Question

`\: \: \: \: \: \: \: \: \: \: \:`​​

Answer 1

Find the critical points of
`f_1` and
`f_2`.

By the fundamental theorem of calculus,

`{f_1}'(x) = \displaystyle \prod_(j=1)^(21) (x-j)^j = (x-1) (x-2)^2 (x-3)^3 \cdots (x-21)^(21)`

and
`{f_1}'=0` for
`x\in\{1,2,3,\ldots,21\}`. The stationary points at odd values have odd multiplicity, while the even ones have even multiplicity. At the points of odd multiplicity,
`{f_1}'` passes through the
`x`-axis and the sign of
`{f_1}'` changes, while at the points of even multiplicity,
`{f_1}'` is tangent the to
`x`-axis. We also have

`\displaystyle \lim_(x\to-\infty) {f_1}'(x) = (-1)^(1+3+5+\cdots+21) \infty = (-1)^(11^2) \infty = -\infty`

`\displaystyle \lim_(x\to\infty) {f_1}'(x) = \infty`

so the plot of
`{f_1}'` looks more or less like the attached curve. (Not drawn to scale)

Wherever the sign of
`{f_1}'` changes from negative to positive, as it does in the case of
`x\in\{1,5,9,13,17,21\}`, there is a local minimum, so
`m_1 = 6`. Similarly, wherever the sign changes from negative to positive, as it does when
`x\in\{3,7,11,15,19\}`, there is a local maximum, so
`n_1 = 5`.

Repeat for
`f_2`.

`{f_2}'(x) = 4900 (x-1)^(49) - 29400 (x-1)^(48) = 4900 (x-1)^(48) (x-7)`

has roots at
`x=1` and
`x=7`, but only the latter is a critical point due to odd multiplicity. A rough sketch of the plot of
`{f_2}'` is also attached. The sign of
`{f_2}'` changes from negative to positive at
`x=7`, so
`m_2 = 1` and
`n_2=0`.

We conclude that

Q.9)
`2m_1+3n_1+m_1n_1 = \boxed{57}`

Q.10)
`6m_2+4n_2+8m_2n_2 = \boxed{6}`