OCR MEI Further Numerical Methods 2021 November — Question 1 5 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2021
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule applied to real-world data
DifficultyStandard +0.3 This question tests understanding of relative error calculation and floating-point representation in spreadsheets. Part (a) involves straightforward application of the relative error formula, while part (b) requires explaining that spreadsheet display vs. stored values differ due to rounding. The conceptual insight needed is modest—recognizing that C1 stores the exact value 2 (from SQRT then squaring) rather than the displayed approximation. This is slightly easier than average as it's primarily about careful reasoning rather than complex calculation or novel problem-solving.

1
    1. Determine the relative error when
      \begin{table}[h]
      ABC
      121.4142142
      \captionsetup{labelformat=empty} \caption{Fig. 1}
      \end{table} The formula in cell B1 is = SQRT (A1)
      and the formula in cell C 1 is \(\quad = \mathrm { B } 1 \wedge 2\).
      Ben evaluates \(1.414214 ^ { 2 }\) on his calculator and obtains 2.000001238 . He states that this shows that the value displayed in cell C1 is wrong. Explain whether Ben is correct.

Question 1:
AnswerMarks Guidance
1(a) i
2
1.414214−√2 1.414214 −2
0.00√02000309449 is2w
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0.000000618898 iswM1
A1
AnswerMarks
A11.1a
1.1
AnswerMarks
1.1ignore modulus signs
to 2 sf or more
to 2 sf or more
[3]
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1(a) ii
error oeB1 2.2a
[1]
AnswerMarks Guidance
1(b) Ben is wrong because the spreadsheet stores
1.414214 to a higher precision than is displayed
(and so when the square of this number is
AnswerMarks Guidance
calculated, 2 is returned) iswB1 2.4
so 1.4142142 ≠ 2 oe
√2
[1]
1
Question 1:
1 | (a) | i | or oe soi
2
1.414214−√2 1.414214 −2
0.00√02000309449 is2w
0.000000618898 isw | M1
A1
A1 | 1.1a
1.1
1.1 | ignore modulus signs
to 2 sf or more
to 2 sf or more
[3]
1 | (a) | ii | the second relative error is double the first relative
error oe | B1 | 2.2a
[1]
1 | (b) | Ben is wrong because the spreadsheet stores
1.414214 to a higher precision than is displayed
(and so when the square of this number is
calculated, 2 is returned) isw | B1 | 2.4 | or 1.414214 is an approximation to
so 1.4142142 ≠ 2 oe
√2
[1]
1
1
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the relative error when

\begin{itemize}
  \item 1.414214 is used to approximate $\sqrt { 2 }$,
  \item $1.414214 ^ { 2 }$ is used to approximate 2.
\item Write down the relationship between your answers to part (a)(i).
\item Fig. 1 shows some spreadsheet output.
\end{itemize}

\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
 & A & B & C &  \\
\hline
1 & 2 & 1.414214 & 2 &  \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{table}

The formula in cell B1 is = SQRT (A1)\\
and the formula in cell C 1 is $\quad = \mathrm { B } 1 \wedge 2$.\\
Ben evaluates $1.414214 ^ { 2 }$ on his calculator and obtains 2.000001238 . He states that this shows that the value displayed in cell C1 is wrong.

Explain whether Ben is correct.
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2021 Q1 [5]}}