Mathematics · May / June 2023 · Paper 41 · Question 1
YMA01
As printed in the paper
Show the answer and mark scheme
1(a)[5]
3
1
1
−
2
3
8 1 2
(
)
−
−
2 x
=
−
x
2
4
(
) (
(
)(
) (
−
−−
1
−
−−
1
−−
2
3
3
3
3
3
3
(
)
(
)(
)
)
)
1 2
−
x
−
= + −
1
−
2
x
+
2
2
−
2
x
2
+
2
2
2
−
2
x
3
+
...
2
3
2
2!
3!
3
−
1
1
2
−
2 x
= +
8
24
x
+
60
x
2
+
140
x
3
+
...
4
1(b)[1]
n = 2
1(c)[3]
2
1
1
1
1
1
−
x
=
−
x
+
x
2
4
2
16
4
4
1
1
1
1
1
1
(
)
2
−
x
=
−
x
+
x
2
8
+
24
x
+
60
x
2
+
140
x
3
+
...
4
2
16
4
4
1
1
1
1
1
1
= ×
8
+
24
x
×
+
60
x
2
×
−×
8
x
−
24
x
×
x
+ ×
8
x
2
16
16
16
4
4
4
1
1
1
x
x
2
...
=
−
−
+
2
2
4
(a)
3
−
B1: Takes out the correct and simplified common factor to obtain (
)
8 1 ...
±
2
Implied by an expansion 8 + ...
3
M1: Attempts the binomial expansion of (
)
1
±
kx
−
to get the third and/or fourth term with an acceptable structure.
2
The correct binomial coefficient must be combined with the correct power of x.
(
) (
(
)(
) (
−
3
−−
3
1
−
3
−−
3
1
−−
3
2
2
2
)
2
2
2
2
)
3
Look for
±
kx
or
±
kx
with or without the brackets on the kx
2!
3!
Even allow with
k =
1
15
35
3
A1: Correct simplified or unsimplified expansion for (
)
−
1 2x
−
. (NB simplified is
1 3
+
x
+
x
2
+
2
2
A1: Two correct and simplified terms of
8
+
24
x
+
60
x
2
+
140
x
3
Ignore extra terms of x4 and above
A1: All correct and simplified
8
+
24
x
+
60
x
2
+
140
x
3
(b)
B1: Correct value, n = 2. Do NOT allow incomplete answers such as 4
2
(c) Hence
1
1
2
B1: Correct expansion of
−
2 x
, simplified or unsimplified
4
M1: Attempts correct strategy to find expansion.
Terms do not need to be collected. There may be other terms as well, for example terms in x3
Look for an attempt to find 4 or more terms from the following (condoning slips)
(
)(
)
2
2
3
2
2
2
A
+
Bx
+
Cx
E
+
Fx
+
Gx
+
Hx
+
...
=
AE
+
AFx
+
AGx
+
BEx
+
BFx
+
CEx
+
...
A1: Correct simplified expansion. Ignore extra terms of x3 and above
(c) Otherwise: Applies binomial expansion to
−
2 x
4
1
1
1
1 1 2
B1: Correct simplified expression
2
1
x
(
x
)
−
=
−
2
4
2
2
but allow (
) (
M1: Correct structure for 3rd term: (
) (
−
1
1
1
−
1
1
1
)
)
2
2
2
±
kx
2
2
2
−
2
x
with or without brackets. Even
2!
2!
allow with
k =
1
1
1
1
A1:
−
x
−
x
2
+
...
Ignore extra terms of x3 and above
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