Mathematics · May / June 2025 · Paper 41 · Question 4
YMA01
As printed in the paper
The question as text
4. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. x x 2 f(x) = 5 17 10 x > 1 x x x ( 1 )( 2 1 )
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[3]
+
−
=
+
−
+
+
+
−
e.g.
5 17
x
10
x
2
A
(2
x
1)(1
x
)
Bx
(2
x
1)
Cx
(1
x
)
A
=
...,
B
=
....,
C
=
....
5
4
8
A
=
5,
B
=
4,
C
=
8
f( )
x
=
+
+
x
(1
−
x
)
(2
x
+
1)
4(b)[5]
5
4
8
(
)
+
+
=
−
−
+
+
+
d
x
5ln
x
4ln 1
x
4ln 2
x
1
c
−
+
x
(1
x
)
(2
x
1)
4
5ln
x
−
4ln 1
−
x
+
4ln 2
x
+
1
2
3
(
)
=
−
+
−
−
+
=
+
5ln 4
4ln3
4ln9
(5ln 2
4ln1
4ln5)
5ln 2
4ln 5
(a)
+
−
=
+
−
+
+
+
−
M1:
Sets
5 17
x
10
x
2
A
(2
x
1)(1
x
)
Bx
(2
x
1)
Cx
(1
x
)
condoning slips on the coefficients of
+
−
5 17
x
10
x
2
only and finds values for A, B and C. The righthand side must be correct. Do not be
concerned how the values are formed, e.g. via substitution or simultaneous equations. Just look for a
suitable identity followed by values for A, B and C.
A1:
Two (out of three) correct values of
A
=
5,
B
=
4,
C
=
8
= or 5
4
8
but condone 5
4
A1:
Accept either
A
=
5,
B
=
4,
C
8
+
+
−
+
x
(1
−
x
)
(2
x
+
1)
x
(
x
−
1)
(2
correct (but not in the form requested).
(b)
Attempts to integrate their "5"
"4"
"8"
M1:
+
+
.
x
(1
−
x
)
(2
x
+
1)
Score for any one "5"
"4"
"8"
→
α
ln
x
→
β
ln 1
−
x
→
ln 2
x
+
1
x
(1
−
x
)
(2
x
+
1)
(Condone invisible modulus signs or invisible brackets if subsequent work implies them)
β
−
→
β
−
Condone
ln 1
x
ln
x
1
throughout (b)
7
7
7 ln(4
C =
→
x
+
2)
Be aware that if they incorrectly find C, e.g.
, then
is a correct follow through
+
2
(4
x
2)
4
−
−
+
+
A1ft: Two of 5ln
x
4ln 1
x
4ln 2
x
1
follow through their A, B and C.
Allow use of brackets instead of modulus signs or allow further work to imply them.
−
−
+
+
A1:
5ln
x
4ln 1
x
4ln 2
x
1
with or without the +c
Allow use of brackets instead of modulus signs or allow further work to imply them.
dM1: Substitutes in the limits, subtracts either way round and applies laws of logs correctly to achieve either an
3
p
ln2
term or a
q
ln 5
term.
It is dependent on the previous M and having integrated to a form
α
ln
x
+
β
ln 1
−
x
α
+
β
−
+
+
or
ln
x
ln
x
1
ln 2
x
1
This is most likely going to be scored via the ln2 term.
The term in ln2 must be found from
α
ln4
−
α
ln2
=
α
ln2
so the value of p and αmust be the same
3
The term in
ln 5
must be found from
β
ln3
+
2 ln3
−
ln5
3
Condone calculations such as
ln( 3)
−
−
ln( 5)
−
=
ln 5
3
+
A1:
CSO:
5ln 2
4ln 5
It cannot be scored following incorrect integration. ISW after sight of this
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