| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Proof by contradiction |
| Difficulty | Moderate -0.5 This is a straightforward proof by contradiction requiring students to assume the terms form a GP, set up the common ratio equation (1+2k)/k = (3+3k)/(1+2k), cross-multiply to get a quadratic, and show it has no real solutions using the discriminant. While it requires understanding of GP definition and proof structure, the algebraic manipulation is routine and the question explicitly tells students to use proof by contradiction, removing any strategic thinking about proof method. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Assume the sequence is geometric | B1 | States an appropriate assumption to set up the contradiction |
| So \(r = \frac{1+2k}{k} = \frac{3+3k}{1+2k}\) | M1 | Uses the assumption to set up an equation in \(k\) only. Allow equivalent work e.g. \(kr = 1+2k\), \(kr^2 = 3+3k\). Allow use of \(\neq\) for \(=\) |
| \(\Rightarrow (1+2k)^2 = k(3+3k) \Rightarrow k^2+k+1=0\) | A1 | Reaches a correct quadratic in \(k\); terms need not all be on one side but terms in \(k\) and \(k^2\) should be collected |
| But \(k^2+k+1 = \left(k+\frac{1}{2}\right)^2 - \frac{1}{4} + 1 \geqslant \frac{3}{4} > 0\) since \(\left(k+\frac{1}{2}\right)^2 \geqslant 0\) for all real \(k\) | dM1 | Completes the square, considers discriminant or other valid means to reach a contradiction. E.g. \(b^2-4ac = 1^2-4(1)(1) = -3 < 0\). Depends on previous M |
| This is a contradiction and hence the original assumption is not true. The sequence is not geometric. | A1 | Correct work leading to contradiction with deduction and conclusion given. Available even if B0 given at start, so 01111 is possible |
## Question 6:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Assume the sequence is geometric | B1 | States an appropriate assumption to set up the contradiction |
| So $r = \frac{1+2k}{k} = \frac{3+3k}{1+2k}$ | M1 | Uses the assumption to set up an equation in $k$ only. Allow equivalent work e.g. $kr = 1+2k$, $kr^2 = 3+3k$. Allow use of $\neq$ for $=$ |
| $\Rightarrow (1+2k)^2 = k(3+3k) \Rightarrow k^2+k+1=0$ | A1 | Reaches a correct quadratic in $k$; terms need not all be on one side but terms in $k$ and $k^2$ should be collected |
| But $k^2+k+1 = \left(k+\frac{1}{2}\right)^2 - \frac{1}{4} + 1 \geqslant \frac{3}{4} > 0$ since $\left(k+\frac{1}{2}\right)^2 \geqslant 0$ for all real $k$ | dM1 | Completes the square, considers discriminant or other valid means to reach a contradiction. E.g. $b^2-4ac = 1^2-4(1)(1) = -3 < 0$. Depends on previous M |
| This is a contradiction and hence the original assumption is not true. The sequence is not geometric. | A1 | Correct work leading to contradiction with deduction and conclusion given. Available even if B0 given at start, so 01111 is possible |
---\begin{enumerate}
\item Three consecutive terms in a sequence of real numbers are given by
\end{enumerate}
$$k , 1 + 2 k \text { and } 3 + 3 k$$
where $k$ is a constant.
Use proof by contradiction to show that this sequence is not a geometric sequence.\\
\hfill \mbox{\textit{Edexcel P4 2022 Q6 [5]}}