Mathematics · Oct / Nov 2022 · Paper 41 · Question 4
YMA01
As printed in the paper
The question as text
4. 1 g (x) = 4 −x 2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[5]
=
4
−
x
=
4
2
1 ...
−
−
x
2
4
1
3
1
3
5
−
×−
−
×−
×−
1
−
2
3
2
2
2
2
2
M1, A1
1
2
1
1
1
1
2
2
2
2
1
−
x
= + −
1
−
x
+
−
x
+
−
x
4
2
4
2
4
3!
4
1
1
1
3
5
2
4
6
=
+
x
+
x
+
x
2
16
256
2048
4
−
x
2
4(b)[1]
x <
2
4(c)[2]
Substitutes an appropriate value of x in both sides with LHS in terms of
3
2048
or 3543
=
E.g. with
x =
1
3
2048
1181
(a) Note that the first mark in (a) is M1 on epen. We are now marking it as B1
B1: Correct constant term/ factor.
1
−
or 1
2
Accept as
4
2 as the constant term or as the factor with the bracket starting (1....
2
M1: Correct attempt at the third or the fourth term of binomial expansion of form
1
+
ax
Look for a correct binomial coefficient with a correct power of x
1
1
1
1
1
−
×−
−
1
−
×−
−
1
×−
−
2
2
2
2
2
2
2
3
±
±
2
2
ax
ax
where a could be
E.g.
or
2
3!
1
A1: Correct unsimplified
For example
1
1
1
1
1
1
−
×−
−
1
−
×−
−
1
×−
−
2
−
2
2
2
2
2
2
3
1
2
1
1
1
1
2
2
2
2
1
−
x
= + −
1
−
x
+
−
x
+
−
x
4
2
4
2
4
3!
4
1
1
3
5
2
4
6
A1: Two correct and simplified terms of
+
x
+
x
+
x
2
16
256
2048
1
1
3
5
2
4
6
A1:
+
x
+
x
+
x
. Accept as a list, ignore any terms with indices greater than
2
16
256
2048
these.
(b)
B1: States a correct range. E.g
x <
o.e.
2
or 2
−<
x
<
2
(c) For example: See **
M1: Substitutes an exact value of x into both sides of their expansion that enables
3 to be found
.
1
1
1
3
5
2
4
6
E.g.
x =
1
=
+
×
1
+
×
1
+
×
1
2
16
256
2048
4 1
−
2
2
Note that
x = −and
x = all work with the same results
1
1
1
=
Alternatively correctly solves
3
o.e. and substitutes this into their expansion
4
−
x
2
2048
or 3543
A1: With
x =
=
2048 . This can only be scored from a correct expansion
1
3
1181
**There are lots of values of x that work, all more difficult than the one above.
8
1
3
11
1
x
2
x
2
E.g
=
=
and
=
=
3
3
2
3
4
−
x
2
4
−
x
2
In both these cases the fractions become much more difficult to find.
8
43
11
55687
x =
2
=
x =
2
=
For
3
and for
3
3
27
3
55296
Alt I (a) By direct expansion
1
3
1
3
−
×−
−
×−
1
−
1
3
1
2
2
5
2
2
2
1
−
1
−
−
2
2
2
2
=
4
−
x
=
4
2
+ −
×
4
2
−
x
+
×
4
2
−
x
+
2
2
3!
−
x
2
4
1
−
2
+
B1: For
4
2x
M1: A correct attempt at the third or fourth terms condoning sign slips on the
−
A1: Correct and unsimplified expansion. See above
1
1
3
5
2
4
6
+
x
+
x
+
x
A1: Two correct and simplified terms of
2
16
256
2048
1
1
3
5
2
4
6
A1:
+
x
+
x
+
x
. Accept as a list
2
16
256
2048
Alt II (a) By difference of two squares
1
1
(
)
1
(
)
1
−
−
−
x
x
2
=
2
−
2
×
2
+
2
=
4
+
......
4
−
x
2
1
−
2
B1: For
4
+
1
1
−
−
1
(
)
1
(
)
1
x
2
x
2
−
−
x
2
x
2
M1: For correctly writing
=
2
−
×
2
+
=
... 1
−
×
1
+
=
2
2
4
−
x
2
with a correct attempt at the third of fourth term in either expansion, followed by an attempt
to combine both the expansions.
2
4
6
A1: One correct term in
x
,
x
or
x
1
1
3
5
+
x
2
+
x
4
+
x
6
A1: Two correct and simplified terms of
with no terms in
2
16
256
2048
3
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