Mathematics · Oct / Nov 2024 · Paper 11 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
O
Fully correct
Correct equation of vertical
asymptote and y intercept
1
Sets
kx
4
(
kx
4
)(
2
x
)
1
(b) (i)
−
=
−
−
=
2
−
x
(
)
kx
2
+ −−
4
2
k x
+
9
=
0
(
)
2
Attempts use
b
2
−
4
ac
…
0
−−
4
2
k
−
4
k
×…
9
0
4
k
2
−
20
k
+
16
…
0
k
2
−
5
k
+
4
… *
0
(
)(
)
k
−
1
k
−
4
…
0
k
„
1,
k
…
4
(ii)
(a)
M1: Correct shape for a 1
x type graph. Look for
or
Must have two branches, be generous with curves that bend back on themselves if the intent
is clear.
A1: Fully correct shape and position. Do not accept curves that clearly bend back on
themselves, but “pen flicks” at the end of a curve can be condoned.
B1: Correct equation (must be an equation) of asymptote and y intercept. The graph must be
asymptotic here and in the correct places (e.g. the equation must match the placement),
though it need not be fully correct – as long as these features are shown. Information
stated on the graph takes precedence but if coordinates or equation are missing, allow if
they are stated in the text. Accept ½ labelled on the yaxis for the intercept. Condone
(½ , 0) if it is in the correct position on the axis.
(b)(i)
1
(
)(
)
M1:
Sets
kx
−
4
=
kx
−
4
2
−
x
=
1
2
−
x
(
)
A1:
kx
2
+ −−
4
2
k x
+
9
=
0
with terms collected. This may be implied by later work, e.g. a = ..., b
=…, c =… listed or used. The “=0” may be implied.
(
)
dM1: Attempts use
b
2
−
4
ac
…
0
−−
4
2
k
2
−
4
k
×… with their coefficients to set up a
9
0
quadratic inequality in k. Must be the correct direction of inequality but allow > or …. Use of
b2 – 4ac = 0 will score dM0 unless clearly recovered before stating that answer.
2
5
+ … with no errors and sufficient working shown –
4
0
A1*: Achieves the given answer of
k
−
k
the bracket should be expanded before the final answer is reached. A0 if x used in place of k.
(b)(ii)
M1: Solves to find the critical values – may be implied by sight of 1 and 4 used in their
answer but must have two values from a valid attempt to solve. Allow from attempts at
their
"
b
2
−
4
ac
"...0
with any inequality or equality.
A1:
k
„
1,
k
… Accept with “and” or “or” used. Accept interval notation
4
(
) (
]
[
)
k
∈
−
∞
,
1
∪
4,
∞ must be correct union if set notation used. Must be the final
answer, but allow if recovery from “incorrect” inequalities when solving for the critical
values. A0 if x used instead of k. Note 4
„
„
1
k
is A0.
Some examples of graphs:
M1A0B0 intent clear for first M. M0A0B1 Only one M1A0B0 this intent is acceptable
Wrong position, A0, incorrect branch, but asymptote
for the shape, but not fully
asymptote and intercept, B0. and intercept correct.
correct as “overlapping” so A0.
Intercept/asymptotes incorrect.
M1A0B0 Intent clear for the M1A0B1 Intent clear for
M1A1B1 Shape and position
First M, but clearly bending first M, clearly bends down correct, the slight “pen flick”
back on itself, so A0. No in fourth quadrant, A0.
is condoned. Intercept and
intercept given .
Asymptote and intercept
vertical asymptote correctly
correct.
labelled.
(
)
P = −
2 , 4
π
−
6(b)[6]
O
Fully correct
Correct equation of vertical
asymptote and y intercept
1
Sets
kx
4
(
kx
4
)(
2
x
)
1
(b) (i)
−
=
−
−
=
2
−
x
(
)
kx
2
+ −−
4
2
k x
+
9
=
0
(
)
2
Attempts use
b
2
−
4
ac
…
0
−−
4
2
k
−
4
k
×…
9
0
4
k
2
−
20
k
+
16
…
0
k
2
−
5
k
+
4
… *
0
(
)(
)
k
−
1
k
−
4
…
0
k
„
1,
k
…
4
(ii)
(a)
M1: Correct shape for a 1
x type graph. Look for
or
Must have two branches, be generous with curves that bend back on themselves if the intent
is clear.
A1: Fully correct shape and position. Do not accept curves that clearly bend back on
themselves, but “pen flicks” at the end of a curve can be condoned.
B1: Correct equation (must be an equation) of asymptote and y intercept. The graph must be
asymptotic here and in the correct places (e.g. the equation must match the placement),
though it need not be fully correct – as long as these features are shown. Information
stated on the graph takes precedence but if coordinates or equation are missing, allow if
they are stated in the text. Accept ½ labelled on the yaxis for the intercept. Condone
(½ , 0) if it is in the correct position on the axis.
(b)(i)
1
(
)(
)
M1:
Sets
kx
−
4
=
kx
−
4
2
−
x
=
1
2
−
x
(
)
A1:
kx
2
+ −−
4
2
k x
+
9
=
0
with terms collected. This may be implied by later work, e.g. a = ..., b
=…, c =… listed or used. The “=0” may be implied.
(
)
dM1: Attempts use
b
2
−
4
ac
…
0
−−
4
2
k
2
−
4
k
×… with their coefficients to set up a
9
0
quadratic inequality in k. Must be the correct direction of inequality but allow > or …. Use of
b2 – 4ac = 0 will score dM0 unless clearly recovered before stating that answer.
2
5
+ … with no errors and sufficient working shown –
4
0
A1*: Achieves the given answer of
k
−
k
the bracket should be expanded before the final answer is reached. A0 if x used in place of k.
(b)(ii)
M1: Solves to find the critical values – may be implied by sight of 1 and 4 used in their
answer but must have two values from a valid attempt to solve. Allow from attempts at
their
"
b
2
−
4
ac
"...0
with any inequality or equality.
A1:
k
„
1,
k
… Accept with “and” or “or” used. Accept interval notation
4
(
) (
]
[
)
k
∈
−
∞
,
1
∪
4,
∞ must be correct union if set notation used. Must be the final
answer, but allow if recovery from “incorrect” inequalities when solving for the critical
values. A0 if x used instead of k. Note 4
„
„
1
k
is A0.
Some examples of graphs:
M1A0B0 intent clear for first M. M0A0B1 Only one M1A0B0 this intent is acceptable
Wrong position, A0, incorrect branch, but asymptote
for the shape, but not fully
asymptote and intercept, B0. and intercept correct.
correct as “overlapping” so A0.
Intercept/asymptotes incorrect.
M1A0B0 Intent clear for the M1A0B1 Intent clear for
M1A1B1 Shape and position
First M, but clearly bending first M, clearly bends down correct, the slight “pen flick”
back on itself, so A0. No in fourth quadrant, A0.
is condoned. Intercept and
intercept given .
Asymptote and intercept
vertical asymptote correctly
correct.
labelled.
(
)
P = −
2 , 4
π
−
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