Mathematics · Oct / Nov 2020 · Paper 41 · Question 7
YMA01
As printed in the paper
Show the answer and mark scheme
7(i)[6]
∫
5
3
x
Chooses a suitable substitution for
d
x
2
x −
1
1
E.g.
u
=
2
x
−
1
E.g.
u
=
2
x
−
1
d
x
d
x
1
u
=
2
x
−⇒
1
=
u
u
=
2
x
−⇒
1
=
d
u
d
u
2
∫
∫
3
x
3
x
d
x
d
x
2
x
−
1
2
x
−
1
∫
3
3
∫
3
1
3
1
−
u
2
+
d
u
=
=
u
2
+
u
2
d
u
2
2
4
4
1
3
3
1
1
3
=
u
3
+
u
=
u
2
+
u
2
2
2
2
2
9
3
1
3
3
1
1
3
Limits
=
u
3
+
u
=
...
=
u
2
+
u
2
=
...
2
2
2
2
1
1
1
3
1
3
1
3
3
1
1
3
×
3
3
+
×
3
−
×
1
3
+
×
1
=
16
×
9
2
+
×
9
2
−
× +
1
×
1
=
16
2
2
2
2
2
2
2
2
7(ii)[6]
6
x
2
−
16
3
.....
=
+
(
)(
)
x
+
1
2
x
−
3
6
x
2
−
16
2
1
=
3
+
−
(
)(
)
x
+
1
2
x
−
3
x
+
1
2
x
−
3
∫
6
x
2
−
16
∫
2
1
d
x
=
3
+
−
d
x
(
)(
)
x
+
1
2
x
−
3
x
+
1
2
x
−
3
1
=
3
x
+
2ln
x
+
1
−
ln 2
x
−
3
+
c
M1 A1ft A1
2
(i)
B1: States or uses a suitable substitution. That is one that will work.
M1: Attempts to change the integral in x to an integral in 'u'
Expect to see all aspects changed including the dx
∫
For
u
=
2
x
−expect to see
1
Au
2+ d
B u
dM1: For correct method of integration. Allow for correct powers. Dependent upon previous M
A1: See scheme. Correct integration which may be left unsimplified
M1: Uses correct limits for their valid substitution.
A1: CSO 16. Look for 18 2 or such before the 16.
16 with no working scores no marks. This is evidence of use of a calculator
(ii) Note that epen scoring is M1 M1 A1 M1 A1 A1
B1: For a quotient of 3.
(
)(
)
(
)
(
)
This can be scored by setting
6
x
2
−
16
=
A x
+
1
2
x
−
3
+
B x
+
1
+
C
2
x
−
3
and reaching
or by division with 3 in the ''correct place''
M1: A full attempt to find A, B and C.
(
)(
)
(
)
(
)
Either sets
6
x
2
−
16
=
A x
+
1
2
x
−
3
+
B x
+
1
+
C
2
x
−
3
condoning slips leading to values of A, B and C
or divides and uses partial fractions on the remainder to form an expression of the correct form.
6
x
2
−
16
2
1
A1: Correct partial fractions seen or implied. (
=
3
+
−
)(
)
(
)
(
)
x
+
1
2
x
−
3
x
+
1
2
x
−
3
∫
∫
B
C
M1: Two of
A
→
...
x
,
d
x
→
...ln(
x
+
1)
d
x
→
...ln(2
x
−
3)
(
x
1
)
(
2
x
3
)
+
−
∫
∫
B
C
C
A1ft: Two correct of A
→
Ax
d
x
→
B
ln(
x
+
1)
and
d
x
→
ln(2
x
−
(
)
(
)
x
+
1
2
x
−
3
2
∫
1
1
Please note that answers such as
d
x
→
ln(4
x
−
6)
are correct
(
)
2 2
x
−
3
2
1
1
A1:
3
x
+
2ln
x
+
1
−
ln 2
x
−
3
+ but condone
c
3
x
+
2ln(
x
+
1)
−
ln(2
x
−
3)
+ The +c is required.
c
2
2
You should isw after a correct answer.
..................................................................................................................................................................................
Note that it is possible to do part (i) by parts. This does not satisfy the demand of the question but can score
some marks
∫
3
x
∫
d
3
2
1
3
2
1d
B1:
x
=
x
x
−−
x
−
x
2
x
−
1
∫
3
x
3
(
)
M1:
d
x
=
ax
2
x
−−
1
b
2
x
−
1
FYI: Correct when
a
2
=
2
x
1
−
∫
5
3
x
third M1 (5th mark on epen): applies limits to a valid expression
d
x
=
ax
2
x
−−
1
b
2
x
−
1
1
∫
5
3
x
5
second A1 (6th mark on epen)
3
o.e.
(
)
(
) (
)
d
x
=
3
x
2
x
−−
1
2
x
−
1
2
=
45
−
27
−
3 1
−
=
16
2
x
−
1
1
1
to score on eepen 110011
.....................................................................................................................................................................
Note that in part (ii) some candidates will write
4
3
2
2
4
+
3
λ
= +
2
2
µ
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