Mathematics · Oct / Nov 2022 · Paper 31 · Question 2
YMA01
As printed in the paper
The question as text
2. The functions f and g are defined by 4 f(x) = 5 − x 0 x + 3 2 x π g(x) = 4sin 3 + x ∈ 6
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Show the answer and mark scheme
2(a)[2]
x <
3
„
f( )
x <
5
4
4
4
2(b)[3]
(i)
y
=
5
−
=
5
−
y
=
3
x
+
2
x
=
...
3
x
+
2
3
x
+
2
5
−
y
1
4
(
)
4
2
or 2
x
−
6
f
−
1
( )
x
=
−
2
oe such as
x −
3 5
−
x
15
3
3
15
3
x
−
−
(ii) Domain is 3
„
x <
5
π
π
π
2(c)[2]
(
)
( )
fg
−
π
=
f
4sin
−
+
=
f
4sin
−
=
f 2
=
...
3
6
6
4
9
=
5
−
=
6
+
2
2
Notes:
(a)
M1: One correct end of range, though allow with
„
or <
and … or > (in correct direction) at the respective
ends, and accept with y instead of f(x), or even with x for the M1.
A1: Correct range, allow with y instead of f(x) or with other correct set notation (but use of x is A0 unless in
formal set notation). Accept as two separate inequalities.
(b)(i)
M1: Attempts to make x or a swapped y the subject of the equation. Allow sign slips, but there should be a
correct order of operations.
A1: Correct rule, must be in terms of x, accept equivalents and isw after a correct answer is seen. Do not be
concerned with the lhs (accept y = … and even condone f(x) =...)
(b)(ii)
B1ft: Follow through on their answer to (a). Accept intervals or set notation answers. Do not accept e.g.
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