Question

If $F(-2)=0$, then the number of possible values of ordered pair $(a,b)$ so that point $P(-2,0)$ becomes the point of minima for $F\left(x\right)=\frac{x^3}{3}+\frac{3x^2}{2}+ax+b$is/are

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$F\left(x\right)=\frac{x^3}{3}+\frac{3x^2}{2}+ax+b$
So $F(-2)=0\Rightarrow 2a-b=\frac{10}{3}$
$F^{\prime }\left(x\right)=x^2+3x+a$
Since, $F^{\prime }(-2)=0$
$\Rightarrow a=2$
So,$F^{\prime }\left(x\right)=x^2+3x+2=(x+1)(x+2)$
$x=-1,-2$
$F^{\prime \prime }\left(x\right)=2x+3$
$F^{\prime \prime }\left(x\right)<0$ at $x=-2$
So maxima occurs at $x=-2$
Hence, there are no possible values of $(a,b)$ for which the minima occurs at P.