Question 3 - A-Level Maths

OCR MEI Further Numerical Methods 2024 June — Question 3 6 marks

Exam Board	OCR MEI
Module	Further Numerical Methods (Further Numerical Methods)
Year	2024
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Fixed Point Iteration
Type	Apply iteration to find root (pure fixed point)
Difficulty	Standard +0.8 This is a Further Maths numerical methods question requiring multiple iterations with hyperbolic functions, understanding of convergence/divergence conditions, and knowledge of relaxation methods. While the calculations are systematic, it demands conceptual understanding beyond standard A-level (explaining why iteration fails, applying relaxation with negative λ), placing it moderately above average difficulty.
Spec	1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

3 The equation $x ^ { 2 } - \cosh ( x - 2 ) = 0$ has two roots, $\alpha$ and $\beta$, such that $\alpha < \beta$.

Use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) \text { where } g \left( x _ { n } \right) = \sqrt { \cosh \left( x _ { n } - 2 \right) } \text {, }$$ starting with $x _ { 0 } = 1$, to find $\alpha$ correct to $\mathbf { 3 }$ decimal places. The diagram shows the part of the graphs of $\mathrm { y } = \mathrm { x }$ and $\mathrm { y } = \mathrm { g } ( \mathrm { x } )$ for $0 \leqslant x \leqslant 7$. \includegraphics[max width=\textwidth, alt={}, center]{83a06341-74e9-4f47-9104-e8e0259e7dfa-3_760_657_753_246}
Explain why the iterative formula used to find $\alpha$ cannot successfully be used to find $\beta$, even if $x _ { 0 }$ is very close to $\beta$.
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 $\lambda = - 0.21$ and $x _ { 0 } = 6.4$, to find $\beta$ correct to $\mathbf { 3 }$ decimal places. In part (c) the method of relaxation was used to convert a divergent sequence of approximations into a convergent sequence.
State one other application of the method of relaxation.

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks	Guidance
3	(a)	[1]

(𝑥 =)1.24(220796762)

1 (𝑥 =)1.14(067093071)

2 (𝑥 =)1.1689(9332106)

7 (𝑥 =)1.1685(9725396)

8 (𝑥 =) 1.168754891

9 (𝑥 =) 1.168692142

10 (𝑥 =)1.168717118

11 (𝑥 =) 1.168707177

12

Answer	Marks
[α =] 1.169	M1

A1

Answer	Marks
[2]	1.1
2.2a	must see 1.24…, 1.14…rounded or truncated to 2 dp or more

NB

1.1800(4699025)

1.1642(2624512)

1.1704(9925209)

1.1679(9852762)

for A1, must see two consecutive iterates from 𝑥 to 𝑥 or better

7 12

eg 𝑥 and 𝑥 rounded or truncated to 4 dp or more

9 10

NB

(𝑥 =) 1.168711134

13 (𝑥 =) 1.168709559

14 (𝑥 =) 1.168710186

15 (𝑥 =) 1.168709936

16

Answer	Marks	Guidance
3	(b)	because g′(𝛽) > 1 oe

[1]

Answer	Marks	Guidance
3	(c)	[6.4]

(𝑥 =)6.4037529

1 (𝑥 =)6.4057774

2 (𝑥 =)6.4077693

5 (𝑥 =)6.4079387

6 (𝑥 =)6.4080298

7 (𝑥 =)6.4080788

8 (𝑥 =)6.4081051

9 (𝑥 =)6.4081192

10

Answer	Marks
[𝛽 = ]6.408	M1
A1	1.1
2.2b	must see 6.403…and 6.405… rounded or truncated to 3 dp or more

NB

6.4068675

6.4074539

must see two consecutive iterates from 𝑥 to 𝑥 or better

5 12

eg 𝑥 and 𝑥 rounded or truncated to 4 dp or more

8 9

NB

(𝑥 =)6.4081268

11 (𝑥 =)6.4081309

12 (𝑥 =)6.4081331

13 ( 𝑥 =)6.4081343

14 [2]

Answer	Marks	Guidance
3	(d)	to accelerate the convergence oe (of a sequence
which already converges)	B1	1.2

[1]

Question 3:
3 | (a) | [1]
(𝑥 =)1.24(220796762)
1
(𝑥 =)1.14(067093071)
2
(𝑥 =)1.1689(9332106)
7
(𝑥 =)1.1685(9725396)
8
(𝑥 =) 1.168754891
9
(𝑥 =) 1.168692142
10
(𝑥 =)1.168717118
11
(𝑥 =) 1.168707177
12
[α =] 1.169 | M1
A1
[2] | 1.1
2.2a | must see 1.24…, 1.14…rounded or truncated to 2 dp or more
NB
1.1800(4699025)
1.1642(2624512)
1.1704(9925209)
1.1679(9852762)
for A1, must see two consecutive iterates from 𝑥 to 𝑥 or better
7 12
eg 𝑥 and 𝑥 rounded or truncated to 4 dp or more
9 10
NB
(𝑥 =) 1.168711134
13
(𝑥 =) 1.168709559
14
(𝑥 =) 1.168710186
15
(𝑥 =) 1.168709936
16
3 | (b) | because g′(𝛽) > 1 oe | B1 | 2.4 | must refer to gradient of g(x) at β.
[1]
3 | (c) | [6.4]
(𝑥 =)6.4037529
1
(𝑥 =)6.4057774
2
(𝑥 =)6.4077693
5
(𝑥 =)6.4079387
6
(𝑥 =)6.4080298
7
(𝑥 =)6.4080788
8
(𝑥 =)6.4081051
9
(𝑥 =)6.4081192
10
[𝛽 = ]6.408 | M1
A1 | 1.1
2.2b | must see 6.403…and 6.405… rounded or truncated to 3 dp or more
NB
6.4068675
6.4074539
must see two consecutive iterates from 𝑥 to 𝑥 or better
5 12
eg 𝑥 and 𝑥 rounded or truncated to 4 dp or more
8 9
NB
(𝑥 =)6.4081268
11
(𝑥 =)6.4081309
12
(𝑥 =)6.4081331
13
( 𝑥 =)6.4081343
14
[2]
3 | (d) | to accelerate the convergence oe (of a sequence
which already converges) | B1 | 1.2 | must refer to convergence
[1]

Show LaTeX source

3 The equation $x ^ { 2 } - \cosh ( x - 2 ) = 0$ has two roots, $\alpha$ and $\beta$, such that $\alpha < \beta$.
\begin{enumerate}[label=(\alph*)]
\item Use the iterative formula

$$x _ { n + 1 } = g \left( x _ { n } \right) \text { where } g \left( x _ { n } \right) = \sqrt { \cosh \left( x _ { n } - 2 \right) } \text {, }$$

starting with $x _ { 0 } = 1$, to find $\alpha$ correct to $\mathbf { 3 }$ decimal places.

The diagram shows the part of the graphs of $\mathrm { y } = \mathrm { x }$ and $\mathrm { y } = \mathrm { g } ( \mathrm { x } )$ for $0 \leqslant x \leqslant 7$.\\
\includegraphics[max width=\textwidth, alt={}, center]{83a06341-74e9-4f47-9104-e8e0259e7dfa-3_760_657_753_246}
\item Explain why the iterative formula used to find $\alpha$ cannot successfully be used to find $\beta$, even if $x _ { 0 }$ is very close to $\beta$.
\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 $\lambda = - 0.21$ and $x _ { 0 } = 6.4$, to find $\beta$ correct to $\mathbf { 3 }$ decimal places.

In part (c) the method of relaxation was used to convert a divergent sequence of approximations into a convergent sequence.
\item State one other application of the method of relaxation.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2024 Q3 [6]}}

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