(a) Differentiate \(y = k \sqrt{4x + 1} - x + 5\) with respect to \(x\):

\(\frac{dy}{dx} = k \cdot \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 - 1\)

\(\frac{dy}{dx} = \frac{2k}{\sqrt{4x+1}} - 1\)

(b) Set \(\frac{dy}{dx} = 0\) for stationary points:

\(\frac{2k}{\sqrt{4x+1}} - 1 = 0\)

\(\frac{2k}{\sqrt{4x+1}} = 1\)

\(\sqrt{4x+1} = 2k\)

\(4x + 1 = 4k^2\)

\(x = \frac{4k^2 - 1}{4}\)

(c) Given \(k = 10.5\), substitute into \(\frac{dy}{dx} = 2\):

\(\frac{2 \times 10.5}{\sqrt{4x+1}} - 1 = 2\)

\(\frac{21}{\sqrt{4x+1}} = 3\)

\(\sqrt{4x+1} = 7\)

\(4x + 1 = 49\)

\(x = 12\)

Find \(y\) at \(x = 12\):

\(y = 10.5 \sqrt{4 \times 12 + 1} - 12 + 5\)

\(y = 66.5\)

The slope of the normal is \(-\frac{1}{2}\) (perpendicular to \(2\)).

Equation of the normal: \(y - 66.5 = -\frac{1}{2}(x - 12)\)

The equation of the curve is \(y = 3x - \frac{4}{x}\).

The derivative is \(\frac{dy}{dx} = 3 + \frac{4}{x^2}\).

The gradient of \(AB\) is \(m_{AB} = 4\).

Set \(\frac{dy}{dx} = 4\) to find \(C\) and \(D\):

\(3 + \frac{4}{x^2} = 4\)

\(\frac{4}{x^2} = 1\)

\(x^2 = 4\)

\(x = \pm 2\)

For \(x = 2\), \(y = 3(2) - \frac{4}{2} = 6 - 2 = 4\), so \(C(2, 4)\).

For \(x = -2\), \(y = 3(-2) - \frac{4}{-2} = -6 + 2 = -4\), so \(D(-2, -4)\).

The midpoint \(M\) of \(CD\) is \((0, 0)\).

The gradient of \(CD\) is \(m_{CD} = 2\).

The perpendicular gradient is \(-\frac{1}{2}\).

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

First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{(3x + 1)^2}\).

Using the chain rule, let \(u = 3x + 1\), then \(y = 4u^{-2}\).

The derivative \(\frac{dy}{du} = -8u^{-3}\).

Since \(u = 3x + 1\), \(\frac{du}{dx} = 3\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -8u^{-3} \cdot 3 = -24(3x + 1)^{-3}\).

When \(x = -1\), \(y = \frac{4}{(3(-1) + 1)^2} = \frac{4}{1} = 1\).

Also, \(\frac{dy}{dx} = -24(3(-1) + 1)^{-3} = 3\).

The equation of the tangent line at \(x = -1\) is given by the point-slope form: \(y - y_1 = m(x - x_1)\).

Substitute \(y_1 = 1\), \(m = 3\), and \(x_1 = -1\):

\(y - 1 = 3(x + 1)\).

Simplifying gives \(y = 3x + 4\).

(i) To find the equation of the tangent, we first differentiate the curve \(y = \sqrt{x^4 + 4x + 4}\).

The derivative is \(\frac{dy}{dx} = \left[ \frac{1}{2} (x^4 + 4x + 4)^{-\frac{1}{2}} \right] \times [4x^3 + 4]\).

At \(x = 0\), \(\frac{dy}{dx} = \frac{1}{2} \times 1 \times 4 = 2\).

The equation of the tangent is \(y - 2 = 2(x - 0)\), which simplifies to \(y - 2 = x\).

(ii) For the intersection, set \(x + 2 = \sqrt{x^4 + 4x + 4}\).

Squaring both sides gives \((x + 2)^2 = x^4 + 4x + 4\).

This simplifies to \(x^2 - x^4 = 0\), or \(x^2(1 - x^2) = 0\).

Thus, \(x = 0, \pm 1\).

(i) Differentiate the curve \(y = (6x + 2)^{\frac{1}{3}}\) to find the gradient at \(A\):

\(\frac{dy}{dx} = 6 \times \frac{1}{3}(6x + 2)^{-\frac{2}{3}} = 2(6x + 2)^{-\frac{2}{3}}\)

At \(x = 1\), \(\frac{dy}{dx} = 2(8)^{-\frac{2}{3}} = \frac{1}{2}\).

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

The gradient of the normal is \(-2\) (negative reciprocal of \(\frac{1}{2}\)).

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

(ii) The coordinates of \(B\) are \((0, \frac{1}{2})\) and \(C\) are \((2, 0)\).

Distance \(BC = \sqrt{(2 - 0)^2 + (0 - \frac{1}{2})^2} = \frac{5}{2}\).

(iii) The equation of \(BC\) is \(y - \frac{1}{2} = -\frac{3}{4}(x - 0)\) or \(y = -\frac{3}{4}x + \frac{1}{2}\).

Intersection \(E\) is found by solving \(-\frac{3}{4}x + \frac{1}{2} = 2x\).

\(x = \frac{6}{11}, \; y = \frac{12}{11}\).

The mid-point of \(OA\) is \(\left( \frac{1}{2}, 1 \right)\), so \(E\) is not the mid-point.