Let f: R rightarrow R and g: R rightarrow R b | vedclass

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(C) Given $f(x) = \frac{x}{1+x^2}$ and $g(x) = \frac{x^2}{1+x^2}$ for $x \in R$.For $f(x)$,$f'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2

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Both $f$ and $g$ are neither one-one nor onto

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(C) Given $f(x) = \frac{x}{1+x^2}$ and $g(x) = \frac{x^2}{1+x^2}$ for $x \in R$.For $f(x)$,$f'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2}$.Since $f'(x)$ changes sign at $x = \pm 1$,$f(x)$ is not monotonic,hence not one-one.The range of $f(x)$ is $[-\frac{1}{2}, \frac{1}{2}]$,which is not equal to the codomain $R$,so $f(x)$ is not onto.For $g(x)$,$g(-x) = \frac{(-x)^2}{1+(-x)^2} = \frac{x^2}{1+x^2} = g(x)$,so $g(x)$ is an even function,hence not one-one.The range of $g(x)$ is $[0, 1)$,which is not equal to the codomain $R$,so $g(x)$ is not onto.Therefore,both $f$ and $g$ are neither one-one nor onto.

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be the functions defined by $f(x) = \frac{x}{1+x^2}$ and $g(x) = \frac{x^2}{1+x^2}$. Then,the correct statement$(s)$ among the following is/are:

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