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

$f\left(x\right)=\alpha \log |x|+\beta x^2+x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\alpha }{x}+2\beta x+1$
At $x=-1,f^{\prime }\left(1\right)=-\alpha -2\beta +1=0$
At $x=2,f^{\prime }\left(2\right)=\frac{\alpha }{2}+4\beta +1=0$
On solving (1) and (2)
$\alpha =2,\beta =-\frac{1}{2}$