<p>The real function \(f(x) = \cos^{-1}\sqrt{x^2 + 3x + 1} + \cos^{-1}\sqrt{x^2 + 3x}\) is defined on the set</p>
<p>(a) \(\{0, 3\}\)</p>
<p>(b) interval \((0, 3)\)</p>
<p>(c) \(\{0, -3\}\)</p>
<p>(d) interval \([-3, 0]\)</p>
Step-by-Step Solution
Key Concept: For inverse cosine to be defined, the argument must satisfy 0 ≤ argument ≤ 1. Both √(x² + 3x + 1) and √(x² + 3x) must be real and ≤ 1 simultaneously, requiring the expressions under the square roots to be non-negative and the square roots themselves to not exceed 1.
<p><strong>Step 1:</strong> For f(x) to be defined, we need both inverse cosine arguments to be in [0,1].</p><p><strong>Step 2:</strong> For √(x² + 3x + 1): We need x² + 3x + 1 ≥ 0. Discriminant = 9 - 4 = 5, so roots are x = (-3 ± √5)/2. This gives x ≤ (-3 - √5)/2 or x ≥ (-3 + √5)/2.</p><p><strong>Step 3:</strong> For √(x² + 3x + 1) ≤ 1: We need x² + 3x + 1 ≤ 1, so x² + 3x ≤ 0, giving x(x + 3) ≤ 0, thus -3 ≤ x ≤ 0.</p><p><strong>Step 4:</strong> For √(x² + 3x) ≥ 0: We need x² + 3x ≥ 0, so x(x + 3) ≥ 0, giving x ≤ -3 or x ≥ 0.</p><p><strong>Step 5:</strong> For √(x² + 3x) ≤ 1: We need x² + 3x ≤ 1, so x² + 3x - 1 ≤ 0. Roots are x = (-3 ± √13)/2, giving (-3 - √13)/2 ≤ x ≤ (-3 + √13)/2.</p><p><strong>Step 6:</strong> Taking intersection of all conditions: From Step 3: [-3, 0]. From Step 4: (-∞, -3] ∪ [0, ∞). Their intersection gives x = -3 or x = 0. Verify with Step 5: both satisfy it.</p><p>∴ Answer: C (The domain is {-3, 0} or a finite discrete set)</p>
Correct Answer: C