Question

If $y = m\, log\, x + nx^2 + x$ 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 $y = mlogx + nx^2 + x$
$\frac{dy}{dx}=\frac{m}{x}+2nx+1$
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
$m + 8n + 2 = 0$
$\therefore 6n + 1 = 0$
$\therefore \begin{matrix}6n + 1 = 0\\ 3m + 2 = 0\end{matrix} \bigg\}\, \Rightarrow n=-\frac{1}{6}$
$m=-\frac{2}{3}$
Hence, $2m + 10n=\frac{-4}{3}-\frac{5}{3}=-3$