Maths Edexcel AS paper 1
pure answers 2023
Edexcel MATHS PURE PAPER 1 2020
Edexcel=MATHS-PURE-PAPER=1 2020
student=friendly mark scheme
M1 – method
... [Show More] mark. This mark is generally given for an appropriate method in the
context of the question. This mark is given for showing your working and may be
awarded even if working is incorrect.
A1 – accuracy mark. This mark is generally given for a correct answer following
correct Guidance working.
B1 – working mark. This mark is usually given when working and the answer cannot
easily be separated.
Some questions require all working to be shown; in such questions, no marks will
be given for an answer with no working (even if it is a correct answer).
on the use of codes within this document
Edexcel MATHS PURE PAPER 1 2020 studentfriendly mark scheme
GCE AS Mathematics (8MA0) – Paper 1
Pure Mathematics
Summer 020 student-friendly mark scheme
Please note that this mark scheme is not the one used by
examiners for making scripts. It is intended more as a guide to
good practice, indicating where marks are given for model
solutions. As such, it doesn’t show follow-through marks (marks
that are awarded despite errors being made) or special cases.
It should also be noted that for many questions, there may be
alternative methods of finding correct solutions that are not
shown here – they will be covered in the formal mark scheme.
2
Question 1 (Total 4 marks)
Part Working or answer an examiner might
expect to see
Mark Notes
(a) y = –
1
x +
3
2 4
M1 This mark is given for a method to
rearrange to find an equation for l1 in
terms of y =
m = 2 A1 This mark is given for deducing the
gradient of the perpendicular line l2
(b) Substituting y = 2x + 7 into 2x + 4y – 3 = 0
gives
2x + 4(2x + 7) – 3 = 0
M1 This mark is given for a method to
substitute to form and solve an equation
in a single variable.
10x + 25
x = –2.5
A1 This mark is given for solving to find
the value of the x-coordinate of the
point P.
Question 2 (Total 8 marks)
Part Working or answer an examiner might
expect to see
Mark Notes
(i) 2 16a
2
16a = 2√a so = 1
1
2a
2
3 3
8a 2 = 1 so a
2 =
1
8
M1 This mark is given for a method to find
an equation to solve with the terms in a
on one side
3
a =
1 2
∣8 𝖩
∣
M1 This mark is given for finding a way to
deal with the indices when solving the
equation
a =
1
4
A1 This mark is given for finding one
correct solution to the equation
a = 0 is also a solution B1 This mark is given for deducing that
a = 0 is also a solution
(ii) b
4 + 7b
2
– 18 = 0 factorises to
(b
2 + 9)(b
2
– 2) = 0
M1 This mark is given for factorising the
equation given
b
2 = –9, 2 A1 This mark is given for finding two
correct solutions for b
2
For real solutions, b
2 = 2 only M1 This mark is given for recognising that
b = √–9 is not a real solution
b = √2, –√2 A1 This mark is given for finding the two
real solutions to the equation
3
Question 3 (Total 6 marks)
Part Working or answer an examiner might
expect to see
Mark Notes
(a)
-3 –2 1
4x + kx dx = –2x +
2
kx
M1 This mark is given for recognising that x
n
becomes x
n + 1 when integrating
A1 This mark is given for two correctly
integrated terms (without c)
2 kx 2
– + + c
x 2 2
A1 This mark is given for a full answer with
a constant (in any correct form)
(b) 2 k
2
∣-
x
2 + 2
x
2
∣
=
0.5
2 4k 2 0.52 k
∣- 2 + ∣ – ∣- 2 + ∣ = 8
2 2 𝖩 0.5 2 𝖩
M1 This mark is given for substituting the
limits 2 and 0.5 and setting equal to 8
∣ -
1
+ 2k ∣ – ∣ - 8 +
k
∣ = 8
2 𝖩 8 𝖩
7.5 +
15 k = 8
8
M1 This mark is given for a method to solve
a linear equation in k
k =
4
15
A1 This mark is given for finding a correct
value for k
Question 4 (Total 5 marks)
Part Working or answer an examiner might
expect to see
Mark Notes
M1 This mark is given for using the
information to create a model of the form
H = mt + b where m is the rate of growth
and b is the original height of the tree
0.93 = 3m,
m = 0.31
M1 This mark is given for finding a value
for m
H = 0.31m so b = 1.42
H = 0.31m + 1.42
A1 This mark is given for finding a value
for b
(b) b represents the original height of the tree B1 This mark is given for recognising what b
represents
(a) 2.35 =
3m + b
3.28 =
6m + b [Show Less]