Mathematics · May / June 2023 · Paper 31 · Question 3
YMA01
As printed in the paper
Show the answer and mark scheme
3(i)[2]
ln sin 3
2
x
=
×
2sin3
x
×
3cos3
x
=
6cot3
x
d
x
sin
2
3
x
(
)
(
)
(ii) (a)
d
6
5
3
x
2
−
4
=
36
x
3
x
2
−
4
d
x
(
)
(
)
1
5
6
2
2
x
3
x
−
4 d
x
=
3
x
−
4
(b)
36
2
(
)
(
)
2
1
1
1
5
6
2
2
( )
6
(
)
6
x
3
x
−
4 d
x
=
3
x
−
4
=
2
−
−
4
= −
112
36
36
36
0
0
Notes
(i)
(
)
d
1
2
M1
Attempts to differentiate a ln function. Award for
ln sin 3
x
=
×
...
where ... could be 1
d
x
2
sin 3
x
(
)
d
d
1
2
(
)
An alternative could be
ln sin 3
x
=
2ln sin3
x
=
(2
×
)
×
...
or
d
x
d
x
sin
3
x
d
1 cos6
2
−
x
=
ln
×
...
d
x
2
1 cos6
−
x
6
cos3
6
Accept also 6sin 6
x
A1
6cot 3x o.e. such as
or
or 6(tan3 )
x
− but not
1
6tan
−
1
3x
sin3
x
ta
n3
x
1 cos6
−
3sin 6
x
and isw after a suitably simplified answer.
sin 3
2
x
Constant terms must be gathered and no uncancelled common factors in numerator and denominator.
(ii) (a)
(
)
(
)
d
6
5
2
2
M1
Achieves
3
x
−
4
=
Ax
3
x
−
4
where A is a constant which may be 1.
d
x
(
)
(
)
d
6
5
2
2
A1
3
x
−
4
=
36
x
3
x
−
4
oe. Need not be simplified. Isw after a correct answer.
d
x
(ii) (b)
(
)
(
)
(
)
1
1
5
6
6
B1ft
x
3
x
2
−
4 d
x
=
3
x
2
−
4
o
r
3
x
2
−
4
following through on their (a) provided it is of
36
A
(
)
(
)
d
6
5
the form
3
x
2
−
4
=
Ax
3
x
2
−
4
This may arise from attempts via substitution and can be scored
d
x
from a restart if (ii)(a) was incorrect. Need not be simplified and isw if simplified incorrectly. Condone
notation errors such as unneeded integral signs mark the expression that is their attempt at the integration.
M1
Substitutes in both limits and subtracts (either way round) into an expression of the form
(
)
6
3
2
4
where D is a constant but allow slips such as a missing power if the intention is clear.
D
x −
Sight of the subtraction is sufficient. Implied by the correct answer for their integral if substitution
not seen. If using integration by substitution they must be substituting the correct limits for their
variable.
Note: Answer only with no working at all shown scores no marks. Correct integral must be seen.
Note: Attempts at integration by parts are unlikely to succeed, but if done correctly and achieve the
correct form of the answer may score the relevant marks.
Note (ii) may be completed by expansion.
(a)
12
10
8
6
4
2
M1 Requires expansion to form
ax
+
bx
+
cx
+
dx
+
ex
+
fx
+
g
followed by an attempt to
integrate each term (power decreased by 1)
11
9
7
5
3
A1 Requires correct derivative.
8748
x
−
58320
x
+
155520
x
−
207360
x
+
138240
x
−
36864
(b)
B1ft Correct answer from a restart, which may be via expansion
12
81
x
10
8
6
4
2
−
162
x
+
540
x
−
960
x
+
960
x
−
512
x
3(ii)[5]
ln sin 3
2
x
=
×
2sin3
x
×
3cos3
x
=
6cot3
x
d
x
sin
2
3
x
(
)
(
)
(ii) (a)
d
6
5
3
x
2
−
4
=
36
x
3
x
2
−
4
d
x
(
)
(
)
1
5
6
2
2
x
3
x
−
4 d
x
=
3
x
−
4
(b)
36
2
(
)
(
)
2
1
1
1
5
6
2
2
( )
6
(
)
6
x
3
x
−
4 d
x
=
3
x
−
4
=
2
−
−
4
= −
112
36
36
36
0
0
Notes
(i)
(
)
d
1
2
M1
Attempts to differentiate a ln function. Award for
ln sin 3
x
=
×
...
where ... could be 1
d
x
2
sin 3
x
(
)
d
d
1
2
(
)
An alternative could be
ln sin 3
x
=
2ln sin3
x
=
(2
×
)
×
...
or
d
x
d
x
sin
3
x
d
1 cos6
2
−
x
=
ln
×
...
d
x
2
1 cos6
−
x
6
cos3
6
Accept also 6sin 6
x
A1
6cot 3x o.e. such as
or
or 6(tan3 )
x
− but not
1
6tan
−
1
3x
sin3
x
ta
n3
x
1 cos6
−
3sin 6
x
and isw after a suitably simplified answer.
sin 3
2
x
Constant terms must be gathered and no uncancelled common factors in numerator and denominator.
(ii) (a)
(
)
(
)
d
6
5
2
2
M1
Achieves
3
x
−
4
=
Ax
3
x
−
4
where A is a constant which may be 1.
d
x
(
)
(
)
d
6
5
2
2
A1
3
x
−
4
=
36
x
3
x
−
4
oe. Need not be simplified. Isw after a correct answer.
d
x
(ii) (b)
(
)
(
)
(
)
1
1
5
6
6
B1ft
x
3
x
2
−
4 d
x
=
3
x
2
−
4
o
r
3
x
2
−
4
following through on their (a) provided it is of
36
A
(
)
(
)
d
6
5
the form
3
x
2
−
4
=
Ax
3
x
2
−
4
This may arise from attempts via substitution and can be scored
d
x
from a restart if (ii)(a) was incorrect. Need not be simplified and isw if simplified incorrectly. Condone
notation errors such as unneeded integral signs mark the expression that is their attempt at the integration.
M1
Substitutes in both limits and subtracts (either way round) into an expression of the form
(
)
6
3
2
4
where D is a constant but allow slips such as a missing power if the intention is clear.
D
x −
Sight of the subtraction is sufficient. Implied by the correct answer for their integral if substitution
not seen. If using integration by substitution they must be substituting the correct limits for their
variable.
Note: Answer only with no working at all shown scores no marks. Correct integral must be seen.
Note: Attempts at integration by parts are unlikely to succeed, but if done correctly and achieve the
correct form of the answer may score the relevant marks.
Note (ii) may be completed by expansion.
(a)
12
10
8
6
4
2
M1 Requires expansion to form
ax
+
bx
+
cx
+
dx
+
ex
+
fx
+
g
followed by an attempt to
integrate each term (power decreased by 1)
11
9
7
5
3
A1 Requires correct derivative.
8748
x
−
58320
x
+
155520
x
−
207360
x
+
138240
x
−
36864
(b)
B1ft Correct answer from a restart, which may be via expansion
12
81
x
10
8
6
4
2
−
162
x
+
540
x
−
960
x
+
960
x
−
512
x
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