Mathematics · May / June 2022 · Paper 41 · Question 5
YMA01
As printed in the paper
The question as text
5. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[4]
−
1 1
π
1
=
2sin
u
p
=
sin
=
2
6
d
x
x
=
2sin
u
=
2cos
u
oe
d
u
3
x
+
2
6sin
u
+
2
6sin
u
+
2
d
x
=
2cos
u u
d
=
2cos
u u
d
3
3
3
(
)
(
)
(
)
4
−
x
2
4
−
4sin
2
u
4cos
2
u
2
2
2
π
12sin
u
2
3
1
6
=
+
d
u
=
sec tan
u
u
+
sec
2
u
d
u
*
8cos
2
u
4cos
2
u
2
2
0
5(b)[4]
3
1
3
1
sec tan
u
u
+
sec
2
u
d
u
=
sec
u
+
tan
u
2
2
2
2
π
3
1
"
"
3
1
3
1
π
π
6
sec
u
+
tan
u
=
sec "
"
+
tan "
"
−
sec0
+
tan0
=
...
2
2
2
6
2
6
2
2
0
3
3
7 3
3
7 3
9
−
=
3
+
−
=
−
=
6
2
6
2
6
Notes
(a)
π
B1:
p
=
Allow if seen anywhere, even in (b). p = 30 is B0.
6
d
x
M1:
x
=
2sin
u
=
±
...cos
u
or any rearrangement of this equation.
d
u
M1:
Full substitution from an integral in terms of x to an integral in terms of u and uses the identity
sin
2
u
+
cos
2
u
= in the denominator. Do not be concerned with the limits for this mark.
1
A1*:
Achieves given answer include du (with their p) with no errors and at least one intermediate step with
the fractional power simplified. Condone missing du in intermediate lines.
(b)
3
1
M1:
sec tan
u
u
+
sec
2
u
d
u
=
...sec
u
+
...tan
u
2
2
3
1
A1:
sec
u
+
tan
u
ignore any constant c
2
2
M1:
Depends on having one term of the correct form, attempts to substitute in their p (≠ 1) and 0, subtracting
either way round. The substitution must be seen or clearly implied, e.g. by correct values for each term
in an intermediate step before the answer (allowing missing 0’s).
7 3
3
or exact equivalent eg 7 3
−
9
A1:
−
Allow if p = 30º was used.
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