The rate of increase of the number, N, of fish in a lake is modelled by the differential equation dN/dt = (k - t)(5000 - N) / t t > 0, 0 < N < 5000 In the given equation, the time t is measured in years from the start of January 2000 and k is a positive constant - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 8

From the problem, at t = 1 (start of January 2001), N = 1200:

$1200 = 5000 - Ae^{-k}$

Thus, we can find A:

$A = 5000 - 1200 e^{k} = 3800 e^{k}$

At t = 2 (start of January 2002), N = 1800:

$1800 = 5000 - 3800 e^{-2k}$

From this, we get:

$3800 e^{-2k} = 3200$

Solving the above, we can derive values for A and k:

$A = 9025$ $k = 9025 / 3800$. Therefore, using the logarithmic properties, we can finalize both constants.

To find the number of fish after five years (t = 5), substitute in our found values:

$N = 5000 - 9025 e^{-5k}$

Calculating e^{-5k} and plugging into the equation allows us to obtain:

$N \approx 4402.828401...$

Rounding to the nearest hundred, we find:

N = 4400 fish.