Mathematics · Oct / Nov 2025 · Paper 41 · Question 10
YMA01
As printed in the paper
Show the answer and mark scheme
10(a)[2]
Requires a minimum of
2
2
2
x
=
(2 )
m
=
4
m
which is even so contradiction, so proven10(b)[4]
2
−
=
2
=
+
e.g.
a
4
b
27
a
27
4
b
which is odd
e.g. If
a is odd then a is odd and can be written in the
2
form
a
=
2
m
+
1
e.g. (
)
+
2
=
+
=
2
m
1
27
4
b
b
...
Requires a full explanation and conclusion e.g.
13
b
=
m
2
+
m
−
which is not an integer
2
so contradiction, hence proven *(a)
2
B1:
States that
2
=
x
4
m
dB1:
Dependent on the previous B mark. Explains or concludes that
(or condone not
odd) and either states contradiction/so proven or
rewrites statement X (condone omission of some of the words provided
the general conclusion is clear). Do not allow if the contradiction is not
based on the assumption that
x is odd and
2
4m is even.
2
(b)
Note that there many ways of showing this – send to review if you think
it might be credit worthy
M1:
a
2
−
4
b
=
27
a
2
=
27
+
4
b
oe with an indication that this is odd. Allow
writing in the form 2(2
+ to imply this.
b +
13) 1
2
A1:
States
a is odd (then a is odd) and writes in the form
a
=
2
m
+
1
dM1: Attempts (
)
+
2
=
+
and rearranges to
b =
or 2
b =
or a form
2
m
1
27
4
b
...
...
13
where a contradiction can be deduced such as
+
=
+
where
m
2
m
b
2
13
and m will both be integers but
b +
is not. It is dependent on the
2
previous method mark. Expect to see the brackets multiplied out and
terms collected condoning slips.
A1*:
cso requires a correct statement showing that there is a contradiction
with a conclusion
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