Statement -1: `sin 1 lt sin 2 lt sin3`
Statement-2: `sin3 lt sin2 lt sin1`
Statement-3: `sinx_(1) lt sinx_(2), x_(1) , x_(2) in (0,pi/2)`

Options

A. TFT
B. FTT
C. FTF
D. TFF

Answer

To solve the problem, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement 1:** \( \sin(1) < \sin(2) < \sin(3) \)

- The sine function is increasing in the interval \( (0, \frac{\pi}{2}) \).
- Since \( 1, 2, \) and \( 3 \) are all in radians and fall within this interval, we can conclude that:
\[ \sin(1) < \sin(2) < \sin(3) \]
- Therefore, **Statement 1 is true**.

### Step 2: Analyze Statement 2
**Statement 2:** \( \sin(3) < \sin(2) < \sin(1) \)

- From our analysis of Statement 1, we know that \( \sin(1) < \sin(2) < \sin(3) \).
- This means that \( \sin(3) \) is actually greater than both \( \sin(2) \) and \( \sin(1) \).
- Therefore, **Statement 2 is false**.

### Step 3: Analyze Statement 3
**Statement 3:** \( \sin(x_1) < \sin(x_2) \) for \( x_1, x_2 \in (0, \frac{\pi}{2}) \)

- The sine function is strictly increasing in the interval \( (0, \frac{\pi}{2}) \).
- If \( x_1 < x_2 \), then it follows that \( \sin(x_1) < \sin(x_2) \).
- However, the statement does not specify that \( x_1 < x_2 \). If \( x_1 \) is greater than \( x_2 \), then \( \sin(x_1) \) could be greater than \( \sin(x_2) \).
- Thus, **Statement 3 is false** because it is not universally true for all \( x_1, x_2 \) in the interval.

### Conclusion
- Statement 1 is true.
- Statement 2 is false.
- Statement 3 is false.

The overall pattern is true, false, false.

### Final Answer
The correct option is option number 4.

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