Mathematics · Oct / Nov 2025 · Paper 31 · Question 1
YMA01
As printed in the paper
The question as text
1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The functions f and g are defined by f (x) = ln (x 2 + 3) x ∈ g (x) = 3 + 5 x x ∈ x > –2 x + 2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
1(a)[1]
The answer is a drawing: see the picture of this mark scheme row.
1(b)[3]
3
−
2
x
g ( )
−
1
x
=
or
oe
−
x
5
x <
5
1(c)[2]
3
3
21
g(0)
=
f
=
ln
2
2
4
1(d)[3]
2
a
3+5e
=
4
5e
2
a
+
3
=
4e
2
a
+
8
e
2
a
=
...
e
2
a
+
2
2e
a
=
5
2
a
=
ln5
a
=
...
1 ln5
a =
2
(a)
B1:
o.e. Accept with y in place of f(x).
(b)
M1:
Attempts to rearrange to find the inverse. Score for achieving the form
±±
3
2
x
±±
3
2
y
= ±±
or
= ±±
. If they use
score for reaching
y
x
x
5
y
5
oe for y =….
−
3
2
x
A1:
g ( )
−
1
x
=
o.e. (e.g. see scheme). Allow with just
or y or even
x
−
5
B1:
x <
5
(c)
M1:
Attempts to find fg(0) . May find the function fg( )
x and substitute 0 into
this. Condone slips. May be implied by awrt 1.7 if no substitution is shown.
21
A1:
or exact simplified equivalent and isw.
ln
4
(d)
2
a
3+5e
M1:
Sets
=
4
and proceeds to make
e a (or condone
2
A
e a
2
e
2
a
+
2
They must achieve the 4 although this may appear later in working.
(
)
−
5
Alternatively, sets
−
=
−
−
=
−
=
=
g g(e )
1
2
a
g f
1
e
4
3
e
2
a
g (4)
1
−
1
dM1: Takes valid lns of both sides (must be lns of positive numbers) and
proceeds to find a value for a. It is dependent on the previous method
mark.
) 1 ln5
A1:
(
or exact equivalent and isw
a =
2
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