Mathematics · May / June 2021 · Paper 31 · Question 3
YMA01
As printed in the paper
Show the answer and mark scheme
3(i)[2]
∫
12
−
6
)(
)
d
x
=
+
c
(
)
2
(
2
x
−
1
2
x
−
1
(ii)(a)
(a) 4
x
+
3
5
4
=
−
x
+
2
x
+
2
(ii)(b)
∫
∫
4
x
+
3
5
(
)
(b)
d
x
=
4
−
d
x
=
4
x
−
5ln
x
+
2
+
c
x
+
2
x
+
2
∫
−
5
4
x
+
3 d
(
) (
)
−
20 5ln3
−
−−
32 5ln6
−
Hence
x
=
x
+
2
−
8
12
5ln 2
=
+
(i)
A
(
)
1
M1: Achieves (
or equivalent e.g.
A
2
x
−
1
−
)
2
x −
1
−
6
(
)
−
1
A1: Achieves (
or
−
6 2
x
−
1
. There is no requirement for the +c
)
2
x
−
1
(ii)(a)
M1: For using division or any other suitable method to find either A or B
4
Using division look for
x
+
2 4
x
+
3
Using an identity look for 4x + 3 = A(x + 2) + B followed by a value for either A or B
It may be implied by either A = 4 or B = −5
5
−
5
A1: Correct value for A and B or a correct expression
or
4
−
4
+
x
+
2
x
+
2
(ii)(b)
B
B
B
M1: Integrates
...ln
x
2 .
Condone
...ln
(
x
2
)
and condone
...ln
→
+
→
+
→
x
+
2
x
+
2
x
+
2
B
B
A1ft: Integrates
Condone
(
)
A
+
→
Ax
+
B
ln
x
+
2 .
A
+
→
Ax
+
B
ln
x
+
2
2
2
x
+
x
+
But do not condone missing brackets unless they are implied by subsequent work.
− into their integrated function containing ln
x +
2
dM1: Substitutes 5
− and 8
or ln(x + 2)
A1: 12
+
5ln 2
or equivalent exact answer such as 12
+
ln32
SC: For candidates who proceed to 12
5ln 2
+
or equivalent exact answer
(
)
(
)
(
)
(
)
via
−
20
−
5ln
−
3
−−
32
−
5ln
−
6
without sight of modulus signs withhold just the final A1
(
)
3(ii)[6]
∫
12
−
6
)(
)
d
x
=
+
c
(
)
2
(
2
x
−
1
2
x
−
1
(ii)(a)
(a) 4
x
+
3
5
4
=
−
x
+
2
x
+
2
(ii)(b)
∫
∫
4
x
+
3
5
(
)
(b)
d
x
=
4
−
d
x
=
4
x
−
5ln
x
+
2
+
c
x
+
2
x
+
2
∫
−
5
4
x
+
3 d
(
) (
)
−
20 5ln3
−
−−
32 5ln6
−
Hence
x
=
x
+
2
−
8
12
5ln 2
=
+
(i)
A
(
)
1
M1: Achieves (
or equivalent e.g.
A
2
x
−
1
−
)
2
x −
1
−
6
(
)
−
1
A1: Achieves (
or
−
6 2
x
−
1
. There is no requirement for the +c
)
2
x
−
1
(ii)(a)
M1: For using division or any other suitable method to find either A or B
4
Using division look for
x
+
2 4
x
+
3
Using an identity look for 4x + 3 = A(x + 2) + B followed by a value for either A or B
It may be implied by either A = 4 or B = −5
5
−
5
A1: Correct value for A and B or a correct expression
or
4
−
4
+
x
+
2
x
+
2
(ii)(b)
B
B
B
M1: Integrates
...ln
x
2 .
Condone
...ln
(
x
2
)
and condone
...ln
→
+
→
+
→
x
+
2
x
+
2
x
+
2
B
B
A1ft: Integrates
Condone
(
)
A
+
→
Ax
+
B
ln
x
+
2 .
A
+
→
Ax
+
B
ln
x
+
2
2
2
x
+
x
+
But do not condone missing brackets unless they are implied by subsequent work.
− into their integrated function containing ln
x +
2
dM1: Substitutes 5
− and 8
or ln(x + 2)
A1: 12
+
5ln 2
or equivalent exact answer such as 12
+
ln32
SC: For candidates who proceed to 12
5ln 2
+
or equivalent exact answer
(
)
(
)
(
)
(
)
via
−
20
−
5ln
−
3
−−
32
−
5ln
−
6
without sight of modulus signs withhold just the final A1
(
)
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