Statement-1: The equation `e^(x-1) +x-2=0` has only one real root.
Statement-2 : Between any two root of an equation f(x)=0 there is a root of its derivative f'(x)=0

Options

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

Explanation

To analyze the statements given in the question, we will break down the problem step by step.

### Step 1: Analyzing Statement 2
**Statement 2:** "Between any two roots of an equation \( f(x) = 0 \), there is a root of its derivative \( f'(x) = 0 \)."

This statement is a consequence of the Mean Value Theorem. If \( f(x) \) has two distinct roots, say \( \alpha \) and \( \beta \), then by the Mean Value Theorem, there exists at least one point \( c \) in the interval \( (\alpha, \beta) \) such that:
\[ f'(c) = \frac{f(\beta) - f(\alpha)}{\beta - \alpha} = 0 \]
Since \( f(\alpha) = 0 \) and \( f(\beta) = 0 \), it follows that \( f'(c) = 0 \). Thus, Statement 2 is **true**.

### Step 2: Analyzing Statement 1
**Statement 1:** "The equation \( e^{(x-1)} + x - 2 = 0 \) has only one real root."

Let us define the function:
\[ f(x) = e^{(x-1)} + x - 2 \]
We need to determine the number of real roots of this function.

### Step 3: Finding the Derivative
To analyze the behavior of \( f(x) \), we will find its derivative:
\[ f'(x) = e^{(x-1)} + 1 \]
Since \( e^{(x-1)} > 0 \) for all \( x \), it follows that:
\[ f'(x) > 0 \quad \text{for all } x \]
This means that \( f(x) \) is an **increasing function**.

### Step 4: Evaluating the Function at a Point
Now, we will evaluate \( f(x) \) at a specific point to check for roots. Let's evaluate \( f(1) \):
\[ f(1) = e^{(1-1)} + 1 - 2 = 1 + 1 - 2 = 0 \]
Thus, \( x = 1 \) is a root of the equation.

### Step 5: Conclusion on the Number of Roots
Since \( f(x) \) is an increasing function and we have found that \( f(1) = 0 \), it implies that there can be no other roots. Therefore, the equation \( e^{(x-1)} + x - 2 = 0 \) has only one real root, which is \( x = 1 \).

### Final Conclusion
Both statements are true:
- Statement 1 is true because the equation has only one real root.
- Statement 2 is true and correctly explains the behavior of the roots of \( f(x) \).

Thus, the correct answer is that both statements are true, and Statement 2 is the correct explanation of Statement 1.