Mathematics · Oct / Nov 2024 · Paper 21 · Question 5
YMA01
As printed in the paper
Show the answer and mark scheme
5(a)[3]
6
6
(
)
6
6
5
(
)
4
(
)
2
2
+
ax
=
2
+
2
ax
+
2
ax
+
...
1
2
=
64 192
+
ax
+
240
a x
2
2
+
...
5(b)[4]
2
1
6
1
3
3
1
3
+
=
9
+
+
or
9
+
+
+
2
2
x
x
x
x
x
x
6
1
(
)
( )
f
x
=
9
+
+
64 192
+
ax
+
240
a x
2
2
+
...
=
...
x
x
2
Constant term is
9 64
×
+ ×
6 192
a
+
240
a
2
2
2
576 1152
+
a
+
240
a
=
576
1152
a
+
240
a
=
0
1152
+
240
a
=
0
a
=
...
24
a = −
5
(a)
M1:
Attempts the binomial expansion up to at least the second term to obtain the correct
structure for the 2nd or 3rd term i.e. the correct binomial coefficient (allow alternative
notation) with the correct power of 2 and the correct power of ax.
The binomial coefficients do not have to be evaluated but must be correct if they are.
If awarding this mark for the x2 term you can condone missing brackets e.g.
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