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 $f(x)=a{x}^3+b{x}^2+cx+d$
$\therefore (x+1)f(x)-x=ax(x-1)(x-2)(x-3)$
$(x+1)\left(a{x}^3+b{x}^2+cx+d\right)-x=\mathrm{ax}(x-1)(x-2)(x-3)$
Comparing both side, get
$a=\frac{1}{24},b=\frac{-7}{24},c=\frac{3}{4},d=0$
$\therefore f(x)=\frac{1}{24}{x}^3-\frac{7}{24}{x}^2+\frac{3}{4}x$
$f(4)=1$