2 In this question you must show detailed reasoning.
Solve the equation \(2 \cosh ^ { 2 } x + 5 \sinh x - 5 = 0\) giving each answer in the form \(\ln ( p + q \sqrt { r } )\) where \(p\) and \(q\) are rational numbers, and \(r\) is an integer, whose values are to be determined.
You are given that the matrix \(\mathbf { A } = \left( \begin{array} { c c c } 1 & 0 & 0
0 & \frac { 2 a - a ^ { 2 } } { 3 } & 0
0 & 0 & 1 \end{array} \right)\), where \(a\) is a positive constant, represents the transformation R which is a reflection in 3-D.
- State the plane of reflection of \(R\).
- Determine the value of \(a\).
- With reference to R explain why \(\mathbf { A } ^ { 2 } = \mathbf { I }\), the \(3 \times 3\) identity matrix.
- By using Euler's formula show that \(\cosh ( \mathrm { iz } ) = \cos z\).
- Hence, find, in logarithmic form, a root of the equation \(\cos z = 2\). [You may assume that \(\cos z = 2\) has complex roots.]
A swing door is a door to a room which is closed when in equilibrium but which can be pushed open from either side and which can swing both ways, into or out of the room, and through the equilibrium position. The door is sprung so that when displaced from the equilibrium position it will swing back towards it.
The extent to which the door is open at any time, \(t\) seconds, is measured by the angle at the hinge, \(\theta\), which the plane of the door makes with the plane of the equilibrium position. See the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-03_317_954_497_255}
In an initial model of the motion of a certain swing door it is suggested that \(\theta\) satisfies the following differential equation.
$$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 25 \theta = 0$$ - Write down the general solution to (\textit{).
- With reference to the behaviour of your solution in part (a)(i) explain briefly why the model using (}) is unlikely to be realistic.
In an improved model of the motion of the door an extra term is introduced to the differential equation so that it becomes
$$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + \lambda \frac { \mathrm { d } \theta } { \mathrm {~d} t } + 25 \theta = 0$$
where \(\lambda\) is a positive constant.
- In the case where \(\lambda = 16\) the door is held open at an angle of 0.9 radians and then released from rest at time \(t = 0\).
- Find, in a real form, the general solution of ( \(\dagger\) ).
- Find the particular solution of ( \(\dagger\) ).
- With reference to the behaviour of your solution found in part (b)(ii) explain briefly how the extra term in ( \(\dagger\) ) improves the model.
- Find the value of \(\lambda\) for which the door is critically damped.
\section*{Total Marks for Question Set 1: 37}
\section*{Resource Materials}
1(a)
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-04_344_621_212_794}
1(b)(i)
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-04_355_556_648_685}
1(b)(ii)
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-04_355_556_1160_685}
\section*{Mark scheme}
\section*{Marking Instructions}
a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available.
\section*{M}
A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly.
\section*{A}
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded.
\section*{B}
Mark for a correct result or statement independent of Method marks.
\section*{E}
A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result.
Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
- When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
- When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:
- If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
- If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
- if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero.
\begin{table}[h]
\captionsetup{labelformat=empty}
\caption{Abbreviations}
| Abbreviations used in the mark scheme | Meaning |
| dep* | Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark |
| cao | Correct answer only |
| ое | Or equivalent |
| rot | Rounded or truncated |
| soi | Seen or implied |
| www | Without wrong working |
| AG | Answer given |
| awrt | Anything which rounds to |
| BC | By Calculator |
| DR | This question included the instruction: In this question you must show detailed reasoning. |
| Question | Answer | Marks | AO | Guidance |
| 1 | (a) | | \includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_395_674_200_575} | | | 4 lines drawn as shown to complete a parallelogram | Or 2 lines drawn to form a triangle which is either the upper or lower half of the parallelogram (split by the leading diagonal). eg |
| (b) | (i) | \includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_374_630_712_568} | | | | \(z _ { 3 }\) and \(z _ { 4 }\) approximately correctly positioned and labelled. | | Approximate correct length (eg \(z _ { 4 }\) length increased by 50\%) and angle (about a quarter of the way round the \(2 ^ { \text {nd } }\) quadrant). |
| | If no labels shown then B1B1 can only follow if there is no ambiguity between points (eg magnitudes shown). | | \(r = 1.8 , \theta = \frac { 5 } { 8 } \pi\) |
|
| (b) | (ii) | \includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_385_634_1154_578} | | | | \(z _ { 3 }\) and \(z _ { 4 }\) approximately correctly positioned and labelled. | | Approximate correct length (eg \(z _ { 3 }\) length halved) and either the same angle as part (b)(i) or about a quarter of the way round the \(2 ^ { \text {nd } }\) quadrant. |
| If no labels shown then B1B1 can only follow if there is no ambiguity between points (eg magnitudes shown). \(r = 0.35 , \theta = \frac { 5 } { 8 } \pi\) |
| Question | Answer | Marks | AO | Guidance |
| 2 | | \(\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1\) | M1 | 1.1a | Reduction to 3 term quadratic in \(\sinh x\) or \(\cosh ^ { 2 } x\) | |
| | | A1 | 1.1 | | |
| | | M1 | 1.1 | Use of \(\ln\) formula for \(\sinh ^ { - 1 }\) or \(\cosh ^ { - }\) 1 | |
| | \(\begin{aligned} | \sinh x = 1 / 2 \text { or } - 3 |
| x = \ln \left( \frac { 1 } { 2 } + \sqrt { \frac { 5 } { 4 } } \right) |
| x = \ln \left( \frac { 1 } { 2 } + \frac { 1 } { 2 } \sqrt { 5 } \right) \end{aligned}\) | | | Must be in the correct form but | |
| | \(x = \ln ( - 3 + \sqrt { 10 } )\) | A1 | 1.1 | | |
| | | [6] | | | |
| 3 | (a) | The \(x\) - \(z\) plane | | 2.2a | or \(y = 0\) | |
| (b) | \(\begin{aligned} | \frac { 2 a - a ^ { 2 } } { 3 } = - 1 |
| a ^ { 2 } - 2 a - 3 = 0 \Rightarrow a = - 1,3 |
| a > 0 \Rightarrow a = 3 \end{aligned}\) | B1 | 1.1 | \multirow{3}{*}{BC. Rearranging the quadratic equation and solving. discarding \(a = - 1\)} | \multirow{3}{*}{} |
| | | М1 | 3.1a | | |
| | | A1 [3] | 2.3 | | |
| \multirow[t]{3}{*}{(c)} | \multirow[t]{3}{*}{Any reflection is self-inverse... oe
\(\text { …so } \mathbf { A } ^ { 2 } = \mathbf { A } \mathbf { A } ^ { - 1 } = \mathbf { I }\)} | B1 | 2.4 | eg "If you do a reflection twice it gets back to where it started" | |
| | | B1 | 2.4 | | |
| | | [2] | | | |
| Question | Answer | Marks | AO | Guidance |
| 4 | (a) | | \(\begin{aligned} | \cosh ( \mathrm { i } z ) = \frac { \mathrm { e } ^ { \mathrm { i } z } + \mathrm { e } ^ { - \mathrm { i } z } } { 2 } |
| = \frac { \cos z + \mathrm { i } \sin z + \cos z - \mathrm { i } \sin z } { 2 } |
| = \frac { 2 \cos z } { 2 } = \cos z \quad \mathbf { A G } \end{aligned}\) | | | | Use of correct exponential form for cosh | | Correct use of Euler's formula at least once | | Both \(\mathbf { M }\) marks must be awarded. Must have \(\cosh ( \mathrm { i } z ) =\) or LHS = |
| Proof must be complete for A1 |
| (b) | | \(\begin{aligned} | \cos z = 2 = > \cosh ( \mathrm { i } z ) = 2 = > z = \left( \cosh ^ { - 1 } 2 \right) / \mathrm { i } |
| = - \mathrm { i } \ln ( 2 + \sqrt { 3 } ) \end{aligned}\) | | 3.1a 1.1 | ± inside or outside the \(\ln\) (ie allow eg \(i \ln ( 2 + \sqrt { 3 } )\) or \(i \ln ( 2 - \sqrt { 3 } )\) and condone eg \(\pm \mathrm { i } \ln ( 2 + \sqrt { } 3 )\) www) | or \(2 \pi n \pm \mathrm { i } \ln ( 2 + \sqrt { } 3 )\) for any integer \(n\) |
| 5 | (a) | (i) | | \(\frac { \mathrm { d } ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } = - \left( \frac { 5 } { 2 } \right) ^ { 2 } \theta\) | | \(\theta = A \cos \omega t + B \sin \omega t\) or \(R \cos ( \omega t + \phi )\) with any positive value for \(\omega\) \(\theta = A \cos \frac { 5 } { 2 } t + B \sin \frac { 5 } { 2 } t \text { or } R \cos \left( \frac { 5 } { 2 } t + \phi \right)\) |
| | | | If \(\mathbf { M 0 }\) then \(\mathbf { S C 1 }\) for \(\theta = A \cos \frac { 5 } { 2 } t\) or \(\theta = A \sin \frac { 5 } { 2 } t\) |
| (a) | (ii) | The model predicts infinite oscillations of the same amplitude; in practice the amplitude must decrease over time. | | 3.5b | | |
| Question | Answer | Marks | AO | Guidance |
| (b) | (i) | AE: \(4 m ^ { 2 } + 16 m + 25 = 0\) | M1 | 1.1a | Writing down the AE correctly or using \(\theta = A \mathrm { e } ^ { m t }\) and substituting into (*) to derive a three term quadratic AE. | |
| | \(\begin{aligned} | - 2 \pm \frac { 3 } { 2 } \mathrm { i } |
| \theta = \mathrm { e } ^ { - 2 t } \left( A \cos \frac { 3 } { 2 } t + B \sin \frac { 3 } { 2 } t \right) \end{aligned}\) | A1ft | 1.1 | Their \(\mathrm { e } ^ { p t } ( A \cos q t + B \sin q t )\) for solution of \(\mathrm { AE } = p \pm q \mathrm { i }\) | |
| | | [3] | | | |
| (b) | (ii) | \(\begin{aligned} | t = 0 , \theta = 0.9 \Rightarrow A = 0.9 |
| \frac { \mathrm {~d} \theta } { \mathrm {~d} t } = - 2 \mathrm { e } ^ { - 2 t } \left( A \cos \frac { 3 } { 2 } t + B \sin \frac { 3 } { 2 } t \right) |
| + \mathrm { e } ^ { - 2 t } \left( - \frac { 3 } { 2 } A \sin \frac { 3 } { 2 } t + \frac { 3 } { 2 } B \cos \frac { 3 } { 2 } t \right) |
| t = 0 , \frac { \mathrm {~d} \theta } { \mathrm {~d} t } = 0 \Rightarrow - 2 A + \frac { 3 } { 2 } B = 0 |
| B = 1.2 |
| \theta = \mathrm { e } ^ { - 2 t } \left( 0.9 \cos \frac { 3 } { 2 } t + 1.2 \sin \frac { 3 } { 2 } t \right) \end{aligned}\) | B1 M1 | 3.4 1.1 a | Attempt to differentiate using the product and chain rules ( \(A\) may be replaced by a number). | |
| | | | | Substituting \(t = 0\) into \(\frac { \mathrm { d } \theta } { \mathrm { d } t }\) to derive an equation in ( \(A\) and) \(B\) | |
| | | A1 | 1.1 | | |
| | | [4] | | | |
| (b) | (iii) | In the modified model \(\theta \rightarrow 0\) as \(t \rightarrow \infty\) oe This is the behaviour we would expect to observe with a real swing door and so the model is an improvement. | B1 B1 | 3.5a 3.5c | ie the amplitude decays etc | |
| (c) | | | Need \(4 m ^ { 2 } + \lambda m + 25 = 0\) to have repeated solutions so \(\lambda ^ { 2 } - 4 \times 4 \times 25 = 0\) | | \(\lambda > 0 \Rightarrow \lambda = 20\) |
| M1 | 3.5c | Using " \(b ^ { 2 } - 4 a c\) " \(= 0\) directly | or \(\left( 2 m + \frac { \lambda } { 4 } \right) ^ { 2 } + 25 - \frac { \lambda ^ { 2 } } { 16 } = 0\) and equating part outside brackets to 0 |
| | | A1 | 3.3 | Not -20 or \(\pm 20\) | |
| | | [2] | | | |