| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for infinite series |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring students to relate rectangular approximations to an integral to derive bounds for ln(N!). Part (a) involves recognizing that the sum of rectangle areas gives a lower bound for the integral, requiring the insight that ln(N!) = sum of ln(k). Part (b) mirrors this with upper rectangles. While it requires some problem-solving and the connection between factorials and logarithms isn't immediately obvious, the rectangular approximation method is standard in numerical integration, and the question provides significant scaffolding through the diagram and structure. More challenging than typical A-level integration but not exceptionally difficult for Further Maths. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | ln2+ln3++lnN | M1 |
| Answer | Marks |
|---|---|
| 1 | M1 |
| Answer | Marks |
|---|---|
| 1 1 1 | M1 A1 |
| lnN!> NlnN −N +1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(b) | ln1+ln2++ln(N −1) (or ln2++ln(N −1)) | M1 |
| Answer | Marks |
|---|---|
| 1 2 | M1 |
| Answer | Marks |
|---|---|
| (or ln N!< (N +1)ln N − N +2(1−ln2)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | ln2+ln3++lnN | M1
> N lnxdx
1 | M1
N lnxdx= [ xlnx ]N − N 1dx=NlnN−N+1
1 1 1 | M1 A1
lnN!> NlnN −N +1 | A1
5
Question | Answer | Marks
--- 4(b) ---
4(b) | ln1+ln2++ln(N −1) (or ln2++ln(N −1)) | M1
< N lnxdx (or < N lnxdx)
1 2 | M1
ln N!< N ln N − N +1+ln N = (N +1)ln N − N +1
(or ln N!< (N +1)ln N − N +2(1−ln2)) | A1
3
Question | Answer | Marks\includegraphics{figure_4}
The diagram shows the curve with equation $y = \ln x$ for $x \geqslant 1$, together with a set of $(N-1)$ rectangles of unit width.
\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that
$$\ln N! > N \ln N - N + 1.$$
[5]
\item Use a similar method to find, in terms of $N$, an upper bound for $\ln N!$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q4 [8]}}