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 step-by-step, we start with the function given:

\[ f(x) = \sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}} \]

where \( x > 0 \) and we need to find the value of \( \alpha \) such that the least value of \( f(x) \) is \( 2\sqrt{3} \).

### Step 1: Analyze the Function
We can rewrite the function in a more manageable form. Let:

\[ y = \sqrt{x^2 + x} \]

Then, we have:

\[ f(x) = y + \frac{\tan^2 \alpha}{y} \]

### Step 2: Apply the AM-GM Inequality
According to the Arithmetic Mean-Geometric Mean (AM-GM) inequality:

\[ \frac{y + \frac{\tan^2 \alpha}{y}}{2} \geq \sqrt{y \cdot \frac{\tan^2 \alpha}{y}} \]

This simplifies to:

\[ \frac{y + \frac{\tan^2 \alpha}{y}}{2} \geq \tan \alpha \]

Thus, we have:

\[ y + \frac{\tan^2 \alpha}{y} \geq 2 \tan \alpha \]

### Step 3: Substitute Back
Since \( f(x) \geq 2 \tan \alpha \), we can conclude that the least value of \( f(x) \) is \( 2 \tan \alpha \).

### Step 4: Set the Least Value
We know from the problem statement that the least value of \( f(x) \) is \( 2\sqrt{3} \). Therefore, we set:

\[ 2 \tan \alpha = 2\sqrt{3} \]

### Step 5: Solve for \( \tan \alpha \)
Dividing both sides by 2 gives:

\[ \tan \alpha = \sqrt{3} \]

### Step 6: Determine \( \alpha \)
The angle \( \alpha \) for which \( \tan \alpha = \sqrt{3} \) is:

\[ \alpha = \frac{\pi}{3} \]

### Conclusion
Thus, the value of \( \alpha \) is:

\[ \alpha = \frac{\pi}{3} \]