| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for definite integral |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on numerical integration using rectangles. Part (a) requires summing a geometric series to derive a given formula, parts (b-c) involve similar calculations with upper bounds, and part (d) applies the results. While it involves geometric series and some algebraic manipulation, the question is highly scaffolded with the formula given in part (a), and the techniques are standard for Further Maths. It's moderately above average difficulty due to the algebraic manipulation of exponentials and series, but not exceptionally challenging. |
| Spec | 1.08g Integration as limit of sum: Riemann sums4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\int_0^1\left(\frac{1}{2}\right)^x dx >\right]\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{1}{N}}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{2}{N}}+\cdots+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N-1}{N}}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N}{N}}\) | M1A1 | Forms the sum of the areas of the rectangles. M1 for correct number of rectangles |
| \(=\frac{1}{N}\sum_{n=1}^{N}\left(\frac{1}{2}\right)^{\frac{n}{N}} = \frac{\left(\frac{1}{2}\right)^{\frac{1}{N}}\left(\frac{1}{2}-1\right)}{N\left(\frac{1}{2}\right)^{\frac{1}{N}}-N} = \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}}{N-N\left(\frac{1}{2}\right)^{\frac{1}{N}}} = \frac{1}{2N\left(2^{\frac{1}{N}}-1\right)}\) | M1A1 | Applies \(\sum_{n=1}^{N}r^n = \frac{r(r^N-1)}{r-1}\), AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1\left(\frac{1}{2}\right)^x dx < \frac{1}{N}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{1}{N}}+\cdots+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N-2}{N}}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N-1}{N}}\) | M1A1 | Forms the sum of the areas of appropriate rectangles |
| \(\frac{1}{N}\sum_{n=0}^{N-1}\left(\frac{1}{2}\right)^{\frac{n}{N}} = \frac{\frac{1}{2}-1}{N\left(\frac{1}{2}\right)^{\frac{1}{N}}-N} = \frac{1}{2N\left(1-\left(\frac{1}{2}\right)^{\frac{1}{N}}\right)} = \frac{2^{\frac{1}{N}}}{2N\left(2^{\frac{1}{N}}-1\right)}\) | M1A1 | Applies \(\sum_{n=0}^{N-1}r^n = \frac{r^N-1}{r-1}\) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2^{\frac{1}{N}}}{2N\left(2^{\frac{1}{N}}-1\right)}-\frac{1}{2N\left(2^{\frac{1}{N}}-1\right)}=\frac{1}{2N}\leq 10^{-3} \Rightarrow 2N\geq 10^3\) | M1 | Simplifies \(U_N - L_N\) to \(\frac{c}{N}\) |
| Least value of \(N\) is \(500\) | A1 | CWO |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(L_N < \frac{1}{2\ln 2} < U_N \Rightarrow \frac{1}{2U_N} < \ln 2 < \frac{1}{2L_N}\) | M1A1FT | Forms inequality. FT on their \(U_N\) |
| \(0.693 = \frac{1}{2U_N} < \ln 2 < \frac{1}{2L_N} = 0.694\) | A1A1 | CWO. Must have used \(N=500\) |
| 4 |
# Question 6:
## Part 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\int_0^1\left(\frac{1}{2}\right)^x dx >\right]\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{1}{N}}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{2}{N}}+\cdots+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N-1}{N}}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N}{N}}$ | M1A1 | Forms the sum of the areas of the rectangles. M1 for correct number of rectangles |
| $=\frac{1}{N}\sum_{n=1}^{N}\left(\frac{1}{2}\right)^{\frac{n}{N}} = \frac{\left(\frac{1}{2}\right)^{\frac{1}{N}}\left(\frac{1}{2}-1\right)}{N\left(\frac{1}{2}\right)^{\frac{1}{N}}-N} = \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}}{N-N\left(\frac{1}{2}\right)^{\frac{1}{N}}} = \frac{1}{2N\left(2^{\frac{1}{N}}-1\right)}$ | M1A1 | Applies $\sum_{n=1}^{N}r^n = \frac{r(r^N-1)}{r-1}$, AG |
| | **4** | |
## Part 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1\left(\frac{1}{2}\right)^x dx < \frac{1}{N}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{1}{N}}+\cdots+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N-2}{N}}+\frac{1}{N}\left(\frac{1}{2}\right)^{\frac{N-1}{N}}$ | M1A1 | Forms the sum of the areas of appropriate rectangles |
| $\frac{1}{N}\sum_{n=0}^{N-1}\left(\frac{1}{2}\right)^{\frac{n}{N}} = \frac{\frac{1}{2}-1}{N\left(\frac{1}{2}\right)^{\frac{1}{N}}-N} = \frac{1}{2N\left(1-\left(\frac{1}{2}\right)^{\frac{1}{N}}\right)} = \frac{2^{\frac{1}{N}}}{2N\left(2^{\frac{1}{N}}-1\right)}$ | M1A1 | Applies $\sum_{n=0}^{N-1}r^n = \frac{r^N-1}{r-1}$ |
| | **4** | |
## Part 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2^{\frac{1}{N}}}{2N\left(2^{\frac{1}{N}}-1\right)}-\frac{1}{2N\left(2^{\frac{1}{N}}-1\right)}=\frac{1}{2N}\leq 10^{-3} \Rightarrow 2N\geq 10^3$ | M1 | Simplifies $U_N - L_N$ to $\frac{c}{N}$ |
| Least value of $N$ is $500$ | A1 | CWO |
| | **2** | |
## Part 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $L_N < \frac{1}{2\ln 2} < U_N \Rightarrow \frac{1}{2U_N} < \ln 2 < \frac{1}{2L_N}$ | M1A1FT | Forms inequality. FT on their $U_N$ |
| $0.693 = \frac{1}{2U_N} < \ln 2 < \frac{1}{2L_N} = 0.694$ | A1A1 | CWO. Must have used $N=500$ |
| | **4** | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-12_533_1532_278_264}
The diagram shows the curve with equation $y = \left( \frac { 1 } { 2 } \right) ^ { x }$ for $0 \leqslant x \leqslant 1$, together with a set of $N$ rectangles each of width $\frac { 1 } { N }$.
\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that $\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x > L _ { N }$, where
$$L _ { N } = \frac { 1 } { 2 N \left( 2 ^ { \frac { 1 } { N } } - 1 \right) }$$
\includegraphics[max width=\textwidth, alt={}, center]{374b91df-926d-4f7f-a1d3-a54c70e8ff0e-12_2717_38_109_2009}
\item Use a similar method to find, in terms of $N$, an upper bound $U _ { N }$ for $\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x$.
\item Find the least value of $N$ such that $U _ { N } - L _ { N } \leqslant 10 ^ { - 3 }$.
\item Given that $\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x = \frac { 1 } { 2 \ln 2 }$ ,use the value of $N$ found in part(c)to find upper and lower bounds for $\ln 2$ .
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q6 [10]}}