Question

Consider a polynomial function $f(x)=x^5+a x^4+b x^3+c x^2+d x+e$. 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 $\int\limits_2^6 f(x) dx$ 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(5 x-11) $

$\Rightarrow x=\frac{11}{5}$ is a point of local minimum.
$f(2)=-6, f(3)=-2$
$\int\limits_2^6 f(x) d x=\int\limits_2^6\left((x-1)^2(x-4)^3+2\right) d x$
Put $x -4= t$
$=\int\limits_{-2}^2\left( t ^3( t +3)^2+2\right) dt =\int\limits_{-2}^2\left( t ^5+9 t ^3+6 t ^4+2\right) dt =2 \int\limits_0^2\left(6 t ^4+2\right) dt =2\left(\frac{6.32}{5}+4\right)=\frac{424}{5}$