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Edexcel IALMathematicsGeneral Applications7 Marks

The population of a species of insects in a laboratory is modelled by the equation P = (800e^(0.1t))/(1 + 4e^(0.1t)), t ≥ 0, where P is the population and t is the time in days.

  1. (a)Find the population when t = 0.[1]
  2. (b)Find the time when the population first reaches 150.[4]
  3. (c)Show that dP/dt = (80e^(0.1t))/(1 + 4e^(0.1t))².[2]

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

  1. (a) B1 for P = 800e⁰ / (1 + 4e⁰) = 800/5 = 160.
  2. (b) M1 for setting P=150: 150 = (800e^(0.1t))/(1 + 4e^(0.1t)).
  3. (b) M1 for rearranging to make e^(0.1t) the subject, e.g., 150 + 600e^(0.1t) = 800e^(0.1t).
  4. (b) A1 for 200e^(0.1t) = 150 or e^(0.1t) = 0.75.
  5. (b) A1 for t = (1/0.1)ln(0.75) = 10ln(0.75) = -2.877 (3 d.p.).
  6. (c) M1 for using the quotient rule correctly: d/dt (800e^(0.1t)) = 80e^(0.1t) and d/dt (1 + 4e^(0.1t)) = 0.4e^(0.1t).
  7. (c) A1 for correct application and simplification to (80e^(0.1t)(1 + 4e^(0.1t)) - 800e^(0.1t)(0.4e^(0.1t))) / (1 + 4e^(0.1t))² = (80e^(0.1t) + 320e^(0.2t) - 320e^(0.2t)) / (1 + 4e^(0.1t))² = (80e^(0.1t))/(1 + 4e^(0.1t))².

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Topic
General Applications, Edexcel IAL Mathematics. Structured question, 7 marks.

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