CAIE P2 2022 November — Question 6 9 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2022
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas Between Curves
TypeArea with Exponential or Logarithmic Curves
DifficultyStandard +0.3 This is a straightforward area-between-curves problem requiring integration of standard functions (rational function and exponential). Students must set up the integral correctly, integrate ln and exponential terms using standard formulas, and simplify to the given form. While it involves multiple steps and algebraic manipulation, it's a routine application of A-level integration techniques with no novel problem-solving required.
Spec1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration
6 \includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-08_616_531_269_799} The diagram shows the curves \(y = \frac { 6 } { 3 x + 2 }\) and \(y = 3 \mathrm { e } ^ { - x } - 3\) for values of \(x\) between 0 and 4. The shaded region is bounded by the two curves and the lines \(x = 0\) and \(x = 4\). Find the exact area of the shaded region, giving your answer in the form \(\ln a + b + c \mathrm { e } ^ { d }\).
Question 6:
AnswerMarks Guidance
AnswerMark Guidance
Integrate \(\frac{6}{3x+2}\) to obtain form \(k_1\ln(3x+2)\)\*M1 for any constant \(k_1\)
Obtain correct \(2\ln(3x+2)\)A1
Apply limits correctlyDM1
Obtain \(2\ln14 - 2\ln2\) and hence \(\ln 49\)A1 at this stage or later
Integrate \(3e^{-x} - 3\) to obtain form \(k_2e^{-x} + k_3x\)M1 for any non-zero constants \(k_2\), \(k_3\)
Obtain correct \(-3e^{-x} - 3x\)A1
Apply limits to obtain \(-3e^{-4} - 12 + 3\)A1 OE; implied if \(\int(y_1 - y_2)\,dx\) approach used
Use correct procedure to find exact total areaM1
Obtain \(\ln 49 + 9 + 3e^{-4}\)A1 
## Question 6:

| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate $\frac{6}{3x+2}$ to obtain form $k_1\ln(3x+2)$ | \*M1 | for any constant $k_1$ |
| Obtain correct $2\ln(3x+2)$ | A1 | |
| Apply limits correctly | DM1 | |
| Obtain $2\ln14 - 2\ln2$ and hence $\ln 49$ | A1 | at this stage or later |
| Integrate $3e^{-x} - 3$ to obtain form $k_2e^{-x} + k_3x$ | M1 | for any non-zero constants $k_2$, $k_3$ |
| Obtain correct $-3e^{-x} - 3x$ | A1 | |
| Apply limits to obtain $-3e^{-4} - 12 + 3$ | A1 | OE; implied if $\int(y_1 - y_2)\,dx$ approach used |
| Use correct procedure to find exact total area | M1 | |
| Obtain $\ln 49 + 9 + 3e^{-4}$ | A1 | |

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6\\
\includegraphics[max width=\textwidth, alt={}, center]{68f4b2dc-a05d-4061-aaf0-de15cfe186a9-08_616_531_269_799}

The diagram shows the curves $y = \frac { 6 } { 3 x + 2 }$ and $y = 3 \mathrm { e } ^ { - x } - 3$ for values of $x$ between 0 and 4. The shaded region is bounded by the two curves and the lines $x = 0$ and $x = 4$.

Find the exact area of the shaded region, giving your answer in the form $\ln a + b + c \mathrm { e } ^ { d }$.\\

\hfill \mbox{\textit{CAIE P2 2022 Q6 [9]}}
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