Mathematics · May / June 2023 · Paper 11 · Question 9
YMA01
As printed in the paper
Show the answer and mark scheme
9(i)(a)[2]
4
13
9
4
The following examples would score both marks,
8
9
45
,
10
2 ,
y
x y
−
x
<
x
9
4
13
8
9
,
10
2 ,
5
y
x y
−
x
<
x
<
9
4
Note that some candidates are also including limits on y and this is acceptable provided the
values are in the correct range. Allow strict or non strict inequalities e.g.
11
11
2,
2, 2
, 2
but e.g.
<
2,
2
scores A0
y
>
y
<
y
<
y
y
y
2
2
Question 8 – General Guidance for marking
1
1
− as
Note that some candidates are misreading the
x
2
x in part (a) and/or part (b)
2
In the majority of cases it will be clear if this is the case as candidates often write down the
expression before they differentiate in part (a) or integrate in part (b)
1
1
If it is clear that a candidate thinks the expression is
3
−
8
in either or both parts then FULL
4 x
x
2
marks should be awarded for correct work – details are below. We are condoning the fact that
1
1
P
(4,12)
does not lie on
y
=
x
3
−
8
x
2
4
If candidates do not write down the original expression (in (a) or (b)) then evidence of misreading
or not can be taken from their derivative or integral.
Note:
If candidates integrate in (a) and differentiate in (b) then the maximum marks available
are:
(a) M0A0M1dM1A0 (b) M0A0dM0A0
If parts are not labelled then you should assume that the first attempt is part (a) and the
second attempt is part (b).
8(a)
1
1
d
1
1
3
y
−
−
=
3
−
8
=
×
3
2
−
8
×−
y
x
x
2
x
x
2
4
d
4
2
x
d
25
y
=
...
d
x
2
=
4
x
25
(
)
y
−
12
=
"
"
x
−
4
2
25
x
−
2
y
−
76
=
0
oe e.g. 25
−
x
+
2
y
+
76
=
0
1
1
1
1
x
4
8
x
2
(
)
−
f
(
)
=
3
−
8
d
=
−
(
+
)
x
x
x
2
x
c
4
4 4
1
2
(
)
f(4)
=
12
16
−
32
+
c
=
12
c
=
... 28
4
x
So (
)
f
( )
x
=
−
1
6
x
+
"28
"
16
Notes:
M1: Correct method of differentiation, at least one power reduced by 1.
1
3
1
−
−
Award for
3
→
. .
.
2
or
−
8
2
→
. .
.
2
x
x
x
x
4
A1: Correct differentiation, need not be simplified.
M1: Finds d
y
x at x = 4. Requires the substitution of x = 4 into a “changed” function to find a value.
d
Note that some candidates are using the “12” in some way to find the gradient e.g.
1
1
3
( )
2
( )
−
12
=
×
3 4
−
8
×
−
4
+
c
c
=
2
m
=
2
and this scores M0
2
4
2
dM1: Depends on previous M. Correct method for the tangent not the normal. Uses their d
x = 4 with y = 12 and x = 4 correctly placed. If using y = mx + c they must proceed as far as
finding a value for c.
A1: Correct equation in the required form including “= 0” or any nonzero integer multiple of it.
M1: Attempts to integrate f(x), look for power increased by 1 on at least one term.
1
1
1
1
−→
Award for
x
3
→
. .
.
x
4
or
−
x
2
.
..
x
2
4
8
A1: Correct integration, need not be simplified. (no need for constant for this mark).
dM1: Depends on first M. Sets f (4) = 12 and proceeds to find a value for c must have a constant of
integration to score this mark.
A1: Correct value for c found.
A1ft: Correct answer following through their c only i.e. the algebraic part must be correct.
x . Depends on the
4
Accept with fractional index rather than square root. Allow
0.0625x for
4
16
previous method mark.
The “f (x) =” is not required so just look for the correct expression. Apply isw if necessary, e.g.
4
1
1
x
correct work leading to (
)
followed by (
)
f
(
x
)
=
−
1
6
x +
2
2
8
f( )x
=
x
4
−
256
x
2
+
448
can score full
16
marks in (b).
=
x
4
1
Condone poor notation e.g. leaving the final answer as (
f( )
)
16
28
x
−
x
2
+
16
8(a)
1
1
d
1
1
1
y
−
=
3
−
8
=
×
3
2
−
8
×
y
x
x
2
x
x
2
4
d
4
2
x
d
y
(
)
=
... 10
d
x
=
4
x
(
)
y
−
12
=
"10"
x
−
4
10
28
0
oe e.g. 10
28
0
x
−
y
−
=
−
x
+
y
+
=
(b)
3
1
1
1
4
8
x
x
2
(
)
f
( )
x
=
x
3
−
8
x
2
d
x
=
−
(
+
c
)
3
4
4
4
2
128
116
f(4)
=
12
16
−
+
c
=
12
c
=
...
3
3
x
4
16
3
116
So (
f(
)
)
"
"
x
=
−
x
2
+
16
3
3
Notes:
(a)
M1: Correct method of differentiation, at least one power reduced by 1.
1
1
1
−
Award for
x
3
→
. .
.
x
2
or
−
8
x
2
→
.
..
x
2
4
A1: Correct differentiation, need not be simplified. Ignore any spurious “= 0”.
M1: Finds d
y
x at x = 4. Requires the substitution of x = 4 into a “changed” function to find a value.
d
Note that some candidates are using the “12” in some way to find the gradient e.g.
1
1
1
( )
2
( )
−
12
=
×
3 4
−
8
×
−
4
+
c
c
=
2
m
=
2
and this scores M0
2
4
2
dM1: Depends on previous M. Correct method for the tangent not the normal. Uses their d
x = 4 with y = 12 and x = 4 correctly placed. If using y = mx + c they must proceed as far as
finding a value for c.
A1: Correct equation in the required form including “= 0” or any nonzero integer multiple of it.
(b)
M1: Attempts to integrate f (x), look for power increased by 1 on at least one term.
1
1
1
3
Award for
x
3
→
. .
.
x
4
or
−
x
2
→
..
.
x
2
4
8
A1: Correct integration, need not be simplified. (no need for constant for this mark).
dM1: Depends on first M. Sets f (4) = 12 and proceeds to find a value for c must have a constant
of integration to score this mark.
A1: Correct value for c found.
A1ft: Correct answer following through their c only i.e. the algebraic part must be correct. Accept
x and equivalent mixed
4
with fractional index rather than square root. Allow
0.0625x for
4
16
fractions for 16
3 and 116
3 . Depends on the previous method mark.
The “f (x) =” is not required so just look for the correct (or correct ft) expression. Apply isw if
3
x
4
16
116
necessary e.g. correct work leading to (
)
f( )
=
−
+
followed by
x
x
2
16
3
3
3
(
)
4
f( )
x
=
3x
−
256
x
2
+
1856
can score full marks in (b).
=
4
1
6
3
1
16
x
Condone poor notation e.g. leaving the final answer as (
)
f
( )
−
+
x
x
2
16
3
3
9(i)(a)
(
)
( )
=
3cos
y
x
(b)
Same shape translated left
or right
All x intercepts labelled
correctly.
Correct y intercept 3 2
2
(ii)(a)
(
)sin(2 )
=
y
x
(b)
Same shape translated
down below the xaxis.
Correct y intercept 2
−
labelled.
Notes:
(i)(a) Remember to check the first page – (i)(a) is often attempted against the question.
(
)
M1: Identifies the curve as a cosine function of the form
α
cos
β
.
x
π
Also allow for a sine function of the form
sin
oe.
α
β
x
+
2
A1: For a fully correct expression as shown or any equivalent. There is no requirement for y = … or
f (x) = … Allow e.g. 3cosθ
(b) There must be a sketch to score the marks in (i)(b) which may be done on Figure 3
B1: Applies a horizontal translation y = f(x), either direction. The maximum at (0, 3) must have
moved to the left or to the right of the yaxis. There should also be one minimum to the left of the
yaxis and one minimum to the right of the yaxis. Ignore any scale on the yaxis.
B1: All x intercepts labelled correctly. These must be the only x intercepts and the curve must pass
through these points. Allow even if there is still a maximum on the yaxis. Allow as shown or as
7
π
7
π
coordinate pairs e.g.
,0
and allow e.g.
0,
as long as the intercepts are in the correct
−
−
4
4
positions. Condone 3sf decimal answers: Awrt −5.50(or −5.5), −2.36, 0.785, 3.93
Condone use of degrees: −315°, −135°, 45°, 225° (with or without the degrees symbol)
3 2
and allow e.g. 3 2 ,0
B1: Correct exact y intercept labelled. Allow as shown or as e.g.
0,
2
as long as it is in the correct position. It must not be a clear maximum.
(ii)(a) Remember to check the first page – (ii)(a) is often attempted against the question
(
)
M1: Identifies the curve as a sine function of the form
α
sin
β
.
x
π
Allow also for a cosine function of the form
α
cos
β±
x
oe
2
A1: Any correct expression. There is no requirement for y = … or g (x) = … Allow e.g. sin 2θ
(b) There must be a sketch to score the marks in (ii)(b) which may be done on Figure 4
B1: Applies a vertical translation down to y = g(x). It must have the same number of cycles as g (x)
and look the same as g (x) and must lie entirely below the xaxis. Do not allow a clear “zigzag”
shape with cusps. Ignore any scale on the xaxis, correct or otherwise.
B1: Correct y intercept. Allow as shown or as e.g. (
)
and allow e.g. (
)
as long as it is in
0,
−
2
−
2,0
the correct position and allow it to be shown by the intercept passing through the line with equation
y = −2.
9(i)(b)[3]
1
1
1
1
x
4
8
x
2
(
)
−
f
(
)
=
3
−
8
d
=
−
(
+
)
x
x
x
2
x
c
4
4 4
1
2
(
)
f(4)
=
12
16
−
32
+
c
=
12
c
=
... 28
4
x
So (
)
f
( )
x
=
−
1
6
x
+
"28
"
16
Notes:
9(ii)(a)[2]
M1: Correct method of differentiation, at least one power reduced by 1.
1
3
1
−
−
Award for
3
→
. .
.
2
or
−
8
2
→
. .
.
2
x
x
x
x
4
A1: Correct differentiation, need not be simplified.
M1: Finds d
y
x at x = 4. Requires the substitution of x = 4 into a “changed” function to find a value.
d
Note that some candidates are using the “12” in some way to find the gradient e.g.
1
1
3
( )
2
( )
−
12
=
×
3 4
−
8
×
−
4
+
c
c
=
2
m
=
2
and this scores M0
2
4
2
dM1: Depends on previous M. Correct method for the tangent not the normal. Uses their d
x = 4 with y = 12 and x = 4 correctly placed. If using y = mx + c they must proceed as far as
finding a value for c.
A1: Correct equation in the required form including “= 0” or any nonzero integer multiple of it.
9(ii)(b)[2]
M1: Attempts to integrate f(x), look for power increased by 1 on at least one term.
1
1
1
1
−→
Award for
x
3
→
. .
.
x
4
or
−
x
2
.
..
x
2
4
8
A1: Correct integration, need not be simplified. (no need for constant for this mark).
dM1: Depends on first M. Sets f (4) = 12 and proceeds to find a value for c must have a constant of
integration to score this mark.
A1: Correct value for c found.
A1ft: Correct answer following through their c only i.e. the algebraic part must be correct.
x . Depends on the
4
Accept with fractional index rather than square root. Allow
0.0625x for
4
16
previous method mark.
The “f (x) =” is not required so just look for the correct expression. Apply isw if necessary, e.g.
4
1
1
x
correct work leading to (
)
followed by (
)
f
(
x
)
=
−
1
6
x +
2
2
8
f( )x
=
x
4
−
256
x
2
+
448
can score full
16
marks in (b).
=
x
4
1
Condone poor notation e.g. leaving the final answer as (
f( )
)
16
28
x
−
x
2
+
16
8(a)
1
1
d
1
1
1
y
−
=
3
−
8
=
×
3
2
−
8
×
y
x
x
2
x
x
2
4
d
4
2
x
d
y
(
)
=
... 10
d
x
=
4
x
(
)
y
−
12
=
"10"
x
−
4
10
28
0
oe e.g. 10
28
0
x
−
y
−
=
−
x
+
y
+
=
(b)
3
1
1
1
4
8
x
x
2
(
)
f
( )
x
=
x
3
−
8
x
2
d
x
=
−
(
+
c
)
3
4
4
4
2
128
116
f(4)
=
12
16
−
+
c
=
12
c
=
...
3
3
x
4
16
3
116
So (
f(
)
)
"
"
x
=
−
x
2
+
16
3
3
Notes:
(a)
M1: Correct method of differentiation, at least one power reduced by 1.
1
1
1
−
Award for
x
3
→
. .
.
x
2
or
−
8
x
2
→
.
..
x
2
4
A1: Correct differentiation, need not be simplified. Ignore any spurious “= 0”.
M1: Finds d
y
x at x = 4. Requires the substitution of x = 4 into a “changed” function to find a value.
d
Note that some candidates are using the “12” in some way to find the gradient e.g.
1
1
1
( )
2
( )
−
12
=
×
3 4
−
8
×
−
4
+
c
c
=
2
m
=
2
and this scores M0
2
4
2
dM1: Depends on previous M. Correct method for the tangent not the normal. Uses their d
x = 4 with y = 12 and x = 4 correctly placed. If using y = mx + c they must proceed as far as
finding a value for c.
A1: Correct equation in the required form including “= 0” or any nonzero integer multiple of it.
(b)
M1: Attempts to integrate f (x), look for power increased by 1 on at least one term.
1
1
1
3
Award for
x
3
→
. .
.
x
4
or
−
x
2
→
..
.
x
2
4
8
A1: Correct integration, need not be simplified. (no need for constant for this mark).
dM1: Depends on first M. Sets f (4) = 12 and proceeds to find a value for c must have a constant
of integration to score this mark.
A1: Correct value for c found.
A1ft: Correct answer following through their c only i.e. the algebraic part must be correct. Accept
x and equivalent mixed
4
with fractional index rather than square root. Allow
0.0625x for
4
16
fractions for 16
3 and 116
3 . Depends on the previous method mark.
The “f (x) =” is not required so just look for the correct (or correct ft) expression. Apply isw if
3
x
4
16
116
necessary e.g. correct work leading to (
)
f( )
=
−
+
followed by
x
x
2
16
3
3
3
(
)
4
f( )
x
=
3x
−
256
x
2
+
1856
can score full marks in (b).
=
4
1
6
3
1
16
x
Condone poor notation e.g. leaving the final answer as (
)
f
( )
−
+
x
x
2
16
3
3
9(i)(a)
(
)
( )
=
3cos
y
x
(b)
Same shape translated left
or right
All x intercepts labelled
correctly.
Correct y intercept 3 2
2
(ii)(a)
(
)sin(2 )
=
y
x
(b)
Same shape translated
down below the xaxis.
Correct y intercept 2
−
labelled.
Notes:
(i)(a) Remember to check the first page – (i)(a) is often attempted against the question.
(
)
M1: Identifies the curve as a cosine function of the form
α
cos
β
.
x
π
Also allow for a sine function of the form
sin
oe.
α
β
x
+
2
A1: For a fully correct expression as shown or any equivalent. There is no requirement for y = … or
f (x) = … Allow e.g. 3cosθ
(b) There must be a sketch to score the marks in (i)(b) which may be done on Figure 3
B1: Applies a horizontal translation y = f(x), either direction. The maximum at (0, 3) must have
moved to the left or to the right of the yaxis. There should also be one minimum to the left of the
yaxis and one minimum to the right of the yaxis. Ignore any scale on the yaxis.
B1: All x intercepts labelled correctly. These must be the only x intercepts and the curve must pass
through these points. Allow even if there is still a maximum on the yaxis. Allow as shown or as
7
π
7
π
coordinate pairs e.g.
,0
and allow e.g.
0,
as long as the intercepts are in the correct
−
−
4
4
positions. Condone 3sf decimal answers: Awrt −5.50(or −5.5), −2.36, 0.785, 3.93
Condone use of degrees: −315°, −135°, 45°, 225° (with or without the degrees symbol)
3 2
and allow e.g. 3 2 ,0
B1: Correct exact y intercept labelled. Allow as shown or as e.g.
0,
2
as long as it is in the correct position. It must not be a clear maximum.
(ii)(a) Remember to check the first page – (ii)(a) is often attempted against the question
(
)
M1: Identifies the curve as a sine function of the form
α
sin
β
.
x
π
Allow also for a cosine function of the form
α
cos
β±
x
oe
2
A1: Any correct expression. There is no requirement for y = … or g (x) = … Allow e.g. sin 2θ
(b) There must be a sketch to score the marks in (ii)(b) which may be done on Figure 4
B1: Applies a vertical translation down to y = g(x). It must have the same number of cycles as g (x)
and look the same as g (x) and must lie entirely below the xaxis. Do not allow a clear “zigzag”
shape with cusps. Ignore any scale on the xaxis, correct or otherwise.
B1: Correct y intercept. Allow as shown or as e.g. (
)
and allow e.g. (
)
as long as it is in
0,
−
2
−
2,0
the correct position and allow it to be shown by the intercept passing through the line with equation
y = −2.
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