Question

Let $y=f(x)$ be a polynomial function of degree 4 , which is symmetrical about the line $x=1$ and touches the $x$-axis and straight line $y+2=0$ and distance between both absolute minima is 4 .
The value of $f(0)+f(1)+f(2)$ is equal to

• A. -7
• B. $\frac{-7}{4}$
• C. $\frac{-7}{8}$
• D. $\frac{-9}{4}$

Answer 1

Answer: (B)

$f(x)=(x-1{)}^2\left(a{x}^2+bx+c\right)$
$f^{\prime }(x)=(x-1{)}^2(2ax+b)+\left(a{x}^2+bx+c\right)2(x-1)$
$f^{\prime }(-1)=f^{\prime }(3)=0\&f(-1)=f(3)=-2$
$f^{\prime }(3)=4(6a+b)+4(9a+3b+c)=0$
$\Rightarrow 15a+4b+c=0$....(1)
$f(3)=4(9a+3b+c)=-2$....(2)
$f(-1)=4(a-b+c)=-2$....(3)

From (1), (2) and (3)
$a=\frac{1}{8},b=-\frac{1}{4},c=-\frac{7}{8}$
$\therefore f(x)=(x-1{)}^2\left(\frac{1}{8}{x}^2-\frac{1}{4}x-\frac{7}{8}\right)$
$=\frac{1}{8}(x-1{)}^2\left((x-1{)}^2-8\right)$
$=\frac{1}{8}\left[(x-1{)}^4-8(x-1{)}^2\right]$
$f(0)+f(1)+f(2)=2f(0)=-\frac{7}{4}$