The equation `x log x = 3-x` has, in the interval (1,3) :

Options

A. exactly one root
B. at most one root
C. at least one root
D. no root

Answer

To solve the equation \( x \log x = 3 - x \) in the interval (1, 3), we can follow these steps:

### Step 1: Define the Functions
Let:
- \( y_1 = x \log x \)
- \( y_2 = 3 - x \)

### Step 2: Analyze the Functions
We need to analyze the behavior of both functions in the interval (1, 3).

### Step 3: Evaluate \( y_1 \) and \( y_2 \) at the Endpoints
1. **At \( x = 1 \)**:
- \( y_1(1) = 1 \log 1 = 1 \cdot 0 = 0 \)
- \( y_2(1) = 3 - 1 = 2 \)

2. **At \( x = 3 \)**:
- \( y_1(3) = 3 \log 3 \) (approximately \( 3 \cdot 1.0986 \approx 3.2958 \))
- \( y_2(3) = 3 - 3 = 0 \)

### Step 4: Determine the Behavior of \( y_1 \) and \( y_2 \)
- The function \( y_1 = x \log x \) is continuous and increasing for \( x > 0 \).
- The function \( y_2 = 3 - x \) is a linear function with a negative slope, decreasing from 2 to 0 as \( x \) goes from 1 to 3.

### Step 5: Find Intersections
Since \( y_1(1) = 0 \) and \( y_2(1) = 2 \), and \( y_1(3) \approx 3.2958 \) while \( y_2(3) = 0 \), we can conclude that:
- At \( x = 1 \), \( y_1 < y_2 \)
- At \( x = 3 \), \( y_1 > y_2 \)

### Step 6: Apply the Intermediate Value Theorem
Since \( y_1 \) is continuous and increases from 0 to approximately 3.2958, while \( y_2 \) decreases from 2 to 0, there must be at least one point in the interval (1, 3) where \( y_1 = y_2 \).

### Conclusion
Thus, the equation \( x \log x = 3 - x \) has exactly one root in the interval (1, 3).

### Final Answer
The answer is **Option 1: Exactly one root**.

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