Mathematics · Oct / Nov 2021 · 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. A curve C has equation y = f(x) where 2 f(x) = 2(x + 1)(x – 3)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
Positive cubic shape anywhere with
18
1 maximum and 1 minimum
Positive cubic shape that at least
reaches the xaxis at x = −1 and with
a minimum on the xaxis at
x = 3
y intercept at 18. Must correspond
−
1
3
O
with their sketch
For the intercepts allow as numbers as above or allow as coordinates e.g. (18, 0),
(0, −1), (0, 3) as long as they are marked in the correct place.
6(b)[3]
E.g.
(2
x
+
2)(
x
2
−
6
x
+
9)
=
...
2
x
3
−
10
x
2
+
6
x
+
18
=
6(c)[4]
(
)
f ( )
′
x
=
6
x
2
−
20
x
+
6
1
1
2
1
f
′
=
6
−
20
+
6
3
3
3
1
′
f
=
0
3
512
y =
27
(a)
3
M1
Correct shape for a
y
= +
x
graph. Do not be too concerned if the “ends” become
vertical or even go beyond the vertical slightly. Condone with no axes and condone
cusp like appearance for the turning points.
A1
y
= +
x
3
shape, intersects (or at least reaches the xaxis) at −1, minimum at x = 3 but
must not stop or cross at x = 3
B1
y intercept at 18
You can ignore the position of the maximum i.e. it may be to the left of or right of or on
the yaxis.
(b)
Mark (b) and (c) together.
M1
Attempts to multiply out.
2
E.g. Look for an attempt to square
(
x
−
3) to obtain
x
±
6
x
±
9
and then an attempt
to multiply by (
)1
or (
)
or an attempt to multiply (
)1
or (
)
x +
2
x +
2
x +
2
x +
2
by (x – 3) and then multiply the result by (x – 3)
Condone slips e.g. attempting (2x + 1)(x – 3)2 but expect to see an expression of the
required form
2
x
3
−
10
x
2
+
6
x
+
18
A1
Two correct terms of
2
x
3
−
10
x
2
+
6
x
+
18
A1
Fully correct
. (Ignore any spurious “= 0”)
2
x
3
10
x
2
6
x
18
Special case: if they obtain
−
+
+
but then attempt to “simplify” as e.g.
f
( )
x
x
3
5
x
2
3
x
9
=
−
+
+
then score A1A0 but note that all marks are available in
(c) in such cases.
(c)
B1ft
Correctly differentiates their
2
x
3
−
10
x
2
+
6
x
+
18.
Allow follow through but only
from a 4 term cubic.
Allow use of product rule e.g.
2
2
f ( )
x
2(
x
1)(
x
3)
f ( )
′
x
2(
x
3)
4(
x
1)(
x
3)
=
+
−
→
=
−
+
+
−
You can condone poor notation so just look for the correct or correct ft expression.
1
M1
Attempts
f
′
3
1
A1
Correctly achieves
f
′
=
0.
Must follow a correct derivative but allow this mark if
3
they have e.g. divided their derivative by 2 before substituting or e.g. if they have
divided their expanded f(x) by 2 before differentiating so they have
f ( )
′
x
=
3
x
2
−
10
x
+
3
or if only their constant in their expansion in (b) is incorrect
1
1
2
1
The working may be minimal so allow e.g.
f
′
=
6
−
20
+
6
=
0
or even
3
3
3
1
f
′
0
=
on its own following a correct derivative. Just look for the answer of 0 (e.g.
3
they may call it y).
512
A1
y =
or exact equivalent. All previous marks must have been scored in (c).
27
1
Acceptable alternative to show
f
′
=
0:
3
1
(
)(
)
f ( )
′
x
=
6
x
2
−
20
x
+
6
=
0
⇒
6
x
−
2
x
−
3
=
0
⇒
f ( )
′
x
=
0 at
x
=
⇒
f
′
3
Score M1 for attempting to solve quadratic (usual rules) and A1 for deducing
1
f
′
=
0
3
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