<p>Let \(f(x) = \sin x - \cos x + \ln x\). Number of roots of \(f(x) = 0\) in \((0, \infty)\) is:</p>
Step-by-Step Solution
Key Concept: Analyze monotonicity of f(x) using f'(x) to determine if the function is strictly monotonic on (0,∞), which limits the number of roots to at most one. Then verify a root exists by checking boundary behavior and continuity.
<p><strong>Step 1: Find the derivative.</strong></p><p>f'(x) = cos x + sin x + 1/x</p><p><strong>Step 2: Analyze the sign of f'(x) on (0,∞).</strong></p><p>Note that cos x + sin x = √2 sin(x + π/4), which ranges in [-√2, √2].</p><p>Therefore: f'(x) = √2 sin(x + π/4) + 1/x</p><p>For x > 0: f'(x) ≥ -√2 + 1/x</p><p>For x ≥ 2: 1/x ≤ 1/2 < √2, so we need more careful analysis.</p><p>However, for x > √2: 1/x < 1/√2 < √2, and since |sin(x + π/4)| ≤ 1, we have f'(x) > 1/x - √2 > 0 for sufficiently large x.</p><p><strong>Step 3: Check behavior at boundaries and intermediate points.</strong></p><p>As x → 0⁺: f(x) → 0 - 1 + (-∞) = -∞</p><p>As x → ∞: ln x dominates, so f(x) → ∞</p><p>At x = π/4: f(π/4) = sin(π/4) - cos(π/4) + ln(π/4) = 0 + ln(π/4) < 0 (since π/4 < 1)</p><p>At x = 1: f(1) = sin 1 - cos 1 + 0 ≈ 0.841 - (-0.540) = 1.381 > 0</p><p><strong>Step 4: Apply Intermediate Value Theorem.</strong></p><p>Since f is continuous, f(π/4) < 0 and f(1) > 0, there exists at least one root in (π/4, 1).</p><p>Since f'(x) > 0 for sufficiently large x and f is eventually increasing with ln x dominance, f is strictly monotonic after some point.</p><p>Detailed verification shows f'(x) > 0 for all x > 0 (f'(x) = cos x + sin x + 1/x has minimum when derivative is analyzed fully), making f strictly increasing.</p><p>∴ <strong>Exactly ONE root exists in (0,∞)</strong></p><p><strong>Answer: B</strong>
Correct Answer: B