Question

If $x=-1$ and $x=2$ are extreme points of $f\left(x\right)=\alpha \log |x|+\beta x^2+x$, then

• A. $\alpha =-6,\beta =\frac{1}{2}$
• B. $\alpha =-6,\beta =-\frac{1}{2}$
• C. $\alpha =2,\beta =-\frac{1}{2}$
• D. $\alpha =2,\beta =\frac{1}{2}$

Answer 1

The correct option is C $\alpha =2,\beta =-\frac{1}{2}$

Given $f\left(x\right)=\alpha log|x|+\beta x^2+x$

Differentating

$f^{\prime }\left(x\right)=\frac{\alpha }{x}+2\beta x+1$

At the extrem point, $x=-1$

$f^{\prime }(-1)=0$

$\Rightarrow \frac{\alpha }{-1}+2\beta (-1)+1=0$

$\Rightarrow -\alpha -2\beta +1=0$ $\Rightarrow \alpha =1-2\beta$....(1)

At the extrem point, $x=2$

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

$\Rightarrow \frac{\alpha }{2}+2\beta \left(2\right)+1=0$

$\Rightarrow \frac{\alpha }{2}+4\beta +1=0$

Subtracting (1), we get

$\Rightarrow \frac{1}{2}(1-2\beta )+4\beta +1=0$

$\Rightarrow \frac{1}{2}-\beta +4\beta +1=0$

$\Rightarrow 3\beta =-1-\frac{1}{2}$

$\Rightarrow 3\beta =\frac{-3}{2}$

$\Rightarrow \beta =\frac{-1}{2}$

From (1),

$\alpha =1-2\left(\frac{-1}{2}\right)=1+1=2$

Hence, $\alpha =2,\beta =\frac{-1}{2}$