| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 11 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Non-linear or complex iterative formula convergence |
| Difficulty | Challenging +1.8 This question requires proving properties of recurrence relations including monotonicity conditions and convergence analysis. Part (i) is routine calculation, but parts (ii)-(iv) demand algebraic manipulation to establish inequalities (showing x_{n+1} > x_n), understanding of limits, and generalizing to prove a condition involving parameters m and k. The final part requires completing the square or discriminant analysis to derive m² < 4k, which is non-trivial problem-solving beyond standard A-level fare, though the multi-part scaffolding provides guidance. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) If \(x_{n+1} > x_n\) where \( | x_n | < 1\) then \((x_{n+1})^2 > x_n^2\) where \( |
**(i)** Obtain to 3 s.f. or better 0.75, 0.854, 0.910, 0.943 B1 **[1]**
**(ii)** Obtain $x_{n+1} - x_n = \frac{x_n^2 - 3x_n + 2}{3}$ B1
Attempt to solve $\frac{x_n^2 - 3x_n + 2}{3} > 0$ or $x_n^2 - 3x_n + 2 > 0$ M1
Obtain critical values 2 and 1 or factors $(x-2)$ and $(x-1)$ A1
Attempt to deal with the inequality via a table of signs, graph or perhaps implied M1
Conclude convincingly $x_n > 2$ or $x_n < 1$ AG A1 **[5]**
**(iii)** If $x_{n+1} > x_n$ where $|x_n| < 1$ then $(x_{n+1})^2 > x_n^2$ where $|x_{n+1}| < 1$ M1
Only limit $\leq 1$ is 1 A1 **[2]**
**(iv)** To be strictly increasing, use the correct inequality sign with their $x_{n+1} - x_n$ M1
$\frac{x_n^2 - mx_n + k}{m} > 0$ or $x_n^2 - mx_n + k > 0$ M1
Clearly state that this requires $b^2 - 4ac < 0$ so $m^2 - 4k < 0$ and $m^2 < 4k$ A1 **[3]**11 A sequence of terms $x _ { n }$ generated by a recurrence relation is said to be strictly increasing if, for each $x _ { n } , x _ { n + 1 } > x _ { n }$.\\
(i) Let a recurrence relation be defined by
$$x _ { n + 1 } = \frac { x _ { n } ^ { 2 } + 2 } { 3 } \quad \text { and } \quad x _ { 0 } = \frac { 1 } { 2 } \quad \text { for } n \geq 0$$
Calculate $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$ correct to 3 significant figures where appropriate.\\
(ii) Given the recurrence relation
$$x _ { n + 1 } = \frac { x _ { n } ^ { 2 } + 2 } { 3 }$$
show that the sequence is strictly increasing when $x _ { n } > 2$ or $x _ { n } < 1$.\\
(iii) If $- 1 < x _ { 0 } < 1$, then the sequence $x _ { n } ( n \geq 0 )$ converges to a limit. Explain briefly why this limit is 1 .\\
(iv) Given the recurrence relation
$$x _ { n + 1 } = \frac { x _ { n } ^ { 2 } + k } { m } \text { with } m > 0$$
prove that $x _ { n }$ is a strictly increasing sequence for all $x _ { n }$ if $m ^ { 2 } < 4 k$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q11 [11]}}