If `(x^(2)+ax+3)/(x^(2)+x+a)`, takes all real values for possible real values of x, then

Options

A. `a^(3)-9a+12 le 0`
B. `4a^(3)+39 lt 0`
C. `a^3 -9a lt 1//4`
D. `a ge 1//4`

Answer

To solve the problem where the expression \(\frac{x^2 + ax + 3}{x^2 + x + a}\) takes all real values for possible real values of \(x\), we can follow these steps:

### Step 1: Set up the equation
Let \(y = \frac{x^2 + ax + 3}{x^2 + x + a}\).

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ y(x^2 + x + a) = x^2 + ax + 3 \]
This simplifies to:
\[ yx^2 + yx + ya = x^2 + ax + 3 \]

### Step 3: Rearrange the equation
Rearranging the equation leads to:
\[ yx^2 - x^2 + yx - ax + ya - 3 = 0 \]
This can be rewritten as:
\[ (y - 1)x^2 + (y - a)x + (ya - 3) = 0 \]

### Step 4: Identify coefficients
From the quadratic equation \(Ax^2 + Bx + C = 0\), we have:
- \(A = y - 1\)
- \(B = y - a\)
- \(C = ya - 3\)

### Step 5: Condition for real values
For the quadratic equation to have real solutions for all values of \(x\), the discriminant must be non-negative:
\[ D = B^2 - 4AC \geq 0 \]
Substituting the coefficients:
\[ (y - a)^2 - 4(y - 1)(ya - 3) \geq 0 \]

### Step 6: Expand the discriminant
Expanding the discriminant:
\[ (y - a)^2 - 4[(y - 1)(ya - 3)] \]
This expands to:
\[ (y - a)^2 - 4(y^2a - 3y - ya + 3) \geq 0 \]

### Step 7: Simplify the discriminant
Simplifying further, we get:
\[ (y - a)^2 - 4y^2a + 12y + 4a - 12 \geq 0 \]

### Step 8: Analyze the quadratic in \(y\)
This is a quadratic in \(y\). For it to be non-negative for all \(y\), the coefficient of \(y^2\) must be non-positive, and the discriminant must be non-positive.

### Step 9: Coefficients conditions
1. Coefficient of \(y^2\) must be \(1 - 4a \geq 0\) which gives \(a \leq \frac{1}{4}\).
2. The discriminant of the quadratic must also be less than or equal to zero.

### Step 10: Solve the cubic inequality
The cubic inequality \(a^3 - 9a + 12 \leq 0\) must be solved.

### Conclusion
The conditions derived from the analysis are:
1. \(a \leq \frac{1}{4}\)
2. \(a^3 - 9a + 12 \leq 0\)