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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
Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding
body
...
For further information visit our
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com or www
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co
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Alternatively, you can
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We’ve been involved in education for
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through innovation in education
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pearson
...
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 + 4 y =12x ⟶ 4 y =12x - 3xy
3x + 4
4y
4 y =x(12 - 3y) ⟶
=x
12 - 3y
4x
0„ x ∈4
Hence f - 1(x) =
12 - 3x

M1

1
...
1

A1

2
...
1b

144x
3x + 4
=
36x +12x +16
3x + 4

M1

1
...
1

=

144x
48x + 16

=

9x
3x +1

*

{ x ∈ ,

x …0}

(3)
(d)

7  9x
7

= ⟶ 18x =21x + 7 ⟶ - 3x =7 ⟶ x =
...
1b

A1

2
...
1b

A1

2
...
1b

A1

2
...
g
...
g
...

Correct range using correct notation
...

A fully correct method to find the inverse
...
e
...

9x
Shows ff (x) =
with no errors seen
...


(d)
9x

M1:

Sets

A1:

Finds x =-

3x +1

to

7

and solves to find x =
...

2

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 4 Notes Continued:

(d)
Alt 1
M1:
A1:

7 
∣ 2∣
 𝖩
7
1  7  28
Deduces f (x) =f ∣ ∣ =
and concludes ff (x) = has no solutions because
2
3
2
 𝖩
28
f (x) = lies outside the range 0 „ f (x) ∈ 4
3
-1
Attempts to find f

(d)
Alt 2
M1:
A1:

Evidence that the upper bound of ff (x) is 3
7
7
0 „ ff (x) ∈ 3 and concludes that ff (x) = has no solutions because > 3
2
2

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

5

Scheme

Marks

AOs

B1

1
...
1

2+h- 2

A1

1
...
1b

 4h +11 =11

A1

2
...
So yQ =4(2 + h) - 5(2 + h)

{P(2, 6), Q(2 + h, 4(2 + h)2 - 5(2 + h))}
Gradient PQ =
=
=

4(4 + 4h + h2 ) - 5(2 + h) - 6
2+h- 2
16 +16h + 4h2 - 10 - 5h - 6
2+h- 2
4h +11h
2

=

dy
dx

=

lim
h⟶ 0

(5)
5
Alt 1

Gradient of chord =

4(x + h)2 - 5(x + h) - (4x2 - 5x)
x+h- x

1
...
1

A1

1
...
1b

A1

2
...
g
...
g
...
E
...

deduce that when y =4x 2 - 5x,
 8x + 4h - 5  =8x - 5
h⟶ 0
dx
dy
=11
Finally, deduces that at the point P,
dx
𝛿
x
Note: For Alt 1,
can be used in place of h

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

6 (a)

{ u =e

1x
2

or x =2 ln u ⟶

}

Scheme

1 1x
du 1
dx 2
2
= e 2 or = u or
= or dx = du or 2du =udx , etc
...
1b

B1

1
...
1

du

{

}

a =1, b =e or evidence of 0 ⟶ 1 and 2 ⟶ e
Criteria 2 (dependent on the first B1 mark)
6
6
2
12
dx =

...
1b

u =0 ⟶ A =3; u =- 4 ⟶ B =- 3

A1

1
...
1a

A1ft

1
...
1

u(u + 4)

u

(u + 4)

3
3 
du = ∣ 
∣ du =3ln u - 3ln(u + 4)
 u(u + 4)
 u (u + 4) 𝖩

∫

12



∫

{ So, [ 3ln u - 3ln(u + 4)] }
e

1

=(3ln e - 3ln(e + 4)) - (3ln1 - 3ln 5)
=3ln e - 3ln(e + 4) + 3ln 5
 5e 
=3ln ∣
*
e+4∣

𝖩

(5)
(8 marks)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 6 Notes:

(a)
B1:
B1:

See scheme

B1:

See scheme

See scheme
Note for Criteria 2: Must start from one of
•
•
•

∫y dx,6 with integral sign and dx
∫e + 4 dx, with integral sign and dx
6
dx
dx
du, with integral sign and
du
∫e + 412du
du

and end at
(b)
M1:

1x
2

1x
2

∫u (u + 4 )

du,

du ,

with integral sign and

with no incorrect working

12

A
B
1
P
Q

+
,

+
, o
...
and a complete method for
Writing
o
...
or
u(u + 4) u (u + 4)
u(u + 4) u (u + 4)
finding the values of both their A and their B (or their P and their Q)
" A"
"B"
+
Note: This mark can be implied by writing down
with values stated for their A
u
(u + 4)

A1:

and their B where either their A =3 or their B =- 3
1
Both their A =3 and their B =- 3 (or their P = 4 and their Q =of the integral sign)

M1:

12

∫

1

with a factor of 12 in front

4

du,

Complete strategy for finding
which consists of
u(u + 4)
12
• expressing
in partial fractions
u(u + 4)
12
M
N
 
; M , N , k 0; (i
...
a two-term partial fraction) to
• and integrating
u(u + 4) u (u k )
obtain both ln(𝛼 u) and ln(𝗉 (u k )); , , 𝛼 , 𝗉 0
A1ft:
A1*:

Integration of both terms is correctly followed through from their M and their N
Applies limits of e and 1 in u (or applies limits of 2 and 0 in x), subtracts the correct way round and
 5e 
uses laws of logarithms to correctly obtain 3ln ∣
with no errors seen
...
1b

Either R =5 or 𝛼 =awrt 53
...
1b

5sin(𝜃 - 53
...
1b

3sin𝜃 - 4cos𝜃 R sin(𝜃 - 𝛼 ); R > 0, 0 ∈ 𝛼 ∈ 90
tan𝛼 =

4

o
...


3

(3)
(b)(i)

Gmax =17 + "5" =22 (C )

B1ft

3
...
13")
sin(15t - "53
...
13") =

3
...
1b

M1

3
...
2a

3

"5"
"5"
After midday solution ⟶ 15t - "53
...
86989
...
1301
...
13"
⟶ t=
15
⟶ t =13
...
⟶ Time =6 : 05 p
...
or 18 : 05

M1

(4)
(8 marks)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 7 Notes:

(a)
4
3
4
or tan𝛼 = or tan𝛼 = or
3
4
3

M1:

For either tan𝛼 =

tan𝛼 =-

B1:

At least one of either R =5 (condone R = 25 ) or 𝛼 =awrt 53
...
13 )

3
4

(b)(i)
B1ft:

Either 22 or follow through “17 + their R from part (a)”

(b)(ii)
M1:

M1:

M1:

A1:

Realisation that the model G =17 + 3sin(15t) - 4cos(15t) can be rewritten as
G =17 + "5"sin(15t - "53
...
13") =
or sin(𝜃 - "53
...
13" =36
...
or 143
...

• 𝜃 - "53
...
86989
...
1301
...
g
...
m
...
13" =180 - 36
...
and so rearranges to give t =
...
m
...
m
...
g
...
1

A1

2
...
This could be e
...

• when x =- 1, e- 3 ∈ e- 2 or e- 3 is not greater than or equal to e- 2

•

M1

2
...
4

when x ∈ 0, e3 x ∈ e 2 x or e3 x is not greater than or equal to e2 x

Followed by an explanation or statement to show when (Bobby’s) claim
e3 x …e2 x is true
...
g
...
E
...

• (Bobby’s) claim is sometimes true

(2)
(ii)
Alt 1

3x
2x
3x
2x
Assuming e …e , then ln(e ) …ln(e ) ⟶ 3x …2x ⟶ x …0

M1

2
...
4

Assume that n2 is even and n is odd
...


M1

2
...

So (Elsa’s) claim is true
...
4

3x
2x
Correct algebra, using logarithms, leading from e …e to x …0

and a correct conclusion
...
g
...
g
...

This could be e
...

• 𝑢 , 9 + 𝑢 ; sum =𝑢 + 9 + 𝑢 =2𝑢 + 9 is irrational

M1

2
...
4

and a correct conclusion
...
g
...


(ii)
M1:

See scheme

A1:

See scheme

(ii)
Alt 1
M1:

Assumes e3 x …e2 x , takes logarithms and rearranges to make x the subject of their inequality

A1:

See scheme

(iii)
M1:

Begins the proof by negating Elsa’s claim and attempts to define n as an odd number

A1:

Shows n2 =4k 2 + 4k +1, where n is correctly defined and gives a correct conclusion

(iv)
M1:

See scheme

A1:

See scheme

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

Scheme

sin x

9 (a)

1 - cos x

+

Marks

AOs

M1

2
...
1b

M1

1
...
1

1 - cos x
sin x

sin x + (1 - cos x)2
2

=
=
=
=

(1 - cos x) sin x
sin2 x + 1 - 2 cos x + cos2 x
(1 - cos x)sin x
1 + 1 - 2 cos x
(1 - cos x)sin x
2 - 2cos x
2(1 - cos x)
=
=
(1 - cos x) sin x
(1 - cos x) sin x

2
= 2cosec x
sin x

{k =2}

(4)
(b)

1 - cos x
 sin x

+
= 1
...
6 ⟶ cosec x =0
...


B1

2
...
6 ⟶  2cosec x =1
...
25

1 - cos x
sin x


As sin x is only defined for - 1 „ sin x „ 1

B1

2
...


(1)
(5 marks)
Question 9 Notes:

(a)
M1:

Begins proof by applying a complete method of rationalising the denominator
sin2 x
Note:

A1:
M1:
A1:

(1 - cos x)sin x

is a valid attempt at rationalising the denominator
sin2 x + 1- 2 cos x + cos2 x
(1 - cos x)sin x

Evidence of applying the identity sin2 x + cos2 x 1
sin x
1 - cos x
+
2cosec x with no errors seen
Uses sin2 x + cos2 x 1 to show that
1 - cos x
sin x
See scheme

(b)
Alt 1
B1:

(1- cos x)sin x

Expands (1 - cos x)2 to give the correct result

(b)
B1:

+

(1 - cos x)2

See scheme

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

Scheme

10

V =4𝑢h(h + 6) =4𝑢 h2 + 24𝑢 h

(a)

Time =

0 „ h „ 25 ;

dV

Marks

AOs

B1 *

3
...
1a

160 =h(h + 6) ⟶ h + 6h - 160 =0 ⟶ (h +16)(h - 10) =0 ⟶ h =
...
1b

{h =- 16, reject}, h =10

A1

1
...
1b

A1

1
...
1a

M1

3
...
1b

2

dV
= 8𝑢 h + 24𝑢
dh
dh
dV dh dV 
⟶ (8𝑢 h + 24𝑢 )
=80𝑢
 × =
dh
dt
dt
dt


80𝑢
 dh dV dh 
 80𝑢 
When h =10, dt = dt  dt =  (8𝑢 (10) + 24𝑢 )  = 124𝑢 




dh 10
=
(cm s- 1) or awrt 0
...
g
...
769
13
dh

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

11 (i)

Scheme

{ y =a
dy

x

⟶}

ln y =ln ax ⟶ ln y = x ln a ⟶

=ln a

y dx
dy

= y ln a ⟶

dx

1 dy

dx

=ax ln a *

Marks

AOs

M1

2
...
1b

(2)
(i)

{ y =a

Alt 1
⟶

x

⟶}

dy
dx

y =exln a ⟶

dy

= (ln a )exln a

dx

=ax ln a *

M1

2
...
1b

(2)
(ii)

d

2 tan y  = 2sec2 y

dy

{ x =2 tan y ⟶ }

dx
=2sec2 y
dy
dx

=2(1 + tan2 y)

dy
dx

or

1 =(2sec2 y)

or

1 =2(1 + tan2 y)

dy
  x 2 

dx
x2 
dx
x2
=2 ∣ 1+ ∣ ∣ ∣ ⟶
=2 ∣ 1 +
=2+
E
...

∣ ⟶
dy
2
dy
dy
2
 4 𝖩
  𝖩 𝖩
2
dx
4+x
dy
2
⟶
=
⟶
=
dy
2
dx
4 + x2

dy
dx

M1

1
...
1b

M1

1
...
1

dx

(4)
(ii)
Alt 1

dy
1
1 
=
×∣ ∣
2
dx 
 2
 𝖩
x
1
+
∣
∣
∣ 2 𝖩 ∣

𝖩
dy
1
dy
1
dy
1
⟶
=
⟶
=
⟶
=
2
2
dx
 x 
dx 
dx  4 + x2 
x 
2∣1 + ∣
∣2 + 2 ∣
∣ 2 ∣
4𝖩


𝖩

𝖩
dy
2
⟶
=
dx
4 + x2

{ x =2 tan y ⟶ }

x
y =arctan ∣ ∣ ⟶
2
 𝖩

M1

1
...
1b

A1

1
...
1

(4)
(6 marks)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 11 Notes:

(i)
M1:

Applies the natural logarithm to both sides of y =a , applies the power law of logarithms and
x

applies implicit differentiation to the result
A1*:

Shows

dy x
=a ln a , with no errors seen
dx

(i)
Alt 1
M1:
A1*:

x
x ln a
Rewrites y =a as y =e
and writes

dy

= c exlna , where c can be 1

dx
Shows

dy x
=a ln a , with no errors seen
dx

(ii)
M1:

Evidence of 2 tan y being differentiated to 2sec2 y

A1:

Differentiates correctly to show that x =2 tan y gives

M1:

Applies sec y =1+ tan y to their differentiated expression

A1:

Shows that

dx

=2sec2 y or 1 =(2sec2 y)

dy
2

dy
dx

2

dy
2
=
, with no errors seen
dx
4 + x2

(ii)
Alt 1
M1:

Evidence of arctan(x); 0 being differentiated to ∣



1
∣ ; , 0
2
∣ 1 + x  ∣

𝖩

Note:  can be 1 for this mark
1

M1:

Differentiates y =arctan  x  ;  0,  1 to give an expression of the form 1 + x 2 × 
 



dy
A1:

Differentiates y =arctan

A1:
Shows that

x
∣ 2∣
 𝖩

correctly to give dx

dy
2
=
, with no errors seen
dx
4 + x2

=

1
× ,
∣ ∣
 x 
2 o
...

 𝖩
∣1 +
∣
∣ ∣ 2∣ ∣
  𝖩 𝖩
1

2



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

12 (a)

Scheme

y =ax2 + c
x =0, y =4 ⟶ c =4
20
1
x =50, y =24 ⟶ 24 =a(50)2 + 4 ⟶ a =
=
or 0
...
3

M1

3
...
1b

(3)
(a)
Alt 1

y =ax2 +bx + c
x =0, y =4 ⟶ c =4

M1

3
...
4

A1

1
...
008
2500
1

y=

x2 + 4

125

{- 50 „ x „ 50}

125

(3)
(b)

x =50 - 19 =31 ⟶ y =

1

(31)2 + 4
125

y =11
...
4

A1

2
...
6227766
...
6227766
...
3772234
...
4

A1

2
...
g
...
g
...
5b

(1)
(6 marks)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 12 Notes:

(a)
M1:
M1:

A1:

2
Attempts to use a model of the form y =ax + c (containing no x term)
Uses the constraints x =0, y =4 and x =50, y =24 (or x =- 50, y =24 ) to find the

values for their c and for their a
1 2
y=
x + 4 (Ignore - 50 „ x „ 50 )
125

(a)
Alt 1
M1:

Attempts to use a model of the form y =ax2 + bx + c and finds or deduces that b =0

M1:

Uses the constraints x =0, y =4; x =50, y =24 and x =- 50, y =24 to find the

A1:

values for their c, for their b and for their a
1 2
y=
x + 4 (Ignore - 50 „ x „ 50 )
125

(b)
M1:

Substitutes x =50 - 19 { =31} or x =- 50 +19 { =- 31} into their quadratic model

A1:

Obtains y =awrt 11
...


A1:

Obtains distance from tower as awrt 18
...
+ 11ln p
11
1
= (2 ln p + (11 - 1) ln p) or
(11)(12) ln p
2
2

M1

3
...
1b

11

ln( p ) = ln p +ln p
n

2

+ ln p3 +
...
+ ln 8 p11

n =1

M1

1
...
1

=11ln 8 + 66ln p
e
...

• 11ln 8 + 66 ln p =11ln 23 + 66 ln p = 33ln 2 + 66ln p
=33(ln 2 + 2 ln p) =33(ln 2 + ln p2 ) =33ln(2 p2 ) *
• 11ln 8 + 66 ln p =11ln 2 + 66 ln p = 33ln 2 + 66ln p
3

=ln(233 p66 ) =ln (2 p2 )33  =33ln(2 p2 ) *

(2)
(c)

S ∈ 0 ⟶ 33ln(2 p2 ) ∈ 0 ⟶ ln(2 p2 ) ∈ 0
2
so either 0 ∈ 2 p ∈ 1 or

2 p2 ∈ 1

1
1
and p > 0 ⟶ 0 ∈ p ∈
2
2
1 

In set notation, e
...
p : 0 ∈ p ∈


2


M1

2
...
5

⟶ p2 ∈

(2)
(6 marks)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 13 Notes:

(a)
11

M1:
Attempts to find

ln( p )
n

by using a complete strategy of

n =1

• applying the power law of logarithms
followed by either
• applying the correct formula for the sum to n terms of an arithmetic series
1
• applying the correct formula n(n +1) ln p
2
• summing the individual terms to give 66 ln p
A1:

66 ln p from correct working

(b)
11

M1:

Deduces S or

ln(8 p ) =11ln 8 + (their answer to part (a))
n

n =1

A1*:

2
and produces a logical argument to correctly show that S =33ln(2 p ) with no errors seen

(c)
M1:

Applies S ∈ 0 to give ln(2 p2 ) ∈ 0 and deduces {e
...
by considering the graph of y =ln x }
that either
• 0 ∈ 2 p2 ∈ 1
•

A1:

2 p2 ∈ 1

Correct answer using set notation
...
g
...
1

A1

1
...
1b

M1

3
...
1

A1

1
...
1b

M1

2
...
1a

6

A1

1
...

4
4

{ y =0 ⟶ x =1 - 16e }
-4

∫4xe

- 2x

-2x

dx = - 2xe
= - 2xe

- 2x

∫- 2e

- e

- 2x

dx

- 2x

Criteria
1

•  - 2xe - 2 x - e - 2 x  = - 2e - 2 - e - 2  -  0 - 1 { =1 - 3e- 2 }
0
1
-6
• Area triangle = 16e - 4   4e - 2  { =32e }
2
- 2

Area(R) =1 - 3e

-6

- 32e

or

e

(10)
(10 marks)

A-Level Edexcel Pure Mathematics Paper 1 Mark Scheme
2022
Question 14 Notes:

M1:

A1:
M1:

Begins the process to find where l intersects the x-axis by differentiating y =4xe- 2 x using the
product rule
dy
=4e- 2x - 8xe- 2x , which can be simplified or un-simplified
dx
A correct method to find the value for the gradient of the normal using mN =

-1
their mT

M1:

Complete strategy to find where l intersects the x-axis
i
...
Applying y - 4e- 2 =m N( x - 1), (where m Ntheir m T) followed by setting y =0 and
rearranging to give x =
...
E
...
1 - 3e - 32e or

- 2x

which can be simplified or un-simplified

e6 - 3e4 - 32
e6

, o
...


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

Scheme

 - 3 ⃗
 3 ⃗
 0 ⃗
 2

∣
∣
∣
∣
∣
∣
∣
OA = 2 , OB = - 1 , BC = 6 , AD = 5 ∣ ;
∣ ∣
∣ ∣
∣ ∣
∣ ∣
∣ 7∣
∣ p∣
∣ - 7∣
∣ - 4∣
 𝖩
 𝖩
 𝖩
 𝖩

15

∣ ⃗ ⃗ ⃗
-3
2

∣  2 ∣ + ∣ 5 ∣ ⟶ ∣ ⃗OD
OD
=OA
+
AD
=


∣ ∣ ∣ ∣
∣
∣ 7 ∣ ∣ - 4∣
∣
 𝖩  𝖩



(a)

 - 1
=∣ 7 ∣
∣ ∣
∣ 3∣
 𝖩

Marks

AOs

B1

1
...
1a

A1

1
...
5 DC ⟶ p - 7 =1
...
1a

p - 7 =1
...
5 p ⟶ p =16

A1

1
...
g

---