O-LevelAdditional MathematicsLogarithmic and exponential functions4037/22 · Oct/Nov 2016 · Q47 Marks
The number of bacteria, N, present in a culture can be modelled by the equation N = 7000+2000e-0.05t, where t is measured in days. Find
- (i)the number of bacteria when t = 10,[1]
- (ii)the value of t when the number of bacteria reaches 7500,[3]
- (iii)the rate at which the number of bacteria is decreasing after 8 days.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.
- [B1] (i) t=10→N=7000+2000e-0.5 = 8213 or 8210 Do not accept non integer responses.
- [M1] (ii) N = 7500→7500 = 7000 + 2000e-0.05t insert and make e-0.05t subject
- e-0.05t = 500/2000
- [M1] -0.05t = ln0.25 → t = ln 0.25 / -0.05 take logs and make t the subject
- [A1] = 27.7 (days) awrt 27.7
- [M1] (iii) dN/dt = -100e-0.05t ke^-0.05t where k is a constant
- [A1] k=-100 or -0.05x2000
- [A1] t=8→ dN/dt = ±67 (.0) awrt ±67 mark final answer
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