Mathematics · Oct / Nov 2021 · Paper 31 · Question 5
YMA01
As printed in the paper
Show the answer and mark scheme
5(i)[4]
d
x
=
+c
(
)
3
(
)
2
2
x
−
3
2
x
−
3
4
∫
4
8
−
2
2
48
d
x
=
= −
+
2
=
(
)
3
(
)
2
25
25
2
x
−
3
2
x
−
3
2
2
(
)
(
)
∫
7
1
8
2
2
5(ii)[2]
x x
+
3
d
x
=
x
+
3
+
c
16
Alt(i)
Let
u
=
2
x
−
3
∫
8
1
2
(
)
×
2 u
d
= −
+c
u
3
u
2
4
5
∫
8
2
2
48
d
x
2
= −
= −
+
=
(
)
3
2
25
25
2
x
−
3
u
2
1
Alt(ii)
Let
u
=
x
2
+
3
(
)
7
8
(
)
∫
7
∫
u
u
1
8
2
2
x x
+
3
d
x
=
d
u
=
+
c
=
x
+
3
+
c
2
16
16
(i)
A
A
M1
Achieves
or equivalent or in the alternative method
(
)
2
2
x −
3
u
2
−
2
2
A1
Achieves
or in the alternative method
−
(which may be unsimplified
(
2
3
)
2
u
2
x
−
but the indices must be processed). There is no requirement for the +c
A
A
dM1
Substitutes 2 and 4 into
or equivalent or 1 and 5 into
and subtracts
(
)
2
x −
3
2
2
u
either way round. May be implied but M1 must have been scored.
48
A1
25 or 1.92 isw after a correct answer
(ii)
Achieves (
)
8
8
M1
k x +
2
3
or equivalent or in the alternative method k
u . Alternatively
multiplies out the expression and integrates achieving an expression of the form
±
...
x
16
±
...
x
14
±
...
x
12
±
...
x
10
±
...
x
8
±
...
x
6
±
...
x
4
±
...
x
2
(
)
1
8
16 x
2
3
c
A1
+
+
Must be in terms of x and the + c must be present
16
3
14
63
12
189
10
2835
8
1701
6
5103
4
2187
2
x
x
x
x
x
x
x
x
Allow
+
+
+
+
+
+
+
+
c or
16
2
4
2
8
2
4
2
simplified equivalent
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