CAIE P2 2021 June — Question 7 11 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2021
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeIterative method for parameter value
DifficultyStandard +0.3 This is a standard parametric equations question with routine iterative method application. Part (a) is simple substitution, part (b) is straightforward evaluation, part (c) follows a standard iterative procedure with given initial value, and part (d) requires dy/dx = (dy/dt)/(dx/dt). All techniques are textbook exercises with no novel insight required, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

7 \includegraphics[max width=\textwidth, alt={}, center]{61df367d-741f-4906-8ab9-2f32e8711aa6-10_465_785_260_680} The diagram shows the curve with parametric equations $$x = 4 t + \mathrm { e } ^ { 2 t } , \quad y = 6 t \sin 2 t$$ for \(0 \leqslant t \leqslant 1\). The point \(P\) on the curve has parameter \(p\) and \(y\)-coordinate 3 .
  1. Show that \(p = \frac { 1 } { 2 \sin 2 p }\).
  2. Show by calculation that the value of \(p\) lies between 0.5 and 0.6 .
  3. Use an iterative formula, based on the equation in part (a), to find the value of \(p\) correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures.
  4. Find the gradient of the curve at \(P\).
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Question 7(a):
AnswerMarks Guidance
AnswerMark Guidance
Equate \(y\) to 3 and confirm \(p = \dfrac{1}{2\sin 2p}\)B1 AG
Total: 1
Question 7(b):
AnswerMarks Guidance
AnswerMark Guidance
Consider sign of \(p - \dfrac{1}{2\sin 2p}\) or equivalent for 0.5 and 0.6M1
Obtain \(-0.09...\) and \(0.06...\) or equivalents and justify conclusionA1 AG
Total: 2
Question 7(c):
AnswerMarks Guidance
AnswerMark Guidance
Use iteration process correctly at least onceM1 Need to see \(0.55494...\)
Obtain final answer \(0.557\) onlyA1 Allow recovery. Allow if iterations are to 4sf. Allow if insufficient iterations seen.
Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval \([0.5565,\ 0.5575]\)A1 If not starting at \(0.55\) then max marks M1A1A0
Total: 3
Question 7(d):
AnswerMarks Guidance
AnswerMark Guidance
Obtain \(\dfrac{dx}{dt} = 4 + 2e^{2t}\)B1
Use product rule to find \(\dfrac{dy}{dt}\)M1 Must be of the form \(p\sin 2t + qt\cos 2t\)
Obtain \(6\sin 2t + 12t\cos 2t\)A1 Allow unsimplified.
Divide to obtain \(\dfrac{dy}{dx}\) using *their* \(\dfrac{dy}{dt}\) and \(\dfrac{dx}{dt}\) correctlyDM1 Must have either B1 or previous M1.
Obtain \(0.826\)A1 AWRT
Total: 5
## Question 7(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Equate $y$ to 3 and confirm $p = \dfrac{1}{2\sin 2p}$ | **B1** | AG |
| | **Total: 1** | |

## Question 7(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Consider sign of $p - \dfrac{1}{2\sin 2p}$ or equivalent for 0.5 and 0.6 | **M1** | |
| Obtain $-0.09...$ and $0.06...$ or equivalents and justify conclusion | **A1** | AG |
| | **Total: 2** | |

## Question 7(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| Use iteration process correctly at least once | **M1** | Need to see $0.55494...$ |
| Obtain final answer $0.557$ only | **A1** | Allow recovery. Allow if iterations are to 4sf. Allow if insufficient iterations seen. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval $[0.5565,\ 0.5575]$ | **A1** | If not starting at $0.55$ then max marks M1A1A0 |
| | **Total: 3** | |

## Question 7(d):

| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\dfrac{dx}{dt} = 4 + 2e^{2t}$ | **B1** | |
| Use product rule to find $\dfrac{dy}{dt}$ | **M1** | Must be of the form $p\sin 2t + qt\cos 2t$ |
| Obtain $6\sin 2t + 12t\cos 2t$ | **A1** | Allow unsimplified. |
| Divide to obtain $\dfrac{dy}{dx}$ using *their* $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ correctly | **DM1** | Must have either B1 or previous M1. |
| Obtain $0.826$ | **A1** | AWRT |
| | **Total: 5** | |
7\\
\includegraphics[max width=\textwidth, alt={}, center]{61df367d-741f-4906-8ab9-2f32e8711aa6-10_465_785_260_680}

The diagram shows the curve with parametric equations

$$x = 4 t + \mathrm { e } ^ { 2 t } , \quad y = 6 t \sin 2 t$$

for $0 \leqslant t \leqslant 1$. The point $P$ on the curve has parameter $p$ and $y$-coordinate 3 .
\begin{enumerate}[label=(\alph*)]
\item Show that $p = \frac { 1 } { 2 \sin 2 p }$.
\item Show by calculation that the value of $p$ lies between 0.5 and 0.6 .
\item Use an iterative formula, based on the equation in part (a), to find the value of $p$ correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures.
\item Find the gradient of the curve at $P$.\\

If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2021 Q7 [11]}}
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