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A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022

Mark Scheme

Pearson Edexcel GCE A Level Mathematics
Pure Mathematics 1 (9MA0/01)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022

Edexcel and BTEC Qualifications
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pearson
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Examiners must

mark the last candidate in exactly the same way as they mark
the first
...
Candidates must be

rewarded for what they have shown they can do rather than
penalised for omissions
...


• All the marks on the mark scheme are designed to be awarded
...
e
...
Examiners should also
be prepared to award zero marks if the candidate’s response
is not worthy of credit according to the mark scheme
...


and

• When examiners

are in doubt regarding the application of the
mark scheme to a candidate’s response, a senior examiner
must be consulted before a mark is awarded
...


A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
PEARSON EDEXCEL GCE MATHEMATICS
General Instructions for Marking
1
...
These 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
...

B marks are unconditional accuracy marks (independent of M marks)

•

Marks should not be subdivided
...
Abbreviations
These are some of the traditional marking abbreviations that will appear
in the mark schemes
...
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

•

o
...
– 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

will be used for correct ft

4
...

A marks are ‘correct answer only’ (cao
...
After a misread
however, the subsequent A marks affected are treated as A ft, but answers
that don’t logically make sense e
...
if an answer given for a probability is >1 or
<0, should never be awarded A marks
...
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
...
Where a candidate has made multiple responses and indicates which response
they wish to submit, examiners should mark this response
...

7
...

8
...
Where appropriate,
alternatives answers are provided in the notes
...
If no such alternative
answer is provided but the response is deemed to be valid, examiners
must escalate the response for a senior examiner to review
...
5 × 1 + 2(e0
...
2 + e 0
...
8 

Marks

AOs

B1

1
...
1b

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
 1

= ×10
...
=2
...
=2
...
73" =10
...
1b

(3)
B1ft

2
...
2a

∣

1 ( x- 1) 2

1

A1

dx  ="2
...
5 or

0
...
25 or

2

1
4

[
...
g
...
73

B1ft:

10
...
6900… which is found directly from integration

(b)(ii)
B1ft:

2
...
6900… or 2
...
52 + 8
...
5)(8
...
2) { ⟶ BC =13
...
}
Arc length AB =7
...
2)

{⟶

Arc length AB =9}

Perimeter AOCBA =7
...
5 + 13
...
+ 9

{ =38
...
}

= 38
...
1b

A1

1
...
1a

M1

3
...
1b

(5)
(5 marks)
Question 2 Notes:

M1:

Application of cosine rule for BC 2 or BC with any angle

A1:

Correct application of cosine rule for BC 2 or BC using 𝑢 - 1
...
5(1
...
2 cao

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question

Scheme

3

3x2 + k =5x + 2
E
...
3x2 - 5x + k - 2 =0 or - 3x2 + 5x + 2 - k =0

{ "b

2

- 4ac" ∈ 0 ⟶

}

25 - 4(3)(k - 2) ∈ 0

Marks

AOs

M1

1
...
1b

B1

1
...
1

25 - 12k + 24 ∈ 0 ⟶ - 12k + 49 ∈ 0
Critical value obtained of

49

o
...


12
k>

49

o
...


12

(4)
(4 marks)
Question 3 Notes:

M1:
M1:

Forms a one-sided quadratic equation or gathers all terms into a single quadratic expression
Understands that the given equation has no real roots by applying "b2 - 4ac" ∈ 0 to their one-sided
quadratic equation or to their one-sided quadratic expression {=0}

B1:

See scheme

A1:

Complete process leading to the correct answer, e
...

49
• k>
12
49
•
∈k
12
49
• k : k > 


12 

with no errors seen in their mathematical argument

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question

Scheme

f (x) =

4

12x
3x + 4

(a)

Marks

AOs

M1

1
...
1b

x ∈, x …0

0 „ f (x) ∈ 4

(2)
(b)

y =

12x
⟶ y(3x + 4) =12x ⟶ 3xy +
Correct process to obtain the gradient of the chord PQ as 𝛼h + 𝗉 ; 𝛼 , 𝗉 0

M1:
A1:

Correctly shows that the gradient of PQ is 4h +11 and applies a limiting argument to deduce that at
lim
dy
the point P on y =4x 2 - 5x,
=11 E
...

 4h +11 =11
h⟶ 0
dx
Note: 𝛿 x can be used in place of h

Alt 1
B1:

4(x + h)2 - 5(x + h), seen or implied

M1:

Begins the proof by attempting to write the gradient of the chord in terms of x and h

A1:

Correct expression for the gradient of the chord in terms of x and h
Correct process to obtain the gradient of the chord as 𝛼 x + 𝗉 h + ; 𝛼 , 𝗉 , 0

M1:
A1:

Correctly shows that the gradient of the chord is 8x + 4h - 5 and applies a limiting argument to
lim
dy
=8x - 5
...
g
...

dx 2
dx 2
du u
u
1 (2)
0
Criteria 1 x =0 ⟶ a =e and x =2 ⟶ b =e 2

Marks

AOs

B1

1
...
1b

B1

2
...
du =
du
1
x
(u + 4 ) u
u (u + 4)
(e2 + 4)

∫

∫

∫

Either Criteria 1 or Criteria 2
Both Criteria 1 and Criteria 2

e

12

and correctly achieves the result

∫u(u + 4)

du

1

(3)
(b)

12



A

+

B

⟶ 12 A(u + 4) + Bu

M1

1
...
1b



M1

3
...
1b

A1*

2
...
e
...
e
...
e
...

e+4∣

𝖩

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question

7
(a)

Scheme

Marks

AOs

M1

1
...
13

B1

1
...
13)

A1

1
...
e
...
4

(1)
(b)(ii)

G =17 + 3sin(15t) - 4cos(15t) ; 0 „ t „ 17
20 =17 + "5"sin(15t - "53
...
13") =

3

or sin(𝜃 - "53
...
4

M1

1
...
1b

A1

3
...
13" =180 - 36
...

143
...
+ "53
...
0840

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