If x is real, the maximum value of `(3x^(2)+9x+17)/(3x^(2)+9x+7)` is :

Options

A. 41
B. 1
C. `17//7`
D. `1//4`

Answer

To find the maximum value of the expression

\[ y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}, \]

we will follow these steps:

### Step 1: Set up the equation
We start by rewriting the equation as follows:

\[ y(3x^2 + 9x + 7) = 3x^2 + 9x + 17. \]

### Step 2: Rearranging the equation
Rearranging gives us:

\[ 3yx^2 + 9yx + 7y - 3x^2 - 9x - 17 = 0. \]

### Step 3: Combine like terms
Now, we can combine the terms:

\[ (3y - 3)x^2 + (9y - 9)x + (7y - 17) = 0. \]

### Step 4: Factor out common terms
Factoring out common terms gives us:

\[ 3(y - 1)x^2 + 9(y - 1)x + (7y - 17) = 0. \]

### Step 5: Identify coefficients for the quadratic equation
From this, we can identify \(a = 3(y - 1)\), \(b = 9(y - 1)\), and \(c = 7y - 17\).

### Step 6: Use the discriminant condition
For \(x\) to have real values, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0. \]

Substituting in our values:

\[ (9(y - 1))^2 - 4(3(y - 1))(7y - 17) \geq 0. \]

### Step 7: Simplify the discriminant
Calculating the discriminant:

\[ 81(y - 1)^2 - 12(y - 1)(7y - 17) \geq 0. \]

### Step 8: Factor the discriminant
Factoring out \((y - 1)\):

\[ (y - 1)(81(y - 1) - 12(7y - 17)) \geq 0. \]

### Step 9: Expand and simplify
Expanding gives:

\[ 81(y - 1) - 84y + 204 \geq 0. \]

This simplifies to:

\[ -3y + 123 \geq 0 \quad \Rightarrow \quad 3y \leq 123 \quad \Rightarrow \quad y \leq 41. \]

### Step 10: Determine the interval for \(y\)
Since \(y - 1 \geq 0\) implies \(y \geq 1\), we have:

\[ 1 \leq y \leq 41. \]

### Step 11: Conclusion
The maximum value of \(y\) is therefore:

\[ \boxed{41}. \]