CAIE P2 2023 June — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeArea under curve using substitution
DifficultyStandard +0.3 This is a straightforward area calculation requiring integration of 1/(2x+3) using substitution u=2x+3, finding intersection points, and subtracting a rectangular area. The substitution is standard, the algebra is routine, and the question clearly guides the student to the final form. Slightly above average due to requiring multiple steps (finding intersection, setting up correct limits, algebraic manipulation) but no novel insight needed.
Spec1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits

3 \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-04_458_892_269_614} The diagram shows part of the curve \(y = \frac { 6 } { 2 x + 3 }\). The shaded region is bounded by the curve and the lines \(x = 6\) and \(y = 2\). Find the exact area of the shaded region, giving your answer in the form \(a - \ln b\), where \(a\) and \(b\) are integers.

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
Integrate to obtain the form \(k\ln(2x+3)\)*M1 Condone lack or misuse of brackets if recovered later
Obtain correct \(3\ln(2x+3)\)A1 Allow unsimplified
Apply limits 0 and 6 correctly to obtain \(k\ln 15 - k\ln 3\)*DM1 Allow unsimplified
Apply relevant logarithm properties correctly to obtain form \(\ln b\)DM1
Obtain \(12 - \ln 125\)A1
Total5
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain the form $k\ln(2x+3)$ | *M1 | Condone lack or misuse of brackets if recovered later |
| Obtain correct $3\ln(2x+3)$ | A1 | Allow unsimplified |
| Apply limits 0 and 6 correctly to obtain $k\ln 15 - k\ln 3$ | *DM1 | Allow unsimplified |
| Apply relevant logarithm properties correctly to obtain form $\ln b$ | DM1 | |
| Obtain $12 - \ln 125$ | A1 | |
| **Total** | **5** | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-04_458_892_269_614}

The diagram shows part of the curve $y = \frac { 6 } { 2 x + 3 }$. The shaded region is bounded by the curve and the lines $x = 6$ and $y = 2$.

Find the exact area of the shaded region, giving your answer in the form $a - \ln b$, where $a$ and $b$ are integers.\\

\hfill \mbox{\textit{CAIE P2 2023 Q3 [5]}}