| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Sketch graphs to show root existence |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on iterative methods requiring standard techniques: sketching y=ln(x+1) and y=40-x³ to show intersection, sign change verification, applying a given iteration formula (no rearrangement needed), and a simple substitution y=ln(x+1) to deduce the final root. All parts are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch increasing curve with correct curvature passing through origin, for \(x \geq 0\) | B1 | |
| Recognisable sketch of \(y = 40 - x^3\), with equation stated, for \(x > 0\) | B1 | |
| Indicate in some way the one intersection, dependent on both curves being roughly correct and both existing for some \(x < 0\) | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Consider signs of \(x^3 + \ln(x+1) - 40\) at \(3\) and \(4\) or equivalent or compare values of relevant expressions for \(x = 3\) and \(x = 4\) | M1 | |
| Complete argument correctly with correct calculations (\(-11.6\) and \(25.6\)) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(3.377\) | A1 | |
| Show sufficient iterations to justify accuracy to 3 d.p. or show sign change in interval (3.3765, 3.3775) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt value of \(\ln(x+1)\) | M1 | |
| Obtain \(1.48\) | A1 | [2] |
**(i)**
Sketch increasing curve with correct curvature passing through origin, for $x \geq 0$ | B1 |
Recognisable sketch of $y = 40 - x^3$, with equation stated, for $x > 0$ | B1 |
Indicate in some way the one intersection, dependent on both curves being roughly correct and both existing for some $x < 0$ | B1 | [3]
**(ii)**
Consider signs of $x^3 + \ln(x+1) - 40$ at $3$ and $4$ or equivalent or compare values of relevant expressions for $x = 3$ and $x = 4$ | M1 |
Complete argument correctly with correct calculations ($-11.6$ and $25.6$) | A1 | [2]
**(iii)**
Use the iterative formula correctly at least once | M1 |
Obtain final answer $3.377$ | A1 |
Show sufficient iterations to justify accuracy to 3 d.p. or show sign change in interval (3.3765, 3.3775) | A1 | [3]
**(iv)**
Attempt value of $\ln(x+1)$ | M1 |
Obtain $1.48$ | A1 | [2]9 (i) Sketch the curve $y = \ln ( x + 1 )$ and hence, by sketching a second curve, show that the equation
$$x ^ { 3 } + \ln ( x + 1 ) = 40$$
has exactly one real root. State the equation of the second curve.\\
(ii) Verify by calculation that the root lies between 3 and 4 .\\
(iii) Use the iterative formula
$$x _ { n + 1 } = \sqrt [ 3 ] { } \left( 40 - \ln \left( x _ { n } + 1 \right) \right)$$
with a suitable starting value, to find the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.\\
(iv) Deduce the root of the equation
$$\left( \mathrm { e } ^ { y } - 1 \right) ^ { 3 } + y = 40$$
giving the answer correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P3 2014 Q9 [10]}}