OCR MEI Further Numerical Methods 2024 June — Question 2 6 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule applied to real-world data
DifficultyStandard +0.3 This is a straightforward application of relative error formulas to hyperbolic functions. Part (a) requires calculating tanh values and their rounded approximations, then applying the standard relative error formula. Part (b) extends this to a quotient. Part (c) asks for observation of the relationship R_C ≈ R_A - R_B. While it involves Further Maths content (hyperbolic functions), the numerical methods are routine and calculator-based with no conceptual challenges or proof required. Slightly easier than average due to its computational nature.

2 You are given that \(a = \tanh ( 1 )\) and \(b = \tanh ( 2 )\). \(A\) is the approximation to \(a\) formed by rounding \(\tanh ( 1 )\) to 1 decimal place. \(B\) is the approximation to \(b\) formed by rounding \(\tanh ( 2 )\) to 1 decimal place.
  1. Calculate the following.

Question 2:
AnswerMarks Guidance
2(a) 0.8−tanh (1) 1−tanh (2)
or
tanh (1) tanh (2)
may be implied by
0.8−0.761… 1−0.964…
eg or
0.761… 0.964…
0.0504 ≤ R ≤ 0.05043
𝐴
0.0373 ≤ R ≤ 0.037315
AnswerMarks
𝐵M1
A1
AnswerMarks
A11.1
1.1
AnswerMarks
1.1allow signs reversed in numerator; may be implied by one correct
answer or the negative of one correct answer
if A0A0 allow SC B1 for awrt ‒0.0504 and awrt ‒0.0373
or SC B1 for 0.050 and 0.037
[3]
AnswerMarks Guidance
2(b) 0.8÷1−0.76159÷0.96403
soi
0.76159÷0.96403
0.0126 ≤ 𝑅 ≤ 0.012642
AnswerMarks
𝐶M1
A11.1
1.1allow signs reversed in numerator; may be implied by ‒ 0.0126
0.8−0.7900128
NB
0.7900128
[2]
AnswerMarks Guidance
2(c) R ≈ R −R oe
𝐶 𝐴 𝐵B1 2.2b
𝐶 𝐴 𝐵
[1]
AnswerMarks Guidance
20.639735 0.80627
Question 2:
2 | (a) | 0.8−tanh (1) 1−tanh (2)
or
tanh (1) tanh (2)
may be implied by
0.8−0.761… 1−0.964…
eg or
0.761… 0.964…
0.0504 ≤ R ≤ 0.05043
𝐴
0.0373 ≤ R ≤ 0.037315
𝐵 | M1
A1
A1 | 1.1
1.1
1.1 | allow signs reversed in numerator; may be implied by one correct
answer or the negative of one correct answer
if A0A0 allow SC B1 for awrt ‒0.0504 and awrt ‒0.0373
or SC B1 for 0.050 and 0.037
[3]
2 | (b) | 0.8÷1−0.76159÷0.96403
soi
0.76159÷0.96403
0.0126 ≤ 𝑅 ≤ 0.012642
𝐶 | M1
A1 | 1.1
1.1 | allow signs reversed in numerator; may be implied by ‒ 0.0126
0.8−0.7900128
NB
0.7900128
[2]
2 | (c) | R ≈ R −R oe
𝐶 𝐴 𝐵 | B1 | 2.2b | allow 𝑅 = 𝑅 −𝑅 ; allow equivalent in words
𝐶 𝐴 𝐵
[1]
2 | 0.639735 | 0.80627 | 0.69525
2 You are given that $a = \tanh ( 1 )$ and $b = \tanh ( 2 )$.\\
$A$ is the approximation to $a$ formed by rounding $\tanh ( 1 )$ to 1 decimal place.\\
$B$ is the approximation to $b$ formed by rounding $\tanh ( 2 )$ to 1 decimal place.
\begin{enumerate}[label=(\alph*)]
\item Calculate the following.

\begin{itemize}
  \item The relative error $\mathrm { R } _ { \mathrm { A } }$ when $A$ is used to approximate $a$.
  \item The relative error $\mathrm { R } _ { \mathrm { B } }$ when $B$ is used to approximate $b$.
\item Calculate the relative error $\mathrm { R } _ { \mathrm { C } }$ when $\mathrm { C } = \frac { \mathrm { A } } { \mathrm { B } }$ is used to approximate $\mathrm { c } = \frac { \mathrm { a } } { \mathrm { b } }$.
\item Comment on the relationship between $R _ { A } , R _ { B }$ and $R _ { C }$.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2024 Q2 [6]}}