Mathematics · Oct / Nov 2020 · Paper 11 · Question 7
YMA01
As printed in the paper
The question as text
7. The curve C has equation 1 y = 2 −x
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[3]
x = 2
y
B1 Negative reciprocal shape
1
B1 Intercept at
0, 2
(0, ½)
y = 0
x
O
B1
x
2,
y
0
=
=
7(b)[4]
1
(
)(
)
(
)
4
x
+
k
=
⇒
4
x
+
k
2
−
x
= ⇒
1
8
x
+
2
k
−
4
x
2
−
kx
−
1
=
0
oe
2
−
x
(
)
4
x
2
+
k
−
8
x
+ −
1 2
k
=
0
a
=
4,
b
=
k
−
8,
c
= −
1 2
k
or
a
= −
4,
b
=
8
−
k c
,
=
2
k
−
1
(
)
2
(
)(
)
k
−
8
−×
4 4 1 2
−
k
>
0
oe
2
16
64 16
32
0
2
16
48
0*
k
−
k
+
−
+
k
>
⇒
k
+
k
+
>
7(c)[4]
k
2
+
16
k
+
48
=
0
⇒
(
k
+
12
)(
k
+
4
)
=
0
⇒
k
=
...
k = −
12,
−
4
k < −
12
or
k > −
4
Notes
(a)
B1
Negative reciprocal shape (top left/bottom right sections) in any position on a set of axes with no
clear vertical or horizontal ‘overlaps’.
Condone slips of the pen as long as there is not a clear turning point, but it should not curve back
on itself. If there are several attempts, then mark the last one. If there are contradictions between
points stated in the text and any on the graph, then the graph takes precedence.
Eg this can score B1 for the correct shape. The ends of their
curve are regarded as a slip of the pen
1
and no others. Allow 1
2 to be marked on the yaxis and condone 1 ,0
B1 Intercept at
0, 2
intercept is in the correct place. Cannot be awarded without a sketch.
B1
x = 2 and y = 0 and no other asymptotes. Do not allow “xaxis” as an asymptote. Do not award
without a sketch.
(b)
Mark (b) and (c) together
M1 Sets line = curve, multiplies by (2
−
)x
, attempts to expand and collects terms on one side of the
equation. Condone the absence of =0.
A1 Correct equation unsimplified with all terms on one side, condone the absence of =0. This may
be implied by later work such as their values of a, b and c substituted in to
b
2
−
4
ac
and full
marks can be awarded in this question if =0 is never seen.
M1 Attempts discriminant in terms of k using their a, b and c from their 3TQ. Sight of these terms
embedded in
b
2
−
4
ac
is sufficient for this mark. Condone invisible brackets and ignore the use
of >, < or = for this mark.
A1* Correctly multiplies out their expression for the discriminant, sets
>
0
and proceeds to the given
answer with no errors including any omission of brackets in their work.
They must have used a correct inequality on at least the previous line before their final answer or
stated
b
2
−
4
ac
>
0
somewhere in their working.
(c)
M1 Solves the given quadratic to obtain 2 values for k. See guidance on solving quadratics. They
may just write the roots down from their calculator which is acceptable.
A1
−
12,
−
4
M1 Attempts ‘outside regions’ so this may include
≤
or
≥ are condoned for this mark.
−<
4
k
< −
12
would imply this mark. Ignore if in terms of x for this mark. Cannot be scored
from a diagram.
A1
12,
> −’, '
4
12 or
4'
, {
}
Accept any of ‘
k
< −
k
k
< −
k
> −
'
k k
:
< −
12
∪
k
> −
4 '
,
{
}
'
k
:
−∞<
k
< −
12
∪−<
4
k
< ∞
'
or their equivalent expressions
Must be in terms of k. (Not x or any other variable)
k < −
12
k > −’ ‘ 12
4
DO NOT ALLOW ‘
and
−
>
k
> −’ , ‘ 4
4
−<
k
< −
12
’
Note: These inequalities without any working score full marks in (c)
)(
)
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