Mathematics · May / June 2024 · Paper 11 · Question 6
YMA01
As printed in the paper
The question as text
6. y C l Q P R O x Figure 3 In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. Figure 3 shows • the line l with equation y – 5x = 75 • the curve C with equation y = 2x 2 + x – 21 The line l intersects the curve C at the points P and Q, as shown in Figure 3.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[4]
y
−
5
x
=
75,
y
=
2
x
2
+ −
x
21
2
2
21
5
75
x
+
x
−
=
x
+
or e.g.
x
2
−
2
x
−
48
=
0
2
x
2
−
4
x
−
96
=
0
or
2
y
−
75
y
−
75
y
=
2
+
−
21
5
5
or e.g.
2
160
5175
0
2
y
2
−
320
y
+
10350
=
0
y
−
y
+
=
(
)(
)
x
2
−
2
x
−
48
=
0
x
−
8
x
+
6
=
0
x
= −
6, 8
x
= −
6
y
=
45
or
x
=
8
y
=
115
(
)
(
)
P
−
6, 45
and
Q
8, 115
6(b)[3]
e.g.
2
2
21
,
y
−
5
x
75
,
y
0
,
x
−
3.5
y
x
+
x
−
where 3.5
−
a
<
3
x
a
(
or
where
15,
3.5
3
)
a
x
b
a
−
−
b
Notes:
(a) Note that solutions relying on calculator technology are not acceptable
so a complete method must be shown.
M1: Equates the line and the curve and rearranges to obtain a 3TQ in x
Alternatively eliminates x to obtain a 3TQ in y
dM1: Solves their 3TQ to obtain at least one value for x or y by an acceptable method e.g. factorisation,
formula or completing the square. They cannot just state the roots without seeing a correct line of
intermediate working.
(
)
2
( )(
)
4
±
−
4
−
4 2
−
96
We will condone e.g.
2
x
2
−
4
x
−
96
=
0
x
=
= −
6, 8
2 2
×
4
±
−
4
2
−
4 2
( )(
−
96
)
But e.g.
2
2
4
96
0
6, 8
scores M0 as it suggests an incorrect
x
−
x
−
=
x
=
= −
2 2
×
( )(
)
4
±
−
4
2
−
4 2
−
96
4
±
784
formula but allow recovery e.g.
2
x
2
−
4
x
−
96
=
0
x
=
=
= −
2 2
×
4
(
)
2
( )
4
±
−
4
−
4 2 ( 96
−
condone a missing trailing bracket e.g.
x
=
= −
6, 8
2 2
×
(
)(
)
If they factorise do not accept
2
x
2
−
4
x
−
96
=
0
x
+
6
x
−
8
=
0
x
= −
6, 8
Depends on the previous method mark.
dM1: Uses at least one x value to find at least one y value (or vice versa) using either of the given equations.
Depends on the FIRST method mark and depends on having solved a 3TQ in x or y.
May be implied by their values. You may need to check (for both equations).
A1: All correct. Must be correctly paired but ignore any labelling regarding P and Q, just look for the correct
paired values. This may be explicit as a coordinate pair or implied by e.g. x = 8, y = 115.
Condone coordinates written without brackets e.g. 6, 45
−
Depends on all previous marks so 1011 is not possible.
Correct answers with no working scores no marks.
(b) Allow strict or nonstrict inequalities and allow a mixture for the first 2 marks.
M1: Obtains at least 2 of the required inequalities e.g. 2 of:
y
2
x
2
+
x
−
21,
y
−
5
x
75
,
y
0,
x
a
where 3.5
−
a
<
3
Allow equivalents e.g.
y
5
x
75
<
+
Allow e.g.
0
2
2
21
or e.g. 0
<
y
<
5
x
+
75
each of which would count as 2 correct
y
x
+
x
−
inequalities.
Do not allow e.g.
R
2
x
2
+
x
−
21
A1: Any 3 of the 4 required inequalities.
A1: All correct and consistent. For consistency, the inequalities for y must be either all strict or all nonstrict.
The restriction on x will depend on their chosen x value. For example
y
0,
y
5
x
+
75,
y
2
x
2
+
x
−
21,
x
<
0
scores full marks
y
0,
y
5
x
+
75,
y
2
x
2
+
x
−
21,
x
<−
3.5
scores M1A1A0
If there are extra incorrect inequalities then score A0
(
)
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