If y=a log |x|+b x^2+x has its extremum values at x = – 1 and x = 2,…

$y=a \log |x|+b x^2+x$

$\log |x|=\left\{\begin{array}{l} \log x ~~~~~\quad \text { if } x>0 \\ \log (-x) \quad \text { if } x<0 \end{array}\right.$

∴  $\frac{d y}{d x}=\left\{\begin{array}{l} \frac{a}{x}+2 b x+1 \text { if } x>0 \\ \frac{a}{x}+2 b x+1 \text { if } x<0 \end{array}\right.$

Now, $\frac{d y}{d x}=0$ at x = -1, 2

∴  $-a-2 b+1=0, \frac{a}{2}+4 b+1=0$

$\Rightarrow a+2 b=1, a+8 b=-2$

Solving, a = 2, b = $-\frac{1}{2}$.