| Date | May 2018 | Marks available | 2 | Reference code | 18M.1.sl.TZ2.10 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 10 | Adapted from | N/A |
Question
The following function models the growth of a bacteria population in an experiment,
P(t) = A × 2t, t ≥ 0
where A is a constant and t is the time, in hours, since the experiment began.
Four hours after the experiment began, the bacteria population is 6400.
Find the value of A.
Interpret what A represents in this context.
Find the time since the experiment began for the bacteria population to be equal to 40A.
Markscheme
6400 = A × 24 (M1)
Note: Award (M1) for correct substitution of 4 and 6400 in equation.
(A =) 400 (A1) (C2)
[2 marks]
the initial population OR the population at the start of experiment (A1) (C1)
[1 mark]
40A = A × 2t OR 40 × 400 = 400 × 2t (M1)
Note: Award (M1) for correct substitution into equation. Follow through with their A from part (a).
40 = 2t (M1)
Note: Award (M1) for simplifying.
5.32 (5.32192…) (hours) OR 5 hours 19.3 (19.3156…) minutes (A1) (C3)
[3 marks]
