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Mark Scheme (Results)

Summer 2022

Pearson Edexcel GCE

In Mathematics (9MA0)

Paper 02 Pure Mathematics 2

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Summer 2022

Question Paper Log Number P69602A_*

Publications Code 9MA0_02_2206_MS*

All the material in this publication is copyright

© Pearson Education Ltd 2022

EDEXCEL GCE MATHEMATICS

General Instructions for Marking

1. The total number of marks for the paper is 100.
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.

• bod – benefit of doubt
• ft – follow through
• 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)
• dep – dependent
• indep – independent
• dp decimal places
• sf significant figures
•  The answer is printed on the paper
• The second mark is dependent on gaining the first mark

4. 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.

5. Where a candidate has made multiple responses and indicates which response they wish

to submit, examiners should mark this response.

If there are several attempts at a question which have not been crossed out, examiners

should mark the final answer which is the answer that is the most complete.

6. Ignore wrong working or incorrect statements following a correct answer.
7. Mark schemes will firstly show the solution judged to be the most common

response expected from candidates. Where appropriate, alternatives answers are

provided in the notes. If examiners are not sure if an answer is acceptable, they will

check the mark scheme to see if an alternative answer is given for the method used.

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

2

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

2

( ax + 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

Solving

2 x + bx + c = 0 :

2

b x q c q

 ±^  ±^ ±^ =^ ≠

, leading to x =...

Method marks for differentiation and integration:

1. Differentiation

Power of at least one term decreased by 1.

1 ( )

n n x x

− →

2. Integration

Power of at least one term increased by 1.

1 ( )

n n x x

→

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.

Alternative solution via squaring

M1 : Attempts to square both sides. Condone poor squaring e.g. (^) ( )

2 2 2 3 − 2 x = 9 ± 4 x or 9 ± 2 x

A1 : Correct quadratic equation

2

3 x − 26 x − 40 = 0.The “= 0” may be implied by their attempt to

solve. Terms must be collected but not necessarily all on one side so allow e.g.

2

3 x − 26 x = 40

dM1 : Correct attempt to solve a 3 term quadratic. See general guidance for solving a quadratic

equation. The roots can be written down from a calculator so the method may be implied by

their values. Depends on the first method mark.

A1 : For both

x = − and^ x = 10 with no extra solutions and neither clearly rejected but ignore any

attempts to find the y coordinates and do not count e.g. x = ˗7 (from where y = 7 + x intersects the

x -axis) or x = 1.5(from finding the value of x at the vertex) as “extras”. Allow exact equivalents

for

− e.g.

− or 1.

− but not e.g. ˗1.

Isw if necessary e.g. ignore subsequent attempts to put the values in an inequality e.g.

− < x <

But if e.g.

x = − is obtained and a candidate states

x = − then score A

Question Scheme Marks AOs

2 (a)

Correct shape or correct

intercept – see notes B1 1.

Fully correct – see notes B1 1.1b

(b) 4

4 100 log 100

x = ⇒ x =

or

e.g.

log log 4 log log 4

x = ⇒ x =

M1 1.1b

⇒ (^) ( x =) awrt 3.32 A1 1.1b

( 4 marks)

Notes:

Note that B0B1 is not possible in part (a)

(a) Axes do not need to be labelled. No sketch is no marks.

B1 : Correct shape or correct intercept_._

Shape : A positive exponential curve in quadrants 1 and 2 only, passing through a point on the

positive y -axis. Must “level out” in quadrant 2 but not necessarily asymptotic to the

x -axis and allow if the curve bends up slightly for x < 0 but do not allow a clear “U”

shape. It must not clearly “stop” on the x -axis to the left of the y -axis.

OR

Intercept : The intercept can be marked as 1 or (0, 1) or y = 1 or (1, 0) as long as it is in the

correct place. May also be seen away from the sketch but must be seen as (0, 1) or

possibly these coordinates in a table but it must correspond to the sketch. If there is

any ambiguity, the sketch takes precedence.

B1 : Fully correct.

Shape : A positive exponential curve in quadrants 1 and 2 only, passing through a point on the

positive y -axis. The curve must appear to be asymptotic to the x -axis and it must level

out at least half way below the intercept. Allow if the curve bends up slightly for x < 0

but do not allow a clear “U” shape. The curve must not bend back on itself on the rhs of

the y -axis. There must be no suggestion that the curve approaches another horizontal

asymptote other than the x -axis e.g. a horizontal dotted line that the curve approaches.

AND

Intercept : As above

See practice items and below for some examples:

O

Question Scheme Marks AOs

3 (a)(i)

(ii)

1

a = 3 , 2

a = 5 , 3

a = 3 ... (^) B1 1.1b

2 B1 1.1b

(b)

( )

85

1

n n

a

=

= × + +

∑

o.e. (^) M1 3.1a

= 339 A1 1.1b

( 4 marks)

Notes:

(a)(i) Mark (a)(i) and (a)(ii) together.

B1 : States the values of at least 2

a = 5 and 3

a = 3.This is sufficient but if more terms are given they

must be correct. There is no need to see e.g. 2 3

a = ..., a = ...just look for values.

Allow an algebraic approach e.g. (^) ( ) 1 2

n n n n n

a a a a a

A conclusion is not needed.

(a)(ii)

B1 : States that the order of the periodic sequence is 2

Allow “second order”, “it repeats every 2 numbers” or equivalent statements that convey the

idea of the period being 2.

Note that ±2 is B

(b)

M1 : Attempts a correct method to find

85

1

n n

a

=

∑

For example ( )

85

1

n n

a

=

= × + +

∑

85

1

n n

a

=

= × + × +

∑

or ( )

85

1

n n

a

=

= × + −

∑

or

85

1

n n

a

=

= × + ×

∑

or

85

1

n 2 n

a

=

= × −

∑

There may be other methods e.g. “Chunking”: 5×(3 + 5) = 40, 40 × 8 = 320, 320 + 3×3 +2×5 = 339

A1 : 339. Correct answer only scores both marks.

Attempts to use an AP formula score M

Question Scheme Marks AOs

( )

(^2 ) 2 2 ...

x h x

h

= M1^ 2.

( )

(^2 2 ) 2 x h 2 x (^) 4 xh 2 h

h h

= A1^ 1.1b

( )

2

0 0

d 4 2 lim lim 4 2 4 * d h^ h

y xh h x h x x →^ h →

= = + = A1* 2.

( 3 marks)

Notes:

Throughout the question allow the use of δ x for h or any other letter e.g. α if used consistently.

If δ x is used then you can condone e.g.

2 δ x for

2 δ x as well as condoning e.g. poorly formed δ's

M1 : Begins the process by writing down the gradient of the chord and attempts to expand

the correct bracket – you can condone “poor” squaring e.g. (^) ( )

(^2 2 ) x + h = x + h.

Note that

( )

(^2 ) 2 2 ...

x h x

h

is also a possible approach.

A1 : Reaches a correct fraction oe with the x

2 terms cancelled out.

E.g.

2 2 4 2 2 ,

xh h x

h

2 2

• 4 xh + 2 h − 2 x , 4 x 2 h h

A1 *: Completes the process by applying a limiting argument and deduces that

d 4 d

y x x

= with no

errors seen. The

d " " d

y

x

= doesn’t have to appear but there must be something equivalent e.g.

"f ′ (^) ( x )= "or “Gradient =” which can appear anywhere in their working. If f ′ (^) ( x )is used then

there is no requirement to see f ( x ) defined first. Condone e.g.

d 4 d

y x x

→ or f ′^ ( x (^) )→4. x

Condone missing brackets so allow e.g.

2

0 0

d 4 2 lim lim 4 2 4 d h^ h

y xh h x h x x →^ h →

= = + =

Do not allow h = 0 if there is never a reference to h 0

e.g. (^) ( )

2

0 0

d 4 2 lim lim 4 2 0 4 d h^ h

y xh h x x x →^ h →

= = + = is acceptable

but e.g. (^) ( )

2 d 4 2 4 2 4 2 0 4 d

y xh h x h x x x h

= = + = + = is not if there is no h0 seen.

The h0 does not need to be present throughout the proof e.g. on every line.

They must reach 4 x + 2 h at the end and not

2 4 xh 2 h

h

(without the h ’s cancelled) to complete

the limiting argument.

(b)(ii)

M1 : Shows correct log work to relate the given question to part (a)

Must reach as far as e.g. (^) [ ]

9 9

3 3 3

2 x + log 2 x d x =...

with correct use of limits on (^) [ ]

9

3

2 x which

may be implied or equivalent work e.g. finds the area of the rectangle as 2 × 6

A1ft : Correct working followed by awrt 25.3 but ft on their 13.3 so allow for 12 + their answer to

part (a) following correct work as shown.

Note that

9

3 3

log 18 x d x =25.32414...

so a correct method must be seen to award marks.

Some examples of an acceptable method are:

( )

9 9 9

3 3 3 3 3 3

log 18 x d x = log 9 × 2 x d x = 2 + log 2 x d x = 6 × 2 + "13.3" =25.

( )

9 9 9

3 3 3 3 3 3

log 18 x d x = log 9 × 2 x d x = 2 + log 2 x d x = 12 + "13.3" =25.

( ) [ ]

9 9 9 9 9

3 3 3 3 3 3 3 3 3

log 18 x d x = log 9 × 2 x d x = 2 + log 2 x d x = 2 x + log 2 x d x =25.

BUT just 12 + "13.3" = 25.3 scores M

Attempts to apply the trapezium rule again in any way score M0 as the instruction in the

question was to use the answer to part (a).

Question Scheme Marks AOs

6(a)

( )

f ( ) 4 cos 3 2

x x

M

A

1.1b

1.1b

Sets

f ( ) 4 cos 3 0 2

x x x

dM1 3.1a

x = 14.0 Cao A1 3.2a

(b) Explains that f (4) > 0 , f (5) < 0

and the function is continuous

B1 2.

(c) Attempts 1

8sin 2.5 15 9 5 "4 cos 2.5 3"

x

(NB f (5) = −1.212… and f (5)′^ = −6.204…)

M1 1.1b

1

x = awrt 4.80 A1 1.1b

( 7 marks)

Notes:

(a)

M1 : Differentiates to obtain

cos 2

k x

± α    

where α is a constant which may be zero and

no other terms. The brackets are not required.

A1 : Correct derivative

f ( ) 4 cos 3. 2

x x

Allow unsimplified e.g.

f ( ) 8cos 3 2 2

x x x

′ = × −

There is no need for f ( )′^ x = ...or

d ... d

y

x

= just look for the expression and the brackets are not

required.

dM1 : For the complete strategy of proceeding to a value for x.

Look for

f ( ) cos 0, , 0 2

x a x b a b

• Correct method of finding a valid solution to

cos 0 2

a x b

Allow for ( )

cos 0 cos 2 cos 2 2

a x b x k x k

   ^ −

  +^ =^ ⇒^  = ±^ ⇒^ =^ ±

where k < 1

If this working is not shown then you may need to check their value(s).

For example

4 cos 3 0 1.4... 2

x x

 −^ =^ ⇒^ =

or 11.1… (or 82.8… or 637.… or 803 in

degrees) would indicate this method.

A1 : Selects the correct turning point x = 14.0and not just 14 or unrounded e.g. 14.011…

Must be this value only and no other values unless they are clearly rejected or 14.0 clearly

selected. Ignore any attempts to find the y coordinate.

Correct answer with no working scores no marks.

(b)

B1 : See scheme. Must be a full reason, (e.g. change of sign and continuous)

Accept equivalent statements for f (4) > 0 , f (5) < 0 e.g. f (4) × f (^) ( ) 5 < 0,”there is a change of

sign”, “one negative one positive”. A minimum is “change of sign and continuous” but do not

allow this mark if the comment about continuity is clearly incorrect e.g. “because x is

continuous” or “because the interval is continuous”

Question Scheme Marks AOs

7 (a) ( )

1

4 − 9 x = 2 1 ± ...^2 B1^ 1.1b

2

1 2

x

x

×  −  − 

 −^  =^ +

or

3 1 1 3 9 " " 2 2 2 4 ... 3!

   ^ x  × − × − −          

M1 1.1b

2 3 1 1 9 1 1 3 9

x x

x

×  −  −  ×  −  ×  −  − 

+ × − + +

A1 1.1b

2 3 9 81 729 4 9 2 4 64 512

x x x − x = − − − A1^ 1.1b

(b) States that the approximation will be an overestimate since all

terms (after the first one) in the expansion are negative

(since x  0 )

B1 3.2b

(5 marks)

Notes:

(a)

B1 : Takes out a factor of 4 and writes (^) ( )

1

4 − 9 x = 2 1 ± ...^2 or ( )

1

4 1 ± ...^2 or ( )

1 1 2 4 1 ±...^2

M1 : For an attempt at the binomial expansion of (^) ( )

1

2 1 + ax a ≠ 1 to form term 3 or term 4 with the

correct structure. Look for the correct binomial coefficient multiplied by the corresponding

power of x e.g.

( )( ) ( )

1 1 2 2 1 2 ...

2!

x

or

( )( )( ) ( )

1 1 1 2 2 1 2 2 3 ... where... 1

3!

x

Condone missing or incorrect brackets around the x terms but the binomial coefficients must be

correct. Allow 2! and/or 3! or 2 and/or 6. Ignore attempts to find more terms.

Do not allow notation such as

unless these are interpreted correctly.

A1 : Correct expression for the expansion of

1

9 2 1 4

^ x  −    

e.g.

2 3 1 1 9 1 1 1 9 1 1 2 1 9 2 2 4 2 2 2 4 1 2 4 2! 3!

x x

x

×  −  ±  ×  −  ×  −  − 

+ ×  − + +

which may be left unsimplified as shown but the bracketing must be correct unless any missing

brackets are implied by subsequent work. If the 2 outside this expansion is only partially applied

to this expansion then score A0 but if it is applied to all terms this A1 can be implied.

OR at least 2 correct simplified terms for the final expansion from,

2 3 9 81 729 , , 4 64 512

x x x − − −

A1 :

2 3 9 81 729 4 9 2 4 64 512

x x x − x = − − − oe and condone e.g.

2 3 9 81 729 2 4 64 512

− x x x

• − −

Allow equivalent mixed numbers and/or decimals for the coefficients e.g.:

Ignore any extra terms if found. Allow terms to be “listed” and apply isw once a correct expansion is

seen. Allow recovery if applicable e.g. if an “ x ” is lost then “reappears”.

Direct expansion in (a) can be marked in a similar way:

( ) ( )

( ) ( )

1 1 3 2 5 3 1 2 2 1 2 2 2

x x x x x

− = − = +   × − +   −  × +   −  −  ×

B1 : For 2 or 4 or

1 2 4 as the constant term in the expansion.

M1 : Correct form for term 3 or term 4.

E.g.

( )

2 1 1 ...

  ^ x   −^ ×   

or

( )

3 1 1 3 ... where... 1 2 2 2 3!

   ^ x   −^  −^ ×^ ≠    

Condone missing brackets around the x terms but the binomial coefficients must be correct.

Allow 2! and/or 3! or 2 and/or 6. Ignore attempts to find more terms.

Do not allow notation such as

unless these are interpreted correctly.

A1 : Correct expansion (unsimplified as above)

OR at least 2 correct simplified terms from,

2 3 9 81 729 , , 4 64 512

x x x − − −

A1 :

2 3 9 81 729 4 9 2 4 64 512

x x x − x = − − − oe and condone e.g.

2 3 9 81 729 2 4 64 512

− x x x

• − −

Allow equivalent mixed numbers and/or decimals for the coefficients e.g.:

Ignore any extra terms if found. Allow terms to be “listed” and apply isw once a correct expansion is

seen. Allow recovery if applicable e.g. if an “ x ” is lost then “reappears”.

(b)

B1 : States that the approximation will be an overestimate due to the fact that all terms (after the first

one) in the expansion are negative or equivalent statements e.g.

• Overestimate because the terms are negative
• Overestimate as the terms are being taken away (from 2)

Condone “overestimate as every term is negative”

If you think a response is worthy of credit but are unsure then use Review.

This mark depends on having obtained an expansion in (a) of the form

2 3 k − px − qx − rx k p q r , , , > 0 but note that if e.g. one of the algebraic terms is zero or was “lost” or

there are extra negative terms this mark is still available.

A1 : Correct working shown leading to

− but also allow^

− or exact equivalents

Award this mark once one of these forms is reached and isw

See overleaf for integration by parts and integration by substitution.

Integration by parts:

( )( ) ( )( ) ( )( ) ( )

1 1 1 (^2 4 1 2 1 2 ) d 2 4 d 2 4 2 6 d 4 4 2 2

x x x x x x x x x x x x x x

− − − = − − = − − − −

M

A

1.1b

1.1b

( )( ) ( ) ( )( )

1 1 1 3 1 (^1 2 1 2 12 2 ) 2 4 2 6 d 2 4 3 d 2 2 2

x − x − x − x − x x = x − x − x − x − x x

( )( )

1 5 3 (^1 2 22 ) 2 4 2 2 5

= x − x − x − x + x

Or e.g. (^) ( )( ) ( )

1 3 5 (^1 2 1 2 ) 2 4 2 6 2 3 15

= x − x − x − x x − + x

dM

A

3.1a

1.1b

Deduces limits of integral are 2 and 4 and applies to their

( )( ) ( )

1 3 5 (^1 2 1 2 ) 2 4 2 6 2 3 15

x − x − x − x x − + x

M

2.2a

Area R

or 2 5 5

A1 2.

Notes:

M1 : Applies integration by parts and reaches the form (^) ( )( ) ( )

1 1 2 2

α x − 2 x − 4 x ± px + q x d x α, p ≠ 0

oe e.g. ( ) ( )

1 1 (^2 2 ) α x − 6 x + 8 x ± px + q x d x α , p ≠ 0

A1 : Correct first application of parts in any form

dM1 : Attempts their (^) ( )

1 2 px + q x d x

by expanding and integrating or may attempt parts again.

E.g. (^) ( )

1 3 1 2 2 2 2 x 6 x d x 2 x 6 x d x ...

or e.g. (^) ( ) ( )

1 3 3 2 2 2 4 2 2 6 d 2 6 d 3 3

x − x x = x x − − x x

If they expand then at least one term requires

n n 1 x x

→ or if parts is attempted again, the structure

must be correct.

A1 : Fully correct integration in any form

M1 : Substitutes the limits 4 and 2 to their (^) ( )( )

1 5 3 (^1 2 22 ) 2 4 2 2 5

= x − x − x − x + x and subtracts

either way round. There is no requirement to evaluate but 4 and 2 must be substituted either way

round with evidence of subtraction, condoning omission of brackets.

E.g. condone

This is an independent mark but the limits must be applied to a “changed” function.

A1 : Correct working shown leading to

− but also allow

− or exact equivalents

Attempts at integration by parts “the other way round” should be sent to review.