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 determine the complete range of \( a \) such that the function \( f(x) = \frac{x^2 + 2x + a}{x^2 + 4x + 3a} \) is surjective, we will follow these steps:

### Step 1: Understand Surjectivity
A function is surjective (onto) if every element in the codomain (in this case, all real numbers \( \mathbb{R} \)) is mapped by at least one element from the domain. This means for every \( y \in \mathbb{R} \), there exists an \( x \in D \) such that \( f(x) = y \).

**Hint:** To check for surjectivity, we often need to analyze the behavior of the function and its limits.

### Step 2: Rewrite the Function
We can rewrite the function \( f(x) \) in a more manageable form. We can express it as:
\[ f(x) = 1 - \frac{2x + 2a}{x^2 + 4x + 3a} \]

**Hint:** Simplifying the function helps us analyze its behavior more easily.

### Step 3: Set Up the Condition for Surjectivity
For the function to be surjective, the term \( \frac{2x + 2a}{x^2 + 4x + 3a} \) must be able to take on all real values. This requires that the denominator does not equal zero for any \( x \) in the domain.

**Hint:** The denominator \( x^2 + 4x + 3a \) must not have real roots for the function to be defined for all \( x \).

### Step 4: Analyze the Denominator
The quadratic \( x^2 + 4x + 3a \) will have real roots if its discriminant is non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 3a = 16 - 12a \]
For the quadratic to have no real roots, we need:
\[ D < 0 \implies 16 - 12a < 0 \implies 12a > 16 \implies a > \frac{4}{3} \]

**Hint:** Ensure that the discriminant is negative to avoid real roots.

### Step 5: Analyze the Numerator
Now, we also need to ensure that the numerator \( 2x + 2a \) can take all real values. This is a linear function and will always cover all real numbers as \( x \) varies.

**Hint:** Linear functions are surjective over \( \mathbb{R} \).

### Step 6: Combine Conditions
From our analysis, we found that for \( f(x) \) to be surjective:
1. The condition \( a > \frac{4}{3} \) must hold for the denominator to not have real roots.

**Hint:** This is the only condition we need to ensure surjectivity.

### Conclusion
Thus, the complete range of \( a \) such that the function \( f(x) \) is surjective is:
\[ \boxed{( \frac{4}{3}, \infty )} \]