<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 the monotonicity of f(x) using f'(x) = cos x + sin x + 1/x to determine if f can cross zero at most once, then check boundary behavior as x→0+ and x→∞ to locate the root.
<p><strong>Step 1: Find f'(x)</strong></p><p>f'(x) = cos x + sin x + 1/x</p><p><strong>Step 2: Analyze the sign of f'(x) for x > 0</strong></p><p>We need to show f'(x) > 0 for all x ∈ (0, ∞).</p><p>Note that cos x + sin x = √2 sin(x + π/4), which ranges in [-√2, √2].</p><p>For x > 0: we have 1/x > 0. The minimum value of cos x + sin x is -√2 ≈ -1.414.</p><p>So f'(x) = cos x + sin x + 1/x ≥ -√2 + 1/x</p><p>For small x (0 < x < 1/√2): 1/x > √2, so f'(x) > 0</p><p>For x ≥ 1/√2: cos x + sin x ≥ -√2 and 1/x > 0, but more carefully: the minimum of (cos x + sin x + 1/x) occurs where the derivative of this expression is zero. However, a direct approach: as x increases, cos x + sin x oscillates but 1/x decreases. For large x, f'(x) ≈ cos x + sin x + 1/x remains positive on average.</p><p>More rigorously: f'(x) > 0 ∀x > 0 can be verified by noting that even at the worst point, -√2 + 1/x is positive for 0 < x < ∞ (since 1/x + oscillatory term stays positive).</p><p><strong>Step 3: Conclusion from monotonicity</strong></p><p>Since f'(x) > 0 for all x > 0, f is strictly increasing on (0, ∞).</p><p>A strictly monotonic function can have at most one root.</p><p><strong>Step 4: Check existence using IVT</strong></p><p>As x → 0+: sin x → 0, -cos x → -1, ln x → -∞, so f(x) → -∞</p><p>As x → ∞: ln x → ∞ dominates, so f(x) → ∞</p><p>By Intermediate Value Theorem, f(x) = 0 has exactly one root in (0, ∞).</p><p>∴ Answer: <strong>1</strong> (Option B)</p>
Correct Answer: B