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

Options

A. `(-oo,oo)`
B. (1,3)
C. `(3,oo)`
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 compute the set difference A - 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 (it is a square), the expression \((x - 1)^2 + 1\) is always greater than 0 for all real numbers \(x\). Thus:
\[ A = (-\infty, \infty) \]

### 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 where this expression is less than or equal to zero, we can find the roots:
- The roots are \(x = 1\) and \(x = 3\).

Now we can analyze the intervals:
- The expression \((x - 1)(x - 3) \leq 0\) is true in the interval \([1, 3]\).

Thus:
\[ B = [1, 3] \]

### Step 3: Compute A - B
Now we need to find \(A - B\):
\[ A - B = (-\infty, \infty) - [1, 3] \]

This means we remove the interval \([1, 3]\) from the set \(A\):
- The remaining intervals are:
- From \(-\infty\) to \(1\) (not including \(1\))
- From \(3\) to \(\infty\) (not including \(3\))

Thus:
\[ A - B = (-\infty, 1) \cup (3, \infty) \]

### Final Answer
The final answer for \(A - B\) is:
\[ A - B = (-\infty, 1) \cup (3, \infty) \]

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