Mathematics · Oct / Nov 2025 · Paper 41 · Question 9
YMA01
As printed in the paper
The question as text
y 9. C R x O ln 9 ln 15 Figure 3 In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. Figure 3 shows a sketch of the curve C with parametric equations 1 x t y t ln 2 5 > 1 t 1 A point P lies on C. Given that the gradient of C at P is –4
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[6]
(
)
2
+
(
)
2
d
t
t
+
1
d
t
2
t
5
d
x
2
t
+
1
2
t
+
5
Sets
−
= −
2
+
+
=
4
8
t
14
t
3
0
(
)
2
t
+
1
2
1
4
(
)(
)
(
)
4
t
+
1
2
t
+
3
=
0
= −
t
y
=
4
3
=
=
(b) (i)
a
2,
b
5
5
5
d
x
1
2
2
Area =
y
d
t
=
×
d
t
=
d
t
(
)(
)
d
t
t
+
1
2
t
+
5
t
+
1 2
t
+
5
2
2
2
A
B
+
=
=
(ii)
(
A
...,
B
...
)(
)
(
)
(
)
t
+
1 2
t
+
5
t
+
1
2
t
+
5
2
4
3
3
=
−
(
)
(
)
+
+
t
1
2
t
5
2
4
2
2
3
3
−
d
t
=
ln
t
+
1
−
ln 2
t
+
5
(
)
(
)
+
+
t
1
2
t
5
3
3
5
2
2
2
2
2
d
t
=
ln6
−
ln15
−
ln3
−
ln9
(
)(
)
t
+
1 2
t
+
5
3
3
3
3
2
2
6
1
36
=
=
ln
ln
3
5
3
25
(a)
Attempts to find d
y
x by dividing their d
y
t by their d
x
M1:
t . Both must be
d
d
d
attempts at differentiation, however poor. Do not allow e.g.
2
→
...ln(2
t
+
5)
as this is clearly
integration.
2
t
+
5
+
d
y
2
t
5
A1:
= −
which may be unsimplified and may be implied by further
(
)
2
d
x
2
t
+
1
work
+
2
t
5
M1:
Sets their
and proceeds to a 3TQ in t. Do not be
"
−
"
= ±
4
(
)
+
2
2
t
1
concerned by the mechanics of the rearrangement.
7
3
2
A1:
8
t
+
14
t
+ =
3
0
o.e. e.g.
2
+
+
=
The = 0 may be implied by further
t
t
0
4
8
work if the three
terms are on the same side of the equation.
dM1: Dependent upon both previous M's, score for solving their 3TQ AND
finding a value for y. Allow any valid attempt to solving the quadratic
including via a calculator. You may need to
check this if just a value for t
is stated. Condone a y value being found from a value of t such that
t <−.
1
) 4
A1:
(
oe only (ignore the x coordinate if found)
y =
3
See alt below – send to review if unsure
2
d
y
−
2e
x
=
=
9 (a)
y
M1A1
2
x
−
d
x
x
−
e
3
(e
3)
−
x
2e
2
Sets
= −
4
2(e )
x
−
13e
x
+
18
=
0
=
0
M1A1
2
(e
x
−
3)
(
)(
)
9
1
4
2e
x
−
9
e
x
−
2
=
0
e
x
=
= −
t
y
=
dM1A1
2
4
3
(b)(i)
B1:
Correct values for a and b. Must be correctly paired or indicated on the
integral. If there is a contradiction between the two then the values on the
integral sign take precedence.
y t with their d
d
x
x
k
M1: For attempting
t in the form 2
where k is a constant.
t +
d
d
5
d
x
Must see either the general expression
y t stated or written as e.g.
d
1
k
×
oe. There needs to be an intermediate stage of working before
t
+
1
2
t
+
5
a
writing (
.
)(
)
t
+
1 2
t
+
5
b
2
It cannot be implied by
d
t
(
)(
)
t
+
1 2
t
+
5
a
b
2
A1:
Area R =
d
t
with or without correct values for a and b. The dt
(
)(
)
t
+
1 2
t
+
5
a
only needs to be present in the final answer. Allow if seen at the start of
(b)(ii). Allow slips in working to be recovered provided it is before the final
answer. This mark cannot be scored without the previous M mark being
scored i.e. we need to see an intermediate stage of working.
b
−
2
Note allow
only if their limits are clearly stated as being
d
t
(
)(
)
t
+
1 2
t
+
5
a
the other way round.
(b)(ii)
k
A
B
M1: Attempts to write (
as (
+
. May be implied by a
)(
)
)
(
)
t
+
1 2
t
+
5
t
+
1
2
t
+
5
correct value for A or B (but not e.g.
=
= without a valid method seen)
A
1,
B
1
May have taken a factor outside of the integral.
2
4
2
3
3
A1:
−
. May have taken a factor outside of the
(
)(
)
(
)
(
)
+
+
+
+
t
1 2
t
5
t
1
2
t
5
1
2
3
3
−
integral so allow e.g. (
oe to score. Must see the partial
)
(
)
t
+
1
2
t
+
5
fractions written or implied by their integration.
M1: Correct attempt at integration. Look for
±α
ln
t
+
1
±β
ln 2
t
+
5
oe where
αβ≠
They must have separated into partial fractions first. Condone
,
0
brackets instead of moduli and condone the omission of brackets to be
implied by further work. Allow this mark to be scored if they miscopy their
partial fractions (e.g. a sign slip or a coefficient slip) so they will still
integrate to a similar form.
2
4
2
2
3
3
A1ft:
−
d
t
=
ln
t
+
1
−
ln 2
t
+
5
oe so allow
(
)
(
)
+
+
t
1
2
t
5
3
3
1
1
ln
t
+
1
−
ln 2
t
+
5
oe but ft on their A and B. May have taken a factor of
3
3
2 outside of the integral. Condone brackets instead of moduli and invisible
brackets to be implied by further work. Ignore any constant of integration if
present.
dM1: Attempts to use their limits and subtract either way round. They must use at
least one law to combine at least two terms. They cannot proceed directly
to the final answer. It is dependent on the previous method mark.
1
36
o.e. in the required form 1 ln
A1:
ln
a where a is a rational number. Do not
3
25
3
2
1
6
accept
ln
3
5
9(b)[9]
(
)
2
+
(
)
2
d
t
t
+
1
d
t
2
t
5
d
x
2
t
+
1
2
t
+
5
Sets
−
= −
2
+
+
=
4
8
t
14
t
3
0
(
)
2
t
+
1
2
1
4
(
)(
)
(
)
4
t
+
1
2
t
+
3
=
0
= −
t
y
=
4
3
=
=
(b) (i)
a
2,
b
5
5
5
d
x
1
2
2
Area =
y
d
t
=
×
d
t
=
d
t
(
)(
)
d
t
t
+
1
2
t
+
5
t
+
1 2
t
+
5
2
2
2
A
B
+
=
=
(ii)
(
A
...,
B
...
)(
)
(
)
(
)
t
+
1 2
t
+
5
t
+
1
2
t
+
5
2
4
3
3
=
−
(
)
(
)
+
+
t
1
2
t
5
2
4
2
2
3
3
−
d
t
=
ln
t
+
1
−
ln 2
t
+
5
(
)
(
)
+
+
t
1
2
t
5
3
3
5
2
2
2
2
2
d
t
=
ln6
−
ln15
−
ln3
−
ln9
(
)(
)
t
+
1 2
t
+
5
3
3
3
3
2
2
6
1
36
=
=
ln
ln
3
5
3
25
(a)
Attempts to find d
y
x by dividing their d
y
t by their d
x
M1:
t . Both must be
d
d
d
attempts at differentiation, however poor. Do not allow e.g.
2
→
...ln(2
t
+
5)
as this is clearly
integration.
2
t
+
5
+
d
y
2
t
5
A1:
= −
which may be unsimplified and may be implied by further
(
)
2
d
x
2
t
+
1
work
+
2
t
5
M1:
Sets their
and proceeds to a 3TQ in t. Do not be
"
−
"
= ±
4
(
)
+
2
2
t
1
concerned by the mechanics of the rearrangement.
7
3
2
A1:
8
t
+
14
t
+ =
3
0
o.e. e.g.
2
+
+
=
The = 0 may be implied by further
t
t
0
4
8
work if the three
terms are on the same side of the equation.
dM1: Dependent upon both previous M's, score for solving their 3TQ AND
finding a value for y. Allow any valid attempt to solving the quadratic
including via a calculator. You may need to
check this if just a value for t
is stated. Condone a y value being found from a value of t such that
t <−.
1
) 4
A1:
(
oe only (ignore the x coordinate if found)
y =
3
See alt below – send to review if unsure
2
d
y
−
2e
x
=
=
9 (a)
y
M1A1
2
x
−
d
x
x
−
e
3
(e
3)
−
x
2e
2
Sets
= −
4
2(e )
x
−
13e
x
+
18
=
0
=
0
M1A1
2
(e
x
−
3)
(
)(
)
9
1
4
2e
x
−
9
e
x
−
2
=
0
e
x
=
= −
t
y
=
dM1A1
2
4
3
(b)(i)
B1:
Correct values for a and b. Must be correctly paired or indicated on the
integral. If there is a contradiction between the two then the values on the
integral sign take precedence.
y t with their d
d
x
x
k
M1: For attempting
t in the form 2
where k is a constant.
t +
d
d
5
d
x
Must see either the general expression
y t stated or written as e.g.
d
1
k
×
oe. There needs to be an intermediate stage of working before
t
+
1
2
t
+
5
a
writing (
.
)(
)
t
+
1 2
t
+
5
b
2
It cannot be implied by
d
t
(
)(
)
t
+
1 2
t
+
5
a
b
2
A1:
Area R =
d
t
with or without correct values for a and b. The dt
(
)(
)
t
+
1 2
t
+
5
a
only needs to be present in the final answer. Allow if seen at the start of
(b)(ii). Allow slips in working to be recovered provided it is before the final
answer. This mark cannot be scored without the previous M mark being
scored i.e. we need to see an intermediate stage of working.
b
−
2
Note allow
only if their limits are clearly stated as being
d
t
(
)(
)
t
+
1 2
t
+
5
a
the other way round.
(b)(ii)
k
A
B
M1: Attempts to write (
as (
+
. May be implied by a
)(
)
)
(
)
t
+
1 2
t
+
5
t
+
1
2
t
+
5
correct value for A or B (but not e.g.
=
= without a valid method seen)
A
1,
B
1
May have taken a factor outside of the integral.
2
4
2
3
3
A1:
−
. May have taken a factor outside of the
(
)(
)
(
)
(
)
+
+
+
+
t
1 2
t
5
t
1
2
t
5
1
2
3
3
−
integral so allow e.g. (
oe to score. Must see the partial
)
(
)
t
+
1
2
t
+
5
fractions written or implied by their integration.
M1: Correct attempt at integration. Look for
±α
ln
t
+
1
±β
ln 2
t
+
5
oe where
αβ≠
They must have separated into partial fractions first. Condone
,
0
brackets instead of moduli and condone the omission of brackets to be
implied by further work. Allow this mark to be scored if they miscopy their
partial fractions (e.g. a sign slip or a coefficient slip) so they will still
integrate to a similar form.
2
4
2
2
3
3
A1ft:
−
d
t
=
ln
t
+
1
−
ln 2
t
+
5
oe so allow
(
)
(
)
+
+
t
1
2
t
5
3
3
1
1
ln
t
+
1
−
ln 2
t
+
5
oe but ft on their A and B. May have taken a factor of
3
3
2 outside of the integral. Condone brackets instead of moduli and invisible
brackets to be implied by further work. Ignore any constant of integration if
present.
dM1: Attempts to use their limits and subtract either way round. They must use at
least one law to combine at least two terms. They cannot proceed directly
to the final answer. It is dependent on the previous method mark.
1
36
o.e. in the required form 1 ln
A1:
ln
a where a is a rational number. Do not
3
25
3
2
1
6
accept
ln
3
5
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