Find the critical points of [tex]f_1[/tex] and [tex]f_2[/tex].
By the fundamental theorem of calculus,
[tex]{f_1}'(x) = \displaystyle \prod_{j=1}^{21} (x-j)^j = (x-1) (x-2)^2 (x-3)^3 \cdots (x-21)^{21}[/tex]
and [tex]{f_1}'=0[/tex] for [tex]x\in\{1,2,3,\ldots,21\}[/tex]. The stationary points at odd values have odd multiplicity, while the even ones have even multiplicity. At the points of odd multiplicity, [tex]{f_1}'[/tex] passes through the [tex]x[/tex]-axis and the sign of [tex]{f_1}'[/tex] changes, while at the points of even multiplicity, [tex]{f_1}'[/tex] is tangent the to [tex]x[/tex]-axis. We also have
[tex]\displaystyle \lim_{x\to-\infty} {f_1}'(x) = (-1)^{1+3+5+\cdots+21} \infty = (-1)^{11^2} \infty = -\infty[/tex]
[tex]\displaystyle \lim_{x\to\infty} {f_1}'(x) = \infty[/tex]
so the plot of [tex]{f_1}'[/tex] looks more or less like the attached curve. (Not drawn to scale)
Wherever the sign of [tex]{f_1}'[/tex] changes from negative to positive, as it does in the case of [tex]x\in\{1,5,9,13,17,21\}[/tex], there is a local minimum, so [tex]m_1 = 6[/tex]. Similarly, wherever the sign changes from negative to positive, as it does when [tex]x\in\{3,7,11,15,19\}[/tex], there is a local maximum, so [tex]n_1 = 5[/tex].
Repeat for [tex]f_2[/tex].
[tex]{f_2}'(x) = 4900 (x-1)^{49} - 29400 (x-1)^{48} = 4900 (x-1)^{48} (x-7)[/tex]
has roots at [tex]x=1[/tex] and [tex]x=7[/tex], but only the latter is a critical point due to odd multiplicity. A rough sketch of the plot of [tex]{f_2}'[/tex] is also attached. The sign of [tex]{f_2}'[/tex] changes from negative to positive at [tex]x=7[/tex], so [tex]m_2 = 1[/tex] and [tex]n_2=0[/tex].
We conclude that
Q.9) [tex]2m_1+3n_1+m_1n_1 = \boxed{57}[/tex]
Q.10) [tex]6m_2+4n_2+8m_2n_2 = \boxed{6}[/tex]
