Mathematics · Oct / Nov 2020 · Paper 21 · Question 1
YMA01
As printed in the paper
Show the answer and mark scheme
1(a)[4]
10
2
3
x
10
x
10
x
10
x
2
−
=
2
10
+
2
9
−
+
2
8
−
+
2
7
−
+
...
4
1
4
2
4
3
4
Attempts the binomial expansion to get the third and/or fourth term with an acceptable
x
structure. The correct binomial coefficient must be combined with the correct power of 4
and the correct power of 2 but condone omission of brackets.
You can ignore the signs between the terms and allow the terms to be listed.
10 2
2
3
x
x
Allow for e.g.
±
8
±
or
±
10
C 2
7
±
but condone omission of brackets.
2
4
3
4
NB 10
C
45,
10
C
120
=
=
2
3
NB 10
C
=
10
C
and
10
C
=
10
C
2
8
3
7
Alternative:
x
10
x
10
10
x
10 9
×
x
2
10 9 8
× ×
x
3
2
−
=
2
10
1
−
=
2
10
1
−
+
−
+
−
+
...
4
8
8
2
8
3!
8
2
3
10 9
×
x
10 9 8
× ×
x
Score M1 for
or
2
10
...
±
−
+
...
2
10
...
±
−
+
...
2
8
3!
8
1024
−
1280x
=
1024 1280
−
x
+
720
x
2
−
240
x
3
720x2 or −240x3
720x2 and −240x3
Note that if any of the “−”’s are “+ −“’s then penalise once on the first occurrence
1024, 1280 ,720
x
x
2
,
240
x
3
Allow the terms to be listed e.g.
−
−
Apply isw once a correct answer is seen.
Ignore any extra terms
1(b)[3]
2
1
6
1
3
3
1
Correct expansion. May be implied by their
3
−
=
9
−
+
9
−
−
+
or
work to find the constant.
x
x
x
2
x
x
x
2
2
10
1
x
6
1
(
)
(
)
3
−
2
−
=
9
−
+
1024 1280
−
x
+
720
x
2
−
240
x
3
+
...
x
4
x
x
2
6
1
(
)
constant term
= ×
9 1024
−
−
1280
x
+
720
x
2
x
x
2
B
C
2
1
This mark depends on having obtained an expression of the form
A
+
+
for
3
−
x
x
2
x
and at least a 3term quadratic expression from part (a) so award for:
A
×
"1024"
+
B
× −
" 1280"
+
C
×
"720"
A, B, C nonzero.
Allow 1 sign error on their values.
May be seen as part of a complete expansion but there must be an attempt to calculate the
value of the constant term with the above conditions.
For reference, true value calculation is: 9216 + 7680 + 720
Correct value. Must be “extracted” if a
= 17616
complete expansion is found above.
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