Mathematics · May / June 2024 · Paper 11 · Question 2
YMA01
As printed in the paper
Show the answer and mark scheme
2(i)(a)[1]
2
n
3
2
n
×
2
3
=
8
m
2(i)(b)[2]
n =
(
)
3
n
16
3
2
4
(
)
12
=
2
12
n
=
2
n
=
m
12
Notes:
(i)(a)
B1: For 8m (Condone 8M). Do not allow
32
m
.
(b)
16 n correctly as an expression involving a power of 2.
M1: For writing
3
Examples: (
)
(
)
(
)
(
)
(
)
3
n
6
n
4
12
n
2
4
, 2
2
, 2
3
n
, 2
n
,
2
12
, 2
12
n
A1: For
m (Condone
12
M )
12
Correct answer only scores both marks.
(
)
2(ii)[3]
x
3
−=
3
x
+
3
x
3
−
x
= +
3
3
x
3 1
−
= +
3
3
x
=
...
(
)
3
1
±
+
3
+
3
=
×
(
)
3
−
1
±
3
+
1
(
)
±
4 3
+
6
3
2 3
=
=
+
(
)
±
3 1
−
Notes:
In this part, the A1 depends on both M marks
(ii)
M1: Collects x terms to one side, factorises and makes x the subject
α
+
β
3 ,
It is for processing the given equation and reaching
x
,
, ,
0
=
αβγ
≠
γ
+
3
M1: Correct attempt to rationalise their denominator.
...
...
γ
−
3
Not formally dependent but requires
=
×
or equivalent.
γ
+
3
γ
+
3
γ
−
3
Sight of e.g. 3
+
3
3
+
1
×
is sufficient for this mark but may be implied by e.g.
3
−
1
3
+
1
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