The function

Given : $f(x)=y=a \log x+b x^{2}+x$

$\Rightarrow f^{\prime}(x)=\frac{a}{x}+2 b x+1$

Since, $f^{\prime}(x)$ has extreme value $s$ at $x=1$ and $x=2, f^{\prime}(1)=0 .$

$\Rightarrow \frac{a}{1}+2 b(1)+1=0$

$\Rightarrow a=-1-2 b$      .....(1)

$f^{\prime}(2)=0$

$\Rightarrow \frac{a}{2}+2 b(2)+1=0$

$\Rightarrow a+8 b=-2$

$\Rightarrow a=-2-8 b$   .....(2)

From eqs. (1) and (2), we get

$-2-8 b=-1-2 b$

$\Rightarrow 6 b=-1$

$\Rightarrow b=\frac{-1}{6}$

Substituting $b=\frac{-1}{6}$ in eq. (1), we get

$a=-1+\frac{1}{3}=\frac{-2}{3}$