Question

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

• A. at $x=\frac{11}{5},f(x)$ has a local minimum.
• B. at $x=\frac{11}{5},f(x)$ has a local maximum.
• C. the value of $f(2)+f(3)$ equals -8 .
• D. the value of $\underset{2}{\overset{6}{\int }}f(x)dx$ is $\frac{424}{5}$.

Answer 1

Answer: (A), (C), (D)

Clearly $f(x)=2\Rightarrow f(x)-2=0$
$\therefore f(x)-2=(x-1{)}^2(x-4{)}^5$
$\therefore f(x)=(x-1{)}^2(x-4{)}^3+2$
$\therefore f^{\prime }(x)=2(x-1)(x-4{)}^3+3(x-1{)}^2(x-4{)}^2$
$=(x-1)(x-4{)}^2[2(x-4)+3(x-1)]$
$=(x-1)(x-4{)}^2(5x-11)$

$\Rightarrow x=\frac{11}{5}$ is a point of local minimum.
$f(2)=-6,f(3)=-2$
$\underset{2}{\overset{6}{\int }}f(x)dx=\underset{2}{\overset{6}{\int }}\left((x-1{)}^2(x-4{)}^3+2\right)dx$
Put $x-4=t$
$=\underset{-2}{\overset{2}{\int }}\left(t^3(t+3{)}^2+2\right)dt=\underset{-2}{\overset{2}{\int }}\left(t^5+9t^3+6t^4+2\right)dt=2\underset{0}{\overset{2}{\int }}\left(6t^4+2\right)dt=2\left(\frac{6.32}{5}+4\right)=\frac{424}{5}$