Mathematics · Oct / Nov 2024 · Paper 41 · Question 4
YMA01
As printed in the paper
The question as text
4. y P O x C Figure 2 Figure 2 shows a sketch of the curve C with equation 2 2 x 3 x 2 y 4 xy 8 11 0 The point P has coordinates (1, 2).
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[1]
x
=
1,
y
=
2
3
x
2
+
2
y
2
−
4
xy
+
8
x
−
11
= + −+ −
3 8 8 8 11
=
0*
4(b)[5]
8
x
8 ln8
x
→
d
y
d
y
or
4
xy
→
4
x
+
4
y
y
2
→
2
y x
d
x
d
d
d
y
y
o.e.
6
x
+
4
y
−
4
y
−
4
x
+
8 ln8
x
=
0
d
x
d
x
d
y
d
y
(
4
y
−
4
x
)
=
4
y
−
6
x
−
8 ln8
x
=
...
d
x
d
x
x
x
x
ln8
d
y
4
y
−
6
x
−
8 ln8
o.e. e.g d
y
6
x
+
8 ln8 4
−
y
d
y
4
y
−
6
x
−
e
ln8
=
=
=
or e.g
d
x
4
y
−
4
x
d
x
4
x
−
4
y
d
x
4
y
−
4
x
4(c)[3]
d
y
8
−−
6
8ln8
4
(
1, 2
)
→
=
y
−
2
=
(
x
−
1
)
d
x
8
−
4
8ln8
−
2
4
(
)
y
=
0
0
−
2
=
x
−
1
x
=
...
8ln8
−
2
x =
2 12ln2
−
(a)
B1*: Substitutes x = 1 and y = 2 and obtains 0 with no errors.
Minimum required without evaluating each term e.g.
3 1 2 2
×+ ×
2
−××+
4 1 2
8
1
−
11
= . We just
0
need a correct equation which evaluates as 0.
Alternatively, substitutes in x = 1, forms a quadratic in y i.e.
2
y
2
−
4
y
=
0
o.e. and
solves to
find y via any method:
3
+
2
y
2
−
4
y
+ −
8 11
=
0
2
y
2
−
4
y
=
0
y
=
2
(ignore the other solution 0)
They may at this stage substitute in y = 2 to get 0.
In any of the approaches a conclusion is not required.
(b)
B1: Correct differentiation of 8x . Award for sight of ln8 8x
×
or
e
x
ln8
×
ln8
. Do not be concerned
with the use of modulus signs around ln8
d
d
y
y
M1: For
or
y
2
→
ky
4
xy
→
Ax
+
By
d
x
d
x
A1: Fully correct differentiation. Condone the omission of = 0 or may be implied by further
d
y
d
y
work. Accept
6
x
4
y
4
y
4
x
e
x
ln8
ln8
0
+
−
−
+
=
d
x
d
x
Ignore if they start their differentiated line with e.g. d
y
x =
d
M1: Attempts to make d
y
x the subject with 2 terms in d
y
x coming from the correct places. i.e. they
d
d
d
y
d
y
where ,
must have two terms (one of each) of the form
α
x x
and
β
y x
αβ are nonzero
d
d
constants.
A1: Any correct expression for d
y
x . Allow
y′=
...
o.e. isw after a correct answer is seen
d
Condone poor dividing lines which do not go quite far enough provided an incorrect method
was not seen. E.g. 4
y
−
6
x
−
8 ln8
x
4
y
−
4
x
(c) Note that an incorrect derivative in part (b) can score maximum 110 in part (c)
M1: Attempts to use x = 1 and y = 2 in their d
y
x to find the gradient at P (condone if this is a
d
decimal) and attempts to form the equation of the normal at P using the negative reciprocal
gradient. Condone one sign slip when substituting in the coordinates into the straight line
equation.
4
y =
0
(
)
They may set
at this stage such that
−=
2
2 x
−
1
which is acceptable.
8ln8
−
If they use y
=
mx
+ they must proceed as far as
c
c =
...
M1: States y = 0 or substitutes y = 0 into their straightline equation using a changed gradient from
their d
y
x and rearranges to find x. Do not be concerned by the mechanics of their
d
rearrangement.
A1: Correct expression in the required form
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