Question

If $y=a\log \left|x\right|+b{x}^2+x$ has its extremum values at $x=-1$ and $x=2$ then find value of $a+2b$ .

Answer 1

Answer: 1

$\frac{dy}{dx}=\frac{a}{x}+2bx+1$ or $\frac{dy}{dx}=0$ at $x=-1,2$
$\frac{a}{-1}+2b\left(-1\right)+1=0$ or $-a-2b+1=0$
$\frac{a}{2}+4b+1=0$ or $a+8b+2=0$
solving we get $;a=2,b=-\frac{1}{2}$
Hence, $a+2b=1$