Mathematics · May / June 2024 · Paper 21 · Question 5
YMA01
As printed in the paper
The question as text
5. In this question you must show detailed reasoning.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[4]
x
−
y
=
x
3
−
3
x y
+
3
xy
−
y
3
(
)
3
3
3
2
2
x
−
y
>
x
−
y
−
3
x y
+
3
xy
>
0
(
)
3
xy y
−
x
>
0
As x and y are positive numbers (
)
−
>
0
>
*
y
x
y
x
5(b)[2]
Chooses a suitable counter example. E.g.
x
=
3,
y
= −
1
Shows that (
)
3
( )
3
(
)
3
3
−−
1
>
3
−−
1
as 64
>
28
(BUT 1
−<
3
)
(a)
Attempts (
)
3
3
2
2
3
M1:
x
−
y
=
x
±
Ax y
±
Bxy
±
y
but may be unsimplified. Condone at most
one slip in an index as long as the middle terms have at least xy in them.
A1:
Correct simplified inequality, cubed terms cancelled, other terms may not all be
gathered.
E.g
−
3
2
+
3
2
>
0
o.e. Terms need not be all on one side, e.g.
3
2
>
x y
xy
xy
dM1: Takes out a common factor of kxy or divides by kxy. Must be clear division or stated
as dividing, but dividing and rearranging to the given answer in one step from
2
2
−
3
x y
+
3
xy
>
0
to y > x is dM0 if method/intermediate step is not stated.
A1*: Requires
correct working (no incorrect lines) before achieving y>
x
reasoning given in the correct/appropriate place,(
)
(
)
o e.g. as x and y are positive so
3
−
>
0
−
>
0
xy y
x
y
x
Alt
(
)
(
2
2
3
3
2
M1:
Attempts (
)
3
(
)
(
)
x
−
y
=
x
−
y
x
+
axy
+
y
and
x
−
y
=
x
−
y
x
+
bxy
may be unsimplified
(
)
2
2
A1:
Both expressions correct (
−
)
3
=
(
−
)
−
2
+
x
y
x
y
x
xy
y
and
(
)
3
3
2
2
x
−
y
=
(
x
−
y
)
x
+
xy
+
y
dM1: Takes to one side, takes out a common factor (
−
)
(dividing by this term is M0)
x
y
and cancels square terms.
A1*: Requires
correct working (no incorrect lines) before achieving y>
x
reasoning given in the correct/appropriate place,(
)
(
)
o e.g. as x and y are positive so
3
xy y
−
x
>
0
y
−
x
>
0
(b)
M1:
Chooses a suitable counter example. E.g.
=
3,
= −
1
x
y
Any positive x and negative y will work. Substitution is not needed for this mark.
Note that if they choose x and y both positive it is M0.
A1:
Shows that the result is not true for their values. As a minimum accept substitution
into both sides of the inequality.
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