Explanation

(b)
Given, $f(0) = 1,\ \ \ \ f(1) = 5,\ \ \ \ \ f(2) = 11$
Let the second degree equation be
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = ax^{2} + bx + c$
$\therefore\ \ \ f(0) = 0 + 0 + c\ \Rightarrow \ c = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$\ \ \ \ \ \ f(1) = a + b + c\ \ \Rightarrow \ \ 5 = a + b + 1$
$\Rightarrow \ \ \ \ a + b = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$\ \ \ \ \ f(2) = 4a + 2b + c\ \Rightarrow 4a + 2b + 1 = 11$
$\Rightarrow \ \ \ \ \ 2a + b = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
On solving Eqs. (ii) and (iii), we get
$\ \ \ a = 1,\ \ \ \ \ \ b = 3$
$\therefore$ The required equation is
$\ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = x^{2} + 3x + 1$