Questions - A-Level Maths

OCR Further Pure Core 1 2021 June Q1

1 Indicate by shading on an Argand diagram the region $$\{ z : | z | \leqslant | z - 4 | \} \cap \{ z : | z - 3 - 2 i | \leqslant 2 \} .$$

OCR Further Pure Core 1 2021 June Q2

2 In this question you must show detailed reasoning. You are given that $x = 2 + 5 \mathrm { i }$ is a root of the equation $x ^ { 3 } - 2 x ^ { 2 } + 21 x + 58 = 0$.
Solve the equation.

OCR Further Pure Core 1 2021 June Q3

3 The diagram shows part of the curve $y = 5 \cosh x + 3 \sinh x$.
\includegraphics[max width=\textwidth, alt={}, center]{ef967953-70b5-4dd1-a342-ad488b5fa79f-02_426_661_906_260}

Solve the equation $5 \cosh x + 3 \sinh x = 4$ giving your solution in exact form.
In this question you must show detailed reasoning. Find $\int _ { - 1 } ^ { 1 } ( 5 \cosh x + 3 \sinh x ) \mathrm { d } x$ giving your answer in the form $a \mathrm { e } + \frac { b } { \mathrm { e } }$ where $a$ and $b$ are integers to be determined.

OCR Further Pure Core 1 2021 June Q4

4 You are given that $y = \tan ^ { - 1 } \sqrt { 2 x }$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
Show that $\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi$ where $k$ is a number to be determined in exact form.

OCR Further Pure Core 1 2021 June Q5

5 The function $\operatorname { sech } x$ is defined by $\operatorname { sech } x = \frac { 1 } { \cosh x }$.

Show that $\operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 }$.
Using a suitable substitution, find $\int \operatorname { sech } x \mathrm {~d} x$.

OCR Further Pure Core 1 2021 June Q6

27 marks

6 In this question you must show detailed reasoning.
You are given the complex number $\omega = \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi$ and the equation $z ^ { 5 } = 1$.

Show that $\omega$ is a root of the equation.
Write down the other four roots of the equation.
Show that $\omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = - 1$.
Hence show that $\left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = 0$.

Hence determine the value of $\cos \frac { 2 } { 5 } \pi$ in the form $a + b \sqrt { c }$ where $a , b$ and $c$ are rational numbers to be found. Total Marks for Question Set 4: 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.
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. 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.
g 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
oe	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

\multirow[t]{2}{*}{3

}

\(\begin{aligned}	5 \cosh x + 3 \sinh x = 4
	\Rightarrow 5 \left( \frac { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } { 2 } \right) + 3 \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { 2 } \right) = 4
	\Rightarrow 4 \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } = 4 \end{aligned}\)
Alternatively: \(\begin{aligned}	5 \cosh x + 3 \sinh x \equiv R \cosh ( x + \alpha )
	\text { where } R = \sqrt { 25 - 9 } = 4
	\tanh \alpha = \frac { 3 } { 5 } \Rightarrow \alpha = \tanh ^ { - 1 } \frac { 3 } { 5 } = \frac { 1 } { 2 } \ln \left( \frac { 1 + \frac { 3 } { 5 } } { 1 - \frac { 3 } { 5 } } \right) = \frac { 1 } { 2 } \ln 4 = \ln 2 \text { M1 A1 }
	\Rightarrow 4 \cosh ( x + \alpha ) = 4 \Rightarrow \cosh ( x + \alpha ) = 0
	\Rightarrow x = - \alpha = - \ln 2 \end{aligned}\)

М1

3.1a

Use of exponentials

Multiply by $\mathrm { e } ^ { x }$

Alternatively make cosh the subject, square and use Pythagoras to give quadratic in cosh

Alternatively use compound angle formula

[4]

(b)

DR \(\begin{aligned}

\int _ { - 1 } ^ { 1 } ( 5 \cosh x + 3 \sinh x ) \mathrm { d } x = [ 5 \sinh x + 3 \cosh x ] _ { - 1 } ^ { 1 }

= \left( 5 \frac { \mathrm { e } ^ { 1 } - \mathrm { e } ^ { - 1 } } { 2 } + 3 \frac { \mathrm { e } ^ { 1 } + \mathrm { e } ^ { - 1 } } { 2 } \right) - \left( 5 \frac { \mathrm { e } ^ { - 1 } - \mathrm { e } ^ { 1 } } { 2 } + 3 \frac { \mathrm { e } ^ { - 1 } + \mathrm { e } ^ { 1 } } { 2 } \right)

= \left( 4 \mathrm { e } ^ { 1 } - \mathrm { e } ^ { - 1 } \right) - \left( 4 \mathrm { e } ^ { - 1 } - \mathrm { e } ^ { 1 } \right)

= 5 \mathrm { e } - \frac { 5 } { \mathrm { e } } \end{aligned}\)

М1

1.1

Attempt at integral (i.e. one function changed)

Alternatively:

M1 convert (including possibly using result from (a))

M1 integrate and use limits correctly

\includegraphics[max width=\textwidth, alt={}]{ef967953-70b5-4dd1-a342-ad488b5fa79f-09_1333_2525_159_242}

Question

Answer

Marks

AO

Guidance

4

(b)

\(\begin{aligned}

\int _ { 1 / 6 } ^ { 1 / 2 } \frac { 1 } { ( 1 + 2 x ) \sqrt { x } } \mathrm {~d} x = \sqrt { 2 } \int _ { 1 / 6 } ^ { 1 / 2 } \frac { 1 } { ( 1 + 2 x ) \sqrt { 2 x } } \mathrm {~d} x

= \sqrt { 2 } \left[ \tan ^ { - 1 } \sqrt { 2 x } \right] _ { 1 / 6 } ^ { 1 / 2 } = \sqrt { 2 } \left( \tan ^ { - 1 } 1 - \tan ^ { - 1 } \frac { 1 } { \sqrt { 3 } } \right)

= \sqrt { 2 } \left( \frac { \pi } { 4 } - \frac { \pi } { 6 } \right) = \frac { \sqrt { 2 } } { 12 } \pi

\text { So } k = \frac { \sqrt { 2 } } { 12 } \end{aligned}\)

M1

A1

М1

A1

[4]

3.1a

1.1

Get into form of (a). Ignore limits

Correct form

Use (a) and correct limits in correct order.

ое

Alternatively: \(\begin{aligned}

\text { Let } u = \sqrt { x }

\mathrm {~d} u = \frac { 1 } { 2 \sqrt { x } } \mathrm {~d} x \Rightarrow \mathrm {~d} x = 2 \sqrt { x } \mathrm {~d} u = 2 u \mathrm {~d} u

\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = \int _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } } \frac { u } { \left( u ^ { 2 } + 2 u ^ { 4 } \right) } 2 u \mathrm {~d} u = 2 \int _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } } \frac { 1 } { \left( 1 + 2 u ^ { 2 } \right) } \mathrm { d } u

= \sqrt { 2 } \left[ \tan ^ { - 1 } u \sqrt { 2 } \right] _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } }

= \sqrt { 2 } \left[ \tan ^ { - 1 } \sqrt { 2 x } \right] _ { x = \frac { 1 } { 6 } } ^ { x = \frac { 1 } { 2 } } = \sqrt { 2 } \left( \tan ^ { - 1 } 1 - \tan ^ { - 1 } \frac { 1 } { \sqrt { 3 } } \right) = \sqrt { 2 } \left( \frac { \pi } { 4 } - \frac { \pi } { 6 } \right)

= \frac { \pi \sqrt { 2 } } { 12 } \end{aligned}\)

Make a substitution

Get into correct form

Use standard result with correct limits in correct order

Question

Answer

Marks

AO

Guidance

5

(a)

\(\begin{aligned}

\cosh x = \frac { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } { 2 } = \frac { \mathrm { e } ^ { 2 x } + 1 } { 2 \mathrm { e } ^ { x } }

\Rightarrow \operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } \quad \text { AG } \end{aligned}\)

М1

A1

[2]

1.1

2.1

Use of $\cosh x$ in exponentials

(b)

\(\begin{aligned}

u = \mathrm { e } ^ { x } \Rightarrow \mathrm {~d} u = \mathrm { e } ^ { x } \mathrm {~d} x

\Rightarrow \mathrm {~d} x = \frac { \mathrm { d } u } { u }

\Rightarrow \int \operatorname { sech } x \mathrm {~d} x = \int \left( \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } \right) \mathrm { d } x

= \int \frac { 2 u } { u ^ { 2 } + 1 } \cdot \frac { \mathrm {~d} u } { u }

= 2 \tan ^ { - 1 } ( u ) + c = 2 \tan ^ { - 1 } \left( \mathrm { e } ^ { x } \right) + c

\text { Alternatively: }

u = \sinh x \Rightarrow \mathrm {~d} u = \cosh x \mathrm {~d} x

\Rightarrow \int \operatorname { sech } x \mathrm {~d} x = \int \frac { 1 } { \cosh x } \cdot \frac { \mathrm {~d} u } { \cosh x } = \int \frac { \mathrm { d } u } { \cosh ^ { 2 } x }

= \int \frac { \mathrm { d } u } { 1 + \sinh ^ { 2 } x } = = \int \frac { \mathrm { d } u } { 1 + u ^ { 2 } }

= \tan ^ { - 1 } u + c

= \tan ^ { - 1 } ( \sinh x ) + c

\text { Alternatively: }

\int \operatorname { sech } x \mathrm {~d} x = \int \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } \mathrm {~d} x

\text { Let } \mathrm { e } ^ { x } = \tan u \Rightarrow \mathrm { e } ^ { x } \mathrm {~d} x = \sec ^ { 2 } u \mathrm {~d} u \Rightarrow \mathrm {~d} x = \frac { \sec ^ { 2 } u } { \tan u } \mathrm {~d} u

\Rightarrow \int \operatorname { sech } x \mathrm {~d} x = \int \frac { 2 \tan u } { \tan ^ { 2 } u + 1 } \cdot \frac { \sec ^ { 2 } u } { \tan u } \mathrm {~d} u = 2 \int \mathrm {~d} u

= 2 u + c

= 2 \tan ^ { - 1 } \left( \mathrm { e } ^ { x } \right) + c \end{aligned}\)

М1

A1

М1

A1

3.1a

Must include $c$

Allow absence of $\mathrm { d } u$

Question

Answer

Marks

AO

Guidance

6

(a)

DR \(\begin{aligned}

\omega = \cos \frac { 2 \pi } { 5 } + \mathrm { i } \sin \frac { 2 \pi } { 5 }

\Rightarrow \omega ^ { 5 } = \left( \cos \frac { 2 \pi } { 5 } + \mathrm { i } \sin \frac { 2 \pi } { 5 } \right) ^ { 5 } = \cos 2 \pi + \mathrm { i } \sin 2 \pi = 1 + 0 \mathrm { i } = 1 \end{aligned}\)

M1 A1

[2]

2.1 1.1

Finding $\omega ^ { 5 }$ AG

Use of exponentials is satisfactory Could be argued backwards

(b)

$\omega ^ { 2 } , \omega ^ { 3 } , \omega ^ { 4 } , 1$

B1

[1]

1.1

Alternative:

Roots are $\cos \frac { 2 k \pi } { 5 } + \mathrm { i } \sin \frac { 2 k \pi } { 5 }$ for $k = 2,3,4$

and 1 ( or $k = 5$ )

Exponentials satisfactory

(c)

DR \(\begin{aligned}	\omega ^ { 5 } - 1 = 0
	\Rightarrow ( \omega - 1 ) \left( \omega ^ { 4 } + \omega ^ { 3 } + \omega ^ { 2 } + \omega + 1 \right) = 0
	\Rightarrow \omega ^ { 4 } + \omega ^ { 3 } + \omega ^ { 2 } + \omega = - 1 \end{aligned}\)
Alternatively: $1 + \omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = \frac { 1 - \omega ^ { 5 } } { 1 - \omega } = \frac { 0 } { 1 - \omega }$
M1
or $\omega + \omega ^ { 2 } + \omega ^ { 3 } + \omega ^ { 4 } = \omega \left( \frac { 1 - \omega ^ { 4 } } { 1 - \omega } \right) = \left( \frac { \omega - \omega ^ { 5 } } { 1 - \omega } \right) = \left( \frac { \omega - 1 } { 1 - \omega } \right) = - 1$ A1
Alternatively:
sum of roots $= = - \frac { b } { a }$ where $b = 0 \mathrm { M } 1 -$ needs
explanation - i.e. coefficient of $z ^ { 4 }$ term $= 0$

Use equation and $\omega$

AG

Question

Answer

Marks

AO

Guidance

6

(d)

\(\begin{aligned}

\mathbf { A G }

\left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = \omega ^ { 2 } + 2 + \frac { 1 } { \omega ^ { 2 } } + \omega + \frac { 1 } { \omega } - 1

= \frac { 1 } { \omega ^ { 2 } } \left( \omega ^ { 4 } + \omega ^ { 2 } + 1 + \omega ^ { 3 } + \omega \right) = 0

\text { Since } \frac { 1 } { \omega ^ { 2 } } \neq 0 , \omega ^ { 4 } + \omega ^ { 2 } + 1 + \omega ^ { 3 } + \omega = 0

\text { or from part (c) } \end{aligned}\)

M1

A1

2.1

1.1

2.2a

Multiply out

Alternatively: \(\begin{aligned}

\omega ^ { 4 } + \omega ^ { 3 } + \omega ^ { 2 } + \omega + 1 = 0

\Rightarrow \omega ^ { 2 } \left( \omega ^ { 2 } + \omega + 1 + \frac { 1 } { \omega } + \frac { 1 } { \omega ^ { 2 } } \right) = 0

\Rightarrow \omega ^ { 2 } \left( \left( \omega ^ { 2 } + 2 + \frac { 1 } { \omega ^ { 2 } } \right) + \left( \omega + \frac { 1 } { \omega } \right) + 1 - 2 \right) = 0

\Rightarrow \omega ^ { 2 } \left( \left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 \right) = 0

\text { Since } \omega ^ { 2 } \neq 0 , \left( \omega + \frac { 1 } { \omega } \right) ^ { 2 } + \left( \omega + \frac { 1 } { \omega } \right) - 1 = 0 \quad \mathbf { A } \mathbf { 1 } \end{aligned}\)

[3]

For extraction of $\omega ^ { 2 }$

For dealing with the 2

Question		Answer	Marks	AO	Guidance
\multirow[t]{4}{*}{6}	(e)	\(\begin{aligned}	\frac { 1 } { \omega } = \cos \frac { 2 \pi } { 5 } - i \sin \frac { 2 \pi } { 5 }
	\Rightarrow \left( \omega + \frac { 1 } { \omega } \right) = 2 \cos \frac { 2 \pi } { 5 } \end{aligned}\)	B1	3.1a	$\omega + \frac { 1 } { \omega }$ may be seen in (d)
		\(\begin{aligned}	\text { From (iii) solving quadratic: } \left( \omega + \frac { 1 } { \omega } \right) = \frac { - 1 \pm \sqrt { 5 } } { 2 }
	\Rightarrow 2 \cos \frac { 2 \pi } { 5 } = \frac { \sqrt { 5 } - 1 } { 2 } \Rightarrow \cos \frac { 2 \pi } { 5 } = \frac { \sqrt { 5 } - 1 } { 4 }
	= - \frac { 1 } { 4 } + \frac { \sqrt { 5 } } { 4 } \text { or } - \frac { 1 } { 4 } + \frac { 1 } { 4 } \sqrt { 5 } \text { or } - 0.25 + 0.25 \sqrt { 5 } \end{aligned}\)	М1	3.1a	BC
			A1	2.3	For taking the valid value and presenting in correct form ое	No other forms acceptable
			[4]

OCR Further Pure Core 1 2021 June Q2

2 In this question you must show detailed reasoning.

Determine the square roots of 25 i in the form $r \mathrm { e } ^ { \mathrm { i } \theta }$, where $0 \leqslant \theta < 2 \pi$.
Illustrate the number 25 i and its square roots on an Argand diagram.

OCR Further Pure Core 1 2021 June Q3

3 By expanding $\left( z ^ { 2 } + \frac { 1 } { z ^ { 2 } } \right) ^ { 3 }$, where $z = \mathrm { e } ^ { \mathrm { i } \theta }$, show that $4 \cos ^ { 3 } 2 \theta = \cos 6 \theta + 3 \cos 2 \theta$.

OCR Further Pure Core 1 2021 June Q4

4 The equations of two non-intersecting lines, $l _ { 1 }$ and $l _ { 2 }$, are
$l _ { 1 } : \mathbf { r } = \left( \begin{array} { c } 1
2
- 1 \end{array} \right) + \lambda \left( \begin{array} { c } 2
1
- 2 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { c } 2
2
- 3 \end{array} \right) + \mu \left( \begin{array} { c } 1
- 1
4 \end{array} \right)$.
Find the shortest distance between lines $l _ { 1 }$ and $l _ { 2 }$.

OCR Further Pure Core 1 2021 June Q5

5 Prove by induction that the sum of the cubes of three consecutive positive integers is divisible by 9 .

OCR Further Pure Core 1 2021 June Q6

6 You are given that the cubic equation $2 x ^ { 3 } + p x ^ { 2 } + q x - 3 = 0$, where $p$ and $q$ are real numbers, has a complex root $\alpha = 1 + i \sqrt { 2 }$.

Write down a second complex root, $\beta$.
Determine the third root, $\gamma$.
Find the value of $p$ and the value of $q$.
Show that if $n$ is an integer then $\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } = 2 \times 3 ^ { \frac { 1 } { 2 } n } \times \cos n \theta + \frac { 1 } { 2 ^ { n } }$ where $\tan \theta = \sqrt { 2 }$.

OCR Further Pure Core 1 2021 June Q7

38 marks

7 A curve has cartesian equation $x ^ { 3 } + y ^ { 3 } = 2 x y$.
$C$ is the portion of the curve for which $x \geqslant 0$ and $y \geqslant 0$. The equation of $C$ in polar form is given by $r = \mathrm { f } ( \theta )$ for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.

Find $f ( \theta )$.
Find an expression for $\mathrm { f } \left( \frac { 1 } { 2 } \pi - \theta \right)$, giving your answer in terms of $\sin \theta$ and $\cos \theta$.
Hence find the line of symmetry of $C$.
Find the value of $r$ when $\theta = \frac { 1 } { 4 } \pi$.

By finding values of $\theta$ when $r = 0$, show that $C$ has a loop. \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. \section*{Abbreviations}

Question

Answer

Marks

AO

Guidance

7

(a)

\(\begin{aligned}

x = r \cos \theta , y = r \sin \theta \Rightarrow ( r \cos \theta ) ^ { 3 } + ( r \sin \theta ) ^ { 3 } = 2 r \cos \theta \cdot r \sin \theta

\Rightarrow r \left( \cos ^ { 3 } \theta + \sin ^ { 3 } \theta \right) = 2 \cos \theta \sin \theta

\Rightarrow r = \frac { 2 \cos \theta \sin \theta } { \cos ^ { 3 } \theta + \sin ^ { 3 } \theta } \end{aligned}\)

M1

A1

3.1a

1.1

Substitution

May see "or $r = 0$ " but not required.

[2]

(b)

\(\begin{aligned} f \left( \frac { 1 } { 2 } \pi - \theta \right)

= \frac { 2 \cos \left( \frac { 1 } { 2 } \pi - \theta \right) \sin \left( \frac { 1 } { 2 } \pi - \theta \right) } { \cos ^ { 3 } \left( \frac { 1 } { 2 } \pi - \theta \right) + \sin ^ { 3 } \left( \frac { 1 } { 2 } \pi - \theta \right) }

= \frac { 2 \sin \theta \cos \theta } { \sin ^ { 3 } \theta + \cos ^ { 3 } \theta } \end{aligned}\)

М1

A1

1.1a

1.1

Correct substitution into their $\mathrm { f } ( \theta )$

[2]

(c)

So the line of symmetry is $\theta = \frac { \pi } { 4 }$

B1

2.2a

Allow $y = x$.

Must have $\theta =$

[1]

(d)

$r = \mathrm { f } \left( \frac { 1 } { 4 } \pi \right) = \sqrt { 2 }$

B1

1.1

BC

[1]

(e)

$r = 0$ when $\theta = 0$.

$r = 0$ also when $\theta = \frac { \pi } { 2 }$

In range $0 < \theta < \frac { \pi } { 2 } , r > 0$ and is continuous

So there is a loop

B1

3.1a

For both, ignore extras.

Conclusion - both statements for $r$ need to be mentioned

[2]

OCR Further Pure Core 1 2021 June Q1

1 Find an expression for $1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + 3 \times 4 ^ { 2 } + \ldots + n ( n + 1 ) ^ { 2 }$ in terms of $n$. Give your answer in fully factorised form.

OCR Further Pure Core 1 2021 June Q2

2
You are given the matrix $\mathbf { A } = \left( \begin{array} { c c c } 1 & 0 & 0
0 & 0 & 1
0 & - 1 & 0 \end{array} \right)$.

Find $\mathbf { A } ^ { 4 }$.
Describe the transformation that A represents. The matrix $\mathbf { B }$ represents a reflection in the plane $x = 0$.
Write down the matrix $B$. The point $P$ has coordinates $( 2,3,4 )$. The point $P ^ { \prime }$ is the image of $P$ under the transformation represented by $\mathbf { B }$.
Find the coordinates of $P ^ { \prime }$.

OCR Further Pure Core 1 2021 June Q3

3

Using exponentials, show that $\cosh 2 u \equiv 2 \sinh ^ { 2 } u + 1$.
By differentiating both sides of the identity in part (a) with respect to $u$, show that $\sinh 2 u \equiv 2 \sinh u \cosh u$.
Use the substitution $x = \sinh ^ { 2 } u$ to find $\int \sqrt { \frac { x } { x + 1 } } \mathrm {~d} x$. Give your answer in the form $a \sinh ^ { - 1 } b \sqrt { x } + \mathrm { f } ( x )$ where $a$ and $b$ are integers and $\mathrm { f } ( x )$ is a function to be determined.
Hence determine the exact area of the region between the curve $y = \sqrt { \frac { x } { x + 1 } }$, the $x$-axis, the line $x = 1$ and the line $x = 2$. Give your answer in the form $p + q \ln r$ where $p , q$ and $r$ are numbers to be determined.

OCR Further Pure Core 1 2021 June Q4

4 A particle of mass 0.5 kg is initially at point $O$. It moves from rest along the $x$-axis under the influence of two forces $F _ { 1 } \mathrm {~N}$ and $F _ { 2 } \mathrm {~N}$ which act parallel to the $x$-axis. At time $t$ seconds the velocity of the particle is $v \mathrm {~ms} ^ { - 1 }$.
$F _ { 1 }$ is acting in the direction of motion of the particle and $F _ { 2 }$ is resisting motion.
In an initial model

$F _ { 1 }$ is proportional to $t$ with constant of proportionality $\lambda > 0$,
$F _ { 2 }$ is proportional to $v$ with constant of proportionality $\mu > 0$.
1. Show that the motion of the particle can be modelled by the following differential equation.

$$\frac { 1 \mathrm {~d} v } { 2 \mathrm {~d} t } = \lambda t - \mu v$$

Solve the differential equation in part (a), giving the particular solution for $v$ in terms of $t$, $\lambda$ and $\mu$. You are now given that $\lambda = 2$ and $\mu = 1$.

Find a formula for an approximation for $v$ in terms of $t$ when $t$ is large. In a refined model

$F _ { 1 }$ is constant, acting in the direction of motion with magnitude 2 N ,
$F _ { 2 }$ is as before with $\mu = 1$.
Write down a differential equation for the refined model.
Without solving the differential equation in part (d), write down what will happen to the velocity in the long term according to this refined model.

OCR Further Pure Core 1 2021 June Q5

31 marks

5
Show that $\int _ { 0 } ^ { \frac { 1 } { \sqrt { 3 } } } \frac { 4 } { 1 - x ^ { 4 } } \mathrm {~d} x = \ln ( a + \sqrt { b } ) + \frac { \pi } { c }$ where $a , b$ and $c$ are integers to be determined. \section*{Total Marks for Question Set 6: 37} \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. \section*{Abbreviations}

OCR Further Pure Core 2 2021 June Q1

1

The Argand diagram below shows the two points which represent two complex numbers, $z _ { 1 }$ and $z _ { 2 }$.
\includegraphics[max width=\textwidth, alt={}, center]{20816f61-154d-4491-9d2d-4c62687bf81e-02_321_592_276_347} On the copy of the diagram in the Resource Materials.
- draw an appropriate shape to illustrate the geometrical effect of adding $z _ { 1 }$ and $z _ { 2 }$,
- indicate with a cross $( \times )$ the location of the point representing the complex number $z _ { 1 } + z _ { 2 }$.
- You are given that $\arg z _ { 3 } = \frac { 1 } { 4 } \pi$ and $\arg z _ { 4 } = \frac { 3 } { 8 } \pi$.
In each part, sketch and label the points representing the numbers $z _ { 3 } , z _ { 4 }$ and $z _ { 3 } z _ { 4 }$ on the diagram in the Resource Materials. You should join each point to the origin with a straight line.
(i) $\left| z _ { 3 } \right| = 1.5$ and $\left| z _ { 4 } \right| = 1.2$
(ii) $\left| z _ { 3 } \right| = 0.7$ and $\left| z _ { 4 } \right| = 0.5$

OCR Further Pure Core 2 2021 June Q2

35 marks

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. $$4 \frac { \mathrm {~d} ^ { 2 } \theta } { \mathrm {~d} t ^ { 2 } } + 25 \theta = 0$$
1. Write down the general solution to (\textit{).
2. 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 $t = 0$.
1. Find, in a real form, the general solution of ( $\dagger$ ).
2. Find the particular solution of ( $\dagger$ ).
3. 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.

\end{table}

Question

Answer

Marks

AO

Guidance

1

(a)

\includegraphics[max width=\textwidth, alt={}]{20816f61-154d-4491-9d2d-4c62687bf81e-08_395_674_200_575}

B1

[2]

2.2a

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}

B1

[2]

1.1

2.2a

$z _ { 3 }$ 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}

B1

[2]

1.1

2.2a

$z _ { 3 }$ 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

B1

[1]

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}\)

M1

М1

A1

[3]

2.1

1.1

2.1

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}\)

M1

A1

[2]

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 $\omega$ $\theta = A \cos \frac { 5 } { 2 } t + B \sin \frac { 5 } { 2 } t \text { or } R \cos \left( \frac { 5 } { 2 } t + \phi \right)$

М1

A1

[2]

1.1

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.

E1

[1]

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]

OCR Further Pure Core 2 2021 June Q1

1 In this question you must show detailed reasoning.
S is the 2-D transformation which is a stretch of scale factor 3 parallel to the $x$-axis. A is the matrix which represents S .

Write down $\mathbf { A }$.
By considering the transformation represented by $\mathbf { A } ^ { - 1 }$, determine the matrix $\mathbf { A } ^ { - 1 }$. Matrix $\mathbf { B }$ is given by $\mathbf { B } = \left( \begin{array} { c c } 0 & - 1
- 1 & 0 \end{array} \right)$. T is the transformation represented by $\mathbf { B }$.
Describe T.
Determine the matrix which represents the transformation S followed by T .
Demonstrate, by direct calculation, that $( \mathbf { B A } ) ^ { - 1 } = \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }$.

OCR Further Pure Core 2 2021 June Q2

2

Find the shortest distance between the point $( - 6,4 )$ and the line $y = - 0.75 x + 7$. Two lines, $l _ { 1 }$ and $l _ { 2 }$, are given by
$l _ { 1 } : \mathbf { r } = \left( \begin{array} { c } 4
3
- 2 \end{array} \right) + \lambda \left( \begin{array} { c } 2
1
- 4 \end{array} \right)$ and $l _ { 2 } : \mathbf { r } = \left( \begin{array} { c } 11
- 1
5 \end{array} \right) + \mu \left( \begin{array} { c } 3
- 1
1 \end{array} \right)$.
Find the shortest distance between $l _ { 1 }$ and $l _ { 2 }$.
Hence determine the geometrical arrangement of $l _ { 1 }$ and $l _ { 2 }$. Three matrices, $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$, are given by $\mathbf { A } = \left( \begin{array} { c c } 1 & 2
a & - 1 \end{array} \right) , \mathbf { B } = \left( \begin{array} { c c } 2 & - 1
4 & 1 \end{array} \right)$ and $\mathbf { C } = \left( \begin{array} { c c } 5 & 0
- 2 & 2 \end{array} \right)$ where $a$ is a
constant.
Using $\mathbf { A } , \mathbf { B }$ and $\mathbf { C }$ in that order demonstrate explicitly the associativity property of matrix multiplication.
Use $\mathbf { A }$ and $\mathbf { C }$ to disprove by counterexample the proposition 'Matrix multiplication is commutative'. For a certain value of $a , \mathbf { A } \binom { x } { y } = 3 \binom { x } { y }$.
Find
- $y$ in terms of $x$,
- the value of $a$.
  \includegraphics[max width=\textwidth, alt={}, center]{570dba92-2c81-43e8-a0a8-741c40718626-3_586_1024_187_404}
The figure shows part of the graph of $y = ( x - 3 ) \sqrt { \ln x }$. The portion of the graph below the $x$-axis is rotated by $2 \pi$ radians around the $x$-axis to form a solid of revolution, $S$. Determine the exact volume of $S$.

OCR Further Pure Core 2 2021 June Q5

33 marks

5
$C$ is the locus of numbers, $z$, for which $\operatorname { Im } \left( \frac { z + 7 i } { z - 24 } \right) = \frac { 1 } { 4 }$.
By writing $z = x + \mathrm { i } y$ give a complete description of the shape of $C$ on an Argand diagram. \section*{Total Marks for Question Set 2: 38} Mark scheme

Question

Answer

Marks

AO

Guidance

1

(a)

\(\mathbf { A } = \left( \begin{array} { l l } 3

0

1 \end{array} \right)\)

B1

[1]

1.1

(b)

Stretch scale factor 1/3 parallel to $x$-axis \(\mathbf { A } ^ { - 1 } = \left( \begin{array} { l l } \frac { 1 } { 3 }

0

1 \end{array} \right)\)

M1

A1

[2]

1.1

2.2a

Must be complete description (except no need to specify 2-D)

(c)

Reflection in the line $y = - x$

B1

[1]

1.2

(d)

\(\begin{aligned}

\mathbf { B A } = \left( \begin{array} { c c } 0

- 1

0 \end{array} \right) \left( \begin{array} { l l } 3

0

1 \end{array} \right) = \ldots

\ldots = \left( \begin{array} { c c } 0

- 1

- 3

0 \end{array} \right) \end{aligned}\)

M1

A1

[2]

1.1a

1.1

For understanding that the matrix representing successive transformations is the product in the correct order. ie $\mathbf { B A }$, not $\mathbf { A B }$

(e)

\(( \mathbf { B A } ) ^ { - 1 } = - \frac { 1 } { 3 } \left( \begin{array} { l l } 0

1

3

0 \end{array} \right)\) \(\left. \begin{array} { r l } \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }

= \left( \frac { 1 } { 3 } \right.

0

1 \end{array} \right) \left( \begin{array} { c c } 0

- 1

0 \end{array} \right) = \left( \begin{array} { c c } 0

- \frac { 1 } { 3 }

- 1

0 \end{array} \right) ~ \left( \begin{array} { l l } 0

1

3

0 \end{array} \right) = ( \mathbf { B } \mathbf { A } ) ^ { - 1 }\)

A1

1.1a

For carrying out the procedure for inverting the matrix found in (d) (or BA worked out from scratch)

OR $\mathbf { M 1 }$ find $\mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 }$ from scratch A1 demonstrate that $( \mathbf { B A } ) \left( \mathbf { A } ^ { - 1 } \mathbf { B } ^ { - 1 } \right)$ is equal to I

Question

Answer

Marks

AO

Guidance

2

(a)

\(\begin{aligned}

3 x + 4 y = 28 \text { so } a = 3 , b = 4 , c = 28 \text { (or any non- }

\text { zero multiples) so } D = \frac { | 3 \times - 6 + 4 \times 4 - 28 | } { \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } }

D = 6 \end{aligned}\)

M1

A1

1.1

Identifying $a , b$ and $c$ and substituting $a , b , c$ and $\left( x _ { 1 } , y _ { 1 } \right)$ correctly into distance formula

Alternative solution $y - 4 = \frac { 4 } { 3 } ( x + 6 ) \text { oe so }$ $- 0.75 x + 7 - 4 = \frac { 4 } { 3 } ( x + 6 ) \Rightarrow x = - 2.4 , y = 8.8$

$D = \sqrt { ( - 2.4 - - 6 ) ^ { 2 } + ( 8.8 - 4 ) ^ { 2 } } = 6$

М1

A1

Finding equation of perpendicular line through $( - 6,4 )$ and solving simultaneously to find foot of perpendicular

[2]

(b)

\(\left( \begin{array} { c } 2
1
- 4 \end{array} \right) \times \left( \begin{array} { c } 3
- 1
1 \end{array} \right) = \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right)\)
\(D = \frac { \left. \left\lvert \, \left( \begin{array} { c } 11
- 1
5 \end{array} \right) - \left( \begin{array} { c } 4
3
- 2 \end{array} \right) \right. \right) \left. \cdot \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\rvert \, } { \left\| \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\| }\) or \(\frac { \left\| \left( \begin{array} { c } 7
- 4
7 \end{array} \right) \cdot \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\| } { \left\| \left( \begin{array} { c } - 3
- 14
- 5 \end{array} \right) \right\| }\)
$D = 0$

B1

М1

A1

1.1a

1.1

Correctly finding a mutual perpendicular BC

Correct substitution into distance formula

Alternative solution \(\begin{aligned}

4 + 2 \lambda = 11 + 3 \mu , 3 + \lambda = - 1 - \mu \text { and } - 2 - 4 \lambda = 5

+ \mu

\lambda = - 1 , \mu = - 3

\text { eg } - 2 - 4 ( - 1 ) = 2 = 5 + - 3 \text { so lines intersect so } D

= 0 \end{aligned}\)

М1

A1

Looking for a PoI so all 3 ( $3 ^ { \text {rd } }$ might be seen later)

Correctly solving any 2 equations Must be checked in the unsolved equation.

Value of each side must be found, not just equality asserted.

[3]

(c)

There are two points, one on each line, such that the distance between the points is $0 . .$.

E1ft

3.1a

If $D$ found to be non-zero in (b) then allow "Because there are not two points..."

Question

Answer

Marks

AO

Guidance

...and so the lines must intersect.

E1ft

2.4

A convincing demonstration that the two direction vectors are not parallel and "...and so the lines must be skew"

[2]

3

(a)

\(\begin{aligned}

\mathbf { A B } = \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \left( \begin{array} { c c } 2

- 1

4

1 \end{array} \right) = \left( \begin{array} { c c } 10

1

2 a - 4

- a - 1 \end{array} \right)

( \mathbf { A B } ) \mathbf { C } = \left( \begin{array} { c c } 10

1

2 a - 4

- a - 1 \end{array} \right) \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right)

= \left( \begin{array} { c c } 48

2

12 a - 18

- 2 a - 2 \end{array} \right)

\mathbf { B C } = \left( \begin{array} { c c } 2

- 1

4

1 \end{array} \right) \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right) = \left( \begin{array} { c c } 12

- 2

18

2 \end{array} \right)

\mathbf { A } ( \mathbf { B C } ) = \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \left( \begin{array} { c c } 12

- 2

18

2 \end{array} \right)

= \left( \begin{array} { c c } 48

2

12 a - 18

- 2 a - 2 \end{array} \right) = ( \mathbf { A B } ) \mathbf { C } \text { (which }

\text { demonstrates associativity of matrix }

\text { multiplication) } \end{aligned}\)

M1

A1

M1

A1

[4]

3.1a

2.1

1.1

2.1

Finding $\mathbf { A B }$ (or $\mathbf { B C }$ )

Finding (AB)C (or $\mathbf { A }$ (BC))

Finding $\mathbf { B C }$ (or $\mathbf { A B }$ )

Correct final matrix and statement of equality

(b)

\(\begin{aligned}

\mathbf { A } \mathbf { C } = \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right) = \left( \begin{array} { c c } 1

4

5 a + 2

- 2 \end{array} \right)

\mathbf { C A } = \left( \begin{array} { c c } 5

0

- 2

2 \end{array} \right) \left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) = \left( \begin{array} { c c } 5

10

2 a - 2

- 6 \end{array} \right) \neq \mathbf { A C } \text { (so }

\text { matrix multiplication is not commutative) } \end{aligned}\)

M1

A1

[2]

1.1

2.1

Finding AC (or CA)

Finding the other and statement of non-equality

(c)

\(\begin{aligned}

\left( \begin{array} { c c } 1

2

a

- 1 \end{array} \right) \binom { x } { y } = \binom { x + 2 y } { a x - y }

x + 2 y = 3 x = > y = x

a x - y = 3 y \text { and } y = x = > a = 4 \end{aligned}\)

M1 >

A1

$[ 3 ]$

3.1a

2.2a

Multiplying the vector into the matrix using the correct procedure

\begin{displayquote}

A1

$[ 3 ]$

\end{displayquote}

Question

Answer

Marks

AO

Guidance

\multirow[t]{5}{*}{4}

\(\begin{aligned}

V = \pi \int _ { 1 } ^ { 3 } ( ( x - 3 ) \sqrt { \ln x } ) ^ { 2 } \mathrm {~d} x = \pi \int _ { 1 } ^ { 3 } ( x - 3 ) ^ { 2 } \ln x \mathrm {~d} x

V = \pi \left( \left[ \frac { 1 } { 3 } ( x - 3 ) ^ { 3 } \ln x \right] _ { 1 } ^ { 3 } - \int _ { 1 } ^ { 3 } \frac { 1 } { 3 } ( x - 3 ) ^ { 3 } \frac { 1 } { x } \mathrm {~d} x \right)

\frac { 1 } { x } ( x - 3 ) ^ { 3 } = x ^ { 2 } - 9 x + 27 - \frac { 27 } { x } \text { soi } \end{aligned}\)

B1

3.1a

\multirow{2}{*}{

Correct substitution into formula (ignore limits) and simplification to integrable (by parts) form Integration by parts with $( x - 3 ) ^ { 2 }$ (may be expanded) being integrated.

May come implicitly from previously expanded form

}

\multirow{3}{*}{ie from $\int _ { 1 } ^ { 3 } x ^ { 2 } \ln x - 6 x \ln x + 9 \ln x \mathrm {~d} x$ integrated by parts term by term}

A1

1.1

A1

1.1

Completing the integral. NB $\left[ ( x - 3 ) ^ { 3 } \ln x \right] _ { 1 } ^ { 3 } = 0$ so may be omitted provided it is seen earlier

\(V = \frac { \pi } { 3 } \left( \left[ \begin{array} { l } ( x - 3 ) ^ { 3 } \ln x - \frac { x ^ { 3 } } { 3 } +

\frac { 9 x ^ { 2 } } { 2 } + 27 x - 27 \ln x \end{array} \right] _ { 1 } ^ { 3 } \right)\)

dep *M1

3.2a

Correctly dealing with limits

[7]

Question	Answer	Marks	AO	Guidance
5	$\frac { z + 7 \mathrm { i } } { z - 24 } = \frac { x + \mathrm { i } y + 7 \mathrm { i } } { x - 24 + \mathrm { i } y } \times \frac { x - 24 - \mathrm { i } y } { x - 24 - \mathrm { i } y }$	M1	3.1a	Substituting $z = x + \mathrm { i } y$ into $\frac { z + 7 \mathrm { i } } { z - 24 }$
\multirow{8}{*}{}	$\operatorname { Im } \frac { z + 7 \mathrm { i } } { z - 24 } = \frac { - x y + ( y + 7 ) ( x - 24 ) } { ( x - 24 ) ^ { 2 } + y ^ { 2 } } = \frac { 1 } { 4 }$	M1	2.1	conjugate of bottom
	$28 x - 96 y - 672 = x ^ { 2 } - 48 x + 576 + y ^ { 2 }$	M1	1.1	Multiplying out to get horizontal
	$0 = ( x - 38 ) ^ { 2 } - 1444 + ( y + 48 ) ^ { 2 } - 2304 + 1248$	M1	1.1	Completing both squares with half signed coefficients of $x$ and $y$
	$( x - 38 ) ^ { 2 } + ( y + 48 ) ^ { 2 } = 2500$	A1	2.2a
	So the shape of $C$ is a circle...	E1	3.2a
	...centre 38 - 48i, radius 50	E1	3.2a	Or $( 38 , - 48 )$
	function $\operatorname { Im } \left( \frac { z + 7 i } { z - 24 } \right)$ is undefined at this point on the circle)			Do not penalise either lack of
		[7]

OCR Further Pure Core 2 2021 June Q2

2 A 2-D transformation $T$ is a shear which leaves the $y$-axis invariant and which transforms the object point $( 2,1 )$ to the image point $( 2,9 )$. $A$ is the matrix which represents the transformation $T$.

Find A .
By considering the determinant of A , explain why the area of a shape is invariant under T .

OCR Further Pure Core 2 2021 June Q3

3 A particle of mass 2 kg moves along the $x$-axis. At time $t$ seconds the velocity of the particle is $v \mathrm {~ms} ^ { - 1 }$. The particle is subject to two forces.

One acts in the positive $x$-direction with magnitude $\frac { 1 } { 2 } t \mathrm {~N}$.
One acts in the negative $x$-direction with magnitude $v \mathrm {~N}$.
1. Show that the motion of the particle can be modelled by the differential equation

$$\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 1 } { 2 } v = \frac { 1 } { 4 } t$$ The particle is at rest when $t = 0$.

Find $v$ in terms of $t$.

Find the velocity of the particle when $t = 2$. When $t = 2$ the force acting in the positive $x$-direction is replaced by a constant force of magnitude $\frac { 1 } { 2 } \mathrm {~N}$ in the same direction.

Refine the differential equation given in part (a) to model the motion for $t \geqslant 2$.

Use the refined model from part (d) to find an exact expression for $v$ in terms of $t$ for $t \geqslant 2$.

OCR Further Pure Core 2 2021 June Q4

32 marks

4 In this question you must show detailed reasoning.

By writing $\sin \theta$ in terms of $\mathrm { e } ^ { \mathrm { i } \theta }$ and $\mathrm { e } ^ { - \mathrm { i } \theta }$ show that $$\sin ^ { 6 } \theta = \frac { 1 } { 32 } ( 10 - 15 \cos 2 \theta + 6 \cos 4 \theta - \cos 6 \theta ) .$$
Hence show that $\sin \frac { 1 } { 8 } \pi = \frac { 1 } { 2 } \sqrt [ 6 ] { 20 - 14 \sqrt { 2 } }$.
Use differentiation to find the first two non-zero terms of the Maclaurin expansion of $\ln \left( \frac { 1 } { 2 } + \cos x \right)$.

By considering the root of the equation $\ln \left( \frac { 1 } { 2 } + \cos x \right) = 0$ deduce that $\pi \approx 3 \sqrt { 3 \ln \left( \frac { 3 } { 2 } \right) }$. \section*{Total Marks for Question Set 3: 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. \section*{Abbreviations}

Question

Answer

Marks

AO

Guidance

4

(a)

DR \(\begin{aligned}

\sin \theta = \frac { \mathrm { e } ^ { \mathrm { i } \theta } - \mathrm { e } ^ { - \mathrm { i } \theta } } { 2 \mathrm { i } }

\sin ^ { 6 } \theta = \left( \frac { \mathrm { e } ^ { \mathrm { i } \theta } - \mathrm { e } ^ { - \mathrm { i } \theta } } { 2 i } \right) ^ { 6 } = - \frac { 1 } { 64 } \left( \mathrm { e } ^ { \mathrm { i } \theta } - \mathrm { e } ^ { - \mathrm { i } \theta } \right) ^ { 6 }

\left( e ^ { i \theta } - e ^ { - i \theta } \right) ^ { 6 } =

\mathrm { e } ^ { 6 \mathrm { i } \theta } - 6 \mathrm { e } ^ { 4 \mathrm { i } \theta } + 15 \mathrm { e } ^ { 2 \mathrm { i } \theta } - 20 + 15 \mathrm { e } ^ { - 2 \mathrm { i } \theta } - 6 \mathrm { e } ^ { - 4 \mathrm { i } \theta } + \mathrm { e } ^ { - 6 \mathrm { i } \theta } \end{aligned}\) \(\begin{aligned}

\mathrm { e } ^ { 6 \mathrm { i } \theta } + \mathrm { e } ^ { - 6 \mathrm { i } \theta } - 6 \left( \mathrm { e } ^ { 4 \mathrm { i } \theta } + \mathrm { e } ^ { - 4 \mathrm { i } \theta } \right) + 15 \left( \mathrm { e } ^ { 2 \mathrm { i } \theta } + \mathrm { e } ^ { - 2 \mathrm { i } \theta } \right) - 20

= 2 \cos 6 \theta - 6 \times 2 \cos 4 \theta + 15 \times 2 \cos 2 \theta - 20

\therefore \sin ^ { 6 } \theta =

- \frac { 1 } { 64 } ( 2 \cos 6 \theta - 12 \cos 4 \theta + 30 \cos 2 \theta - 20 )

= \frac { 1 } { 32 } ( 10 - 15 \cos 2 \theta + 6 \cos 4 \theta - \cos 6 \theta ) \end{aligned}\)

M1

dep*A1

[5]

1.1a

2.1

Genuine attempt to use binomial expansion with correct evaluated binomial coefficients. Condone sign errors

Collecting terms and using $\mathrm { e } ^ { \mathrm { i } \phi } + \mathrm { e } ^ { - \mathrm { i } \phi } = 2 \cos \phi$ at least once.

AG. Fully correct argument

Condone $2 \mathrm { i } \sin \theta = \mathrm { e } ^ { \mathrm { i } \theta } - \mathrm { e } ^ { - \mathrm { i } \theta }$

Allow use of $\sin \theta = \frac { e ^ { i \theta } + e ^ { - i \theta } } { 2 i }$ for $1 ^ { \text {st } }$ two M marks only

If i omitted from denominator their expression for $\sin \theta$ then only this M mark can still be awarded

(b)

DR $\theta = \frac { \pi } { 8 } \text { and } \mathrm { eg } \cos 2 \theta = \frac { \sqrt { 2 } } { 2 }$

$\sin ^ { 6 } \frac { \pi } { 8 } = \frac { 1 } { 32 } \left( 10 - 15 \times \frac { \sqrt { 2 } } { 2 } - \frac { - \sqrt { 2 } } { 2 } ( + 6 ( 0 ) ) \right)$

$\sin \frac { \pi } { 8 } = \sqrt [ 6 ] { \frac { 1 } { 64 } ( 20 - 15 \sqrt { 2 } + \sqrt { 2 } ) }$

$= \frac { 1 } { 2 } \sqrt [ 6 ] { 20 - 14 \sqrt { 2 } }$

*M1

dep*M1

2.2a

Choice of $\theta$ soi and calculation of at least one cos term.

Substitution and calculation of all cos terms

Terms must be shown distinct either in this line or in the form of $\cos n \frac { \pi } { 8 }$

Question

Answer

Marks

AO

Guidance

5

(a)

$f ( 0 ) = \ln \left( \frac { 1 } { 2 } + \cos 0 \right) = \ln \left( \frac { 3 } { 2 } \right)$ $\frac { \mathrm { d } ^ { 2 } \ln \left( \frac { 1 } { 2 } + \cos x \right) } { \mathrm { d } x ^ { 2 } } = \frac { - \cos x \left( \frac { 1 } { 2 } + \cos x \right) + \sin x ( - \sin x ) } { \left( \frac { 1 } { 2 } + \cos x \right) ^ { 2 } }$ $\ldots \Rightarrow f ^ { \prime \prime } ( 0 ) = - \frac { 2 } { 3 }$

$\ln \left( \frac { 1 } { 2 } + \cos x \right) = \ln \left( \frac { 3 } { 2 } \right) - \frac { x ^ { 2 } } { 3 } + \ldots$

B1

3.1a

Differentiating using chain rule (or rule for $\ln ( \mathrm { f } ( x ) )$ and evaluating when $x = 0$

Allow sign error in numerator

A1

1.1

Differentiating again using quotient (or product/chain) rule.

\multirow[t]{3}{*}{

NB Simplifies to $- \frac { \frac { 1 } { 2 } \cos x + 1 } { \left( \frac { 1 } { 2 } + \cos x \right) ^ { 2 } }$

If zero scored then SC1 for correct expansion

}

[4]

\multirow{4}{*}{}

(b)

\multirow{4}{*}{}

B1

1.1

Finding either $\pm \pi / 3$ as a root. Allow $60 ^ { \circ }$ for B 1 . Ignore other roots

\multirow{4}{*}{Or equating their expression (approximately) to 0 and rearranging for $x$ : $\ln \left( \frac { 3 } { 2 } \right) - \frac { x ^ { 2 } } { 3 } \approx 0 \Rightarrow x \approx \sqrt { 3 \ln \left( \frac { 3 } { 2 } \right) }$}

\(\begin{aligned}

\ln \left( \frac { 1 } { 2 } + \cos x \right) = 0 \Rightarrow x = \frac { \pi } { 3 } \left( \text { or } - \frac { \pi } { 3 } \right)

\therefore \ln \left( \frac { 3 } { 2 } \right) - \frac { \left( \frac { \pi } { 3 } \right) ^ { 2 } } { 3 } \approx 0 \end{aligned}\)

M1

3.1a

Substituting their root, in radians, into their Maclaurin series and equating (approximately) to 0 .

$\ln \left( \frac { 3 } { 2 } \right) - \frac { \pi ^ { 2 } } { 27 } \approx 0 \Rightarrow \pi \approx \sqrt { 27 \ln \left( \frac { 3 } { 2 } \right) } = 3 \sqrt { 3 \ln \left( \frac { 3 } { 2 } \right) }$

A1

3.2a

Could see ± but must be removed by final conclusion.

Must use approximately equals symbol (not just equals symbol)

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