Mathematics · May / June 2025 · Paper 41 · Question 8
YMA01
As printed in the paper
Show the answer and mark scheme
8(a)[4]
1
−
1
−
1
x
2
=
+
4
2
1
4
+
x
4
1
1
−
−
−
1
1
−
2
x
1
x
2
2
x
2
+
= + −
+
+
1
1
...
4
2
4
2!
4
1
1
3
−
x
+
x
2
+
...
2
16
256
8(b)[1]
1
1
3
+
+
+
x
x
2
...
2
16
256
8(c)[2]
1
1
3
1
1
3
1
1
−
x
+
x
2
+
x
+
x
2
=
+
x
2
2
16
256
2
16
256
4
128
8(d)[3]
1
1
= 1
1
33
×
" " "
+
"
=
"
"
4
128
128
5
3
128
128
×=
"
" 3
(see alt (d) for alternative answer)
33
11
Alt(d)
1
1
= 1
1
33
×
" " "
+
"
=
"
"
4
128
128
5
3
1
15
15
33
33
1485
×
=
≈
"
"
135
=
3 15
≈×
3 15 "
×
"
=
15
128
128
128
15
15
(a) It is scored B, M, A, A on epen. We are scoring it M, M, A, A
1
1
−
−
−
1
x
2
1
x
2
M1:
Takes out a factor of 4 proceeding to
4
2
1
+
o.e such as
×
1
+
4
2
4
M1:
Attempts the binomial expansion.
Look for the correct binomial coefficient for the third term with the correct power of x. So, award for
1
1
−
−
−
1
3
2
2
(
)
2
×
...
x
or
...
x
2
where … could even be 1. If there is no intermediate working it is
2!
8
1
−
3
x
3
2
2
+
2
implied by the correct coefficient of
128 x in the expansion of
1
or
256 x in the expansion of
4
1
(
)
−
+
4
x
2
1
1
3
−
+
2
+
A1:
Two correct and simplified of
x
x
...
So 1,0,1, 0 is possible
2
16
256
1
1
3
−
+
+
A1:
All three correct and simplified of
x
x
2
...
2
16
256
Accept as a list. Condone any extra terms even if incorrect
1
3
−
×−
1
1
1
3
2
2
5
1
1
Direct expansion: (
)
−
−
−
+
−
=
+ −
×
+
×
2
+
=
−
+
4
x
4
2
4
2
x
4
2
x
...
x
2
2
2
2
16
256
− or 1
1
M1:
For a constant term of
4
2
2
1
3
−
×−
5
2
2
×
−
M1: For a correct third term which may be unsimplified
4
x
2
2
2
3
256 x for the term in x2
2
It is implied if they get
1
1
3
−
+
+
A1:
Two correct and simplified of
x
x
2
...
2
16
256
1
1
3
A1:
All three correct and simplified of
−
x
+
x
2
+
...
2
16
256
Accept as a list. Condone any extra terms even if incorrect
(b)
1
1
3
+
+
+
B1ft: For a correct answer
x
x
2
...
which can be scored following a restart.
2
16
256
Allow as a list. Condone any extra terms even if incorrect.
α
−
β
+
+
αβ are positive to (b)
α
Or follow through on their (a)
x
x
2
...
where , ,
Answers must be seen in (b).
(c)
1
1
3
1
1
3
M1:
Attempts to multiply out the correct
−
x
+
x
2
+
x
+
x
2
or follow through using their
2
16
256
2
16
256
answers to parts (a) and (b) and attempts to collect terms. It must be an attempt on
(
) (
)
αβ are positive leading to
α
+
α
−
β
x
+
x
2
×
α
+
β
x
+
x
2
where , ,
2
2x
. Condone extra terms
where the index is greater than 2, such as a term in x4
1
1
1
−
Alternatively, attempts the expansion (
)
−
1
1
−
2
16
−
x
2
=
16
1
−
x
2
leading to
±
2
2
16
4
1
1
A1:
+
128 x
2
ONLY Any extra terms would be A0. Condone = for ≈
4
(d) This cannot be scored from made up values of a and b.
x = into both sides of their part (c) to achieve 1
1
1
1
M1:
Substitutes
1
×
=
''
'' ''
+
''
4
128
5
3
= ×
dM1: Uses the fact that 135
3
15
and finds a rational approximation for 135
Method 1: 1
1
1
1
15
1
1
1
1
×
=
''
'' ''
+
''
=
''
'' ''
+
''
135
=
''
'' ''
+
' '×45
4
128
15
4
128
4
128
5
3
Method 2: 1
1
1
1
1
1
1
3
×
=
''
'' ''
+
''
=
''
'' ''
+
''
135
=
5
3
4
128
15
4
128
1
1
''
'' ''
+
'
4
128
1
1
The method marks M1, dM1 may be implied, e.g. by the calculation
''
'' ''
+
' ' 45
×
without sight of 135
4
128
Either 128
) or 1485
A1:
11 (or
11
7
(or
11
77
) (or accept both if found)
11
128
128
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