Mathematics · May / June 2024 · Paper 11 · Question 4
YMA01
As printed in the paper
The question as text
4. The curve C1 has equation 2 + kx – 9 y = x and the curve C2 has equation 2 – 5x + k y = –3x where k is a constant. Given that C1 and C2 meet at a single point P
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[3]
(
)
x
2
+
kx
−
9
= −
3
x
2
−
5
x
+
k
4
x
2
+
kx
+
5
x
−−
9
k
=
0
(
)
2
(
)
b
2
−
4
ac
=
0
k
+
5
−×
4 4
−−
9
k
=
0
k
2
+
26
k
+
169
=
0*
4(b)[3]
k
2
+
26
k
+
169
=
0
k
= −
13
k
= −
13
4
x
2
−
8
x
+
4
=
0
x
=
...
(
1,
−
21
)
Notes:
Mark (a) and (b) together
(a)
M1: Equates both curves and collects terms to one side. Allow slips but there must be some attempt to collect
terms to one side. The “ = 0” may be implied by their attempt at the discriminant.
2
2
M1: Attempts to find the discriminant
"
b
−
4
ac
"
of a 3TQ. May be seen as an attempt at e.g.
b
−
equivalent e.g.
b
2
=
4
ac
or may be seen embedded in an attempt at the quadratic formula.
It requires b and c both of the form
pk
+
q
,
p q
,
≠
0
and requires a as a constant.
If clearly a wrong formula/expression is used e.g.
"
b
2
+
4
ac
"
this scores M0
A1*: Achieves the printed answer with no errors and sufficient working shown.
The “ = 0” must appear at least once before the printed answer unless they start with
b
2
=
4
ac
(b)
B1: Correct value of k. Condone x = −13 if it is subsequently used as a value for k.
M1: Either:
substitutes their value of k into their equation obtained by equating the curves and collecting terms frompart (a) and solves a 3TQ for x by any correct method including a calculator or
substitutes their k into the 2 given equations, equates, collects terms to one side and solves a 3TQ for xby any correct method including a calculator
A1: Correct coordinates. Allow as e.g. x = 1, y = –21
Correct answer only in (b) scores 3/3
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