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^

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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.

List-$I$	List-$II$
$(A) f(-5)+f(0)+f(-1)$	$(I) 16$
$(B) f(f(5)+10f(-3))$	$(II) 40$
$(C) f(f(-4))$	$(III) -31$
$(D) f(f(f(1)))$	$(IV) -12$
	$(V) 19$

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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