Given,
$$
\begin{aligned}
f(x) & =x^2+2 x+2 \\
& =x^2+2 x+1+1=(x+1)^2+1
\end{aligned}
$$
Here, $f(x)$
$$
\begin{aligned}
& \in[1, \infty) \text { and } g(x)=-x^2+2 x-1 \\
& =-\left(x^2-2 x+1\right)=-(x-1)^2
\end{aligned}
$$
Here, $g(x) \in(-\infty, 0]$
Now, $\frac{f}{g}(x)=\frac{x^2+2 x+2}{-x^2+2 x-1}=y$
$$
\begin{array}{ll}
\Rightarrow & x^2+2 x+2=-y x^2+2 x y-y \\
\Rightarrow & x^2+y x^2+2 x-2 x y+2+y=0 \\
\Rightarrow & x^2(1+y)+(2-2 y) x+2+y=0 \\
\because \quad & D \geq 0 \\
\therefore \quad & (2-2 y)^2-4(2+y)(1+y) \geq 0 \\
& 4+4 y^2-8 y-4(2+y)(1+y) \geq 0 \\
& 1+y^2-2 y-(2+y)(1+y) \geq 0 \\
& 1+y^2-2 y-\left(2+2 y+y+y^2\right) \geq 0 \\
& 1+y^2-2 y-2-3 y-y^2 \geq 0 \\
& -5 y-1 \geq 0 \\
& y \leq-\frac{1}{5}
\end{array}
$$
So, $\quad \frac{f}{g}(x) \in\left(-\infty,-\frac{1}{5}\right]$
So, $\quad a=1, b=0$ and $c=-\frac{1}{5}$
Hence,
$$
\begin{aligned}
a+2 b+5 c+4 & \\
& =1+0+5\left(-\frac{1}{5}\right)+4 \\
& =1-1+4=4
\end{aligned}
$$