If 'x' is real, then greatest value of the expression, `(x+2)/(2x^(2)+3x +6)` is :

Options

A. `1//13`
B. `-1//13`
C. `3`
D. `1//3`

Explanation

To find the greatest value of the expression \( y = \frac{x+2}{2x^2 + 3x + 6} \), we will follow these steps:

### Step 1: Set up the expression
We start with the expression:
\[ y = \frac{x + 2}{2x^2 + 3x + 6} \]

### Step 2: Rearrange the equation
We can rearrange the equation to eliminate the fraction:
\[ y(2x^2 + 3x + 6) = x + 2 \]
This simplifies to:
\[ 2xy^2 + 3xy + 6y - x - 2 = 0 \]

### Step 3: Rearrange into standard quadratic form
Rearranging gives us a quadratic in terms of \( x \):
\[ 2xy^2 + (3y - 1)x + (6y - 2) = 0 \]

### Step 4: Apply the discriminant condition
For \( x \) to have real solutions, the discriminant \( D \) of this quadratic must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = 2y \), \( b = 3y - 1 \), and \( c = 6y - 2 \). Thus, we have:
\[ (3y - 1)^2 - 4(2y)(6y - 2) \geq 0 \]

### Step 5: Expand and simplify the discriminant
Expanding the discriminant:
\[ (3y - 1)^2 = 9y^2 - 6y + 1 \]
And for the second term:
\[ 4(2y)(6y - 2) = 48y^2 - 16y \]
Combining these gives:
\[ 9y^2 - 6y + 1 - (48y^2 - 16y) \geq 0 \]
This simplifies to:
\[ -39y^2 + 10y + 1 \geq 0 \]

### Step 6: Rearranging the inequality
Rearranging gives:
\[ 39y^2 - 10y - 1 \leq 0 \]

### Step 7: Find the roots of the quadratic
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 39, b = -10, c = -1 \):
\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 39 \cdot (-1)}}{2 \cdot 39} \]
Calculating the discriminant:
\[ 100 + 156 = 256 \]
Thus:
\[ y = \frac{10 \pm 16}{78} \]
This gives us two roots:
\[ y_1 = \frac{26}{78} = \frac{1}{3}, \quad y_2 = \frac{-6}{78} = -\frac{1}{13} \]

### Step 8: Determine the intervals
The quadratic \( 39y^2 - 10y - 1 \) opens upwards, so it is less than or equal to zero between the roots:
\[ -\frac{1}{13} \leq y \leq \frac{1}{3} \]

### Step 9: Identify the greatest value
The greatest value of \( y \) occurs at the upper bound:
\[ \text{Greatest value of } y = \frac{1}{3} \]

### Conclusion
Thus, the greatest value of the expression \( \frac{x + 2}{2x^2 + 3x + 6} \) is:
\[ \boxed{\frac{1}{3}} \]