Let `f : D toR` bge defined as `:f (x) = (x^(2) +2x+a)/(x ^(2)+ 4x+3a)` where D and R denote the domain of f and the set of all the real numbers respectively. If f is surjective mapping. Then the complete range of a is :

Options

A. `0 lt a le 1`
B. `0 lt a le 1`
C. `0le a lt 1`
D. `0 lt a lt 1`

Answer

To solve the problem, we need to determine the complete range of \( a \) such that the function \( f(x) = \frac{x^2 + 2x + a}{x^2 + 4x + 3a} \) is a surjective mapping from its domain \( D \) to the set of real numbers \( \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding Surjectivity**:
A function \( f: D \to \mathbb{R} \) is surjective if for every \( y \in \mathbb{R} \), there exists at least one \( x \in D \) such that \( f(x) = y \). This means the function must cover all real values.

2. **Rewriting the Function**:
We can rewrite the function as:
\[ f(x) = \frac{x^2 + 2x + a}{x^2 + 4x + 3a} \]
We need to analyze the behavior of this function to determine the conditions under which it is surjective.

3. **Identifying the Denominator**:
The denominator \( x^2 + 4x + 3a \) must not equal zero for any \( x \) in the domain \( D \). Therefore, we need to ensure that this quadratic does not have any real roots.

4. **Finding Roots of the Denominator**:
The roots of the quadratic \( x^2 + 4x + 3a \) can be found using the discriminant:
\[ \Delta = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 3a = 16 - 12a \]
For the quadratic to have no real roots, we require:
\[ 16 - 12a < 0 \implies 12a > 16 \implies a > \frac{4}{3} \]

5. **Analyzing the Numerator**:
Next, we need to ensure that the function can take all real values. This can be checked by analyzing the behavior of the function as \( x \) approaches the roots of the denominator.

6. **Behavior of the Function**:
We can express \( f(x) \) in terms of another variable \( g(x) \):
\[ g(x) = \frac{x + 2 + \frac{a - 2}{x^2 + 4x + 3a}}{1} \]
For \( g(x) \) to be surjective, we need to ensure that the limits of \( f(x) \) as \( x \to \pm \infty \) cover all real numbers.

7. **Finding Conditions on \( a \)**:
We can analyze the limits:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2 + 2x + a}{x^2 + 4x + 3a} = \frac{1}{1} = 1 \]
\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x^2 + 2x + a}{x^2 + 4x + 3a} = \frac{1}{1} = 1 \]
This indicates that the function approaches 1 as \( x \) goes to both positive and negative infinity.

8. **Conclusion on the Range of \( a \)**:
For \( f(x) \) to be surjective, we need to ensure that the function can take values less than and greater than 1. This can be achieved by ensuring that the quadratic in the denominator does not allow \( f(x) \) to stabilize at a single value.

9. **Final Range of \( a \)**:
From the analysis, we find that:
\[ a \in (0, 1) \]
This is the complete range of \( a \) for which the function \( f \) is surjective.