Higher November 2020 Paper 4 Q15
15 Here are two pieces of work.
For each one, describe the error made and give the complete correct solution.
(a)
Question:
Solve by factorisation.
\(3x^2 - 2x - 5 = 0\)
Solution:
\((3x + 5)(x - 1) = 0\)
Therefore \(x = {}^-5/3\) or \(x = 1\)
Error:
Correct solution: [3]
(b)
Question:
Solve, giving your answers correct to 3 significant figures.
\(2x^2 - 8x + 3 = 0\)
Solution:
\(x = -({}^-8) \pm \dfrac{\sqrt{({}^-8)^2 - 4 \times 2 \times 3}}{2 \times 2}\)
Therefore \(x = 6.42\) or \(x = 9.58\)
Error:
Correct solution: [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct reason e.g. the factors give \(+2x\) or factors are \((3x - 5)(x + 1)\) or the signs are the wrong way round | B1 | See appendix | |
| \((3x - 5)(x + 1)\) and \(-1\) and \(\frac{5}{3}\) oe | B2 | B1 for the correct factorisation or two correct solutions FT from their incorrect factorisation | |
Appendix: Exemplar responses for Q15(a)
| Response | Mark |
|---|---|
| the factors give +2x | 1 |
| factors are \((3x - 5)(x + 1)\) | 1 |
| the signs are the wrong way round | 1 |
| Factorised incorrectly | 1 BOD |
| \(-\frac{5}{3}\) should be positive | 1 BOD |
| \(x - 1\) should be \(x + 1\) | 1 |
| the symbols are the wrong way round | 1 BOD |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Correct reason e.g. the \(-b\) term should be in the numerator or \(\dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 2 \times 3}}{2 \times 2}\) | B1 | allow [+] 8 for \(-(-8)\) throughout this part see appendix | |
| \(\dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 2 \times 3}}{2 \times 2}\) and 0.419 and 3.58 | B2 | B1 for \(\dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 2 \times 3}}{2 \times 2}\) or 0.419 and 3.58 or 0.4188..., 0.4189 or 0.419 and 3.58[1...] | |
Appendix: Exemplar responses for Q15(b)
| Response | Mark |
|---|---|
| the −b term should be in the numerator | 1 |
| dividing line [in fraction] is not long enough | 1 BOD |