The following functions are continuous on `(0, pi)`

Options

A. tan x
B. `int_(0)^(x) sin 1/t dt`
C. `{{:(1, 0 lt x le (3pi)/4),(2 sin (x/3), (3pi)/4 lt x lt pi):}`
D. `{{:(x sin x, 0 lt x le pi/2),(pi/2sin(2pi+x), pi/2 lt x lt pi):}`

Answer

To determine which of the given functions are continuous on the interval \( (0, \pi) \), we will analyze each option step by step.

### Step 1: Analyze the first function \( f(x) = \tan x \)

**Continuity Check:**
- The function \( \tan x \) is continuous wherever it is defined.
- However, \( \tan x \) is undefined at \( x = \frac{\pi}{2} \).
- Since \( \frac{\pi}{2} \) lies within the interval \( (0, \pi) \), \( \tan x \) is discontinuous.

**Conclusion:**
- \( f(x) = \tan x \) is **not continuous** on \( (0, \pi) \).

### Step 2: Analyze the second function \( f(x) = \int_0^x \sin\left(\frac{1}{t}\right) dt \)

**Continuity Check:**
- The function defined by the integral is continuous as long as the integrand \( \sin\left(\frac{1}{t}\right) \) is continuous on the interval \( (0, x) \).
- The function \( \sin\left(\frac{1}{t}\right) \) is continuous for \( t > 0 \).
- Therefore, \( f(x) \) is continuous for \( x \in (0, \pi) \).

**Conclusion:**
- \( f(x) = \int_0^x \sin\left(\frac{1}{t}\right) dt \) is **continuous** on \( (0, \pi) \).

### Step 3: Analyze the third function \( f(x) = \begin{cases} 0 & \text{if } x < \frac{3\pi}{4} \\ 2\sin\left(\frac{x}{3}\right) & \text{if } x \geq \frac{3\pi}{4} \end{cases} \)

**Continuity Check:**
- We need to check the continuity at the point \( x = \frac{3\pi}{4} \).
- Calculate \( f\left(\frac{3\pi}{4}\right) \):
- For \( x < \frac{3\pi}{4} \), \( f(x) = 0 \).
- For \( x \geq \frac{3\pi}{4} \), \( f(x) = 2\sin\left(\frac{3\pi/4}{3}\right) = 2\sin\left(\frac{\pi}{4}\right) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \).
- The left-hand limit as \( x \to \frac{3\pi}{4}^- \) is \( 0 \) and the right-hand limit as \( x \to \frac{3\pi}{4}^+ \) is \( \sqrt{2} \).
- Since \( 0 \neq \sqrt{2} \), the function is discontinuous at \( x = \frac{3\pi}{4} \).

**Conclusion:**
- \( f(x) \) is **not continuous** on \( (0, \pi) \).

### Step 4: Analyze the fourth function \( f(x) = \begin{cases} x \sin x & \text{if } x \in (0, \frac{\pi}{2}) \\ \pi/2 \sin(2\pi + x) & \text{if } x \in (\frac{\pi}{2}, \pi) \end{cases} \)

**Continuity Check:**
- We need to check the continuity at \( x = \frac{\pi}{2} \).
- Calculate \( f\left(\frac{\pi}{2}\right) \):
- For \( x \in (0, \frac{\pi}{2}) \), \( f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \).
- For \( x \in (\frac{\pi}{2}, \pi) \), \( f(x) = \frac{\pi}{2} \sin(2\pi + x) = \frac{\pi}{2} \sin(x) \).
- The left-hand limit as \( x \to \frac{\pi}{2}^- \) is \( \frac{\pi}{2} \) and the right-hand limit as \( x \to \frac{\pi}{2}^+ \) is \( \frac{\pi}{2} \).
- Since both limits are equal and match the function value, \( f(x) \) is continuous at \( x = \frac{\pi}{2} \).

**Conclusion:**
- \( f(x) \) is **continuous** on \( (0, \pi) \).

### Final Summary:
- The functions that are continuous on \( (0, \pi) \) are:
- **Option 2:** \( \int_0^x \sin\left(\frac{1}{t}\right) dt \)
- **Option 4:** \( \begin{cases} x \sin x & \text{if } x \in (0, \frac{\pi}{2}) \\ \frac{\pi}{2} \sin(2\pi + x) & \text{if } x \in (\frac{\pi}{2}, \pi) \end{cases} \)