6. Let $g(x)$ be a strictly increasing function defined for all $x \geq 0$. It is known that the range of $t$ satisfying
$$
\mathrm{g}\left(2 t^{2}+t+5\right)<\mathrm{g}\left(t^{2}-3 t+2\right)
$$
is $b<t<a$. Find $a-b$.
6. Answer: 2
Solution. Note that $2 t^{2}+t+5=2\left(t+\frac{1}{4}\right)^{2}+\frac{39}{8}>0$. Hence $\mathrm{g}\left(2 t^{2}+t+5\right)<\mathrm{g}\left(t^{2}-3 x+2\right)$ is true if and only if
$$
2 t^{2}+t+5<t^{2}-3 t+2 \text {, }
$$
which is equivalent to $(t+3)(t+1)<0$. Hence the range of $t$ satisfying the given inequality is $-3<t<-1$, which yields $a-b=(-1)-(-3)=2$.