Mathematics · May / June 2024 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. The current, x amps, at time t seconds after a switch is closed in a particular electric circuit is modelled by the equation x d k 3 x t d where k is a constant. Initially there is zero current in the circuit.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[6]
3
t =
k
−
x
d
x
=
d
t
oe
d
k
−
3
x
1 ln
(
)
(
)
−
k
−
3
x
=
t
+
c
3
1 n
0,
0
l
t
=
x
=
−
k
=
c
3
k
−
3
x
3
(
)
−
ln
k
−
3
x
=
3
t
−
ln
k
ln
= −
3
t
−
1
x
=
e
−
3
t
x
=
...
k
k
k
(
)
x
=
1 e
−
−
3
t
3
7(b)[2]
k
k
t
→∞
e
−
3
t
0
x
→
=
7
k
=
...
3
3
k = 21
(
)
7(c)[3]
x
=
5
7 1 e
−
−
3
t
=
5
e
−
3
t
=
...
1
2
= −
t
ln
3
7
Awrt 0.42 seconds
Alt (a): Use the same criteria for the first 3 marks. See notes
x
t
7(a)
d
x
1
t =
k
−
3
x
d
d
oe
x
=
t
d
k
−
3
x
0
0
x
1 ln
(
)
t
k
3
x
t
−
−
=
3
0
0
1
(
)
1
−
ln
k
−
3
x
+
ln
k
=
t
3
3
(
)
k
k
l
n
k
−
ln
k
−
3
x
=
3
t
ln
=
3
t
=
e
3
t
k
=
k
e
3
t
−
3
x
e
3
t
x
=
..
k
−
3
x
k
−
3
x
k
(
)
x
=
1 e
−
−
3
t
3
Mark as one complete question. For example, it is possible to get (b) without doing (a)
(a)
M1: Separates the variables correctly, with or without the integration signs shown. The dx and dt must be
shown in the correct positions
1
(
)
M1: Attempts the integration of the LHS with
a
l
n
k
3
x
A missing bracket should be
x →±
−
k
−
3
penalised unless recovered later in the solution
A1: Fully correct integration, with or without the + c for this mark.
A missing bracket should be penalised unless recovered later in the solution
M1: Attempts to find the constant of integration following any attempt at integration to an equation
involving a constant. This involves
t
=
0,
x
=
0
c
=
..
.
where c is non zero. For a definite integral we
would require the x = 0 to be clearly substituted to obtain a constant term.
M1: Correct log work to make x the subject. May be scored before the previous M if the constant is found
later. Allow if the constant is zero, or is missing. Condone slips on signs.
You may see versions of the following showing correct exponential work:
1
k
−
(
)
(
)
−
ln
k
−
3
x
= +
t
c
ln
k
−
3
x
= −
3
t
+
d
k
−
3
x
=
e
−
3
t d
+
k
−
3
x
=
A
e
−
3
t
x
=
3
(
)
k
e
3
t
−
1
A1: Correct answer or exact equivalent. E.g.
.
x
=
3
3e
t
(b)
M1: Correctly identifies
e
−→
3
t
0
(which may be implied) and deduces the longterm limit for their
expression and sets their longterm limit to 7 and solves for k. The equation must be of the form
x
=
p
−
q −
e
3
t
with p in terms of k and q > 0. Alternatively, if they have an equation of the form
(
)
k
e
3
t
−
β
k
x
they would need to solve 7
= to find the value of k.
=
3
e
t
Condone work on equations of the same form where they have lost the 3 and used another positive value,
(
)
k
e
t
−
β
usually 1. So, score for equations
x
=
p
−
q
e
−
t
o
r
x
=
>
0
with the same conditions.
e
t
A1: k = 21 following a correct equation.
Correct answer following correct equation with no working scores both marks
Alt (b)
M1: Correctly identifies d
x
t =
k
−
3
x
=
0
and substitutes in x = 7 leading to a value for k
d
A1: k = 21
(c)
M1: Substitutes x = 5 and their value for k into their equation from (a), then rearranges to
This is dependent upon having a solvable equation of the form
x
= ±±
p
q −
e
3
t
o.e such as
Condone work on equations of the same form where they have lost the 3 and used another positive value,
±
μ
e
t
±
usually 1. So, score for equations
x
= ±
p
±
q
e
−
t
o
r
x
=
>
0
with the same conditions.
e
t
dM1: Solves an equation of the form
e
±=
3
t
k k
,
>
0
to find a value for t. Condone work in solving
equations of the form
e
±
t
=
k k
,
>
0
following the loss of the 3
A1: cso 0.42 (seconds)
Alt (c)
(
)
M1: Substitutes x = 5 into their solvable equation of the form
±
β
ln
k
−
3
x
= ±
t
ln
k
with their k
dM1: ...And finds a value for t
A1: cso awrt 0.42 (seconds)
4
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