Mathematics · May / June 2025 · Paper 31 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The temperature, θ°C, of a computer processor, t minutes after the computer is switched off, is modelled by the equation –kt θ = 21 + Ae where A and k are positive constants. Given that the temperature of the processor was 75°C when the computer was switched off,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[2]
θ=
75,
t
=
0
75
=
21
+
A
A
=
...
A =
54
6(b)[3]
21
54e
−
kt
25
21
54e
−
5
k
θ
=
+
=
+
2
2
5
k
5
k
54e
−
=
4
e
−
=
−
5
k
=
ln
k
=
...
27
27
1
2
k = −
ln
=
0.521
5
27
Notes:
(a)
M1: Substitutes θ = 75 and t = 0 into the given equation, uses
0e
= and proceeds to find a value for A.
1
Look for 75
=
21
+
A
×
1
A
=
...
A1:
For A = 54
Sight of 54 with no incorrect working scores both marks.
(b)
M1: Uses θ = 25 and t = 5 with their A in
θ
=
21
+
A
e kt
±
so e.g.
25
=
21 54e
+
±
5
k
is all that is required.
M1: Proceeds from an equation of the form
e
±
5
k
=
α
where
α>
0
to 5
±
k
=
ln
α
k
=
...
to obtain a value
for k. Condone slips but the log work must be correct. Must be using ln and not e.g.
log which for
10
reference gives k = 0.226
May be implied by the correct exact value for k following through their A.
A1:
For awrt 0.521 from correct and sufficient work.
1
2
ln
Allow the exact value e.g.
−
and apply isw if subsequently rounded incorrectly.
5
27
May be seen embedded e.g.
θ
=
21 54e
+
−
0.521
t
1
4
25
=
21 54e
+
−
5
k
k
= −
ln
Minimum for full marks in (b) is:
5
54
4
25
=
21 54e
+
−
5
k
e
−
5
k
=
k
=
0.521
scores M1dM0A0
54
25
21 54e
−
5
k
k
0.521
scores M1dM0A0
=
+
=
Note that some candidates are using θ = 50 and t = 5 leading to k = 0.124 and this scores a maximum
of M0M1A0 in (b)
6(c)[3]
d
θ
θ
=
21 54e
+
−
0.521
t
= −
28.1...e
−
0.521
t
d
t
9
9
−
28.1...e
−
0.521
T
= −
9
e
−
0.521
T
=
−
0.521
T
=
ln
28.1...
28.1...
9
T
=
ln
÷−
0.521
28.1...
=
2.19
(c)
M1: Differentiates to obtain d
θ
e
−
kt
=
α
where α ≠ A
d
t
dM1: Depends on the previous mark.
d
θ
Sets their d
θ = − 9 or sets
−
their
=
9
and proceeds to obtain a value for t (or T).
dt
dt
You do not need to be concerned about the processing as long as they reach a value.
For this mark they must have obtained d
θ
=
α
e
−
kt
with
and k > 0
α<
0
d
t
Do not condone work where candidates ignore/cross out signs to make their equation solvable.
A1:
For awrt 2.19
Just look for this value so ignore any units, correct or incorrect, and apply isw once awrt 2.19 is seen.
Must follow M1dM1 previously.
Note candidates who used θ = 50 and t = 5 leading to k = 0.124 in (b) who proceed correctly in (c)
will obtain T = −2.38 and can score a maximum of M1dM1A0 in (c)
Note correct answer with no working scores no marks in (c).
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