Mathematics · Oct / Nov 2022 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(i)[7]
u
=
x
−
=
x
e
3
e
d
x
(
)
3
(
)
2
3
x
4
u
+
3
4
u
+
3
4e
d
x
=
d
u
=
d
u
(
)
u
x
u u
+
3
e
−
3
=
4
u
2
+
24
u
+
36 d
u
u
36
2
=
u
+
+
u
=
u
+
u
+
u
4
24
d
2
24
36ln
u
u
=
4
2
=
2
u
+
24
u
+
36ln
u
u
=
2
=
72
+
36ln 2
3
3
x
x
x
7(ii)[5]
3e cos2 d
x x
=
e sin 2
x
−
e sin 2 d
x x
2
2
3
3
3
x
x
x
=
x
−−
x
+
x x
e sin 2
e cos2
e cos2 d
2
4
4
Collect terms 15
3
3
x
x
x
e cos2 d
x x =
x
+
x
e sin 2
e cos2
4
2
4
6
3
(
)
x
x
x
x x =
x
+
x
+
c
3e cos2 d
e sin 2
e cos2
5
5
(i) The 4 must be present to score the A marks. No misreads
B1: States or uses d
u
o.e.such as d
x
1
x =
x
=
This may be seen within the integrand.
e
d
d
u
u
+
3
M1: A ''correct'' attempt to write the integrand in terms of u.
(
)
3
3
x
P u
±
3
4e
Look for
d
x
=
d
u
which does not need to be simplified.
u u
(
)
x
±
3
e
−
3
Condone a missing du but dx is M0
(
)
u
+
2
4
3
d
u
. Condone a missing du (cannot be dx) .
A1: Correct
u
dM1: Integrates by attempting to multiply out and integrating each term.
P
Look for two ''correct'' terms, one of which must be
d
u
=
P
ln
u
.
u
(
)
2
P u
±
3
It is dependent upon having achieved
d
u
u
2
u
+
u
+
u
A1: Correct
2
24
36ln
o.e.
M1: Uses limits 2 and 4 within their attempted integral and subtracts . Condone poor attempts at the
integration
Alternatively converts their answer in u back to x's using the correct substitution and uses the
given limits.
u =
x
−at some point.
A1: 72
+
or 36ln 2
+
36ln 2
72
(ii) Condone missing dx's here
x
x
x
x x
=
A
x
±
B
x x
M1: Attempts to integrate by parts
3e cos2 d
e sin 2
e sin 2 d
3
3
x
x
x
A1:
3e cos2 d
x x
=
e sin 2
x
−
e sin 2 d
x x
Condone missing dx's
2
2
The 3 may have been ''removed''/factorised out and replaced at the final answer.
If so you can this would be awarded at the sight of a correct answer.
dM1: Attempts to integrate by parts again. See below of what to look for
x
x
x
x
Look for
3e cos2 d
x x
=
A
e sin 2
x
±
B
e cos2
x
±
C
e cos2 d
x x
ddM1: Dependent upon both previous M's
x
It is awarded for attempting to collect both terms in
e cos2 d
x x
6
3
(
)
x
x
x
x x =
x
+
x
+
c
condone the omission of the + c.
A1:
3e cos2 d
e sin 2
e cos2
5
5
Alternative method for (ii) . This can be scored in a similar way to the above.
x
x
x
(ii)
x x
=
x
+
x x
3e cos2 d
3e cos2
6e sin 2 d
−
x
x
x
=
x
+
x
x x
3e cos2
6e sin 2
12e cos2 d
x
x
x
15
e cos2 d
x x =
6e sin 2
x
+
3e cos2
x
6
3
(
)
x
x
x
x x =
x
+
x
+
c
3e cos2 d
e sin 2
e cos2
5
5
x
x
x
1stM1:
3e cos2 d
x x
=
P
e cos2
x
±
Q
e sin 2 d
x x
x
x
x
x
x x
=
P
x
±
Q
x
±
R
x x
2ndM1:
3e cos2 d
e cos2
e sin 2
e cos2 d
(
)
(
)
x
−
y
=
x
+
y
=
3
25 and
1
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.