AS June 2019 Paper 1 Q5
5 In this question you must show detailed reasoning.
You are given that \(\alpha\), \(\beta\) and \(\gamma\) are the roots of the equation \(5x^3 - 2x^2 + 3x + 1 = 0\).
| Scheme | Marks | AO |
|---|---|---|
| M1 | 1.1a | |
| \(\alpha\beta\gamma = -\dfrac{1}{5},\ \sum\alpha\beta = \dfrac{3}{5},\ \sum\alpha = \dfrac{2}{5}\) | A1 | 1.1 |
| \((\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum\alpha^2\beta^2 + \sum 2\alpha\beta\gamma\alpha\) | M1 | 3.1a |
| \((\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum\alpha^2\beta^2 + 2\alpha\beta\gamma\sum\alpha\) | A1 | 1.1 |
| \(\sum\alpha^2\beta^2 = \dfrac{13}{25}\) | A1 | 1.1 |
| [5] |
Notes
DR: Detailed reasoning required for this question
See appendix for Special Cases for this question (below).
M1: (1st) Any one of \(\alpha\beta\gamma\), \(\alpha\beta + \beta\gamma + \gamma\alpha\), \(\alpha + \beta + \gamma\)
A1: (1st) All three correct
M1: (2nd) Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in \(\alpha\), \(\beta\) and \(\gamma\). Might have missing 2
A1: (2nd) Correct, useful form
Could see early substitution (see alternative scheme)
Alternate
| Scheme | Marks |
|---|---|
| Substitution \(x = \sqrt{u}\) or \(x^2 = u\) used | B1 |
| \(5u\sqrt{u} - 2u + 3\sqrt{u} + 1 = 0\) | M1* |
| \(\left((5u + 3)\sqrt{u}\right)^2 = (2u - 1)^2\) | M1* |
| \(25x^3 + 26x^2 + 13x - 1 = 0\) | A1 |
| \(\sum\alpha^2\beta^2 = \dfrac{13}{25}\) | A1ft dep(*) |
B1: Soi by correct substitution – must be used
M1*: (1st) Substitution and dealing with \(\left(\sqrt{u}\right)^3\) term
M1*: (2nd) Rearrangement with \(\sqrt{u}\) terms collected on one side and squaring. LHS does not need to be factorised before squaring
A1: Correct expansion and rearranging. Could be in terms of “\(u\)”. Or any non-zero integer multiple. Must be an equation.
Must see the correct entire cubic for this mark if using this method.
A1ft dep(*): Correct coefficient ratio identified. Must gain method marks and have a cubic.
Cubic might not be completely correct but coefficients of \(x^3\) and \(x\) must be.
Appendix: Special cases for question 5 using the substitution method
Working is shown in part (b) and no working shown in part (a).
1. Working in part (b) is fully correct. Then award:
5/5 if a reference to the working in part (b) is made, the correct cubic equation is written down, there is a comment stating that the roots of the new cubic are \(\alpha^2, \beta^2, \gamma^2\) and the correct answer is given.
4/5 as above but no comment about the roots of the new cubic equation being \(\alpha^2, \beta^2, \gamma^2\) or there is no reference to the working in part (b) (i.e. one element of explanation is missing).
3/5 if the correct cubic equation is written down and correct answer of \(\dfrac{13}{25}\)
3/5 if a reference is made to working in part (b) but the cubic equation is not re-written and no comment about the roots being \(\alpha^2, \beta^2, \gamma^2\) (and correct answer of \(\dfrac{13}{25}\) is given)
2. Working in part (b) is not fully correct.
3/5 if a reference to the working in part (b) is made, the same cubic equation from part (b) is written down, there is a comment stating that the roots of the new cubic are \(\alpha^2, \beta^2, \gamma^2\) and the correct follow through coefficient ratio is given.
2/5 as above but no comment about the roots of the new cubic equation being \(\alpha^2, \beta^2, \gamma^2\) or there is no reference to the working in part (b) (i.e. one element of explanation is missing).
1/5 if same cubic equation as in part (b) is written down and the correct follow through coefficient ratio is given.
3. If \(\dfrac{13}{25}\) appears as an answer in part (a) with no supporting working or comments then 0/5
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = \dfrac{1}{25}\) | B1ft | 2.2a |
| \((\alpha + \beta + \gamma)^2 = \sum\alpha^2 + 2\sum\alpha\beta\) | M1 | 3.1a |
| \(\sum\alpha^2 = -\dfrac{26}{25}\) | A1 | 1.1 |
| \(25x^3 + 26x^2 + 13x - 1 = 0\) | A1 | 1.1 |
| [4] |
Notes
See appendix for Special Cases for this question (below).
M1: Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in \(\alpha\), \(\beta\) and \(\gamma\). Might have 2 missing.
\(\sum\alpha^2 = \left(\tfrac{2}{5}\right)^2 - 2 \times \tfrac{3}{5}\)
A1: (1st) Seen or implied
A1: (2nd) Or any non-zero integer multiple. Must be = 0. CAO
Alternate
| Scheme | Marks |
|---|---|
| Substitution \(x = \sqrt{u}\) or \(x^2 = u\) used | B1 |
| \(5u\sqrt{u} - 2u + 3\sqrt{u} + 1 = 0\) | M1 |
| \(\left((5u + 3)\sqrt{u}\right)^2 = (2u - 1)^2\) | M1 |
| \(25x^3 + 26x^2 + 13x - 1 = 0\) | A1 |
B1: Soi by correct substitution - must be used
M1: (1st) Substitution and dealing with \(\left(\sqrt{u}\right)^3\) term
M1: (2nd) Rearrangement with \(\sqrt{u}\) terms collected on one side and squaring. LHS does not need to be factorised before squaring
A1: Correct expansion and rearranging. Could be in terms of “\(u\)”. Or any non-zero integer multiple. Must be = 0. CAO
Appendix: Special cases for question 5 using the substitution method
Working is shown in part (a). If working shown in part (b) mark as mark scheme above. If little or no working then mark part (b) as:
1. Working in (a) is fully correct. Then award:
4/4 If correct cubic equation is re-written and there is a reference to the working in part (a) such as “See working above” or arrows etc.
3/4 If correct cubic equation re-written and there is no reference to the working in part (a)
0/4 If an incorrect cubic is written down, or “=0” missing
2. Working in (a) is not fully correct. Then award:
0/4 if B1 M1 M1 is not awarded in part (a) (i.e. at least three marks given for (a))
3/4 if at least three marks awarded in part (a) and reference is made to the working in part (a) and the same cubic as in part (a) is re-written in part (b) with and “=0”
2/4 if at least three marks awarded in part (a) and the same cubic as in part (a) is re-written in part (b) with and “=0” but no reference is made to the working in part (a)