$$f\left( x \right) = {x \over {1 + {x^2}}}$$

$$ \therefore $$ $$f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)$$

$$ \therefore $$ f(x) is many-one function.

Now let y = f(x) = $${x \over {1 + {x^2}}}$$

$$ \Rightarrow $$ y + x²y = x

$$ \Rightarrow $$ yx - x + y = 0

As x $$ \in $$ R

$$ \therefore $$ (-1)² - 4(y)(y) $$ \ge $$ 0

$$ \Rightarrow $$ 1 - 4y² $$ \ge $$ 0

$$ \Rightarrow $$ y $$ \in $$ $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$

$$ \therefore $$ Range = Codomain = $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$

So, f(x) is surjective.

$$ \therefore $$ f(x) is surjective but not injective