Statement-1 : The function f defined as `f(x) = a^(x)` satisfies the inequality `f(x_(1)) lt f(x_(2))` for `x_(1) gt x_(2)` when `0 lt a lt 1`.
and
Statement-2 : The function f defined as f(x) `= a^(x)` satisfies the inequality `f(x_(1)) lt f(x_(2))` for `x_(1) lt x_(2)` when `a gt 1`.

Options

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

Answer

To analyze the given statements, we will evaluate each statement step by step.

### Step 1: Analyze Statement 1
The function defined is \( f(x) = a^x \) where \( 0 < a < 1 \). We need to check if the inequality \( f(x_1) < f(x_2) \) holds true when \( x_1 > x_2 \).

1. **Understanding the function**:
- Since \( a \) is between 0 and 1, the function \( f(x) = a^x \) is a decreasing function. This means as \( x \) increases, \( f(x) \) decreases.

2. **Evaluate the inequality**:
- If \( x_1 > x_2 \), then since \( f(x) \) is decreasing, we have:
\[ f(x_1) < f(x_2) \]
- Therefore, the statement \( f(x_1) < f(x_2) \) is true when \( x_1 > x_2 \).

**Conclusion for Statement 1**: The statement is **true**.

### Step 2: Analyze Statement 2
Now, we analyze the second statement where \( a > 1 \). We need to check if the inequality \( f(x_1) < f(x_2) \) holds true when \( x_1 < x_2 \).

1. **Understanding the function**:
- Since \( a \) is greater than 1, the function \( f(x) = a^x \) is an increasing function. This means as \( x \) increases, \( f(x) \) increases.

2. **Evaluate the inequality**:
- If \( x_1 < x_2 \), then since \( f(x) \) is increasing, we have:
\[ f(x_1) < f(x_2) \]
- Therefore, the statement \( f(x_1) < f(x_2) \) is true when \( x_1 < x_2 \).

**Conclusion for Statement 2**: The statement is **true**.

### Final Conclusion
- Statement 1 is **true**.
- Statement 2 is **true**.

### Summary of Results:
- Statement 1: True
- Statement 2: True