Mathematics · Oct / Nov 2025 · Paper 11 · Question 5
YMA01
As printed in the paper
The question as text
5. y C l P R M x O Figure 1 In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[3]
2
2
2
−
+
−
+
±
±
2
x
16
x
50
2
x
8
x
25
2(
x
...)
...
2
2(
x
−
4)
+
18
Notes
(a)
2
2
2
B1:
Writes
2
x
−
16
x
+
50
in the form
2(
x ±
...)
±
...
or e.g.
±
±
2 (
x
...)
...
. Allow e.g. 16
8
2
2
M1: For
x −
±
or e.g.
−
±
−
or
−
for −4
2(
4)
...
2 (
x
4)
...
4
2
2
x −
+
A1:
For
2(
4)
18
x −
2
+
=
Condone any spurious “ = 0” e.g.
2(
4)
18
0
5(b)[5]
18
Eqn of l is
y
=
''
''
x
4
or
d
y
2
y
=
2
x
−
16
x
+
50
=
4
x
−
16
d
x
18
At M
−
=
=
=
=
4
x
16
0
x
4,
y
18
y
x
4
9
41
2
−
+
=
2
−
+
=
or e.g.
2
+
=
2
x
16
x
50
x
4
x
41
x
100
0
2
x
50
x
2
2
2
(
)(
)
4
x
−
41
x
+
100
=
0
x
−
4
4
x
−
25
=
0
x
=
...
25
x =
o.e.
4
Notes
(b)
18
c
=
= −
M1:
Deduces that equation of l is
y
''
''
x
o.e. Follow through on
y
''
''
x
from (a)
4
b
Alternatively finds the coordinates of M (p, q) by using calculus and forms the equation
2
=
−
+
= ±
M1: Equates
y
2
x
16
x
50
and their l of the form y
mx
and collects terms to form a 3 term
=
2
−
+
quadratic equation. Condone a miscopy of
y
2
x
16
x
50
as long as the intention is clear.
A1:
Correct 3 term quadratic equation. The terms do not have to all be on the same side but must be
2
collected so e.g.
2
x
+
50
=
41
x
is acceptable.
2
If the terms are all one side, the “= 0” may be implied by their attempt to solve.
ddM1: Correct non calculator method of solving their 3TQ.
Depends on both previous method marks.
Do not allow for quoting the quadratic formula quoted followed by values e.g.
2
−±
b
b
−
4
ac
25
2
4
x
−
41
x
+
100
=
0
x
=
=
, 4
2
a
4
2
41
±
41
−××
4 4 100
2
But allow e.g.
4
x
−
41
x
+
100
=
0
x
=
2 4
×
Do not allow obvious use of a calculator where the factors do not correspond with their 3TQ
25
25
2
(
)
e.g.
4
x
−
41
x
+
100
=
0
x
−
x
−
4
=
0
x
=
, 4
4
4
41
25
2
(
)(
)
or e.g.
2
x
−
x
+
50
=
0
4
x
−
25
x
−
4
=
0
x
=
, 4
2
4
25
2
(
)(
)
−
+
=
−
−
=
=
or e.g.
4
x
41
x
100
0
4
x
25
4
x
16
0
x
, 4
4
25
25
2
(
)
but allow e.g.
4
x
−
41
x
+
100
=
0
4
x
−
x
−
4
=
0
x
=
, 4
4
4
2
−
+
=
(
−
)(
−
)
=
(
−
)(
−
)
=
and allow e.g.
4
x
41
x
100
0
4
x
25
4
x
16
4
x
25
x
4
0
x
25
25
A1:
x =
only. If x = 4 is also seen, it must clearly be rejected or the
x =
clearly selected.
4
4
Ignore any attempts to find y.
9
5(c)[2]
≤
2
−
+
≥
y
2
x
16
x
50,
y
'' 2 x
''
9
2
≤
≤
−
+
and 0
≤
x
≤
4
Fully defined region E.g.
2 x
y
2
x
16
x
50
Notes
Some candidates are answering (c) at the bottom of the first page of the question so please check there.
Allow use of f (x) for y.
2
R
≤
2
x
−
16
x
+
50
Do not allow use of R for y e.g.
or e.g. y > line
B1ft: Uses C and l with inequalities in the correct directions and ignore any other inequalities for this
mark.
Allow strict or nonstrict inequalities and condone a mix of both for this mark.
Follow through on their equation for l and their possibly incorrect completed square form for C.
Examples:
9
9
<
2
−
+
2
y
2
x
16
x
50
and
y
≥
''
x
''
or e.g.
"
2 x
"
<
y
<
2
x
−
16
x
+
50
2
B1ft: Correct and fully consistently defined region and no incorrect inequalities.
Follow through on their equation for l and their possibly incorrect completed square form for C.
<
<
>
For the restriction on x, allow e.g. 0
x
4
or written separately as e.g.
x
0,
x
25
<
<
a <
0
x
a
where
4
If extra inequalities are given that do not contradict then they can be ignored (see example below)
Examples of full marks in (c):
9
2
y
≤
2
x
−
16
x
+
50,
y
≥
''
x
''
,
x
≥
0,
x
≤
4
2
9
2
<
<
−
+
<
<
"
2 x
"
y
2
x
16
x
50, 0
x
4
9
<
<
2
−
+
<
<
>
"
2 x
"
y y
,
2
x
16
x
50, 0
x x
,
6.25,
y
0
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