(a) To find the stationary point \(R\), we need to find \(\frac{dy}{dx}\) and set it to zero. The derivative is \(\frac{dy}{dx} = -\frac{8}{(2x-1)^3} + 1\).

Substitute \(x = \frac{3}{2}\) into the derivative: \(\frac{dy}{dx} = -\frac{8}{8} + 1 = 0\). Hence, the \(x\)-coordinate of \(R\) is \(\frac{3}{2}\).

When \(x = \frac{3}{2}\), \(y = \frac{2}{4} + \frac{3}{2} = 2\). Thus, the \(y\)-coordinate of \(R\) is 2.

(b) The \(y\)-coordinate of \(P\) is 3, and the \(y\)-coordinate of \(Q\) is \(\frac{20}{9}\).

The area under the curve from \(x = 1\) to \(x = 2\) is given by:

\(\int_{1}^{2} \left( \frac{2}{(2x-1)^2} + x \right) \, dx = \left[ \frac{1}{3} x^3 + 2x - \frac{1}{2x-1} \right]_{1}^{2}\).

Evaluating this integral gives \(\frac{13}{6}\).

The area of the trapezium formed by the lines \(x = 1\), \(x = 2\), and the line \(y = x\) is \(\frac{1}{2} \left( 3 + \frac{20}{9} \right) = \frac{47}{18}\).

Thus, the exact value of the area of the shaded region is \(\frac{47}{18}\).

(a) To find the points of intersection, set the equations equal: \((3x - 2)^{\frac{1}{2}} = \frac{1}{2}x + 1\).

Square both sides: \(3x - 2 = \left( \frac{1}{2}x + 1 \right)^2\).

Expand the right side: \(3x - 2 = \frac{1}{4}x^2 + x + 1\).

Rearrange to form a quadratic equation: \(\frac{1}{4}x^2 - 2x + 3 = 0\).

Multiply through by 4 to clear the fraction: \(x^2 - 8x + 12 = 0\).

Factorize: \((x - 6)(x - 2) = 0\).

Thus, \(x = 2\) or \(x = 6\).

Substitute back to find \(y\):

For \(x = 2\), \(y = \frac{1}{2}(2) + 1 = 2\).

For \(x = 6\), \(y = \frac{1}{2}(6) + 1 = 4\).

Coordinates are \((2, 2)\) and \((6, 4)\).

(b) The area between the curve and the line is given by the integral:

\(\text{Area} = \int_{2}^{6} \left( (3x - 2)^{\frac{1}{2}} - \left( \frac{1}{2}x + 1 \right) \right) \, dx\).

Integrate each part:

\(\int (3x - 2)^{\frac{1}{2}} \, dx = \frac{2}{9}(3x - 2)^{\frac{3}{2}}\).

\(\int \left( \frac{1}{2}x + 1 \right) \, dx = \frac{1}{4}x^2 + x\).

Evaluate from 2 to 6:

\(\left[ \frac{2}{9}(3x - 2)^{\frac{3}{2}} \right]_{2}^{6} - \left[ \frac{1}{4}x^2 + x \right]_{2}^{6}\).

Calculate each part:

\(\frac{2}{9}(16)^{\frac{3}{2}} - \frac{2}{9}(4)^{\frac{3}{2}} - \left( \frac{1}{4}(36) + 6 - \left( \frac{1}{4}(4) + 2 \right) \right)\).

\(= \frac{4}{9}\).

(a) To find the area between the curves, we calculate the definite integral of the difference between the two functions from \(x = \frac{1}{4}\) to \(x = 4\):

\(\int_{1/4}^{4} \left( \frac{5}{2} - x^{-\frac{1}{2}} - x^{-\frac{1}{2}} \right) \, dx\)

\(= \int_{1/4}^{4} \left( \frac{5}{2} - 2x^{-\frac{1}{2}} \right) \, dx\)

Integrating term by term, we have:

\(\frac{5}{2}x - \frac{2}{3}x^{\frac{3}{2}} \bigg|_{1/4}^{4}\)

Substituting the limits:

\(\left( \frac{5}{2} \times 4 - \frac{2}{3} \times 4^{\frac{3}{2}} \right) - \left( \frac{5}{2} \times \frac{1}{4} - \frac{2}{3} \times \left( \frac{1}{4} \right)^{\frac{3}{2}} \right)\)

\(= \left( 10 - \frac{16}{3} \right) - \left( \frac{5}{8} - \frac{1}{12} \right)\)

\(= \frac{9}{8}\)

(b) The derivative of \(y = x^{-\frac{1}{2}}\) is \(\frac{dy}{dx} = -\frac{1}{2}x^{-\frac{3}{2}}\).

At \(x = 1\), the gradient \(m = -\frac{1}{2}\).

The equation of the normal at \((1, 1)\) is:

\(y - 1 = 2(x - 1)\)

Substituting \(x = 0\) to find \(p\):

\(y - 1 = 2(0 - 1)\)

\(y - 1 = -2\)

\(y = -1\)

Thus, \(p = -1\).

(a) Differentiate the curve: \(\frac{dy}{dx} = \frac{1}{2} + \frac{1}{3(x-2)^{\frac{4}{3}}}\).

Evaluate \(\frac{dy}{dx}\) at \(x = 3\): \(\frac{1}{2} + \frac{1}{3(3-2)^{\frac{4}{3}}} = \frac{5}{6}\).

The gradient of the normal is the negative reciprocal: \(-\frac{6}{5}\).

Equation of the normal: \(y - \frac{6}{5} = -\frac{6}{5}(x - 3)\).

Simplify: \(y = -\frac{6}{5}x + \frac{24}{5}\).

Set \(y = 0\) to find \(x\): \(0 = -\frac{6}{5}x + \frac{24}{5} \Rightarrow x = 4\).

(b) Integrate the curve from \(x = 2.5\) to \(x = 3\):

\(\int \left( \frac{1}{2}x + \frac{7}{10} - \frac{1}{(x-2)^{\frac{1}{3}}} \right) \, dx\).

Calculate the definite integral: \(\left[ \frac{1}{4}x^2 + \frac{7}{10}x - 3(x-2)^{\frac{2}{3}} \right]_{2.5}^{3}\).

Evaluate: \(\left( \frac{9}{4} + 2.1 - \frac{3}{2} \right) - \left( \frac{6.25}{4} + 1.75 - 3 \times 0.5^{\frac{2}{3}} \right) = 0.48\).

Area of triangle: \(0.6\).

Total area: \(0.48 + 0.6 = 1.08\).

(a) Differentiate \(y = x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}\) to find \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}k^2 x^{-\frac{3}{2}}\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(\frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2}k^2 x^{-\frac{3}{2}}\).

Simplify to get \(x = k^2\). Substitute back into the original equation to find \(y = 2k\).

Thus, the minimum point is \((k^2, 2k)\).

(b) At \(x = 4k^2\), find \(\frac{dy}{dx} = \frac{1}{4k} - \frac{1}{16k} = \frac{3}{16k}\).

The y-coordinate at \(x = 4k^2\) is \(y = 2k + \frac{k}{2} = \frac{5k}{2}\).

The equation of the tangent is \(y - \frac{5k}{2} = \frac{3}{16k}(x - 4k^2)\).

When \(x = 0\), \(y = \frac{7k}{4}\).

(c) Integrate \(\int (x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}) \, dx\) to get \(\frac{2}{3}x^{\frac{3}{2}} + 2k^2 x^{\frac{1}{2}}\).

Apply limits from \(x = \frac{9}{4}k^2\) to \(x = 4k^2\).

Calculate the definite integral: \(\left[ \frac{16k^3}{3} + 4k^3 \right] - \left[ \frac{9k^3}{4} + 3k^3 \right] = \frac{49k^3}{12}\).