| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find coordinate from gradient condition |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on fixed-point iteration. Part (i) requires basic differentiation and algebraic manipulation (product rule, rearranging). Part (ii) is simple substitution to verify a sign change. Part (iii) applies a given iterative formula repeatedly—a routine procedure requiring no insight. All steps are standard A-level techniques with clear guidance, making this slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | Use product rule on a correct expression | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain correct derivative in any form | A1 | dy ( ) x |
| Answer | Marks | Guidance |
|---|---|---|
| ln 8−x | A1 | Given answer: check carefully that it follows from correct |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | Calculate values of a relevant expression or pair of relevant | |
| expressions at x = 2.9 and x = 3.1 | M1 | 8 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Complete the argument correctly with correct calculated values | A1 | Note: valid to consider gradient at 2.9 (1.06..) and 3.1 (0.95..) |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(iii) | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| SR: Clear successive use of 0, 1, 2, 3 etc., or equivalent, scores M0. | M1 | ( ) |
| Answer | Marks |
|---|---|
| Obtain final answer 3.02 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| or show there is a sign change in the interval (3.015, 3.025) | A1 | Must have two consecutive values rounding correctly to 3.02 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | Use product rule on a correct expression | M1 | Condone with + x unless there is clear evidence of incorrect
8−x
product rule.
Obtain correct derivative in any form | A1 | dy ( ) x
=ln 8−x −
dx 8−x
8
Equate derivative to 1 and obtain x=8−
( )
ln 8−x | A1 | Given answer: check carefully that it follows from correct
working
Condone the use of a for x throughout
3
--- 5(ii) ---
5(ii) | Calculate values of a relevant expression or pair of relevant
expressions at x = 2.9 and x = 3.1 | M1 | 8 8
8− =3.09>2.9, 8− =2.97<3.1
ln5.1 ln4.9
Clear linking of pairs needed for M1 by this method
(0.19 and -0.13)
Complete the argument correctly with correct calculated values | A1 | Note: valid to consider gradient at 2.9 (1.06..) and 3.1 (0.95..)
and comment on comparison with 1
2
Question | Answer | Marks | Guidance
--- 5(iii) ---
5(iii) | 8
Use the iterative process x = 8 – correctly to find at
n+1 ( )
ln 8−x
n
least two successive values.
SR: Clear successive use of 0, 1, 2, 3 etc., or equivalent, scores M0. | M1 | ( )
3,3.0293,3.0111,3.0225,3.0154, 3.0198
2.9,3.0897,2.9728,3.0460,3.0006,3.290,3.0113,3.0223,3.0155
3.1,2.9661,3.0501,2.9980,3.0305,3.0103,3.0229,3.0151
Allow M1 if values given to fewer than 4 dp
Obtain final answer 3.02 | A1
Show sufficient iterations to at least 4 d.p. to justify 3.02 to 2 d.p.,
or show there is a sign change in the interval (3.015, 3.025) | A1 | Must have two consecutive values rounding correctly to 3.02
3
Question | Answer | Marks | GuidanceThe equation of a curve is $y = x \ln(8 - x)$. The gradient of the curve is equal to 1 at only one point, when $x = a$.
\begin{enumerate}[label=(\roman*)]
\item Show that $a$ satisfies the equation $x = 8 - \frac{8}{\ln(8 - x)}$. [3]
\item Verify by calculation that $a$ lies between 2.9 and 3.1. [2]
\item Use an iterative formula based on the equation in part (i) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2018 Q5 [8]}}