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

Options

A. `(0, (pi)/(4))`
B. `[0, (pi)/(3)]`
C. `[-(pi)/(4), (pi)/(4)]`
D. `(-(pi)/(4), (pi)/(4))`

Answer

To determine the interval for which the function \( f(x) = \tan^{-1} \left( \frac{2x}{1+x^2} \right) \) is both one-one and onto, we will follow these steps:

### Step 1: Understand the function
The function is defined as:
\[ f(x) = \tan^{-1} \left( \frac{2x}{1+x^2} \right) \]
We need to analyze the inner function \( g(x) = \frac{2x}{1+x^2} \) first.

### Step 2: Find the range of \( g(x) \)
To find the range of \( g(x) \), we will consider it as a function of \( x \) and determine its maximum and minimum values.

1. **Differentiate \( g(x) \)**:
\[ g'(x) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} \]
Setting \( g'(x) = 0 \):
\[ 2 - 2x^2 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \]

2. **Evaluate \( g(x) \) at critical points and endpoints**:
- \( g(1) = \frac{2(1)}{1+1^2} = \frac{2}{2} = 1 \)
- \( g(-1) = \frac{2(-1)}{1+(-1)^2} = \frac{-2}{2} = -1 \)
- As \( x \to \infty \), \( g(x) \to 0 \).
- As \( x \to -\infty \), \( g(x) \to 0 \).

Thus, the range of \( g(x) \) is \( [-1, 1] \).

### Step 3: Determine the range of \( f(x) \)
Since \( f(x) = \tan^{-1}(g(x)) \) and the range of \( g(x) \) is \( [-1, 1] \):
- The minimum value of \( f(x) \) occurs at \( g(-1) \):
\[ f(-1) = \tan^{-1}(-1) = -\frac{\pi}{4} \]
- The maximum value of \( f(x) \) occurs at \( g(1) \):
\[ f(1) = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 4: Conclusion about the function
The function \( f(x) \) is strictly increasing because \( g(x) \) is strictly increasing in the interval \( (-\infty, \infty) \). Therefore, \( f(x) \) is one-one.

Since the range of \( f(x) \) is \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \), it is onto when the codomain \( B \) is taken as this interval.

### Final Result
Thus, the function \( f(x) \) is both one-one and onto when \( B \) is in the interval:
\[ B = \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]