Number of real roots of the equation ` e^(x-1)-x=0` is

Options

A. 1
B. 2
C. 3
D. 0

Answer

To find the number of real roots of the equation \( e^{x-1} - x = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation in a more manageable form:
\[ e^{x-1} = x \]
This can also be expressed as:
\[ f(x) = e^{x-1} - x = 0 \]

### Step 2: Analyze the functions
We define two functions:
- \( f(x) = e^{x-1} \)
- \( g(x) = x \)

We are interested in finding the points where these two functions intersect.

### Step 3: Find a point of intersection
Let's check if \( x = 1 \) is a solution:
\[ f(1) = e^{1-1} = e^0 = 1 \]
\[ g(1) = 1 \]
Since \( f(1) = g(1) \), we have found that \( x = 1 \) is indeed a point of intersection.

### Step 4: Determine the behavior of the functions
Next, we need to analyze the behavior of \( f(x) \) and \( g(x) \):

1. **Calculate the derivative of \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(e^{x-1}) = e^{x-1} \]
Since \( e^{x-1} > 0 \) for all \( x \), \( f(x) \) is always increasing.

2. **Calculate the derivative of \( g(x) \)**:
\[ g'(x) = 1 \]
The function \( g(x) \) is a straight line with a constant slope of 1.

### Step 5: Compare the slopes at the intersection point
At \( x = 1 \):
- \( f'(1) = e^{1-1} = 1 \)
- \( g'(1) = 1 \)

Both functions have the same slope at \( x = 1 \). Since \( f(x) \) is increasing and touches \( g(x) \) at this point, it indicates that they do not intersect again.

### Step 6: Conclusion
Since \( f(x) \) is increasing and only touches \( g(x) \) at \( x = 1 \), we conclude that the equation \( e^{x-1} - x = 0 \) has exactly one real root.

Thus, the number of real roots of the equation is:
\[ \boxed{1} \]