If x `in` R then `(x^(2)+2x+a)/(x^(2)+4x+3a)` can take all real values if

Options

A. `(1,2)`
B. `[0,1]`
C. `(0,1)`
D. `(-1,0)`

Answer

To solve the problem, we need to determine the conditions under which the expression \( \frac{x^2 + 2x + a}{x^2 + 4x + 3a} \) can take all real values for \( x \in \mathbb{R} \).

### Step 1: Set the expression equal to a constant \( k \)

We start by setting the expression equal to a constant \( k \):

\[ \frac{x^2 + 2x + a}{x^2 + 4x + 3a} = k \]

### Step 2: Rearrange the equation

Cross-multiplying gives us:

\[ x^2 + 2x + a = k(x^2 + 4x + 3a) \]

Rearranging this leads to:

\[ x^2 - kx^2 + 2x - 4kx + a - 3ka = 0 \]

This simplifies to:

\[ (1 - k)x^2 + (2 - 4k)x + (a - 3ka) = 0 \]

### Step 3: Identify coefficients

From the quadratic equation \( Ax^2 + Bx + C = 0 \), we identify:

- \( A = 1 - k \)
- \( B = 2 - 4k \)
- \( C = a - 3ka \)

### Step 4: Condition for real values

For the quadratic equation to have real solutions, the discriminant must be non-negative:

\[ D = B^2 - 4AC \geq 0 \]

Substituting our coefficients into the discriminant:

\[ D = (2 - 4k)^2 - 4(1 - k)(a - 3ka) \geq 0 \]

### Step 5: Expand the discriminant

Expanding \( D \):

\[ D = (2 - 4k)^2 - 4(1 - k)(a - 3ka) \]

Calculating \( (2 - 4k)^2 \):

\[ = 4 - 16k + 16k^2 \]

Calculating \( -4(1 - k)(a - 3ka) \):

\[ = -4(a - 3ka) + 4k(a - 3ka) = -4a + 12ka + 4ka - 12k^2a \]

Combining these gives:

\[ D = 4 - 16k + 16k^2 - 4a + 16ka - 12k^2a \]

### Step 6: Set the discriminant greater than or equal to zero

Now, we want:

\[ 4 + 16k^2 - 16k - 4a + 16ka - 12k^2a \geq 0 \]

### Step 7: Analyze the quadratic in \( k \)

This is a quadratic in \( k \). For this quadratic to be non-negative for all \( k \), the discriminant of this quadratic must be less than or equal to zero.

### Step 8: Calculate the discriminant of the quadratic in \( k \)

The discriminant \( D_k \) of the quadratic \( 16k^2 + (-16 + 16a)k + (4 - 4a) \):

\[ D_k = (-16 + 16a)^2 - 4 \cdot 16 \cdot (4 - 4a) \]

Setting \( D_k \leq 0 \) gives us the conditions for \( a \).

### Step 9: Solve for \( a \)

After simplifying the discriminant condition, we find:

\[ 4 - 3a > 0 \implies a < \frac{4}{3} \]

### Step 10: Determine the range of \( a \)

To ensure the quadratic can take all real values, we also need \( a \geq 0 \).

Thus, the final range for \( a \) is:

\[ 0 \leq a < \frac{4}{3} \]

### Final Answer

The values of \( a \) for which the expression can take all real values are:

\[ a \in [0, \frac{4}{3}) \]