Mathematics · May / June 2022 · Paper 31 · Question 2
YMA01
As printed in the paper
The question as text
2. The functions f and g are defined by − x f(x) = 5 x ∈ , x ≠ – 2 x + 3 2 3 x ∈ g(x) = 2x – 7
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Show the answer and mark scheme
2(a)[2]
5 "3"
2
−
fg(5)
=
f(3)
=
,
=
3 "3" 2
×
+
11
2(b)[3]
5
−
2
x
f
−
1
( )
x
=
3
x
+
1
1
,
x
∈
x
≠−
3
2(c)[4]
5
−
1
a
=
2(
a
+
3)
−
7
3
×+
1
2
a
5
−
1
a
=
2(
a
+
3)
−
7
4
a
2
−
a
−
2
=
0
3
×+
1
2
a
) 1
±
33
(
a
=
8
(a)
M1
Correct order of operations applying g before f on 5
5
−
(2
x
−
7)
Allow an attempt to substitute
x =
5
into
fg( )
x
=
condoning slips. They must
3(2
x
−
7)
+
2
proceed to find a value.
This may be completed in two steps so allow for an attempt at g(5)
=
A
→
f( )
A
=
...
Condoning arithmetical slips and bracket errors/omissions.
2
or exact equivalent. Accept 0.18
A1
but not eg 0.18 or 0.18….
11
(b)
5
−
x
...
±
2
y
5
−
y
...
±
2
x
x
=
→
y
=
M1
Changes the subject of
y
=
→
x
=
or
3
y
2
3
x
...
3
x
+
2
±
3
y
±
...
+
±
±
5
−
2
x
2
x
−
5
5
f
−
1
( )
x
=
f
−
1
( )
x
= −
f
−
1
( )
x
=
A1
or
or exact equivalent expressions such as
3
1
3
1
x
+
x
−
3
.
5
−
2
x
Accept
f
−
1
:
x
→
Condone
f
−=
1
...
(or
f
−=
1
y
=
...
) but do not allow just
y =
...
3
x
+
1
f
−
1
:
y
=
...
1
B1
x ≠−
. The omission of x∈
is condoned.
3
5
2
1
−
t
Note: It is also acceptable to define
f −in any variable eg as
1
f
−
1
( )
t
=
t ≠−
as long as the
3
t
+
1
3
variable is used consistently to score M1A1B1. If another variable is used other than x it must be
fully defined eg
f
−
1
( )
t
=
...
not just
f
−=
1
...
(c)
5
−
1
1
1
"5" "2"(2
−
a
−
1)
f
−
1
(2
a
1)
B1ft
Sets
a
=
2(
a
+
3)
−
7
or their
=
−
=
3
×+
1
2
a
a
"3"(2
a
−
1)
+
"1"
a
(may be in terms of x). Follow through on their
f
−
1
( )
x
.
May be implied by later work if they
have invisible brackets.
M1
They must start with an equation of one of these forms (which may be in x):
1
A
±
5
−
1
a
=
Da
+
E
or equivalent so typically look for
a
=
2(
a
+
3)
−
7
1
3
×+
2
1
B
×
±
C
a
a
Aa
±
1
(may be manipulated to
=
Da
+
E
)
B
±
Ca
1
Aa
±
B
=
or equivalent
a
Ca
±
D
Rearranges their equation of an allowable form to a 3TQ. They must be attempting to
1
=
solve
f
g(
a
3)
so it cannot be awarded for an attempt to solve ( )
f
a
g( )
a
or
+
=
a
1
1
f
=
g
a
a
Do not be concerned with the mechanics of their arrangement and condone invisible
brackets.
A1
4
2
2
0
or equivalent equation involving 3 terms. Eg
4
2
2
(may be in terms
a
−
a
−
=
a
−
a
=
of x). May be implied by awrt
±
0.843
and the = 0 may be implied by their attempt to solve
their quadratic equation.
) 1
±
33
1
±
(
a
=
x
=
A1
or exact equivalent. Do not accept rounded decimals. Condone
8
8
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