OCR MEI Further Numerical Methods 2019 June — Question 2 8 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSign Change & Interval Methods
TypeExplain Sign Change Method Failure
DifficultyStandard +0.3 This is a straightforward conceptual question about numerical methods. Part (a) requires recognizing that false position fails when there's no sign change in the interval (both f(0) and f(1) are positive based on the graph). Part (b) is simple interpretation of given spreadsheet data. This is below average difficulty as it tests basic understanding rather than calculation or problem-solving.
Spec1.09a Sign change methods: locate roots
2 Fig. 2.1 shows the graph of \(y = x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-4_757_545_315_248} \captionsetup{labelformat=empty} \caption{Fig. 2.1}
\end{figure} There are three roots of the equation \(x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5 = 0\). The roots are \(\alpha , \beta\) and \(\gamma\), where \(\alpha < \beta < \gamma\).
  1. Explain why it is not possible to use the method of false position with \(x _ { 0 } = 0\) and \(x _ { 1 } = 1\) to find \(\beta\) and \(\gamma\). The graph of the function indicates that the root \(\gamma\) lies in the interval [0.6, 0.8]. Fig. 2.2 shows some spreadsheet output using the method of false position using these values as starting points. \begin{table}[h]
    ABCDEF
    1af(a)bf(b)approx
    20.6-0.104760.80.4699410.636457-0.07876
    30.636457-0.078760.80.4699410.659931-0.04748
    40.659931-0.047480.80.4699410.672783-0.0249
    50.672783-0.02490.80.4699410.679184-0.01211
    60.679184-0.012110.80.4699410.682218-0.00567
    70.682218-0.005670.80.4699410.683623-0.00261
    80.683623-0.002610.80.4699410.684266-0.00119
    90.684266-0.001190.80.4699410.684559-0.00054
    100.684559-0.000540.80.4699410.684692-0.00025
    110.684692-0.000250.80.4699410.684753-0.00011
    120.684753-0.000110.80.4699410.68478\(- 5.1 \mathrm { E } - 05\)
    \captionsetup{labelformat=empty} \caption{Fig. 2.2}
    \end{table}
  2. Without doing any further calculation, write down the smallest possible interval which is certain to contain \(\gamma\).
  3. State what is being calculated in column F. The formula in cell A3 is \(\quad = \operatorname { IF } ( \mathrm { F } 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )\).
  4. Explain the purpose of this formula in the application of the method of false position. The method of false position uses the same formula for obtaining new approximations as the secant method.
  5. Explain how the method of false position differs from the secant method.
  6. Give one advantage and one disadvantage of using the method of false position instead of the secant method.
Question 2:
AnswerMarks Guidance
2(a) f(0) and f(1) are both positive – a sign change
is needed for this method oeB1
[1]3.1a allow the chord does not cut the
x-axis oethere are 2 roots oe is
insufficient
AnswerMarks Guidance
2(b) 0.68478 < γ < 0.8
[1]2.2a
2(c) the value of f(x ) oe
newB1
[1]1.1 must allude to the value of the
functionallow eg f(approx.)
2(d) the formula represents a logical test for a
change of sign oe
if the value in F2 is negative, the value in E2
will be copied into A3; if it is not negative, the
AnswerMarks
value in A2 will be copied into A3B1
B1
AnswerMarks
[2]3.1a
2.4allow eg establishing the interval
for the next iteration
must consider both possibilities and
must not refer to value in column C
AnswerMarks Guidance
2(e) the secant method does not (necessarily)
incorporate a sign change whereas the method
of false position does require a change of sign
AnswerMarks Guidance
oeB1
[1]1.2 must mention both methods
2(f) eg false position always straddles the root so
always converges
eg converges more slowly
eg harder to implement because it incorporates
AnswerMarks
the programming of a sign checkB1
B1
AnswerMarks
[2]2.4
2.4advantage
disadvantage
Question 2:
2 | (a) | f(0) and f(1) are both positive – a sign change
is needed for this method oe | B1
[1] | 3.1a | allow the chord does not cut the
x-axis oe | there are 2 roots oe is
insufficient
2 | (b) | 0.68478 < γ < 0.8 | B1
[1] | 2.2a
2 | (c) | the value of f(x ) oe
new | B1
[1] | 1.1 | must allude to the value of the
function | allow eg f(approx.)
2 | (d) | the formula represents a logical test for a
change of sign oe
if the value in F2 is negative, the value in E2
will be copied into A3; if it is not negative, the
value in A2 will be copied into A3 | B1
B1
[2] | 3.1a
2.4 | allow eg establishing the interval
for the next iteration
must consider both possibilities and
must not refer to value in column C
2 | (e) | the secant method does not (necessarily)
incorporate a sign change whereas the method
of false position does require a change of sign
oe | B1
[1] | 1.2 | must mention both methods
2 | (f) | eg false position always straddles the root so
always converges
eg converges more slowly
eg harder to implement because it incorporates
the programming of a sign check | B1
B1
[2] | 2.4
2.4 | advantage
disadvantage
2 Fig. 2.1 shows the graph of $y = x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-4_757_545_315_248}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}

There are three roots of the equation $x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5 = 0$. The roots are $\alpha , \beta$ and $\gamma$, where $\alpha < \beta < \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Explain why it is not possible to use the method of false position with $x _ { 0 } = 0$ and $x _ { 1 } = 1$ to find $\beta$ and $\gamma$.

The graph of the function indicates that the root $\gamma$ lies in the interval [0.6, 0.8]. Fig. 2.2 shows some spreadsheet output using the method of false position using these values as starting points.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
 & A & B & C & D & E & F \\
\hline
1 & a & f(a) & b & f(b) & approx &  \\
\hline
2 & 0.6 & -0.10476 & 0.8 & 0.469941 & 0.636457 & -0.07876 \\
\hline
3 & 0.636457 & -0.07876 & 0.8 & 0.469941 & 0.659931 & -0.04748 \\
\hline
4 & 0.659931 & -0.04748 & 0.8 & 0.469941 & 0.672783 & -0.0249 \\
\hline
5 & 0.672783 & -0.0249 & 0.8 & 0.469941 & 0.679184 & -0.01211 \\
\hline
6 & 0.679184 & -0.01211 & 0.8 & 0.469941 & 0.682218 & -0.00567 \\
\hline
7 & 0.682218 & -0.00567 & 0.8 & 0.469941 & 0.683623 & -0.00261 \\
\hline
8 & 0.683623 & -0.00261 & 0.8 & 0.469941 & 0.684266 & -0.00119 \\
\hline
9 & 0.684266 & -0.00119 & 0.8 & 0.469941 & 0.684559 & -0.00054 \\
\hline
10 & 0.684559 & -0.00054 & 0.8 & 0.469941 & 0.684692 & -0.00025 \\
\hline
11 & 0.684692 & -0.00025 & 0.8 & 0.469941 & 0.684753 & -0.00011 \\
\hline
12 & 0.684753 & -0.00011 & 0.8 & 0.469941 & 0.68478 & $- 5.1 \mathrm { E } - 05$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{table}
\item Without doing any further calculation, write down the smallest possible interval which is certain to contain $\gamma$.
\item State what is being calculated in column F.

The formula in cell A3 is $\quad = \operatorname { IF } ( \mathrm { F } 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$.
\item Explain the purpose of this formula in the application of the method of false position.

The method of false position uses the same formula for obtaining new approximations as the secant method.
\item Explain how the method of false position differs from the secant method.
\item Give one advantage and one disadvantage of using the method of false position instead of the secant method.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2019 Q2 [8]}}
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