Let `f:R rarr B`, be a function defined `f(x)=tan^(-1).(2x)/(sqrt3(1+x^(2)))`, then f is both one - one and onto when B, is the interval

Options

A. `(0, (x)/(6))`
B. `[0, (pi)/(6))`
C. `[-(pi)/(6), (pi)/(6)]`
D. `(-(pi)/(6),(pi)/(6))`

Answer

To solve the problem, we need to determine the interval \( B \) such that the function \( f(x) = \tan^{-1}\left(\frac{2x}{\sqrt{3}(1+x^2)}\right) \) is both one-one and onto.

### Step 1: Understanding the Function
The function \( f(x) \) is defined as:
\[ f(x) = \tan^{-1}\left(\frac{2x}{\sqrt{3}(1+x^2)}\right) \]
We need to analyze the behavior of this function to find its range.

### Step 2: Finding the Range of the Inner Function
The expression inside the arctangent function is:
\[ y = \frac{2x}{\sqrt{3}(1+x^2)} \]
To find the range of \( f(x) \), we first need to find the maximum and minimum values of \( y \).

### Step 3: Differentiate to Find Critical Points
We differentiate \( y \) with respect to \( x \):
\[ y = \frac{2x}{\sqrt{3}(1+x^2)} \]
Using the quotient rule:
\[ y' = \frac{(1+x^2)(2) - 2x(2x)}{\sqrt{3}(1+x^2)^2} = \frac{2(1+x^2 - 2x^2)}{\sqrt{3}(1+x^2)^2} = \frac{2(1-x^2)}{\sqrt{3}(1+x^2)^2} \]
Setting \( y' = 0 \) gives:
\[ 1 - x^2 = 0 \implies x = \pm 1 \]

### Step 4: Evaluate \( y \) at Critical Points and Endpoints
Now we evaluate \( y \) at \( x = 1 \) and \( x = -1 \):
- For \( x = 1 \):
\[ y(1) = \frac{2(1)}{\sqrt{3}(1+1^2)} = \frac{2}{\sqrt{3} \cdot 2} = \frac{1}{\sqrt{3}} \]
- For \( x = -1 \):
\[ y(-1) = \frac{2(-1)}{\sqrt{3}(1+(-1)^2)} = \frac{-2}{\sqrt{3} \cdot 2} = -\frac{1}{\sqrt{3}} \]

### Step 5: Analyze the Behavior as \( x \to \pm \infty \)
As \( x \to \infty \) or \( x \to -\infty \):
\[ y \to 0 \]

### Step 6: Determine the Range of \( y \)
From the evaluations:
- The maximum value of \( y \) is \( \frac{1}{\sqrt{3}} \).
- The minimum value of \( y \) is \( -\frac{1}{\sqrt{3}} \).

Thus, the range of \( y \) is:
\[ -\frac{1}{\sqrt{3}} \leq y \leq \frac{1}{\sqrt{3}} \]

### Step 7: Find the Range of \( f(x) \)
Now we find the range of \( f(x) \):
\[ f(x) = \tan^{-1}(y) \]
The range of \( \tan^{-1}(y) \) when \( y \) is in the interval \( \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \) is:
\[ f(-1/\sqrt{3}) = -\frac{\pi}{6}, \quad f(1/\sqrt{3}) = \frac{\pi}{6} \]
Thus, the range of \( f(x) \) is:
\[ \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \]

### Conclusion
For \( f \) to be both one-one and onto, the codomain \( B \) must be the interval:
\[ B = \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \]