Explanation

(d)
Given, $f(x)$ is a polynomial of degree 3.
$\therefore f(x) = x^{3} + ax^{2} + bx + c$
But $f(x) = x^{3} + x^{2} + f^{'}(1) + xf^{''}(2) + f^{'''}(3)$
$\therefore a = f^{'}(1),\ b = f^{''}(2),c = f^{'''}(3)$
Now, $f^{'}(x) = 3x^{2} + 2ax + b$
$f^{''}(x) = 6x + 2a$
$f^{'''}(x) = 6$
$\therefore a = f^{'}(1) = 3 + 2a + b$
$\Rightarrow \ - a - b = 3$ …(i)
$b = 12 + 2a$
$\Rightarrow - 2a + b = 12$ …(ii)
and $c = 6$ …(iii)
On solving Eqs. (i), (ii) and (iii), we get
$a = - 5,\ b = 2$ and $c = 6$
$\therefore f(x) = x^{3} - 5x^{2} + 2x + 6$
Thus, $f(0) = 6,\ f(1) = 4,\ f(2) = - 2,\ f(3) = - 6$