(i) To find \(\frac{dy}{dx}\), differentiate \(y = 1 - \frac{4}{(2x+1)^2}\) using the chain rule:

\(\frac{dy}{dx} = -\frac{d}{dx} \left( \frac{4}{(2x+1)^2} \right) = -\frac{8}{(2x+1)^3}\).

To find \(\int y \, dx\), integrate \(y = 1 - \frac{4}{(2x+1)^2}\):

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

(ii) At \(A\), \(x = \frac{1}{2}\). The gradient of the normal is the negative reciprocal of \(\frac{dy}{dx}\) at \(x = \frac{1}{2}\), which is \(-\frac{1}{2}\).

The equation of the normal is \(y - 0 = -\frac{1}{2}(x - \frac{1}{2})\), simplifying to \(y = -\frac{1}{2}x + \frac{1}{4}\).

Thus, \(B(0, \frac{1}{4})\).

(iii) The area of the shaded region is found by integrating \(y\) from 0 to \(\frac{1}{2}\):

\(\int_0^{\frac{1}{2}} \left( 1 - \frac{4}{(2x+1)^2} \right) dx = \left[ x + \frac{1}{2x+1} \right]_0^{\frac{1}{2}} = \frac{1}{2} + \frac{1}{2} - 0 - 1 = -\frac{1}{4}\).

The area of the triangle above the x-axis is \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{16}\).

Total area of shaded region = \(\frac{9}{16}\).

(a) To find the coordinates of \(A\) and \(B\), equate the two expressions for \(y\):

\(2x^{\frac{1}{2}} + 13x^{-\frac{1}{2}} = 3x^{-\frac{1}{2}} + 12\)

Rearrange to isolate terms:

\(2x^{\frac{1}{2}} + 13x^{-\frac{1}{2}} - 3x^{-\frac{1}{2}} - 12 = 0\)

\(2x^{\frac{1}{2}} + 10x^{-\frac{1}{2}} - 12 = 0\)

Multiply through by \(x^{\frac{1}{2}}\) to clear the fraction:

\(2x + 10 - 12x^{\frac{1}{2}} = 0\)

Rearrange to form a quadratic in \(x^{\frac{1}{2}}\):

\((x^{\frac{1}{2}} - 1)(x^{\frac{1}{2}} - 25) = 0\)

Solving gives \(x = 1\) and \(x = 25\).

Substitute back to find \(y\):

For \(x = 1\), \(y = 2(1)^{\frac{1}{2}} + 13(1)^{-\frac{1}{2}} = 15\).

For \(x = 25\), \(y = 2(25)^{\frac{1}{2}} + 13(25)^{-\frac{1}{2}} = \frac{128}{5}\).

(b) To find the area of the shaded region, integrate the difference between the curves from \(x = 1\) to \(x = 25\):

\(\text{Area} = \int_{1}^{25} \left( 2x^{\frac{1}{2}} + 13x^{-\frac{1}{2}} - (3x^{-\frac{1}{2}} + 12) \right) \, dx\)

\(= \int_{1}^{25} \left( 2x^{\frac{1}{2}} + 10x^{-\frac{1}{2}} - 12 \right) \, dx\)

Integrate each term:

\(\int 2x^{\frac{1}{2}} \, dx = \frac{4}{3}x^{\frac{3}{2}}\)

\(\int 10x^{-\frac{1}{2}} \, dx = 20x^{\frac{1}{2}}\)

\(\int -12 \, dx = -12x\)

Evaluate from 1 to 25:

\(\left[ \frac{4}{3}(25)^{\frac{3}{2}} + 20(25)^{\frac{1}{2}} - 12(25) \right] - \left[ \frac{4}{3}(1)^{\frac{3}{2}} + 20(1)^{\frac{1}{2}} - 12(1) \right]\)

\(= \frac{128}{3}\) or approximately 42.7.

(i) Differentiate \(y = (3x + 4)^{\frac{1}{2}}\) to find the gradient: \(\frac{1}{2}(3x + 4)^{-\frac{1}{2}} \times 3 = \frac{3}{2}(3x + 4)^{-\frac{1}{2}}\).

At \(x = 4\), \(\frac{dy}{dx} = \frac{3}{8}\).

The equation of the tangent is \(y - 3 = \frac{3}{8}(x - 4)\), which simplifies to \(y = \frac{3}{8}x + \frac{5}{2}\).

(ii) The area under the line is \(\frac{1}{2} \times 4 \times 3 = 3\).

The area under the curve is \(\int_0^4 (3x + 4)^{\frac{1}{2}} \, dx = \left[ \frac{2}{3}(3x + 4)^{\frac{3}{2}} \right]_0^4 = 13\).

The area of the shaded region is \(13 - 12 = \frac{5}{9}\) (or 0.556).

(iii) Given \(\frac{dy}{dx} = \frac{1}{2}\), solve \(\frac{3}{2}(3x + 4)^{-\frac{1}{2}} = 1\).

This gives \((3x + 4)^{\frac{1}{2}} = 3\), so \(3x + 4 = 9\), leading to \(x = \frac{5}{3}\).

(i) To find \(\frac{dy}{dx}\), differentiate \(y = \sqrt{4x+1} + \frac{9}{\sqrt{4x+1}}\).

\(\frac{dy}{dx} = \frac{1}{2}(4x+1)^{-\frac{1}{2}} \times 4 + \frac{9}{2}(4x+1)^{-\frac{3}{2}} \times (-4)\)

\(= \frac{2}{\sqrt{4x+1}} - \frac{18}{(4x+1)^{\frac{3}{2}}}\)

To find \(\int y \, dx\), integrate \(y = \sqrt{4x+1} + \frac{9}{\sqrt{4x+1}}\).

\(\int y \, dx = \int (4x+1)^{\frac{1}{2}} \, dx + \int \frac{9}{(4x+1)^{\frac{1}{2}}} \, dx\)

\(= \frac{(4x+1)^{\frac{3}{2}}}{6} + \frac{9}{2}(4x+1)^{\frac{1}{2}} + C\)

(ii) To find the coordinates of \(M\), set \(\frac{dy}{dx} = 0\).

\(\frac{2}{\sqrt{4x+1}} - \frac{18}{(4x+1)^{\frac{3}{2}}} = 0\)

\(4x+1 = 9\) or \((4x+1)^2 = 81\)

\(x = 2, y = 6\) so \(M = (2, 6)\)

(iii) The area is \(\int y \, dx\) from 0 to 2.

\(\left[ \frac{(4x+1)^{\frac{3}{2}}}{6} + \frac{9}{2}(4x+1)^{\frac{1}{2}} \right]_0^2\)

\(= [4.5 + 13.5] - [\frac{1}{6} + 4.5] = \frac{4}{3}\) or 1.33

(i) To find the \(x\)-coordinate of \(Q\), we first find the derivative of the curve \(y = \frac{3}{\sqrt{1 + 4x}}\). The derivative is \(\frac{dy}{dx} = 3 \times -\frac{1}{2} \times (1 + 4x)^{-\frac{3}{2}} \times 4\).

At \(x = 2\), the gradient \(m\) is \(-\frac{2}{9}\). The perpendicular gradient (normal) is \(\frac{9}{2}\).

The equation of the normal is \(y - 1 = \frac{9}{2}(x - 2)\).

Setting \(y = 0\) to find \(Q\), we solve \(0 - 1 = \frac{9}{2}(x - 2)\), giving \(x = \frac{16}{9}\).

(ii) The area under the curve from \(x = 0\) to \(x = 2\) is \(\int_{0}^{2} \frac{3}{\sqrt{1 + 4x}} \, dx = \frac{3}{2} \sqrt{1 + 4x} \bigg|_0^2 \times \frac{1}{4}\).

Evaluating gives \(3\).

The area of the triangle is \(\frac{1}{2} \times 1 \times \frac{2}{9} = \frac{1}{9}\).

The shaded area is \(3 - \frac{1}{9} = \frac{8}{9}\).