Mathematics · Oct / Nov 2021 · Paper 11 · Question 8
YMA01
As printed in the paper
The question as text
y 8. l1 M C O x Figure 4 Figure 4 shows a sketch of the curve C with equation y = 4 + 12x – 3x 2 The point M is the maximum turning point on C.
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Show the answer and mark scheme
8(a)[5]
(
)
2
(
)
2
4 12
+
x
−
3
x
=
a
±
3
x
+
c
or
a
+
b x
±
2
2
16
−
3
(
x
−
2
)
2
a
16,
b
3,
c
2
Two of
or two of
=
= −
= −
(
)
2
16
−
3
x
−
2
(ii)
Coordinates
M =
(
2,16
)
(b)
States or implies that 2l has equation
y
=
''8''
x
+
k
Sets
4 12
+
x
−
3
x
=
''8 ''
x
+
k
and proceeds to 3TQ
2
Correct 3TQ
3
x
−
4
x
+
k
−
4
=
0
2
2
Attempts to use
b
−
4
ac
=
0
to find k
16
16
k
=
⇒
y
=
8
x
+
3
3
(a)(i)
M1
For attempting to complete the square. Look for
b = ± or
3
c = ±
2
(
)
2
A1
Two correct constants or two correct integers from
16
−
3
x
−
2
(
)
2
(
)
2
A1
16
−
3
x
−
2
(
16
−
3 2
−
x
scores M1A1A0)
Alternative by comparing coefficients:
(
)
(
)
2
a
+
b x
+
c
=
a
+
b x
2
+
2
xc
+
c
2
=
bx
2
+
2
bcx
+
a
+
bc
2
bx
2
+
2
bcx
+
a
+
bc
2
≡
4 12
+
x
−
3
x
2
b = −
3
2
bc
=
12
⇒
c
= −
2
a
−
12
=
4
⇒
a
=
16
(
)
2
Score M1 for expanding
a
+
b x
+
c
and compare x2 coefficients to find a value for b
(NB this can be deduced directly and would score the M mark for
b = ± as above)
3
A1: Continues the process and compares x coefficients to find both b = −3 and c = −2
A1: a = 16
(a)(ii)
(
)
2
B1ft
Either
x =
2
or
y =
16
but follow through on their
16
−
3 x
−
2
where a ≠ 0
(
)
2
B1ft
Both
x =
2
and
y =
16
but follow through on their (− c, a) from
a
+
b x
+
c
where
b ≠±
1
For correct or correct ft coordinates the wrong way round e.g. (16, 2) score SC B1 B0
but apply isw if the correct or correct ft answers are seen as x = …, y = …
M1
States or implies that 2l has equation
y
=
''8''
x
+
k
,
k
≠
0
a
y
coordinateof their
M
Follow through on their
y
=
''
''
x
+
k
or on
y
=
x
c
x
coordinateof their
M
dM1
Sets
4 12
+
x
−
3
x
2
=
''8 ''
x
+
k
and proceeds to 3TQ
A1
Correct 3TQ
3
x
2
−
4
x
+
k
−
4
=
0
(The “= 0” may be implied by subsequent work)
b
2
−
4
ac
=
0
ddM1 Attempts to use
to find k.
16
16
16
A1
k =
⇒
y
=
8
x
+
. Condone just
k =
if
y
=
8
x
+
k
was mentioned as the
3
3
3
equation for 2l
Alternative for part (b)
M1
Attempts to differentiate
4 12
+
x
−
3
x
2
(xn → xn1at least once) and sets equal to their 8
dM1
Solves for x and proceeds to find the coordinates of point of contact
Tangent meets curve at 2 32
A1
,
o.e.
3
3
ddM1 Substitutes their 2 32
,
in their
y
=
''8''
x
+
k
to find k.
3
3
16
y
8
x
A1
=
+
3
8(b)[5]
M1
States or implies that 2l has equation
y
=
''8''
x
+
k
,
k
≠
0
a
y
coordinateof their
M
Follow through on their
y
=
''
''
x
+
k
or on
y
=
x
c
x
coordinateof their
M
dM1
Sets
4 12
+
x
−
3
x
2
=
''8 ''
x
+
k
and proceeds to 3TQ
A1
Correct 3TQ
3
x
2
−
4
x
+
k
−
4
=
0
(The “= 0” may be implied by subsequent work)
b
2
−
4
ac
=
0
ddM1 Attempts to use
to find k.
16
16
16
A1
k =
⇒
y
=
8
x
+
. Condone just
k =
if
y
=
8
x
+
k
was mentioned as the
3
3
3
equation for 2l
Alternative for part (b)
M1
Attempts to differentiate
4 12
+
x
−
3
x
2
(xn → xn1at least once) and sets equal to their 8
dM1
Solves for x and proceeds to find the coordinates of point of contact
Tangent meets curve at 2 32
A1
,
o.e.
3
3
ddM1 Substitutes their 2 32
,
in their
y
=
''8''
x
+
k
to find k.
3
3
16
y
8
x
A1
=
+
3
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