Question

If $x=-1$ and $x=2$ are the extremum 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 $\alpha =2,\beta =-\frac{1}{2}$

$f\left(x\right)=\alpha \log |x|+\beta x^2+x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\alpha }{x}+2\beta x+1$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2\beta x^2+x+\alpha }{x}$
Given that $f\left(x\right)$ has extrema at $x=-1,2$
From $f^{\prime }(-1)=0$
$\Rightarrow 2\beta +\alpha -1=0\dots \left(1\right)$
And from $f^{\prime }\left(2\right)=0$
$\Rightarrow 8\beta +\alpha +2=0\dots \left(2\right)$
On solving $\left(1\right)$ and $\left(2\right)$, we get
$\beta =-\frac{1}{2},\alpha =2$