C