<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 function from [0,∞) to A, the range of f must equal A. Since tan⁻¹x ∈ [0, π/2) for x ∈ [0,∞), we need to find the range by analyzing the function g(t) = √t + √(π - t) where t = tan⁻¹x ∈ [0, π/2).
<p><strong>Step 1:</strong> Let t = tan⁻¹x. Since x ∈ [0,∞), we have t ∈ [0, π/2).</p><p><strong>Step 2:</strong> Rewrite f as g(t) = √t + √(π - t) for t ∈ [0, π/2).</p><p><strong>Step 3:</strong> Find critical points: g'(t) = 1/(2√t) - 1/(2√(π-t)) = 0<br/>This gives √(π-t) = √t, so t = π/2.</p><p><strong>Step 4:</strong> Evaluate at boundaries:<br/>• At t = 0: g(0) = √0 + √π = √π<br/>• At t = π/2: g(π/2) = √(π/2) + √(π/2) = 2√(π/2) = √(2π)<br/>• Since t ∈ [0, π/2), maximum √(2π) is approached but not attained</p><p><strong>Step 5:</strong> Check monotonicity: g'(t) > 0 for t < π/2, so g is strictly increasing on [0, π/2).<br/>Therefore, range = [√π, √(2π))</p><p><strong>Step 6:</strong> For f to be onto, A = [√π, √(2π)).</p><p>∴ Answer: B,C</p>
Correct Answer: B,C