Question 6 - A-Level Maths

OCR MEI Further Numerical Methods 2021 November — Question 6 12 marks

Exam Board	OCR MEI
Module	Further Numerical Methods (Further Numerical Methods)
Year	2021
Session	November
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Fixed Point Iteration
Type	Apply iteration to find root (pure fixed point)
Difficulty	Challenging +1.2 This is a structured Further Maths numerical methods question covering Newton-Raphson and fixed point iteration. While it requires multiple techniques (Newton-Raphson, understanding convergence failure, relaxed iteration), each part is guided with clear instructions. The relaxed iteration in part (f) is given explicitly with the λ value provided, making it computational rather than requiring deep insight. Slightly above average difficulty due to Further Maths content and multi-part nature, but the scaffolding keeps it accessible.
Spec	1.09d Newton-Raphson method

6 The equation $0.5 \ln x - x ^ { 2 } + x + 1 = 0$ has two roots $\alpha$ and $\beta$, such that $0 < \alpha < 1$ and $1 < \beta < 2$.

Use the Newton-Raphson method with $x _ { 0 } = 1$ to obtain $\beta$ correct to $\mathbf { 6 }$ decimal places. Fig. 6.1 shows part of the graph of $y = 0.5 \ln x - x ^ { 2 } + x + 1$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
\end{figure}
On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain $x _ { 1 }$ from $x _ { 0 } = 1$. Beth is trying to find $\alpha$ correct to 6 decimal places.
Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration. Beth tries to find $\alpha$ using the Newton-Raphson method with a starting value of $x _ { 0 } = 0.5$. Her spreadsheet output is shown in Fig. 6.2. \begin{table}[h]
$r$ $\mathrm { x } _ { \mathrm { r } }$
0 0.5
1 - 0.40343
2 \#NUM!
\captionsetup{labelformat=empty} \caption{Fig. 6.2}
\end{table}
Explain how the display \#NUM! has arisen in the cell for $x _ { 2 }$. Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
Determine the outcome when Beth uses this formula with $x _ { 0 } = 0.5$.
Use the relaxed iteration $\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)$ with $\lambda = - 0.041$ and $x _ { 0 } = 0.5$ to obtain $\alpha$ correct to $\mathbf { 6 }$ decimal places.

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6	(a)	1

−2x+1 seen

2x

0.5ln(x )−x2 + x +1

x =x − n n n oe soi

n+1 n 1

−2x +1

n

2x

Answer	Marks	Guidance
n	M1
M1	2.1
1.1	may be implied by correct iterates	condone omission of

subscripts

1

3 2.0791668

1.7783346

1.7360141

1.7351281

1.7351277

Answer	Marks
1.735128	M1
A1	1.1
1.1	at least three further correct iterates

derived from starting at 1

if M0 allow SC1 for 1.735128 from

N-R method used with different

and at least 3 correct iterates shown

Answer	Marks
𝑥𝑥0	correct to at least 5 sf

where appropriate

[4]

Answer	Marks	Guidance
6	(b)	M1
A1	2.4
1.1	tangent at (1,1)

(1,1) to (3,0)

[2]

Answer	Marks	Guidance
6	(c)	N-R generally has 2nd order convergence whereas

fixed point iteration generally has 1st order

Answer	Marks	Guidance
convergence	B1	2.4

allow eg fixed point iteration more

likely to fail oe

[1]

Answer	Marks	Guidance
6	(d)	ln(-0.403) is undefined (so the spreadsheet
cannot compute a value )	B1	2.2a

[1]

Answer	Marks	Guidance
6	(e)	0.5

1.0739769

1.4524673

1.6245304

1.6932631

1.7194743

1.7293015

Answer	Marks
converges to β	M1
A1	2.1
2.2a	need to see at least 3 iterates correct to

at least 5 sf

[2]

0.5 1.0739769

1.4524673

1.6245304

1.6932631

1.7194743

1.7293015

Answer	Marks	Guidance
6	(f)	0.5

0.4764669

0.4528879

0.4293074

0.4057756

0.3823498

…

0.1116318

0.1111278

0.1110835

0.1110821

Answer	Marks
0.111082	M1
A1	1.1
2.2a	at least 3 correct iterates derived from

starting at 0.5

if M0 allow SC1 for 0.111082 from

relaxation method used with different

and at least 3 correct iterates shown

Answer	Marks
𝑥𝑥0	iterates correct to at least

5 sf

[2]

Question 6:
6 | (a) | 1
−2x+1 seen
2x
0.5ln(x )−x2 + x +1
x =x − n n n oe soi
n+1 n 1
−2x +1
n
2x
n | M1
M1 | 2.1
1.1 | may be implied by correct iterates | condone omission of
subscripts
1
3
2.0791668
1.7783346
1.7360141
1.7351281
1.7351277
1.735128 | M1
A1 | 1.1
1.1 | at least three further correct iterates
derived from starting at 1
if M0 allow SC1 for 1.735128 from
N-R method used with different
and at least 3 correct iterates shown
𝑥𝑥0 | correct to at least 5 sf
where appropriate
[4]
6 | (b) | M1
A1 | 2.4
1.1 | tangent at (1,1)
(1,1) to (3,0)
[2]
6 | (c) | N-R generally has 2nd order convergence whereas
fixed point iteration generally has 1st order
convergence | B1 | 2.4 | allow eg N-R converges faster
allow eg fixed point iteration more
likely to fail oe
[1]
6 | (d) | ln(-0.403) is undefined (so the spreadsheet
cannot compute a value ) | B1 | 2.2a
[1]
6 | (e) | 0.5
1.0739769
1.4524673
1.6245304
1.6932631
1.7194743
1.7293015
converges to β | M1
A1 | 2.1
2.2a | need to see at least 3 iterates correct to
at least 5 sf
[2]
0.5
1.0739769
1.4524673
1.6245304
1.6932631
1.7194743
1.7293015
6 | (f) | 0.5
0.4764669
0.4528879
0.4293074
0.4057756
0.3823498
…
0.1116318
0.1111278
0.1110835
0.1110821
0.1110821
0.111082 | M1
A1 | 1.1
2.2a | at least 3 correct iterates derived from
starting at 0.5
if M0 allow SC1 for 0.111082 from
relaxation method used with different
and at least 3 correct iterates shown
𝑥𝑥0 | iterates correct to at least
5 sf
[2]

Show LaTeX source

6 The equation $0.5 \ln x - x ^ { 2 } + x + 1 = 0$ has two roots $\alpha$ and $\beta$, such that $0 < \alpha < 1$ and $1 < \beta < 2$.
\begin{enumerate}[label=(\alph*)]
\item Use the Newton-Raphson method with $x _ { 0 } = 1$ to obtain $\beta$ correct to $\mathbf { 6 }$ decimal places.

Fig. 6.1 shows part of the graph of $y = 0.5 \ln x - x ^ { 2 } + x + 1$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
\item On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain $x _ { 1 }$ from $x _ { 0 } = 1$.

Beth is trying to find $\alpha$ correct to 6 decimal places.
\item Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration.

Beth tries to find $\alpha$ using the Newton-Raphson method with a starting value of $x _ { 0 } = 0.5$. Her spreadsheet output is shown in Fig. 6.2.

\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | l | }
\hline
$r$ & \multicolumn{1}{|c|}{$\mathrm { x } _ { \mathrm { r } }$} \\
\hline
0 & 0.5 \\
\hline
1 & - 0.40343 \\
\hline
2 & \#NUM! \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{table}
\item Explain how the display \#NUM! has arisen in the cell for $x _ { 2 }$.

Beth decides to use the iterative formula

$$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
\item Determine the outcome when Beth uses this formula with $x _ { 0 } = 0.5$.
\item Use the relaxed iteration $\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)$ with $\lambda = - 0.041$ and $x _ { 0 } = 0.5$ to obtain $\alpha$ correct to $\mathbf { 6 }$ decimal places.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2021 Q6 [12]}}

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Q1 5 Q2 6 Q3 7 Q4 6 Q5 9 Q6 12 Q7 15

\(r\)	\(\mathrm { x } _ { \mathrm { r } }\)
0	0.5
1	- 0.40343
2	\#NUM!