If `x` is a real, then the maximum value `(x^2+14x+9)/(x^2+2x+3)` `(i)2 (ii)4 (iii)6 (iv) 8`

Options

A. 3, 1
B. -5,4
C. `0, -oo`
D. `oo, -oo`

Explanation

To find the maximum value of the expression \(\frac{x^2 + 14x + 9}{x^2 + 2x + 3}\), we can follow these steps:

### Step 1: Define the Expression
Let \(y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}\).

### Step 2: Cross Multiply
Cross-multiplying gives us:
\[ y(x^2 + 2x + 3) = x^2 + 14x + 9 \]
This simplifies to:
\[ yx^2 + 2yx + 3y = x^2 + 14x + 9 \]

### Step 3: Rearrange the Equation
Rearranging the equation leads to:
\[ (y - 1)x^2 + (2y - 14)x + (3y - 9) = 0 \]

### Step 4: Apply the Discriminant Condition
Since \(x\) is real, the discriminant of this quadratic equation must be greater than or equal to zero. The discriminant \(D\) is given by:
\[ D = (2y - 14)^2 - 4(y - 1)(3y - 9) \geq 0 \]

### Step 5: Simplify the Discriminant
Calculating the discriminant:
\[ D = (2y - 14)^2 - 4(y - 1)(3y - 9) \]
Expanding this gives:
\[ D = 4y^2 - 56y + 196 - 4[(3y^2 - 9y) - (3y - 9)] \]
\[ = 4y^2 - 56y + 196 - (12y^2 - 36y + 36) \]
\[ = 4y^2 - 56y + 196 - 12y^2 + 36y - 36 \]
\[ = -8y^2 - 20y + 160 \]

### Step 6: Set the Discriminant Greater Than or Equal to Zero
Now we need to solve:
\[ -8y^2 - 20y + 160 \geq 0 \]
Dividing the entire inequality by -1 (and flipping the inequality):
\[ 8y^2 + 20y - 160 \leq 0 \]

### Step 7: Factor the Quadratic
To factor \(8y^2 + 20y - 160\), we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 8 \cdot (-160)}}{2 \cdot 8} \]
Calculating the discriminant:
\[ = \frac{-20 \pm \sqrt{400 + 5120}}{16} = \frac{-20 \pm \sqrt{5520}}{16} \]
Finding the roots gives us the critical points.

### Step 8: Find the Range of \(y\)
After solving, we find the roots and determine the intervals where the quadratic is less than or equal to zero. This leads us to:
\[ -5 \leq y \leq 4 \]

### Step 9: Determine the Maximum Value
The maximum value of \(y\) occurs at the upper bound:
\[ \text{Maximum value of } y = 4 \]

### Conclusion
Thus, the maximum value of \(\frac{x^2 + 14x + 9}{x^2 + 2x + 3}\) is \(\boxed{4}\).

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