Mathematics · Oct / Nov 2023 · Paper 31 · Question 4
YMA01
As printed in the paper
The question as text
4. A new mobile phone is released for sale. The total sales N of this phone, in thousands, is modelled by the equation –0.109t t 0 N = 125 – Ae where A is a constant and t is the time in months after the phone was released for sale. Given that when t = 0 , N = 32
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[1]
A =
93
4(b)[3]
100
=
125 "93"e
−
−
0.109
T
A
e
−
0.109
T
=
...
B
ln
A
A
ekT
=
B
T
=
k
T =
12.05
4(c)[2]
d
N
=
0.109 "93"e
×
−
0.109 7
×
d
t
4 730 (total sales per month)
4(d)[1]
The limit is 125 000 / the model has a limit below 150 000
Notes
(a)
B1:
93
(b)
M1:
Sets
100
125 "93"e
0.109
T
and proceeds to
e
0.109
T
...
May use t instead of T. Condone
=
−
−
A −
=
slips.
dM1: Uses correct order of operations and correct log work from an equation of form
A
e
kT
=
B AB
,
>
0
to proceed to a value for T (must be a solvable equation). Implied by a correct
answer for their equation.
A1:
12.05 cao
Note: Candidates who use 100 000 in the equation will score no marks in this part.
Note: Answer only with no working send to review, but 12.05 following a correct equation seen score
M1dM1A1.
(c)
M1:
Attempts to differentiate to form
λ
e
−
0.109
×, λ constant, and substitutes in
t
t =
7
. Award for an
expression of the form
×if no incorrect working is seen. The substitution of t = 7 may be
...e−
0.109 7
implied by a correct value for their derivative of correct form.
A1:
awrt 4 730 (total sales per month). Accept equivalent forms for the answer, e.g. 4.73 thousand,
but just 4.73 is A0. “Sales per month” may be omitted, but score A0 if an incorrect unit is given.
Note: Answer only scores no marks.
(d)
B1:
A correct reason given – any factual statements must be correct. E.g The limit is 125 000 / the
model has a limit below 150 000 oe. Accept maximum value of N is 125, or N cannot reach 150
as asymptote at 125. Also allow attempts to substitute in 150 and find that they cannot take the
log a negative number, but score B0 if a “negative time” is reached and used as reason.
Attempts using e.g 150 000 in the equation score B0.
There must be a reference to the information given in the answer, just “number is too large” or
similar is B0.
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