The function \[f:\mathbb{R} \to \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]\] is defined as \[f(x) = \dfrac{{\left[ x \right]}}{{\left[ {1 + {x^2}} \right]}}\], is
A. injective but not surjective.
B. surjective but not injective.
C. neither injective nor surjective.
D. invertible.

Hint: First differentiate the given function with respect to x, then identify if the function is only greater than zero or only less than zero in the given domain, then conclude whether the function is injective or not.
After that, suppose y is the range of the given function, then form a quadratic equation of x and calculate the discriminant to obtain the range of y. Then observe if the range of y is equal to the co-domain or not.

Formula Used: \[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x)\dfrac{d}{{dx}}f(x) - f(x)\dfrac{d}{{dx}}g(x)}}{{{{\left( {g(x)} \right)}^2}}};g(x) \ne 0\]

Complete step by step solution:
The given function is \[f(x) = \dfrac{{\left[ x \right]}}{{\left[ {1 + {x^2}} \right]}}\].
Differentiate the given function with respect to x.
\[f'(x) = \dfrac{{(1 + {x^2}).1 - x.(0 + 2x)}}{{{{(1 + {x^2})}^2}}}\]
          =\[\dfrac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}\]
Here we can see that,
\[f'(x) > 0 \Rightarrow 1 - {x^2} > 0\]
That is, when \[x < \pm 1\]
Similarly, \[f'(x) < 0 \Rightarrow 1 - {x^2} < 0\]
That is, when \[x > \pm 1\]
We know that if \[f'(x) > 0{\rm{ or f'(x) < 0 }}\forall {\rm{x}} \in {\rm{domain}}\], then f(x) is one-one.
Here we can conclude that \[f'(x)\] is not always greater than zero and not always less than zero.
It means f(x) is not a one-one or injective function.
Let us suppose that \[y = \dfrac{x}{{1 + {x^2}}}\].
Now, let us construct a quadratic equation of x from \[y = \dfrac{x}{{1 + {x^2}}}\].
Therefore, the equation is \[{x^2}y - x + y = 0\].
Now, this equation should have real roots.
Therefore, the discriminant is,
\[{( - 1)^{^2}} - 4(y)(y) \ge 0\]
\[ \Rightarrow 1 - 4{y^2} \ge 0\]
\[ \Rightarrow 4{y^2} \le 1\]
\[ \Rightarrow {y^2} \le \dfrac{1}{4}\]
\[ \Rightarrow y \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]\], which is the co-domain of the given function.
As the co-domain and the range of the given function are equal, hence the given function is onto or surjective.

Hence the correct option is B.

Note: The given question contains box function, let us quickly recall the box function. The box function is the greatest integer function, that is, \[\left[ {1.2} \right] = 1\].