If f(x) is a polynomial satisfying $f(x) = x^3+x^2 f'(1)+x f''(2)+f'''(3)$ for all $x \in R$. Then,

all the above

We have,

$f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$           ....(i)

$\Rightarrow f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$           ....(ii)

$\Rightarrow f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$           ....(iii)

$\Rightarrow f^{\prime \prime \prime}(x)=6$           ....(iv)

Putting $x=1,2,3$ in (ii), (iii), and (iv) respectively, we get

$f^{\prime}(1)=3+2 f^{\prime}(1)+f^{\prime \prime}(2)$

$f^{\prime \prime}(2)=12+2 f^{\prime}(1), f^{\prime \prime \prime}(3)=6$

Solving these equations, we get

$f^{\prime}(1)=-5, f^{\prime \prime}(2)=2$ and $f^{\prime \prime \prime}(3)=6$

∴  $f(x)=x^3-5 x^2+2 x+6$

$\Rightarrow f(0)=6, f(1)=4, f(2)=8-20+4+6=-2$ and, f(3) = -6

Clearly, these values satisfy options (1), (2) and (3).