Mathematics · Oct / Nov 2024 · Paper 11 · Question 2
YMA01
As printed in the paper
The question as text
2. 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
2(i)[3]
=
2
4
y
4
x y
16
=
a
27
+
4
2(ii)[4]
3
+
1
States or uses
27
=
3 3
3
−
1
16
Correct attempt at rationalising seen e.g.
×
−
3
+
1
3
1
Correct attempt to make ''a'' the subject and make progress
towards the form
p
3
+
q
e.g.
8 3
12
8 3
−
8
=
3
a
3
+
4
3
a
3
=
8 3
−
12
a
=
−
3 3
3 3
4
8
a = −
3
+
3
3
(i)
x or 1
− for 1
10
1
B1: For any one of 6,
y Allow
y
y may be seen separately, and allow for
one correctly simplified term even if a spurious extra term in the same variable is
present.
B1: For any two of the above as the only term in that variable within a single expression.
3
3
E.g.
2
x
10
×
is fine for B1B1 but
2
x
10
+
is B1B0.
y
y
10
6x
10
− must be one expression. Allow with y1 in the denominator.
1
B1: For
or
6x y
y
(ii)
B1: States or uses
. May be implied by working as long as it is seen before the
27
=
3 3
final answer.
M1: Correct attempt at rationalisation seen anywhere. This can be on any expression with a
two term surd expression in a denominator during their working. The denominator may
be simplified without the multiplication being seen (and need not be correct), but the
evidence of the process must be clear.
M1: Attempts to make ''a'' the subject and make progress towards the form
p
(
)
,p q ≠
0
must reach a two term expression for a (but may be over common
denominator), though the
27 may not have been simplified. E.g. accept for
8 3
−
12
a =
. Can be scored if a calculator was used to rationalise the surd expression
27
but there must be an intermediate step before their final answer to show the method.
4
8
A1cso:
a = −
3
+
but isw after a correct expression. Must have score all previous marks,
3
3
showing the relevant steps as outline above before the final answer is given but
condone if “invisible brackets” are recovered. Accept with terms in the other order, and
1 8
allow (
)
−
4 3
3
Note that there are many ways of attempting (b). The above scheme can still be applied
A common alternative method is;
(
)(
)
16
=
a
27
+
4
16
=
a
27
+
4
3
+
1
=
9
a
+
4
+
4 3
+
3 3
a
for B1
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.