(i) Separate variables and integrate:

\(\int \frac{1}{x} \, dx = -k \int \frac{1}{\sqrt{t}} \, dt\)

\(\ln x = -2k\sqrt{t} + C\)

Use initial condition \(x = 100\) when \(t = 0\):

\(\ln 100 = C\)

Thus, \(\ln x = -2k\sqrt{t} + \ln 100\)

(ii) Given \(t = 25\) when \(x = 80\):

\(\ln 80 = -2k\sqrt{25} + \ln 100\)

\(\ln 80 = -10k + \ln 100\)

\(10k = \ln 100 - \ln 80\)

\(k = \frac{\ln 100 - \ln 80}{10}\)

Find \(t\) when \(x = 40\):

\(\ln 40 = -2k\sqrt{t} + \ln 100\)

\(2k\sqrt{t} = \ln 100 - \ln 40\)

\(\sqrt{t} = \frac{\ln 100 - \ln 40}{2k}\)

Substitute \(k\):

\(\sqrt{t} = \frac{\ln 100 - \ln 40}{2 \times \frac{\ln 100 - \ln 80}{10}}\)

\(\sqrt{t} = \frac{10(\ln 100 - \ln 40)}{2(\ln 100 - \ln 80)}\)

\(t = \left( \frac{10(\ln 100 - \ln 40)}{2(\ln 100 - \ln 80)} \right)^2\)

\(t \approx 64.1\)

(i) Separate variables: \(\frac{1}{x(4-x)} \frac{dx}{dt} = k\).

Integrate both sides: \(\int \frac{1}{x(4-x)} \, dx = \int k \, dt\).

Use partial fractions: \(\frac{1}{x(4-x)} = \frac{A}{x} + \frac{B}{4-x}\).

Find \(A = \frac{1}{4}\) and \(B = \frac{1}{4}\).

Integrate: \(\frac{1}{4} \ln x - \frac{1}{4} \ln(4-x) = kt + C\).

Use initial conditions \(t = 0, x = 0.4\) and \(t = 2, x = 2\) to find \(C\) and \(k\).

\(\ln \left( \frac{x}{4-x} \right) = 4kt - 8k\).

Substitute \(t = 2, x = 2\): \(\ln 1 = 8k - 8k\).

Substitute \(t = 0, x = 0.4\): \(\ln \left( \frac{0.4}{3.6} \right) = -8k\).

Solve for \(k\): \(k = \frac{1}{4} \ln 3\).

(ii) When 90% of the area is infected, \(x = 3.6\).

Substitute \(x = 3.6\) into \(\ln \left( \frac{x}{4-x} \right) = 4kt - 8k\).

Solve for \(t\): \(t = 4\).

(i) Separate variables and integrate:

\(\int \frac{1}{\sqrt{M}} \, dM = \int k \cos(0.02t) \, dt\)

Integrating both sides gives:

\(2\sqrt{M} = 50k \sin(0.02t) + C\)

Using initial condition \(M = 100\) when \(t = 0\), we find \(C = 20\).

Thus, \(2\sqrt{M} = 50k \sin(0.02t) + 20\).

(ii) Given \(M = 196\) when \(t = 50\), substitute to find \(k\):

\(2\sqrt{196} = 50k \sin(1) + 20\)

\(28 = 50k \sin(1) + 20\)

\(8 = 50k \sin(1)\)

\(k = \frac{8}{50 \sin(1)} \approx 0.190\)

(iii) Express \(M\) in terms of \(t\):

\(2\sqrt{M} = 50 \times 0.190 \sin(0.02t) + 20\)

\(\sqrt{M} = 4.75 \sin(0.02t) + 10\)

\(M = (4.75 \sin(0.02t) + 10)^2\)

The least possible number of micro-organisms occurs when \(\sin(0.02t) = -1\):

\(M = (4.75(-1) + 10)^2 = (5.25)^2 = 27.5625\)

Thus, the least possible number is approximately 28 or 27.5 or 27.6.

(i) Separate variables and integrate:

\(\int \frac{1}{x} \, dx = \int \frac{e^{-t}}{k + e^{-t}} \, dt\)

\(\ln x = -\ln(k + e^{-t}) + C\)

Given \(x = 10\) when \(t = 0\), substitute to find \(C\):

\(\ln 10 = -\ln(k + 1) + C\)

Thus, \(C = \ln 10 + \ln(k + 1)\).

Therefore, the solution is:

\(\ln x - \ln 10 = -\ln(k + e^{-t}) + \ln(k + 1)\)

(ii) Substitute \(x = 20\), \(t = 1\) into the solution:

\(\ln 20 - \ln 10 = -\ln(k + e^{-1}) + \ln(k + 1)\)

Simplify and solve for \(k\):

\(\ln 2 = \ln \left( \frac{k + 1}{k + \frac{1}{e}} \right)\)

\(2(k + \frac{1}{e}) = k + 1\)

\(2k + \frac{2}{e} = k + 1\)

\(k = 1 - \frac{2}{e}\)

(iii) As \(t \to \infty\), \(e^{-t} \to 0\):

\(\ln x = -\ln k + \ln(k + 1)\)

\(x = \frac{k + 1}{k}\)

Substitute \(k = 1 - \frac{2}{e}\):

\(x = \frac{1 - \frac{2}{e} + 1}{1 - \frac{2}{e}} = \frac{2 - \frac{2}{e}}{1 - \frac{2}{e}}\)

\(x \to 48\) as \(t \to \infty\).

(i) Separate variables and integrate:

\(\int \frac{1}{R} \, dR = \int \left( \frac{1}{x} - 0.57 \right) \, dx\)

\(\ln R = \ln x - 0.57x + C\)

Exponentiate both sides:

\(R = e^{\ln x - 0.57x + C} = e^C \cdot xe^{-0.57x}\)

When \(x = 0.5, R = 16.8\):

\(16.8 = e^C \cdot 0.5e^{-0.285}\)

\(e^C = \frac{16.8}{0.5e^{-0.285}} = 33.6\)

Thus, \(R = 33.6xe^{(0.285 - 0.57x)}\).

(ii) To find the maximum \(R\), set \(\frac{dR}{dx} = 0\):

\(\frac{1}{x} - 0.57 = 0\)

\(x = 0.57^{-1} \approx 1.75\)

Substitute back to find \(R\):

\(R = 33.6 \times 1.75 \times e^{(0.285 - 0.57 \times 1.75)} \approx 28.8\)