Mathematics · Oct / Nov 2024 · Paper 41 · Question 3
YMA01
As printed in the paper
The question as text
3. y O x Figure 1 Figure 1 shows a sketch of the curve C with parametric equations π π 3 x 3sin θ y 1 cos2 θ θ 2 2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[3]
d
x
o.e.
x
=
3sin
3
θ
=
9sin
2
θ
cos
θ
d
θ
d
2sin 2
y
−
θ
=
d
x
9sin
2
θ
cos
θ
d
y
−
4sin
θ
cos
θ
4 cosec
=
= −
θ
d
x
9sin
2
θ
cos
θ
9
3(b)[3]
3
π
d
y
8
1
1
θ
=
= −
"
",
x
=
3
=
...,
y
= +
1
=
...
6
d
x
9
2
2
3
8
3
y
−
= −
x
−
2
9
8
16
x
+
18
y
−
33
=
0
3(c)[3]
x
2
3
x
e.g.
x
=
3sin
θ
sin
θ
=
3
3
y
= +
1
cos2
θ
= + −
1 1
2sin
2
θ
=
2
−
2 3
3
2
2
x
x
*
3
3
(
)
3
y
=
2
−
2
2
=
2
−
y
8
x
2
=
9 2
−
y
3
3
q = 3
Alt(c)
(
)
(
)
(
)
(
)
3
e.g.
8
x
2
=
72sin
6
θ
,
9 2
−
y
3
=
9 2
−
cos2
θ
−
1
3
=
9 1
−
1 2sin
−
2
θ
(
)
3
72sin
6
θ
= ×
9 8sin
6
θ
8
x
2
=
9 2
−
y
*
q = 3
(a)
B1: Correct expression for d
θ which may be part of their expression for d
x
y
x . Condone poor
d
d
labelling/notation provided the intention is clear. Must be in terms of θ
1 cos2
−
θ
May use identities before differentiating using the product rule e.g.
sin
θ
2
x . They must be attempting d
y
d
θ
M1: Correct method for d
y
o.e. condoning copying slips and
θ×
d
d
d
x
condoning poor differentiation, labelling /notation provided the intention is clear.
A1: Achieves d
y
4 cosec
with no errors but condone invisible brackets and poor notation
= −
θ
d
x
9
4
. Note d
y
being recovered. Allow equivalent fractions to
−
x = may be seen in an earlier line
9
d
of their solution which is acceptable.
Alt (a) Implicit differentiation – note that some candidates may find the cartesian equation
and differentiate implicitly or explicitly.
B1: A correctly differentiated cartesian equation. (Do not be concerned with where the cartesian
equation has come from for this mark)
d
y
e.g.
16
x
= ××−
9 3 ( 1) (2
×
−
y
)
2
×
d
x
M1: Substitutes in for x and y into their differentiated equation and rearranges to achieve an
expression for d
y
x in terms of θ. Condone poor differentiation.
d
A1: Achieves d
4 cosec
y
with no errors but condone invisible brackets and poor notation
= −
θ
d
x
9
4
being recovered. Allow equivalent fractions to
−
. They must have shown correctly how
9
they found the cartesian equation. i.e. they cannot just use the given result in (c).
(b)
M1: Attempts to find d ,
y x
π
and y when
. Must achieve values for all three which may be
θ=
d
6
x
implied by further work. Condone slips but if values are only stated then their d
y
x must be
d
correct for their k (look for 2× their k) and one of x or y must be correct.
16
x
Note that they may use their d
y
x in terms of x and y
= −
"
"
by either finding the
27(2
−
y
)
2
d
differentiated equation in (a) or by converting in (b).
dM1: Complete method for the tangent using their values. It is dependent on the previous method
mark. Condone one sign error when substituting in the coordinates. If they use y
=
mx
+
they must proceed as far as
c =
...
A1: 16
or any integer multiple of this equation. Must have = 0.
x
+
18
y
−
33
=
0
(c)
M1: A complete method to eliminate θ. Any identities used must be correct when substituted in.
Do not allow the parametric equations to be expressed as θ = and equated. May also be
awarded if seen in (a).
A1*: Correct proof with no errors including brackets although condone missing trailing brackets if
it does not affect the expression. Must be seen in (c).
B1: Correct value of q. Do not accept
x = Can be scored for the domain 3
−„
x
„
3
which must
3
be in terms of x
Alt(c)
M1: A complete method to express lhs and rhs both in terms of sin θ only or both in terms cos 2θ
only. Any identities used must be correct when substituted in. They cannot proceed from
8
x
2
=
72sin
6
θ
9(2
−
y
)
3
=
72sin
6
θ
without an intermediate stage but condone
A1*: Fully correct work showing lhs = rhs and gives a conclusion. There should be no errors
including brackets but condone missing trailing brackets if it does not affect the expression.
B1: As above in the notes for (c)
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