CAIE P3 2020 Specimen — Question 4 9 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeIterative method for parameter value
DifficultyStandard +0.3 This is a straightforward iterative method question requiring standard parametric differentiation to derive the equation, then mechanical application of an iteration formula. The derivation involves routine dy/dx calculation and algebraic rearrangement, while the iteration is purely computational with no conceptual challenges—slightly easier than average A-level.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t - 3 } , \quad y = 4 \ln t$$ where \(t > 0\). When \(t = a\) the gradient of the curve is 2 .
  1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 2 } ( 3 - \ln a )\).
  2. Verify by calculation that this equation has a root between 1 and 2 .
  3. Use the iterative formula \(a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

Question 4(a):
AnswerMarks Guidance
AnswerMarks Guidance
State or imply \(\frac{dx}{dt} = 2e^{2t-3}\) or \(\frac{dy}{dt} = \frac{4}{t}\)B1
Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\)M1
Obtain \(\frac{dy}{dx} = \frac{4}{2te^{2t-3}}\), or equivalentA1
Set \(t = a\), equate gradient to 2 and obtain the given answerA1 AG
Total4
Question 4(b):
AnswerMarks Guidance
AnswerMarks Guidance
Calculate \(a - \frac{1}{2}(3 - \ln a)\) when \(a = 1\) and \(a = 2\), or equivalentM1
Complete the argument by considering the signs of the correct calculated valuesA1
Total2
Question 4(c):
AnswerMarks Guidance
AnswerMarks Guidance
Use the iterative formula correctly at least onceM1
Obtain final answer 1.35A1
Show sufficient iterations to 4 dp to justify 1.35 to 2 dp, OR show there is a sign change in the interval (1.345, 1.355)A1
Total3
## Question 4(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $\frac{dx}{dt} = 2e^{2t-3}$ or $\frac{dy}{dt} = \frac{4}{t}$ | B1 | |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 | |
| Obtain $\frac{dy}{dx} = \frac{4}{2te^{2t-3}}$, or equivalent | A1 | |
| Set $t = a$, equate gradient to 2 and obtain the given answer | A1 | AG |
| **Total** | **4** | |

## Question 4(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculate $a - \frac{1}{2}(3 - \ln a)$ when $a = 1$ and $a = 2$, or equivalent | M1 | |
| Complete the argument by considering the signs of the correct calculated values | A1 | |
| **Total** | **2** | |

## Question 4(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer 1.35 | A1 | |
| Show sufficient iterations to 4 dp to justify 1.35 to 2 dp, **OR** show there is a sign change in the interval (1.345, 1.355) | A1 | |
| **Total** | **3** | |
4 The parametric equations of a curve are

$$x = \mathrm { e } ^ { 2 t - 3 } , \quad y = 4 \ln t$$

where $t > 0$. When $t = a$ the gradient of the curve is 2 .\\
(a) Show that $a$ satisfies the equation $a = \frac { 1 } { 2 } ( 3 - \ln a )$.\\
(b) Verify by calculation that this equation has a root between 1 and 2 .\\
(c) Use the iterative formula $a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)$ to calculate $a$ correct to 2 decimal places, showing the result of each iteration to 4 decimal places.\\

\hfill \mbox{\textit{CAIE P3 2020 Q4 [9]}}