If x is real, the expression ` (x+2)/(2x^2 + 3x+6)` takes all value in the interval

Options

A. `(1/13,1/3)`
B. `[-1/13,1/3]`
C. `(-1/3,1/13)`
D. none of these

Answer

To solve the problem, we need to find the interval of values that the expression \( \frac{x+2}{2x^2 + 3x + 6} \) can take when \( x \) is a real number.

### Step-by-Step Solution:

1. **Define the Expression**:
Let \( y = \frac{x + 2}{2x^2 + 3x + 6} \).

2. **Rearranging the Equation**:
Rearranging gives:
\[ y(2x^2 + 3x + 6) = x + 2 \]
This can be rewritten as:
\[ 2yx^2 + (3y - 1)x + (6y - 2) = 0 \]
Here, we have a quadratic equation in \( x \).

3. **Discriminant Condition**:
For \( x \) to be real, the discriminant \( D \) of this quadratic must be non-negative:
\[ D = (3y - 1)^2 - 4(2y)(6y - 2) \geq 0 \]

4. **Calculate the Discriminant**:
Expanding the discriminant:
\[ D = (3y - 1)^2 - 8y(6y - 2) \]
\[ = 9y^2 - 6y + 1 - 48y^2 + 16y \]
\[ = -39y^2 + 10y + 1 \]

5. **Set the Discriminant Greater than or Equal to Zero**:
We need:
\[ -39y^2 + 10y + 1 \geq 0 \]
or equivalently,
\[ 39y^2 - 10y - 1 \leq 0 \]

6. **Finding Roots of the Quadratic**:
Use the quadratic formula to find the roots:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 39 \cdot (-1)}}{2 \cdot 39} \]
\[ = \frac{10 \pm \sqrt{100 + 156}}{78} = \frac{10 \pm \sqrt{256}}{78} = \frac{10 \pm 16}{78} \]
The roots are:
\[ y_1 = \frac{26}{78} = \frac{1}{3}, \quad y_2 = \frac{-6}{78} = -\frac{1}{13} \]

7. **Determine the Interval**:
The quadratic \( 39y^2 - 10y - 1 \) opens upwards (as the coefficient of \( y^2 \) is positive). Therefore, it is less than or equal to zero between its roots:
\[ -\frac{1}{13} \leq y \leq \frac{1}{3} \]

8. **Conclusion**:
Thus, the expression \( \frac{x + 2}{2x^2 + 3x + 6} \) takes all values in the interval:
\[ \left[-\frac{1}{13}, \frac{1}{3}\right] \]