Mathematics · Oct / Nov 2022 · Paper 11 · Question 9

YMA01

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The question as text

9. y C l P O x M Figure 3 Figure 3 shows a sketch of the curve C with equation y = 1 2 x2 – 10x + 22

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9(a)[3]
1
1
1
2
−
10
+
22
=
(
±
...)
2
±
...
or states
a =
x
x
x
2
2
2
1
1
1
2
10
22
(
10)
2
...
or states
and
10
x
−
x
+
=
x
±
±
a =
b = ±
2
2
2
1
1
x
2
−
10
x
+
22
=
(
x
−
10)
2
−
28
2
2
9(b)[2]
(
)
"10", " 28"
−
(c)(i)
Gradient of tangent
=
8
d
y
x =
−
10
=
8
=
...
x
x
d
18,
4
x
=
y
=
(c)(ii)
1
25
−
×
"18"
=
"4"
=
k
k
8
4
9(c)[6]
1
1
1
2
−
10
+
22
=
(
±
...)
2
±
...
or states
a =
x
x
x
2
2
2
1
1
1
2
10
22
(
10)
2
...
or states
and
10
x
−
x
+
=
x
±
±
a =
b = ±
2
2
2
1
1
x
2
−
10
x
+
22
=
(
x
−
10)
2
−
28
2
2
(
)
"10", " 28"
−
(c)(i)
Gradient of tangent
=
8
d
y
x =
−
10
=
8
=
...
x
x
d
18,
4
x
=
y
=
(c)(ii)
1
25
−
×
"18"
=
"4"
=
k
k
8
4
25
1
1
x …
"10"
One of
or
y
„
"
"
−
x
or
y
…
x
2
−
10
x
+
22
4
8
2
25
1
1
x …
"10"
Two of
or
y
„
"
"
−
x
or
y
…
x
2
−
10
x
+
22
4
8
2
25
1
1
x …
10
All three of
,
y
„
−
x
and
y
…
x
2
−
10
x
+
22
4
8
2
(a)
If there is a contradiction between the embedded values of a, b and c within their
expression and their stated values then the embedded values take precedence.
1
1
1
B1
Achieves
x
2
−
10
x
+
22
=
(
x
±
...)
2
±
...
or states that
a =
2
2
2
1
M1
Deals correctly with the first two terms of
2
−
10
+
22
2 x
x
1
1
1
Scored for
x
2
−
10
x
+
22
=
(
x
±
10)
2
±
...
or states that
a =
&
b = ±
10
(may be found
2
2
2
using symmetry by finding the roots)
1
1
A1
x
2
−
10
x
+
22
=
(
x
−
10)
2
−
28
(cannot just be the stated values). Do not isw if for
2
2
example they attempt to multiply by 2 to achieve integer values
This may also be done by equating coefficients using the expanded form
a x
(
+
b
)
2
+ =
c
ax
2
+
2
abx
+
ab
2
+ but the marks can be applied in the same way
c
(b)
B1ft One of the coordinates of (
)
or follow through one of their (
)
10,
−
28
−
,b c
from (a).
(
)
10,
−
28
or follow through their (
−
, )
B1ft
b c
from (a). Accept
x
=
"10",
y
= −
" 28"
. Condone
lack of bracketing as long as the intention as to which coordinate is which is clear.
(c)(i)
B1
Gradient of tangent
=
8
(stated or implied)
1
M1
Differentiates
2
−
10
+
22
to achieve a derivative of the form px
and sets equal to 8 .
2 x
x
+
q
They then proceed to find a value for x.
1
Alternatively, sets
8
+
α
=
2
−
10
+
22
, proceeds to a quadratic in x, sets the
x
x
x
2
discriminant = 0 and solves to find α. They then use this value of αin their original
equation to solve and find x
A1
18
x =
A1
y =
4
(c)(ii)
1 "18"
dM1
Sets
k −
×
=
"4"
and proceeds to find a value for k. It is dependent on the method mark
8
in (i)
25
A1
k
=
oe
4
(d)
Ignore any references to OR or AND (or equivalent use of set notation)
1
25
1
Note that
x
2
−
10
x
+
22
„
y
„
"
"
−
x
is acceptable and counts as two of the required
2
4
8
inequalities in part (d)
Use of strict or inclusive inequalities must be consistent for all of their inequalities on the
last mark only
25
1
B1ft
One of
x …
"10"
(or "10 "
„ x „
x
d
where d is at least 18) or
„
"
"
−
or
y
x
4
8
1
…
2
10
22
follow through on their minimum point or their k. Allow if l is still in
y
x
−
x
+
2
terms of k
25
1
x …
"10"
(or 10
„ x „
„
"
"
B1ft
Two of
x
d
where d is at least 18) or
y
−
x
or
4
8
1
y
…
x
2
−
10
x
+
22
follow through on their minimum point or their k. Allow if l is still in
2
terms of k
25
1
B1
All three of
x …
10
(or 10
„ x „
x
d
where d is at least 18),
„
−
and
y
x
4
8
1
…
2
−
10
+
22
. Condone the additional minimum interval (or greater) 28
„
y
x
x
−
y
2
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
9(d)[3]
25
1
1
x …
"10"
One of
or
y
„
"
"
−
x
or
y
…
x
2
−
10
x
+
22
4
8
2
25
1
1
x …
"10"
Two of
or
y
„
"
"
−
x
or
y
…
x
2
−
10
x
+
22
4
8
2
25
1
1
x …
10
All three of
,
y
„
−
x
and
y
…
x
2
−
10
x
+
22
4
8
2
(a)
If there is a contradiction between the embedded values of a, b and c within their
expression and their stated values then the embedded values take precedence.
1
1
1
B1
Achieves
x
2
−
10
x
+
22
=
(
x
±
...)
2
±
...
or states that
a =
2
2
2
1
M1
Deals correctly with the first two terms of
2
−
10
+
22
2 x
x
1
1
1
Scored for
x
2
−
10
x
+
22
=
(
x
±
10)
2
±
...
or states that
a =
&
b = ±
10
(may be found
2
2
2
using symmetry by finding the roots)
1
1
A1
x
2
−
10
x
+
22
=
(
x
−
10)
2
−
28
(cannot just be the stated values). Do not isw if for
2
2
example they attempt to multiply by 2 to achieve integer values
This may also be done by equating coefficients using the expanded form
a x
(
+
b
)
2
+ =
c
ax
2
+
2
abx
+
ab
2
+ but the marks can be applied in the same way
c
(b)
B1ft One of the coordinates of (
)
or follow through one of their (
)
10,
−
28
−
,b c
from (a).
(
)
10,
−
28
or follow through their (
−
, )
B1ft
b c
from (a). Accept
x
=
"10",
y
= −
" 28"
. Condone
lack of bracketing as long as the intention as to which coordinate is which is clear.
(c)(i)
B1
Gradient of tangent
=
8
(stated or implied)
1
M1
Differentiates
2
−
10
+
22
to achieve a derivative of the form px
and sets equal to 8 .
2 x
x
+
q
They then proceed to find a value for x.
1
Alternatively, sets
8
+
α
=
2
−
10
+
22
, proceeds to a quadratic in x, sets the
x
x
x
2
discriminant = 0 and solves to find α. They then use this value of αin their original
equation to solve and find x
A1
18
x =
A1
y =
4
(c)(ii)
1 "18"
dM1
Sets
k −
×
=
"4"
and proceeds to find a value for k. It is dependent on the method mark
8
in (i)
25
A1
k
=
oe
4
(d)
Ignore any references to OR or AND (or equivalent use of set notation)
1
25
1
Note that
x
2
−
10
x
+
22
„
y
„
"
"
−
x
is acceptable and counts as two of the required
2
4
8
inequalities in part (d)
Use of strict or inclusive inequalities must be consistent for all of their inequalities on the
last mark only
25
1
B1ft
One of
x …
"10"
(or "10 "
„ x „
x
d
where d is at least 18) or
„
"
"
−
or
y
x
4
8
1
…
2
10
22
follow through on their minimum point or their k. Allow if l is still in
y
x
−
x
+
2
terms of k
25
1
x …
"10"
(or 10
„ x „
„
"
"
B1ft
Two of
x
d
where d is at least 18) or
y
−
x
or
4
8
1
y
…
x
2
−
10
x
+
22
follow through on their minimum point or their k. Allow if l is still in
2
terms of k
25
1
B1
All three of
x …
10
(or 10
„ x „
x
d
where d is at least 18),
„
−
and
y
x
4
8
1
…
2
−
10
+
22
. Condone the additional minimum interval (or greater) 28
„
y
x
x
−
y
2
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom

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