Mathematics · Oct / Nov 2024 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. Use the substitution u 3 x 1 to show that 5 9 x k 3 3 x x x A c d 2 1 3 x 1 where k and A are constants to be found and c is an arbitrary constant. (5)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
d
u
3
1
2
3
1
2
3
2
u
=
x
+
u
=
x
+
u
=
x
d
x
or
2 d
x
3
2
3
u
=
x
+
1
u
=
x
+
1
2
u
=
3
x
d
u
or
d
u
1
1
(
)
−
u
=
x
3
+
1
=
x
3
+
1
2
×
3
x
2
d
x
2
or
1
d
x
2 (
2
−
(
2
1)
u u
2
1)
x
=
u
−
3
=
−
3
d
u
3
9
x
5
9
x
5
2
u
(
)
e.g.
d
x
=
d
u
=
6
x
3
d
u
=
6
u
2
−
1 d
u
u
3
x
2
x
3
+
1
or
9
x
5
9
x
5
2
x
3
+
1
(
)
e.g.
d
x
=
d
u
=
6
x
3
d
u
=
6
u
2
−
1 d
u
3
x
2
x
3
+
1
x
3
+
1
3
1
(
)
(
)
(
)
6
u
2
−
1 d
u
=
2
u
3
−
6 (
u
+
c
)
=
2
x
3
+
1
2
−
6
x
3
+
1
2
(
+
c
)
1
1
(
)
(
)
(
) (
)
=
2
x
3
+
1
2
x
3
+
1
−
3
+
c
=
2
x
3
+
1
2
x
3
−
2
+
c
B1: Any correct equation involving d
u
x or d
x
u or du and dx separately. May be implied by
d
d
further work.
9
x
5
M1: Attempts to fully change
d
x
into an integral with respect to u. There must be no x’s
x +
3
1
anywhere. Must include an attempt at replacing dx to get du so M0 if there is no du term
present (unless implied by further work) or dx becomes du without an attempt at connecting
them first. (i.e. there must have been an attempt at d
u
x or d
x
u ). Condone no integral sign for
d
d
this mark.
(
)
6
u
2
−
1 d
u
A1: Reaches
o.e. Condone the omission of the integral sign but they must have a du
or implied by further work.
M1: Attempts to integrate their expression in terms of u only (usually
(
Au
2
+
B
)d
u
) and
attempts to reverse the substitution so that the expression is completely in terms of x (with or
without the constant of integration). For the integration look for a correct method to integrate
C
a term i.e.
...
u
n
→
...
u +
n
1
or B
→
Bu
or
u →
ln
u
at least once. Just seeing + c is not an
example of integrating.
A1: Correct expression in the required form including the + c
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