Mathematics · Oct / Nov 2022 · Paper 21 · Question 9

YMA01

As printed in the paper

The question as text

9. In this question you must show detailed reasoning. Solutions relying entirely on calculator technology are not acceptable. y C1 R l A T C2 O x Figure 3 Figure 3 shows • the curve C1 with equation y = x3 – 5x2 + 3x + 14 • the circle C2 with centre T The point T is the minimum turning point of C1 Using Figure 3 and calculus,

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9(a)[3]
d
y
y
=
x
3
−
5
x
2
+
3
x
+
14
3
x
2
10
x
3
0
=
−
+
=
d
x
1
Roots are
3, 3 when
x
=
3,
y
=
"3"
3
−×
5 "3"
2
+ ×
3 "3" 14
+
=
....
Centre is (3, 5)
9(b)[3]
8
At A
y =
r =
2
(2
−
"3")
2
+
("8" "5")
−
2
(=10)
(
x
−
3)
2
+
(
y
−
5)
2
=
10
9(c)[3]
"8" "5"
−
(
)
=
...
−
3
2 "3"
−
1
y
−
"8"
=
" "(
x
−
2)
3
1
22
y
=
x
+
*
3
3
9(d)[3]
1
5
3
2
x
3
−
5
x
2
+
3
x
+
14 d
x
=
...
x
4
−
x
3
+
x
2
+
14
x
4
3
2
0
2
1
5
3
1
22
Area =
x
4
−
x
3
+
x
2
+
14
x
−
×
+
"8"
×
2
=
...
4
3
2
2
3
0
1
5
3
46
×
16
−
×+
8
×+
4 14 2
×−
4
3
2
3
74
46
28
−
=
3
3
3
(a)
M1: Differentiates the equation of the curve and sets equal to 0 which may be implied by their
attempt to solve below. Do not credit this differentiation in any other parts of the question.
For the differentiation, look for at least 2 of
→
x
3
→
...
x
2
,
−
5
x
2
→
... , 3
x
x
3
dM1: Solves a 3 term quadratic equation by any valid means (including a calculator) and substitutes
one of their roots in the original equation to find the y coordinate. Depends on the first mark.
1
A1: Correct coordinates(3, 5)or e.g. x = 3, y = 5 (Ignore any work with e.g.
x =
and ignore any
3
work attempting to prove maximum/minimum points – just look for correct coordinates identified)
(b)
B1: y coordinate at A is 8 (Allow this to score anywhere)
M1: Attempts to find the radius of the circle (or radius2) using (2,"8") and their minimum point.
Score for
(2
−
"3")
2
+
("8" "5")
−
2
or equivalent correct work for their coordinates of T and A.
You can ignore what they call it e.g. condone
radius
=
(2
−
"3")
2
+
("8" "5")
−
2
(
)
2
A1:
(
x
−
3)
2
+
(
y
−
5)
2
=
10
oe e.g.
(
x
−
3)
2
+
(
y
−
5)
2
=
10
,
(
x
−
3)
2
+
(
y
−
5)
2
=
10
(c)
M1: Attempts to find the gradient between A and T using their coordinates. Note that as the
equation of the tangent is given we do not accept −3 just written down.
M1: Attempts to find the equation of the straight line using x = 2, their y value at A and the negative
reciprocal of what they think is the gradient of AT.
1
22
A1*:
y
=
x
+
with no errors and both previous method marks scored.
3
3
d
y
d
y
3
−
x
3
−
2
1
Alternative for first M:
(
x
−
3)
2
+
(
y
−
5)
2
=
10
2(
x
−
3)
+
2(
y
−
5)
=
0
=
=
=
d
x
d
x
y
−
5
8
−
5
3
(d)
M1: Attempts to integrate the curve. Award for a power increasing by 1 on one of the terms.
dM1: A correct method to find the shaded area:
Substitutes in 2 (and 0) into the integrated function and subtracts either way round.
May not see 0 substituted in.
Attempts to find the shaded area by subtracting the area of the trapezium
1
22
"8"
2
×
+
×
from their area under the curve. It is dependent on the previous
2
3
method mark.
It must be a correct method for the area of the trapezium e.g. 1
22
×
+
their
y
at
A
×
2
2
3
22
1
22
Or e.g. rectangle + triangle:
2
×
+
×
2 their
y
at
A
−
3
2
3
2
2
1
22
1
22
2
44
46
Or by integration
2
x
+
d
x
=
x
+
x
=
+
=
3
3
6
3
3
3
3
0
0
A1: 28
3 or exact equivalent cso
Alt(d)
M1: Attempts to integrate (curve – line) or (line – curve).
Award for the power increasing by 1 on one of the terms.
dM1: A correct method to find the shaded area.
Substitutes 2 (and 0) into the integrated function and subtracts either way round:
2
8
20
x
4
5
x
3
4
x
2
20
x
2
4
5 2
×
3
4 2
×
2
20 2
×
2
x
3
−
5
x
2
+
x
+
d
x
=
−
+
+
=
−
+
+
=
...
0
3
3
4
3
3
3
4
3
3
3
0
2
1
22
x
4
5
x
3
3
x
2
x
2
22
x
2
Or e.g.
x
3
−
5
x
2
+
3
x
+
14
−
x
−
d
x
=
−
+
+
14
x
−
−
=
3
3
4
3
2
6
3
0
0
2
4
5 2
×
3
3 2
×
2
2
2
22 2
×
14 2
...
=
−
+
+
×−
−
=
4
3
2
6
3
You can condone slips when collecting terms or for a slip with brackets e.g. attempting:
1
22
2
x
3
5
x
2
3
x
14
x
d
x
−
+
+
−
+
0
3
3
A1: 28
3 or exact equivalent cso
28
With otherwise correct work leading to
−
allow full marks if then made positive.
3

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