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

Options

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

Explanation

To solve the problem, we need to find the sets A and B based on the given conditions and then determine their union.

### Step 1: Define Set A
Set A is defined as:
\[ A = \{ x : x^2 - 2x + 2 > 0 \} \]

To analyze this inequality, we can rewrite the quadratic expression:
\[ x^2 - 2x + 2 = (x - 1)^2 + 1 \]

### Step 2: Analyze Set A
The expression \((x - 1)^2 + 1\) is always greater than 0 for all real numbers \(x\) because the square of a real number is non-negative and we are adding 1 to it. Therefore:
\[ A = (-\infty, \infty) \]

### Step 3: Define Set B
Set B is defined as:
\[ B = \{ x : x^2 - 4x + 3 \leq 0 \} \]

We can factor the quadratic expression:
\[ x^2 - 4x + 3 = (x - 1)(x - 3) \]

### Step 4: Analyze Set B
To find the values of \(x\) for which this expression is less than or equal to 0, we set up the inequality:
\[ (x - 1)(x - 3) \leq 0 \]

The critical points are \(x = 1\) and \(x = 3\). We can test intervals around these points to determine where the product is non-positive.

- For \(x < 1\), both factors are negative, so the product is positive.
- For \(1 \leq x \leq 3\), the product is non-positive (0 at the endpoints).
- For \(x > 3\), both factors are positive, so the product is positive.

Thus, the solution for Set B is:
\[ B = [1, 3] \]

### Step 5: Find the Union of Sets A and B
Now we need to find the union of sets A and B:
\[ A \cup B = (-\infty, \infty) \cup [1, 3] \]

Since \(A\) includes all real numbers, the union will also include all real numbers:
\[ A \cup B = (-\infty, \infty) \]

### Final Answer
Thus, the final answer is:
\[ A \cup B = (-\infty, \infty) \]

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