CAIE P2 2022 June — Question 6 8 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeDerive equation from integral condition
DifficultyStandard +0.8 This question requires setting up and evaluating a definite integral of an exponential function, algebraically manipulating the result to isolate a in a specific form (non-trivial rearrangement), then applying iterative methods. The algebraic manipulation to reach the given form requires careful handling of exponentials and logarithms across multiple steps, which is more demanding than standard integration exercises.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.08d Evaluate definite integrals: between limits1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

6 \includegraphics[max width=\textwidth, alt={}, center]{712be8e6-e1e9-4662-b1f1-51c39c2c9df1-08_542_661_269_731} The diagram shows the curve \(y = 3 \mathrm { e } ^ { 2 x - 1 }\). The shaded region is bounded by the curve and the lines \(x = a , x = a + 1\) and \(y = 0\), where \(a\) is a constant. It is given that the area of the shaded region is 120 square units.
  1. Show that \(a = \frac { 1 } { 2 } \ln \left( 80 + \mathrm { e } ^ { 2 a - 1 } \right) - \frac { 1 } { 2 }\).
  2. Use an iterative formula, based on the equation in part (a), to find the value of \(a\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

Question 6(a):
AnswerMarks Guidance
AnswerMarks Guidance
Integrate to obtain form \(ke^{2x-1}\)M1 any constant \(k\), \(k \neq 6\)
Obtain correct \(\frac{3}{2}e^{2x-1}\)A1
Apply limits correctly to \(ke^{2x-1}\) and equate to 120M1 Allow one slip
Rearrange as far as \(a = \ldots\)M1
Obtain \(a = \frac{1}{2}\ln(80 + e^{2a-1}) - \frac{1}{2}\)A1 AG – necessary detail needed
Total5
Question 6(b):
AnswerMarks Guidance
AnswerMarks Guidance
Use iteration process correctly at least onceM1
Obtain final answer 1.76A1 Answer required to exactly 3 s.f.
Show sufficient iterations to 5 s.f. to justify answer or show sign change in the interval \([1.755, 1.765]\)A1
Total3
## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain form $ke^{2x-1}$ | M1 | any constant $k$, $k \neq 6$ |
| Obtain correct $\frac{3}{2}e^{2x-1}$ | A1 | |
| Apply limits correctly to $ke^{2x-1}$ and equate to 120 | M1 | Allow one slip |
| Rearrange as far as $a = \ldots$ | M1 | |
| Obtain $a = \frac{1}{2}\ln(80 + e^{2a-1}) - \frac{1}{2}$ | A1 | AG – necessary detail needed |
| **Total** | **5** | |

---

## Question 6(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use iteration process correctly at least once | M1 | |
| Obtain final answer 1.76 | A1 | Answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in the interval $[1.755, 1.765]$ | A1 | |
| **Total** | **3** | |

---
6\\
\includegraphics[max width=\textwidth, alt={}, center]{712be8e6-e1e9-4662-b1f1-51c39c2c9df1-08_542_661_269_731}

The diagram shows the curve $y = 3 \mathrm { e } ^ { 2 x - 1 }$. The shaded region is bounded by the curve and the lines $x = a , x = a + 1$ and $y = 0$, where $a$ is a constant. It is given that the area of the shaded region is 120 square units.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = \frac { 1 } { 2 } \ln \left( 80 + \mathrm { e } ^ { 2 a - 1 } \right) - \frac { 1 } { 2 }$.
\item Use an iterative formula, based on the equation in part (a), to find the value of $a$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2022 Q6 [8]}}
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