Mathematics · Oct / Nov 2025 · Paper 41 · 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 entirely on calculator technology are not acceptable. The curve C has equation x 2 y 2 xy x 4 2 24

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Show the answer and mark scheme

4(a)[5]
5
M1:
One correct end. e.g.
oe. Do not be concerned over the
g( )
x „
2
inequality being strict or inclusive.
5
5
A1:
0
<
g( )
x
„
o.e. Condone use of y or g but not x e.g.
0
<
y
„
and
2
2
alternative equivalent
notation
5
5
5
e.g.
or
0, 2
0
<
g( ),
x
…
g( )
x
0
<
g( )
x
∩
…
g( )
x
0
<
g( )AND
x
2
2
do not
accept “OR” or use of ∪
5
Special case:
oe score M1A0
0
<
x
„
2
d
y
2
4 (a)
Attempt at chain rule
y
→
... d
y x
d
y
Attempt at product rule
2
xy
→
α
x
+
β
y
d
x
d
y
d
y
Correct differentiation
oe
8
x
+
2
y
−
2
y
−
2
x
=
24
d
x
d
x
) d
y
Collects terms in (
2
y
−
2
x
=
24
−
8
x
+
2
y
d
x
d
y
12
−
4
x
+
y
oe
=
d
x
y
−
x
−
+
1
Sets 12
4
x
y
or
=
−
(b)
=
2
y
=
12
−
2
x
x
6
y
y
−
x
2
1
=
−
=
−
+
2
−
=
Substitute
y
12
2 or
x
x
6
y
i
n
t
o
4
x
2
y
2
xy
24
x
2
2
−
+
=
oe or
x
8
x
12
0
y
2
−
8
y
=
0
Solves
=
=
or
=
=
x
6,
y
0
x
2,
y
8
Chooses ( )
( )
= or states (
)
=
2, 8 only
a x
2,
b y
8
Allow use of y′ for d
y
(a)
x
d
d
y
M1:
Attempt at chain rule
2
where ... is a constant
y
→
... d
y x
d
y
M1:
Attempt at product rule
→
α
+
β where
α β ... are nonzero
,
2
xy
x
y
d
x
constants.
d
y
Condone
−
2
xy
→±
α
x
±
β
y
d
x
d
y
d
y
A1:
Correct differentiation
oe
8
x
+
2
y
−
2
y
−
2
x
=
24
d
x
d
x
Collects their two different terms in d
y
M1:
x which must have come from
d
2 and 2
differentiating
y
xy (however poorly the differentiation was carried
out) on one side of the equation
with the remaining terms on the
d
y
other side proceeding to
(... ...)
±
=
............
. May be
implied by a
d
x
correct answer.
Ignore a spurious d
y
x e.g. on the left hand side as long as it is not included
d
in the collecting of terms.
−
+
−
−
d
y
12
4
x
y
oe including d
y
4
x
y
12
A1:
=
=
Must be in simplest
−
−
d
x
y
x
d
x
x
y
form and seen in (a)
(b)
May work in terms of a and b which is acceptable
Sets their d
y
M1:
x =
2
and proceeds to a linear equation in x and y. May be
d
d
y
unsimplified. Alternatively, substitutes in
x =
into the given
2
d
equation
(
)
8
x
+
2
y
×−
2
2
y
−
2
x
×=
2
24
4
x
+
2
y
=
24
dM1:
Substitutes their linear equation in x and y in the equation for C and
proceeds to a quadratic equation in x or y. Do not be concerned by the
mechanics of their rearrangement. Terms should be collected but all
terms do not need to be on the same side of the equation. (if all on
one side then = 0 must be seen or may be implied by further work). It is
dependent on the previous method mark.
A1:
Correct quadratic in x or y.
x
2
−
8
x
+
12
=
0
oe or
y
2
−
8
y
=
0
oe terms
should be collected but all terms do not need to be on the same side of
the equation. (if all on one side then = 0 may be implied by further work)
ddM1: Solves their quadratic in x or y and proceeds to a pair of coordinates for P.
Allow any valid method (usual rules apply). If via a calculator to find x then
they must have rearranged to a three term quadratic with all terms on one
side. You may need to check their roots are correct for their quadratic. It is
dependent on both of the previous method marks.
Note when finding a value for b they may substitute into C resulting in
false solutions.
e.g. ("6", "0"), ("6", "6") or ("2", "8"), ("2", " 4")
−
which is acceptable for
this mark.
(
)
A1:
only but allow other forms such as
=
= . Condone if labelled
P
2, 8
a
2,
b
8
as x and y.
If they find the coordinates (6,0) then they must clearly identify (
their chosen answer.
Attempts to solve 2
+ λ= −+ μ and 3
+ μ
4(b)[5]
5
M1:
One correct end. e.g.
oe. Do not be concerned over the
g( )
x „
2
inequality being strict or inclusive.
5
5
A1:
0
<
g( )
x
„
o.e. Condone use of y or g but not x e.g.
0
<
y
„
and
2
2
alternative equivalent
notation
5
5
5
e.g.
or
0, 2
0
<
g( ),
x
…
g( )
x
0
<
g( )
x
∩
…
g( )
x
0
<
g( )AND
x
2
2
do not
accept “OR” or use of ∪
5
Special case:
oe score M1A0
0
<
x
„
2
d
y
2
4 (a)
Attempt at chain rule
y
→
... d
y x
d
y
Attempt at product rule
2
xy
→
α
x
+
β
y
d
x
d
y
d
y
Correct differentiation
oe
8
x
+
2
y
−
2
y
−
2
x
=
24
d
x
d
x
) d
y
Collects terms in (
2
y
−
2
x
=
24
−
8
x
+
2
y
d
x
d
y
12
−
4
x
+
y
oe
=
d
x
y
−
x
−
+
1
Sets 12
4
x
y
or
=
−
(b)
=
2
y
=
12
−
2
x
x
6
y
y
−
x
2
1
=
−
=
−
+
2
−
=
Substitute
y
12
2 or
x
x
6
y
i
n
t
o
4
x
2
y
2
xy
24
x
2
2
−
+
=
oe or
x
8
x
12
0
y
2
−
8
y
=
0
Solves
=
=
or
=
=
x
6,
y
0
x
2,
y
8
Chooses ( )
( )
= or states (
)
=
2, 8 only
a x
2,
b y
8
Allow use of y′ for d
y
(a)
x
d
d
y
M1:
Attempt at chain rule
2
where ... is a constant
y
→
... d
y x
d
y
M1:
Attempt at product rule
→
α
+
β where
α β ... are nonzero
,
2
xy
x
y
d
x
constants.
d
y
Condone
−
2
xy
→±
α
x
±
β
y
d
x
d
y
d
y
A1:
Correct differentiation
oe
8
x
+
2
y
−
2
y
−
2
x
=
24
d
x
d
x
Collects their two different terms in d
y
M1:
x which must have come from
d
2 and 2
differentiating
y
xy (however poorly the differentiation was carried
out) on one side of the equation
with the remaining terms on the
d
y
other side proceeding to
(... ...)
±
=
............
. May be
implied by a
d
x
correct answer.
Ignore a spurious d
y
x e.g. on the left hand side as long as it is not included
d
in the collecting of terms.
−
+
−
−
d
y
12
4
x
y
oe including d
y
4
x
y
12
A1:
=
=
Must be in simplest
−
−
d
x
y
x
d
x
x
y
form and seen in (a)
(b)
May work in terms of a and b which is acceptable
Sets their d
y
M1:
x =
2
and proceeds to a linear equation in x and y. May be
d
d
y
unsimplified. Alternatively, substitutes in
x =
into the given
2
d
equation
(
)
8
x
+
2
y
×−
2
2
y
−
2
x
×=
2
24
4
x
+
2
y
=
24
dM1:
Substitutes their linear equation in x and y in the equation for C and
proceeds to a quadratic equation in x or y. Do not be concerned by the
mechanics of their rearrangement. Terms should be collected but all
terms do not need to be on the same side of the equation. (if all on
one side then = 0 must be seen or may be implied by further work). It is
dependent on the previous method mark.
A1:
Correct quadratic in x or y.
x
2
−
8
x
+
12
=
0
oe or
y
2
−
8
y
=
0
oe terms
should be collected but all terms do not need to be on the same side of
the equation. (if all on one side then = 0 may be implied by further work)
ddM1: Solves their quadratic in x or y and proceeds to a pair of coordinates for P.
Allow any valid method (usual rules apply). If via a calculator to find x then
they must have rearranged to a three term quadratic with all terms on one
side. You may need to check their roots are correct for their quadratic. It is
dependent on both of the previous method marks.
Note when finding a value for b they may substitute into C resulting in
false solutions.
e.g. ("6", "0"), ("6", "6") or ("2", "8"), ("2", " 4")
−
which is acceptable for
this mark.
(
)
A1:
only but allow other forms such as
=
= . Condone if labelled
P
2, 8
a
2,
b
8
as x and y.
If they find the coordinates (6,0) then they must clearly identify (
their chosen answer.
Attempts to solve 2
+ λ= −+ μ and 3
+ μ

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