June 2024 Paper 1 Q1
1. \[\mathrm{g}(x) = 3x^3 - 20x^2 + (k+17)x + k\]
where \(k\) is a constant.
Given that \((x-3)\) is a factor of \(\mathrm{g}(x)\), find the value of \(k\). (3)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{g}(3) = 3(3)^3 - 20(3)^2 + 3(k+17) + k = 0\) | M1 | 3.1a |
| \(4k - 48 = 0 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(\{k=\}\,12\) | A1 | 1.1b |
| (3) | ||
| (3 marks) |
Notes
Note: Ignore any use of \(\mathrm{f}(x)\) in place of \(\mathrm{g}(x)\) throughout.
M1: Attempts \(\mathrm{g}(3) = 0\) to set up a linear equation in \(k\). The = 0 may implied by their value of \(k\).
Expect to see 3 substituted for \(x\) at least twice but condone minor slips copying the function. May be scored for e.g. \(81 - 180 + 3(k+17) + k = 0\)
Missing brackets may be recovered.
Attempting \(\mathrm{g}(-3) = 0\) scores M0 but note that the second M1 is available.
If algebraic division is attempted, they need to achieve a linear remainder in \(k\) only and set =0 Condone slips in their calculations.
As a minimum, expect to see \(3x^2 + \lambda x,\ \lambda \neq 0\) as their quotient leading to a linear remainder in k only set = 0 (the = 0 may be implied by their value for \(k\)).
For reference, the correct division is
\[\begin{array}{rrrrr} & & 3x^2 & -11x & +\;k-16 \\x-3\,\big) & 3x^3 & -20x^2 & +(k+17)x & +\;k \\ & \underline{3x^3} & \underline{-9x^2} & & \\ & & -11x^2 & +(k+17)x & +\;k \\ & & \underline{-11x^2} & \underline{+33x} & \\ & & & (k-16)x & +\;k \\ & & & \underline{(k-16)x} & \underline{-3k+48} \\ & & & & 4k-48=0\end{array}\]You may also see variations on the table below.
Here, the M1 is scored when the sum of both coefficients of \(x\) are equated to \((k+17)\)
| \(3x^2\) | \(-11x\) | \(-\dfrac{k}{3}\) | |
|---|---|---|---|
| \(x\) | \(3x^3\) | \(-11x^2\) | \(-\dfrac{k}{3}x\) |
| \(-3\) | \(-9x^2\) | \(33x\) | \(k\) |
\(33-\dfrac{k}{3}=k+17\) scores M1
M1: Scored for attempting to solve a linear equation in \(k\) having attempted \(\mathrm{g}(\pm 3) = 0\)
Do not be concerned about the process, e.g. \(-81 + 180 - 3(k+17) + k = 0 \rightarrow k = \ldots\) scores M1.
Via division they must have a linear remainder in \(k\) set = 0
The = 0 may be implied by their value for \(k\) in all approaches.
A1: Obtains \(\{k=\}\,12\) only. Do not accept e.g. \(\dfrac{48}{4}\) Allow slips in working to be recovered.
Condone e.g. \(x = 12\) provided it has come from a linear equation in \(k\).
Note that e.g. \(3(3)^3 - 20(3)^2 + 3(k+17) + k\{=0\} \rightarrow k = 12\) and \(81 - 180 + 3k + 51 + k\{=0\} \rightarrow k = 12\) are sufficient to imply M1M1A1.