October 2020 Paper 1 Q6
6.
Given that in the expansion of \((1 + kx)^{10}\) the coefficient \(x^3\) is 3 times the coefficient of \(x\),
| Scheme | Marks | AO |
|---|---|---|
| \((1 + kx)^{10} = 1 + \dbinom{10}{1}(kx)^1 + \dbinom{10}{2}(kx)^2 + \dbinom{10}{3}(kx)^3 \ldots\) | M1 A1 | 1.1b 1.1b |
| \(= 1 + 10kx + 45k^2x^2 + 120k^3x^3 \ldots\) | A1 | 1.1b |
| (3) |
Notes
M1: An attempt at the binomial expansion. This may be awarded for either the second or third term or fourth term. The coefficients may be of the form \({}^{10}\mathrm{C}_1\), \(\dbinom{10}{2}\) etc or eg \(\dfrac{10 \times 9 \times 8}{3!}\)
A1: A correct unsimplified binomial expansion. The coefficients must be numerical so cannot be of the form \({}^{10}\mathrm{C}_1\), \(\dbinom{10}{2}\). Coefficients of the form \(\dfrac{10 \times 9 \times 8}{3!}\) are acceptable for this mark.
The bracketing must be correct on \((kx)^2\) but allow recovery
A1: \(1 + 10kx + 45k^2x^2 + 120k^3x^3 \ldots\) or \(1 + 10(kx) + 45(kx)^2 + 120(kx)^3 \ldots\)
Allow if written as a list.
| Scheme | Marks | AO |
|---|---|---|
| Sets \(120k^3 = 3 \times 10k\) | B1 | 1.2 |
| \(4k^2 = 1 \Rightarrow k = \ldots\) | M1 | 1.1b |
| \(k = \pm\dfrac{1}{2}\) | A1 | 1.1b |
| (3) | ||
| (6 marks) |
Notes
B1: Sets their \(120k^3 = 3 \times\) their \(10k\) (Seen or implied)
For candidates who haven’t cubed allow \(120k = 3 \times\) their \(10k\)
If they write \(120k^3x^3 = 3 \times\) their \(10kx\) only allow recovery of this mark if \(x\) disappears afterwards.
M1: Solves a cubic of the form \(Ak^3 = Bk\) by factorising out/cancelling the \(k\) and proceeding correctly to at least one value for \(k\). Usually \(k = \sqrt{\dfrac{B}{A}}\)
A1: \(k = \pm\dfrac{1}{2}\) o.e ignoring any reference to 0