Mathematics · May / June 2021 · Paper 41 · Question 6

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As printed in the paper

The question as text

6. y R O x Figure 3 Figure 3 shows a sketch of the curve C with parametric equations π x = 2 cos 2t y = 4 sin t 0  t  2 The region R, shown shaded in Figure 3, is bounded by the curve, the x‑axis and the y‑axis. π ∫ 4

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Show the answer and mark scheme

6(a)[6]
Area R =
y
t
=
4sin
t
×−
4sin 2 d
t t
d
t
= −∫
2
32sin
t
cos d
t t
π
∫
0
∫
2
4
2
π
= −
32sin
t
cos d
t t
=
32sin
t
cos d
t t
* A1*
and
y
=
0
⇒=
t
0
⇒ Area
x
=
0
⇒=
t
4
π
0
4
π
32
32
2 2
8 2
3
4
(ii)
sin
Area =
t
=
×
=
3
3
8
3
0
2
x
y
2
Attempts to use
cos2
t
= −
1 2sin
t
⇒
= −
1 2
2
4
y
=
8
−
4
x
Range is 0
f
4
(a)(i)
M1: Attempts to multiply y by d
x
t to achieve an integrand of the form
±
k
sin sin 2
t
t
d
∫
d
x
∫
(
)
( )
(
)
Look for area R =
y
d
t
=
4 sin
t
×±
k
sin 2
t
d
t
but condone a missing 4 or dt
d
t
∫
∫
d
x
(
)
(
)
A1: Correct integrand. Achieves area R =
y
d
t
=
4sin
t
×−
4sin 2
t
d
t
d
t
Condone a missing dt and do not be concerned by the limits. Allow unsimplified.
= ±∫
2
dM1: Substitutes sin 2
→ seen or implied
A
sin
t
cos d
t t
t
=
2sin cos
t
t
Dependent upon previous M. Condone a missing dt and do not be concerned by the limits.
π
∫
2
0
→∫
= −∫
4
2
2
A1*: Achieves area
=
y x
d
32sin
t
cos d
t t
32sin
t
cos d
t t
π
0
0
4
This is a given answer so you must see
evidence of the dt in at least one line other than the given answer
0
−∫π
→∫
... dt
correct application of the limits as seen above with as a minimum
4
(a)(ii)
3
M1: Integrates to a form
A
sin
t
with some attempt to apply the limit(s)
You may see a substitution
u
=
sin
t
which is fine. Just look for
Au with some attempt to apply the
3
adapted limits
A1: Correct answer 8 2
3 o.e. following correct algebraic integration.
Cannot be scored without the M mark.
(b)
2
M1: Attempts to use a double angle formula of the form
cos2
t
= ± ±
1
2sin
t
with the parametric equations to
get a Cartesian equation.
If the parametric equations are substituted into the given form of the answer y
=
ax
+
b
only scored when the double angle formula is used
2
x
y
A1: Any correct unsimplified equation
= −
1 2
2
4
On the alternative method, this A mark is scored when the candidate writes down
a
+
2
b
=
0.
−
4
b
=
16
⇒
a
=
...,
b
=
...
A1:
y
=
8
−
4
x
or
y
=
−
4
x
+
8
ONLY
(c)
B1: Range is 0
f
4
( )
and [
]
Allow other acceptable forms such as 0
y
4
,
0
f
x
4
0,4
, [
Examples of unacceptable forms are 0
x
4
0,4
)
4
4
1
=
4
4
2
4
2
2
2
6
4
6(b)[3]
Attempts to use
cos2
t
= −
1 2sin
t
⇒
= −
1 2
2
4
y
=
8
−
4
x
6(c)[1]
Range is 0
f
4
(a)(i)
M1: Attempts to multiply y by d
x
t to achieve an integrand of the form
±
k
sin sin 2
t
t
d
∫
d
x
∫
(
)
( )
(
)
Look for area R =
y
d
t
=
4 sin
t
×±
k
sin 2
t
d
t
but condone a missing 4 or dt
d
t
∫
∫
d
x
(
)
(
)
A1: Correct integrand. Achieves area R =
y
d
t
=
4sin
t
×−
4sin 2
t
d
t
d
t
Condone a missing dt and do not be concerned by the limits. Allow unsimplified.
= ±∫
2
dM1: Substitutes sin 2
→ seen or implied
A
sin
t
cos d
t t
t
=
2sin cos
t
t
Dependent upon previous M. Condone a missing dt and do not be concerned by the limits.
π
∫
2
0
→∫
= −∫
4
2
2
A1*: Achieves area
=
y x
d
32sin
t
cos d
t t
32sin
t
cos d
t t
π
0
0
4
This is a given answer so you must see
evidence of the dt in at least one line other than the given answer
0
−∫π
→∫
... dt
correct application of the limits as seen above with as a minimum
4
(a)(ii)
3
M1: Integrates to a form
A
sin
t
with some attempt to apply the limit(s)
You may see a substitution
u
=
sin
t
which is fine. Just look for
Au with some attempt to apply the
3
adapted limits
A1: Correct answer 8 2
3 o.e. following correct algebraic integration.
Cannot be scored without the M mark.
(b)
2
M1: Attempts to use a double angle formula of the form
cos2
t
= ± ±
1
2sin
t
with the parametric equations to
get a Cartesian equation.
If the parametric equations are substituted into the given form of the answer y
=
ax
+
b
only scored when the double angle formula is used
2
x
y
A1: Any correct unsimplified equation
= −
1 2
2
4
On the alternative method, this A mark is scored when the candidate writes down
a
+
2
b
=
0.
−
4
b
=
16
⇒
a
=
...,
b
=
...
A1:
y
=
8
−
4
x
or
y
=
−
4
x
+
8
ONLY
(c)
B1: Range is 0
f
4
( )
and [
]
Allow other acceptable forms such as 0
y
4
,
0
f
x
4
0,4
, [
Examples of unacceptable forms are 0
x
4
0,4
)
4
4
1
=
4
4
2
4
2
2
2
6
4

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