Mathematics · Oct / Nov 2023 · Paper 41 · Question 3

YMA01

As printed in the paper

The question as text

3. In this question you must show all stages of your working. Solutions based on calculator technology are not acceptable.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

3(i)[5]
1
2
x
2
x
2
x
x
2
e
d
x
=
x
2
e
−
x
e
d
x
2
✓
2
x
1
1
e
2
x
2
x
=
x
2
e
−
x
e
−
d
x
2
2
2
2
x
1
1
e
2
x
2
x
=
x
2
e
−
x
e
+
2
2
4
4
2
x
8
4
1
1
e
e
1
2
x
2
x
2
x
8
8
x
2
e
d
x
=
x
2
e
−
x
e
+
=
8e
−
2e
+
−
2
2
4
4
4
0
0
8
25e
1
=
−
4
4
(i) Condone the omission of “dx” throughout
M1: Attempts to integrate by parts once to achieve
2
x
2
x
Px
2e
−
Q
x
e d ,
x
P Q
,
>
0
2e
x
M1: Attempts to integrate
to achieve
2
x
2
x
x
λ
x
e
±
μ
e
d ,
x
λ
>
0
2
x
1
1
e
2
x
2
x
2
x
2
e
d
2
e
e
A1:
x
x
=
x
−
x
+
which may be unsimplified and isw once a correct answer is seen.
2
2
4
1
1
e
2
x
(
)
x
2
e
2
x
d
x
=
x
e
2
x
−
x
e
2
x
−
+
c
Watch for
2
which scores all 3 marks. ISW after sight of this.
2
2
4
Watch for D & I method which may be seen. You will just see the answer here.
D
I
2x
2e x
+
1
1
1
x
2
e
2
x
x
e
2
x
e
2
x
=
−
+
1 e
−
2
2
4
2x
2
x
2
2
1 e
2
x
2
x
2
x
+
2
x
Score M2 for
Ax
2e
±
Bx
e
±
C
e ,
A
>
0
4
2
x
1
1
e
(
)
0
1 e
2
x
2
x
and then A1 for
x
2
e
−
x
e
+
+
c
2
x
2
2
4
8
2
x
2
x
2
x
M1: Attempts to substitute both limits into an expression of the form
Ax
2e
±
Bx
e
±
C
e ,
8e
subtracts either way round, and simplifies to the form
α
+
βαβ
≠
0
where α and β are numeric.
8e terms are not combined)
(Allow if the
8
25e
1
1
25
1
8
8
−
25e
−
1 ,
e
−
A1:
or exact simplified equivalent e.g.
. Do not condone “+ c” here.
3(ii)[7]
x
x
1.
Allow correct work leading to e.g.
k =
3
The d
x
"
t = must appear at some point in their working for this mark.
"
d
(a) Alt.
or d
S
B1: States or uses
S
6
x
2
(ignore any units associated with either)
=
t =
4
d
may be implied by d
S
Note that
S
=
6
x
2
x =
12
x
d
and d
S
B1: States or uses
(ignore any units associated with either)
S
=
6
x
2
t =
4
d
may be implied by d
S
Note that
S
=
6
x
2
12
x =
x
d
M1: Integrates d
S
t =
4
to obtain S = 4t + c or S = 4t and replaces S in terms of x and differentiates wrt t or
d
d
x
d
t
wrt x to obtain
α
x t
=
β
or
α
x
=
β
d
d
x
1
A1: d
x
1
4
3 , 12
oe e.g.
x from correct work and apply isw once a correct answer is seen.
=
d
t
3
x
x
Note that d
x
1
sometimes comes from incorrect work so check their method.
=
d
t
3
x
This can be scored from obtaining S = 4t + c or S = 4t earlier.
The d
x
"
"
t = must appear at some point in their working for this mark.
d
(b)
d
V
B1: States or uses
x =
3
x
2
Allow this mark to score anywhere in the question.
d
M1: Attempts to use d
V
d
V
d
x
or equivalent with their d
V
x and d
x
t which may be implied.
=
×
d
t
d
x
d
t
d
d
d
V
1
which may be implied by e.g. d
V
1
p
A1:
t =
V
t =
x x
,
=
V
p
=
3
.
d
d
3
The d
V
"
t = must appear at some point in their working for this mark.
"
d
d
S
1
2
−
B1: States or uses
=
4
V
(from
S
=
6
x
2
,
V
=
x
3
S
=
6
V
)
3
3
d
V
M1: Attempts to use d
V
d
V
d
S
or equivalent with their d
V
S and d
S
=
×
t which may be implied.
d
t
d
S
d
t
d
d
1
d
V
which may be implied by e.g. d
V
1
p
A1:
t =
V
t =
x x
,
=
V
p
=
3
d
d
3
The d
V
"
t = must appear at some point in their working for this mark.
"
d
Note that a common incorrect response involves using S = x2 rather than S = 6x2, giving:
2 d
S
d
S
d
S
d
x
d
x
d
x
2
(a)
S
=
x
,
=
4,
=
×
4
=
2
x
×
=
scoring B1B0M1A0
d
t
d
t
d
x
d
t
d
t
d
t
x
1
d
V
d
V
d
V
d
x
d
V
2
(b)
scoring B1M1A0
=
3
x
2
,
=
×
=
3
x
2
×
=
6
x
=
6
V
3
d
x
d
t
d
x
d
t
d
t
x
1
2
x
2
x
2
x
x
2
e
d
x
=
x
2
e
−
x
e
d
x
2
✓
2
x
1
1
e
2
x
2
x
=
x
2
e
−
x
e
−
d
x
2
2
2
2
x
1
1
e
2
x
2
x
=
x
2
e
−
x
e
+
2
2
4
4
2
x
8
4
1
1
e
e
1
2
x
2
x
2
x
8
8
x
2
e
d
x
=
x
2
e
−
x
e
+
=
8e
−
2e
+
−
2
2
4
4
4
0
0
8
25e
1
=
−
4
4
(i) Condone the omission of “dx” throughout
M1: Attempts to integrate by parts once to achieve
2
x
2
x
Px
2e
−
Q
x
e d ,
x
P Q
,
>
0
2e
x
M1: Attempts to integrate
to achieve
2
x
2
x
x
λ
x
e
±
μ
e
d ,
x
λ
>
0
2
x
1
1
e
2
x
2
x
2
x
2
e
d
2
e
e
A1:
x
x
=
x
−
x
+
which may be unsimplified and isw once a correct answer is seen.
2
2
4
1
1
e
2
x
(
)
x
2
e
2
x
d
x
=
x
e
2
x
−
x
e
2
x
−
+
c
Watch for
2
which scores all 3 marks. ISW after sight of this.
2
2
4
Watch for D & I method which may be seen. You will just see the answer here.
D
I
2x
2e x
+
1
1
1
x
2
e
2
x
x
e
2
x
e
2
x
=
−
+
1 e
−
2
2
4
2x
2
x
2
2
1 e
2
x
2
x
2
x
+
2
x
Score M2 for
Ax
2e
±
Bx
e
±
C
e ,
A
>
0
4
2
x
1
1
e
(
)
0
1 e
2
x
2
x
and then A1 for
x
2
e
−
x
e
+
+
c
2
x
2
2
4
8
2
x
2
x
2
x
M1: Attempts to substitute both limits into an expression of the form
Ax
2e
±
Bx
e
±
C
e ,
8e
subtracts either way round, and simplifies to the form
α
+
βαβ
≠
0
where α and β are numeric.
8e terms are not combined)
(Allow if the
8
25e
1
1
25
1
8
8
−
25e
−
1 ,
e
−
A1:
or exact simplified equivalent e.g.
. Do not condone “+ c” here.

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