Mathematics · Oct / Nov 2025 · Paper 11 · Question 4
YMA01
As printed in the paper
The question as text
4. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(i)[3]
Examples:
2
p
2
p
( )
1
(
)
5
3
3
2
2
p
−
5
p
−
−
−
2
p
5
=
2
p
×
5
2
p
=
p
2
2
lhs:
3
3
3
or
or
or 3
3
or 9
or 9
5
243
3
3
1
1
1
3
−
1
=
or 9 ,
27
=
3 3 or 3 ,
2
27
=
rhs:
2
9
9
3
3
Examples:
1
−
−
−
2
p
5
2
p
5
0.5
3
=
27
3
=
3
2
p
−
5
= −
0.5
p
=
2.25
9
2
p
−
1
3
−
2
p
5
0.5
2
p
4.5
3
=
27
=
3
3
=
3
2
p
=
4.5
p
=
2.25
9
5
3
−=
1
1
2
p
5
2
p
−
5
p
3
27
3
×
3
=
27
9
=
81 3
p
=
2.25
9
9
1
1
2
p
−
5
2
p
−
5
2
p
−
5
2
p
−
4
0.5
=
=
×
=
=
3
27
3
3
3 3
3
3
3
9
3
2
p
−
4
=
0.5
p
=
2.25
5
3
−
−=
1
p
5
1
2
p
5
−
1
3
27
9
2
=
9
×
9
4
p
−
= −
p
=
2.25
9
2
4
(i) Notes
This part of the question can be solved in many different ways.
If you are unsure if a particular approach deserves credit, seek advice from your TL
M1: This is effectively a B mark and is for using a correct law of indices on the lhs or rhs of the
equation. This may be seen in isolation i.e. not necessarily within the equation.
See scheme for some examples.
dM1: Reaches a value for p condoning arithmetic slips but with no incorrect index work e.g.
3
3
1
1
1
1
2
×
2
=
×
2
=
2
3
or e.g.
3 3
9 .
9
3
See main scheme for examples of acceptable work.
See also supplementary document for some example responses with marks.
Depends on the first method mark.
o.e. e.g. 9
1
A1:
Fully correct work leading to
p =
2.25
, 2
4 following award of both Method marks.
4
Correct answer with no working scores no marks in (i)
(
)(
)
4(ii)[4]
36
x
2
+
1
=
x
2
+
1
x
2
−
4
=
36
x
4
−
4
x
2
+
x
2
−
4
=
36
M1
x
2
−
4
2
2
x
4
−
3
x
−
40
=
0
or e.g.
x
4
−
3
x
=
40
A1
(
)(
)
x
2
−
8
x
2
+
5
=
0
x
2
=
...
dM1
x = ±
A1
2 2
Notes
M1:
Attempts to cross multiply and expand to obtain a polynomial equation in x or another variable if
36
(
)(
)
y
=
x
2
y
+
1
=
y
+
1
y
−
4
=
36
y
2
−
4
y
+
y
a substitution is used e.g.
−
y
4
x . The terms do not have to all be on the same side but
2
A1:
Correct 3 term quadratic equation in
4
−
2
=
must be collected so e.g.
x
3
x
40
is acceptable.
If the terms are all one side, the “= 0” may be implied by their attempt to solve.
May also be implied by a substitution e.g.
36
(
)(
)
=
2
+
=
+
−
=
2
−
+
−
=
2
−
=
y
x
y
1
y
1
y
4
36
y
4
y
y
4
36
y
3
y
40
y
−
4
dM1: Attempts to solve 3TQ in
x by any noncalculator means leading to values for
2
x
2
=
2
−
−
=
(
−
)(
+
)
=
This may be implied by substitution e.g.
y
x
y
2
3
y
40
0
y
8
y
5
This may also be implied by their values for x e.g. (
)(
)
−
+
=
= ±
x
2
8
x
2
5
0
x
2 2
−
2
−
=
(
−
)(
+
)
=
=
− scores dM0
Note that e.g.
x
4
3
x
40
0
x
8
x
5
0
x
8,
5
(
)(
)
2
4
−
2
−
=
−
+
=
=
− scores dM1 by implication.
but
x
3
x
40
0
x
8
x
5
0
x
8,
5
x = ±
±
8
A1:
2 2
o.e. e.g.
and no other values. Do not allow decimal values.
x = ±
Do not isw if they obtain e.g.
8
and then
x = ± score A0
4
Similarly if they obtain
x = ±
2 2
and then declare e.g. therefore
x =
2 2
score A0
Must clearly be identified as values of x not any other variable e.g. y or t.
(
)
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