| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a straightforward fixed-point iteration question requiring: (i) finding an integer root by substitution, (ii) algebraic rearrangement to isolate x, and (iii) applying a given iterative formula. All steps are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Find \(y\) for \(x = -2\) | M1 | |
| Obtain \(0\) and conclude that \(\alpha = -2\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Find cubic factor by division or inspection or equivalent | M1 | |
| Obtain \(x^3 + 2x - 8\) | A1 | |
| Rearrange to confirm given equation \(x = \sqrt[3]{8-2x}\) | A1 | |
| Or: Derive cubic factor from given equation and form product with \((x-\alpha)\): \((x+2)(x^3+2x-8)\) | M1, A1 | |
| Obtain quartic \(x^4 + 2x^3 + 2x^2 - 4x - 16\,(=0)\) | A1 | |
| Or: Derive cubic factor from given equation and divide the quartic by the cubic \((x^4+2x^3+2x^2-4x-16)\div(x^3+2x-8)\) | M1, A1 | |
| Obtain correct quotient and zero remainder | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use the given iterative formula correctly at least once | M1 | |
| Obtain final answer \(1.67\) | A1 | |
| Show sufficient iterations to at least 4 d.p. to justify answer \(1.67\) to 2 d.p., or show there is a change of sign in interval \((1.665,\, 1.675)\) | A1 | [3] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Find $y$ for $x = -2$ | M1 | |
| Obtain $0$ and conclude that $\alpha = -2$ | A1 | [2] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Find cubic factor by division or inspection or equivalent | M1 | |
| Obtain $x^3 + 2x - 8$ | A1 | |
| Rearrange to confirm given equation $x = \sqrt[3]{8-2x}$ | A1 | |
| **Or:** Derive cubic factor from given equation and form product with $(x-\alpha)$: $(x+2)(x^3+2x-8)$ | M1, A1 | |
| Obtain quartic $x^4 + 2x^3 + 2x^2 - 4x - 16\,(=0)$ | A1 | |
| **Or:** Derive cubic factor from given equation and divide the quartic by the cubic $(x^4+2x^3+2x^2-4x-16)\div(x^3+2x-8)$ | M1, A1 | |
| Obtain correct quotient and zero remainder | A1 | [3] |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the given iterative formula correctly at least once | M1 | |
| Obtain final answer $1.67$ | A1 | |
| Show sufficient iterations to at least 4 d.p. to justify answer $1.67$ to 2 d.p., or show there is a change of sign in interval $(1.665,\, 1.675)$ | A1 | [3] |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{adbef77f-e2ac-40ce-a56b-cf6776534ec1-3_561_732_255_705}
The diagram shows the curve $y = x ^ { 4 } + 2 x ^ { 3 } + 2 x ^ { 2 } - 4 x - 16$, which crosses the $x$-axis at the points $( \alpha , 0 )$ and $( \beta , 0 )$ where $\alpha < \beta$. It is given that $\alpha$ is an integer.\\
(i) Find the value of $\alpha$.\\
(ii) Show that $\beta$ satisfies the equation $x = \sqrt [ 3 ] { } ( 8 - 2 x )$.\\
(iii) Use an iteration process based on the equation in part (ii) to find the value of $\beta$ correct to 2 decimal places. Show the result of each iteration to 4 decimal places.
\hfill \mbox{\textit{CAIE P3 2012 Q6 [8]}}