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To check if the function $f(x) = \frac{2x}{1+x^2}$ is one-one, we need to verify if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Let $f(x_1) = f(x_2)$. Then, $$\frac{2x_1}{1+x_1^2} = \frac{2x_2}{1+x_2^2}$$ $$2x_1(1+x_2^2) = 2x_2(1+x_1^2)$$ $$x_1 + x_1x_2^2 = x_2 + x_2x_1^2$$ $$x_1 - x_2 + x_1x_2^2 - x_2x_1^2 = 0$$ $$(x_1 - x_2) + x_1x_2(x_2 - x_1) = 0$$ $$(x_1 - x_2) - x_1x_2(x_1 - x_2) = 0$$ $$(x_1 - x_2)(1 - x_1x_2) = 0$$ So, either $x_1 = x_2$ or $x_1x_2 = 1$. Since $x_1x_2 = 1$ does not necessarily imply $x_1 = x_2$, the function is not one-one. For example, $f(2) = \frac{4}{5}$ and $f(\frac{1}{2}) = \frac{1}{\frac{5}{4}} = \frac{4}{5}$. Thus, $f(2) = f(\frac{1}{2})$ but $2 \neq \frac{1}{2}$.
To check if the function is onto, we need to determine the range of the function and see if it equals the codomain $R$. Let $y = f(x) = \frac{2x}{1+x^2}$. Then, $$y(1+x^2) = 2x$$ $$yx^2 - 2x + y = 0$$ For $x$ to be real, the discriminant of this quadratic equation must be non-negative. $$D = (-2)^2 - 4(y)(y) \geq 0$$ $$4 - 4y^2 \geq 0$$ $$1 - y^2 \geq 0$$ $$y^2 \leq 1$$ $$-1 \leq y \leq 1$$ Thus, the range of $f(x)$ is $[-1, 1]$. Since the range $[-1, 1]$ is not equal to the codomain $R$, the function is not onto.
For the function $f: R \rightarrow A$ to be onto, the range of $f(x)$ must be equal to $A$. We found that the range of $f(x)$ is $[-1, 1]$. Therefore, $A = [-1, 1]$.
Final Answer: The function is neither one-one nor onto. The set A is [-1, 1].
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