If `A={x:x^(2)-2x+2gt0}andB={x:x^(2)-4x+3le0}`
`AnnB` equals

Options

A. `[1,oo]`
B. [1,3]
C. `(-oo,3]`
D. `(-oo,1)uu(3,oo)`

Explanation

To solve the problem, we need to find the sets \( A \) and \( B \) as defined in the question and then determine the intersection \( A \cap B \).

### Step 1: Define Set A
Set \( A \) is defined as:
\[ A = \{ x : x^2 - 2x + 2 > 0 \} \]
To analyze this inequality, we can complete the square:
\[ x^2 - 2x + 2 = (x - 1)^2 + 1 \]
Since \((x - 1)^2\) is always non-negative for all real \( x \), \((x - 1)^2 + 1\) is always greater than 0. Therefore, set \( A \) includes all real numbers:
\[ A = \mathbb{R} \]

### Step 2: Define Set B
Set \( B \) is defined as:
\[ B = \{ x : x^2 - 4x + 3 \leq 0 \} \]
We can factor the quadratic:
\[ x^2 - 4x + 3 = (x - 1)(x - 3) \]
To find the values of \( x \) for which this expression is less than or equal to 0, we find the roots:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
Next, we test the intervals determined by these roots:
- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 3) = 3 > 0 \)
- For \( 1 \leq x \leq 3 \): Choose \( x = 2 \) → \( (2 - 1)(2 - 3) = -1 \leq 0 \)
- For \( x > 3 \): Choose \( x = 4 \) → \( (4 - 1)(4 - 3) = 3 > 0 \)

Thus, the set \( B \) is:
\[ B = [1, 3] \]

### Step 3: Find the Intersection \( A \cap B \)
Since \( A = \mathbb{R} \) and \( B = [1, 3] \), the intersection is simply:
\[ A \cap B = B = [1, 3] \]

### Final Answer
Thus, \( A \cap B \) is:
\[ A \cap B = [1, 3] \]