$\underline{\textbf{Answer: B}}$
$\underline{\textbf{Explanation:}}$
Given:
$\mathrm {f'(x) = 4x^3 -3x^2 + 2x + k,\; f(0) = 1 \;\text{and}\; f(1) = 4}$
Now, we can integrate the $\mathrm {f'(x)}$ to get the $\mathrm {f(x)}$:
$\text{Let} \;\mathrm{I=f(x) = \int(4x^3-3x^2+2x+k)}$
$\Rightarrow \mathrm{f(x)=\frac{4x^4}{4} -\frac{3x^3}{3}+\frac{2x^2}{2}+kx + C, \;where\;c\;is\;a\;constant}$
$\Rightarrow \mathrm{f(x) =x^4 -x^3+x^2+kx + C}\tag{1}$
Now,
$\mathrm f(0) = 0\;\;\;\text{[Given]}$
$\therefore \mathrm {f(0)} = 0-0+0+\mathrm k\times 0 + \mathrm C = 1 $
This is possible only when $\mathrm C = 1$
$\therefore \bf{C = 1}$
Now,
$\mathrm{f(1) = 1-1+1+k.1+C = 4 \\ \Rightarrow 1 + k + 1 = 4\\ \Rightarrow \bf{k = 2}}$
Now, substitute $\mathrm C = 1,\;\text{and}\; \mathrm k = 2$, we get:
$\mathrm {x^4-x^3+x^2+2x + 1}$
$\therefore \mathbf B$ is the correct option.