| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.8 This question requires finding dy/dx using product and chain rules, setting it to zero, then algebraically manipulating to derive a specific iterative form—requiring careful exponential/logarithm manipulation. The iteration itself is routine, but deriving the rearrangement and verifying convergence elevates this above standard calculus exercises. |
| Spec | 1.07l Derivative of ln(x): and related functions1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct product rule | M1 | |
| Obtain correct derivative in any form: \(\frac{dy}{dx} = -2e^{-2x}\ln(x-1) + \frac{e^{-2x}}{x-1}\) | A1 | |
| Equate derivative to zero and derive \(x = 1 + e^{\frac{1}{2(x-1)}}\) or \(p = 1 + \frac{1}{2(p-1)}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculate values of a relevant expression or pair of relevant expressions at \(x = 2.2\) and \(x = 2.6\): \(f(x) = \ln(x-1) - \frac{1}{2(x-1)} \Rightarrow f(2.2) = -0.234,\ f(2.6) = 0.317\) or \(f(x) = 2e^{-2x}\ln(x-1) + \frac{e^{-2x}}{x-1} \Rightarrow f(2.2) = 0.005\ldots,\ f(2.6) = -0.0017\ldots\) | M1 | |
| Complete the argument correctly with correct calculated values | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the iterative process \(p_{n+1} = 1 + \exp\!\left(\dfrac{1}{2(p_n - 1)}\right)\) correctly at least once | M1 | |
| Obtain final answer 2.42 | A1 | |
| Show sufficient iterations to 4 d.p. to justify 2.42 to 2 d.p., or show there is a sign change in the interval \((2.415,\ 2.425)\) | A1 |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct product rule | M1 | |
| Obtain correct derivative in any form: $\frac{dy}{dx} = -2e^{-2x}\ln(x-1) + \frac{e^{-2x}}{x-1}$ | A1 | |
| Equate derivative to zero and derive $x = 1 + e^{\frac{1}{2(x-1)}}$ or $p = 1 + \frac{1}{2(p-1)}$ | A1 | AG |
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate values of a relevant expression or pair of relevant expressions at $x = 2.2$ and $x = 2.6$: $f(x) = \ln(x-1) - \frac{1}{2(x-1)} \Rightarrow f(2.2) = -0.234,\ f(2.6) = 0.317$ or $f(x) = 2e^{-2x}\ln(x-1) + \frac{e^{-2x}}{x-1} \Rightarrow f(2.2) = 0.005\ldots,\ f(2.6) = -0.0017\ldots$ | M1 | |
| Complete the argument correctly with correct calculated values | A1 | |
## Question 5(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative process $p_{n+1} = 1 + \exp\!\left(\dfrac{1}{2(p_n - 1)}\right)$ correctly at least once | M1 | |
| Obtain final answer 2.42 | A1 | |
| Show sufficient iterations to 4 d.p. to justify 2.42 to 2 d.p., or show there is a sign change in the interval $(2.415,\ 2.425)$ | A1 | |
---5 The curve with equation $y = \mathrm { e } ^ { - 2 x } \ln ( x - 1 )$ has a stationary point when $x = p$.\\
(i) Show that $p$ satisfies the equation $x = 1 + \exp \left( \frac { 1 } { 2 ( x - 1 ) } \right)$, where $\exp ( x )$ denotes $\mathrm { e } ^ { x }$.\\
(ii) Verify by calculation that $p$ lies between 2.2 and 2.6.\\
(iii) Use an iterative formula based on the equation in part (i) to determine $p$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.\\
\hfill \mbox{\textit{CAIE P3 2019 Q5 [8]}}