(i) The gradient of the curve is given by \(\frac{dy}{dx} = kxy\), where \(k\) is a constant. At \((1, 2)\), \(\frac{dy}{dx} = 4\), so \(4 = k \cdot 1 \cdot 2\), giving \(k = 2\). Thus, \(\frac{dy}{dx} = 2xy\).

Separate variables: \(\frac{1}{y} dy = 2x dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int 2x dx\).

This gives \(\ln |y| = x^2 + C\).

Exponentiate: \(y = e^{x^2 + C}\).

At \((1, 2)\), \(2 = e^{1^2 + C}\), so \(2 = e^{1 + C}\), giving \(C = \ln 2 - 1\).

Thus, \(y = e^{x^2 + \ln 2 - 1} = 2e^{x^2 - 1}\).

(ii) The gradient at \((-1, 2)\) is \(\frac{dy}{dx} = 2(-1)(2) = -4\).

(a)(i) The gradient of the normal at \(P\) is \(-\frac{1}{\frac{dy}{dx}}\). The length \(MN\) is the horizontal distance from \(M\) to \(N\), which is \(y \cdot \frac{dy}{dx}\). Therefore, \(\frac{MN}{y} = \frac{dy}{dx}\).

(a)(ii) The area of triangle \(PMN\) is \(\frac{1}{2} \times MN \times y = \frac{1}{2} \times y \cdot \frac{dy}{dx} \times y = \frac{1}{2}y^2 \frac{dy}{dx}\). Given that this area equals \(\tan x\), we have \(\frac{1}{2}y^2 \frac{dy}{dx} = \tan x\).

(b) Separate variables: \(\frac{1}{2}y^2 dy = \tan x \, dx\).

Integrate both sides: \(\int \frac{1}{2}y^2 dy = \int \tan x \, dx\).

This gives \(\frac{1}{6}y^3 = -\ln |\cos x| + C\).

Using the initial condition \(y = 1\) when \(x = 0\), we find \(C = \frac{1}{6}\).

Thus, \(\frac{1}{6}y^3 = -\ln |\cos x| + \frac{1}{6}\).

Rearranging gives \(y^3 = 1 - 6\ln |\cos x|\).

Therefore, \(y = \sqrt[3]{1 - 6\ln \cos x}\).

(i) The differential equation representing the biologist's claim is:

\(\frac{dA}{dt} = k\sqrt{2A - 5}\)

(ii) To solve the differential equation, separate variables and integrate:

\(\int \frac{1}{\sqrt{2A - 5}} \, dA = \int k \, dt\)

Integrating both sides, we obtain:

\((2A - 5)^{1/2} = kt + C\)

Using the initial condition \(t = 0\) and \(A = 7\), we find \(C\):

\((2 \times 7 - 5)^{1/2} = C\)

\(C = 3\)

Using \(t = 10\) and \(A = 27\) to find \(k\):

\((2 \times 27 - 5)^{1/2} = 10k + 3\)

\(7 = 10k + 3\)

\(k = 0.4\)

Substitute \(t = 20\), \(C = 3\), and \(k = 0.4\) to find \(A\):

\((2A - 5)^{1/2} = 0.4 \times 20 + 3\)

\((2A - 5)^{1/2} = 11\)

\(2A - 5 = 121\)

\(2A = 126\)

\(A = 63\)

(i) The rate of formation is proportional to \((20 - x)\), so \(\frac{dx}{dt} = k(20 - x)\). Given \(\frac{dx}{dt} = 1\) when \(t = 0\) and \(x = 0\), we have \(1 = k(20 - 0)\), thus \(k = 0.05\).

(ii) Separate variables: \(\frac{dx}{20 - x} = 0.05 \, dt\). Integrate both sides: \(-\ln|20 - x| = 0.05t + C\). Using \(t = 0\), \(x = 0\), we find \(C = -\ln 20\). Thus, \(\ln(20 - x) = \frac{1}{20}t\).

(iii) Substitute \(t = 10\) into \(\ln(20 - x) = \frac{1}{20}t\): \(\ln(20 - x) = \frac{1}{2}\). Solving gives \(x = 7.9\).

(iv) As \(t\) becomes very large, \(x\) approaches 20.

(i) The rate of increase of the surface area is proportional to the volume, so \(\frac{dA}{dt} = kV\). Given \(A = 4\pi r^2\) and \(V = \frac{4}{3}\pi r^3\), differentiate \(A\) with respect to \(t\):

\(\frac{dA}{dt} = 8\pi r \frac{dr}{dt}\).

Equating \(8\pi r \frac{dr}{dt} = k \frac{4}{3}\pi r^3\), solve for \(\frac{dr}{dt}\):

\(\frac{dr}{dt} = \frac{k}{6} r^2\).

Using \(\frac{dr}{dt} = 2\) and \(r = 5\), find \(k\):

\(2 = \frac{k}{6} \times 25 \Rightarrow k = 0.48\).

Thus, \(\frac{dr}{dt} = 0.08r^2\).

(ii) Separate variables and integrate:

\(\int \frac{1}{r^2} \, dr = \int 0.08 \, dt\).

\(-\frac{1}{r} = 0.08t + C\).

Using \(t = 0, r = 5\), find \(C\):

\(-\frac{1}{5} = C\).

Thus, \(-\frac{1}{r} = 0.08t - \frac{1}{5}\).

Solve for \(r\):

\(r = \frac{5}{1 - 0.4t}\).

(iii) For \(r\) to be positive, \(1 - 0.4t > 0\), so \(t < 2.5\). Therefore, \(0 \leq t < 2.5\).