<p>\(f(x)=\cot^{-1}\!\sqrt{x(x+3)}+\cos^{-1}\!\sqrt{x^2+3x+1}\) is defined on set \(S\). \(S\) equals:</p>
Step-by-Step Solution
<div class="solution"><p><strong>Term 1:</strong> $\sqrt{x(x+3)}\ge 0$ requires $x\in(-\infty,-3]\cup[0,\infty)$.</p><p><strong>Term 2 condition A:</strong> $x^2+3x+1\ge 0$.</p><p><strong>Term 2 condition B:</strong> $\sqrt{x^2+3x+1}\le 1\implies x^2+3x+1\le 1\implies x(x+3)\le 0\implies x\in[-3,0]$.</p><p><strong>Intersection:</strong> $(-\infty,-3]\cup[0,\infty)$ intersected with $[-3,0]$ gives $\{-3,0\}$.</p><p><strong>Answer: (C) $\{0,-3\}$</strong></p><div class="trap-box"><strong>Trap:</strong> Including all of $[-3,0]$ -- the first term's square root condition collapses this to just the endpoints.<div class="key-concept"><strong>Key Concept:</strong> Intersecting multiple domain constraints often collapses to discrete points
Correct Answer: 3