Mathematics · May / June 2022 · Paper 11 · Question 6
YMA01
As printed in the paper
The question as text
6. 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
6(a)[2]
(
)
2
x x
3
−
6x
−
3
x
2
=
5
0
2
x −
4
12
x
2
−
3
x
2
−
50
=
0
2
x
4
−
15
x
2
−
5
0
=
0
*
CSO
(
)(
)
2
x
2
+
5
x
2
−
10
=
0
x
2
=
...
6(b)[5]
So
x =
2
10
(
)
3
y
=
10
−
6
10
=
...
x
=
10
,
y
=
4
10
one solution pair is
x
=
10
,
y
=
4
10
x
= −
10, y
= −
4
10
Solutions are
and
CSO
(a)
M1: Attempts to substitute their expression for y from rearranging the second equation into the first
equation. More difficult approaches are possible, but all should lead to an equation in just x
E.g 1 substituting their expression for y from rearranging the first equation into the second equation
E.g. 2 multiplying the second equation by 2x and subtracting
A1*: Correct work leading to the given answer. Expect to see correct bracketing etc and at least one
intermediate line with the two x2 terms initially uncollected. See scheme
(b) This is a non calculator question so look for evidence of calculations
2x .
M1: Uses a non calculator method to solve the given quadratic in
Allow factorisation, the quadratic formula or completing the square.
u
x
2
You may see a substitution, say
=
. In this case it is for solving to find u
y
=
x
2
x
=
x
2
Condone substitutions
or
if it is clear what is meant.
(
15
)
(
15
)
2
4 2
50
−−
±
−
−××−
5
2
E.g.
2
x
4
−
15
x
2
−
50
=
0
x
=
=
10,
−
scores 0 unless recovered to
x =
10
4
2
x =
10
A1: A correct solution for x or x2 e.g.
x =
2
10
,
u =
10
x = ±
10
,
x =
10
etc but not x = 10
Candidates cannot just write down
2
x −
4
15
x
2
−
50
=
0
x
= ±
10
1
1 1
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