be a real polynomial of degree 3 which vanishes at

Question:

Let $\mathrm{P}(\mathrm{x})$ be a real polynomial of degree 3 which vanishes at $x=-3$. Let $\mathrm{P}(\mathrm{x})$ have local minima

at $x=1$, local maxima at $x=-1$ and $\int_{-1}^{1} \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}=18$, then the sum of all the coefficients of the polynomial $\mathrm{P}(\mathrm{x})$ is equal to

Solution:

Let $p^{\prime}(x)=a(x-1)(x+1)=a\left(x^{2}-1\right)$

$p(x)=a \int\left(x^{2}-1\right) d x+c$

$=a\left(\frac{x^{3}}{3}-x\right)+c$

Now $p(-3)=0$

$\Rightarrow \mathrm{a}\left(-\frac{27}{3}+3\right)+\mathrm{c}=0$

$\Rightarrow-6 a+c=0$

Now $\int_{-1}^{1}\left(a\left(\frac{x^{3}}{3}-x\right)+c\right) \mathrm{dx}=18$

$=2 c=18 \Rightarrow c=9 \quad \ldots(2)$

$\Rightarrow$ from

(1) $\&(2) \Rightarrow-6 a+9=0 \Rightarrow a=\frac{3}{2}$

$\Rightarrow \mathrm{p}(\mathrm{x})=\frac{3}{2}\left(\frac{\mathrm{x}^{3}}{3}-\mathrm{x}\right)+9$

sum of coefficient

$=\frac{1}{2}-\frac{3}{2}+9$

$=8$