O-LevelAdditional MathematicsLogarithmic and exponential functions4037/22 · Oct/Nov 2014 · Q57 Marks
The number of bacteria B in a culture, t days after the first observation, is given by B = 500 + 400e^(0.2t).
- (i)Find the initial number present.[1]
- (ii)Find the number present after 10 days.[1]
- (iii)Find the rate at which the bacteria are increasing after 10 days.[2]
- (iv)Find the value of t when B = 10000.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.
- [B1] (i) B = 900
- [B1] (ii) B = 500 + 400e² = 3455 or 3456 or 3460. 3455.6 scores B0
- [B1] (iii) dB/dt = 80e^(0.2t)
- [B1] (iii) t=10 → dB/dt = 80e² = 591 (/day) awrt
- [M1] (iv) 10000=500+ 400e^(0.2t) → e^(0.2t) = (23.75). e^(0.2t) = k
- [DM1] (iv) 0.2t = ln 23.75. Take logs: 0.2t = ln k
- [A1] (iv) t = 15.8 (days) awrt
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