Question

If $y=mlogx+n{x}^2+x$ has its extreme values at $x=2$ and $x=1,$ then $2m+10n=$

• A. $-1$
• B. $-4$
• C. $-2$
• D. $-3$

Answer 1

Answer: (D)

Let $y=mlogx+n{x}^2+x$
$\frac{dy}{dx}=\frac{m}{x}+2nx+1$
At $x=2,\frac{dy}{dx}=0$
$\therefore \frac{m}{2}+2n\left(2\right)+1=0$
At $x=1,\frac{dy}{dx}=0$
$m+2n+1=0$
Thus, we have
$m+8n+2=0$
$\therefore 6n+1=0$
$\therefore \begin{array}{c}6n+1=0\\ 3m+2=0\end{array}\}\Rightarrow n=-\frac{1}{6}$
$m=-\frac{2}{3}$
Hence, $2m+10n=\frac{-4}{3}-\frac{5}{3}=-3$