Let `a in (0,(pi)/(2))` and `f(x)=sqrt(x^(2)+x)+(tan^(2)alpha)/(sqrt(x^(2)+x)), x gt 0`. If the least value of `f(x)` is `2sqrt3`, then `alpha` is equal to

Options

A. `(pi)/(3)`
B. `(pi)/(8)`
C. `(pi)/(6)`
D. `(pi)/(4)`

Answer

To solve the problem, we need to find the value of \(\alpha\) such that the least value of the function \(f(x) = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}\) is \(2\sqrt{3}\).

### Step-by-step Solution:

1. **Define the function**:
\[ f(x) = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} \]

2. **Apply the AM-GM inequality**:
According to the Arithmetic Mean-Geometric Mean inequality (AM-GM), we can write:
\[ A + B \geq 2\sqrt{AB} \]
where \(A = \sqrt{x^2 + x}\) and \(B = \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}\).

3. **Set up the inequality**:
\[ \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} \geq 2\sqrt{\sqrt{x^2 + x} \cdot \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}} \]
Simplifying the right-hand side:
\[ 2\sqrt{\tan^2 \alpha} = 2\tan \alpha \]

4. **Establish the relationship**:
Thus, we have:
\[ f(x) \geq 2\tan \alpha \]

5. **Given condition**:
We know from the problem that the least value of \(f(x)\) is \(2\sqrt{3}\). Therefore, we can set up the equation:
\[ 2\tan \alpha = 2\sqrt{3} \]

6. **Simplify the equation**:
Dividing both sides by 2:
\[ \tan \alpha = \sqrt{3} \]

7. **Find the angle**:
We know that:
\[ \tan \frac{\pi}{3} = \sqrt{3} \]
Therefore, we conclude:
\[ \alpha = \frac{\pi}{3} \]

### Final Answer:
\[ \alpha = \frac{\pi}{3} \]