If x is real and the expression `(x^(2) + 2x - 11)/( x - 3)` takes all values which do not lie between a and b, then find a and b

Answer

To solve the problem, we need to analyze the expression given and find the values of \( a \) and \( b \) such that the expression takes all values except those between \( a \) and \( b \).

### Step-by-Step Solution:

1. **Define the Expression**:
Let \( y = \frac{x^2 + 2x - 11}{x - 3} \).

2. **Rearranging the Expression**:
Multiply both sides by \( x - 3 \) (assuming \( x \neq 3 \)):
\[ y(x - 3) = x^2 + 2x - 11 \]
This simplifies to:
\[ yx - 3y = x^2 + 2x - 11 \]

3. **Rearranging to Form a Quadratic Equation**:
Rearranging gives us:
\[ x^2 + (2 - y)x + (3y - 11) = 0 \]

4. **Identifying Coefficients**:
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = 2 - y \)
- \( c = 3y - 11 \)

5. **Discriminant Condition**:
For \( x \) to be real, the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values:
\[ (2 - y)^2 - 4(1)(3y - 11) \geq 0 \]

6. **Expanding the Discriminant**:
Expanding gives:
\[ (2 - y)^2 - 12y + 44 \geq 0 \]
This simplifies to:
\[ 4 - 4y + y^2 - 12y + 44 \geq 0 \]
Combining like terms:
\[ y^2 - 16y + 48 \geq 0 \]

7. **Factoring the Quadratic**:
To factor \( y^2 - 16y + 48 \):
We need two numbers that multiply to \( 48 \) and add to \( -16 \). The factors are \( -12 \) and \( -4 \):
\[ (y - 12)(y - 4) \geq 0 \]

8. **Finding the Intervals**:
The critical points are \( y = 4 \) and \( y = 12 \). The expression \( (y - 12)(y - 4) \geq 0 \) is satisfied in the intervals:
- \( y \leq 4 \) or \( y \geq 12 \)

9. **Conclusion**:
The expression takes all values except those between \( 4 \) and \( 12 \). Therefore, the values of \( a \) and \( b \) are:
\[ a = 4, \quad b = 12 \]

### Final Answer:
\[ a = 4, \quad b = 12 \]