| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 16 |
| Topic | Newton-Raphson method |
| Type | Iterative formula from rearrangement |
| Difficulty | Challenging +1.8 This is a substantial multi-part question requiring proof techniques, understanding of cubic behavior, reverse-engineering Newton-Raphson, and convergence analysis. While individual parts use standard A-level techniques (stationary points, intermediate value theorem), part (iii) requires algebraic insight about discriminant conditions, and part (iv) demands working backwards from an iteration formula to identify the original function—both requiring more sophistication than typical textbook exercises. The length and integration of multiple concepts elevates this above average difficulty. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| \(n\) | 1 | 2 |
| \(x_n - a\) | 0.32 | 0.068 |
(i) The function assumes both negative and positive values, so must cross the $x$-axis at some real value of $x$ [B1] **1 mark**
(ii)(a) As the leading term of $p(x)$ is positive, the local maximum is to the left of the local minimum (or a graphical demonstration) $\Rightarrow y_{\max} > 0$ and $y_{\min} < 0$ [M1]
$y = p(x)$ has three changes of sign $\Rightarrow$ 3 real roots [A1 (AG)] **2 marks**
(b) One (and only one) turning point is on the $x$-axis [M1]
There are two (distinct) real roots [A1] **2 marks**
(iii) The cubic has no turning points for any value of $c$ [M1]
Calculation of the discriminant of $p'(x)$: $\Delta = 4a^2 - 12b$ [A1]
Statement that $\Delta \leq 0$ [A1 (AG)] **3 marks**
(iv)(a) If $x_n \to a$ then $a$ satisfies $a = \frac{2a^3+1}{3a^2+1}$ [M1]
Rearrange as the cubic $x^3 + x - 1 = 0$ [A1] **2 marks**
(b) Use of the iterative scheme (as illustrated by the values of the first two iterates):
$x_1 = 1$; $x_2 = 0.75$ [M1]
$x_3 = 0.68604(6511)$; $x_4 = 0.68233(9582)$; $x_5 = 0.68232(7803)$ [A1]
$x_6 = 0.68232(7803)$
$a = 0.6823$ to 4dp, with sufficient iterates to justify [A1] **3 marks**
(c) Attempt to calculate the differences $x_n - a$ [M1]
| $n$ | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $x_n - a$ | 0.32 | 0.068 | 0.002 | 0.00004 |
[A1]
Doubling of number of correct decimal places indicative of a second order scheme [A1] **3 marks**9 The cubic polynomial $x ^ { 3 } + a x ^ { 2 } + b x + c$, where $a , b$ and $c$ are real, is denoted by $\mathrm { p } ( x )$.\\
(i) Give a reason why the equation $\mathrm { p } ( x ) = 0$ has at least one real root.\\
(ii) Suppose that the curve with equation $y = \mathrm { p } ( x )$ has a local minimum point and a local maximum point with $y$-coordinates $y _ { \text {min } }$ and $y _ { \text {max } }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Prove that if $y _ { \text {min } } y _ { \text {max } } < 0$, then the equation $\mathrm { p } ( x ) = 0$ has three real roots.
\item Comment on the number of distinct real roots of the equation $\mathrm { p } ( x ) = 0$ in the case $y _ { \text {min } } y _ { \text {max } } = 0$.\\
(iii) Suppose instead that the equation $\mathrm { p } ( x ) = 0$ has only one real root for all values of $c$. Prove that $a ^ { 2 } \leqslant 3 b$.\\
(iv) The iterative scheme
$$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 1 } { 3 x _ { n } ^ { 2 } + 1 } , \quad x _ { 0 } = 0$$
converges to a root of the cubic equation $\mathrm { p } ( x ) = 0$.\\
(a) Find $\mathrm { p } ( x )$.\\
(b) Find the limit of the iteration, correct to 4 decimal places.
\item Determine the rate of convergence of the iterative scheme.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q9 [16]}}