Question

Let $p\left(x\right)$ be a real polynomial of least degree which has local minima at $x=-2$ and which changes it concavity at $x=2.$ If $p\left(0\right)=1$ and $p\left(1\right)=68,$ then the coefficient of $x^3$ in $p\left(x\right)$ is

• A. $-\frac{67}{41}$
• B. $\frac{47}{41}$
• C. $-3$
• D. $3$

Answer 1

The correct option is A $-\frac{67}{41}$

Let $\frac{d^{2}p}{d{x}^2}=\lambda (x-2)$
$\Rightarrow \frac{dp}{dx}=\frac{\lambda x^2}{2}-2\lambda x+c$
${\frac{dp}{dx}|}_{x=-2}=0$
$\Rightarrow 6\lambda +c=0\cdots \left(i\right)$

$p\left(x\right)=\frac{\lambda x^3}{6}-\lambda x^2+cx+k$
$p\left(0\right)=1$
$\Rightarrow p\left(x\right)=\frac{\lambda x^3}{6}-\lambda x^2+cx+1$
$p\left(1\right)=68$
$\Rightarrow \lambda =-\frac{67\times 6}{41}\left[\text{Using}\right(i\left)\right]$
$\Rightarrow$ Coefficient of $x^3$ is $-\frac{67}{41}$