Consider a polynomial function  $f\left(x\right)=x^5+a{x}^4+b{x}^3+c{x}^2+dx+e.$ if  the line  $y=2$ touches it at  x=1 and  x=4 where  x=4  is also a point of inflection of  $y=f\left(x\right),$  then

(A,C,D)
Clearly  $f\left(x\right)=2\Rightarrow f\left(x\right)-2=0$

$\therefore f\left(x\right)-2={(x-1)}^2{(x-4)}^3 \therefore f\left(x\right)={(x-1)}^2{(x-4)}^3+2$

$\therefore f^1\left(x\right)=2(x-1){(x-4)}^3+3{(x-1)}^2{(x-4)}^2$

$=(x-1){(x-4)}^2\left[2\right(x-4)+3(x-1\left)\right] =(x-1){(x-4)}^2(5x-11)$

$\Rightarrow x=\frac{11}{5} is a po\mathrm{int} of local \min imum$

$$\begin{gathered} f\left(2\right)=-6,f(3=-2) \\ \underset{2}{\overset{6}{\int }}f\left(x\right)dx=\underset{2}{\overset{6}{\int }}({(x-1)}^2{(x-4)}^3+2)dx \\ Put x-4=t \\ =\underset{-2}{\overset{2}{\int }}(t^3{(t+3)}^2+2)dt=\underset{-2}{\overset{2}{\int }}(t^5+9t^3+6t^4+2)dt=2\underset{0}{\overset{2}{\int }}(6t^4+2)dt=2(\frac{6.32}{5}+4)=\frac{424}{5}] \end{gathered}$$