Mathematics · May / June 2025 · Paper 11 · Question 7
YMA01
As printed in the paper
The question as text
7. y C1 x O Figure 3 Figure 3 shows a sketch of part of the curve C1 Given that C1 • has equation y = f (x) where f (x) is a cubic function • touches the x-axis at the origin and cuts the x-axis at x = 4 • passes through the point (10, 120)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[3]
2
+
AC
2
−
2
×
17
×
AC
co
s0.6
AC
=
=
..
.
2
Allow if degrees mode is used. Here angle CAO = 0.680
Watch for cases where the inverse sine is not taken, which give 0.6399. These will score M0dM0A0.
dM1: Full method to find angle COA. As scheme or in the Alt they may proceed to use the sine rule with
their AC (note they should get 1.293… from this method).
A1*: Complete and correct proof. In the Alt the selection of the positive root must be justified (from
diagram angle COA is obtuse so…) and justification of the correct angle seen (ie
π−
1.293...
1.847…). Allow as long as all values are to at least 3dp throughout (if seen).
There may be attempts via verification. Allow
M1: For a correct attempt at the sine (or cosine) rule as a part of their check.
dM1: For a full process to check the angle, e.g.
15sin 0.69459...
angle
CAO
=
π
−
0.6 1.847
−
=
0.69459...
∠
ACO
=
arcsin
=
0.600
✓
17
A1*: If all is correct with a minimal conclusion.
Note: for (b) and (c) if they use their 1.847 rather than the given value in the question, allow the M’s for
correct method with their value.
M1: Correct attempt at one relevant area, allowing the minor sector as relevant. See scheme.
dM1: Complete method to find both areas and adds.
A1: Awrt 622 m2 including the units
M1: Attempts major or minor arc length AB OR length AC or its square, e.g.
rθ=
15
×θ where
(
) or just 1.847
2
2
(
)
θ
=
2
π
−
1.847
OR attempts
15
+
17
−×
2 15 17cos 1.847
×
condoning if it is called
AC.
15sin1.847
There may be alternative approaches to these (e.g. sine rule
AC =
).
sin 0.6
For each approach allow if correct formula is quoted but they subsequently forget to apply the cosine or
sine to the angle.
(
)
dM1: Attempts both major arc length and length AC ie
r =
θ
15
×
2
π−
1.847
AND
2
2
AC =
15
17
2 15 17cos 1.847
(
)
or equivalent approaches. Must correctly apply the trig ratios
+
−×
×
for this mark.
A1: Awrt 94.1 m. Should include units unless the A mark in (b) was penalised only for lack of units (ie
penalise only the first instance of incorrect or missing units).
2
7 (a)
States or implies that
f( )
x
= λ
x
(
x
−
4)
2
Attempts to find λ. E.g.
120
= λ
10 (10
−
4)
λ=
...
2
f( )
x
=
0.2
x
(
x
−
4)
(b)
''Upside down'' parabola M1
Passing through O and +ve x axis > 4 A1
(c)
Sets
1.2 (8
x
−
x
)
=
thei
r
0.2
x x
2
(
−
4
)
2
2
1.2 (8
x
−
x
)
=
0.2
x
(
x
−
4)
x x
2
+
2
x
−
48
=
0
or
x
3
+
2
x
−
48
x
=
0
...,...(,0)
x
=
For x = 8,0,6
OR one of (
8, 153.6 , 6,14.4
) (
)
−
−
−
768
72
✓
All of (
) (
) (
)
(
)
−
8, 153.6 , 6,14.4 , 0,0
−
−
8,
−
, 6,
, 0,0
5
5
(a)
2
M1: States or implies that
f( )
x
= λ
x
(
x
−
4)
. Allow with λ = 1 (or any other constant). Credited when first
seen, but must use this expression to access the dM mark.
2
dM1: Attempts to findλ. E.g.
120
= λ
10 (10
−
4)
λ=
...
A1:
2
f( )
0.2
(
4)
o.e. Allow with(
x −
0
)
2
x
=
x
x
−
3
2
Alt: May see attempts at using
f( )x
=
ax
+
bx
+
cx
+
d
3
2
M1: States or implies
f( )x
=
ax
+
bx
+
cx
+
d
AND uses (0,0) on curve to deduce d = 0
dM1: Full method to find the remaining constants, so f (0)
′
=
0
c
=
0
and
f(4)
=
0
64
a
+
16
b
=
0
b
= −
4
a
then f(10)
=
120
1000
a
+
100
b
=
120
50
a
+
5
b
=
They must set up sufficient equations to be able to solve for all the constants.
3
2
A1: For
f( )
x
=
0.2
x
−
0.8
x
o.e.
Note, do not be concerned if the f(x) = … is missing, mark the expression.
(b)
Note: If there are multiple attempts with no clearly designated answer, Diagram 1 takes precedence if not
crossed out. If Diagram 1 is not used, or is crossed out and no clear statement of answer is given, then an
answer in the answer space takes precedence over an answer on Figure 3.
M1: Correct shape, a ∩shape parabola in any position. Be tolerant with the shape, must at least touch the x
axis.
A1: Correct shape and position, a ∩shape parabola passing through O and the +ve x axis to the right hand
side of x = 4. The graph must clearly extend into quadrants 3 and 4, not stop at the xaxis.
Ignore any coordinates on the graph, and don’t be concerned if the intersection in quadrant 3 is not
shown.
(c)
B1ft: Sets
1.2 (8
x
−
x
)
=
thei
r
0.2
x x
2
(
−
4
)
. It is for a ''correct'' ft equation on their part (a) as long as their
answer to part (a) is a cubic expression.
M1: Valid attempt to solve an equation of the form 1.2 (8
x
−
x
)
=
cubic
with their cubic (and their quadratic
(
)
1.2
x
7(b)[2]
M1: Correct attempt at one relevant area, allowing the minor sector as relevant. See scheme.
dM1: Complete method to find both areas and adds.
A1: Awrt 622 m2 including the units
7(c)[5]
M1: Attempts major or minor arc length AB OR length AC or its square, e.g.
rθ=
15
×θ where
(
) or just 1.847
2
2
(
)
θ
=
2
π
−
1.847
OR attempts
15
+
17
−×
2 15 17cos 1.847
×
condoning if it is called
AC.
15sin1.847
There may be alternative approaches to these (e.g. sine rule
AC =
).
sin 0.6
For each approach allow if correct formula is quoted but they subsequently forget to apply the cosine or
sine to the angle.
(
)
dM1: Attempts both major arc length and length AC ie
r =
θ
15
×
2
π−
1.847
AND
2
2
AC =
15
17
2 15 17cos 1.847
(
)
or equivalent approaches. Must correctly apply the trig ratios
+
−×
×
for this mark.
A1: Awrt 94.1 m. Should include units unless the A mark in (b) was penalised only for lack of units (ie
penalise only the first instance of incorrect or missing units).
2
7 (a)
States or implies that
f( )
x
= λ
x
(
x
−
4)
2
Attempts to find λ. E.g.
120
= λ
10 (10
−
4)
λ=
...
2
f( )
x
=
0.2
x
(
x
−
4)
(b)
''Upside down'' parabola M1
Passing through O and +ve x axis > 4 A1
(c)
Sets
1.2 (8
x
−
x
)
=
thei
r
0.2
x x
2
(
−
4
)
2
2
1.2 (8
x
−
x
)
=
0.2
x
(
x
−
4)
x x
2
+
2
x
−
48
=
0
or
x
3
+
2
x
−
48
x
=
0
...,...(,0)
x
=
For x = 8,0,6
OR one of (
8, 153.6 , 6,14.4
) (
)
−
−
−
768
72
✓
All of (
) (
) (
)
(
)
−
8, 153.6 , 6,14.4 , 0,0
−
−
8,
−
, 6,
, 0,0
5
5
(a)
2
M1: States or implies that
f( )
x
= λ
x
(
x
−
4)
. Allow with λ = 1 (or any other constant). Credited when first
seen, but must use this expression to access the dM mark.
2
dM1: Attempts to findλ. E.g.
120
= λ
10 (10
−
4)
λ=
...
A1:
2
f( )
0.2
(
4)
o.e. Allow with(
x −
0
)
2
x
=
x
x
−
3
2
Alt: May see attempts at using
f( )x
=
ax
+
bx
+
cx
+
d
3
2
M1: States or implies
f( )x
=
ax
+
bx
+
cx
+
d
AND uses (0,0) on curve to deduce d = 0
dM1: Full method to find the remaining constants, so f (0)
′
=
0
c
=
0
and
f(4)
=
0
64
a
+
16
b
=
0
b
= −
4
a
then f(10)
=
120
1000
a
+
100
b
=
120
50
a
+
5
b
=
They must set up sufficient equations to be able to solve for all the constants.
3
2
A1: For
f( )
x
=
0.2
x
−
0.8
x
o.e.
Note, do not be concerned if the f(x) = … is missing, mark the expression.
(b)
Note: If there are multiple attempts with no clearly designated answer, Diagram 1 takes precedence if not
crossed out. If Diagram 1 is not used, or is crossed out and no clear statement of answer is given, then an
answer in the answer space takes precedence over an answer on Figure 3.
M1: Correct shape, a ∩shape parabola in any position. Be tolerant with the shape, must at least touch the x
axis.
A1: Correct shape and position, a ∩shape parabola passing through O and the +ve x axis to the right hand
side of x = 4. The graph must clearly extend into quadrants 3 and 4, not stop at the xaxis.
Ignore any coordinates on the graph, and don’t be concerned if the intersection in quadrant 3 is not
shown.
(c)
B1ft: Sets
1.2 (8
x
−
x
)
=
thei
r
0.2
x x
2
(
−
4
)
. It is for a ''correct'' ft equation on their part (a) as long as their
answer to part (a) is a cubic expression.
M1: Valid attempt to solve an equation of the form 1.2 (8
x
−
x
)
=
cubic
with their cubic (and their quadratic
(
)
1.2
x
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