(i) To find the equation of the normal, first find the derivative of \(y = \sqrt{5x - 1}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2}(5x - 1)^{-\frac{1}{2}} \times 5 = \frac{5}{2\sqrt{5x - 1}}\).

At \(P(2, 3)\), \(\frac{dy}{dx} = \frac{5}{2 \times 3} = \frac{5}{6}\).

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

The equation of the normal is \(y - 3 = -\frac{6}{5}(x - 2)\).

(ii) To find the area of the shaded region, integrate the curve from \(x = \frac{1}{5}\) to \(x = 2\):

\(\int_{\frac{1}{5}}^{2} \sqrt{5x - 1} \, dx = \left[ \frac{2}{15}(5x - 1)^{\frac{3}{2}} \right]_{\frac{1}{5}}^{2} = 3.6\).

The normal crosses the x-axis at \(x = \frac{27}{5}\).

The area of the triangle formed by the normal and the x-axis is \(\frac{1}{2} \times \frac{15}{4} \times 3 = 3.75\).

Total area of the shaded region is \(3.6 + 3.75 = 7.35\) or \(\frac{147}{20}\).

(i) To find the \(x\)-coordinate of \(A\), set \(\frac{dy}{dx} = 0\):

\(2x - \frac{2}{x^3} = 0\)

\(2x = \frac{2}{x^3}\)

\(x^4 = 1\)

Thus, \(x = 1\) at \(A\).

(ii) To find \(f(x)\), integrate \(\frac{dy}{dx} = 2x - \frac{2}{x^3}\):

\(f(x) = \int (2x - \frac{2}{x^3}) \, dx\)

\(f(x) = x^2 + \frac{1}{x^2} + c\)

Using the point \(\left(4, \frac{189}{16}\right)\):

\(\frac{189}{16} = 16 + \frac{1}{16} + c\)

\(c = -\frac{17}{4}\)

Thus, \(f(x) = x^2 + \frac{1}{x^2} - \frac{17}{4}\).

(iii) To find the \(x\)-coordinates of \(B\) and \(C\), solve \(f(x) = 0\):

\(x^2 + \frac{1}{x^2} - \frac{17}{4} = 0\)

Multiply by \(4x^2\):

\(4x^4 - 17x^2 + 4 = 0\)

\((4x^2 - 1)(x^2 - 4) = 0\)

\(x = \frac{1}{2}, 2\)

(iv) To find the area of the shaded region, integrate \(f(x)\) from \(x = \frac{1}{2}\) to \(x = 2\):

\(\int_{1/2}^{2} \left(x^2 + \frac{1}{x^2} - \frac{17}{4}\right) \, dx\)

\(= \left[ \frac{x^3}{3} - \frac{1}{x} - \frac{17x}{4} \right]_{1/2}^{2}\)

\(= \left( \frac{8}{3} - \frac{1}{2} - \frac{17}{2} \right) - \left( \frac{1}{24} - 2 - \frac{17}{8} \right)\)

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

(a) Let \(u = 2x - 3\). Then the equations become \(y = 2u^4\) and \(y = u^2 + 1\).

Equating the two: \(2u^4 = u^2 + 1\).

Rearrange to form: \(2u^4 - u^2 - 1 = 0\).

Factor or solve using the quadratic formula: \((2u^2 + 1)(u^2 - 1) = 0\).

Thus, \(u = \pm 1\).

For \(u = 1\), \(2x - 3 = 1\) gives \(x = 2\).

For \(u = -1\), \(2x - 3 = -1\) gives \(x = 1\).

Coordinates are \((1, 2)\) and \((2, 2)\).

(b) Integrate the functions: \(\int_{1}^{2} \frac{(2x-3)^3}{3 \times 2}\) and \(\int_{1}^{2} \frac{2(2x-3)^5}{5 \times 2}\).

Calculate: \(\left[ \frac{1}{6} + 2 \right] - \left[ \frac{1}{5} - \frac{1}{5} \right]\).

Subtract the two areas: \(\frac{4}{3} - \frac{2}{5}\).

Final area: \(\frac{14}{15}\).

(i) To find the coordinates of \(A\), solve \(y = (2x - 1)^2\) and \(y^2 = 1 - 2x\) simultaneously. At \(A\), \(y = 0\), so substitute into \(y^2 = 1 - 2x\):

\(0^2 = 1 - 2x\)

\(0 = 1 - 2x\)

\(2x = 1\)

\(x = \frac{1}{2}\)

Thus, \(A = \left( \frac{1}{2}, 0 \right)\).

(ii) To find the area of the shaded region, integrate the difference between the curves from \(x = 0\) to \(x = \frac{1}{2}\).

First, integrate \(y^2 = 1 - 2x\):

\(\int (1 - 2x)^{1/2} \, dx = \left[ \frac{(1 - 2x)^{3/2}}{3/2} \right] \div (-2)\)

Evaluate from \(0\) to \(\frac{1}{2}\):

\(\left[ 0 - \left(-\frac{1}{3}\right) \right]\)

Next, integrate \(y = (2x - 1)^2\):

\(\int (2x - 1)^2 \, dx = \left[ \frac{(2x - 1)^3}{3} \right] \div 2\)

Evaluate from \(0\) to \(\frac{1}{2}\):

\(\left[ 0 - \left(-\frac{1}{6}\right) \right]\)

The area of the shaded region is:

\(\frac{1}{3} - \frac{1}{6} = \frac{1}{6}\).

(i) To find the \(x\)-coordinate of \(P\), set \(y = 0\) in the equation \(y = \frac{1}{16}(3x-1)^2\). Solving \(\frac{1}{16}(3x-1)^2 = 0\) gives \(3x-1 = 0\), so \(x = \frac{1}{3}\).

(ii) Differentiate \(y = \frac{1}{16}(3x-1)^2\) to find the gradient at \(Q\). \(\frac{dy}{dx} = \frac{2}{16}(3x-1) \times 3 = \frac{3}{8}(3x-1)\). At \(x = 3\), \(\frac{dy}{dx} = 3\). The equation of the tangent at \(Q\) is \(y - 4 = 3(x - 3)\). Setting \(y = 0\) to find \(R\), \(0 - 4 = 3(x - 3)\) gives \(x = \frac{5}{3}\).

(iii) The area under the curve from \(x = \frac{1}{3}\) to \(x = 3\) is \(\int_{1/3}^{3} \frac{1}{16}(3x-1)^2 \, dx\). This evaluates to \(\frac{1}{16 \times 3}[(3x-1)^3]_{1/3}^{3} = \frac{32}{9}\). The area of triangle \(\Delta\) is \(\frac{1}{2} \times (3 - \frac{1}{3}) \times 4 = \frac{8}{3}\). The shaded area \(PQR\) is \(\frac{32}{9} - \frac{8}{3} = \frac{8}{9}\) (or 0.889).