If `f(0)=1,f(1)=5" and "f(2)=11`, then the equation of polynomial of degree two is

Options

A. `x^(2)+1=0`
B. `x^(2)+3x+1=0`
C. `x^(2)-2x+1=0`
D. None of these

Answer

B

Explanation

Given, `f(0)=1, f(1)=5" and "f(2)=11`
Let the second degree polynomial be `f(x)=ax^(2)+bx+c`.
`:." "f(0)=0+0+c`
`implies" "c=1" "…(i)`
`f(1)=a+b+c`
`implies" "5=a+b+1`
`implies" "a+b=4" "...(ii)`
`f(2)=4a+2b+c`
`implies" "4a+2b+1=11`
`implies" "2a+b=5" "...(iii)`
On solving Eqs. (ii) and (iii), we get
a = 1, b = 3
So, the required equation is `x^(2)+3x+1=0`.