Mathematics · May / June 2023 · Paper 11 · Question 1
YMA01
As printed in the paper
The question as text
1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. Solve the inequality 2 – 3x + 7 4x + 9 4x (4)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4
x
2
−
3
x
+
7
4
x
+
9
(
)(
)
4
7
2
...0
4
1
2
..
.0
..
.
x
2
−
x
−
x
+
x
−
x
=
or
(
)
(
)
2
( )(
)
−−
7
±
−
7
−
4 4
−
2
4
7
2
...0
...
x
2
−
x
−
x
=
x
=
2 4
×
or
2
2
7
1
7
7
1
4
2
−
7
−
2
...0
4
2
−
−
...0
4
−
−
−
...0
=
...
x
x
x
x
x
x
4
2
8
8
2
1 ,2
x = −
4
1
"
" ,
"
2"
x
−
x
4
1
2
x
−
or
x
oe
4
Notes:
M1: Gathers terms to one side and solves their 3TQ which is not
4
3
7
0
to obtain at least
x
2
−
x
+
=
one critical value for x. Allow by factorisation, the quadratic formula or completing the square. Do
not be concerned about the inequality symbol for the first two marks and the “…0” can be implied.
1
(
)
Note that
4
2
−
7
−
2
...0
+
−
2
...
0
=
...
scores M0
x
x
x
x
x
4
For completing the square, the conditions for their
4
x −
2
7
x
−
2
...0
are as follows:
(
)
Either
4
x
2
−
7
x
−
2
...0
4
x
2
+
bx
+
c
...0
then as general guidance
2
7
Or
4
x
2
−
7
x
−
2
...0
2
x
−
±
q
±
2...0
x
=
...
4
A1: Both correct critical values. Allow unsimplified but must be single fractions e.g. allow
These values may be seen embedded in their final answer but the first M1 must have been scored.
M1: Chooses the outside region for their 2 critical values. Allow with < and > for this mark.
1
1
1
Condone poor notation e.g.
"
−
"
x
"2"
or e.g.
"
−
"
>
x
>
"2"
or e.g.
"2"
<
x
<−
"
4
4
4
1 ,
A1: Correct solution only – depends on all previous marks. Allow e.g.
x
−
x
2
;
4
1
1
1
1 and
)
)
)
−
2
;
−
∞
,
−
,
2,
∞
;
−
∞
,
−
o
r
2,
∞
;
−∞
,
−
and 2,
∞
x
x
4
4
4
4
1
1
1
)
Do not allow
;
;
,
2,
−
x
2
x
−
∩
x
2
−∞
−
∩
∞
4
4
4
2 16
Allow unsimplified constants but must be single fractions e.g. allow
−
,
as above.
8
8
1 ,
Apply isw if possible if there is no contradiction so e.g. correct work leading to
x
−
x
4
1
followed by
−
2
can score full marks. If several different answers are offered, mark the
x
4
final answer.
Particular cases and how to mark them:
Case 1
4
2
−
3
+
7
4
+
9
4
2
−
7
−
2
0
x
x
x
x
x
1
1
= −
, 2
−
,
2
x
x
x
4
4
Scores M0A0 M1A0
Case 2
4
x
2
−
3
x
+
7
4
x
+
9
4
x
2
−
7
x
−
2
0
1 ,
x
−
x
2
4
Scores M0A0M0A0
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