Let `A = MINIMUM (x^2-2x+7),x in R and B = MINIMUM ( x^2-2x+7),x in [2,oo),` then: `(log)_((B-A))(A+B)` is not defined `A+B=13` `(log)_((2B-A))A<1` (d) `(log)_((2A-B))A >1`

Options

A. `log_((B-A))(A+B)` is not defined
B. `A+B=13`
C. `log_((2B-A))A lt 1`
D. `log_((2A-B))A gt 1`

Answer

To solve the problem, we need to find the minimum values \( A \) and \( B \) from the given quadratic expression and then evaluate the logarithmic expressions.

### Step 1: Find \( A \)
We need to find the minimum value of the function \( f(x) = x^2 - 2x + 7 \) for \( x \in \mathbb{R} \).

1. **Differentiate the function**:
\[ f'(x) = 2x - 2 \]
2. **Set the derivative to zero to find critical points**:
\[ 2x - 2 = 0 \implies x = 1 \]
3. **Evaluate the function at the critical point**:
\[ A = f(1) = 1^2 - 2 \cdot 1 + 7 = 1 - 2 + 7 = 6 \]

### Step 2: Find \( B \)
Now we need to find the minimum value of the same function but for \( x \in [2, \infty) \).

1. **Evaluate the function at the endpoint \( x = 2 \)**:
\[ B = f(2) = 2^2 - 2 \cdot 2 + 7 = 4 - 4 + 7 = 7 \]

### Step 3: Calculate \( A + B \)
Now we can calculate \( A + B \):
\[ A + B = 6 + 7 = 13 \]

### Step 4: Evaluate the logarithmic expressions
1. **Check if \( \log_{(B-A)}(A+B) \) is defined**:
\[ B - A = 7 - 6 = 1 \quad \text{(logarithm base cannot be 1)} \]
Therefore, \( \log_{(B-A)}(A+B) \) is not defined.

2. **Check \( \log_{(2B-A)}(A) < 1 \)**:
\[ 2B - A = 2 \cdot 7 - 6 = 14 - 6 = 8 \]
\[ \log_{8}(6) < 1 \quad \text{(since 8 is greater than 6)} \]

3. **Check \( \log_{(2A-B)}(A) > 1 \)**:
\[ 2A - B = 2 \cdot 6 - 7 = 12 - 7 = 5 \]
\[ \log_{5}(6) > 1 \quad \text{(since 5 is less than 6)} \]

### Summary of Results
- \( A = 6 \)
- \( B = 7 \)
- \( A + B = 13 \)
- \( \log_{(B-A)}(A+B) \) is not defined.
- \( \log_{(2B-A)}(A) < 1 \) is true.
- \( \log_{(2A-B)}(A) > 1 \) is true.