Questions - A-Level Maths

Determine the probability that the total capacitance of a randomly chosen capacitor of Type B and two randomly chosen capacitors of Type A is at least 16 units.
Determine the probability that the capacitance of a randomly chosen capacitor of Type C is within 1 unit of the total capacitance of four randomly chosen capacitors of Type B. When the manufacturer gets a new batch of 1000 capacitors from the supplier, a random sample of 10 of them is tested to check the capacitances. For a new batch of Type C capacitors, summary statistics for the capacitances, $x$ units, of the random sample are as follows. $n = 10$ $$\sum x = 299.6 \quad \sum x ^ { 2 } = 8981.0$$ You should assume that the capacitances of the sample come from a Normally distributed population, but you should not assume that the standard deviation is 0.64 as for previous Type C capacitors.
In this question you must show detailed reasoning. Carry out a hypothesis test at the $5 \%$ significance level to check whether it is reasonable to assume that the capacitors in this batch have the specified mean capacitance for Type C of 30.2 units.

OCR MEI Further Statistics Major 2021 November Q6

14 marks Standard +0.3

6 Cosmic rays passing through the upper atmosphere cause muons, and other types of particle, to be formed. Muons can be detected when they reach the surface of the earth. It is known that the mean number of muons reaching a particular detector is 1.7 per second. The numbers of muons reaching this detector in 200 randomly selected periods of 1 second are shown in Fig. 6.1. \begin{table}[h]

Number of muons	0	1	2	3	4	5	6	$\geqslant 7$
Frequency	34	65	55	24	14	6	2	0

\captionsetup{labelformat=empty} \caption{Fig. 6.1}

\end{table}

Use the values of the sample mean and sample variance to discuss the suitability of a Poisson distribution as a model. The screenshot in Fig. 6.2 shows part of a spreadsheet to assess the goodness of fit of the distribution Po(1.7). \begin{table}[h]

	A	B	C	D	E
1	Number of muons	Observed frequency	Poisson probability	Expected frequency	Chi-squared contribution
2	0	34	0.1827	36.5367	0.1761
3	1	65
4	2	55	0.2640	52.7955	0.0920
5	3	24	0.1496	29.9175	1.1704
6	4	14			0.1299
7	$\geqslant 5$	8	0.0296	5.9230	0.7284

\captionsetup{labelformat=empty} \caption{Fig. 6.2}

\end{table}

Calculate the missing values in each of the following cells.
- C3
- D3
- E3
- Explain why the numbers for 5, 6 and at least 7 muons have been combined into the single category of at least 5 muons, as shown in Fig. 6.2.
- In this question you must show detailed reasoning.
Carry out the test at the 5\% significance level.

OCR MEI Further Statistics Major 2021 November Q7

10 marks Standard +0.3

7 A physiotherapist is investigating hand grip strength in adult women under 30 years old. She thinks that the grip strength of the dominant hand will be on average 2 kg higher than the grip strength of the non-dominant hand. The physiotherapist selects a random sample of 12 adult women under 30 years old and measures the grip strength of each of their hands. She then uses software to produce a $95 \%$ confidence interval for the mean difference in grip strength between the two hands (dominant minus nondominant), as shown in Fig. 7. \begin{table}[h]

}


Result
T Estimate of a Mean
Mean	2.79
s	3.92
SE	1.13161
N	12
df	11
Lower Limit	0.29935
Upper Limit	5.28065
Interval	$2.79 \pm 2.49065$

\captionsetup{labelformat=empty} \caption{Fig. 7} \end{table}

Explain why the physiotherapist used the same people for testing their dominant and nondominant grip strengths.
State any assumptions necessary in order to construct the confidence interval shown in Fig. 7.
Explain whether the confidence interval supports the physiotherapist's belief.
The physiotherapist then finds some data which have previously been collected on grip strength using a sample of 100 adult women. A 95\% confidence interval, based on this sample and calculated using a Normal distribution, for the mean difference in grip strength between the two hands (dominant minus non-dominant) is (1.94, 2.84).
1. For this sample, find
  - the mean difference
the standard deviation of the differences.
(ii) Explain what you would need to know about the nature of this sample if you wanted to draw conclusions about the mean difference in grip strength in the population of adult women.

OCR MEI Further Statistics Major 2021 November Q8

16 marks Standard +0.3

$\mathrm { VO } _ { 2 \max }$ is a measure of athletic fitness. Since $\mathrm { VO } _ { 2 \max }$ is fairly time-consuming and expensive to measure, an exercise scientist wants to predict $\mathrm { VO } _ { 2 _ { \text {max } } }$ from data such as times for running different distances. The scientist uses these data for a random sample of 15 athletes to predict their $\mathrm { V } \mathrm { O } _ { 2 \text { max } }$ values, denoted by $y$, in suitable units. She also obtains accurate measurements of the $\mathrm { V } \mathrm { O } _ { 2 \text { max } }$ values, denoted by $x$, in the same units. The scatter diagram in Fig. 8.1 shows the values of $x$ and $y$ obtained, together with the equation of the regression line of $y$ on $x$ and the value of $r ^ { 2 }$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce557137-f9eb-4c09-a7e3-e4ec626109dc-08_750_1324_660_317} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
\end{figure}
1. Use the regression line to estimate the predicted $\mathrm { VO } _ { 2 \text { max } }$ of an athlete whose accurately measured $\mathrm { VO } _ { 2 \text { max } }$ is 50 .
2. Comment on the reliability of your estimate.
3. The equation of the regression line of $x$ on $y$ is $x = 0.7565 y + 10.493$. Find the coordinates of the point at which the two regression lines meet.
4. State what the point you found in part (iii) represents.
It is known that there is negative correlation between $\mathrm { VO } _ { 2 \text { max } }$ and marathon times in very good runners (those whose best marathon times are under 3 hours). The exercise scientist wishes to know whether the same applies to runners who take longer to run a marathon. She selects a random sample of 20 runners whose best marathon times are between $3 \frac { 1 } { 2 }$ hours and $4 \frac { 1 } { 2 }$ hours and accurately measures their $\mathrm { VO } _ { 2 \text { max } }$. Fig. 8.2 is a scatter diagram of accurately measured $\mathrm { VO } _ { \text {2max } }$, $v$ units, against best marathon time, $t$ hours, for these runners. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce557137-f9eb-4c09-a7e3-e4ec626109dc-09_671_1064_648_319} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
\end{figure}
1. Explain why the exercise scientist comes to the conclusion that a test based on Pearson's product moment correlation coefficient may be valid. Summary statistics for the 20 runners are as follows. $$\sum t = 80.37 \quad \sum v = 970.86 \quad \sum t ^ { 2 } = 324.71 \quad \sum v ^ { 2 } = 47829.24 \quad \sum t v = 3886.53$$
2. Find the value of Pearson's product moment correlation coefficient.
3. Carry out a test at the $5 \%$ significance level to investigate whether there is negative correlation between accurately measured $\mathrm { VO } _ { 2 _ { \text {max } } }$ and best marathon time for runners whose best marathon times are between $3 \frac { 1 } { 2 }$ hours and $4 \frac { 1 } { 2 }$ hours.

OCR MEI Further Statistics Major 2021 November Q9

6 marks Standard +0.3

9 The discrete random variable $X$ has a uniform distribution over the set of all integers between $- n$ and $n$ inclusive, where $n$ is a positive integer.

Given that $n$ is odd, determine $\mathrm { P } \left( \mathrm { X } > \frac { 1 } { 2 } \mathrm { n } \right)$, giving your answer as a single fraction in terms of $n$.
Determine the variance of the sum of 10 independent values of $X$, giving your answer in the form $\mathrm { an } ^ { 2 } + \mathrm { bn }$, where $a$ and $b$ are constants.

OCR MEI Further Statistics Major 2021 November Q10

9 marks Standard +0.3

10 Sarah takes a bus to work each weekday morning and returns each evening. The times in minutes that she has to wait for the bus in the morning and evening are modelled by uniform distributions over the intervals $[ 0,10 ]$ and $[ 0,6 ]$ respectively. The times in minutes for the bus journeys in the morning and evening are modelled by $\mathrm { N } ( 25,4 )$ and $\mathrm { N } ( 28,16 )$ respectively. You should assume that all of the times are independent. The total time in minutes that she takes for her two journeys, including the waiting times, is denoted by the random variable $T$. The spreadsheet below shows the first 20 rows of a simulation of 500 return journeys. It also shows in column H the numbers of values of $T$ that are less than or equal to the corresponding values in column G. For example, there are 156 out of the 500 simulated values of $T$ which are less than or equal to 58 minutes. All of the times have been rounded to 2 decimal places.

	A	B	C	D	E	F	G	H
1	Waiting time morning	Journey time morning	Waiting time evening	Journey time evening	Total time		Total time t	Number $\leqslant \mathbf { t }$
2	0.89	20.78	1.88	26.30	49.86		46	0
3	3.55	21.24	1.04	29.61	55.44		48	4
4	2.13	21.83	2.40	28.64	55.00		50	13
5	5.12	25.04	3.13	24.30	57.60		52	35
6	4.03	27.49	2.19	30.81	64.52		54	57
7	2.47	20.54	4.32	34.61	61.93		56	104
8	3.21	26.93	3.78	27.66	61.58		58	156
9	9.72	24.15	0.63	27.53	62.03		60	218
10	1.59	28.45	0.08	35.87	65.99		62	288
11	7.34	23.04	4.02	24.77	59.17		64	357
12	1.04	24.69	1.66	31.95	59.33		66	408
13	7.17	22.16	2.55	25.39	57.28		68	441
14	5.20	26.97	2.41	30.05	64.62		70	475
15	5.01	26.84	1.88	36.21	69.93		72	490
16	3.76	26.03	2.21	30.96	62.96		74	496
17	0.96	23.72	2.55	29.36	56.59		76	500
18	8.64	24.97	2.82	26.39	62.82
19	0.59	20.82	4.57	31.41	57.38
20	9.85	23.68	5.54	29.92	68.99
01

Use the spreadsheet output to estimate each of the following.
- $\mathrm { P } ( T \leqslant 56 )$
- $\mathrm { P } ( T > 61 )$
- The random variable $W$ is Normally distributed with the same mean and variance as $T$. Find each of the following.
- $\mathrm { P } ( W \leqslant 56 )$
- $\mathrm { P } ( W > 61 )$
- Explain why, if many more journeys were used in the simulation, you would expect $\mathrm { P } ( T > 61 )$ to be extremely close to $\mathrm { P } ( W > 61 )$.

OCR MEI Further Statistics Major 2021 November Q11

11 marks Challenging +1.8

11 The continuous random variable $X$ has probability density function given by $f ( x ) = \begin{cases} a x ^ { 2 } & 0 \leqslant x < 2 , \\ b ( 3 - x ) ^ { 2 } & 2 \leqslant x \leqslant 3 , \\ 0 & \text { otherwise } \end{cases}$ where $a$ and $b$ are positive constants.

Given that $\mathrm { E } ( X ) = 2$, determine the values of $a$ and $b$.
Determine the median value of $X$.
A random sample of 50 observations of $X$ is selected. Given that $\operatorname { Var } ( X ) = 0.2$, determine an estimate of the probability that the mean value of the 50 observations is less than 1.9.

OCR MEI Further Numerical Methods 2019 June Q1

6 marks Moderate -0.5

1 Fig. 1 shows some spreadsheet output concerning the values of a function, $\mathrm { f } ( x )$. \begin{table}[h]

	A	B	C
1	$x$	$\mathrm { f } ( x )$
2	1	0.367879441	0.367879441
3	2	0.018315639	0.38619508
4	3	0.00012341	0.38631849
5	4	$1.12535 \mathrm { E } - 07$	0.386318602
6	5	$1.38879 \mathrm { E } - 11$	0.386318602

\captionsetup{labelformat=empty} \caption{Fig. 1}

\end{table} The formula in cell B2 is ==EXP(-(A2\^{}2)) and equivalent formulae are in cells B3 to B6. The formula in cell C 2 is $= \mathrm { B } 2$.
The formula in cell C3 is $\quad = \mathrm { C } 2 + \mathrm { B } 3$. Equivalent formulae are in cells C4 to C6.

Use sigma notation to express the formula in cell C5 in standard mathematical notation.
Explain why the same value is displayed in cells C 5 and C 6. Now suppose that the value in cell C2 is chopped to 3 decimal places and used to approximate the value in cell C2.
Calculate the relative error when this approximation is used. Suppose that the values in cells B4, B5 and B6 are chopped to 3 decimal places and used as approximations to the original values in cells B4, B5 and B6 respectively.
Explain why the relative errors in these approximations are all the same.

OCR MEI Further Numerical Methods 2019 June Q2

8 marks Standard +0.3

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 $\alpha < \beta < \gamma$.

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]

	A	B	C	D	E	F
1	a	f(a)	b	f(b)	approx
2	0.6	-0.10476	0.8	0.469941	0.636457	-0.07876
3	0.636457	-0.07876	0.8	0.469941	0.659931	-0.04748
4	0.659931	-0.04748	0.8	0.469941	0.672783	-0.0249
5	0.672783	-0.0249	0.8	0.469941	0.679184	-0.01211
6	0.679184	-0.01211	0.8	0.469941	0.682218	-0.00567
7	0.682218	-0.00567	0.8	0.469941	0.683623	-0.00261
8	0.683623	-0.00261	0.8	0.469941	0.684266	-0.00119
9	0.684266	-0.00119	0.8	0.469941	0.684559	-0.00054
10	0.684559	-0.00054	0.8	0.469941	0.684692	-0.00025
11	0.684692	-0.00025	0.8	0.469941	0.684753	-0.00011
12	0.684753	-0.00011	0.8	0.469941	0.68478	$- 5.1 \mathrm { E } - 05$

\captionsetup{labelformat=empty} \caption{Fig. 2.2}

\end{table}

Without doing any further calculation, write down the smallest possible interval which is certain to contain $\gamma$.
State what is being calculated in column F. The formula in cell A3 is $\quad = \operatorname { IF } ( \mathrm { F } 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$.
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.
Explain how the method of false position differs from the secant method.
Give one advantage and one disadvantage of using the method of false position instead of the secant method.

OCR MEI Further Numerical Methods 2019 June Q3

11 marks Standard +0.3

3 In the first week of an outbreak of influenza, 9 patients were diagnosed with the virus at a medical practice in Pencaster. Records were kept of $y$, the total number of patients diagnosed with influenza in week $n$. The data are shown in Fig. 3. \begin{table}[h]

$n$	1	2	3	4	5
$y$	9	32	63	96	125

\captionsetup{labelformat=empty} \caption{Fig. 3}

\end{table}

Complete the difference table in the Printed Answer Booklet.
Explain why a cubic model is appropriate for the data.
Use Newton's method to find the interpolating polynomial of degree 3 for these data. In both week 6 and week 7 there were 145 patients in total diagnosed with influenza at the medical practice.
Determine whether the model is a good fit for these data.
Determine the maximum number of weeks for which the model could possibly be valid.

OCR MEI Further Numerical Methods 2019 June Q4

13 marks Standard +0.8

4 Fig. 4 shows the graph of $y = x ^ { 5 } - 6 \sqrt { x } + 4$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-6_867_700_317_246} \captionsetup{labelformat=empty} \caption{Fig. 4}

\end{figure} There are two roots of the equation $x ^ { 5 } - 6 \sqrt { x } + 4 = 0$. The roots are $\alpha$ and $\beta$, such that $\alpha < \beta$.

Show that $0 < \alpha < 1$ and $1 < \beta < 2$.
Obtain the Newton-Raphson iterative formula $$x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { \frac { 11 } { 2 } } - 6 x _ { n } + 4 \sqrt { x _ { n } } } { 5 x _ { n } ^ { \frac { 9 } { 2 } } - 3 }$$
Use the iterative formula found in part (b) with a starting value of $x _ { 0 } = 1$ to obtain $\beta$ correct to 6 decimal places.
Use the iterative formula found in part (b) with a starting value of $x _ { 0 } = 0$ to find $x _ { 1 }$.
Give a geometrical explanation of why the Newton-Raphson iteration fails to find $\alpha$ in part (d).
Obtain the iterative formula $$x _ { n + 1 } = \left( \frac { x _ { n } ^ { 5 } + 4 } { 6 } \right) ^ { 2 }$$
Use the iterative formula found in part (f) with a starting value of $x _ { 0 } = 0$ to obtain $\alpha$ correct to 6 decimal places.

OCR MEI Further Numerical Methods 2019 June Q5

12 marks Standard +0.8

5 Fig. 5 shows spreadsheet output concerning the estimation of the derivative of a function $\mathrm { f } ( x )$ at $x = 2$ using the forward difference method. \begin{table}[h]

	A	B	C	D
1	h	estimate	difference	ratio
2	0.1	6.3050005
3	0.01	6.0300512	-0.274949
4	0.001	6.0030018	-0.027049	0.098379
5	0.0001	6.0003014	-0.0027	0.099835
6	0.00001	6.0000314	-0.00027	0.099983
7	0.000001	6.0000044	$- 2.7 \mathrm { E } - 05$	0.099994
8	1E-07	6.0000016	$- 2.71 \mathrm { E } - 06$	0.100352
9	1E-08	6.0000013	$- 3.02 \mathrm { E } - 07$	0.111457
10	1E-09	6.0000018	$4.885 \mathrm { E } - 07$	-1.61765
11	1E-10	6.0000049	$3.109 \mathrm { E } - 06$	6.363636
12	1E-11	6.0000005	$- 4.44 \mathrm { E } - 06$	-1.42857
13	1E-12	6.0005334	0.0005329	-120
14	1E-13	5.9952043	-0.005329	-10
15	1E-14	6.1284311	0.1332268	-25
16	1E-15	5.3290705	-0.799361	-6
17	1E-16	0	-5.329071	6.666667

\captionsetup{labelformat=empty} \caption{Fig. 5}

\end{table}

Write down suitable cell formulae for
- cell C3,
- cell D4.
- Explain what the entries in cells D4 to D8 tell you about the order of the convergence of the forward difference method.
- Write the entry in cell A10 in standard mathematical notation.
- Explain what the values displayed in cells D10 to D17 suggest about the values in cells B10 to B16.
- Write down the value of the derivative of $\mathrm { f } ( x )$ at $x = 2$ to an accuracy that seems justified, explaining your answer.
The formula in cell B2 is $\quad = ( \mathrm { LN } ( \mathrm { SQRT } ( \mathrm { SINH } ( ( 2 + \mathrm { A } 2 ) \wedge 3 ) ) ) - \mathrm { LN } ( \mathrm { SQRT } ( \mathrm { SINH } ( 2 \wedge 3 ) ) ) ) / \mathrm { A } 2$ and equivalent formulae are entered in cells B3 to B17.
Write $\mathrm { f } ( x )$ in standard mathematical notation. The value displayed in cell B17 is zero, even though the calculation results in a non-zero answer.
Explain how this has arisen.

OCR MEI Further Numerical Methods 2019 June Q6

10 marks Standard +0.3

6 The spreadsheet output in Fig. 6 shows approximations to $\int _ { 0 } ^ { 1 } x ^ { - \sqrt { x } } \mathrm {~d} x$ found using the midpoint rule, denoted by $M _ { n }$, and the trapezium rule, denoted by $T _ { n }$. \begin{table}[h]

	A	B	C
1	$n$	$M _ { n }$	$T _ { n }$
2	1	1.632527	1
3	2	1.641461	1.316263
4	4	1.623053	1.478862
5	8	1.610295	1.550957
6	16	1.604132	1.580626
7	32	1.601505	1.592379

\captionsetup{labelformat=empty} \caption{Fig. 6}

\end{table}

Write down an efficient spreadsheet formula for cell C3.
By first completing the table in the Printed Answer Booklet using the Simpson's rule, calculate the most accurate estimate of $\int _ { 0 } ^ { 1 } x ^ { - \sqrt { x } } \mathrm {~d} x$ that you can, justifying the precision quoted. \section*{END OF QUESTION PAPER}

OCR MEI Further Numerical Methods 2022 June Q2

7 marks Moderate -0.5

2 The table shows some values of $x$ and the associated values of $y = f ( x )$.

$x$	2.75	3	3.25
$\mathrm { f } ( x )$	0.920799	1	1.072858

Calculate an estimate of $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 3$ using the forward difference method, giving your answer correct to $\mathbf { 5 }$ decimal places.
Calculate an estimate of $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 3$ using the central difference method, giving your answer correct to $\mathbf { 5 }$ decimal places.
Explain why your answer to part (b) is likely to be closer than your answer to part (a) to the true value of $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 3$. When $x = 5$ it is given that $y = 1.4645$ and $\frac { \mathrm { dy } } { \mathrm { dx } } = 0.1820$, correct to 4 decimal places.
Determine an estimate of the error when $\mathrm { f } ( 5 )$ is used to estimate $\mathrm { f } ( 5.024 )$.

OCR MEI Further Numerical Methods 2022 June Q3

7 marks Standard +0.3

3 The equation $\mathrm { f } ( x ) = \sin ^ { - 1 } ( x ) - x + 0.1 = 0$ has a root $\alpha$ such that $- 1 < \alpha < 0$.
Alex uses an iterative method to find a sequence of approximations to $\alpha$. Some of the associated spreadsheet output is shown in the table.

	C	D	E
4	$r$	$\mathrm { x } _ { \mathrm { r } }$	$\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } } \right)$
5	0	- 1	- 0.4707963
6	1	- 0.8	- 0.0272952
7	2	- 0.787691	- 0.0193610
8	3	- 0.7576546	- 0.0020574
9	4	- 0.7540834	- 0.0001740
10	5
11	6

The formula in cell D7 is $$= ( \mathrm { D } 5 * \mathrm { E } 6 - \mathrm { D } 6 * \mathrm { E } 5 ) / ( \mathrm { E } 6 - \mathrm { E } 5 )$$ and equivalent formulae are in cells D8 and D9.

State the method being used.
Use the values in the spreadsheet to calculate $x _ { 5 }$ and $x _ { 6 }$, giving your answers correct to 7 decimal places.
State the value of $\alpha$ as accurately as you can, justifying the precision quoted. Alex uses a calculator to check the value in cell D9, his result is - 0.7540832686 .
Explain why this is different to the value displayed in cell D9. The value displayed in cell E11 in Alex's spreadsheet is $- 1.4629 \mathrm { E } - 09$.
Write this value in standard mathematical notation.

OCR MEI Further Numerical Methods 2022 June Q4

8 marks Standard +0.8

4 Fig. 4.1 shows part of the graph of $y = e ^ { x } - x ^ { 2 } - x - 1.1$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_805_789_299_274} \captionsetup{labelformat=empty} \caption{Fig. 4.1}

\end{figure} The equation $\mathrm { e } ^ { x } - x ^ { 2 } - x - 1.1 = 0$ has a root $\alpha$ such that $1 < \alpha < 2$.
Ali is considering using the Newton-Raphson method to find $\alpha$. Ali could use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$.

Without doing any calculations, explain whether Ali should use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$, or whether using either starting value would work equally well. Ali is also considering using the method of fixed point iteration to find $\alpha$. Ali could use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$. Fig. 4.2 shows parts of the graphs of $y = x$ and $y = \ln \left( x ^ { 2 } + x + 1.1 \right)$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_819_1011_1818_255} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
\end{figure}

Without doing any calculations, explain whether Ali should use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$ or whether either starting value would work equally well. Ali used one of the above methods to find a sequence of approximations to $\alpha$. These are shown, together with some further analysis in the associated spreadsheet output in Fig. 4.3. \begin{table}[h]

		M	N	O
	$r$	$\mathrm { X } _ { \mathrm { r } }$
4	0	2
5	1	1.879008	- 0.121
6	2	1.858143	- 0.021	0.172
7	3	1.857565	$- 6 \mathrm { E } - 04$	0.028
8	4	1.857564	$- 4 \mathrm { E } - 07$	$8 \mathrm { E } - 04$
9	5	1.857564	$- 2 \mathrm { E } - 13$	$6 \mathrm { E } - 07$

\captionsetup{labelformat=empty} \caption{Fig. 4.3}

\end{table} The formula in cell N5 is =M5-M4
and the formula in cell O6 is =N6/N5
equivalent formulae are in cells N6 to N9 and O7 to O9 respectively.

State what is being calculated in the following columns of the spreadsheet.
1. Column N
2. Column O
Explain whether the values in column O suggest that Ali used the Newton-Raphson method or the iterative formula $\mathrm { x } _ { \mathrm { n } + 1 } = \ln \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } + \mathrm { x } _ { \mathrm { n } } + 1.1 \right)$ to find this sequence of approximations to $\alpha$.

OCR MEI Further Numerical Methods 2022 June Q5

9 marks Standard +0.3

5 Kai uses the midpoint rule, trapezium rule and Simpson's rule to find approximations to $\int _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { f } ( \mathrm { x } ) \mathrm { dx }$, where $a$ and $b$ are constants. The associated spreadsheet output is shown in the table. Some of the values are missing.

	F	G	H	I
3	$n$	$\mathrm { M } _ { \mathrm { n } }$	$\mathrm { T } _ { \mathrm { n } }$	$\mathrm { S } _ { 2 \mathrm { n } }$
4	1	0.2436699	0.1479020
5	2	0.2306967

Write down a suitable spreadsheet formula for cell H 5 .
Complete the copy of the table in the Printed Answer Booklet, giving the values correct to 7 decimal places.
Use your answers to part (b) to determine the value of $\int _ { a } ^ { b } f ( x ) d x$ as accurately as you can, justifying the precision quoted.

OCR MEI Further Numerical Methods 2022 June Q6

11 marks Standard +0.8

6 Charlie uses fixed point iteration to find a sequence of approximations to the root of the equation $\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0$. Charlie uses the iterative formula $\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)$, where $\mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = \sin \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } - 1 \right)$.
Two sections of the associated spreadsheet output, showing $x _ { 0 }$ to $x _ { 6 }$ and $x _ { 102 }$ to $x _ { 108 }$, are shown in Fig. 6.1.

$r$	$\mathrm { x } _ { \mathrm { r } }$	difference	ratio
0	0
1	-0.841471	-0.84147
2	-0.287798	0.553673	-0.65798
3	-0.793885	-0.50609	-0.91405
4	-0.361379	0.432507	-0.85461
5	-0.763945	-0.40257	-0.93078
6	-0.404459	0.359486	-0.89299

\begin{table}[h]

102	-0.596302	0.004626	-0.95886
103	-0.600738	-0.00444	-0.95911
104	-0.596484	0.004254	-0.95887
105	-0.600564	-0.00408	-0.95910
106	-0.596652	0.003912	-0.95888
107	-0.600404	-0.00375	-0.95909
108	-0.596806	0.003598	-0.95889

\captionsetup{labelformat=empty} \caption{Fig. 6.1}

\end{table}

Use the information in Fig. 6.1 to find the value of the root as accurately as you can, justifying the precision quoted. The relaxed iteration $\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)$, with $\lambda = 0.51$ and $x _ { 0 } = 0$, is to be used to find the root of the equation $\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0$.
Complete the copy of Fig. 6.2 in the Printed Answer Booklet, giving the values of $\mathrm { x } _ { \mathrm { r } }$ correct to 7 decimal places and the values in the difference column and ratio column correct to 3 significant figures. \begin{table}[h]
$r$ $\mathrm { x } _ { \mathrm { r } }$ difference ratio
0 0
1
2
3
4 -0.000192
5 $- 1.99 \times 10 ^ { - 7 }$ 0.00103
6 $- 1.82 \times 10 ^ { - 10 }$ 0.000914
\captionsetup{labelformat=empty} \caption{Fig. 6.2}
\end{table}
Write down the value of the root correct to 7 decimal places.
Explain why extrapolation could not be used in this case to find an improved approximation using this sequence of iterates. In this case the method of relaxation has been used to speed up the convergence of an iterative scheme.
Name another application of the method of relaxation.

OCR MEI Further Numerical Methods 2022 June Q7

14 marks Standard +0.8

7 Sam decided to go on a high-protein diet. Sam's mass in $\mathrm { kg } , M$, after $t$ days of following the diet is recorded in Fig. 7.1. \begin{table}[h]

$t$	0	10	20	30
$M$	88.3	80.05	78.7	78.85

\captionsetup{labelformat=empty} \caption{Fig. 7.1}

\end{table} A difference table for the data is shown in Fig. 7.2. \begin{table}[h]

$t$	$M$	$\Delta M$	$\Delta ^ { 2 } M$	$\Delta ^ { 3 } M$
0	88.3

10	80.05

20	78.7

30	78.85

\captionsetup{labelformat=empty} \caption{Fig. 7.2}

\end{table}

Complete the copy of the difference table in the Printed Answer Booklet. Sam's doctor uses these data to construct a cubic interpolating polynomial to model Sam's mass at time $t$ days after starting the diet.
Find the model in the form $\mathrm { M } = \mathrm { at } ^ { 3 } + \mathrm { bt } ^ { 2 } + \mathrm { ct } + \mathrm { d }$, where $a , b , c$ and $d$ are constants to be determined. Subsequently it is found that when $\mathrm { t } = 40 , \mathrm { M } = 78.7$ and when $\mathrm { t } = 50 , \mathrm { M } = 80.05$.
Determine whether the model is a good fit for these data.
By completing the extended copy of Fig. 7.2 in the Printed Answer Booklet, explain why a quartic model may be more appropriate for the data.
Refine the doctor's model to include a quartic term.
Explain whether the new model for Sam's mass is likely to be appropriate over a longer period of time.

OCR MEI Further Numerical Methods 2023 June Q1

7 marks Standard +0.8

1 You are given that $\left( x _ { 1 } , y _ { 1 } \right) = ( 0.9,2.3 )$ and $\left( x _ { 2 } , y _ { 2 } \right) = ( 1.1,2.7 )$.
The values of $x _ { 1 }$ and $x _ { 2 }$ have been rounded to $\mathbf { 1 }$ decimal place.

Determine the range of possible values of $x _ { 2 } - x _ { 1 }$. The values of $y _ { 1 }$ and $y _ { 2 }$ have been chopped to $\mathbf { 1 }$ decimal place.
Determine the range of possible values of $y _ { 2 } - y _ { 1 }$. You are given that $m = \frac { y _ { 2 } - y _ { 1 } } { x _ { 2 } - x _ { 1 } }$.
Determine the range of possible values of $m$.
Explain why your answer to part (c) is much larger than your answer to part (a) and your answer to part (b).

OCR MEI Further Numerical Methods 2023 June Q2

8 marks Standard +0.3

2 A car tyre has a slow puncture. Initially the tyre is inflated to a pressure of 34.5 psi . The pressure is checked after 3 days and then again after 5 days. The time $t$ in days and the pressure, $P$ psi, are shown in the table below. You are given that the pressure in a car tyre is measured in pounds per square inch (psi).

$t$	0	3	5
$P$	34.5	29.4	27.0

The owner of the car believes the relationship between $P$ and $t$ may be modelled by a polynomial.

Explain why it is not possible to use Newton's forward difference interpolation method for these data.
Use Lagrange's form of the interpolating polynomial to find an interpolating polynomial of degree 2 for these data. The car owner uses the polynomial found in part (b) to model the relationship between $P$ and $t$.
Subsequently it is found that when $t = 6 , P = 26.0$ and when $t = 10 , P = 24.4$.
Determine whether the owner's model is a good fit for these data.
Explain why the model would not be suitable in the long term.

OCR MEI Further Numerical Methods 2023 June Q3

6 marks Standard +0.3

3 The diagram shows the graph of $y = f ( x )$ for values of $x$ from 1 to 3.5. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-03_945_1248_312_244} The table shows some values of $x$ and the associated values of $y$.

$x$	1.5	2	2.5
$y$	1.682137	2.094395	2.318559

Use the forward difference method to calculate an approximation to $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 2$.
Use the central difference method to calculate an approximation to $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 2$.
On the copy of the diagram in the Printed Answer Booklet, show how the central difference method gives the approximation to $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 2$ which was found in part (b).
Explain whether your answer to part (a) or your answer to part (b) is likely to give a better approximation to $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 2$.

OCR MEI Further Numerical Methods 2023 June Q4

4 marks Moderate -0.8

4 A spreadsheet is used to approximate $\int _ { a } ^ { b } f ( x ) d x$ using the midpoint rule with 1 strip. The output is shown in the table below.

	B	C	D
3	$x$	$\mathrm { f } ( x )$	$\mathrm { M } _ { 1 }$
4	1.5	1.3103707	0.65518535

The formula in cell C4 is $= \mathrm { B } 4 \wedge ( 1 / \mathrm { B } 4 )$.
The formula in cell D4 is $= 0.5 ^ { * } \mathrm { C } 4$.

Write the integral in standard mathematical notation. A graph of $y = f ( x )$ is included in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-04_789_1004_1199_235}
Explain whether 0.65518535 is an over-estimate or an under-estimate of $\int _ { a } ^ { b } f ( x ) d x$.

OCR MEI Further Numerical Methods 2023 June Q5

6 marks Standard +0.8

5 The equation $3 - 2 \ln x - x = 0$ has a root near $x = 1.8$.
A student proposes to use the iterative formula $\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = 3 - 2 \ln \mathrm { x } _ { \mathrm { n } }$ to find this root.
The diagram shows the graphs of $\mathrm { y } = \mathrm { x }$ and $\mathrm { y } = \mathrm { g } ( \mathrm { x } )$ for values of $x$ from - 2 to 6 . \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-05_913_917_502_233}

With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near $x = 1.8$.
Use the relaxed iteration $\mathrm { x } _ { \mathrm { n } + 1 } = \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) + ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } }$, with $\lambda = 0.475$ and $x _ { 0 } = 2$, to determine the root correct to $\mathbf { 6 }$ decimal places. A student uses the same relaxed iteration with the same starting value. Some analysis of the iterates is carried out using a spreadsheet, which is shown in the table below.
$r$ difference ratio
0
1 - 0.1834898
2 - 0.0049137 0.02678
3 $- 6.44 \mathrm { E } - 06$ 0.00131
4 $- 3.862 \mathrm { E } - 09$ 0.0006
5 $- 2.313 \mathrm { E } - 12$ 0.0006
Explain what the analysis tells you about the order of convergence of this sequence of approximations.

OCR MEI Further Numerical Methods 2023 June Q6

6 marks Standard +0.8

1. Calculate the relative error when $\pi$ is chopped to $\mathbf { 2 }$ decimal places in approximating $$\pi ^ { 2 } + 2 .$$
2. Without doing any calculation, explain whether the relative error would be the same when $\pi$ is chopped to 2 decimal places when approximating $( \pi + 2 ) ^ { 2 }$. The table shows some spreadsheet output. The values of $x$ in column A are exact.
  A B C
  1 $x$ $10 ^ { x }$ $\log _ { 10 } 10 ^ { x }$
  2 $1 \mathrm { E } - 12$ 1 $1.00001 \mathrm { E } - 12$
  3 $1 \mathrm { E } - 11$ 1 $9.99998 \mathrm { E } - 12$
  The formula in cell B2 is $= 10 ^ { \wedge } \mathrm { A } 2$.
  This has been copied down to cell B3.
  The formula in cell C2 is $\quad =$ LOG(B2) .
  This formula has been copied down to cell C3.
1. Write the value displayed in cell C 2 in standard mathematical notation.
2. Explain why the values in cells C 2 and C 3 are neither zero nor the same as the values in cells A2 and A3 respectively.

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