OCR Further Pure Core 2 (Further Pure Core 2) 2021 June

2 The equations of two intersecting lines \(l _ { 1 }\) and \(l _ { 2 }\) are
\(l _ { 1 } : \mathbf { r } = \left( \begin{array} { l } 1
0
a \end{array} \right) + \lambda \left( \begin{array} { r } 2
1
- 3 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } 7
9
- 2 \end{array} \right) + \mu \left( \begin{array} { r } - 1
1
2 \end{array} \right)\)
where \(a\) is a constant.
The equation of the plane \(\Pi\) is
r. \(\left( \begin{array} { l } 1
5
3 \end{array} \right) = - 14\).
\(l _ { 1 }\) and \(\Pi\) intersect at \(Q\).
\(\zeta _ { 2 }\) and \(\Pi\) intersect at \(R\).

1. Verify that the coordinates of \(R\) are \(( 13,3 , - 14 )\).
2. Determine the exact value of the length of \(Q R\).

3 A capacitor is an electrical component which stores charge. The value of the charge stored by the capacitor, in suitable units, is denoted by \(Q\). The capacitor is placed in an electrical circuit. At any time \(t\) seconds, where \(t \geqslant 0 , Q\) can be modelled by the differential equation \(\frac { \mathrm { d } ^ { 2 } Q } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} Q } { \mathrm {~d} t } - 15 Q = 0\). Initially the charge is 100 units and it is given that \(Q\) tends to a finite limit as \(t\) tends to infinity.

1. Determine the charge on the capacitor when \(t = 0.5\).
2. Determine the finite limit of \(Q\) as \(t\) tends to infinity. The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { r r } 0.6 & 2.4
   - 0.8 & 1.8 \end{array} \right)\).
3. Find \(\operatorname { det } \mathbf { A }\). The matrix \(\mathbf { A }\) represents a stretch parallel to one of the coordinate axes followed by a rotation about the origin.
4. By considering the determinants of these transformations, determine the scale factor of the stretch.
5. Explain whether the stretch is parallel to the \(x\)-axis or the \(y\)-axis, justifying your answer.
6. Find the angle of rotation.

5 Two thin poles, \(O A\) and \(B C\), are fixed vertically on horizontal ground. A chain is fixed at \(A\) and \(C\) such that it touches the ground at point \(D\) as shown in the diagram. On a coordinate system the coordinates of \(A , B\) and \(D\) are \(( 0,3 ) , ( 5,0 )\) and \(( 2,0 )\).
\includegraphics[max width=\textwidth, alt={}, center]{420598e3-4531-44da-8ae3-088e433f4c05-03_696_1338_1011_262} It is required to find the height of pole \(B C\) by modelling the shape of the curve that the chain forms.
Jofra models the curve using the equation \(y = k \cosh ( a x - b ) - 1\) where \(k , a\) and \(b\) are positive constants.

1. Determine the value of \(k\).
2. Find the exact value of \(a\) and the exact value of \(b\), giving your answers in logarithmic form. Holly models the curve using the equation \(y = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 3\).
3. Write down the coordinates of the point, \(( u , v )\) where \(u\) and \(v\) are both non-zero, at which the two models will agree.
4. Show that Jofra's model and Holly's model disagree in their predictions of the height of pole \(B C\) by 3.32 m to 3 significant figures. \section*{Total Marks for Question Set 5: 38} \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.

   \end{table}

   Question		Answer	Marks	AO	Guidance
   1		DR \(z = \frac { - - 20 \pm \sqrt { ( - 20 ) ^ { 2 } - 4 \times 4 \times 169 } } { 2 \times 4 }\) \(z = \frac { 5 \pm 12 \mathrm { i } } { 2 }\) \(r = \sqrt { \left( \frac { 5 } { 2 } \right) ^ { 2 } + \left( \frac { 12 } { 2 } \right) ^ { 2 } } = \frac { 13 } { 2 } \mathrm { oe }\) \(\theta = \tan ^ { - 1 } \frac { 6 } { 2.5 }\) oe \(\frac { 13 } { 2 } ( \cos ( - 1.18 ) + \mathrm { i } \sin ( - 1.18 ) )\)	M1	1.1	Term by term substituting into formula. If formula quoted, allow one slip ... Or correctly completes the square	Condone anything correct of the form \(\frac { p \pm \sqrt { q } } { r }\) eg \(4 \left( \left( z - \frac { 5 } { 2 } \right) ^ { 2 } - \frac { 25 } { 4 } \right) + 169 = 0\)
   			B1ft	1.1		Ft workings from complex conjugate distinct pair (with real component)
   			M1	1.1	Attempting to find argument using trigonometry
   			A1	2.5	Angle must be in radians. Argument could be 5.11 but both angles must be the same.	Not 5.10 (rounding error) Not e.g. \(\cos ( - 1.18 ) + \mathrm { i } \sin ( 5.11 )\)

   Question		Answer	Marks	AO	Guidance
   \multirow[t]{3}{*}{2}	\multirow[t]{3}{*}{(a)}	\(\begin{aligned}	\left( \begin{array} { c } 13
   3
   - 14 \end{array} \right) \cdot \left( \begin{array} { l } 1
   5
   3 \end{array} \right) = 13 + 15 - 42 = - 14 ( \text { so } R \text { is on } \Pi )
   	\operatorname { eg } 7 - \mu = 13 \Rightarrow \mu = - 6 \Rightarrow
   	\mathbf { r } = \left( \begin{array} { c } 7
   9
   - 2 \end{array} \right) - 6 \left( \begin{array} { c } - 1
   1
   2 \end{array} \right) = \left( \begin{array} { c } 13
   3
   - 14 \end{array} \right) \quad \left( \text { so } R \text { is also on } l _ { 2 } \right) \end{aligned}\)	B1 B1	1.1 1.1	AG. Intermediate working must be seen AG. Or \(9 + \mu = 3\) or \(- 2 + 2 \mu = - 14\) but must be checked in other two equations.
   		Alternate method \(\begin{aligned}	\left( \begin{array} { l } 1
   5
   3 \end{array} \right) \cdot \left( \left( \begin{array} { c } 7
   9
   - 2 \end{array} \right) + \mu \left( \begin{array} { c } - 1
   1
   2 \end{array} \right) \right) = 46 + 10 \mu = - 14 \Rightarrow \mu = - 6
   	\mu = - 6 \Rightarrow
   	\mathbf { r } = \left( \begin{array} { c } 7
   9
   - 2 \end{array} \right) - 6 \left( \begin{array} { c } - 1
   1
   2 \end{array} \right) = \left( \begin{array} { c } 13
   3
   - 14 \end{array} \right) \text { so } R \text { is } ( 13,3 , - 14 ) \end{aligned}\)	М1 A1		AG. Substituting in expression of the point into the equation of the plane to find a value for \(\mu\) AG.	Answer in vector form is acceptable.
   			[2]
   	(b)	\(\text { Since lines intersect } \left( \begin{array} { l } 1 0 a \end{array} \right) + \lambda \left( \begin{array} { c } 2 1 - 3 \end{array} \right) = \left( \begin{array} { c } 7 9 - 2 \end{array} \right) + \mu \left( \begin{array} { c } - 1 1 2 \end{array} \right)\) for some \(\lambda\) and \(\mu\) \(\begin{aligned} \text { so } 1 + 2 \lambda = 7 - \mu \lambda = 9 + \mu ( a - 3 \lambda = - 2 + 2 \mu ) \end{aligned}\) \(\Rightarrow \lambda = 5 , \mu = - 4\) so \(a + 5 \times ( - 3 ) = - 2 + ( - 4 ) \times 2 \Rightarrow a = 5\)	M1	3.1a	Equating the lines and deriving 2 useful equations. Ignore attempts at \(z\) coefficient equation	Can be BC

   \includegraphics[max width=\textwidth, alt={}]{420598e3-4531-44da-8ae3-088e433f4c05-09_1607_2571_86_239}

   Question			Answer	Marks	AO	Guidance
   4	(a)		\(\operatorname { det } \mathbf { A } ( = 0.6 \times 1.8 - - 0.8 \times 2.4 ) = 3\)	B1 [1]	1.1
   	(b)		Determinant of rotation \(= 1\) Determinant of rotation × determinant of stretch \(= 1 \times \mathrm { sf } = 3 \Rightarrow \mathrm { sf } = 3\)	B1 B1 [2]	1.1 2.2a
   	(c)		Since the second column of A contains entries bigger than 1 (in magnitude) the stretch must be parallel to the \(y\)-axis.	B1 [1]	2.4	Or any correct, complete explanation.	\(\begin{gathered} \left( \begin{array} { c c } \cos \theta	- \sin \theta
   \sin \theta	\cos \theta \end{array} \right) \left( \begin{array} { l l } 1	0
   0	3 \end{array} \right)
   = \left( \begin{array} { c c } \cos \theta	- 3 \sin \theta
   \sin \theta	3 \cos \theta \end{array} \right) \end{gathered}\)
   	(d)		\(\sin \theta = - 0.8\) and \(\cos \theta = 0.6\) oe awrt \(- 53 ^ { \circ }\) (or - 0.93 rads)	M1 A1 [2]	2.2a 1.1	Condone if only one equation or \(53 ^ { \circ }\) ( 0.93 rads) clockwise or \(307 ^ { \circ }\) (5.36 rads) (anticlockwise).
   5	(a)		Min value of cosh is 1 (and point on ground is at the minimum) \(( \text { so } 0 = k \times 1 - 1 \Rightarrow ) k = 1\)	M1 A1 [2]	2.2a 2.2a	Using minimum point of curve and knowledge of cosh graph	Could be derived by differentiation If zero scored then sc1 for \(\mathrm { k } = 1\) www
   	(b)		\(\begin{aligned}	\text { Passes through } ( 0,3 ) = > 3 = \cosh ( - b ) - 1
   	= > b = - \cosh ^ { - 1 } ( 3 + 1 )
   	b = ( \pm ) \ln \left( 4 + \sqrt { } \left( 4 ^ { 2 } - 1 \right) \right)
   	= > b = \ln ( 4 + \sqrt { } 15 )
   	\text { Passes through } ( 2,0 ) = > 0 = \cosh ( 2 a - b ) - 1
   	\Rightarrow b = 2 a
   	= > a = 1 / 2 \ln ( 4 + \sqrt { } 15 ) \end{aligned}\)	*M1 dep*M1 A1 M1 A1 [5]	3.3 3.1a 1.1 3.3 1.1	Use of ( 0,3 ) to derive an expression for \(b\) Correct numerical use of formula Use of \(( 2,0 )\) to derive \(b = 2 a\)	\(\operatorname { accept } \cosh ( - b ) = \frac { 4 } { k }\) Or rearranges. Could be from (a). Allow ft
   	(c)		(By symmetry of both;) (4,3)	B1 [1]	2.2a
   	(d)		Holly's model; \(d _ { \mathrm { H } } = 6.75\) Jofra's model: \(d _ { \mathrm { J } } = \cosh ( 5 a - b ) - 1\) AG \(d _ { \mathrm { J } } - d _ { \mathrm { H } } = 10.067 \ldots - 6.75 = 3.32 ( 3 \mathrm { sf } )\)	B1 M1 A1 [3]	3.4 3.4 1.1	Use of \(x = 5\) with their values of \(a\) and \(b\) to predict \(d\). Must have - 1 From correct values	Condone 27/4 \(a = 1.0317 \ldots , b = 2.0634 \ldots , d _ { J } = 10.067 \ldots\) Condone 10.07 only if clear evidence of production