8/2/2019 Test Mark Scheme - Logs and Exponents (2012) v2
4/5
4
Paper 2: Calculator Allowed Date Block
25 marks, 25 minutes allowed
Full marks are not necessarily awarded for a correct answer with
no working. Answers must be supported
by working and/or explanations. In particular, solutions found
from a graphic display calculator should be
supported by suitable working, e.g. if graphs are used to find a
solution, you should sketch these as part of
your answer. Where an answer is incorrect, some marks may be
given for a correct method, provided this
is shown by written working. You are therefore advised to show
all working.
(8) (a) eln(x + 2)
= e3
(M1)
x + 2 = e3
(A1)
x = e3 2 (= 18.1) A1 N3
(b) log10 (102x
) = log10 500 (accept lg and log for log10) (M1)
2x = log10 500 (A1)
x = log10 500 A1 N3
Note: In both parts (a) and (b), if candidates usea graphical
approach, awardM1 for a
sketch, A1 for indicating appropriate points
of intersection, and A1 for the answer.[6]
(9) A population of ladybugs rapidly multiplies so that the
population tdays from now is given by
() .
a. What is the present ladybug population?b. How many complete
days will it take for the population to reach 3218 ladybugs?
a) 3000 A1b) Solve 0.013218 3000 te (M1)
0.013218
3000
te
3218ln 0.013000
t
M1 A1
7.01477...t (A1) The population will reach 3218 ladybugs on the
8th day. A1[Accept 8t days. Do not accept 7t ]
(Total 6 marks)
2
1
1.35
100log
500log
8/2/2019 Test Mark Scheme - Logs and Exponents (2012) v2
5/5
5
(10) (a) (i) f(a) = 1 A1 N1
(ii) f(1) = 0 A1 N1
(iii) f(a4) = 4 A1 N1
(b)
A1A1A1 N3
Note: Award A1 for approximate reflection of
f in y = x, A1 for y intercept at 1, and
A1 for curve asymptotic to x axis.[6]
(11) Note: A reminder that a candidate is penalized only once in
this question
for not giving answers to 3 sf
(a) V(5) = 10000 (0.9335) = 7069.8
= 7070 (3 sf) (A1) 1
(b) We want twhen V= 5000 (M1)
5000 = 10000 (0.933)t
0.5 = 0.933t
(A1)
9.9949 = tAfter 10 minutes 0 seconds, to nearest second (or 600
seconds). (A1) 3
(c) 0.05 = 0.933t
(M1)
= t= 43.197 minutes (M1)(A1)
3/4 hour (AG) 3
2
2
1
1
1
1
2
2 0
f
f1y
x
(0.933)ln
(0.5)lnor
)933.0(log
)5.0(logt
)933.0(log
)05.0(log
