Question 6 - A-Level Maths

CAIE Further Paper 2 2024 November — Question 6 10 marks

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 multi-part question on Riemann sums with clear guidance at each step. Part (a) requires summing a geometric series to verify a given formula, parts (b-c) follow similar methods, and part (d) is straightforward substitution. While it involves geometric series manipulation and requires careful algebraic work, the question provides significant scaffolding and uses standard Further Maths techniques without requiring novel insight.
Spec	1.08g Integration as limit of sum: Riemann sums 4.08c Improper integrals: infinite limits or discontinuous integrands

6 \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-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 $\frac { 1 } { N }$.

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]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-12_2717_38_109_2009}
Use a similar method to find, in terms of $N$, an upper bound $U _ { N }$ for $\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x$.
Find the least value of $N$ such that $U _ { N } - L _ { N } \leqslant 10 ^ { - 3 }$.
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 $\ln 2$ .

Show mark scheme Show mark scheme source

Question 6(a):

Answer	Marks	Guidance
$\left[\int_0^1\left(\frac{1}{2}\right)^x dx >\right]\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{2}{N}}+\cdots+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-1}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N}{N}}$	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.

Total: 4

Question 6(b):

Answer	Marks	Guidance
$\int_0^1\left(\frac{1}{2}\right)^x dx < \left(\frac{1}{N}\right)+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}+\cdots+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-2}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-1}{N}}$	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 $\sum_{n=0}^{N-1}r^n = \frac{r^N-1}{r-1}$.

Total: 4

Question 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.

Total: 2

Question 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 $N=500$.

Total: 4

## Question 6(a):

$\left[\int_0^1\left(\frac{1}{2}\right)^x dx >\right]\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{2}{N}}+\cdots+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-1}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N}{N}}$ | 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.

**Total: 4**

---

## Question 6(b):

$\int_0^1\left(\frac{1}{2}\right)^x dx < \left(\frac{1}{N}\right)+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}+\cdots+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-2}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-1}{N}}$ | 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 $\sum_{n=0}^{N-1}r^n = \frac{r^N-1}{r-1}$.

**Total: 4**

---

## Question 6(c):

$\frac{2^{\frac{1}{N}}}{2N\left(2^{\frac{1}{N}}-1\right)}-\frac{1}{2N\left(2^{\frac{1}{N}}-1\right)} = \frac{1}{2N} \leq 10^{-3} \Rightarrow 2N \geq 10^3$ | M1 | Simplifies $U_N - L_N$ to $\frac{c}{N}$.

Least value of $N$ is $500$ | A1 | CWO.

**Total: 2**

---

## Question 6(d):

$L_N < \frac{1}{2\ln 2} < U_N \Rightarrow \frac{1}{2U_N} < \ln 2 < \frac{1}{2L_N}$ | 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 $N=500$.

**Total: 4**

Show LaTeX source

6\\
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-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]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-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]}}

This paper (7 questions)

View full paper

Q2 6 Q3 12 Q4 9 Q5 10 Q6 10 Q7 10 Q8 14

Answer	Marks	Guidance
\(\left[\int_0^1\left(\frac{1}{2}\right)^x dx >\right]\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{2}{N}}+\cdots+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-1}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N}{N}}\)	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.

Answer	Marks	Guidance
\(\int_0^1\left(\frac{1}{2}\right)^x dx < \left(\frac{1}{N}\right)+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{1}{N}}+\cdots+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-2}{N}}+\left(\frac{1}{N}\right)\left(\frac{1}{2}\right)^{\frac{N-1}{N}}\)	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 \(\sum_{n=0}^{N-1}r^n = \frac{r^N-1}{r-1}\).

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.

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 \(N=500\).