| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule with stated number of strips |
| Difficulty | Moderate -0.8 Part (a) is a routine trapezium rule application with clear ordinates and straightforward arithmetic—standard C2 bookwork requiring only formula recall and calculator work. Part (b) tests basic understanding of transformations (stretch parallel to x-axis means replacing x with x/3), which is also standard curriculum content with minimal problem-solving required. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(h = 0.5\); ordinates at \(x = 0, 0.5, 1.0, 1.5\) | B1 | Correct strip width |
| \(y\) values: \(2, \sqrt{27(0.125)+4}, \sqrt{31}, \sqrt{27(3.375)+4}\) i.e. \(2, 2.179, 5.568, 9.813\) | B1 | At least 3 correct ordinates |
| \(\approx \frac{0.5}{2}[2 + 9.813 + 2(2.179 + 5.568)]\) | M1 | Correct trapezium rule structure |
| \(= 6.15\) | A1 | 3 significant figures |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Stretch parallel to \(x\)-axis scale factor 3 replaces \(x\) with \(\frac{x}{3}\) | M1 | Correct reasoning |
| \(g(x) = \sqrt{27\left(\frac{x}{3}\right)^3 + 4} = \sqrt{x^3 + 4}\) | A1 | cao |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $h = 0.5$; ordinates at $x = 0, 0.5, 1.0, 1.5$ | B1 | Correct strip width |
| $y$ values: $2, \sqrt{27(0.125)+4}, \sqrt{31}, \sqrt{27(3.375)+4}$ i.e. $2, 2.179, 5.568, 9.813$ | B1 | At least 3 correct ordinates |
| $\approx \frac{0.5}{2}[2 + 9.813 + 2(2.179 + 5.568)]$ | M1 | Correct trapezium rule structure |
| $= 6.15$ | A1 | 3 significant figures |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Stretch parallel to $x$-axis scale factor 3 replaces $x$ with $\frac{x}{3}$ | M1 | Correct reasoning |
| $g(x) = \sqrt{27\left(\frac{x}{3}\right)^3 + 4} = \sqrt{x^3 + 4}$ | A1 | cao |
---4
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with four ordinates (three strips) to find an approximate value for $\int _ { 0 } ^ { 1.5 } \sqrt { 27 x ^ { 3 } + 4 } \mathrm {~d} x$, giving your answer to three significant figures.
\item The curve with equation $y = \sqrt { 27 x ^ { 3 } + 4 }$ is stretched parallel to the $x$-axis with scale factor 3 to give the curve with equation $y = \mathrm { g } ( x )$. Write down an expression for $\mathrm { g } ( x )$.\\
(2 marks)
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{1c06ba04-575c-4eb8-b4aa-0a7510838cd2-05_1988_1717_719_150}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q4 [6]}}