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**Hint:**The function $F(x)$ is said to be injective, that is $F(a) = F(b)$ if and only if $a = b$. So, here we have to examine whether the given function is injective or not by putting values of x at respective points and then equating it. The function $F(x)$ is said to be surjective if for every y in the co-domain there exists x in the domain such that $F(x) = y$.

**Complete step-by-step answer:**

Firstly, we will check the given function is injective or not:

Let $F(a) = F(b)$ $\forall a,b \in R$

\[ \Rightarrow \dfrac{{{a^2} + a + 1}}{{{a^2} - a + 1}} = \dfrac{{{b^2} + b + 1}}{{{b^2} - b + 1}}\]

Since it cannot be further solved, so, the given function is not injective.

Take $a = 1$ and $b = - 1$ then $F(1) \ne F( - 1)$, then also it is not injective.

Secondly, we will check that given function is surjective or not:

Let \[\dfrac{{{x^2} + x + 1}}{{{x^2} - x + 1}} = t\] (say)

On further solving it, we will get the equation:

\[{x^2} + x + 1 = t({x^2} - x + 1)\]

\[ \Rightarrow (t - 1){x^2} + (t + 1)x + (t - 1) = 0\]

Now, finding discriminant of the above, we will get:

\[(3t - 1)(3 - t) \geqslant 0\]

So, it means \[t \in \left[ {\dfrac{1}{3},3} \right]\], which is the range of $F(x)$.

The co-domain and range of F is not the same, hence, the given function is not surjective.

**Since, the given function is neither surjective nor injective, hence, the correct answer is option C.**

**Note:**1) Any function is said to be injective if and only if $F(a) = F(b)$ whenever $a = b$. If this is not the case, the given function is not injective.

2) Any function is said to be surjective if and only if $F(x) = y$ for all x, y belonging to the given domain.

3) Another way to find if a function is surjective or not is that if the co-domain and range of the function is the same, it is surjective.

4) A function is called bijective if it is both one-one and onto. 5) Injective function is also called as one one function.

6) Surjective function is also called as onto function.

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