Mathematics · Oct / Nov 2023 · Paper 31 · Question 2
YMA01
As printed in the paper
The question as text
2. The function f is defined by f (x) = x 3 4 x ∈ , x ≠ 4 x
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[2]
9
"
" 3
+
9
2
ff(6)
=
f " "
=
=
...;
=
15
2
9
"
" 4
−
2
2(b)[3]
4
x
+
3
f
−
1
( )
x
=
1
x
−
x
∈
,
x
≠
1
2(c)[3]
2
x
+
3
a
+
3
+
E.g.
or
5
=
7
=
(
±
)
7
−
5
x
−
4
a −
4
x
2
−
22
x
+
23
=
0
x
=
..
.
or (
a
+
3)
= ±
( ) 2(
a
−
4)
a
=
...
(
)11 7 2
a =
+
oe
Notes
(a)
M1:
Substitutes
x =
6
into f and substitutes the result back into f to find a value for ff(6)
x
+
3
+
3
x
−
4
into ff( )x =
Alternatively substitutes
x =
6
condoning slips.
x
+
3
−
4
x
−
4
A1:
15
(b)
x
+
3
y
+
3
...
x
±
3
...
y
±
3
x
x
M1:
Changes the subject of
y
=
or
=
to
y
=
or
=
4
...
1
x
−
4
y
−
...
x
±
1
y
±
4
x
+
3
A1:
Achieves
f
−
1
( )
x
=
. Accept
or y =… instead of
f
1( )
x
=
...
.
f
−=
1
...
−
x
1
−
B1:
x ≠
1
. The omission of x∈
is condoned. Accept alternative notations.
(c) Note this is M1A1A1 on epen but is being marked as M1dM1A1.
x
3
2
+
+
M1:
Correct attempt to set up an equation in x or a. This will usually be
5
=
7
x
−
4
f( )
a
=
g (7)
may be used first. Score when a suitable equation is set up and allow if a minor
−
1
slip or miscopy is made if the intention is clear.
dM1: Proceeds to solve for x or a, e.g. forms a 3TQ and solves, or take the 5 across, square roots,
cross multiplies and makes a the subject, condoning e.g. sign slips when rearranging.
Alternatively, uses (
)
2
( )
f( )
x
+
5
=
7
f( )
x
= ±
2
and their part (b) leading to
(
)
"4 2
+
3"
a
=
f
−
1
" 2"
=
. In this method both M’s may be scored together.
" 2
−
1"
or 4 2
+
3
−
4 2
−
3
A1
11 7 2
+
Accept 11
+
98
or
. Common factors should be cancelled.
2
−
1
1
−
2
Ignore labelling of x or a. Condone e.g. if 11 7 2
−
is also given as a solution but A0 if any
other extra solutions are included.
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