AS June 2023 Paper 1 Q1
1 The roots of the equation \(4x^4 - 2x^3 - 3x + 2 = 0\) are \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\). By using a suitable substitution, find a quartic equation whose roots are \(\alpha + 2\), \(\beta + 2\), \(\gamma + 2\) and \(\delta + 2\) giving your answer in the form \(at^4 + bt^3 + ct^2 + dt + e = 0\), where \(a\), \(b\), \(c\), \(d\), and \(e\) are integers. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(u = x + 2\) | B1 | 1.1 |
| \((u - 2)^3 = u^3 - 3 \times 2u^2 + 3 \times 2^2u - 2^3\) | M1 | 1.1 |
| \((u - 2)^4 = u^4 - 4 \times 2u^3 + 6 \times 2^2u^2 - 4 \times 2^3u + 2^4\) | M1 | 1.1 |
| \(4(u - 2)^4 - 2(u - 2)^3 - 3(u - 2) + 2\ (= 0)\) | M1 | 1.1 |
| \(\therefore 4(u^4 - 8u^3 + 24u^2 - 32u + 16) - 2(u^3 - 6u^2 + 12u - 8) - 3(u - 2) + 2 = 0\) \(\therefore 4u^4 - 34u^3 + 108u^2 - 155u + 88 = 0\) | A1 | 1.1 |
| [5] |
Notes
B1: Correct substitution stated or used
M1: Attempt to expand their \((u - 2)^3\) following a linear substitution. 4 terms using \(\binom{n}{r}2^ru^{n-r}\)
\(u^3 - 6u^2 + 12u - 8\)
Binomial coefficients might not be correct or evaluated. Might be by expanding brackets.
M1: Attempt to expand their \((u - 2)^4\) following a linear substitution. 5 terms using \(\binom{n}{r}2^ru^{n-r}\)
\(u^4 - 8u^3 + 24u^2 - 32u + 16\)
Binomial coefficients might not be correct or evaluated. Might be by expanding brackets.
M1: Forming (LHS of) equation in \(u\) (could be using their expansions)
A1: Final answer can be in terms of \(x\)
Final answer needs to be an equation with coefficients simplified.
ISW attempts to cancel down following correct equation