Mathematics · May / June 2022 · Paper 31 · Question 4
YMA01
As printed in the paper
The question as text
4. log10 N (5, 3.85) (0, 3.08) O t Figure 1 The number of subscribers to an online video streaming service, N, is modelled by the equation N = ab t where a and b are constants and t is the number of years since monitoring began. The line in Figure 1 shows the linear relationship between t and log10 N The line passes through the points (0, 3.08) and (5, 3.85) Using this information,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[2]
3.85
−
3.08
(=0.154 or 77
Gradient
=
)
5
−
0
500
log
N
=
3.08
+
0.154
t
oe
10
4(b)[3]
a =
10
3.08
=
awrt 1200
log
b =
"0.154"
or
b =
10
"0.154"
10
b =
awrt1.43
4(c)[2]
500000
log 500000
3.08
0.154
T
T
...
T
=
log
=
...
=
+
=
or
10
"1.43"
"1200"
(
)awrt 17
T =
Mark (a) and (b) together
(a)
M1
A correct attempt to find the gradient between the given coordinates. This may be implied by
or awrt 1.43 or 77
. The values embedded in the expression 3.85 3.08
or 3.08 3.85
−
−
is
500
5
−
0
−
5
sufficient. Check Figure 1 and may be seen in (b)
A1
log
N
=
3.08
+
0.154
t
Must be in terms of
log
N and t (may be seen in (b))
10
10
Condone log N or lg N for
log
N but do not accept ln N
10
77
77
Accept equivalent equations such as
log
N
=
+
t
or
500log
N
=
1540
+
77
t
or
10
25
500
10
500
1540
log
t
=
N
−
10
77
77
Do not accept
N =
1200 1.43t
×
(b)
B1
awrt 1200
M1
A correct equation to find b using their gradient in (a). Can be awarded for
log
b =
"0.154"
10
or
b =
10
"0.154"
(may be implied by awrt 1.43). Allow log or lg but not ln.
A1
awrt 1.43
(c)
N
ab
t
M1
Substitutes in N = 500 000 into their
=
using their values for a and b and uses a valid
method to find a value or expression for T (can be left as a logarithm)
This can be either by:
using correct inverse laws
using a calculator and proceeding directly to a value for t
500000
log "
1250
"
"1200" "1.43"
×
T
=
500000
"1.43"
T
=
T
=
10
=
...
Typically,
3
"1200"
log "1.43"
10
500000
500000
"1.43"
T
=
T
=
log
=
...
"1200"
"1.43"
"1200"
Alternatively uses their
log
N
=
"3.08" "0.154"
+
t
and rearranges to find a value for T
10
allowing sign slips only:
log 500000 "3.08"
−
log 500000
"3.08" "0.154"
T
T
...
=
+
=
10
=
10
"0.154"
May be implied by a correct answer.
(
)awrt 17
T =
A1
isw after a correct answer is seen (cannot be left as a logarithm)
This mark can only be scored provided EITHER their equation in (a) is correct OR their values
for a and b are correct in (b).
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