Mathematics · May / June 2024 · Paper 21 · Question 1
YMA01
As printed in the paper
Show the answer and mark scheme
1(a)[3]
x
x
x
x
6
6
2
6
3!
6
3
7
2
3
= −
1
x
+
x
−
x
2
18
3
7
2
3
(
10
3
)
1
1(b)[2]
x
+
−
x
+
x
−
x
2
18
7
53
3x =
10 1 3
×+ ×−
=
Coefficient of
18
6
(a)
M1:
For an attempt at the binomial expansion. Score for a correct attempt at any of term 2,
3 or 4.
1
1
1
2
1
3
9
9
9
Accept sight of
C
±
6 x
or
C
±
6 x
or
C
±
6 x
condoning the omission
1
2
3
of brackets.
Coefficients may appear from Pascal's triangle but should be correct for the term.
3
7
2
3
−
+
−
A1:
For any two simplified terms of
x
x
x
2
18
3
7
2
3
1
−
+
−
A1:
For
x
x
x
ignoring terms with greater powers. Allow as a list, allow
2
18
3
+ −
etc but the fractions must be simplified. Accept 1x2. ISW following a correct
2 x
expression (e.g. if multiply through by 2). If given as decimals they must be exact
(recurring if appropriate).
(b)
M1:
For attempting to find 10
3
combined to a single term for their
b
±
c
)(
)
(
2
3
10
+
3 1
+
+
+
+
x
ax
bx
cx
cao for 53
53
3
A1:
6 but condone
6 x and isw. Accept embedded in an expansion with other
terms as long as simplified to correct single x3 term. If given as a decimal must be
exact recurring.
10 2
5
2
and
+
9
= −
80
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