Statement-1`f(x)=(sin x)/(x) lt 1 for 0 lt x lt pi/2`
Statement -2 `f(x)=(sin x)/(x)` is decreasing function on `(0,pi//2)`

Options

A. Statement-1 True statement -1 is True,Statement -2 is True statement -2 is a correct explanation for Statement-4
B. Statement-1 True statement -1 is True,Statement -2 is True statement -2 is not a correct explanation for Statement-4
C. Statement-1 True statement -1 is True,Statement -2 is False
D. Statement-1 is False ,Statement -2 is True

Explanation

To solve the problem, we need to analyze both statements regarding the function \( f(x) = \frac{\sin x}{x} \) for \( 0 < x < \frac{\pi}{2} \).

### Step 1: Analyze Statement 1
**Statement 1:** \( f(x) = \frac{\sin x}{x} < 1 \) for \( 0 < x < \frac{\pi}{2} \).

To prove this statement, we can evaluate the behavior of \( \sin x \) and \( x \) in the interval \( (0, \frac{\pi}{2}) \).

1. **Behavior of \( \sin x \)**: The function \( \sin x \) is increasing in the interval \( (0, \frac{\pi}{2}) \) and takes values from \( 0 \) to \( 1 \).
2. **Behavior of \( x \)**: The function \( x \) is also increasing in the interval \( (0, \frac{\pi}{2}) \) and takes values from \( 0 \) to \( \frac{\pi}{2} \).

Since \( \sin x < x \) for \( 0 < x < \frac{\pi}{2} \), we can conclude:

\[ \frac{\sin x}{x} < 1 \quad \text{for } 0 < x < \frac{\pi}{2}. \]

Thus, Statement 1 is true.

### Step 2: Analyze Statement 2
**Statement 2:** \( f(x) = \frac{\sin x}{x} \) is a decreasing function on \( (0, \frac{\pi}{2}) \).

To determine if \( f(x) \) is decreasing, we can find the derivative \( f'(x) \):

1. **Differentiate \( f(x) \)** using the quotient rule:
\[ f'(x) = \frac{x \cos x - \sin x}{x^2}. \]

2. **Analyze the numerator**: We need to check the sign of \( x \cos x - \sin x \) in the interval \( (0, \frac{\pi}{2}) \).
- At \( x = 0 \), \( f'(0) \) is not defined, but we can analyze the limit.
- As \( x \) approaches \( 0 \), \( \sin x \) behaves like \( x \), and \( \cos x \) approaches \( 1 \). Thus, \( \sin x < x \) implies \( x \cos x - \sin x < 0 \).
- For \( 0 < x < \frac{\pi}{2} \), \( \cos x \) is positive, and since \( \sin x < x \), we have \( x \cos x < \sin x \).

Thus, \( x \cos x - \sin x < 0 \) implies \( f'(x) < 0 \), indicating that \( f(x) \) is decreasing on \( (0, \frac{\pi}{2}) \).

### Conclusion
Both statements are true:
- Statement 1 is true: \( f(x) < 1 \) for \( 0 < x < \frac{\pi}{2} \).
- Statement 2 is true: \( f(x) \) is a decreasing function on \( (0, \frac{\pi}{2}) \).

### Final Answer
Both statements are true, and Statement 2 provides the correct explanation for Statement 1.