If $f\left(x\right)=a\log \left|x\right|+b{x}^2+x$ has its extremum values  at $x=-1$ and $x=2$, then

Solution:

We have,

$f\left(x\right)=a\log \left|x\right|+b{x}^2+x\Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{a}{x}+2bx+1$

Since $f\left(x\right)$ attains its extremum values at $x=-1, 2$

$\begin{array}{l}\therefore f^{\mathrm{\prime }}(-1)=0\text{ and }{f}^{\mathrm{\prime }}\left(2\right)=0\\ \Rightarrow -a-2b+1=0\text{ and }\frac{a}{2}+4b+1=0\\ \Rightarrow a=2\text{ and }b=-1/2.\end{array}$