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 determine the interval for which the function \( f(x) = \frac{2x}{\sqrt{3}(1+x^2)} \) is both one-one and onto, we will analyze the function step by step.

### Step 1: Understanding the Function
The function \( f(x) \) is defined as:
\[ f(x) = \frac{2x}{\sqrt{3}(1+x^2)} \]
This is a rational function where the numerator is a linear function of \( x \) and the denominator is a quadratic function.

### Step 2: Finding the Range of \( f(x) \)
To find the range of \( f(x) \), we will set \( y = f(x) \):
\[ y = \frac{2x}{\sqrt{3}(1+x^2)} \]
Rearranging gives:
\[ y \sqrt{3}(1+x^2) = 2x \]
This simplifies to:
\[ \sqrt{3}y + \sqrt{3}yx^2 - 2x = 0 \]
This is a quadratic equation in \( x \):
\[ \sqrt{3}yx^2 - 2x + \sqrt{3}y = 0 \]

### Step 3: Applying the Discriminant Condition
For \( f(x) \) to be one-one, the quadratic equation must have real roots. Therefore, the discriminant must be non-negative:
\[ D = b^2 - 4ac \]
Here, \( a = \sqrt{3}y \), \( b = -2 \), and \( c = \sqrt{3}y \):
\[ D = (-2)^2 - 4(\sqrt{3}y)(\sqrt{3}y) = 4 - 12y^2 \]
Setting the discriminant \( D \geq 0 \):
\[ 4 - 12y^2 \geq 0 \]
This simplifies to:
\[ 12y^2 \leq 4 \implies y^2 \leq \frac{1}{3} \implies -\frac{1}{\sqrt{3}} \leq y \leq \frac{1}{\sqrt{3}} \]

### Step 4: Conclusion on the Range
Thus, the function \( f(x) \) is both one-one and onto when \( y \) lies in the interval:
\[ \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \]

### Final Answer
The interval \( B \) for which \( f \) is both one-one and onto is:
\[ B = \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \]