| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find equation satisfied by limit |
| Difficulty | Standard +0.3 This question involves standard A-level techniques: (i) differentiation using product and chain rules to find a maximum, (ii) understanding that fixed points satisfy x = f(x) and algebraic manipulation to show a point lies on the curve, (iii) straightforward iteration. While it requires multiple steps and careful algebra, all techniques are routine for P3 level with no novel problem-solving insight required. |
| Spec | 1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | ||
| Use correct product or quotient rule and use chain rule at least once | M1 | |
| Obtain derivative in any correct form | A1 | |
| Equate derivative to zero and solve an equation with at least two non-zero terms for real \(x\) | M1 | |
| Obtain answer \(x = \frac{2}{\sqrt{2}}\), or exact equivalent | A1 | [4] |
| (ii) | ||
| State a suitable equation, e.g. \(\alpha = \sqrt{(\ln(4+8\alpha^2))}\) | B1 | |
| Rearrange to reach \(e^{\alpha^2} = 4 + 8\alpha^2\) | B1 | |
| Obtain \(\frac{1}{2} = e^{-\frac{1}{2}\alpha}\sqrt{(1+2\alpha^2)}\), or work vice versa | B1 | [3] |
| (iii) | ||
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer 1.86 | A1 | |
| Show sufficient iterations to 4 d.p. to justify 1.86 to 2 d.p., or show there is a sign change in the interval (1.855, 1.865) | A1 | [3] |
**(i)** | |
Use correct product or quotient rule and use chain rule at least once | M1 |
Obtain derivative in any correct form | A1 |
Equate derivative to zero and solve an equation with at least two non-zero terms for real $x$ | M1 |
Obtain answer $x = \frac{2}{\sqrt{2}}$, or exact equivalent | A1 | [4]
**(ii)** | |
State a suitable equation, e.g. $\alpha = \sqrt{(\ln(4+8\alpha^2))}$ | B1 |
Rearrange to reach $e^{\alpha^2} = 4 + 8\alpha^2$ | B1 |
Obtain $\frac{1}{2} = e^{-\frac{1}{2}\alpha}\sqrt{(1+2\alpha^2)}$, or work vice versa | B1 | [3]
**(iii)** | |
Use the iterative formula correctly at least once | M1 |
Obtain final answer 1.86 | A1 |
Show sufficient iterations to 4 d.p. to justify 1.86 to 2 d.p., or show there is a sign change in the interval (1.855, 1.865) | A1 | [3]8\\
\includegraphics[max width=\textwidth, alt={}, center]{7fe27759-d014-4bc6-8391-342d9df8280e-3_397_750_255_699}
The diagram shows the curve $y = \mathrm { e } ^ { - \frac { 1 } { 2 } x ^ { 2 } } \sqrt { } \left( 1 + 2 x ^ { 2 } \right)$ for $x \geqslant 0$, and its maximum point $M$.\\
(i) Find the exact value of the $x$-coordinate of $M$.\\
(ii) The sequence of values given by the iterative formula
$$x _ { n + 1 } = \sqrt { } \left( \ln \left( 4 + 8 x _ { n } ^ { 2 } \right) \right) ,$$
with initial value $x _ { 1 } = 2$, converges to a certain value $\alpha$. State an equation satisfied by $\alpha$ and hence show that $\alpha$ is the $x$-coordinate of a point on the curve where $y = 0.5$.\\
(iii) Use the iterative formula to determine $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\hfill \mbox{\textit{CAIE P3 2012 Q8 [10]}}