Mathematics · Oct / Nov 2022 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. y C O x P Figure 3 Figure 3 shows a sketch of the curve C with parametric equations π t 3 π x = 1 + 3 tan t y = 2 cos 2t – 6 The curve crosses the x-axis at point P, as shown in Figure 3.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[5]
t = π or x coordinate at P
For correct parameter
x =
4
4
y
d
y
−
t
d
t
4sin 2
d
=
=
x
d
x
d
2
t
3sec
d
t
(
)
2
2
8
Equation of tangent
y
−
0
= −
x
−
4
y
= −
x
+
3
3
3
k = −
6(b)[1]
1
3
6(c)[2]
−„
1
f
„
2
(a)
B1: Correct parameter or x coordinate at P.
The correct x coordinate
x =
x =
° )
4
(can even be scored following
45
y
d
M1: Attempts to find d
y
t
d
x using
. Condone poor differentiation.
x t
d
d d
y
−
t
d
4sin 2
A1: Correct
=
. You may see other forms for this following use of identities.
x
d
3sec
2
t
d
y
−
4sin 2
t
d
y
−
4sin 2
=
=
If you see
then only allow if they subsequently use this as
d
x
x
d
x
3sec
2
3sec
2
−
8sin cos
t
t
−
t
3
t
8sin cos
Other possible correct gradients are
and
3sec
2
t
3
ISW after you see a correct answer. Quite often we see candidates incorrectly adapting a
correct answer.
It is also possible for candidates to find d
y
t = π, d
x
t = π and divide the two
t at
''
''
t at
''
''
d
4
d
4
values.
t = π
dM1: Attempts to find the equation of the tangent at
4
It is dependent upon the previous M and use of (
)
t = π
4,0 and
4
Allow substitution into their adapted d
y
x
d
2
8
y
= −
x
+
A1:
3
3
(b)
k = −
B1:
1
3
(c)
M1: Either end correct. Allow strict inequalities here. E.g f < 2 scores M1 but f > 2 is M0
−
A1: 1
−„
f
„
2
o.e such as
1,2
and 1
−„
y
„
2
d
u
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