If f(x) is continuous in [0,2] and f(0)=f(2). Then the equation f(x)=f(x+1) has

Options

A. no real root in [0,2]
B. at least one real root in [0,1]
C. at least one real root in [0,2]
D. at least one real root in [1,2]

Explanation

To solve the problem, we need to analyze the function \( g(x) = f(x) - f(x+1) \) and determine the roots of this equation, specifically where \( g(x) = 0 \).

### Step 1: Define the function
We define the function:
\[ g(x) = f(x) - f(x+1) \]

### Step 2: Evaluate \( g(0) \) and \( g(1) \)
Next, we evaluate \( g(0) \) and \( g(1) \):
\[ g(0) = f(0) - f(1) \]
\[ g(1) = f(1) - f(2) \]

Since \( f(0) = f(2) \) (given in the problem), we can substitute \( f(2) \) with \( f(0) \):
\[ g(1) = f(1) - f(0) \]

### Step 3: Analyze the signs of \( g(0) \) and \( g(1) \)
Now we have:
- \( g(0) = f(0) - f(1) \)
- \( g(1) = f(1) - f(0) \)

Notice that:
- If \( f(0) > f(1) \), then \( g(0) > 0 \) and \( g(1) < 0 \).
- If \( f(0) < f(1) \), then \( g(0) < 0 \) and \( g(1) > 0 \).

In both cases, \( g(0) \) and \( g(1) \) have opposite signs.

### Step 4: Apply the Intermediate Value Theorem
Since \( g(x) \) is continuous on the interval \([0, 1]\) (because \( f(x) \) is continuous), and \( g(0) \) and \( g(1) \) have opposite signs, by the Intermediate Value Theorem, there exists at least one root \( c \) in the interval \( (0, 1) \) such that:
\[ g(c) = 0 \]

### Step 5: Conclusion about the roots
Since we have found at least one root in the interval \( (0, 1) \), we can conclude that:
- There is at least one root of the equation \( f(x) = f(x+1) \) in the interval \( (0, 1) \).
- Consequently, there is also at least one root in the larger interval \( (0, 2) \).

### Final Answer
Thus, the equation \( f(x) = f(x+1) \) has at least one root in the interval \( (0, 1) \) and also in \( (0, 2) \).

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