| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule with stated number of strips |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C2 techniques: solving an exponential equation using logarithms (routine), applying the trapezium rule with given intervals (mechanical calculation), and understanding whether trapezium rule over/underestimates for convex curves (bookwork). All parts are textbook exercises with no problem-solving required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.25^x = 2 \Rightarrow x\log 1.25 = \log 2\) | B1 | For correct initial use of logs |
| M1 | For correct log expression for \(x\) | |
| A1 | 3 | For correct numerical value |
| Hence \(x = \frac{\log 2}{\log 1.25} = 3.11\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}\left[1.25^0 + 2(1.25^1 + 1.25^2 + 1.25^3) + 1.25^4\right]\) | B1 | For correct recognition of \(h = 1\) |
| M1 | For any use of values \(1.25^x\) for \(x = 0, \ldots, 4\) | |
| M1 | For use of correct formula | |
| A1 | 4 | For correct answer |
| Area is 6.49 |
| Answer | Marks | Guidance |
|---|---|---|
| The trapezia used in (ii) extend above the curve. Hence the trapezium rule overestimates the area. | M1 | For stating or sketching trapezia above curve |
| A1 | 2 | For stating overestimate with correct reason |
| Answer | Marks | Guidance |
|---|---|---|
| Use more trapezia, with a smaller value of \(h\) | B1 | 1 |
| Total: 10 |
## Part (i)
$1.25^x = 2 \Rightarrow x\log 1.25 = \log 2$ | B1 | For correct initial use of logs
| M1 | For correct log expression for $x$
| A1 | 3 | For correct numerical value
Hence $x = \frac{\log 2}{\log 1.25} = 3.11$ | |
## Part (ii)
$\frac{1}{2}\left[1.25^0 + 2(1.25^1 + 1.25^2 + 1.25^3) + 1.25^4\right]$ | B1 | For correct recognition of $h = 1$
| M1 | For any use of values $1.25^x$ for $x = 0, \ldots, 4$
| M1 | For use of correct formula
| A1 | 4 | For correct answer
Area is 6.49 | |
## Part (iii)
The trapezia used in (ii) extend above the curve. Hence the trapezium rule overestimates the area. | M1 | For stating or sketching trapezia above curve
| A1 | 2 | For stating overestimate with correct reason
## Part (iv)
Use more trapezia, with a smaller value of $h$ | B1 | 1 | For stating that more trapezia should be used
| **Total: 10** | |\includegraphics{figure_8}
The diagram shows the curve $y = 1.25^x$.
\begin{enumerate}[label=(\roman*)]
\item A point on the curve has y-coordinate 2. Calculate its x-coordinate. [3]
\item Use the trapezium rule with 4 intervals to estimate the area of the shaded region, bounded by the curve, the axes, and the line $x = 4$. [4]
\item State, with a reason, whether the estimate found in part (ii) is an overestimate or an underestimate. [2]
\item Explain briefly how the trapezium rule could be used to find a more accurate estimate of the area of the shaded region. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q8 [10]}}