Official solution
\[f\left( x \right) = {x \over {1 + {x^2}}}\]
\[ \therefore \] \[f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)\]
\[ \therefore \] 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