Question

The curve $f\left(x\right)=a^2{x}^3-0.5a{x}^2-2x-b$ has its local minima at $x=\frac{1}{3}$. If $f\left(\frac{1}{3}\right)>0,$ then

• A. $a=-2$ and $b<-\frac{11}{27}$
• B. $a=3$ and $b<-\frac{1}{2}$
• C. $a=-2$ and $b>-\frac{1}{2}$
• D. $a=3$ and $b<-\frac{3}{4}$

Answer 1

Answer: (A), (B)

The correct options are
A $a=-2$ and $b<-\frac{11}{27}$
B $a=3$ and $b<-\frac{1}{2}$

$f\left(x\right)=a^2{x}^3-0.5a{x}^2-2x-b\Rightarrow f^{\prime }\left(x\right)=3a^2{x}^2-ax-2$
$x=\frac{1}{3}$ is a critical point.
$\therefore f^{\prime }\left(\frac{1}{3}\right)=0$
$\Rightarrow \frac{a^2}{3}-\frac{a}{3}-2=0\Rightarrow (a+2)(a-3)=0$
$\Rightarrow a=-2,3$
For $a=-2,3,f\left(\frac{1}{3}\right)>0,$ so both values are acceptable

When $a=-2$
$f\left(\frac{1}{3}\right)=\frac{4}{3^3}+\frac{1}{3^2}-\frac{2}{3}-b>0\Rightarrow b<-\frac{11}{27}$

When $a=3$
$f\left(\frac{1}{3}\right)=\frac{9}{3^3}-\frac{3}{2\times 3^2}-\frac{2}{3}-b>0\Rightarrow b<-\frac{1}{2}$