Question

Let $f ( x )$ be a polynomial of degree three with $f (0)=0, f (1)=\frac{1}{2}, f (2)=\frac{2}{3}$ and $f (3)=\frac{3}{4}$, then value of $[ f (4)]$ is
[Note: [k] denote greatest integer function less than or equal to $k$.]

• A. 0
• B. 1
• C. 3
• D. 5

Answer 1

Answer: (B)

$\Theta f(x)=\frac{x}{x+1}$ for $x=0,1,2,3$
$\Rightarrow( x +1) f ( x )- x =0 \text { have roots } x =0,1,2,3$
$\Theta f ( x )$ is cubic
$\therefore( x +1) f ( x )- x$ will be biquadratic
Consider $f ( x )= ax ^3+ bx ^2+ cx + d$
$\therefore( x +1) f ( x )- x = ax ( x -1)( x -2)( x -3) $
$( x +1)\left( ax ^3+ bx ^2+ cx + d \right)- x =\operatorname{ax}( x -1)( x -2)( x -3)$
Comparing both side, get
$ a =\frac{1}{24}, b =\frac{-7}{24}, c =\frac{3}{4}, d =0 $
$\therefore f ( x )=\frac{1}{24} x ^3-\frac{7}{24} x ^2+\frac{3}{4} x $
$ f (4)=1 $