Question

If $f\left(x\right)=x^3+a{x}^2+bx+c$ is minimum at $x=3$ and maximum at $x=-1$

• A. $a=-3,b=-9,c=0$
• B. $a=3,b=9,c=0$
• C. $a=-3,b=-9,c\in R$
• D. none of these

Answer 1

The correct option is C $a=-3,b=-9,c\in R$

$f\left(x\right)=x^3+a{x}^2+bx+c$
$f^{\prime }\left(x\right)=3x^2+2ax+b$
$f^{\prime }\left(x\right)=0$
$3x^2+2ax+b=0$
$x^2+\frac{2ax}{3}+\frac{b}{3}=0......\left(1\right)$
Now, given $x=3$ is the point of minimum and $x=-1$ is a point for maximum
So, $(x-3)(x+1)=0$
$x^2-2x-3=\mathrm{0.......}\left(2\right)$
Since, equation $\left(1\right)$ and $\left(2\right)$ are same, so on comparing we get
$\frac{2a}{3}=-2$; $\frac{b}{3}=-3$
$a=-3$; $b=-9$
and $c\in R$
so $a=-3$; $b=-9$, $c\in R$