Mathematics · Oct / Nov 2025 · Paper 21 · Question 7
YMA01
As printed in the paper
The question as text
7. y C x O R N T Figure 3 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 3 shows a sketch of part of the curve C with equation 5 1 3 2 2 y x x kx x x 0 4 28 44 where k is a positive constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[2]
2
2
2
2
y
=
x
3
−
4
x
−
kx
+
28
x
−
44
=
3
x
2
−
10
x
−
x
+
28
d
x
2
7(b)[2]
into d
y
k
Subs
x =
9
x and sets = 0
×
−
×
−
+
=
3 81 10 27
28
0
d
6
k
k
243
−
−
+
=
=
=
*
270
28
0
1
k
6
6
6
5
1
7
3
1
8
7(c)[6]
3
−
2
−
2
+
−
=
4
−
2
−
2
+
2
−
x
4
x
6
x
28
x
44 d
x
x
x
4
x
14
x
44
x
4
7
Correct value of
y = −
at T
53
9
7
3
1
8
4
2
Shaded area =
''53'' 9
×+
x
−
x
2
−
4
x
2
+
14
x
−
44
x
4
7
0
= 248
(a)
M1: Attempts to differentiate. Look for two correct terms, one of which could be 44
Allow the coefficients to be unsimplified but the indices must be processed
3
1
d
y
=
k
−
A1:
2
−
2
−
2
+
or other simplified equivalent. ISW after a correct answer
3
x
10
x
x
28
d
x
2
(b)
into their d
y
M1: Substitutes
x =
9
x and sets = 0. Look for embedded values, or 4 simplified
d
terms set = 0
into their d
y
For the M1 only, allow a substitution of
x =
x which develops into an equation that
9
d
3
1
implies d
y
d
y
k
−
k
k =
x = 0. E.g.
=
×
2
−
×
2
−
×
2
+
=
− So
3 9
10
9
9
28
1
1
d
d
x
2
6
6
into their d
y
x and finds the value of d
Alternatively substitutes
x =
and
9
k =
6
d
d
conditions as above.
A1*: Shows that
k =
6
Via the main method you must see
1
an unsimplified equation such as
×
−
−
+
=
3 81
270
6 k
28
0
k = o.e.
a simplified equation such as
1
6
and the given answer
k =
6
Via the alternative method you must see
an unsimplified expression for d
y
1
x =
3 81
×
−
270
−
×
6
+
28
d
6
before proceeding to d
y
x =
0
d
followed by a conclusion stating that there is a stationary point when
(c)
Main method: Finds both areas and subtracts
M1:
Attempts to integrate. Look for two correct indices which must be processed
A1:
Two correct terms but allow the coefficients to be left unsimplified
7
3
1
8
A1:
4
−
2
−
2
+
2
−
allowing coefficients to be left unsimplified.
x
x
4
x
14
x
44
x
4
7
Condone an additional +c
B1:
y = −
at T May be awarded at any point in the question, even from the diagram
53
dM1:
Look for a FULL method that involves (area of rectangle −area under curve) either
way around to find a value for the area of R. Condone slips but it is dependent
upon the previous M.
The area of the rectangle must be found using 9× (an attempt at the y value of the
curve at x = 9), and not for instance using 9 44
×
as 44 is the intersection of the
curve with the y axis.
7
3
1
8
4
2
You don’t need to see working for the evaluation of
−
2
−
2
+
−
x
x
4
x
14
x
44
x
4
7
AWRT 248. Allow the exact answer 6939
A1:
28 Note that 248
−
is A0
(c) Alt method via integration of line – curve (or curve −line)
This could also be attempted by translating the curve upwards by 53 units
The following example shows a correct line – curve approach
5
1
5
1
3
2
2
3
2
2
' 53'
−
−
x
−
4
x
−
6
x
+
28
x
−
44 d
x
=
−−
9
x
+
4
x
+
6
x
−
28 d
x x
7
3
1
8
2
−
9
x
−
x
4
+
x
2
+
4
x
2
−
14
x
4
7
7
3
1
8
6939
2
Shaded area =
−×
9
9
−
×
9
4
+
×
9
2
+
4
×
9
2
−
14
×
9
=
4
7
28
5
1
3
2
2
M1: Attempts to integrate an expression of the form
±−
x
+
4
x
+
6
x
−
28
x
±
p
Look for two correct indices which must be processed. Condone bracketing errors
Allow even with p = 97 or 0
7
3
1
8
2
2
2
A1: Two correct terms of
±−
x
4
+
x
+
4
x
−
14
x
±
px
but allow the coefficients to be
4
7
left unsimplified. Follow through on their value of p which may be 97 or 0
7
3
1
8
2
A1: Correct integration. So look for
±−
x
4
+
x
2
+
4
x
2
−
14
x
±
px
, allowing the
4
7
coefficients to be left unsimplified. Follow through on their value of p which may be 97 or
0. Condone an additional + c
B1: For
y = −
at T. May be awarded at any point in the question, even from the diagram
53
9
7
3
1
8
2
dM1: Attempts the value of
±−
4
+
2
+
2
−
−
. No intermediate working is
x
x
4
x
14
x
9
x
4
7
0
required.
This is dependent upon the first M and also obtaining ‘9x’
AWRT 248. Allow the exact answer 6939
A1:
28
Note that final answers of 248
−
and
−
are A0
248
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