Mathematics · Oct / Nov 2025 · Paper 11 · Question 1
YMA01
As printed in the paper
The question as text
1. The curve C has equation 5 y 6 x 2 3 x x > 0 8
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
1(a)[3]
5
=
2
+
+
y
6
x
3
x
8
d
y
3
1
−
=
+
12
x
x
2
d
x
2
Notes
(a)
M1: Attempts to differentiate achieving one correct index or for 5
8 →
0
1
−
(oe) or for 5
Award for
6
x
2
→
...
x
or
3
x
→
...
x
2
8 →
0
1 1
−
2
→
x −
2 1
→
Allow indices to be unprocessed for this mark e.g.
6
x
...
or
3
x
...
x
2
3
1
− oe e.g.
3
1.5
A1:
One correct simplified algebraic term. E.g. 12x or
2 x
2
,
2 x
1
x
2
1
Condone
12x for 12x for this mark.
1
1
3
1
−
−
×
Do not allow
3
2 x
2
for
2 x
2
Award as soon as a correct simplified term is seen.
d
y
3
1
−
1.5
=
+
A1:
12
x
x
2
or simplified equivalent such as
12x
+
.
d
x
2
x
1
Do not allow
12x for 12x for this mark.
There is no need for d
y
"
x = so just look for the correct expression and apply isw once a correct
"
d
answer is seen.
d
y
3
1
d
y
3
1
−
−
=
12
x
+
x
2
+
c
scores A0 here as does
=
12
x
+
x
2
+
0
d
x
2
d
x
2
1(b)[3]
1
d
y
1
3
4
( )
x =
=
×
+
×
=
12
... 6
4
d
x
4
2
1
1 5
5
1
−
=
−
,
y
''6''
x
2
4
4 2
or e.g.
5
1
=
+
= ×
+
=
y
''6''
x
c
6
c
c
...
2
4
=
+
y
6
x
1
Notes
(b)
1
into their d
y
x and finds its value. Their d
y
x =
M1:
Substitutes
x cannot be the same as ''y''.
4
d
d
1
substituted consistently into their d
y
x =
Score for sight of
x followed by a value or it may
4
d
be implied by a correct value for their d
y
x if no substitution is seen.
d
Condone a miscopy of their d
y
x if the intention is clear.
d
Do not award this mark for work such as
1
−
1
d
y
3
1
3 1
2
5
−
=
12
x
+
x
2
+
c
=
12
+
+
c
=
0 or
c
=
...
d
x
2
4
2 4
2
dM1: Correct method for finding the equation of the tangent (not the normal). Look for the correct use
of their nonzero d
y
1
and the point 1
5
x =
x at
,
with the values correctly placed.
d
4
4
2
+ is used they must proceed as far as c = ... e.g. 5
1
=
=
+
If the form y
mx
c
"6"
c
2
4
Depends on the first method mark.
=
+ Cso
A1:
y
6
x
1
Note if they do not use part (a) for part (b) e.g. just use a calculator they will need to obtain a
gradient of 6 for the first mark, then apply the scheme.
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