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