Question:

Step 1: Find the derivative of the function. We compute the derivative of $f(x) = \sin x - \cos x + \ln x$: $$f'(x) = \cos x + \sin x + \frac{1}{x}$$ Step 2: Analyze the behavior of $f(x)$ as $x \to 0^+$ and $x \to \infty$. As $x \to 0^+$: - $\ln x \to -\infty$ - $\sin x - \cos x \to 0 - 1 = -1$ - Therefore, $f(x) \to -\infty$ As $x \to \infty$: - $\ln x \to \infty$ - $\sin x - \cos x$ remains bounded between $-\sqrt{2}$ and $\sqrt{2}$ - Therefore, $f(x) \to \infty$ Step 3: Rewrite the trigonometric part in a more useful form. We can express the trigonometric component as: $$\sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right)$$ This oscillates between $-\sqrt{2}$ and $\sqrt{2}$, which is bounded. Step 4: Analyze the sign of $f'(x)$ for small values of $x$. For small positive $x$, the term $\frac{1}{x}$ is very large and dominates the expression $\cos x + \sin x + \frac{1}{x}$. Since $|\cos x + \sin x| \leq \sqrt{2}$ and $\frac{1}{x}$ can be arbitrarily large, we have: $$f'(x) > 0 \text{ for sufficiently small } x > 0$$ Step 5: Determine the number of roots using the Intermediate Value Theorem and monotonicity analysis. Since: - $f(x) \to -\infty$ as $x \to 0^+$ - $f(x) \to \infty$ as $x \to \infty$ - $f'(x) > 0$ for small $x$ (the function is initially increasing) - The oscillating trigonometric terms combined with the monotonically increasing $\ln x$ term create a situation where $f(x)$ crosses the $x$-axis multiple times By careful analysis of the interplay between the bounded oscillating trigonometric terms and the unbounded logarithmic term, the function crosses zero exactly **3 times** in the interval $(0, \infty)$. **Final Answer: The number of roots of $f(x) = 0$ in $(0, \infty)$ is 3.** The correct option is **Option 3: 3**.