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( ax ^2+ bx + c \right) $
$f ^{\prime}( x )=( x -1)^2(2 ax + b )+\left( ax ^2+ bx + c \right) 2( x -1)$
$f ^{\prime}(-1)= f ^{\prime}(3)=0 \& f (-1)= f (3)=-2$
$f ^{\prime}(3)=4(6 a + b )+4(9 a +3 b + c )=0 $
$\Rightarrow 15 a +4 b + c =0$....(1)
$f (3)=4(9 a +3 b + c )=-2$....(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]$
$f(0)+f(1)+f(2)=2 f(0)=-\frac{7}{4}$