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A-LEVEL EDEXCEL PAPER 2 PURE MATHEMATICS June 2024 MARK SCHEME (9MA0/02) £6.49
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A-LEVEL EDEXCEL PAPER 2 PURE MATHEMATICS June 2024 MARK SCHEME (9MA0/02)

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A-LEVEL EDEXCEL PAPER 2 PURE MATHEMATICS June 2024 MARK SCHEME (9MA0/02)

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  • January 17, 2025
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  • 2024/2025
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Mark Scheme (Results)


Summer 2024




Pearson Edexcel GCE Mathematics

Pure 2 Paper 9MA0/02

, General Principles for Pure Mathematics Marking

(But note that specific mark schemes may sometimes override these general
principles)



Method mark for solving 3 term quadratic:
1. Factorisation

( x 2 + bx + c) = ( x + p )( x + q ), where pq = c , leading to x = ...


(ax 2 + bx + c) = (mx + p )(nx + q ), where pq = c and mn = a , leading to x = ...

2. Formula
Attempt to use the correct formula (with values for a, b and c)



3. Completing the square
2
 b
Solving x + bx + c = 0 :  x    q  c = 0, q  0 , leading to
2
x = ...
 2
Method marks for differentiation and integration:
1. Differentiation

Power of at least one term decreased by 1. (x) →
n x −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 errors 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.

,Question Scheme Marks AOs

1(a)(i)  dy  M1 1.1b
y = 4 x 3 − 7 x 2 + 5 x − 10   = 12 x 2 − 14 x + 5
 dx  A1 1.1b
(ii)  d2 y 
 2 =  24 x − 14 A1ft 1.1b
 dx 
(3)
(b) 24 x − 14 = 0  x = ... M1 1.1b
7 oe e.g. 14
x= x= A1 1.1b
12 24
(2)
(5 marks)
Notes
(a)(i) If “+ c” is included with either derivative penalise it only once on the first occurrence.
M1: Award for x3 → x 2 or x 2 → x or 5 x → 5 or −10 → 0
Indices may be unprocessed e.g. x3 → x3−1 or x 2 → x 2−1 or 5 x → 5 x 0
A1: Correct simplified expression with indices processed 12 x 2 − 14 x + 5.
Do not allow x1 for x or 5x0 for 5.
Apply isw if necessary once a correct answer is seen.
dy
The “ = ” is not required.
dx
(ii)
A1ft: Correct simplified second derivative 24 x − 14 or follow through their first derivative.
Must be simplified so do not allow e.g. x1 for x or x0 for 1 as above.
Apply isw if necessary once a correct answer is seen.
d2 y
The “ = ” is not required.
dx 2
(b)
M1: Sets their second derivative of the form ax + b, a, b  0 equal to 0 and proceeds to a
value for x. Condone slips in rearranging as long as a value for x is obtained.
This may be implied by their value of x or may be implied by their working e.g.
 d2 y 
 2 =  24 x − 14 → 24 x = 14  x = ...
 dx 
Condone one slip in copying their second derivative.
 d2 y 
Also condone if they “cancel” e.g.  2 =  24 x − 14 → 12 x − 7 = 0  x = ...
 dx 
A1: Correct value from correct work and a correct second derivative but allow recovery if they
“cancel” their second derivative to obtain e.g. 12 x − 7.
14
Allow exact equivalents e.g. but not rounded decimals e.g. 0.583
24

Allow recurring decimal if clearly indicated e.g. 0.583
Correct answer only from a correct second derivative (or correctly cancelled second
derivative) scores both marks.
Isw after a correct answer is seen.

, Question Scheme Marks AOs

2(a) ( u12 = ) 400 + 11 −10 = 290 * or e.g. ( u12 = ) 400 − 110 = 290*
or e.g. B1* 1.1b
( u12 = ) 400 + (12 − 1)  −10 = 290 * or e.g. ( u12 = ) 410 + 12  −10 = 290*
(1)
Alternative 1:
400 + ( n − 1)  −10 = 290
B1* 1.1b
 400 − 10n + 10 = 290  10n = 120  n = 12*
Alternative 2:
290 = 400 + (12 − 1) d  11d = −110  d = −10* B1* 1.1b

N ( 2  400 + ( N − 1)  −10 )
(b) 1
8100 =
2
or e.g. M1 1.1b
N ( 400 + 400 + ( N − 1)  −10 )
1
8100 =
2
8100 = N ( 2  400 + ( N − 1)  −10 )
1
2
A1* 2.1
 16200 = 800 N − 10 N + 10 N or e.g.  8100 = 400 N − 5 N 2 + 5 N
2


 N 2 − 81N + 1620 = 0*
(2)
(c) N − 81N + 1620 = 0  ( N − 45 )( N − 36 ) = 0  N = 45,36
2
M1 1.1b
(N =) 36 A1 2.3
(2)
(5 marks)
Notes
(a)
B1*: Correct working to obtain 290. Must be a correct calculation so do not condone missing
brackets unless they are recovered. E.g. ( u12 = ) 400 + 12 − 1 −10 = 290 scores B0 unless
followed by = 400 + 11 −10 = 290 . Condone ( u12 = ) 400 + (12 − 1) − 10 = 290
The “£” symbol is not required but the “290” must appear.
Alternative 1:
B1*: Correct working using the 290 to obtain n = 12.
There must be at least one intermediate line after setting up the equation and must be
correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as 12 is correctly obtained.
Alternative 2:
B1*: Correct working using the 290 and 400 to obtain d = −10.
There must be at least one intermediate line after setting up the equation and must be
correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as −10 is correctly obtained.

Allow candidates to list terms and show the 12th term is 290 e.g.
400, 390, 380, 370, 360, 350, 340, 330, 320, 310, 300, 290
Must list all 12 terms which must be correct and end with 290
Condone if missing 400, 390, 380 as these are given in the question.
(b) Mark (b) and (c) together

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