If `f:R rarrA` defined as `f(x)=tan^(-1)(sqrt(4(x^(2)+x+1)))` is surjective, then A is equal to

Options

A. `(-(pi)/(2), (pi)/(2))`
B. `[0, (pi)/(2))`
C. `[(pi)/(3), (pi)/(2))`
D. `(0, (pi)/(3)]`

Answer

To determine the set \( A \) for which the function \( f: \mathbb{R} \to A \) defined by

\[ f(x) = \tan^{-1}(\sqrt{4(x^2 + x + 1)}) \]

is surjective, we need to find the range of the function \( f(x) \).

### Step-by-Step Solution

1. **Identify the expression inside the function**:
We start with the expression under the square root:
\[ \sqrt{4(x^2 + x + 1)} \]

2. **Simplify the expression**:
The expression \( x^2 + x + 1 \) is a quadratic function. We can analyze its minimum value to understand the behavior of \( f(x) \).

3. **Find the minimum value of \( x^2 + x + 1 \)**:
The vertex of the quadratic \( x^2 + x + 1 \) can be found using the formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{1}{2(1)} = -\frac{1}{2} \]
Now, substituting \( x = -\frac{1}{2} \) into the quadratic:
\[ y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

4. **Calculate the minimum value of \( \sqrt{4(x^2 + x + 1)} \)**:
The minimum value of \( x^2 + x + 1 \) is \( \frac{3}{4} \), thus:
\[ \sqrt{4\left(\frac{3}{4}\right)} = \sqrt{3} \]

5. **Determine the range of \( f(x) \)**:
Since \( \sqrt{4(x^2 + x + 1)} \) can take values from \( \sqrt{3} \) to \( \infty \), we can find the range of \( f(x) \):
\[ f(x) = \tan^{-1}(\sqrt{4(x^2 + x + 1)}) \]
The minimum value of \( f(x) \) occurs when \( \sqrt{4(x^2 + x + 1)} = \sqrt{3} \):
\[ f\left(-\frac{1}{2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
As \( x \) approaches \( \infty \), \( f(x) \) approaches:
\[ \tan^{-1}(\infty) = \frac{\pi}{2} \]

6. **Conclude the range of \( f(x) \)**:
Therefore, the range of \( f(x) \) is:
\[ \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \]

7. **Identify the codomain \( A \)**:
For the function \( f(x) \) to be surjective, the codomain \( A \) must equal the range of \( f(x) \):
\[ A = \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \]

### Final Answer
Thus, the set \( A \) is equal to:
\[ A = \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \]