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Solution:

It is given that, $y=a \log x+b x^{2}+x$ has its extreme values at $x=1$ and $x=2$.

$\therefore \frac{d y}{d x}=0$ at $x=1$ and $x=2$

$y=a \log x+b x^{2}+x$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x}=\frac{a}{x}+2 b x+1$

Now,

$\left(\frac{d y}{d x}\right)_{x=1}=0$

$\Rightarrow a+2 b+1=0$

$\Rightarrow a+2 b=-1$       .....(1)

Also,

$\left(\frac{d y}{d x}\right)_{x=2}=0$

$\left(\frac{d y}{d x}\right)_{x=2}=0$

$\Rightarrow \frac{a}{2}+4 b+1=0$

 

$\Rightarrow a+8 b=-2 \quad \ldots(2)$

Subtracting (1) from (2), we get

$6 b=-1$

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

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

$a+2 \times\left(-\frac{1}{6}\right)=-1$

 

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

Thus, the values of $a$ and $b$ are $-\frac{2}{3}$ and $-\frac{1}{6}$, respectively.

Hence, the ordered pair $(a, b)$ is $\left(-\frac{2}{3},-\frac{1}{6}\right)$

If $y=\operatorname{alog} x+b x^{2}+x$ has its extreme values at $x=1$ and $x=2$, then $(a, b)=$   $\left(-\frac{2}{3},-\frac{1}{6}\right)$