Questions Further Pure Core 2

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)

OCR Further Pure Core 2 2021 June Q2

2 In this question you must show detailed reasoning.
Show that $\int _ { 5 } ^ { \infty } ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = 1$.
$3 A$ is a fixed point on a smooth horizontal surface. A particle $P$ is initially held at $A$ and released from rest. It subsequently performs simple harmonic motion in a straight line on the surface. After its release it is next at rest after 0.2 seconds at point $B$ whose displacement is 0.2 m from $A$. The point $M$ is halfway between $A$ and $B$. The displacement of $P$ from $M$ at time $t$ seconds after release is denoted by $x \mathrm {~m}$.

Sketch a graph of $x$ against $t$ for $0 \leqslant t \leqslant 0.4$.
Find the displacement of $P$ from $M$ at 0.75 seconds after release.

OCR Further Pure Core 2 2021 June Q4

4 In an Argand diagram the points representing the numbers $2 + 3 \mathrm { i }$ and $1 - \mathrm { i }$ are two adjacent vertices of a square, $S$.

Find the area of $S$.
Find all the possible pairs of numbers represented by the other two vertices of $S$.

OCR Further Pure Core 2 2021 June Q5

23 marks

5 In this question you must show detailed reasoning.
The diagram below shows the curve $r = \sqrt { \sin \theta } \mathrm { e } ^ { \frac { 1 } { 3 } \cos \theta }$ for $0 \leqslant \theta \leqslant \pi$.
\includegraphics[max width=\textwidth, alt={}, center]{58789f16-bfc3-4f21-b7c4-abca4f549fc7-03_574_878_276_477}

Find the exact area enclosed by the curve.

Show that the greatest value of $r$ on the curve is $\sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 6 } }$. Total Marks for Question Set 4: 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.
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.

Abbreviations

Question

Answer

Marks

AO

Guidance

\multirow[t]{7}{*}{2}

\multirow{7}{*}{}

DR

\multirow{6}{*}{

B1

М1

A1

B1

A1

}

\multirow{4}{*}{

1.1a

2.1

1.1

}

\multirow[b]{3}{*}{Consideration of a finite upper limit}

\multirow{7}{*}{Can be seen as part of limit of both terms, but must be explicitly shown as zero}

$\int ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = - 2 ( x - 1 ) ^ { - \frac { 1 } { 2 } } ( + c )$

$\int _ { 5 } ^ { N } ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = \left[ - 2 ( x - 1 ) ^ { - \frac { 1 } { 2 } } \right] _ { 5 } ^ { N }$ $- \frac { 2 } { \sqrt { N - 1 } } + \frac { 2 } { \sqrt { 5 - 1 } }$

$\lim _ { N \rightarrow \infty } \frac { 1 } { \sqrt { N - 1 } } = 0$ oe

2.1

Not just eg $\frac { 1 } { \infty } = 0$

$\int _ { 5 } ^ { \infty } ( x - 1 ) ^ { - \frac { 3 } { 2 } } \mathrm {~d} x = \lim _ { N \rightarrow \infty } \left\{ - \frac { 2 } { \sqrt { N - 1 } } + \frac { 2 } { \sqrt { 5 - 1 } } \right\} = 1$

2.2a

AG. Convincing argument equating improper integral to solution

[5]

Question

Answer

Marks

AO

Guidance

3

(a)

\includegraphics[max width=\textwidth, alt={}]{58789f16-bfc3-4f21-b7c4-abca4f549fc7-10_450_671_93_559}

B1

1.2

3.4

3.1b

At least one cycle of $A \cos \omega t$ graph

Amplitude 0.1

Period 0.4

Intersect with the $t$-axis at 0.1 and 0.3 (values must be indicated or implied unambiguously (eg by tickmarks and a single value))

graph must instantaneously horizontal at top/bottom, continuous and not vertical at any point.

Ignore any graph outside [0, 0.4] Non-inverted cos graph can still get 4/4

(b)

So $x = \pm 0.1 \cos ( 5 \pi \mathrm { t } )$ or $x = 0.1 \sin \left( 5 \pi t \pm \frac { 1 } { 2 } \pi \right)$ when $t = 0.75$ or 0.35 $x = - \frac { \sqrt { 2 } } { 20 } ( = - 0.0707 \text { to } 3 \mathrm { sf } )$

M1

A1

[2]

3.1b

3.4

or by argument from sketch

Condone amplitude of 0.2 for M1

4

(a)

\(\begin{aligned}

( 2 + 3 i ) - ( 1 - i ) ( = \pm ( 1 + 4 i ) ) \text { soi }

| 1 + 4 i | = \sqrt { 1 ^ { 2 } + 4 ^ { 2 } }

17 \end{aligned}\)

B1

M1

A1

[3]

1.1

2.2a

1.1

Either way round

Can be implied by vector

Or finding the square of their side

(b)

$( \pm \mathrm { i } ) \times ( \pm ( 1 + 4 \mathrm { i } ) )$

$( 2 + 3 i ) \pm i ( 1 + 4 i )$ and $( 1 - i ) \pm i ( 1 + 4 i )$

So vertices at - 3 and $- 2 + 4 \mathrm { i }$

Or at $5 - 2 \mathrm { i }$ and $6 + 2 \mathrm { i }$

*M1

dep*M1

A1

[4]

3.1a

2.2a

3.2a

Method to find a complex number representing perpendicular side. Can be implied by $\pm ( 4 - \mathrm { i } )$ Method to find both pairs of numbers

Both clearly paired and in complex number form for final A1

Or vector form if take geometric approach

If M1M0A0A0 then add SC1 for any two correct vertices

Question

Answer

Marks

AO

Guidance

5

(a)

DR \(\begin{aligned}

\frac { 1 } { 2 } \int \left( \sqrt { \sin \theta } e ^ { \frac { 1 } { 3 } \cos \theta } \right) ^ { 2 } \mathrm {~d} \theta

\mathrm {~A} = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } \sin \theta \mathrm { e } ^ { \frac { 2 } { 3 } \cos \theta } \mathrm {~d} \theta

= \frac { 1 } { 2 } \times - \frac { 3 } { 2 } \left[ \mathrm { e } ^ { \frac { 2 } { 3 } \cos \theta } \right] _ { 0 } ^ { \pi }

\frac { 3 } { 4 } \left( \mathrm { e } ^ { \frac { 2 } { 3 } } - \mathrm { e } ^ { - \frac { 2 } { 3 } } \right) \end{aligned}\)

M1

*A1

dep*M1

A1

[4]

3.1a

2.1

1.1a

1.1

Correct form, in terms of $\theta$,

Integrand has been squared out. Must include limits (can be seen later)

Might be as result of substitution Allow coefficient error for M1 isw

M1 can be implied by 1.0757 ... BC

eg $\frac { 3 } { 4 } \left[ \mathrm { e } ^ { u } \right] _ { - \frac { 2 } { 3 } } ^ { \frac { 2 } { 3 } }$ or $\frac { 3 } { 4 } \left[ \mathrm { e } ^ { \frac { 2 } { 3 } u } \right] _ { - 1 } ^ { 1 }$ oe

(b)

DR $\frac { \mathrm { d } r } { \mathrm {~d} \theta } = \frac { 1 } { 2 } \cos \theta ( \sin \theta ) ^ { - \frac { 1 } { 2 } } e ^ { \frac { 1 } { 3 } \cos \theta } +$ $( \sin \theta ) ^ { \frac { 1 } { 2 } } \left( - \frac { 1 } { 3 } \sin \theta \right) e ^ { \frac { 1 } { 3 } \cos \theta }$

$\frac { \mathrm { d } r } { \mathrm {~d} \theta } = \frac { 1 } { 6 } ( \sin \theta ) ^ { - \frac { 1 } { 2 } } e ^ { \frac { 1 } { 3 } \cos \theta } \left( 3 \cos \theta - 2 \sin ^ { 2 } \theta \right)$

$\frac { \mathrm { d } r } { \mathrm {~d} \theta } = 0 \Rightarrow 3 \cos \theta - 2 \sin ^ { 2 } \theta = 0$

$2 \cos ^ { 2 } \theta + 3 \cos \theta - 2 = 0$

$\cos \theta = \frac { 1 } { 2 } , - 2$

$\cos \theta \neq - 2$

$\Rightarrow \sin \theta = \frac { \sqrt { 3 } } { 2 } \Rightarrow r = \sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 3 } \times \frac { 1 } { 2 } } = \sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 6 } }$

*M1

A1

dep*M1

M1

*A1

dep*A1

3.1a

1.1

2.2a

2.1

1.1

2.3

2.2a

Attempt to differentiate using product and chain rules.

Setting $r ^ { \prime }$ to zero and factorising/cancelling to produce a quadratic equation in $\cos$ and/or sin Use of $\cos ^ { 2 } + \sin ^ { 2 } = 1$ to find 3 term quadratic equation in $\cos \theta$.

Solving quadratic correctly Explicitly rejecting root

AG. At least one intermediate step must be seen.

Must be in the form $u v ^ { \prime } + u ^ { \prime } v$ with at most one of $u , v , u ^ { \prime }$ or $v ^ { \prime }$ incorrect or omitted

Or could be in $\sin ^ { 2 } \theta$; $4 \sin ^ { 4 } \theta + 9 \sin ^ { 2 } \theta - 9 = 0$

$\sin ^ { 2 } \theta = \frac { 3 } { 4 } , - 3$

Rejects $\sin ^ { 2 } \theta = - 3$ and $\sin \theta =$ $- \frac { \sqrt { 3 } } { 2 }$

Can be awarded even if rejection of root(s) was implicit.

OCR Further Pure Core 2 2021 June Q2

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 $R$.

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

OCR Further Pure Core 2 2021 June Q3

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.

Determine the charge on the capacitor when $t = 0.5$.
Determine the finite limit of $Q$ as $t$ tends to infinity. The matrix $\mathbf { A }$ is given by $\mathbf { A } = \left( \begin{array} { r r } 0.6 & 2.4
- 0.8 & 1.8 \end{array} \right)$.
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.
By considering the determinants of these transformations, determine the scale factor of the stretch.
Explain whether the stretch is parallel to the $x$-axis or the $y$-axis, justifying your answer.
Find the angle of rotation.

OCR Further Pure Core 2 2021 June Q5

19 marks

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.

Determine the value of $k$.
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 $y = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 3$.
Write down the coordinates of the point, $( u , v )$ where $u$ and $v$ are both non-zero, at which the two models will agree.

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

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

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

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

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

OCR Further Pure Core 2 2021 June Q1

1 In this question you must show detailed reasoning.
The roots of the equation $3 x ^ { 3 } - 2 x ^ { 2 } - 5 x - 4 = 0$ are $\alpha , \beta$ and $\gamma$.

Find a cubic equation with integer coefficients whose roots are $\alpha ^ { 2 } , \beta ^ { 2 }$ and $\gamma ^ { 2 }$.
Find the exact value of $\frac { \alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } } { \alpha \beta \gamma }$.

OCR Further Pure Core 2 2021 June Q2

2 In this question you must show detailed reasoning.

Use partial fractions to show that $\sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }$.
Write down the value of $\lim _ { n \rightarrow \infty } \left( \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } \right)$.

OCR Further Pure Core 2 2021 June Q3

3 The equation of a curve in polar coordinates is $r = \ln ( 1 + \sin \theta )$ for $\alpha \leqslant \theta \leqslant \beta$ where $\alpha$ and $\beta$ are non-negative angles. The curve consists of a single closed loop through the pole.

By solving the equation $r = 0$, determine the smallest possible values of $\alpha$ and $\beta$.
Find the area enclosed by the curve, giving your answer to 4 significant figures.
Hence, by considering the value of $r$ at $\theta = \frac { \alpha + \beta } { 2 }$, show that the loop is not circular.

OCR Further Pure Core 2 2021 June Q4

4 In this question you must show detailed reasoning.
The complex number $- 4 + i \sqrt { 48 }$ is denoted by $z$.

Determine the cube roots of $z$, giving the roots in exponential form. The points which represent the cube roots of $z$ are denoted by $A , B$ and $C$ and these form a triangle in an Argand diagram.
Write down the angles that any lines of symmetry of triangle $A B C$ make with the positive real axis, justifying your answer.

OCR Further Pure Core 2 2021 June Q5

26 marks

5 Let $\mathrm { f } ( x ) = \sin ^ { - 1 } ( x )$.

1. Determine $f ^ { \prime \prime } ( x )$.
2. Determine the first two non-zero terms of the Maclaurin expansion for $\mathrm { f } ( x )$.
3. By considering the first two non-zero terms of the Maclaurin expansion for $\mathrm { f } ( \mathrm { x } )$, find an approximation to $\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x$. Give your answer correct to 6 decimal places.

By writing $\mathrm { f } ( x )$ as $\sin ^ { - 1 } ( x ) \times 1$, determine the value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x$. Give your answer in exact form. \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. \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)

DR \(\begin{aligned}

u = x ^ { 2 }

3 ( \sqrt { u } ) ^ { 3 } - 2 ( \sqrt { u } ) ^ { 2 } - 5 \sqrt { u } - 4 ( = 0 ) \end{aligned}\) $3 u \sqrt { u } - 5 \sqrt { u } = 2 u + 4 \Rightarrow u ( 3 u - 5 ) ^ { 2 } = ( 2 u + 4 ) ^ { 2 }$ \(\begin{aligned}

u \left( 9 u ^ { 2 } - 30 u + 25 \right) = 4 u ^ { 2 } + 16 u + 16 = >

9 u ^ { 3 } - 34 u ^ { 2 } + 9 u - 16 = 0 \end{aligned}\)

B1

М1

A1

3.1a

1.1

3.2a

Correct substitution chosen

Oe Attempting to make substitution

Rearranging and squaring bs to remove the square root(s) Rearranging to answer

or preparation for substitution by removing odd powers. $\operatorname { eg } x ^ { 2 } \left( 3 x ^ { 2 } - 5 \right) ^ { 2 } = \left( 2 x ^ { 2 } + 4 \right) ^ { 2 } \ldots$

...and then substituting $u ( 3 u - 5 ) ^ { 2 } = ( 2 u + 4 ) ^ { 2 }$

Equation can be in $x$

Alternative method

DR $\alpha ^ { 2 } \beta ^ { 2 } \gamma ^ { 2 } = ( \alpha \beta \gamma ) ^ { 2 } = \left( - \frac { - 4 } { 3 } \right) ^ { 2 } = \frac { 16 } { 9 }$

$\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }$ $= ( \alpha \beta + \beta \gamma + \gamma \alpha ) ^ { 2 } - 2 \alpha \beta \gamma ( \alpha + \beta + \gamma )$ $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = ( \alpha + \beta + \gamma ) ^ { 2 } - 2 ( \alpha \beta + \beta \gamma + \gamma \alpha )$

$u ^ { 3 } - \left( \left( \frac { 2 } { 3 } \right) ^ { 2 } - 2 \times \frac { - 5 } { 3 } \right) u ^ { 2 } + \left( \left( \frac { - 5 } { 3 } \right) ^ { 2 } - 2 \times \frac { 4 } { 3 } \times \frac { 2 } { 3 } \right) u - \frac { 16 } { 9 }$ $= u ^ { 3 } - \frac { 34 } { 9 } u ^ { 2 } + u - \frac { 16 } { 9 } = 0 \Rightarrow 9 u ^ { 3 } - 34 u ^ { 2 } + 9 u - 16 = 0$

B1

М1

A1

Writing the expression in terms of standard symmetrical forms

Writing the expression in terms of standard symmetrical forms Substituting in and rearranging to answer

Must include one intermediate step $\mathrm { NB } \sum \alpha = \frac { 2 } { 3 } , \sum \alpha \beta = - \frac { 5 } { 3 } , \alpha \beta \gamma = \frac { 4 } { 3 }$

Condone without factorisation of "2"

$\mathrm { NB } \sum \alpha ^ { 2 } = \frac { 34 } { 9 } , \sum \alpha ^ { 2 } \beta ^ { 2 } = 1$

[4]

(b)

DR $\frac { \sum \alpha ^ { 2 } \beta ^ { 2 } } { \alpha \beta \gamma } = \frac { \left( \frac { 9 } { 9 } \right) } { \left( \frac { 4 } { 3 } \right) } \text { or } \frac { 1 } { \left( \frac { 4 } { 3 } \right) }$

$= \frac { 3 } { 4 }$

M1

A1

[2]

3.1a

1.1

Their $\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }$ from part (a) over $\pm \frac { 4 } { 3 }$

Strict ft

Question

Answer

Marks

AO

Guidance

2

(a)

\(\begin{aligned}

\text { DR }

( r + 2 ) ( r - 1 ) \end{aligned}\) \(\begin{aligned}

\frac { A } { r - 1 } + \frac { B } { r + 2 }

A = 1 , B = - 1

= \end{aligned}\) \(\begin{array} { c c c c c c } \frac { 1 } { 4 }

-

\frac { 1 } { 7 }

\cdots

-

\frac { 1 } { n - 1 }

\frac { 1 } { 5 }

-

\frac { 1 } { 8 }

\frac { 1 } { n - 3 }

-

\frac { 1 } { n }

\frac { 1 } { 6 }

-

\frac { 1 } { 9 }

\frac { 1 } { n - 2 }

-

\frac { 1 } { n + 1 }

\frac { 1 } { 7 }

-

\cdots

\frac { 1 } { n - 1 }

-

\frac { 1 } { n + 2 } \end{array}\) \(\begin{aligned}

= \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }

= \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } \end{aligned}\)

A1

1.1

Correct factorisation of denominator soi Correct form for partial fractions

M1 can be ft from any $A , B$ having opposite signs

For M1, condone omission of $\frac { 1 } { 7 }$ or $- \frac { 1 } { n - 1 }$

Alternative Method \(\begin{aligned}

\therefore \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 5 } ^ { n } \frac { 1 } { r + 2 }

= \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 8 } ^ { n + 3 } \frac { 1 } { r - 1 }

= \sum _ { r = 5 } ^ { 7 } \frac { 1 } { r - 1 } - \sum _ { r = n + 1 } ^ { n + 3 } \frac { 1 } { r - 1 } \end{aligned}\)

М1

Using partial fractions, separating into two sums, re-indexing so that the summands have identical form and cancelling central terms.

Might see start and end terms explicitly. eg $\sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 8 } ^ { n + 3 } \frac { 1 } { r - 1 }$

Question

Answer

Marks

AO

Guidance

\multirow{2}{*}{}

\(\begin{aligned}

\therefore \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } =

= \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }

= \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } \end{aligned}\)

A1

AG.

Might see formal substitution of index. eg
Let $R = r + 3 \Rightarrow r + 2 = R - 1$ \(\begin{aligned}	\therefore \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 5 } ^ { n } \frac { 1 } { r + 2 }
	= \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { R = 8 } ^ { n + 3 } \frac { 1 } { R - 1 } \end{aligned}\)

[5]

(b)

$= \frac { 37 } { 60 }$ or awrt 0.617

B1

[1]

2.2a

3

(a)

\(\begin{aligned}

\ln ( 1 + \sin \theta ) = 0 \Rightarrow 1 + \sin \theta = 1 \Rightarrow \sin \theta = 0

\text { so } \alpha = 0 \text { and } \beta = \pi \end{aligned}\)

M1

A1 [2]

1.1a

2.2a

(b)

\(\begin{aligned}

A = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } ( \ln ( 1 + \sin \theta ) ) ^ { 2 } \mathrm {~d} \theta

= 0.4162 ( 4 \mathrm { sf } ) \text { cao } \end{aligned}\)

M1

A1 [2]

1.2

1.1

Correct formula for area with $r$ correctly substituted and their limits. Must be unambiguous but can be implied by correct answer/later work BC

Incorrect formula = M0A0 Condone missing $\mathrm { d } \theta$

(c)

$\theta = \frac { \pi } { 2 } \Rightarrow r = \ln 2 = 0.6931 ( 4 \mathrm { sf } )$ which would be the diameter, $D$, of the circle

But $A = 0.4162 ( 4 \mathrm { sf } ) = > D = 0.7280 ( 4 \mathrm { sf } )$ or $R = 0.3640 ( 4 \mathrm { sf } )$ so the curve is not circular

M1

A1

[2]

3.1a

3.2a

or radius $R = 0.3466 ( 4 \mathrm { sf } )$ condone correct $R$ or $D$ without reasoning

or $R = 0.3466 ( 4 \mathrm { sf } ) ($ or $D = 0.6931 ) \Rightarrow A = 0.3773 ( 4 \mathrm { sf } )$ which is not $0.4162 ( 4 \mathrm { sf } )$

It must be clear that the $r$ value would be the diameter of the circle; the calculation alone is insufficient for M1.

M1 can be implied by area given as $\pi \left( \frac { \ln 2 } { 2 } \right) ^ { 2 }$

Explanation must include comparison of $R$ 's, $D$ 's or $A$ 's and conclusion . Allow correct working to 3 sf .

Question

Answer

Marks

AO

Guidance

4

(a)

DR

\(\begin{aligned}

r ^ { 2 } = ( - 4 ) ^ { 2 } + ( \sqrt { 48 } ) ^ { 2 } \quad \text { or } ( r \cos \theta = - 4 \text { and }

r \sin \theta = \sqrt { 48 } ) \text { or } \tan \theta = - \sqrt { 3 } \text { oe }

r = 8 \left( \mathrm { ie } z = 8 \mathrm { e } ^ { \mathrm { i } \theta } \right) \quad \theta = 2 \pi / 3 \left( \mathrm { ie } z = r \mathrm { e } ^ { \mathrm { i } 2 \pi / 3 } \right)

\sqrt [ 3 ] { 8 } \text { or } 2

\frac { 2 \pi } { 9 } \text { soi }

\frac { 2 \pi } { 3 } + 2 \pi k \text { for } k = 1 \text { and } 2 \text { oe seen }

2 \mathrm { e } ^ { \frac { 2 } { 9 } \pi i } , 2 \mathrm { e } ^ { \frac { 8 } { 9 } \pi i } \text { and } 2 \mathrm { e } ^ { - \frac { 4 } { 9 } \pi i } \end{aligned}\)

A1

2.1

Correct use of relevant formula(e). Some working must be seen.

Correct answer with no working: M0A0 or eg $\theta = 8 \pi / 3$

B1ft

2.1

Argument of (principal) cube root is one third of their argument

M1

2.2a

Considering further arguments at angular distance $2 \pi$

A1

1.1

or eg $2 \mathrm { e } ^ { \frac { 2 } { 9 } \pi i } , 2 \mathrm { e } ^ { \frac { 8 } { 9 } \pi i }$ and $2 \mathrm { e } ^ { \frac { 14 } { 9 } \pi i }$

Must be in exponential form, not just $r =$ and $\theta =$. Do not condone any missing i's.

(b)

DR

The cube roots form an equilateral triangle which has (3) lines of symmetry, (one) through each vertex $\theta = \frac { 2 \pi } { 9 } , \theta = \frac { 8 \pi } { 9 } \text { and } \theta = - \frac { 4 \pi } { 9 } \text { soi }$

B1

2.2a

for one

for all three without extras

ft their angles if $2 \pi / 3$ apart.

If valid alternatives, must come from clear explanation/diagram

Question

Answer

Marks

AO

Guidance

5

(a)

(i)

\(\begin{aligned}

\mathrm { f } ^ { \prime } ( x ) = \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } } \text { from the formula book }

\text { so } \mathrm { f } ^ { \prime \prime } ( x ) = - \frac { 1 } { 2 } \cdot \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \cdot ( - 2 x )

= \frac { x } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \end{aligned}\)

М1

1.1

Formula from the Formula Booklet and attempt differentiation

To within sign error

(a)

(ii)

\(\begin{aligned}

f ( 0 ) = 0 , f ^ { \prime } ( 0 ) = 1 \text { and } f ^ { \prime \prime } ( 0 ) = 0

f ^ { \prime \prime \prime } ( x ) = \frac { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } - x \cdot \frac { 3 } { 2 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \cdot ( - 2 x ) } { \left( 1 - x ^ { 2 } \right) ^ { 3 } }

\text { so } f ^ { \prime \prime \prime } ( 0 ) = 1 \text { and } f ( x ) = x + \frac { 1 } { 6 } x ^ { 3 } + \ldots \end{aligned}\)

B1

M1

A1

[3]

1.1

3.1a

2.1

or $a _ { 0 } = 0 , a _ { 1 } = 1$ and $a _ { 2 } = 0$

Differentiate and simplify far enough to be able to justify value 1

Condone 3! In place of 6

Ignore sign error in $\mathrm { f } ^ { \prime \prime } ( x )$

Either full derivative or "zero term" denoted as such

Not BC. If M0 then SC1 for correct expansion

(a)

(iii)

\(\begin{aligned} \int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x \approx \int _ { 0 } ^ { \frac { 1 } { 2 } } x

+ \frac { 1 } { 6 } x ^ { 3 } \mathrm {~d} x

=

0.127604167 \ldots

=

0.127604 \text { to } 6 \mathrm { dp } \end{aligned}\)

M1

A1

[2]

1.1

Integral of their 2 term cubic with limits

Could be BC

(b)

\(\begin{aligned}

\int 1 \times \sin ^ { - 1 } x \mathrm {~d} x = x \sin ^ { - 1 } x - \int \frac { x } { \sqrt { 1 - x ^ { 2 } } } \mathrm {~d} x

= x \sin ^ { - 1 } x + \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } ( + \mathrm { c } )

\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) = \frac { \pi } { 12 } + \frac { \sqrt { 3 } } { 2 } - 1 \end{aligned}\)

M1

A1

[3]

3.1a

1.1

Attempt integration by parts

ignore limits. Formula for parts must be correct

Questions Further Pure Core 2 (116 questions)

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OCR Further Pure Core 2 2021 June Q4

OCR Further Pure Core 2 2021 June Q2

OCR Further Pure Core 2 2021 June Q4

OCR Further Pure Core 2 2021 June Q5

OCR Further Pure Core 2 2021 June Q2

OCR Further Pure Core 2 2021 June Q3

OCR Further Pure Core 2 2021 June Q5

OCR Further Pure Core 2 2021 June Q1

OCR Further Pure Core 2 2021 June Q2

OCR Further Pure Core 2 2021 June Q3

OCR Further Pure Core 2 2021 June Q4

OCR Further Pure Core 2 2021 June Q5