Question

lf $f\left(x\right)=a\log x+b{x}^2+x$ has extreme values at $x=-1,x=2$ then $a=\dots ..,b=\dots \dots$

• A. $2,\frac{-1}{2}$
• B. $\frac{-1}{2},2$
• C. $\frac{1}{2},2$
• D. $2,\frac{1}{2}$

Answer 1

The correct option is A $2,\frac{-1}{2}$

$f^{\prime }\left(x\right)=\frac{a}{x}+2bx+1$
$f^{\prime }\left(x\right)$ is 0 at $x=-1$ & $x=2$
(1) $0=-a-2b+1$
$a+2b=1$
(2) $0=\frac{a}{2}+4b+1$
$a+8b+2=0$
So $a=2,b=\frac{-1}{2}$