Let $$f(x)=\begin{cases}2x+a;&x\ge-1\\bx^2+3;&x\lt-1\end{cases}$$ and $$g(x)=\begin{cases}x+4;&0\le x\le4\\-3x-2;&-2\lt x\lt0\end{cases}$$ then $g(f(x))$ is not defined if
$(1)\;a\in(10,\infty),b\in(5,\infty)$
$(2)\;a\in(4,10),b\in(5,\infty)$
$(3)\;a\in(10,\infty),b\in(0,1)$
$(4)\;a\in(4,10),b\in(1,5)$
My Attempt:
$$g(f(x))=\begin{cases}f(x)+4;&0\le f(x)\le4\\-3f(x)-2;&-2\lt f(x)\lt0\end{cases}$$
So, $g(f(x))$ will not be defined if $f(x)\gt4$ or $f(x)\le-2$
Also, $f(x)=2x+a$ is an increasing function. So, its minimum value is at $x=-1$ i.e. $a-2$
For $g(f(x))$ to be not defined $a-2\gt4\implies a\gt6$
Also, $f(x)=bx^2+3$ is a parabola. If $b\lt0$ then it's a downward parabola, and its maximum value is $b+3$. For $g(f(x))$ to be not defined, $b+3\le-2\implies b\le-5$ (is this correct?)
If $b\gt0$ then $f(x)$ is an upward parabola and its minimum value is $b+3$. For $g(f(x))$ to be not defined, $b+3\gt4\implies b\gt1$
Accordingly, I think option $(1)$ is correct. But is there a way to find the exhaustive domain of $a?$
In the hint, they have written $-2+a\gt8, b+3\gt8$
Seems like they have calculated the value of $g(x)$ at $x=4$ and then compared it with the values of $f(x)$. But shouldn't we be comparing the domain (and not the range) of $g(x)$ with the range of $f(x)?$