To find the stationary points, we first find the derivative of the curve:

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

Set \(\frac{dy}{dx} = 0\) to find the stationary points:

\(3x^2 + 2x - 8 = 0\).

Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 2\), and \(c = -8\), we find:

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-8)}}{2 \times 3}\).

\(x = \frac{-2 \pm \sqrt{4 + 96}}{6}\).

\(x = \frac{-2 \pm \sqrt{100}}{6}\).

\(x = \frac{-2 \pm 10}{6}\).

This gives the solutions \(x = \frac{8}{6} = \frac{4}{3}\) and \(x = \frac{-12}{6} = -2\).

For the curve to have no stationary points in the interval \(a < x < b\), the interval must not include these solutions. Therefore, the least possible value of \(a\) is \(-2\) and the greatest possible value of \(b\) is \(\frac{4}{3}\).

First, find the derivative of the curve \(y = 2x^3 - 5x^2 - 3x + c\):

\(\frac{dy}{dx} = 6x^2 - 10x - 3\).

Evaluate the derivative at \(x = 2\):

\(\frac{dy}{dx} = 6(2)^2 - 10(2) - 3 = 24 - 20 - 3 = 1\).

Since the line is tangent to the curve, the slope \(a\) of the line is equal to the derivative at \(x = 2\):

\(a = 1\).

Substitute \(x = 2\) and \(y = 6\) into the line equation \(y = ax + b\):

\(6 = 2(1) + b \Rightarrow b = 4\).

Substitute \(x = 2\) and \(y = 6\) into the curve equation \(y = 2x^3 - 5x^2 - 3x + c\):

\(6 = 2(2)^3 - 5(2)^2 - 3(2) + c\).

Simplify to find \(c\):

\(6 = 16 - 20 - 6 + c \Rightarrow c = 16\).

(i) To find \(\frac{dy}{dx}\), differentiate \(y = (2x - 1)^{-1} + 2x\).

Using the chain rule, \(\frac{d}{dx}((2x - 1)^{-1}) = -1(2x - 1)^{-2} \cdot 2 = -2(2x - 1)^{-2}\).

The derivative of \(2x\) is \(2\).

Thus, \(\frac{dy}{dx} = -2(2x - 1)^{-2} + 2\).

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = -2(2x - 1)^{-2} + 2\).

Using the chain rule again, \(\frac{d}{dx}(-2(2x - 1)^{-2}) = -2 \cdot -2(2x - 1)^{-3} \cdot 2 = 8(2x - 1)^{-3}\).

Thus, \(\frac{d^2y}{dx^2} = 8(2x - 1)^{-3}\).

(ii) To find stationary points, set \(\frac{dy}{dx} = 0\).

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

\(-2(2x - 1)^{-2} = -2\)

\((2x - 1)^{-2} = 1\)

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

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

\(2x = 2 \Rightarrow x = 1\)

\(2x = 0 \Rightarrow x = 0\)

For \(x = 0\), \(\frac{d^2y}{dx^2} = 8(2 \cdot 0 - 1)^{-3} = -8\), indicating a maximum.

For \(x = 1\), \(\frac{d^2y}{dx^2} = 8(2 \cdot 1 - 1)^{-3} = 8\), indicating a minimum.

(i) To find stationary points, we set the derivative equal to zero. The derivative is \(\frac{dy}{dx} = 3x^2 - 4x + 5\). The discriminant of this quadratic is \(b^2 - 4ac = (-4)^2 - 4 \times 3 \times 5 = 16 - 60 = -44\), which is negative. Therefore, there are no real roots, and the curve has no stationary points.

(ii) The gradient \(m = 3x^2 - 4x + 5\). To find the stationary value of \(m\), we differentiate \(m\) with respect to \(x\): \(\frac{dm}{dx} = 6x - 4\). Setting \(\frac{dm}{dx} = 0\) gives \(6x - 4 = 0\), so \(x = \frac{2}{3}\). Substituting \(x = \frac{2}{3}\) into \(m = 3x^2 - 4x + 5\), we get \(m = 3 \left( \frac{2}{3} \right)^2 - 4 \left( \frac{2}{3} \right) + 5 = \frac{11}{3}\).

To determine the nature, we find \(\frac{d^2m}{dx^2} = 6\), which is positive, indicating a minimum.

First, find the derivative \(f'(x)\):

\(f(x) = \frac{8}{x-2} + 2\)

\(f'(x) = \frac{-8}{(x-2)^2}\)

Next, find the inverse function \(f^{-1}(x)\):

Let \(y = \frac{8}{x-2} + 2\)

\(y - 2 = \frac{8}{x-2}\)

\(x - 2 = \frac{8}{y-2}\)

\(f^{-1}(x) = \frac{8}{x-2} + 2\)

Substitute into the inequality:

\(6f'(x) + 2f^{-1}(x) - 5 < 0\)

\(6 \left( \frac{-8}{(x-2)^2} \right) + 2 \left( \frac{8}{x-2} + 2 \right) - 5 < 0\)

\(\frac{-48}{(x-2)^2} + \frac{16}{x-2} + 4 - 5 < 0\)

\(\frac{-48}{(x-2)^2} + \frac{16}{x-2} - 1 < 0\)

Multiply through by \((x-2)^2\):

\(-48 + 16(x-2) - (x-2)^2 < 0\)

\(-48 + 16x - 32 - (x^2 - 4x + 4) < 0\)

\(-48 + 16x - 32 - x^2 + 4x - 4 < 0\)

\(x^2 - 20x + 84 < 0\)

Factor the quadratic:

\((x-6)(x-14) < 0\)

Find the intervals where the inequality holds:

\(2 < x < 6, \; x > 14\)

(i) To find the stationary points, set \(\frac{dy}{dx} = 0\). First, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = x - 6x^{\frac{1}{2}} + 8\).

Set \(\frac{dy}{dx} = 0\):

\(x - 6x^{\frac{1}{2}} + 8 = 0\).

Let \(u = x^{\frac{1}{2}}\), then \(u^2 = x\). The equation becomes:

\(u^2 - 6u + 8 = 0\).

Solving the quadratic equation gives \(u = 4\) or \(u = 2\).

Thus, \(x = 16\) or \(x = 4\).

(ii) Differentiate \(\frac{dy}{dx} = x - 6x^{\frac{1}{2}} + 8\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 1 - 3x^{-\frac{1}{2}}\).

(iii) Evaluate \(\frac{d^2y}{dx^2}\) at the stationary points:

When \(x = 16\), \(\frac{d^2y}{dx^2} = 1 - 3 \times 16^{-\frac{1}{2}} = \frac{1}{4} > 0\), indicating a minimum point.

When \(x = 4\), \(\frac{d^2y}{dx^2} = 1 - 3 \times 4^{-\frac{1}{2}} = -\frac{1}{2} < 0\), indicating a maximum point.

(i) To find the stationary points, set \(\frac{dy}{dx} = 0\):

\(-x^2 + 5x - 4 = 0\)

Solving the quadratic equation:

\(x^2 - 5x + 4 = 0\)

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

Thus, \(x = 1\) and \(x = 4\).

(ii) Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -2x + 5\)

Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 1\):

\(\frac{d^2y}{dx^2} = -2(1) + 5 = 3\)

Since \(3 > 0\), \(x = 1\) is a minimum.

Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 4\):

\(\frac{d^2y}{dx^2} = -2(4) + 5 = -3\)

Since \(-3 < 0\), \(x = 4\) is a maximum.

(i) Differentiate \(y = 8\sqrt{x} - 2x\) to find \(\frac{dy}{dx} = 4x^{-\frac{1}{2}} - 2\). Set \(\frac{dy}{dx} = 0\) to find the stationary point: \(4x^{-\frac{1}{2}} - 2 = 0\). Solving gives \(\sqrt{x} = 2\), so \(x = 4\). Substitute back to find \(y = 8 \sqrt{4} - 2 \times 4 = 8\). The stationary point is \((4, 8)\).

(ii) Differentiate again to find \(\frac{d^2y}{dx^2} = -2x^{-\frac{3}{2}}\). At \(x = 4\), \(\frac{d^2y}{dx^2} = -\frac{1}{4}\), which is negative, indicating a maximum.

(iii) Set \(y = 6\) in the curve equation: \(8\sqrt{x} - 2x = 6\). Rearrange to form a quadratic in \(\sqrt{x}\): \(2u^2 - 8u + 6 = 0\) where \(u = \sqrt{x}\). Solve to find \(u = 1\) or \(u = 3\), giving \(x = 1\) or \(x = 9\).

(iv) For the line \(y = k\) not to meet the curve, \(k > 8\) since the maximum value of \(y\) on the curve is 8.

(i) To find \(f'(x)\), use the chain rule. Let \(u = 4x + 1\), then \(f(x) = u^{\frac{3}{2}}\).

\(\frac{du}{dx} = 4\)

\(\frac{df}{du} = \frac{3}{2}u^{\frac{1}{2}}\)

Thus, \(f'(x) = \frac{3}{2}(4x + 1)^{\frac{1}{2}} \times 4 = 6(4x + 1)^{\frac{1}{2}}\).

To find \(f''(x)\), differentiate \(f'(x)\):

\(f'(x) = 6(4x + 1)^{\frac{1}{2}}\)

\(f''(x) = 6 \times \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \times 4 = 12(4x + 1)^{-\frac{1}{2}}\).

(ii) Calculate \(f(2)\), \(f'(2)\), and \(f''(2)\):

\(f(2) = (4(2) + 1)^{\frac{3}{2}} = 9^{\frac{3}{2}} = 27\)

\(f'(2) = 6(4(2) + 1)^{\frac{1}{2}} = 6 \times 3 = 18\)

\(f''(2) = 12(4(2) + 1)^{-\frac{1}{2}} = 12 \times \frac{1}{3} = 4\)

The terms form a geometric progression: \(27, 18, 4k\).

The common ratio \(r\) is \(\frac{18}{27} = \frac{2}{3}\).

Thus, \(\frac{4k}{18} = \frac{2}{3}\).

Solving for \(k\):

\(4k = 12\)

\(k = 3\)

To find the stationary points, we first find the derivative \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = 8 + [-2] [(2x-1)^{-2}]\).

Set \(\frac{dy}{dx} = 0\):

\(0 = 8 - \frac{2}{(2x-1)^2}\).

Rearrange to find \((2x-1)^2 = \frac{1}{4}\).

Solving gives \(x = \frac{1}{4}\) and \(x = \frac{3}{4}\).

Next, find the second derivative \(\frac{d^2y}{dx^2} = 8(2x-1)^{-3}\).

Evaluate \(\frac{d^2y}{dx^2}\) at each stationary point:

When \(x = \frac{1}{4}\), \(\frac{d^2y}{dx^2} = -64\), which is less than 0, indicating a maximum.

When \(x = \frac{3}{4}\), \(\frac{d^2y}{dx^2} = 64\), which is greater than 0, indicating a minimum.

First, find the first derivative \(f'(x)\):

\(f'(x) = \frac{d}{dx}[2x + (x + 1)^{-2}] = 2 - 2(x + 1)^{-3}\).

Next, find the second derivative \(f''(x)\):

\(f''(x) = \frac{d}{dx}[-2(x + 1)^{-3}] = 6(x + 1)^{-4}\).

Evaluate \(f'(x)\) at \(x = 0\):

\(f'(0) = 2 - 2(1)^{-3} = 0\).

This shows that \(x = 0\) is a stationary point.

Evaluate \(f''(x)\) at \(x = 0\):

\(f''(0) = 6(1)^{-4} = 6 > 0\).

Since \(f''(0) > 0\), the function has a minimum at \(x = 0\).

(i) Differentiate \(y = \frac{8}{x} + 2x\) with respect to \(x\):

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

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(-8x^{-2}) = 16x^{-3} = \frac{16}{x^3}\).

(ii) To find stationary points, set \(\frac{dy}{dx} = 0\):

\(-\frac{8}{x^2} + 2 = 0 \Rightarrow 2x^2 - 8 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\).

For \(x = 2\), \(y = \frac{8}{2} + 2(2) = 4 + 4 = 8\).

For \(x = -2\), \(y = \frac{8}{-2} + 2(-2) = -4 - 4 = -8\).

Check the nature of each stationary point using \(\frac{d^2y}{dx^2}\):

At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{16}{2^3} = 2 > 0\), hence a minimum.

At \(x = -2\), \(\frac{d^2y}{dx^2} = \frac{16}{(-2)^3} = -2 < 0\), hence a maximum.

Given \(u = 2x(y - x)\) and \(x + 3y = 12\).

First, express \(y\) in terms of \(x\):

\(x + 3y = 12\)

\(3y = 12 - x\)

\(y = \frac{12 - x}{3}\)

Substitute \(y\) in the expression for \(u\):

\(u = 2x\left(\frac{12 - x}{3} - x\right)\)

\(u = 2x\left(\frac{12 - x - 3x}{3}\right)\)

\(u = 2x\left(\frac{12 - 4x}{3}\right)\)

\(u = \frac{2x(12 - 4x)}{3}\)

\(u = \frac{24x - 8x^2}{3}\)

\(u = 8x - \frac{8x^2}{3}\)

To find the stationary value, differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 8 - \frac{16x}{3}\)

Set \(\frac{du}{dx} = 0\):

\(8 - \frac{16x}{3} = 0\)

\(8 = \frac{16x}{3}\)

\(24 = 16x\)

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

Substitute \(x = \frac{3}{2}\) back into the equation for \(y\):

\(y = \frac{12 - \frac{3}{2}}{3}\)

\(y = \frac{21}{6}\)

\(y = \frac{7}{2}\)

Substitute \(x = \frac{3}{2}\) and \(y = \frac{7}{2}\) back into the expression for \(u\):

\(u = 2 \times \frac{3}{2} \times \left(\frac{7}{2} - \frac{3}{2}\right)\)

\(u = 3 \times 2\)

\(u = 6\)

(i) To find stationary points, set the derivative \(\frac{dy}{dx} = 3x^2 + 2px\) to zero:

\(3x^2 + 2px = 0\)

\(x(3x + 2p) = 0\)

Thus, \(x = 0\) or \(x = -\frac{2p}{3}\).

For \(x = 0\), \(y = 0\), so one stationary point is \((0, 0)\).

For \(x = -\frac{2p}{3}\), substitute back into the equation:

\(y = \left(-\frac{2p}{3}\right)^3 + p\left(-\frac{2p}{3}\right)^2\)

\(y = -\frac{8p^3}{27} + \frac{4p^3}{9} = \frac{4p^3}{27}\)

So the other stationary point is \(\left(-\frac{2p}{3}, \frac{4p^3}{27}\right)\).

(ii) To determine the nature, find the second derivative:

\(\frac{d^2y}{dx^2} = 6x + 2p\)

At \((0, 0)\), \(\frac{d^2y}{dx^2} = 2p > 0\), so it is a minimum.

At \(\left(-\frac{2p}{3}, \frac{4p^3}{27}\right)\), \(\frac{d^2y}{dx^2} = -2p < 0\), so it is a maximum.

(iii) For the curve \(y = x^3 + px^2 + px\), the derivative is:

\(\frac{dy}{dx} = 3x^2 + 2px + p\)

Using the discriminant \(b^2 - 4ac\) for \(3x^2 + 2px + p = 0\):

\((2p)^2 - 4 \cdot 3 \cdot p < 0\)

\(4p^2 - 12p < 0\)

\(4p(p - 3) < 0\)

Thus, \(0 < p < 3\).

To find \(f'(x)\), we use the quotient rule for differentiation. The function \(f(x) = \frac{15}{2x+3}\) can be rewritten as \(f(x) = 15(2x+3)^{-1}\).

The derivative of \(f(x)\) is:

\(f'(x) = -15(2x+3)^{-2} \cdot 2 = \frac{-30}{(2x+3)^2}\).

Since \((2x+3)^2\) is always positive, \(f'(x) < 0\) for all \(x\) in the domain \(0 \leq x \leq 6\). This means \(f(x)\) is strictly decreasing, and therefore, \(f\) has an inverse.

To find the stationary points, we first differentiate the equation \(y = \frac{k^2}{x+2} + x\) with respect to \(x\):

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

Set \(\frac{dy}{dx} = 0\) to find the stationary points:

\(-\frac{k^2}{(x+2)^2} + 1 = 0\).

Solving for \(x\), we get:

\((x+2)^2 = k^2\).

\(x+2 = \pm k\).

Thus, \(x = -2 + k\) or \(x = -2 - k\).

Next, we find the second derivative to determine the nature of the stationary points:

\(\frac{d^2y}{dx^2} = 2k^2(x+2)^{-3}\).

Substitute \(x = -2 + k\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{2}{k}\), which is positive, indicating a minimum.

Substitute \(x = -2 - k\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{-2}{k}\), which is negative, indicating a maximum.

To find the gradient of the curve, we need to differentiate \(y = \frac{2}{\sqrt{5x - 6}}\) with respect to \(x\).

First, rewrite the equation as \(y = 2(5x - 6)^{-\frac{1}{2}}\).

Using the chain rule, the derivative \(\frac{dy}{dx}\) is:

\(\frac{dy}{dx} = 2 \times -\frac{1}{2} \times (5x - 6)^{-\frac{3}{2}} \times 5\).

Simplifying, \(\frac{dy}{dx} = -\frac{5}{(5x - 6)^{\frac{3}{2}}}\).

Substitute \(x = 2\) into the derivative:

\(\frac{dy}{dx} = -\frac{5}{(5(2) - 6)^{\frac{3}{2}}} = -\frac{5}{(10 - 6)^{\frac{3}{2}}} = -\frac{5}{4^{\frac{3}{2}}}\).

Since \(4^{\frac{3}{2}} = 8\), the gradient is \(-\frac{5}{8}\).

We start with the equations:

1. \(u = x^2 y\) 2. \(y + 3x = 9\)

Express \(y\) in terms of \(x\):

\(y = 9 - 3x\)

Substitute \(y\) into the equation for \(u\):

\(u = x^2 (9 - 3x) = 9x^2 - 3x^3\)

Differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 18x - 9x^2\)

Set \(\frac{du}{dx} = 0\) to find the stationary points:

\(18x - 9x^2 = 0\)

\(9x(2 - x) = 0\)

\(x = 0\) or \(x = 2\)

Since \(x\) cannot be zero (as \(x\) is non-zero), we have \(x = 2\).

Substitute \(x = 2\) back into \(y = 9 - 3x\):

\(y = 9 - 3(2) = 3\)

Calculate \(u\):

\(u = (2)^2 (3) = 12\)

To determine if this is a maximum or minimum, find the second derivative:

\(\frac{d^2u}{dx^2} = 18 - 18x\)

Evaluate at \(x = 2\):

\(\frac{d^2u}{dx^2} = 18 - 18(2) = -18\)

Since \(\frac{d^2u}{dx^2} < 0\), the stationary point is a maximum.

(i) Substitute \(u = x^{\frac{1}{2}}\), then \(f'(x) = 3u + \frac{3}{u} - 10 = 0\).

Multiply through by \(u\): \(3u^2 - 10u + 3 = 0\).

Factorize: \((3u - 1)(u - 3) = 0\).

So, \(u = \frac{1}{3}\) or \(u = 3\).

Since \(u = x^{\frac{1}{2}}\), \(x = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\) or \(x = 3^2 = 9\).

(ii) Differentiate \(f'(x)\) to find \(f''(x) = \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{-\frac{3}{2}}\).

At \(x = \frac{1}{9}\), \(f''(x) = \frac{3}{2}(3) - \frac{3}{2}(27) = -36 < 0\), indicating a maximum.

At \(x = 9\), \(f''(x) = \frac{3}{2} \times \frac{1}{3} - \frac{3}{2} \times \frac{1}{27} = \frac{4}{9} > 0\), indicating a minimum.

(i) To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2(3x + 4)^{\frac{3}{2}} - 6x - 8\) with respect to \(x\).

The derivative of \(2(3x + 4)^{\frac{3}{2}}\) is \(2 \times \frac{3}{2} \times (3x + 4)^{\frac{1}{2}} \times 3 = 9(3x + 4)^{\frac{1}{2}}\).

The derivative of \(-6x\) is \(-6\), and the derivative of \(-8\) is \(0\).

Thus, \(\frac{d^2y}{dx^2} = 9(3x + 4)^{\frac{1}{2}} - 6\).

(ii) To verify the stationary point at \(x = -1\), substitute \(x = -1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2(3(-1) + 4)^{\frac{3}{2}} - 6(-1) - 8 = 2(1)^{\frac{3}{2}} + 6 - 8 = 0\).

Since \(\frac{dy}{dx} = 0\), there is a stationary point at \(x = -1\).

To determine the nature, evaluate \(\frac{d^2y}{dx^2}\) at \(x = -1\):

\(\frac{d^2y}{dx^2} = 9(3(-1) + 4)^{\frac{1}{2}} - 6 = 9(1)^{\frac{1}{2}} - 6 = 3\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

To find the stationary point, we first differentiate the equation \(y = 2x + \frac{1}{(x-1)^2}\).

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

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 - 2(2-1)^{-3} = 2 - 2 = 0\).

This confirms a stationary point at \(x = 2\).

Next, we find the second derivative to determine the nature of the stationary point:

\(\frac{d^2y}{dx^2} = 6(x-1)^{-4}\).

Substitute \(x = 2\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 6(2-1)^{-4} = 6\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point at \(x = 2\) is a minimum.

(i) To find where the gradient is less than 5, first find the derivative: \(f'(x) = 3x^2 - 4x + 1\).

Set \(f'(x) < 5\):

\(3x^2 - 4x + 1 < 5\)

\(3x^2 - 4x - 4 < 0\)

Factorize: \((3x + 2)(x - 2) < 0\)

Find the critical points: \(x = -\frac{2}{3}\) and \(x = 2\).

The solution is \(-\frac{2}{3} < x < 2\).

(ii) To find stationary points, set \(f'(x) = 0\):

\(3x^2 - 4x + 1 = 0\)

Factorize: \((3x - 1)(x - 1) = 0\)

Solutions: \(x = \frac{1}{3}\) and \(x = 1\).

Find \(f(x)\) at these points:

\(f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \frac{1}{3} = \frac{4}{27}\)

\(f(1) = 1^3 - 2 \times 1^2 + 1 = 0\)

Determine nature using second derivative: \(f''(x) = 6x - 4\).

\(f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) - 4 = -2\) (negative, so maximum)

\(f''(1) = 6 \times 1 - 4 = 2\) (positive, so minimum)

Maximum at \(\left( \frac{1}{3}, \frac{4}{27} \right)\); Minimum at \((1, 0)\).

To find the gradient of the curve, we need to differentiate the equation \(y = 3x^3 - 6x^2 + 4x + 2\) with respect to \(x\).

The derivative is \(\frac{dy}{dx} = 9x^2 - 12x + 4\).

We need to show that \(9x^2 - 12x + 4 \geq 0\) for all \(x\).

Consider the expression \(9x^2 - 12x + 4\).

This can be rewritten as \((3x - 2)^2\).

Since \((3x - 2)^2\) is a square, it is always non-negative.

Therefore, \(9x^2 - 12x + 4 \geq 0\) for all \(x\), and the gradient of the curve is never negative.

To find \(g'(x)\), differentiate \(g(x) = 2(x-1)^3 + 8\).

Using the chain rule, \(\frac{d}{dx}[(x-1)^3] = 3(x-1)^2 \cdot 1\).

Thus, \(g'(x) = 2 \cdot 3(x-1)^2 = 6(x-1)^2\).

Since \(g'(x) = 6(x-1)^2 > 0\) for \(x > 1\), \(g(x)\) is strictly increasing.

This implies \(g\) is one-to-one (1:1), so \(g\) has an inverse.

(i) Given \(z = 3x + 2y\) and \(xy = 600\), we can express \(y\) in terms of \(x\):

\(y = \frac{600}{x}\).

Substitute \(y\) in the expression for \(z\):

\(z = 3x + 2\left(\frac{600}{x}\right) = 3x + \frac{1200}{x}\).

(ii) To find the stationary value, differentiate \(z\) with respect to \(x\):

\(\frac{dz}{dx} = 3 - \frac{1200}{x^2}\).

Set \(\frac{dz}{dx} = 0\) to find critical points:

\(3 - \frac{1200}{x^2} = 0\).

\(\frac{1200}{x^2} = 3\).

\(x^2 = 400\).

\(x = 20\) (since \(x\) is positive).

Substitute \(x = 20\) back into the expression for \(z\):

\(z = 3(20) + \frac{1200}{20} = 60 + 60 = 120\).

To determine the nature, find the second derivative:

\(\frac{d^2z}{dx^2} = \frac{2400}{x^3}\).

At \(x = 20\), \(\frac{d^2z}{dx^2} = \frac{2400}{20^3} > 0\), indicating a minimum.

(i) To find \(\frac{dy}{dx}\), differentiate \(y = \frac{1}{x-3} + x\).

The derivative of \(\frac{1}{x-3}\) is \(\frac{-1}{(x-3)^2}\) and the derivative of \(x\) is 1.

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

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = \frac{-1}{(x-3)^2} + 1\).

The derivative of \(\frac{-1}{(x-3)^2}\) is \(\frac{2}{(x-3)^3}\) and the derivative of 1 is 0.

Thus, \(\frac{d^2y}{dx^2} = \frac{2}{(x-3)^3}\).

(ii) To find the critical points, set \(\frac{dy}{dx} = 0\):

\(\frac{-1}{(x-3)^2} + 1 = 0 \Rightarrow (x-3)^2 = 1 \Rightarrow x-3 = \pm 1\).

Solving gives \(x = 4\) and \(x = 2\).

For \(x = 4\), substitute into \(y = \frac{1}{x-3} + x\):

\(y = \frac{1}{4-3} + 4 = 1 + 4 = 5\).

For \(x = 2\), substitute into \(y = \frac{1}{x-3} + x\):

\(y = \frac{1}{2-3} + 2 = -1 + 2 = 1\).

Check the second derivative \(\frac{d^2y}{dx^2}\) to determine concavity:

When \(x = 4\), \(\frac{d^2y}{dx^2} = \frac{2}{(4-3)^3} = 2 > 0\), indicating a minimum.

When \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{2}{(2-3)^3} = -2 < 0\), indicating a maximum.

Thus, the maximum point is \(A(2, 5)\) and the minimum point is \(B(4, 1)\).

Given \(y = \frac{9}{2-x}\), we can rewrite it as \(y = 9(2-x)^{-1}\).

Using the chain rule, \(\frac{dy}{dx} = -9(2-x)^{-2} \times (-1)\).

This simplifies to \(\frac{dy}{dx} = \frac{9}{(2-x)^2}\).

For stationary points, \(\frac{dy}{dx} = 0\). However, \(\frac{9}{(2-x)^2} \neq 0\) for any real \(x\), so there are no stationary points.

(i) To find \(\frac{dy}{dx}\), use the chain rule on \(y = (2x - 3)^3 - 6x\):

\(\frac{dy}{dx} = 3(2x-3)^2 \cdot \frac{d}{dx}(2x-3) - 6\)

\(\frac{dy}{dx} = 3(2x-3)^2 \cdot 2 - 6\)

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

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(6(2x-3)^2)\)

\(\frac{d^2y}{dx^2} = 12(2x-3) \cdot \frac{d}{dx}(2x-3)\)

\(\frac{d^2y}{dx^2} = 12(2x-3) \cdot 2\)

\(\frac{d^2y}{dx^2} = 24(2x-3)\)

(ii) Stationary points occur where \(\frac{dy}{dx} = 0\):

\(6(2x-3)^2 - 6 = 0\)

\(6(2x-3)^2 = 6\)

\((2x-3)^2 = 1\)

\(2x-3 = \pm 1\)

\(2x = 4\) or \(2x = 2\)

\(x = 2\) or \(x = 1\)

To determine the nature, use \(\frac{d^2y}{dx^2}\):

For \(x = 2\), \(\frac{d^2y}{dx^2} = 24(2 \cdot 2 - 3) = 24 \cdot 1 = 24\) (positive, minimum)

For \(x = 1\), \(\frac{d^2y}{dx^2} = 24(2 \cdot 1 - 3) = 24 \cdot (-1) = -24\) (negative, maximum)

The equation of the curve is \(y = \frac{k}{x}\). To find the gradient, we differentiate \(y\) with respect to \(x\).

\(\frac{dy}{dx} = -\frac{k}{x^2}\).

We are given that the gradient \(\frac{dy}{dx} = -3\) when \(x = 2\).

Substitute \(x = 2\) into the derivative:

\(-\frac{k}{2^2} = -3\).

Simplify:

\(-\frac{k}{4} = -3\).

Multiply both sides by \(-4\):

\(k = 12\).

To find the gradient of the curve, we need to differentiate \(y = \frac{12}{x^2 - 4x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = 12\) and \(v = x^2 - 4x\).

First, find \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = 2x - 4\).

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(x^2 - 4x)(0) - 12(2x - 4)}{(x^2 - 4x)^2}\)

\(= \frac{-12(2x - 4)}{(x^2 - 4x)^2}\)

\(= -12(x^2 - 4x)^{-2} \times (2x - 4)\)

Now, substitute \(x = 3\) into the derivative:

\(\frac{dy}{dx} = -12(3^2 - 4 \times 3)^{-2} \times (2 \times 3 - 4)\)

\(= -12(9 - 12)^{-2} \times (6 - 4)\)

\(= -12(-3)^{-2} \times 2\)

\(= -12 \times \frac{1}{9} \times 2\)

\(= -\frac{24}{9}\)

\(= -\frac{8}{3}\)

(i) To find \(\frac{dy}{dx}\), differentiate \(y = x^2 + \frac{2}{x}\):

\(\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{2}{x}\right)\)

\(\frac{dy}{dx} = 2x - \frac{2}{x^2}\)

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2x - \frac{2}{x^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(2x) + \frac{d}{dx}\left(-\frac{2}{x^2}\right)\)

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

(ii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

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

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

\(2x^3 = 2\)

\(x^3 = 1\)

\(x = 1\)

Substitute \(x = 1\) into \(y = x^2 + \frac{2}{x}\):

\(y = 1^2 + \frac{2}{1} = 3\)

Stationary point is \((1, 3)\).

To determine the nature, evaluate \(\frac{d^2y}{dx^2}\) at \(x = 1\):

\(\frac{d^2y}{dx^2} = 2 + \frac{4}{1^3} = 6\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

(i) Differentiate \(y = x^3 + 3x^2 - 9x + k\) with respect to \(x\):

\(\frac{dy}{dx} = 3x^2 + 6x - 9\).

(ii) Stationary points occur where \(\frac{dy}{dx} = 0\):

\(3x^2 + 6x - 9 = 0\).

Factorize the quadratic equation:

\(3(x^2 + 2x - 3) = 0\).

\(3(x + 3)(x - 1) = 0\).

Thus, \(x = -3\) or \(x = 1\).

(iii) For the curve to have a stationary point on the \(x\)-axis, \(y = 0\) at the stationary points:

Substitute \(x = -3\) into \(y = x^3 + 3x^2 - 9x + k\):

\((-3)^3 + 3(-3)^2 - 9(-3) + k = 0\).

\(-27 + 27 + 27 + k = 0\).

\(k = -27\).

Substitute \(x = 1\) into \(y = x^3 + 3x^2 - 9x + k\):

\(1^3 + 3(1)^2 - 9(1) + k = 0\).

\(1 + 3 - 9 + k = 0\).

\(k = 5\).

Thus, the values of \(k\) are \(-27\) and \(5\).