Mathematics · Oct / Nov 2023 · Paper 11 · Question 2

YMA01

As printed in the paper

The question as text

2. Given that 1 16 2 a = x b = 64 x n where k and n are simplified constants. express each of the following in the form kx 1

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Show the answer and mark scheme

2(a)[1]
−
6
x−
3
−
7
expression but withhold this mark if there is a constant of integration. Isw once a correct
answer is seen. Do not accept
15
2
6
−
3
7
0
15
2
6
x−
3
7
x
−
x
−
x
or e.g.
x
+ −
−
M1:
Decreases the power by 1 on one of the terms of their changed function. Do not allow this
→
0
mark to be scored for a constant of integration in (a)
in (b)
18
18
A1:
30
x
+
18
x−
4
or simplified equivalent e.g.
30x
+
. Condone e.g.
30x
1
+
. Isw once a
x
4
x
4
correct answer is seen. Withhold this mark if there is a constant of integration and it has not
already been penalised in (a).
Allow this mark to be scored following
15
2
−
6
x−
3
±
7
in (a). Also allow full marks to be
x
scored if they have no constant.
− and (b) 30
x + will score M1A1A0 M1A0
6
Note: (a)
15
2
−
6
x−
1
7
x
2a
1
8 x
b
3
1
256 x
2
4
4
−
−
3
1
1
16
3
1
3
2
×
=
=
16
−
2
x
x
2
x
2 64
8
x
(a)
1
1
1
x but
B1:
8 x or simplified equivalent e.g. 0.125x or
8 x Accept e.g.
1
is B0.
±
x
8
8
Do not withhold this mark unless it is clear that they intend to write 1
8x
(b)
3
3
1
3
x
2
B1:
256 x or simplified equivalent e.g.
0.00390625x condone
2
. Isw after a correct
2
256
answer is seen.
1
1
or 1
(
)
3
Do not accept e.g.
x or
3
256 x x as n is not a simplified constant.
x
256
256
3
1
Do not accept
x ÷
256
or
− (not of the required form)
2
3
256x
2
1
Do not withhold this mark unless it is clear that they intend to write
3
256x
2
(c)
M1:
Attempts to use the index laws and proceeds to either
4
4
−
3
3
3
3
−
...x
2
or
...x
2
8
2
8
2
−×
−
−
...
x
3
...
x
3
(or
...
x
3
÷
...
x
3
)
...x−
2
1
16
x where
n
n ≠−
2
16
Allow
x to score M1.
2
16
Be aware that incorrect understanding of indices such as
is M0A0
=
16x
−
2
x
A1:
16x− (from a correct method)
2
(Correct answer with no incorrect working seen can score M1A1). Isw after a correct
16
answer is seen.
x is A0
2
2(b)[1]
M1:
Decreases the power by 1 on one of the terms of their changed function. Do not allow this
→
0
mark to be scored for a constant of integration in (a)
in (b)
18
18
A1:
30
x
+
18
x−
4
or simplified equivalent e.g.
30x
+
. Condone e.g.
30x
1
+
. Isw once a
x
4
x
4
correct answer is seen. Withhold this mark if there is a constant of integration and it has not
already been penalised in (a).
Allow this mark to be scored following
15
2
−
6
x−
3
±
7
in (a). Also allow full marks to be
x
scored if they have no constant.
− and (b) 30
x + will score M1A1A0 M1A0
6
Note: (a)
15
2
−
6
x−
1
7
x
2a
1
8 x
b
3
1
256 x
2
2(c)[2]
4
4
−
−
3
1
1
16
3
1
3
2
×
=
=
16
−
2
x
x
2
x
2 64
8
x
(a)
1
1
1
x but
B1:
8 x or simplified equivalent e.g. 0.125x or
8 x Accept e.g.
1
is B0.
±
x
8
8
Do not withhold this mark unless it is clear that they intend to write 1
8x
(b)
3
3
1
3
x
2
B1:
256 x or simplified equivalent e.g.
0.00390625x condone
2
. Isw after a correct
2
256
answer is seen.
1
1
or 1
(
)
3
Do not accept e.g.
x or
3
256 x x as n is not a simplified constant.
x
256
256
3
1
Do not accept
x ÷
256
or
− (not of the required form)
2
3
256x
2
1
Do not withhold this mark unless it is clear that they intend to write
3
256x
2
(c)
M1:
Attempts to use the index laws and proceeds to either
4
4
−
3
3
3
3
−
...x
2
or
...x
2
8
2
8
2
−×
−
−
...
x
3
...
x
3
(or
...
x
3
÷
...
x
3
)
...x−
2
1
16
x where
n
n ≠−
2
16
Allow
x to score M1.
2
16
Be aware that incorrect understanding of indices such as
is M0A0
=
16x
−
2
x
A1:
16x− (from a correct method)
2
(Correct answer with no incorrect working seen can score M1A1). Isw after a correct
16
answer is seen.
x is A0
2

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