| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule with stated number of strips |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard P2 techniques: quotient rule differentiation, solving f'(x)=0 for stationary points, applying trapezium rule with given intervals, and reasoning about concavity. All parts are routine applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate using quotient rule (or product rule) | M1* | |
| Obtain \(\frac{5(4x^3+1)-60x^3}{(4x^3+1)^2}\) | A1 | OE |
| Equate first derivative to zero and attempt solution | DM1 | |
| Obtain \(x = \frac{1}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(y\) values \(\frac{5}{5}\), \(\frac{10}{33}\), \(\frac{15}{109}\) or decimal equivalents | B1 | |
| Use correct formula, or equivalent, with \(h=1\) | M1 | |
| Obtain \(\frac{1}{2}\left(1+\frac{20}{33}+\frac{15}{109}\right)\) or equivalent and hence 0.87 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State over-estimate with reference to top of each trapezium above curve | B1 | Or clear equivalent |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate using quotient rule (or product rule) | M1* | |
| Obtain $\frac{5(4x^3+1)-60x^3}{(4x^3+1)^2}$ | A1 | OE |
| Equate first derivative to zero and attempt solution | DM1 | |
| Obtain $x = \frac{1}{2}$ | A1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $y$ values $\frac{5}{5}$, $\frac{10}{33}$, $\frac{15}{109}$ or decimal equivalents | B1 | |
| Use correct formula, or equivalent, with $h=1$ | M1 | |
| Obtain $\frac{1}{2}\left(1+\frac{20}{33}+\frac{15}{109}\right)$ or equivalent and hence 0.87 | A1 | |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State over-estimate with reference to top of each trapezium above curve | B1 | Or clear equivalent |
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\includegraphics[max width=\textwidth, alt={}, center]{9cf008d5-c15f-4491-9e4d-4bd070f896d5-06_446_832_260_653}
The diagram shows part of the curve with equation $y = \frac { 5 x } { 4 x ^ { 3 } + 1 }$. The shaded region is bounded by the curve and the lines $x = 1 , x = 3$ and $y = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and hence find the $x$-coordinate of the maximum point.
\item Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
\item State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2021 Q4 [8]}}