Key Concepts and Formulas

• Inverse Trigonometric Identity: The identity $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ is valid for $x \in (-1, 1)$.
• Range of $\tan^{-1}x$: The principal value range of $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
• Properties of Functions:
  • One-one (Injective): A function $f$ is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. A strictly monotonic function is always one-one.
  • Onto (Surjective): A function $f: A \to B$ is onto if its range is equal to its codomain $B$.

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Step-by-Step Solution

Step 1: Simplify the function $f(x)$ using the inverse trigonometric identity. We are given the function $f: (-1,1) \to B$ defined by $f\left( x \right) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$. The domain of the function is $x \in (-1, 1)$. We know the identity $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, which is valid for $x \in (-1, 1)$. Since the domain of $f(x)$ is precisely $(-1, 1)$, we can directly apply this identity to simplify $f(x)$: $f(x) = 2\tan^{-1}x$ This simplified form is valid for all $x \in (-1, 1)$.

Step 2: Determine the range of the simplified function $f(x)$. We have $f(x) = 2\tan^{-1}x$ and the domain of $x$ is $(-1, 1)$. First, let's find the range of $\tan^{-1}x$ for $x \in (-1, 1)$. As $x$ approaches $-1$ from the right ($x \to -1^+$), $\tan^{-1}x$ approaches $-\frac{\pi}{4}$ from above (${\tan ^{ - 1}}x \to -{\frac{\pi}{4}}^+$). As $x$ approaches $1$ from the left ($x \to 1^-$), $\tan^{-1}x$ approaches $\frac{\pi}{4}$ from below (${\tan ^{ - 1}}x \to {\frac{\pi}{4}}^-$). Therefore, for $x \in (-1, 1)$, the range of $\tan^{-1}x$ is $\left( { - {\frac{\pi}{4}},{\frac{\pi}{4}}} \right)$.

Now, we find the range of $f(x) = 2\tan^{-1}x$ by multiplying the interval by 2: $\text{Range of } f(x) = 2 \times \left( { - {\frac{\pi}{4}},{\frac{\pi}{4}}} \right) = \left( {2 \times \left( { - {\frac{\pi}{4}}} \right), 2 \times \left( {{\frac{\pi}{4}}} \right)} \right) = \left( { - {\frac{\pi}{2}},{\frac{\pi}{2}}} \right)$ So, the range of $f(x)$ is $\left( { - {\frac{\pi}{2}},{\frac{\pi}{2}}} \right)$.

Step 3: Analyze the conditions for $f$ to be one-one and onto. The problem states that $f$ is both one-one and onto.

• One-one: The function $y = \tan^{-1}x$ is strictly increasing over its domain. Multiplying by a positive constant (2) preserves this strictly increasing nature. Thus, $f(x) = 2\tan^{-1}x$ is strictly increasing on $(-1, 1)$, which guarantees that it is one-one.

• Onto: For a function $f: A \to B$ to be onto, its range must be equal to its codomain $B$. We found the range of $f(x)$ to be $\left( { - {\frac{\pi}{2}},{\frac{\pi}{2}}} \right)$. Therefore, for $f$ to be onto, the codomain $B$ must be equal to this range. $B = \left( { - {\frac{\pi}{2}},{\frac{\pi}{2}}} \right)$

Step 4: Select the correct option for B. The codomain $B$ must be $\left( { - {\frac{\pi}{2}},{\frac{\pi}{2}}} \right)$ for the function to be onto. Comparing this with the given options: (A) $\left( {0,{\pi \over 2}} \right)$ (B) $\left[ {0,{\pi \over 2}} \right)$ (C) $\left[ { - {\pi \over 2},{\pi \over 2}} \right]$ (D) $\left( { - {\pi \over 2},{\pi \over 2}} \right)$ The interval $\left( { - {\frac{\pi}{2}},{\frac{\pi}{2}}} \right)$ matches option (D).

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Common Mistakes & Tips

• Domain of Identity: Be extremely careful about the domain for which the identity $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ holds. If the domain of $x$ were outside $(-1, 1)$, a different form of the identity would be required.
• Range of Inverse Functions: Always remember the principal value ranges of inverse trigonometric functions. For $\tan^{-1}x$, it is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
• Onto Condition: The requirement for a function to be onto is that its codomain must exactly match its range. Any mismatch, including differences in open/closed intervals or endpoints, means the function is not onto for that specific codomain.

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Summary

The problem requires us to find the codomain $B$ for which the function $f\left( x \right) = {\tan ^{ - 1}}{{\frac{{2x}}{{1 - {x^2}}}}}$ is both one-one and onto, given its domain is $(-1,1)$. We first simplified the function using the identity $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, valid for $x \in (-1,1)$, to get $f(x) = 2\tan^{-1}x$. We then determined the range of this simplified function over the domain $(-1,1)$ to be $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. For the function to be onto, its codomain $B$ must equal its range. Since $f(x) = 2\tan^{-1}x$ is strictly increasing on $(-1,1)$, it is one-one. Therefore, $B$ must be $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

The final answer is \boxed{\left( { - {\pi \over 2},{\pi \over 2}} \right)}.