Mathematics · Oct / Nov 2020 · Paper 41 · Question 2
YMA01
As printed in the paper
Show the answer and mark scheme
2(a)[4]
1
1
or 1
−
4
2
or
2
1
2
4
1
1
−
−
1
5
x
2
(4
−
5 )
x
2
=
1
−
2
4
1
3
−
× −
2
1
5
x
2
2
5
x
=
... 1
+ −
× −
+
× −
+
...
2
4
2!
4
1
5
75
2
=
+
x
+
x
+
...
2
16
256
2
1
5
75
+
kx
(
)
2
=
2
+
kx
+
x
+
x
+
...
3
2
16
256
(2
−
3 )
x
Compares x terms leading to k = .... E.g. 10
k
3
13
2(b)[2]
+
=
⇒
k
= −
16
2
10
20
75
5
13
49
2x terms leading to m = ... E.g.
2(c)[2]
Compares
m
=
+
× −
'
'
⇒
m
=
128
16
20
128
1
−
1
2
4
or
B1
Alt (a)
1
4
2
1
3
1
1
3
−
× −
5
1
2
−
−
−
2
2
−
(4
5 )
x
2
4
2
4
2
(
5
x
)
4
2
(
5
x
)
−
=
+ −
−
+
−
2
2
1
5
75
2
=
+
x
+
x
+
...
2
16
256
.
(a)
1
1
−
B1: For taking out a factor of
4
or 2
2
1
−+
For a direct expansion look for
or equivalent.
4
2
....
1
M1: For the form of the binomial expansion (
)
1
+
ax
−
where
a ≠ or 5
1
−
2
To score M1 it is sufficient to see either term two or term three. Allow a slip on the sign.
1
3
−
× −
1
2
2
±
So allow for either
(
)
or
(
)
2
−
±
ax
ax
2
2
1
3
3
−
× −
5
1 4
2
−
2
2 4
−
2
(
)
2
(
)
In the alternative version look for
−
−
5
x
or
−
5
x
condoning sign slips
2
2
1
x −
5
2
A1: Any (unsimplified) but correct form of the binomial expansion for .
1
−
4
Ignore the factor preceding the bracket for this mark
1
3
−
× −
2
1
5
x
2
2
5
x
Score for
1
+ −
× −
+
× −
o.e.
2
4
2!
4
1
3
1
1
3
−
× −
5
1
−
−
−
2
2
−
In the alternative version look for
(4
5 )
2
4
2
4
2
(
5
)
4
2
(
−
x
=
+ −
−
x
+
−
2
2
1
5
75
2
A1: cao
+
x
+
x
+
...
This must be simplified
2
16
256
)(
)
3
(
2
2
2
+
kx
P
+
Qx
+
Rx
+
...
= +
1
x
+
mx
10
(b)
M1: For a correct equation in k formed by comparing the x terms. It must lead to a value for k
3
Follow through on their expansion. So look for
Pk
+
2
Q
=
⇒
k
=
...
Condone slips, i.e copying errors.
10
3
Condone
Pkx
+
2
Qx
=
x
as long as it leads to a value for k
10
13
A1:
k = −
20
(c)
2x terms, following through on their expansion and their value for k leading to a
M1: Correctly compares the
value for m . Condone slips, i.e copying errors.
2
2
2
Look for
Qk
+
2
R
=
m
Condone
Qkx
+
2
Rx
=
mx
as long as it leads to a value for m
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