Marking instructions AO Mark Typical solution
1 Circles the correct response 1.1b B1
Total 1
Q Marking instructions AO Mark Typical solution
2 Circles the correct response 1.1b B1
Total 1
Q Marking instructions AO
...
Marking instructions AO Mark Typical solution
1 Circles the correct response 1.1b B1
Total 1
Q Marking instructions AO Mark Typical solution
2 Circles the correct response 1.1b B1
Total 1
Q Marking instructions AO Mark Typical solution
3 Circles the correct response 1.1b B1 6.4 cm2
Total 1
−4log10 ( a )
dd
y kx
ke
x
=
6MARK SCHEME – A-LEVEL MATHEMATICS – 7357/1 – JUNE 2019
Q Marking instructions AO Mark Typical solution
4 Uses negative reciprocal to
obtain equation with correct
gradient
3.1a M1
4 5
1
5 4 17
3
4 1 5 3 11
5 4 11
4 11
5 5
x y k
x
y
y
k
y x
y x
− + =
=
⇒ + =
⇒ =
= − × + × =
− =
= +
Obtains correct x coordinate of
midpoint
Or obtains correct equations of
lines through A and B
perpendicular to AB
5 4 31.5 5 4 9.5 y x y x − = − = −
OE
1.1b B1
Substitutes their mid-point value
of x to obtain value of y
coordinate of midpoint (not in
terms of a or b)
Or
Finds a value for their
2
a b +
Or
Finds k by adding correct
equations of lines through A and
B perpendicular to AB
Or equating intercepts.
1.1a M1
Obtains correct equation ACF
Eg 4 , 2.2
5
y x c c = + =
ISW once correct answer seen.
1.1b A1
Total 4
7MARK SCHEME – A-LEVEL MATHEMATICS – 7357/1 – JUNE 2019
Q Marking instructions AO Mark Typical solution
5(a) Uses =260 for arithmetic
sequence with n=16 to form a
correct equation
PI by 8 2 15 260 ( a d + = )
1.1a M1
Completes rigorous argument
with correct algebraic
manipulation to show required
result
Must see at least one line of
simplification after
8 2 15 260 ( a d + = ) before given
answer.
2.1 R1
5(b) Forms a second equation in a
and d using and
solves simultaneously to find a
or d
3.1a M1
( )
41 ( )
30 2 59 315
20 590 105
20
0.5
41
2 20 40 0.5 410
2
a d
a d
a d S
+ =
+ =
=
= −
= × − × =
Obtains correct a and d 1.1b A1
Uses their a and d to obtain their
value of = 41 820 a d +
Follow through provided one of
their a or d is correct.
1.1b A1F
5(c) Explains that values of are
positive n < 41
Or
Explains that values of are
negative for
Or
Uses quadratic manipulation or
differentiation of formula for Sn
to obtain n = 40.5
CSO
2.4 M1 The terms before the 41st term are
all positive. The terms after the 41st
term are all negative so the sum of
the first 41 terms must be a
maximum value.
Completes a valid argument
explaining all terms positive
before 41 and negative after 41
Or
Completes argument linking
40.5 with the sum to 40 terms
and the sum to 41 terms.
CSO
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