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To check if $f$ is one-one, we assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$. $$ \frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}} $$ Squaring both sides: $$ \frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2} $$ $$ x_1^2(1+x_2^2) = x_2^2(1+x_1^2) $$ $$ x_1^2 + x_1^2x_2^2 = x_2^2 + x_1^2x_2^2 $$ $$ x_1^2 = x_2^2 \implies |x_1| = |x_2| $$ Since the function $f(x)$ is strictly increasing (its derivative $f'(x) = (1+x^2)^{-3/2} > 0$), $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Thus, $f$ is one-one.
To check if $f$ is onto, we examine the range. Let $y = \frac{x}{\sqrt{1+x^2}}$. As $x \to \infty$, $y \to 1$. As $x \to -\infty$, $y \to -1$. Since the function is continuous and strictly increasing, the range is the open interval $(-1, 1)$. Because the codomain is $R$ and the range $(-1, 1) \neq R$, there exist values in the codomain (e.g., $y = 2$) that have no pre-image in the domain. Thus, $f$ is not onto.
Final Answer: The function is one-one because $f(x_1)=f(x_2) \implies x_1=x_2$, and not onto because the range $(-1, 1)$ is a proper subset of the codomain $R$.
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