Mark Scheme (Results)

January 2024

Pearson Edexcel International Advanced Level

in Pure Mathematics P2 (WMA12) Paper 01
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January 2024
Question Paper Log Number P74311A
Publications Code WMA12_01_2401_MS
All the material in this publication is copyright
© Pearson Education Ltd 2024
General Marking Guidance

• All candidates must receive the same treatment. Examiners must

mark the first candidate in exactly the same way as they mark the
last.
• Mark schemes should be applied positively. Candidates must be
rewarded for what they have shown they can do rather than
penalised for omissions.
• Examiners should mark according to the mark scheme not
according to their perception of where the grade boundaries may
lie.
• There is no ceiling on achievement. All marks on the mark scheme
should be used appropriately.
• All the marks on the mark scheme are designed to be awarded.
Examiners should always award full marks if deserved, i.e. if the
answer matches the mark scheme. Examiners should also be
prepared to award zero marks if the candidate’s response is not
worthy of credit according to the mark scheme.
• Where some judgement is required, mark schemes will provide
the principles by which marks will be awarded and exemplification
may be limited.
• When examiners are in doubt regarding the application of the
mark scheme to a candidate’s response, the team leader must be
consulted.
• Crossed out work should be marked UNLESS the candidate has
replaced it with an alternative response.
 EDEXCEL IAL MATHEMATICS

General Instructions for Marking

1. The total number of marks for the paper is 75.

2. The Edexcel Mathematics mark schemes use the following types of marks:
• M marks: Method marks are awarded for ‘knowing a method and attempting to apply
it’, unless otherwise indicated.
• A marks: Accuracy marks can only be awarded if the relevant method (M) marks have
been earned.
• B marks are unconditional accuracy marks (independent of M marks)
Marks should not be subdivided.

3. Abbreviations

These are some of the traditional marking abbreviations that will appear in the mark
schemes and can be used if you are using the annotation facility on ePEN:
• bod – benefit of doubt
• ft – follow through
o the symbol will be used for correct ft
• cao – correct answer only
• cso – correct solution only. There must be no errors in this part of the question to
obtain this mark
• isw – ignore subsequent working
• awrt – answers which round to
• SC – special case
• oe – or equivalent (and appropriate)
• d… or dep – dependent
• indep – independent
• dp – decimal places
• sf – significant figures
•  – The answer is printed on the paper or ag- answer given

• or d… – The second mark is dependent on gaining the first mark

4. All A marks are ‘correct answer only’ (cao), unless shown, for example, as A1 ft to indicate
that previous wrong working is to be followed through. After a misread however, the
 subsequent A marks affected are treated as A ft, but manifestly absurd answers should
never be awarded A marks.

5. For misreading which does not alter the character of a question or materially simplify it,
deduct two from any A or B marks gained, in that part of the question affected. If you are
using the annotation facility on ePEN, indicate this action by ‘MR’ in the body of the
script.

6. If a candidate makes more than one attempt at any question:

a) If all but one attempt is crossed out, mark the attempt which is NOT crossed out.
b) If either all attempts are crossed out or none are crossed out, mark all the attempts
and score the highest single attempt.

7. Ignore wrong working or incorrect statements following a correct answer.

 General Principles for Pure Mathematics Marking
(NB specific mark schemes may sometimes override these general principles)

Method mark for solving 3 term quadratic:

1. Factorisation
( x2 + bx + c) = ( x + p)( x + q), where pq = c leading to x = …
(ax2 + bx + c) = (mx + p)(nx + q), where pq = c and mn = a leading to x = …
2. Formula
Attempt to use correct formula (with values for a, b and c)
3. Completing the square
Solving x 2 + bx + c = 0 : ( x  b2 )  q  c, q  0 leading to x = …
2

Method marks for differentiation and integration:

1. Differentiation
Power of at least one term decreased by 1 ( x n → x n −1 )
2. Integration
Power of at least one term increased by 1 ( x n → x n +1 )

Use of a formula
Where a method involves using a formula that has been learnt, the advice given in recent
examiners’ reports is that the formula should be quoted first.

Normal marking procedure is as follows:

Method mark for quoting a correct formula and attempting to use it, even if there are small
mistakes in the substitution of values.

Where the formula is not quoted, the method mark can be gained by implication from correct
working with values, but may be lost if there is any mistake in the working.

Exact answers
Examiners’ reports have emphasised that where, for example, an exact answer is asked for, or
working with surds is clearly required, marks will normally be lost if the candidate resorts to
using rounded decimals.

Answers without working

The rubric says that these may not gain full credit. Individual mark schemes will give details of
what happens in particular cases. General policy is that if it could be done “in your head”,
detailed working would not be required. Most candidates do show working, but there are
occasional awkward cases and if the mark scheme does not cover this, please contact your team
leader for advice.
 January 2024
WMA12 Mark Scheme

Question
Number Scheme Marks
f ( x) = ax + 3x − 8 x + 2
2
1 3

Sets f (2) = 3  8a + 12 − 16 + 2 = 3 M1

5
 8a = 5  a = M1A1
8
(3)
(3 marks)

M1: Attempts to set f (2) = 3 . The values embedded and the expression set equal to 3 is sufficient. May be
implied by further work. If 2 embedded in the expression is not seen, condone slips in their evaluation
provided there is still evidence that the intention was to substitute in 2.
( )
May also be seen as f (2) − 3 = a(2)3 + 3(2)2 − 8(2) − 1 = 0

M1: Solves a linear equation in a arising from setting f (2) = 3 . Condone slips in their rearrangement
proceeding to a = ... (May be implied by further work).

5
A1: a= or exact equivalent
8
----------------------------------------------------------------------------------------------------------------------------------------
5
Answers of a = may appear with very little or no working, possibly via trial and improvement. If so, then
8
marks can only be allocated if evidence is shown.
5
e.g. f ( x) =  x3 + 3x 2 − 8 x + 2  f (2) = 5 + 3(2) − 8(2) + 2 = 3  dividing by ( x − 2) gives a remainder of 3
2

8
More difficult alternative methods may be seen:
----------------------------------------------------------------------------------------------------------------------------------------
Alt I: e.g. algebraic division (note they may also divide ax3 + 3 x − 8 x − 1 by ( x − 2) and set the remainder
2

equal to 0)

ax 2 + (3 + 2a) x + (−2 + 4a)

x − 2 ax3 + 3x 2 − 8x +2
ax3 − 2ax 2
(3 + 2a) x 2 − 8x
(3 + 2a) x 2 + (−6 − 4a) x
(−2 + 4a) x +2
(−2 + 4a) x + 4 − 8a
5
− 2 + 8a  −2 + 8a = 3  a =
8
M1: Divides the cubic by ( x − 2) , leading to a quadratic quotient where both the coefficient of x and the
constant term are in terms of a. They should then have a linear remainder in a which is then set equal to 3

M1: Solves an equation resulting from setting a linear remainder in a equal to 3 . It is dependent on the first
method mark via this route.
5
A1: Completely correct with a =
8
----------------------------------------------------------------------------------------------------------------------------------------
Alt II: You may also see a grid or an attempt at factorisation of ax3 + 3 x − 8 x − 1 via inspection
2

 1
M1: For an attempt at factorising e.g. ax3 + 3x 2 − 8 x − 1 = ( x − 2)  ax 2 + bx + 
 2
M1: Forms two correct simultaneous equations and proceeds to find a value for a. Condoning slips in their
solving.
x 2 : 3 = −2a + b
17
3−
1 17 4 =5
x : − 8 = − 2b  b =  a =
2 4 −2 8

5
A1: Completely correct with a =
8
Question
Number Scheme Marks
2 3 
12

 + 4 x 
8
 

5 5
3 3 7
( )
12 7 12
7 7
Term in x is C7   4 x or coefficient of x is C7   4 M1 A1
8 8
   

Coefficient is 96228 A1

(3)
(3 marks)

Note that you do not need to see x 7 in their solution

5
3
Combines a correct binomial coefficient with   and either ( 4x ) , 4 7 or x 7
7
M1:
8
 12  12  12  12
Look for the binomial coefficient of the form e.g.   , C7 or 792 o.e. Condone   or C5 .
7 5
   
May be implied by further work.
12 12 12 12 7
3   3  32   3  12 1110  9  8  7  6  32 
Alternatively  + 4 x  =    1 + x       x
8  8 3 8 7!  3 
         
5
12 3 7
Condone invisible brackets for this mark. e.g. C7   4x
8
5 5
3 7 12!  3 
12 7
A1: Correct unsimplified term or coefficient e.g. C7    4 or e.g.    4 . Invisible brackets
8 5!7! 8
   
may be implied by further work. Do not be concerned by the presence or absence of x 7
5 5
3 3
( ) ( )
12 7 12 7
Note C7   4 x or C5   4 x scores M1A1 (may be seen as part of a list or several terms
8 8
   
found in the expansion)

A1: For 96228 but condone 96 228 x 7 isw once a correct answer is seen. The term or coefficient
must be identified if they have more than one term.
A correct answer on its own with no incorrect working seen can score full marks.

 5 
3
Note: C7    4 x = 23.493...   23.5 will score M1A0A0 (where they multiply by 4 instead of
 12 7

 8 
   
 
47 )
Question
Number Scheme Marks
3(a)
( ) ( ) = ...
2 2
Attempts 8 − 3 + 5 − −7 M1

Writes ( x − 3) + ( y − 5 ) = k
2 2
M1

( x − 3) + ( y − 5) = 169
2 2
A1
(3)
(b)
( )
2
Attempts d + 2 22 = 169  d = ...
2
M1

States or uses y = 5 + d( ) dM1

y = 14 A1
(3)
(6 marks)

(a)

M1: Attempts to find the radius or radius 2. Must proceed to find a value.
Condone a sign slip if attempting 5 − −7 if this is seen as 5 − 7 o.e.
May be implied by 13 or 169. Do not be concerned with the labelling of the expression or value as r
or r 2

( x − 3) + ( y − 5 )
2 2
M1: Writes the equation of the circle in the form = k , where k  0 , o.e.

e.g. x 2 + y 2 − 6 x − 10 y + c = 0 where c  0

( ) ( )
2 2 2
Invisible brackets may be implied by further work. Condone x − 3 + y − 5 = r where no
attempt has been made to find the radius or radius2

( x − 3) + ( y − 5 )
2 2
A1: = 169 o.e. e.g. x2 + y 2 − 6 x − 10 y − 135 = 0 isw once a correct unsimplified

( ) ( )
2 2 2
equation for C is found. Condone x − 3 + y − 5 = 13
A correct equation scores M1M1A1 but it must be seen in (a).
(b) Note that (3 + 2 22, 14) as the final answer is A0
M1: Attempts Pythagoras' theorem with “13” (as the hypotenuse) and 2 22 to find d. Not just d 2
(See diagram)

dM1: ( )
States or uses the equation for MN, y = 5 + d but also condone stating/using y = 5 − d ( )
(may be implied by their value for y)

A1: y = 14 only (Must be exactly 14 e.g. 13.99... = 14 is A0). Further work leading to a different
equation of a line is A0.

M N
d

Alternative using their centre and the x coordinates of M and N are equidistant from the centre:

M1: Attempts to find the x coordinates of M or N e.g. 3  2 22 , uses their equation of the circle with one
of x = 3  2 22 and forms a three term quadratic in y (e.g. y 2 − 10 y − 56 = 0 )

dM1: Attempts to solve their quadratic proceeding to a value for y. Usual rules for solving a
quadratic. Accept via a calculator. It is dependent on the previous method mark.

A1: y = 14 only (Must be exactly 14 e.g. 13.99... = 14 is A0) Further work leading to a different
equation of a line is A0.
-------------------------------------------------------------------------------------------------------------------------------
--------
Alternative finding an angle in the triangle MNA, then using the angle to find d
M1: e.g. attempts the cosine rule to find the angle MNA, and then trigonometry to find d

( )
2
132 + 4 22 − 132
cos ( MNA ) =  MNA = 43.8...
2 13  4 22
d = sin 43.8... 13 = ...
dM1: Same as main scheme

A1: y = 14 only (Must be exactly 14 e.g. 13.99... = 14 is A0) Further work leading to a different
equation of a line is A0.
-------------------------------------------------------------------------------------------------------------------------------------------
---------
Note: There may be other more complex attempts (including use of area) Send to review if unsure.
 9 220 − 27 22 
e.g. Attempting to find the line which passes through AN  y = x+  and solving
 2 22 44 
simultaneously with the equation of the circle  y = 14
This typically will score M1 for attempting Pythagoras’ theorem, but can score for attempting the
gradient of the line AN.
The dM1 mark would be for the full attempt to find where the line through AN intersects the equation of
the circle.
Question
Scheme Marks
Number
4(a)

Correct shape
B1

y=4 Correct intercept B1

Correct equation for asymptote B1

(3)
(b) Strip width = 2.5 B1


 − 12 x 
8.5
5
 3 + 4  dx  13 + 4.009 + 2  (6.280 + 4.577 + 4.146 + 4.037)
−4   4 M1 A1
= 68.86125 ( = awrt 69)
(3)


(c)  
8.5 1
− x
(i) 3  dx = 69 − 4  ( 8.5 − −4 ) = awrt 19
2
M1, A1ft
−4  

 
 − 12 x   2x 
8.5 4 1

(ii)  3 + 4  dx +  3 + 4  dx = awrt 138 B1ft

−4   −8.5  
(3)
(9 marks)
(a) If there are multiple attempts then mark the most complete attempt. If coordinates/the
equation of the asymptote are stated by the question and on their diagram, then the sketch
takes precedence.
B1: Correct shape in quadrants 1 and 2 and does not appear in quadrants 3 or 4 or touch the x-
axis.
Do not be concerned about the y intercept or asymptote. Mark the intention to draw an exponentially
decreasing graph – do not penalise parts of the curve which may appear linear.

B1: Intercept at (0, 5). Condone invisible brackets, or may just have 5 marked on the y-axis, or
x = 0, y = 5 . If the point is labelled as (5, 0) then condone provided it is a y intercept which is
above the x-axis.

B1: Graph with asymptote at y = 4 (asymptote does not need to be drawn) Must be an equation.
There must be a graph with the intention of a horizontal asymptote above the x-axis to score the mark. If
more than one horizontal asymptote is given, then B0. Where the asymptote is not drawn (e.g. with
dashed lines) it must be clear that y = 4 refers to the asymptote of the graph and not the y-intercept.
Examples:

B0B0B1 B0B0B1 B1B0B0 B1B1B0

(b)
B1: For a correct strip width of h = 2.5 o.e. May be implied by 1.25 or equivalent in front of the
bracket

M1: Correct application of the trapezium rule with all the y values and their h (which may be 1
and condone if negative)
Condone a missing trailing bracket. Condone other bracketing errors only if followed by a
correct answer for their values. Condone miscopying of the y values from the table but the intended
values must be used in the correct places.
Also award for attempts at forming individual trapezia and adding these together.
e.g.

 13 + 6.280   6.280 + 4.577   4.577 + 4.146   4.146 + 4.037   4.037 + 4.009 

2.5   + 2.5   + 2.5   + 2.5   + 2.5  
 2   2   2   2   2 
A1: awrt 69 (exact answer is 68.86125) isw once a correct answer is seen. awrt 69 following a
correct calculation.


 − 12 x 
8.5

Note that the calculator answer for  3 + 4  dx is 66.367... which scores 0 marks.
−4  

(c) They may use their 69 or a more accurate value of their 69 which is acceptable.

(i) Attempts at redoing the trapezium rule with new values is M0A0


8.5

M1: Attempts "69" 4  (8.5 − −4) or "69" 4 dx or (b)  50

−4

May also be implied by just stating the value of (b)  50

A1ft: awrt 19 but follow through on (b) − 50 .

(c)(ii)
B1ft: awrt 138 but follow through on 2  (b)
-------------------------------------------------------------------------------------------------------------------------------
---------

Part (c) when the answer to part (b) is awrt 66 (where (b) has been done using the calculator)

(c)(i) If a method is shown then M1A1ft is possible e.g. 66  4  (8.5 − −4) or e.g. 66 − 50 for
awrt 16.
awrt 16 with no working is M0A0ft (as this could have been done using the calculator)

(c)(ii) awrt 132 scores B1ft (condoned)

Question
Number Scheme Marks
5(i) r = 0.25 B1

1.5
S = =2 M1 A1
1 − 0.25
(3)
(ii)
(a) (u = 3,) u = 0, u = 32 , u = 3, ...
1 2 3 4
M1

(u = 3) , u = 0, u = 32 , u = 3, ... and e.g. states that u

1 2 3 4 4
= u1 so sequence will repeat (is
A1
periodic)
(2)
(b) (Order) 3 A1
(1)
 3

70

(c) un = 23   3 + 0 +  + 3 = 106.5 M1, A1

 2
n =1  
(2)
(8 marks)
(i)
B1: States or uses r = 0.25 (may be implied by using r = 0.25 in the sum to infinity formula). 6  0.25 or
any other term in the sequence is not sufficient evidence that this is a geometric sequence.
Alternatively, shows an understanding of the sigma notation by writing at least the first three terms in
2 3
1 1 1
the sequence e.g. 6  + 6    + 6    + ...
4 4 4
Condone invisible brackets and may be written as a list.

a
M1: Attempts S = with a = 1.5 (or 6  0.25), r = 0.25 or a = 6, r = 0.25
1− r
a
You may also see S = − a with a = 6, r = 0.25 which is acceptable
1− r
Beware of other variants such as taking a factor of 6 out and using a = 0.25, r = 0.25
The values embedded in the expression is sufficient to score this mark.

A1: 2 cao following evidence of the use of the sum to infinity formula

2 3
1 1 1
Note 6  + 6    + 6    + ... = 2 is B1M0A0 (no evidence of using the sum to infinity
4 4 4
formula)
(ii)(a)
M1: Uses the recurrence relation correctly at least once. If they do not achieve a correct u2 = 0
then check e.g. their u3 is a correct follow through for their u2

( ) 3
A1: u1 = 3 , u2 = 0, u3 = , u4 = 3, ... with a statement or conclusion such as u4 = u1 /repeats/cycles (is
2
periodic)
3
As a minimum accept (3), 0, , 3 (hence) periodic
2
(ii)(b) This mark can only be scored following correct values in (ii)(a) (or a restart in (ii)(b))
A1: (Order is) 3 (do not accept (repeats every) third).
Cannot be for just listing the terms of the sequence.

(ii)(c)

u
70

M1: Establishes a correct method of finding n

for their periodic sequence of order 3 found
n =1

in (a).
If they did not have a periodic sequence of order 3 in (a) and they do not restart then this mark cannot
be scored.
Seeing the values embedded in the expression is sufficient to score this mark.
Attempts to use the sum of a geometric or an arithmetic series is M0A0

213
A1: 106.5 or exact equivalent e.g. ignore any subsequent incorrect attempts to round.
2
Question
Scheme Marks
Number
6 (a)
( ) ( )
2
1
2 log 4 x + 3 = log 4 x + 3 or = log 4 2 o.e B1
2

( ) ( )
2
Combines two terms e.g. 2 log 4 x + 3 + log 4 x = log 4 x x + 3 M1

( ) ( )
2
e.g. x x + 3 = 2 4 x + 2 A1

 2 
e.g. x  x + 6 x + 9  = 8 x + 4  x + 6 x + x − 4 = 0 *
3 2
A1*
 
(4)
(b) (i)
3 2  2

( ) 
x + 6 x + x − 4 = x + 1  x + 5x − 4 

M1

−5  25 + 16 −5  41
x= x= dM1 A1
2 2
(3)
(ii) −5 + 41
x= B1
2
(1)
(8 marks)

(a) Do not penalise the omission of base 4 in their working provided the terms and values are
consistent with base 4.

B1: Writes (or may be implied by further work)

( ) ( )
2
• 2log 4 x + 3 as log 4 x + 3 or
1
• as = log 4 2
2

M1: Correctly combines (at least) two of the original terms. e.g.

( ) ( ) ( ) ( )
2
• 2 log 4 x + 3 + log 4 x = log 4 x x + 3 is M1 but 2log 4 x + 3 + log 4 x = 2log 4 x x + 3 is M0

 4x + 2 
• ( )
log 4 4 x + 2 − log 4 x = log 4 
 x 
is M1
 

• ( 1
) (
log 4 4 x + 2 + = log 4 8 x + 4 is M1
2 )
Do not penalise if they subsequently make errors or apply laws incorrectly. May be implied if they

( )
2
proceed to an equation not involving logs e.g. x x + 3 = 2(4 x + 2) (but not the given answer)

Beware that 2log( x + 3) + log x = 2log x( x + 3) = log x( x + 3)2 is B1M0A0A0

A1: A correct intermediate equation not involving logs (but not the given answer).
log 4 x( x + 3)2 1 x( x + 3) 2
Must follow correct log work. e.g. =  = 2 is A0
log 4 (4 x + 2) 2 (4 x + 2)
 A1*: Correct proof. Expect to see
• a correct equation not involving logs which is not the final answer
• the brackets multiplied out ( x + 3) 2 or ( x 2 + 3x)( x + 3) before proceeding to the given answer.
Do not allow this mark for recovery from incorrect log work but allow invisible brackets to be
“recovered” or implied if the intention is clear.

(b) (i) Answers with no working score 0 marks in (b)(i)

M1: Divides by ( x + 1) . May be implied by a correct quadratic. May be seen as the quotient if
 2
( ) 
attempting algebraic division. If via division look for x + 1  x  5 x....  or via inspection
 

( x + 1)  x 2 
 ....x  4 


dM1: Attempts to find the roots of their quadratic factor (usual rules for solving a quadratic).
Accept solving the quadratic via a calculator (you may need to check this). Condone decimals to 2sf
for this mark. e.g. may see 0.70 or −5.7
Note that if there are multiple attempts to solve the quadratic (including via a calculator so roots may
be stated) then mark the one which scores the most marks).

A1: x =( ) −5 2 41 or exact equivalent e.g. −2.5  10.25 (Do not accept decimal answers but isw if
seen after an exact answer is seen)
Ignore the presence or absence of x = −1

(b)(ii) This mark can only be scored provided x = ( ) −5 +2 41 (or awrt 0.702) is found in (i), even if
this is via solving the cubic on a calculator
−5 + 41
B1: x = or exact equivalent ONLY (accept awrt 0.702 if given in (b)(i)) If both answers are
2
−5 + 41
present then they must indicate they have chosen x = e.g circling or ticking (or putting a
2
cross by the other one)
Question
Scheme Marks
Number
7.(a) Attempts to use 4000 = 300 +11d to find 'd' M1
Uses 300 + 3'' d '' M1
Wheat production in year 4 is awrt 1310 (to the nearest 10) (tonnes) A1
(3)
11
(b) Attempts to use 4000 = 300r to find 'r' M1
Finds r = (1.266...) and MULTIPLIES this by 300 M1
Wheat production in year 2 is awrt 380 (to the nearest 10) (tonnes) A1
(3)
(c)
Attempts
12
300 + 4000 or
(
300 "1.266..." − 1
12
) M1
2 "1.266..."− 1

Finds
12
300 + 4000 −
(
300 "1.266..." − 1
12
)
= ( 25800 − 17935 ) dM1
2 "1.266..."− 1
Difference = 7860 but allow 7870 (tonnes) (not AWRT) A1
(3)
(9 marks)
Condone slips copying 300 and 4000 if they lose or gain an extra 0 in all parts
(a)
M1: Attempts to use the AP formula in an attempt to find 'd'

Accept an attempt at 4000 = 300 +11d resulting in a value for d.

4000 − 300
Accept the calculation proceeding to a value (condoning slips)
11
3700
May be implied by or awrt 336
11

M1: Attempts to find the fourth term (do not allow a misread) by attempting 300 + 3 ''336'' .

You may award this following an ''incorrect'' AP formula with 12d being used instead of 11d.
4000 − 300
e.g. 4000 = 300 +12d or more likely usually leading to an answer of 1225. If no
12
method seen to find d then it must be correct.
A1: awrt 1310 (to the nearest 10) (tonnes) isw once a correct answer is seen.
-------------------------------------------------------------------------------------------------------------------------------
---------
Alt (a)
x − 300 4 −1
M1: Forms the equation = or may be implied by further work
4000 − 300 12 − 1
4 −1
M1: Rearranges the equation to x = 300 +  ( 4000 − 300 ) or allow this mark for
12 − 1
4
x = 300 +  ( 4000 − 300 )
12
A1: awrt 1310 (to the nearest 10) (tonnes) isw once a correct answer is seen.

(b)
M1: Attempts to use the GP formula in an attempt to find 'r'
 11 400011 4000
Look for 4000 = 300r  r =  r = 11 condoning slips. Implied by r = awrt 1.3
300 300
provided they do not start from an incorrect equation.

M1: A correct attempt to find the second term (do not allow a misread) by multiplying 300 by
their "1.266..." .

It is dependent on an attempt to find r from either 4000 = 300r11 or 4000 = 300r12 condoning
slips.
If no method is seen to find r then it must be correct.
12 4000
e.g. following 4000 = 300r or 12 . Condone slips.
300

A1: awrt 380 (to the nearest 10) (tonnes) isw once a correct answer is seen.

(c) Condone use of their d and their r even if they have come from incorrect methods
M1: A correct method to find the sum of either the AP or the GP
12 12
For the AP accept an attempt at either 300 + 4000 or 2  300 + 11 "336"
2 2

For the GP accept an attempt at either

(
300 "1.266..." − 1
12
) or 300 (1 − "1.266..." ) 12

"1.266..."− 1 1 − "1.266..."

dM1: Both formulae must be attempted ''correctly'' (see above) and the difference taken

FYI if d and r are correct, the sums are 25 800 tonnes and 17 935

A1: Difference = 7860 or 7870 (tonnes) This is NOT AWRT.

(Note that use of r = 1.266 will lead to a difference of 7810 which is max M1dM1A0)

-------------------------------------------------------------------------------------------------------------------------------------------
Alt (c) Listing values

Accept listing the values

M1: Attempts to add the 12 terms for the arithmetic or geometric series. Must include 300 and 4000, but
condone one omission of one of the middle terms. May be implied by calculating the difference
between the AP and GP values for each year and adding those together.

dM1: Attempts to add the 12 terms for the arithmetic and geometric series. Must include 300 and
4000 for both sums, but condone one omission of one of the middle terms in each of the
summations. May be implied.

A1: Difference = 7860 or 7870 This is NOT AWRT.

Year AP GP Difference
1 300 300 0
2 636.3636 379.6544 256.7093
3 972.7273 480.4582 492.2691
4 1309.091 608.0268 701.0641
5 1645.455 769.4668 875.9877
6 1981.818 973.7715 1008.047
7 2318.182 1232.322 1085.86
8 2654.545 1559.521 1095.024
9 2990.909 1973.597 1017.312
10 3327.273 2497.616 829.6567
11 3663.636 3160.77 502.8668
12 4000 4000 0

Sum 25800 17935.2 7864.796

Question
Number Scheme Marks
8 (i)
Substitutes a value e.g. n = 6 into n + 3n + 1 where n + 3n + 1 is not prime
2 2
M1

Correct calculation for that value e.g n + 3n + 1 = 55

2
A1
And conclusion ''which is not prime''
(2)
Attempts to find n − 2 for either odds or evens
(ii) 2

(
E.g Attempts 2 p + 1 − 2 or 2 p − 2 ) 2
( ) 2 M1

( ) 2 2
Achieves either 2 p + 1 − 2 = 4 p + 4 p − 1 or 2 p − 2 = 4 p 2 − 2 ( ) 2

and shows or gives a reason why the expression is not a multiple of 4 where required A1
(see notes)
Attempts to find n − 2 for both odds and evens (See above)
2
dM1
Achieves both
( 2 p +1) 2 2
( ) 2
− 2 = 4 p + 4 p − 1 and 2 p − 2 = 4 p − 2 and shows or gives reasons why
2

these are not multiples of 4 where required (see notes) A1*

With a conclusion that they are not multiples of 4. *

(4)
(6 marks)
(i)
M1: Substitutes a value into n + 3n + 1 where n + 3n + 1 is not prime. (Does not have to be evaluated)
2 2

Possible values are n = 6, 11,13,16 etc. Condone slips substituting in or if they have evaluated
incorrectly provided the intention was to substitute in a valid value for n. (Note that n = 0 is not
acceptable)

A1: A correct calculation for the value and a conclusion. There must be some reference to not being
prime or they show that the number is divisible by e.g. 5 and state that it is false/not true
e.g. 62 + 3  6 + 1 = 55 which is not prime (so not true) is M1A1
e.g. when n = 11 then n + 3n + 1 = 155 155  5 = 31 so false is M1A1
2

e.g. 62 + 3  6 + 1 = 55 so false is M1A0

Values of n up to 50 for which n + 3n + 1 is not prime:
2

n + 3n + 1 n + 3n + 1
2 2
n n
6 55 31 1055
11 155 32 1121
13 209 33 1189
16 305 35 1331
17 341 36 1405
21 505 39 1639
22 551 41 1805
24 649 42 1891
26 755 46 2255
28 869 48 2449
50 2651
 (ii) There should be no errors in the algebra but allow e.g. invisible brackets to be “recovered”.
Withhold the final mark if n is used instead of k

Uses algebra to describe odds e.g. n = 2p+1 or evens e.g. n = 2p and attempts n − 2 (allow
2
M1:
equivalent representation of odd or even e.g. n = 2k + 2 or 2n  5 )
Condone arithmetical slips and condone the use of e.g. n = 2n and n = 2n  1

A1: They must

• achieve a correct expanded n − 2 for either odds or evens
2

• show or give an explanation why the expression is not a multiple of 4 (other than the trivial case when
n = 2p.

( ) 2 2
e.g. 2 p + 1 − 2 = 4 p + 4 p − 1 which is 1 less than a multiple of 4 is A1 (explanation)
Do not accept. e.g. “you cannot take 4 out as a common factor” as this is insufficient and should be

( 2
) 2  2
shown. e.g. 2 p + 1 − 2 = 4 p + 4 p − 1 = 4  p +


p  − 1 is A1 (shows)

( ) 2
For the case when n = 2 p  2 p − 2 = 4 p − 2 it does not need to be followed by an explanation
2

as it is clearly not a multiple of 4 but if one is given then it must be correct. e.g. when divided by 4
1
p −
2
gives
2
They must have evaluated numbers with indices so do not accept 22 p 2 − 2

Attempts to find n − 2 for both odds and evens (See above)

2
dM1:
A1: Fully correct proof. They must
• achieve a correct expanded n − 2 for both odds and evens
2

• show or gives a reason as to why each expression is not a multiple of 4 (other than for n = 2 p )
• make a concluding overall statement. “Hence not a multiple of 4 (for all n )_” Accept “hence
proven”, “statement proved”, “QED” if they have stated for each separate case that the expression is
not a multiple of 4.
----------------------------------------------------------------------------------------------------------------------------------------
 Attempts using 4k , 4k + 1, 4k + 2, 4k + 3 should be applied similarly to the main scheme above:
M1: Attempts at least two of the four cases from e.g. 4k , 4k + 1, 4k + 2, 4k + 3 (but could be others e.g.
4k − 2, 4k − 1, 4k , 4k + 1
A1: They must
• achieve two correct expanded n − 2
2

• show or give a reason as to why the two expressions are not a multiple of 4 (same examples as in notes
above for main scheme).
n −2
2

4k − 3 16k 2 − 24k + 7
4k − 2 16k 2 − 16k + 2
4k −1 16k 2 − 8k − 1
4k 16k 2 − 2
4k + 1 16k 2 + 8k − 1
4k + 2 16k 2 + 16k + 2
4k + 3 16k 2 + 24k + 7

Attempts to find n − 2 for all four cases (see above)

2
dM1:

A1: Fully correct proof (see above)

Question
Scheme Marks
Number
9 (i) sin x
States or uses tan x = B1
cos x
sin x tan x = 5  sin x = 5cos x  1 − cos x = 5cos x
2 2
M1A1

(
cos x + 5cos x − 1 = 0  cos x =
2
) −5 2 29  x = awrt 78.9, 281.1 M1dM1A1
(6)
(ii)
(a) A = 5 B1

(1)
3 3
(b) 2 − =   = ... M1
8 2
15
= A1
16
y coordinate Q = −3 (or 2 − " A " ) B1ft
(3)
 3   3  2
(c) Sets 0 = ''5''sin  2 −  + 2  sin  2 −  = M1
 8   8  "5"
   
 3  2  3   2
sin  2 −  =    2 −  = arcsin    = ... dM1
 8  5  8   5
     
One of  = 0.38, 2.4, 3.5, 5.5, 6.7, 8.6, 9.8.... A1
 = awrt 5.51 A1
(4)
Total 14

(i) Do not penalise working in another variable or if the variable is omitted at times in their
working. e.g. sin instead of sin x is condoned as a slip or cos 2 x as cos x 2 .They are required to show
the stages of working but not to the same level of accuracy in their presentation as a “show” or
“prove” question.
sin x
B1: States or uses tan x = (may be seen as sin x tan x  tan 2 x cos x )
cos x
Uses  sin x  cos x = 1 to set up a quadratic equation in cos x only. If terms have been
2 2
M1:
collected on one side condone the omission of = 0. May form the equation tan 2 x cos x = 5 and use
the identity 1  tan 2 x =  sec 2 x proceeding to a quadratic in cos x only
1
e.g.  cos x(sec 2 − 1) = 5  − cos x = 5  1 − cos 2 x = 5cos x or alternatively, square both sides
cos x
and proceed to a quartic in cos x only.
A1: Correct 3TQ (or quartic) in cos x. The three terms do not need to be on the same side of the equation.
The =0 may be implied by further work. e.g. attempting to solve the quadratic. Alternatively award for
the quartic equation cos 4 x − 27 cos 2 x + 1 = 0

M1: Attempts to solve their 3TQ (or quartic) in cos x. Usual rules apply for solving the quadratic.
Accept via a calculator but at least one root must be correct (you may need to check this)
 They cannot proceed from the 3TQ directly to an angle for x; this is M0 and no further marks can be
scored.
Accept either root as a rounded decimal to 2sf e.g. for the correct quadratic it would be awrt 0.19 or
awrt -5.2

dM1: Attempts to find one value for x in the given range using their root. You may need to check
this. It is dependent on the previous method mark. Accept in radians. It is acceptable to proceed from
the roots of their quadratic directly to an angle in the given range to score this mark. May be implied by
awrt 79 or awrt 281 (or in radians awrt 1.4 or awrt 4.9)
A1: awrt x = 78.9, 281.1 (must be in degrees) and no others in the range.
(ii)(a)
B1: A = 5 Check by the question. If there are multiple answers, mark the answer in the main body of the
work.
(ii)(b) Solutions with no working: Note if the two coordinates are given with no working then max
M0A0B1
 15 
 − 3,  (or coordinates given the wrong way round) with no working seen then SC001
16 
 
3 3 3 
M1: Attempts to solve 2 − = or any other minimum value e.g. 2 − =− .
8 2 8 2
3 3

They must proceed using the correct order of operations e.g allow  = 2 8 , which may be
2
implied by their answer. Answer only with no working though is M0A0.
If they have a mixture of radians and degrees within an equation and they do not “recover”
then the method mark cannot be scored.
15  15 
( )
A1:  =
16
must be exact following a correct equation. May be seen as 
 16

, − 3  isw if they


proceed to write the coordinates the wrong way round.
B1ft: y coordinate Q = −3 or follow through on 2 − " A " (may be seen on the diagram or by the
question). If there is a contradiction, then the answer in the main body of the work takes
precedence. If they proceed to write the coordinates the wrong way round then isw. May be seen as a
pair of coordinates.

(ii)(c) In EPEN this is M1A1dM1A1 but we are marking this M1dM1A1A1

Solutions with no working in (c) scores 0 marks.
 3 
M1: Sets 0 = ''5''sin  2 −  + 2 (which may be implied) and proceeds to
 8 
 
 3  2
sin  2 −  =  o.e.
 8 "5"
 
3
May be implied by e.g. 2 − = "− 0.41..." or any other equivalent angle in radians or
8
2 − 67.5 = "− 23.5..." or any other equivalent if working in degrees. You may need to check this. Allow
3
use of X for 2 or 2 −
8
  3  2 3  2
dM1: Proceeds from sin  2 −  = to 2 − = arcsin    = ... which is one of the
 8  5 8  5
   
values below:
3  2
2 − = arcsin  −  = −0.41, 3.6, 5.9, 9.8, 12.2
8  5
3 2
2 − = arcsin   = 0.41, 2.7, 6.7,9.0, 13.0, 15.3
8 5
 2  3
arcsin    +
 2  5 8
May be implied by 2 = arcsin    + 1.17... or allow the expression =  
 5 2
 
The sign slip is only condoned before they take arcsin(…)
A1: Any one of the values in the table provided M1dM1 has been scored. Do not withhold this mark if
other incorrect angles are seen.
Radians (awrt) Degrees (awrt)
 0.38, 2.4, 3.5, 5.5, 6.7, 8.6, 9.8, 11.8, 22, 136, 202, 316, 382, 496, 562, 676, 742
12.9
A1:  = awrt 5.51 only (must be in radians). (Can only be scored from correct working and all
previous marks are scored)
Question
Number Scheme Marks
10 (a) 5
 dy  −

 dx =  x − 2187 x
2
M1, A1
 

( )
5 7
−
= 0  x = 2187 or e.g. x = ( 7 2187 )  x = 9 *
2
Sets x − 2187 x 2 2
dM1A1*
(4)
(b) 1 3
 1

  2 x
− 1 −
e.g. 2
+ 1458 x 2
− 74 dx = x3 − 2916 x 2
− 74 x
 6
or
M1A1ft
1 2 3
 1


− 1 −
 x + 1458x 2
− "94.5" dx = x3 − 2916 x 2 − 94.5 x
2  6

y value at P is 20.5 B1
9
1 −
1

e.g. Area R =  x3 − 2916 x 2 − 74 x  − ( 9 − 4 )  "20.5"
6 4
1 3 −
1
 1 3 −
1
 dM1
=   9 − 2916  9 − 74  9  −   4 − 2916  4 2 − 74  4  − (9 − 4 )  "20.5"
2

6  6 

 1 1 1 1
 −1516 + 1743 − 102  = 124 A1
2 3 2 3
 
(5)
Total 9
(a)
3 3
− − −1
M1: Attempts to differentiate with one index correct e.g. ...x 2 → ...x or x 2
→x 2
but not for
74 → 0 .

A1: Correct differentiation. May be left unsimplified but indices must be processed.
1
dy n
= 0 of the form x − ...x = 0 where m is a fraction via x = A or x = A
m n
dM1: Either solves their
dx
dy
It must be a solvable equation and they must correctly deal with the indices for their . Look out for
dx
dy
attempts where is manipulated before being set equal to zero which is acceptable.
dx
7 7
dy
e.g. = x 2 − 2187 = 0  x 2 = 2187  x = ...
dx
dy
Alternatively substitutes x = 9 into their and finds its value.
dx
A1*: Correct calculations and working leading to prove that the x coordinate of P is 9. It is
5
n −
x = A and proceed directly to x = 9 but e.g. x − 2187 x = 0  x = 9 would be
2
sufficient to achieve
dM0A0.
 5
−
It is also acceptable to proceed from x − 2187 x = 0 to an expression for x which is not the given
2

before achieving x = 9 e.g. x = ( 7 2187 )

2
answer
If using the verification method they should conclude that x = 9 (or have a preamble that if x = 9 is
dy
substituted in then = 0 , then substitute in, achieves 0 followed by e.g. tick, QED etc.)
dx
(b) Note that if no algebraic integration is seen then maximum score is M0A0B1dM0A0
3 3
− − +1
M1: Attempts to integrate with one index correct e.g. ...x 2 → ...x 3 or x 2 → x 2 (indices do not
need to be processed for this mark). Also accept 74 → 74x or "− 94.5" → "− 94.5"x

1
1 3 −
A1ft: Correct integration. x − 2916 x 2 − 74 x or follow through if curve − line is integrated
6
1
1 3 −
proceeding to x − 2916 x 2 − "94.5" x (allowing for an error on the coefficient of x when
6
attempting to subtract before integrating)
May be left unsimplified but indices must be processed (may be implied by further working – not just
the final answer). Do not be concerned with spurious notation e.g. a dx or integral sign still present.

41
B1: y value at P is 20.5 or e.g. . This may be seen as part of a wider calculation or on the diagram or
2
next to the questions (can also be scored for sight in (a))
This mark can be scored for sight of −94.5 or −94.5x when they are attempting to integrate curve −
line.

dM1: Full method to find the area of R. They do not need to proceed as far as finding a value for the
area, the values embedded is sufficient. It is dependent on the previous method mark.
The method for finding y at P must be correct (may be implied by e.g. 20.5 or attempting to
substitute x = 9 into the equation of the curve).
If no integration is seen then this mark cannot be scored.
If they proceed from the integrated expression to the answer without showing some substitution of
the limits (or evidence of this which is not the answer) then dM0A0.
9
1 −
1
 1
e.g.  x3 − 2916 x 2 − 74 x  − ( 9 − 4 )  20.5 = 124 scores dM0A0 (no evidence of limits substituted)
6 4 3
1 1 1 1
e.g. −1516 + 1743 − 102 = 124 scores dM1A1 (evidence of limits substituted)
2 3 2 3

1361 5
(Note = 226 is only the area under the curve so this is not a full method to find the required
6 6
area)

•
1 373
A1: Correct working and calculations seen leading to the answer of 124 o.e. including 124.3 or
3 3
124.3 124.33… but not rounded answers of 124.3
Can only be scored provided all of the previous marks have been scored.
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