Mathematics · Oct / Nov 2023 · Paper 41 · Question 5
YMA01
As printed in the paper
The question as text
5. y y = x P C O x Figure 2 Figure 2 shows a sketch of the curve C with equation 3 2 2 y x 4 x y k where k is a positive constant greater than 1
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[5]
y
3
−
x
2
+
4
x y
=
k
2 d
y
y
3
3
y
→
d
x
2 d
y
2
4
x y
→
8
xy
+
4
x
d
x
d
d
y
y
y
3
−
x
2
+
4
x y
2
=
k
3
y
2
−
2
x
+
8
xy
+
4
x
2
=
0
d
x
d
x
2 d
y
3
y
2
+
4
x
=
2
x
−
8
xy
d
x
d
y
2
x
−
8
xy
=
d
x
2
2
3
y
+
4
x
5(b)[5]
d
y
x = −
1
at P
d
Uses d
y
x = ±
1
and y = x to set up and solve equation in x, y or ‘p’
d
2
p
−
8
p
2
−=
1
p
=
2
3
p
2
+
4
p
2
2
e.g.
k =
"2"
3
−
"2"
+ ×
4 "2"
3
k =
36
(a) Allow equivalent notation for the d
y
x e.g. y′
d
2 d
y
M1: Differentiates
y
3
→
Py
d
x
2 d
y
M1: Uses the product rule to differentiate
4x y and obtains
2
Qxy
Rx
+
d
x
A1: Correct differentiation including the “ = 0” which may be implied by subsequent work.
Note that some candidates have a spurious d
y
x =
...
at the start (as their intention to differentiate) and this
d
d
y
d
y
d
y
2
2
can be ignored for the first 3 marks so condone
=
3
y
−
2
x
+
8
xy
+
4
x
=
0
d
x
d
x
d
x
2
2
Allow versions such as
3
y
d
y
−
2 d
x x
+
8
xy x
d
+
4
x
d
y
=
0
M1: Dependent upon having achieved two different terms in d
y
x , one from each of the terms
3y and
d
) d
d
y
y
Look for (
which may be implied by their working.
... ...
±
=
...
=
...
d
x
d
x
For those candidates who had a spurious d
y
x =
...
at the start, they may incorporate this in their
d
rearrangement in which case they will have 3 terms in d
y
x and so score M0.
d
If they ignore it, then this mark is available for the condition as described above.
d
y
2
x
−
8
xy
d
y
8
xy
−
2
x
A1:
=
or exact equivalent e.g.
=
d
d
x
x
3
y
2
+
4
x
2
−
3
y
2
−
4
x
2
M1: States or uses d
y
x = − which may be implied by their working e.g.
1
d
2
2
2
x
−
8
xy
8
xy
−
2
x
3
y
+
4
x
= −
1,
=
1,
=
1
etc.
8
xy
−
2
x
3
2
4
2
3
2
4
2
y
+
x
y
+
x
Do not award for e.g. “Gradient = −1”
M1: Uses y = x or x = y or y = x = “p” in their d
y
x sets
= ±
1
and attempts to solve to obtain a value for their
d
variable.
2
2
2
2
x
−
8
x
2
y
−
8
y
2
p
−
8
p
e.g.
= ±
1
x
=
...,
= ±
1
y
=
...,
= ±
1
p
=
...,
3
x
2
+
4
x
2
3
y
2
+
4
y
2
3
p
2
+
4
p
2
A1: For x, y or “p” = 2 at P
ddM1: Uses their nonzero value for x, y or “p” to obtain a value for k using the given equation of the
curve.
Depends on both previous method marks.
A1:
k =
36
Correct answer only with no working scores no marks.
(
)
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