EDEXCEL A-Level Pure Mathematics Paper 2 2025 - Mark Scheme

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Mark Scheme (Results)


Summer 2025




Pearson Edexcel GCE
In Mathematics (9MA0)

Paper 02 Pure Mathematics

,ALEVELPAPERZ2025

Question Scheme Marks AOs

1 ( ) (
2 x x 2 − 12 x + 20 or x 2 x 2 − 24 x + 40 ) B1 1.1b
x 2 − 12 x + 20 = ( x − 2 )( x − 10 )
M1 1.1b
or e.g. 2 x 2 − 24 x + 40 = ( 2 x − 4 )( x − 10 )
2 x ( x − 2 )( x − 10 ) A1 1.1b
(3)
(3 marks)
Notes
B1: Takes a factor of 2x or x out of the given cubic correctly.
( )
Dividing by 2 and then achieving x x 2 − 12 x + 20 does not score this mark unless recovered.
M1: Attempts to factorise their quadratic. Invisible brackets may be implied by later work.
Score for ( x ± b )( x ± d ) where bd = 20 coming from x 2 ± 12 x ± 20
or for ( ax ± b )( cx ± d ) where ac = 2 and bd = 40 coming from 2 x 2 ± 24 x ± 40
May be scored if they have divided by x but not from e.g. 2 x 3 − 24 x 2 + 40 x → ( 2 x − 4 )( x − 10 )
without clear indication that they have divided by or taken out a factor of x or 2x.
There may be incorrect intermediate steps such as ( 2 x − 20 )( 2 x − 4 ) which do not score the
mark on their own but may be ignored if they return to e.g. ( 2 x − 20 )( x − 2 ) .
A1: 2 x ( x − 2 )( x − 10 ) or e.g. 2 ( x − 10 )( x − 2 ) x . Do not accept e.g. x ( 2 x − 4 )( x − 10 ) . Ignore = 0
ISW after a fully correct factorisation e.g. 2 x ( x − 2 )( x − 10 ) that becomes x ( x − 2 )( x − 10 )
and ignore any attempt to find roots (before or after factorisation). Allow ( 2 x )( x − 2 )( x − 10 )
Alternative:
B1: Takes a factor of ( x − 2 ) or ( x − 10 ) out correctly i.e. ( x − 2 ) 2 x 2 − 20 x or( )
( x − 10 ) ( 2 x 2 − 4 x ) . Must be seen as a product of factors and not just in a division attempt but
may be implied by later work.
M1: Attempts to factorise their quadratic. Invisible brackets may be implied by later work.
Score for ax ( bx ± c ) where ab = "2" and ac = their "20" or "4" coming from Ax 2 + Bx
A1: As main scheme.
Note: Solutions that solve the cubic = 0 to achieve x = 0, 2 and 10 and then arrive at e.g.
x ( x − 2 )( x − 10 ) score no marks without a prior line of working such as x x 2 − 12 x + 20 ( )
( )
(would score M1) or x 2 x 2 − 24 x + 40 (would score B1).
Some examples:
• 2 x 3 − 24 x 2 + 40 x → x ( x − 2 )( x − 10 ) scores B0M0A0
• 2 x 3 − 24 x 2 + 40 x → ( 2 x − 4 )( x − 10 ) scores B0M0A0
• x 2 − 12 x + 20 → ( x + 2 )( x − 10 ) scores B0M1A0
• x 3 − 12 x 2 + 20 x → x ( x − 2 )( x − 10 ) scores B0M1A0
• ( )
x 2 x 2 − 24 x + 40 → x ( x − 2 )( x − 10 ) scores B1M0A0
• x ( 2x 2
− 24 x + 40 ) → { x ( 2 x + 2 )( 2 x + 40 ) →} x ( 2 x + 2 )( x + 20 ) scores B1M1A0
• 2x ( x 2
− 12 x + 20 ) → 2 x ( x − 1)( x + 20 ) scores B1M1A0
• 2 x ( x − 2 )( x − 10 ) on its own scores B1M1A1

,Question Scheme Marks AOs

2 1 3
M1 1.1a
x 4 → … x5 or x 2 → … x 2 or −3 → … x
1 5 6 32
Any of x or − x or −3x
5 3 A1 1.1b
 
2
1 5 6 32
Any two of x or − x or −3x
5 3 A1 1.1b
 
2
3
1 5
x − 4 x 2 − 3x + c A1 1.1b
5
(4)
(4 marks)
Notes
M1: For increasing any power by one. Score for x n → x n +1 in any term, including, −3 → … x
where … is a constant, but not for + c. Allow the indices to be unprocessed, e.g., x 4+1
A1: One correct term which may be unsimplified and indices may be unprocessed.
1
6 +1
Condone e.g. −3x1 or − x 2 for this mark. Not scored for + c
1 
 + 1
2 
A1: Two correct terms which may be unsimplified but indices must be processed.
Condone −3x1 for this mark. Not scored for + c
A1: cao Requires all terms simplified and + c
Ignore the LHS i.e. ignore what they call their integral.
3
1 3
Allow 0.2x5 for x5 and −4 x3 or −4 x or −4x x or −4x1.5 for −4x 2
5
Do not allow −3x1 for this mark.
3
1 5
Condone spurious integral signs e.g. ⌠  x − 4 x 2
− 3 x + c or dx left in their answer
⌡5
ISW after a correct expression seen e.g. if they multiply through by 5 or e.g. try to solve = 0
3
1 5
Do not allow e.g. x + −4 x 2 + −3 x + c
5

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Question Scheme Marks AOs

3 3x =→
7y y log 7
x log 3 = M1 3.1a
 x  log 7
e.g.  =  A1 1.1b
 y  log 3
(2)
(2 marks)
Notes
x
Note: Condone absence of reference to the value of being undefined when x= y= 0
y

M1: For the key step in attempting to take logarithms with the same base of both sides and
apply the power law to both sides, e.g., x = y log 3 7 or y = x log 7 3 or x ln 3 = y ln 7
Condone e.g. x = log 3 7 y but not x = log 3 ( 7 y )
Alternatively, takes a root of both sides (either x or y), takes logs with the same base of both
x
x y x
sides and applies the power law e.g., 3 =7 → 3 =7 → y
log 3 =log 7
y
May be implied by a correct answer.
 x  log 7 ln 7 1 ln 7
A1:  = or equivalent, e.g, or log 3 7 or Condone e.g.
 y  log 3 ln 3 log 7 3 ln 3

Correct answer only scores both marks provided there is no incorrect log work.
log k 7
Any base k may be used for e.g. provided it is the same base in both logs, although
log k 3
1
 log 3 7  2 log 3 k
watch out for e.g.   or for constant k > 0 which are exceptions and correct.
 log 7 3  log 7 k
x
There is no need to see = but it should be clear what their answer is.
y
log 7 7
Do not ISW if they incorrectly apply log laws e.g. = log =or log 7 − log 3
log 3 3
or log ( 7 − 3) all of which score M1A0.
=
You may ISW after a correct answer if they go on to provide a decimal approximation.
A decimal approximation with no correct log work seen (1.7712…) scores no marks.