Question

If the function $f\left(x\right)=a\log |x|+b{x}^2+x$ has its extremum values at $x=-1$ and $x=2$ then the values of $a$ and $b$.

• A. $a=2b=\frac{1}{2}$
• B. $a=\frac{1}{2}b=2$
• C. $a=2b=-\frac{1}{2}$
  
• D. $a=-2b=-\frac{1}{2}$

Answer 1

The correct option is C $a=2b=-\frac{1}{2}$


$f\left(x\right)$ has its extremum values at $-1$ and $2$, so $f^{\prime }(-1)=0=f^{\prime }\left(2\right)$,
$f^{\prime }\left(x\right)=\frac{a}{x}+2bx+1$

$f^{\prime }\left(2\right)=\frac{a}{2}+4b+1\Rightarrow a+8b=-2\cdots \left(1\right)$
$f^{\prime }(-1)=-a-2b+1\Rightarrow a+2b=1\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$,
$a=2\text{and}b=-\frac{1}{2}$