5. If $a$ is a real constant, the function $f(x)$ for any non-zero real number $x$ satisfies $f\left(\frac{1}{x}\right)=a f(x)-x-1$, and $f(1)=1$, then the range of the function $F(x)=f(x)(x \in D=\{x \mid x$ $\in \mathbf{R}, x \neq 0, f(x) \geqslant x\})$ is $\qquad$
5. $\left(-\infty, \frac{1}{2}-\frac{\sqrt{3}}{4}\right] \cup\left[\frac{1}{2}+\frac{\sqrt{3}}{4},+\infty\right)$.

Let $x=1$, we get $f(1)=a f(1)-2$.
Then, by $f(1)=1$, we get $a=3$.
Thus, $f\left(\frac{1}{x}\right)=3 f(x)-x-1$.
Substitute $x$ with $\frac{1}{x}$ in equation (1), we get
$$
f(x)=3 f\left(\frac{1}{x}\right)-\frac{1}{x}-1 \text {. }
$$

Eliminate $f\left(\frac{1}{x}\right)$ from equations (1) and (2), we get
$$
\begin{array}{l}
f(x)=\frac{3}{8} x+\frac{1}{8} \times \frac{1}{x}+\frac{1}{2} . \\
f(x) \geqslant x \Leftrightarrow 5 x-\frac{1}{x}-4 \leqslant 0 \\
\Leftrightarrow \frac{(5 x+1)(x-1)}{x} \leqslant 0
\end{array}
$$
$\Leftrightarrow 0<x \leqslant 1$ or $x \leqslant-\frac{1}{5}$.
Therefore, $D=\{x \mid 0<x \leqslant 1\} \cup\left\{x \left\lvert\, x \leqslant-\frac{1}{5}\right.\right\}$.
(1) If $0<x \leqslant 1$, then
$$
\begin{array}{l}
F(x)=f(x)=\frac{1}{8}\left(3 x+\frac{1}{x}\right)+\frac{1}{2} \\
\geqslant \frac{1}{8} \times 2 \sqrt{3 x \cdot \frac{1}{x}}+\frac{1}{2}=\frac{\sqrt{3}}{4}+\frac{1}{2},
\end{array}
$$

Equality holds if and only if $3 x=\frac{1}{x}$, i.e., $x=\frac{\sqrt{3}}{3}$.
Additionally, when $x \rightarrow 0^{+}$, $f(x) \rightarrow+\infty$.
Thus, $F(x) \in\left[\frac{1}{2}+\frac{\sqrt{3}}{4},+\infty\right)$.
(2) If $x \leqslant-\frac{1}{5}$, let $x=-x_{1}$, then $x_{1} \geqslant \frac{1}{5}$.
$$
\begin{array}{l}
F(x)=f(x)=-\frac{1}{8}\left(3 x_{1}+\frac{1}{x_{1}}\right)+\frac{1}{2} \\
\leqslant-\frac{1}{8} \times 2 \sqrt{3 x_{1} \cdot \frac{1}{x_{1}}}+\frac{1}{2}=\frac{1}{2}-\frac{\sqrt{3}}{4},
\end{array}
$$

Equality holds if and only if $3 x_{1}=\frac{1}{x_{1}}$, i.e., $x_{1}=\frac{1}{\sqrt{3}}$.
Additionally, when $x \rightarrow-\infty$, i.e., $x_{1} \rightarrow+\infty$, $f(x) \rightarrow-\infty$. Thus, $F(x) \in\left(-\infty, \frac{1}{2}-\frac{\sqrt{3}}{4}\right]$.