October 2021 Paper 1 Q4
4 In this question you must show detailed reasoning.
The cubic polynomial \(\mathrm{f}(x)\) is defined by \(\mathrm{f}(x) = 2x^3 - 3x^2 - 11x + 6\).
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathrm{f}(0.5) = 0.25 - 0.75 - 5.5 + 6 = 0\) | B1 | 2.1 |
| [1] |
Notes
B1: Attempt \(\mathrm{f}(0.5)\) and show equal to 0
Must be using factor theorem so B0 for alternative methods
B0 for just \(\mathrm{f}(0.5) = 0\)
Condone \(2(0.5)^3 - 3(0.5)^2 - 11(0.5) + 6 = 0\)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\mathrm{f}(x) = (2x - 1)(x^2 - x - 6)\) | M1 A1 | 1.1 1.1 |
| \(\mathrm{f}(x) = (2x - 1)(x - 3)(x + 2)\) | A1 | 1.1 |
| [3] |
Notes
M1: Attempt complete division by \((2x - 1)\)
DR so need to see quadratic factor
Allow equivalent complete methods eg coefficient matching / inspection / grid method
Condone slip(s) in otherwise correct method
A1: Obtain correct quadratic factor
Seen in division / correct coeffs eg \(A = 1\) etc / at top of grid
A1: Obtain correct fully factorised \(\mathrm{f}(x)\)
Must be seen as a product of all 3 factors
SC B1 for correct factorisation with no DR
| Scheme | Marks | AO |
|---|---|---|
| DR \(x = 2^y\) | B1 | 3.1a |
| \(2^y = 0.5\), \(y = -1\) \(2^y = 3\), \(y = 1.58\) | M1 | 1.1 |
| \(2^y = -2\), no solutions as \(2^y \gt 0\) for all \(y\) Hence \(y = -1\), \(y = \log_2 3\) | A1 | 2.4 |
| [3] |
Notes
B1: State or imply that \(x = 2^y\)
Could be implied by equating \(2^y\) to at least one of their roots
M1: Attempt to find at least one value of \(y\)
Exact or decimal
A1: Obtain both correct values, and no others
1.58 or better, or \(\dfrac{\log_n 3}{\log_n 2}\) for \(\log_2 3\)
Must give reason for \(2^y = -2\) having no solution
eg cannot log a negative number
\(2^y\) always greater than 0