Mathematics · Oct / Nov 2025 · Paper 31 · Question 8
YMA01
As printed in the paper
The question as text
8. y P C2 C1 R Q x O Figure 2 Figure 2 shows a sketch of the graph C1 with equation y = –2x 3 + 5x 2 + 4x – 3 and a sketch of the graph C2 with equation y = a + |5x + b| where a and b are constants. Graphs C1 and C2 intersect at point P, point Q and point R, as shown in Figure 2. Given that P has coordinates (–2, 25)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[2]
*
8(b)[3]
9
=
a
+
10
+
b
a
=
... or
b
=
....
= −
a
=
7,
b
8
8(c)[2]
8 , 7
5
8(d)[6]
15
−
5
x
= −
2
x
3
+
5
x
2
+
4
x
−
3
2
x
3
−
5
x
2
−
9
x
+
18
=
0
2
x
3
−
5
x
2
−
9
x
+
18
=
(
x
+
2)(2
x
2
−
9
x
+
9)
3
(ignore
x =
)
2
x
2
−
9
x
+
9
=
0
x
=
3
2
3 15
,
2
2
(a)
M1:
Substitutes ( 2, 25)
into
. You must either see a correct
−
y
=
a
−
(5
x
+
b
)
unsimplified equation without the modulus before the answer or
indication of correct branch with attempt to substitute.
Note they may use the cubic and substitute
to achieve the 25.
For methods via squaring send to review if you are unsure if the marks are
deserved.
A1*:
Achieves the given answer with no errors seen following a correct
unsimplified equation.
If they attempt both equations, they must clearly indicate the choice of
solution.
(b)
M1:
Substitutes (2, 9) into
=
+
+
and solves simultaneously with the
y
a
(5
x
b
)
equation from (a) to find a value for a or a value for b. May be implied by a
correct value for a or b.
A1:
One of
=
= −
a
7,
b
8
A1:
Both
a
=
7,
b
= − only
8
Note if there are attempts at squaring they must reach only the correct
values. If unsure, use review.
(c)
8 , 7
B1ft: One of the coordinates of
. Follow through their a and b for this mark
5
ie one of
8 , 7
B1:
Aceept as x = …, y = …
5
(d)
M1:
Sets their
equal to
− and collects terms to
y
=
15
−
5
x
−
2
x
3
+
5
x
2
+
4
x
3
achieve a cubic. Condone a miscopy of the cubic if the intention is clear.
dM1:
Attempts to find a quadratic factor (by division or equating coefficients).
It is dependent on the previous method mark. For factorisation expect to
)(
)
)(
)
see (
or via division (
x
±
2
2
x
2
+
......
±
9
x
±
2
2
x
2
+
...
x
+
...
Valid non
calculator method must be seen, solutions which go direct to a
“factorised” form with no intermediate quadratic factor seen score dM0.
Note they may use another linear factor to divide through by, the scheme
applies the same way, but must be one of the correct factors.
A1:
+ (or correct factor for their linear term).
2
x
2
−
9
x
9
ddM1: Attempts to solve their quadratic via a noncalculator method, leading to
a value for x. It is dependent on the previous two method marks. Valid
noncalculator method must be seen to award this mark.
M1:
Proceeds to find both coordinates of Q, accepting answers where the x
value was found via a calculator. It is not dependent on the previous Ms
but must attempted an initial equation using the cubic and either branch,
set up to show what has been solved on the calculator. Correct answer
with no working at all seen will score no marks.
3 15
A1:
,
(oee) only. A0 if a second set of coordinates is given. Allow for
2
2
solutions where the cubic or quadratic was solved via calculator.
Allow listed as x =…, y = … and isw but must clear clearly seen as the correct
coordinates.
Attempts at squaring, use review if you are unsure how to score unsuccessful
attempts.
NB
If no quadratic factor or suitable method is seen to solve the cubic then
maximum of three mark is available, M1dM0A0ddM0M1A1.
If the quadratic factor is reached but the quadratic is solved via calculator
a maximum of 5 marks may be scored, M1dM1A1ddM0M1A1.
M0dM0A0ddM0M1A1 is also possible if they give the initial equation and
proceed to give the solutions without first gathering terms.
A noncalculator approach must be shown for full marks.
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