<p>The domain of \(f(x) = \sqrt{\log\left|\frac{1}{\sin x}\right|}\) is</p>
<p>(a) \(R\)</p>
<p>(b) \(R - [-\pi, \pi]\)</p>
<p>(c) \(R - \{n\pi, n \in Z\}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: For the square root to be defined, we need the argument ≥ 0, which means log|1/sin x| ≥ 0. This occurs when |1/sin x| ≥ 1, i.e., |sin x| ≤ 1, but more crucially when |sin x| ≤ 1 AND sin x ≠ 0. The logarithm is non-negative when its argument ≥ 1, so |1/sin x| ≥ 1 gives |sin x| ≤ 1.
<p><strong>Step 1:</strong> For √u to be defined, we need u ≥ 0, so log|1/sin x| ≥ 0.</p><p><strong>Step 2:</strong> The logarithm log(a) ≥ 0 when a ≥ 1. Therefore: |1/sin x| ≥ 1.</p><p><strong>Step 3:</strong> This means 1/|sin x| ≥ 1, which gives |sin x| ≤ 1. Since |sin x| ≤ 1 is always true for real x, we need the stronger condition from the logarithm argument being defined: sin x ≠ 0.</p><p><strong>Step 4:</strong> For log|1/sin x| ≥ 0, we need |sin x| ≤ 1 AND equality holds when sin x = ±1. But sin x = 0 makes the log undefined. The condition simplifies to: |sin x| ≤ 1 with sin x ≠ 0, which gives us x ≠ nπ where n ∈ ℤ.</p><p><strong>Step 5:</strong> However, the complete analysis shows we need |sin x| ≤ 1 (always satisfied) AND sin x ≠ 0, giving domain: <strong>ℝ \ {nπ : n ∈ ℤ}</strong> or equivalently <strong>x ∈ (nπ, (n+1)π) for n ∈ ℤ</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C