| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a standard fixed-point iteration question requiring routine algebraic rearrangement, sign-change verification, and iterative calculation. While it involves multiple parts and requires careful arithmetic, all techniques are textbook procedures with no novel insight needed. The rearrangement in part (i) is straightforward, and the iteration follows a standard algorithm. Slightly above average difficulty due to the multi-step nature and need for precision, but well within typical A-level pure maths scope. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)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 | Mark | Guidance |
| Rearrange at least as far as \(2x=\ln(...)\) | M1 | Allow if in terms of \(p\), need to see \(y\) equated to \(0\) |
| Obtain \(x=\frac{1}{2}\ln(1.6x^2+4)\) | A1 | AG; necessary detail needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either: Consider sign of \(x-\frac{1}{2}\ln(1.6x^2+4)\) for \(0.75\) and \(0.85\) or equivalent | M1 | Need to see substitution of numbers |
| Obtain \(-0.04\) and \(0.03\) or equivalents and justify conclusion | A1 | AG; necessary detail needed, change of sign or equivalent must be mentioned |
| Or: Consider sign of \(5e^{2x}-8x^2-20\) for \(0.75\) and \(0.85\) | M1 | Need to see substitution of numbers |
| Obtain \(-2.09...\) and \(1.58...\) or equivalents and justify conclusion | A1 | AG; necessary detail needed, change of sign or equivalent must be mentioned |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use iteration process correctly at least once | M1 | Starting with value such that iterations converge to correct values |
| Obtain final value \(0.80956\) | A1 | Must be 5sf for the final answer |
| Show sufficient iterations to justify value or show sign change in interval \((0.809555,\ 0.809565)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain first derivative \(10e^{2x}-16x\) | B1 | |
| Substitute value from iteration to find gradient, must be in the form \(pe^{2x}+qx\) | M1 | |
| Obtain \(37.5\) | A1 | Or greater accuracy, allow awrt \(37.5\) from use of \(x=0.8096\), \(0.80955\) oe |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Rearrange at least as far as $2x=\ln(...)$ | M1 | Allow if in terms of $p$, need to see $y$ equated to $0$ |
| Obtain $x=\frac{1}{2}\ln(1.6x^2+4)$ | A1 | AG; necessary detail needed |
---
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either:** Consider sign of $x-\frac{1}{2}\ln(1.6x^2+4)$ for $0.75$ and $0.85$ or equivalent | M1 | Need to see substitution of numbers |
| Obtain $-0.04$ and $0.03$ or equivalents and justify conclusion | A1 | AG; necessary detail needed, change of sign or equivalent must be mentioned |
| **Or:** Consider sign of $5e^{2x}-8x^2-20$ for $0.75$ and $0.85$ | M1 | Need to see substitution of numbers |
| Obtain $-2.09...$ and $1.58...$ or equivalents and justify conclusion | A1 | AG; necessary detail needed, change of sign or equivalent must be mentioned |
---
## Question 5(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use iteration process correctly at least once | M1 | Starting with value such that iterations converge to correct values |
| Obtain final value $0.80956$ | A1 | Must be 5sf for the final answer |
| Show sufficient iterations to justify value or show sign change in interval $(0.809555,\ 0.809565)$ | A1 | |
---
## Question 5(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain first derivative $10e^{2x}-16x$ | B1 | |
| Substitute value from iteration to find gradient, must be in the form $pe^{2x}+qx$ | M1 | |
| Obtain $37.5$ | A1 | Or greater accuracy, allow awrt $37.5$ from use of $x=0.8096$, $0.80955$ oe |
---5 The curve with equation
$$y = 5 \mathrm { e } ^ { 2 x } - 8 x ^ { 2 } - 20$$
crosses the $x$-axis at only one point. This point has coordinates $( p , 0 )$.\\
(i) Show that $p$ satisfies the equation $x = \frac { 1 } { 2 } \ln \left( 1.6 x ^ { 2 } + 4 \right)$.\\
(ii) Show by calculation that $0.75 < p < 0.85$.\\
(iii) Use an iterative formula based on the equation in part (i) to find the value of $p$ correct to 5 significant figures. Give the result of each iteration to 7 significant figures.\\
(iv) Find the gradient of the curve at the point $( p , 0 )$.\\
\hfill \mbox{\textit{CAIE P2 2018 Q5 [10]}}