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