(a) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = -\left( \frac{1}{2}x + k \right)^{-3} \cdot \frac{1}{2}\).

Since the curve has a minimum at \(x = 2\), \(f''(2) > 0\):

\(-\left( 1 + k \right)^{-3} > 0\).

This implies \(k < -1\).

(b) Integrate \(f'(x)\) to find \(f(x)\):

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

Using integration,

\(f(x) = \left[ \left( \frac{1}{2}x - 3 \right)^{-1} \cdot \frac{-1}{\frac{1}{2}} \right] - \left[ (-2)^{-1} \cdot x \right] + c\).

Substitute \(x = 2, y = 3\frac{1}{2}\) to find \(c\):

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

\(c = 3\).

Thus, \(f(x) = \frac{-2}{\left( \frac{1}{2}x - 3 \right)} \cdot \frac{x}{4} + 3\).

(c) Set \(f'(x) = 0\) to find the other stationary point:

\(\left( \frac{1}{2}x - 3 \right)^{-2} - (-2)^{-2} = 0\).

\(\left( \frac{1}{2}x - 3 \right)^{-2} = 4\).

\(\frac{1}{2}x - 3 = \pm 2\).

Solving gives \(x = 10\).

Substitute \(x = 10\) into \(f(x)\):

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

Check the nature using \(f''(10)\):

\(f''(10) = -\left( 5 - 3 \right)^{-3} < 0\), indicating a maximum.

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = ax^2 + bx - 4\).

The integral is \(y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x + c\).

Since the curve passes through (0, 11), substitute \(x = 0\) and \(y = 11\) to find \(c\):

\(11 = \frac{1}{3}a(0)^3 + \frac{1}{2}b(0)^2 - 4(0) + c\)

\(c = 11\)

Thus, the equation of the curve is \(y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x + 11\).

(ii) At the stationary point (2, 3), \(\frac{dy}{dx} = 0\).

Substitute \(x = 2\) into \(\frac{dy}{dx} = ax^2 + bx - 4\):

\(4a + 2b - 4 = 0\)

Substitute \(x = 2\) and \(y = 3\) into the integrated equation:

\(3 = \frac{1}{3}a(2)^3 + \frac{1}{2}b(2)^2 - 4(2) + 11\)

\(\frac{8}{3}a + 2b - 8 + 11 = 3\)

Solve the simultaneous equations:

1. \(4a + 2b - 4 = 0\)

2. \(\frac{8}{3}a + 2b + 3 = 3\)

Solving these gives \(a = 3\) and \(b = -4\).

(i) At a stationary point, \(\frac{dy}{dx} = 0\). Given \(\frac{dy}{dx} = ax^2 + a^2 x\), substitute \(x = 3\):

\(0 = 9a + 3a^2\)

Solving gives \(a = -3\).

(ii) Integrate \(\frac{dy}{dx} = -3x^2 + 9x\) to find \(y\):

\(y = \int (-3x^2 + 9x) \, dx = -x^3 + \frac{9x^2}{2} + c\)

Substitute \((3, 9\frac{1}{2})\) to find \(c\):

\(9\frac{1}{2} = -27 + \frac{40}{2} + c\)

\(c = -4\)

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

(iii) Find \(\frac{d^2y}{dx^2}\) to determine the nature of the stationary point:

\(\frac{d^2y}{dx^2} = -6x + 9\)

At \(x = 3\), \(\frac{d^2y}{dx^2} = -9\), which is less than 0, indicating a maximum.

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \sqrt{4x + 1}\).

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

Using the point \((2, 5)\), substitute to find \(C\):

\(5 = \frac{2}{3}(4(2)+1)^{\frac{3}{2}} + C\)

\(5 = \frac{2}{3}(9)^{\frac{3}{2}} + C\)

\(5 = 12 + C\)

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

Thus, the equation is \(y = \frac{2}{3}(4x+1)^{\frac{3}{2}} + \frac{1}{2}\).

(ii) Given \(\frac{dy}{dt} = 0.06\), use the chain rule:

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\)

\(0.06 = \sqrt{4(2) + 1} \cdot \frac{dx}{dt}\)

\(0.06 = 3 \cdot \frac{dx}{dt}\)

\(\frac{dx}{dt} = 0.02\)

(iii) Differentiate \(\frac{dy}{dx} = \sqrt{4x + 1}\) to find \(\frac{d^2y}{dx^2}\):

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

\(\frac{d^2y}{dx^2} = \frac{2}{\sqrt{4x+1}}\)

Then, \(\frac{d^2y}{dx^2} \times \frac{dy}{dx} = \frac{2}{\sqrt{4x+1}} \times \sqrt{4x+1} = 2\)

(i) To find the stationary point, set \(f'(x) = 0\):

\(ax^2 + bx = 0\)

\(x(ax + b) = 0\)

\(x = 0\) or \(x = -\frac{b}{a}\)

The non-zero value is \(x = -\frac{b}{a}\).

To determine the nature, find \(f''(x)\):

\(f''(x) = 2ax + b\)

Substitute \(x = -\frac{b}{a}\) into \(f''(x)\):

\(f''\left(-\frac{b}{a}\right) = 2a\left(-\frac{b}{a}\right) + b = -2b + b = -b\)

Since \(f''\left(-\frac{b}{a}\right) < 0\), it is a maximum point.

(ii) Given \(f'(-2) = 0\) and \(f'(1) = 9\):

\(a(-2)^2 + b(-2) = 0\)

\(4a - 2b = 0\)

\(2a = b\)

\(a(1)^2 + b(1) = 9\)

\(a + b = 9\)

Substitute \(b = 2a\) into \(a + b = 9\):

\(a + 2a = 9\)

\(3a = 9\)

\(a = 3\)

\(b = 6\)

Integrate \(f'(x) = 3x^2 + 6x\) to find \(f(x)\):

\(f(x) = \int (3x^2 + 6x) \, dx = x^3 + 3x^2 + c\)

Given \(f(-2) = -3\):

\(-3 = (-2)^3 + 3(-2)^2 + c\)

\(-3 = -8 + 12 + c\)

\(c = -7\)

Thus, \(f(x) = x^3 + 3x^2 - 7\).

(i) To find \(f'(x)\), integrate \(f''(x) = (4x + 1)^{-\frac{1}{2}}\):

\(f'(x) = \int (4x + 1)^{-\frac{1}{2}} \, dx = \frac{1}{2}(4x + 1)^{\frac{1}{2}} + c\).

Given \(f'(2) = 0\), substitute \(x = 2\):

\(\frac{1}{2}(4(2) + 1)^{\frac{1}{2}} + c = 0 \Rightarrow \frac{3}{2} + c = 0 \Rightarrow c = -\frac{3}{2}\).

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

(ii) Given \(f''(0), f'(0), f(0)\) form an arithmetic progression:

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

Arithmetic progression: \(1, -1, f(0)\).

Common difference \(d = -2\), so \(f(0) = -1 - 2 = -3\).

(iii) Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int \left( \frac{1}{2}(4x + 1)^{\frac{1}{2}} - \frac{3}{2} \right) \, dx\).

\(f(x) = \frac{1}{12}(4x + 1)^{\frac{3}{2}} - \frac{3}{2}x + k\).

Given \(f(0) = -3\), substitute \(x = 0\):

\(\frac{1}{12}(1)^{\frac{3}{2}} - 0 + k = -3 \Rightarrow \frac{1}{12} + k = -3 \Rightarrow k = -\frac{37}{12}\).

Thus, \(f(x) = \frac{1}{12}(4x + 1)^{\frac{3}{2}} - \frac{3}{2}x - \frac{37}{12}\).

Minimum value at \(x = 2\):

\(f(2) = \frac{1}{12}(9)^{\frac{3}{2}} - 3 - \frac{37}{12} = \frac{27}{12} - 3 - \frac{37}{12} = \frac{23}{6}\) or \(-3.83\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 7 - x^2 - 6x\):

\(y = \int (7 - x^2 - 6x) \, dx = 7x - \frac{x^3}{3} - \frac{6x^2}{2} + c\).

Using the point \((3, -10)\), substitute to find \(c\):

\(-10 = 7(3) - \frac{3^3}{3} - \frac{6(3)^2}{2} + c\)

\(-10 = 21 - 9 - 27 + c\)

\(c = 5\)

Thus, the equation is \(y = 7x - \frac{x^3}{3} - \frac{6x^2}{2} + 5\).

(ii) Express \(7 - x^2 - 6x\) in the form \(a - (x + b)^2\):

Complete the square for \(-x^2 - 6x\):

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

Thus, \(7 - x^2 - 6x = 7 + 9 - (x + 3)^2 = 16 - (x + 3)^2\)

So, \(a = 16\) and \(b = 3\).

(iii) Find the set of values of \(x\) for which the gradient is positive:

\(7 - x^2 - 6x > 0\)

Using the completed square form: \(16 - (x + 3)^2 > 0\)

\((x + 3)^2 < 16\)

\(-4 < x + 3 < 4\)

\(-7 < x < 1\)

(i) At \(x = a^2\), \(\frac{dy}{dx} = \frac{2}{a}a^{-1} + a \cdot a^{-3} = \frac{3}{a^2}\). The equation of the tangent is \(y - 3 = \frac{3}{a^2}(x - a^2)\), simplifying to \(y = \frac{3}{a^2}x\).

(ii) Integrate \(\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-\frac{3}{2}}\) to get \(y = \frac{2}{a} \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + a \cdot \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + c = \frac{4x^{\frac{1}{2}}}{a} - 2ax^{-\frac{1}{2}} + c\). Substituting \(x = a^2, y = 3\) gives \(c = 1\), so \(y = \frac{4x^{\frac{1}{2}}}{a} - 2ax^{-\frac{1}{2}} + 1\).

(iii) Substituting \(x = 16, y = 8\) into the curve equation gives \(8 = \frac{4}{a} \times 4 - 2a \times \frac{1}{4} + 1\). Solving \(a^2 + 14a - 32 = 0\) gives \(a = 2\). Points are \(A = (4, 3)\), \(B = (16, 8)\). Distance \(AB = \sqrt{(16-4)^2 + (8-3)^2} = 13\).

(i) To find the \(x\)-coordinate of \(A\), set \(f'(x) = -1\):

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

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

\(3x - \frac{2}{x} = -1\)

Rearrange to form a quadratic equation:

\(3x + 1 = \frac{2}{x}\)

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

Let \(z = x^{\frac{1}{2}}\), then \(z^2 = x\):

\(3z^2 + z - 2 = 0\)

Solving this quadratic gives \(z = \frac{2}{3}\) or \(z = -1\).

Since \(z = x^{\frac{1}{2}}\), \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

(ii) To find the \(y\)-coordinate of \(A\), integrate \(f'(x)\) to find \(f(x)\):

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

\(f(x) = 3 \cdot \frac{2}{3} x^{\frac{3}{2}} - 2 \cdot 2 x^{\frac{1}{2}} + c\)

\(f(x) = 2x^{\frac{3}{2}} - 4x^{\frac{1}{2}} + c\)

Given \(f(4) = 10\):

\(10 = 2(4)^{\frac{3}{2}} - 4(4)^{\frac{1}{2}} + c\)

\(10 = 16 - 8 + c\)

\(c = 2\)

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

Substitute \(x = \frac{4}{9}\) to find \(y\):

\(y = 2 \left(\frac{4}{9}\right)^{\frac{3}{2}} - 4 \left(\frac{4}{9}\right)^{\frac{1}{2}} + 2\)

\(y = 2 \cdot \frac{8}{27} - 4 \cdot \frac{2}{3} + 2\)

\(y = \frac{16}{27} - \frac{8}{3} + 2\)

\(y = \frac{16}{27} - \frac{72}{27} + \frac{54}{27}\)

\(y = -\frac{2}{27}\)

(i) Given \(\frac{dy}{dx} = 6x^2 + \frac{k}{x^3}\) and the gradient at \(P(1, 9)\) is 2, substitute \(x = 1\) into the derivative:

\(6(1)^2 + \frac{k}{(1)^3} = 2\)

\(6 + k = 2\)

Solving for \(k\), we get \(k = -4\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = 6x^2 - \frac{4}{x^3}\):

\(y = \int (6x^2 - \frac{4}{x^3}) \, dx\)

\(y = 2x^3 + 2x^{-2} + c\)

Using the point \(P(1, 9)\), substitute \(x = 1\) and \(y = 9\):

\(9 = 2(1)^3 + 2(1)^{-2} + c\)

\(9 = 2 + 2 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = 2x^3 + 2x^{-2} + 5\).

(a) To find \(f(x)\), integrate \(f'(x) = 2x^2 - 7 - \frac{4}{x^2}\):

\(f(x) = \int (2x^2 - 7 - \frac{4}{x^2}) \, dx = \frac{2}{3}x^3 - 7x + 4x^{-1} + c\).

Given \(f(1) = -\frac{1}{3}\), substitute to find \(c\):

\(-\frac{1}{3} = \frac{2}{3}(1)^3 - 7(1) + 4(1)^{-1} + c\)

\(-\frac{1}{3} = \frac{2}{3} - 7 + 4 + c\)

\(-\frac{1}{3} = -\frac{1}{3} + c\)

\(c = 2\)

Thus, \(f(x) = \frac{2}{3}x^3 - 7x + 4x^{-1} + 2\).

(b) Stationary points occur where \(f'(x) = 0\):

\(2x^2 - 7 - \frac{4}{x^2} = 0\)

\(2x^4 - 7x^2 - 4 = 0\)

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

\(x^2 = 4 \Rightarrow x = \pm 2\)

For \(x = 2\), \(f(2) = \frac{2}{3}(2)^3 - 7(2) + 4(2)^{-1} + 2 = \frac{14}{3}\)

For \(x = -2\), \(f(-2) = \frac{2}{3}(-2)^3 - 7(-2) + 4(-2)^{-1} + 2 = -\frac{26}{3}\)

Coordinates: \((2, \frac{14}{3})\) and \((-2, -\frac{26}{3})\).

(c) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = \frac{d}{dx}(2x^2 - 7 - \frac{4}{x^2}) = 4x + 8x^{-3}\).

(d) Evaluate \(f''(x)\) at the stationary points:

\(f''(2) = 4(2) + 8(2)^{-3} = 9 > 0\), so minimum at \(x = 2\).

\(f''(-2) = 4(-2) + 8(-2)^{-3} = -9 < 0\), so maximum at \(x = -2\).

To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = 2 - 8(3x + 4)^{-\frac{1}{2}}\).

Integrate term by term:

\(\int 2 \, dx = 2x + C_1\)

For the second term, let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

\(\int -8(3x + 4)^{-\frac{1}{2}} \, dx = -8 \int u^{-\frac{1}{2}} \cdot \frac{du}{3}\)

\(= -\frac{8}{3} \int u^{-\frac{1}{2}} \, du\)

\(= -\frac{8}{3} \cdot 2u^{\frac{1}{2}} + C_2\)

\(= -\frac{16}{3} \sqrt{3x+4} + C_2\)

Combine the results:

\(y = 2x - \frac{16}{3} \sqrt{3x+4} + C\)

Given that the curve intersects the y-axis where \(y = \frac{4}{3}\), substitute \(x = 0\) and \(y = \frac{4}{3}\) to find \(C\):

\(\frac{4}{3} = 2(0) - \frac{16}{3} \sqrt{4} + C\)

\(\frac{4}{3} = -\frac{16}{3} \cdot 2 + C\)

\(\frac{4}{3} = -\frac{32}{3} + C\)

\(C = \frac{4}{3} + \frac{32}{3} = 12\)

Thus, the equation of the curve is:

\(y = 2x + \frac{8}{3}\sqrt{3x+4} + 12\)

(i) At \(x = 4\), \(\frac{dy}{dx} = 2\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 2 \times 3 = 6\).

(ii) Integrate \(\frac{dy}{dx} = 1 + 2x^{-\frac{1}{2}}\) to find \(y\).

\(y = x + 4x^{\frac{1}{2}} + c\).

Substitute \(x = 4, y = 6\):

\(6 = 4 + 4(4^{\frac{1}{2}}) + c\)

\(c = -6\)

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

(iii) Equation of the tangent at A is \(y - 6 = 2(x - 4)\) or \(y = 2x - 8\).

The tangent crosses the x-axis at \(B = (1, 0)\).

The gradient of the normal is \(-\frac{1}{2}\), so the equation is \(y = -\frac{1}{2}x + 8\).

The normal crosses the x-axis at \(C = (16, 0)\).

The area of triangle ABC is \(\frac{1}{2} \times 15 \times 6 = 45\).

(i) Given \(f''(x) = \frac{12}{x^3}\), integrate to find \(f'(x)\):

\(f'(x) = \int \frac{12}{x^3} \, dx = -\frac{6}{x^2} + c\).

Since \(f'(2) = 0\), substitute \(x = 2\):

\(0 = -\frac{6}{4} + c \Rightarrow c = \frac{3}{2}\).

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

Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int \left(-\frac{6}{x^2} + \frac{3}{2}\right) \, dx = \frac{6}{x} + \frac{3x}{2} + A\).

Since \(f(2) = 10\), substitute \(x = 2\):

\(10 = \frac{6}{2} + 3 + A \Rightarrow A = 4\).

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

(ii) Set \(f'(x) = 0\):

\(-\frac{6}{x^2} + \frac{3}{2} = 0 \Rightarrow x = \pm 2\).

The other stationary point is \((-2, -2)\).

(iii) Evaluate \(f''(x)\) at the stationary points:

At \(x = 2\), \(f''(2) = \frac{12}{8} = 1.5\), indicating a minimum.

At \(x = -2\), \(f''(-2) = -1.5\), indicating a maximum.

(i) To determine whether the stationary point is a maximum or minimum, evaluate the second derivative at \(x = 3\):

\(f''(3) = 36 \times 3^{-3} = \frac{36}{27} = \frac{4}{3}\).

Since \(f''(3) > 0\), the stationary point at \((3, 7)\) is a minimum.

(ii) To find \(f'(x)\), integrate \(f''(x)\):

\(f''(x) = 36x^{-3}\) implies \(f'(x) = \int 36x^{-3} \, dx = -18x^{-2} + c\).

Given that \(f'(3) = 0\), substitute to find \(c\):

\(0 = -18 \times 3^{-2} + c\)

\(0 = -2 + c\)

\(c = 2\)

Thus, \(f'(x) = -18x^{-2} + 2\).

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

\(f(x) = \int (-18x^{-2} + 2) \, dx = 18x^{-1} + 2x + k\).

Given that \(f(3) = 7\), substitute to find \(k\):

\(7 = 18 \times 3^{-1} + 2 \times 3 + k\)

\(7 = 6 + 6 + k\)

\(k = -5\)

Thus, \(f(x) = 18x^{-1} + 2x - 5\).

(i) At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{24}{2^3} - 4 = 3 - 4 = -1\), which is negative, indicating a maximum point.

(ii) To find \(\frac{dy}{dx}\), integrate \(\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4\):

\(\frac{dy}{dx} = \int \left( \frac{24}{x^3} - 4 \right) dx = \int 24x^{-3} dx - \int 4 dx\)

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

Given \(\frac{dy}{dx} = 0\) at \(x = 2\), solve for \(A\):

\(0 = -12(2)^{-2} - 4(2) + A\)

\(0 = -3 - 8 + A\)

\(A = 11\)

Thus, \(\frac{dy}{dx} = -12x^{-2} - 4x + 11\).

(iii) Integrate \(\frac{dy}{dx} = -12x^{-2} - 4x + 11\) to find \(y\):

\(y = \int (-12x^{-2} - 4x + 11) dx\)

\(= 12x^{-1} - 2x^2 + 11x + c\)

Given \(y = 13\) when \(x = 1\):

\(13 = 12(1)^{-1} - 2(1)^2 + 11(1) + c\)

\(13 = 12 - 2 + 11 + c\)

\(c = -8\)

Thus, \(y = 12x^{-1} - 2x^2 + 11x - 8\).

At \(x = 2\), \(y = 12(2)^{-1} - 2(2)^2 + 11(2) - 8\)

\(y = 6 - 8 + 22 - 8 = 12\)

Therefore, the coordinates of the stationary point \(P\) are \((2, 12)\).

(i) Calculate f'(2):

\(f'(2) = 2(2) - \frac{2}{2^2} = 4 - \frac{1}{2} = \frac{7}{2}\)

The gradient of the normal is the negative reciprocal:

\(-\frac{2}{7}\)

Equation of the normal using point-slope form:

\(y - 6 = -\frac{2}{7}(x - 2)\)

(ii) Integrate f'(x) to find f(x):

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

Substitute point (2, 6):

\(6 = 4 + 1 + c \Rightarrow c = 1\)

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

\((iii) Find stationary points by setting f'(x) = 0:\)

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

Determine nature using second derivative:

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

\(f''(1) = 6\) (which is > 0), hence minimum.

(i) The slope of the normal is \(-\frac{1}{3}\), so the slope of the tangent is \(3\) (since \(m_1 m_2 = -1\)).

Set \(\frac{dy}{dx} = 3 = \frac{12}{\sqrt{4 \times 2 + a}}\).

\(3 = \frac{12}{\sqrt{8 + a}}\)

\(\sqrt{8 + a} = 4\)

\(8 + a = 16\)

\(a = 8\)

(ii) Integrate \(\frac{dy}{dx} = \frac{12}{\sqrt{4x + 8}}\) to find \(y\).

\(y = \int 12(4x + 8)^{-\frac{1}{2}} \, dx\)

\(y = 12 \times 2 \sqrt{4x + 8} + c\)

\(y = 24 \sqrt{4x + 8} + c\)

Using point \(P(2, 14)\):

\(14 = 24 \sqrt{16} + c\)

\(14 = 24 \times 4 + c\)

\(14 = 96 + c\)

\(c = -82\)

Thus, the equation of the curve is \(y = 24 \sqrt{4x + 8} - 82\).

We start with the second derivative \(\frac{d^2y}{dx^2} = 2x - 1\). To find the first derivative \(\frac{dy}{dx}\), integrate:

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

Given the minimum point at \(x = 3\), \(\frac{dy}{dx} = 0\) at this point:

\(0 = 3^2 - 3 + c\) which simplifies to \(c = -6\).

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

Setting \(\frac{dy}{dx} = 0\) to find critical points:

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

Factoring gives \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\).

Since \(x = 3\) is a minimum, \(x = -2\) is a maximum.

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (x^2 - x - 6) \, dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 6x + k\).

Given \(y = -10\) when \(x = 3\), substitute to find \(k\):

\(-10 = \frac{1}{3}(3)^3 - \frac{1}{2}(3)^2 - 6(3) + k\).

\(-10 = 9 - 4.5 - 18 + k\).

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

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

Substitute \(x = -2\) to find \(y\) at the maximum point:

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

\(y = -\frac{8}{3} - 2 + 12 + \frac{3}{2}\).

\(y = 10.8\).

Therefore, the coordinates of the maximum point are \((-2, 10.8)\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\).

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

Substitute the point \((4, \frac{2}{3})\) into the equation:

\(\frac{2}{3} = \frac{2}{3}(4)^{\frac{3}{2}} - 2(4)^{\frac{1}{2}} + c\).

\(\frac{2}{3} = \frac{16}{3} - 4 + c\).

\(c = -\frac{2}{3}\).

Thus, the equation of the curve is \(y = \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} - \frac{2}{3}\).

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

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

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

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

\(x = 1\).

Substitute \(x = 1\) into the equation of the curve:

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

Thus, the stationary point is \((1, -2)\).

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

\(\frac{d^2y}{dx^2} \bigg|_{x=1} = \frac{1}{2}(1)^{-\frac{1}{2}} + \frac{1}{2}(1)^{-\frac{3}{2}} = 1 > 0\).

Hence, the stationary point is a minimum.

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = x + \frac{4}{x^2}\).

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

\(y = \frac{x^2}{2} - \frac{4}{x} + c\)

Using the point \(P(4, 8)\), substitute \(x = 4\) and \(y = 8\):

\(8 = \frac{4^2}{2} - \frac{4}{4} + c\)

\(8 = 8 - 1 + c\)

\(c = 1\)

Thus, the equation of the curve is \(y = \frac{x^2}{2} - \frac{4}{x} + 1\).

(ii) To find when the gradient has a minimum value, find the second derivative:

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

Set \(\frac{d^2y}{dx^2} = 0\) to find critical points:

\(1 - \frac{8}{x^3} = 0\)

\(\frac{8}{x^3} = 1\)

\(x^3 = 8\)

\(x = 2\)

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

\(\frac{dy}{dx} = 2 + \frac{4}{2^2} = 2 + 1 = 3\)

Check the nature of the critical point using the third derivative or sign change:

\(\frac{d}{dx} \left( 1 - \frac{8}{x^3} \right) = \frac{24}{x^4}\)

Since \(\frac{24}{x^4} > 0\) for \(x > 0\), the gradient has a minimum value at \(x = 2\).

Thus, the minimum gradient is 3 when \(x = 2\).

(a) To find the value of \(a\), set \(\frac{dy}{dx} = 0\) at \(x = a\):

\(2(a + 3)^{\frac{1}{2}} - a = 0\)

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

Square both sides:

\(4(a + 3) = a^2\)

\(a^2 - 4a - 12 = 0\)

Factorize:

\((a - 6)(a + 2) = 0\)

\(a = 6\) (since \(a\) is positive)

(b) To determine the nature of the stationary point, find the second derivative:

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

Substitute \(x = a = 6\):

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

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

(c) To find the equation of the curve, integrate \(\frac{dy}{dx} = 2(x + 3)^{\frac{1}{2}} - x\):

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

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

Substitute \(x = 6\) and \(y = 14\):

\(14 = \frac{4}{3}(9)^{\frac{3}{2}} - 18 + c\)

\(14 = 36 - 18 + c\)

\(c = -4\)

Thus, the equation of the curve is:

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

(i) Given \(\frac{d^2y}{dx^2} = -4x\), integrate to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int -4x \, dx = -2x^2 + c\)

Since the curve has a maximum at \(x = 2\), \(\frac{dy}{dx} = 0\) when \(x = 2\):

\(0 = -2(2)^2 + c \Rightarrow c = 8\)

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (-2x^2 + 8) \, dx = -\frac{2x^3}{3} + 8x + C\)

Use the point (2, 12) to find \(C\):

\(12 = -\frac{2(2)^3}{3} + 8(2) + C\)

\(12 = -\frac{16}{3} + 16 + C\)

\(C = \frac{4}{3}\)

Thus, the equation of the curve is \(y = -\frac{2x^3}{3} + 8x + \frac{4}{3}\).

(ii) To find \(\frac{dy}{dt}\), use the chain rule:

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\)

From part (i), \(\frac{dy}{dx} = -2x^2 + 8\).

At \(x = 3\), \(\frac{dy}{dx} = -2(3)^2 + 8 = -18 + 8 = -10\).

Given \(\frac{dx}{dt} = 0.05\),

\(\frac{dy}{dt} = -10 \times 0.05 = -0.5\).

Thus, the \(y\)-coordinate is decreasing at 0.5 units per second.

(i) Since \(P(3, -10)\) is a stationary point, \(f'(3) = 0\). Substituting into \(f'(x) = 2x^2 + kx - 12\), we get:

\(2(3)^2 + 3k - 12 = 0\)

\(18 + 3k - 12 = 0\)

\(3k = -6\)

\(k = -2\)

To find the other stationary point, set \(f'(x) = 0\):

\(2x^2 - 2x - 12 = 0\)

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

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

\(x = 3\) or \(x = -2\)

Thus, the \(x\)-coordinate of \(Q\) is \(-2\).

(ii) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = 4x - 2\)

At \(x = 3\), \(f''(3) = 4(3) - 2 = 10 > 0\), so \(P\) is a minimum.

At \(x = -2\), \(f''(-2) = 4(-2) - 2 = -10 < 0\), so \(Q\) is a maximum.

(iii) Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int (2x^2 - 2x - 12) \, dx\)

\(f(x) = \frac{2}{3}x^3 - x^2 - 12x + c\)

Using \(P(3, -10)\), substitute to find \(c\):

\(-10 = \frac{2}{3}(3)^3 - (3)^2 - 12(3) + c\)

\(-10 = 18 - 9 - 36 + c\)

\(c = 17\)

Thus, \(f(x) = \frac{2}{3}x^3 - x^2 - 12x + 17\).

(i) The gradient of the normal line \(3y + x = 17\) is \(-\frac{1}{3}\). Since the line is normal to the curve, the product of the gradients of the tangent and the normal is \(-1\). Therefore, the gradient of the tangent \(\frac{dy}{dx} = 3\).

Setting \(5 - \frac{8}{x^2} = 3\), we solve for \(x\):

\(5 - \frac{8}{x^2} = 3\)

\(\frac{8}{x^2} = 2\)

\(x^2 = 4\)

\(x = 2\) (since \(x\) is positive)

Substitute \(x = 2\) into the normal line equation to find \(y\):

\(3y + 2 = 17\)

\(3y = 15\)

\(y = 5\)

Thus, the coordinates of \(P\) are \((2, 5)\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = 5 - \frac{8}{x^2}\):

\(y = \int (5 - \frac{8}{x^2}) \, dx\)

\(y = 5x + 8x^{-1} + c\)

Using the point \((2, 5)\) to find \(c\):

\(5 = 5(2) + \frac{8}{2} + c\)

\(5 = 10 + 4 + c\)

\(c = -9\)

Therefore, the equation of the curve is \(y = 5x + \frac{8}{x} - 9\).

(i) To find the stationary point, set the derivative equal to zero:

\(f'(x) = 2x - 6 = 0\)

Solving for \(x\), we get:

\(2x = 6\)

\(x = 3\)

Thus, the value of \(x\) for which \(f(x)\) has a stationary value is 3.

(ii) To find \(f(x)\), integrate \(f'(x)\):

\(f(x) = \int (2x - 6) \, dx\)

\(f(x) = x^2 - 6x + c\)

Given \(f(x) \geq -4\), substitute \(x = 3\) to find \(c\):

\(f(3) = 3^2 - 6(3) + c = -4\)

\(9 - 18 + c = -4\)

\(c = 5\)

Thus, \(f(x) = x^2 - 6x + 5\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{3}{(1 + 2x)^2}\).

Let \(u = 1 + 2x\), then \(du = 2dx\) or \(dx = \frac{du}{2}\).

\(\int \frac{3}{u^2} \cdot \frac{du}{2} = -\frac{3}{2u} + c\).

Substitute back \(u = 1 + 2x\):

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

Using the point \((1, \frac{1}{2})\):

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

\(\frac{1}{2} = -\frac{3}{6} + c\).

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

\(c = 1\).

Thus, the equation of the curve is \(y = \frac{3}{1 + 2x} + 1\).

(ii) We need \(\frac{3}{(1 + 2x)^2} < \frac{1}{3}\).

\(9 < (1 + 2x)^2\).

\((1 + 2x)^2 > 9\).

\(1 + 2x > 3\) or \(1 + 2x < -3\).

\(2x > 2\) or \(2x < -4\).

\(x > 1\) or \(x < -2\).

Thus, the set of values of \(x\) is \(x > 1, x < -2\).

(i) At \(x = 2\), the tangent has gradient \(3\). Therefore, the normal has gradient \(-\frac{1}{3}\) since \(m_1 m_2 = -1\).

The equation of the normal is \(y - 11 = -\frac{1}{3}(x - 2)\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{6}{\sqrt{3x - 2}}\).

Integrate: \(y = 6 \int \frac{1}{\sqrt{3x - 2}} \, dx\).

This gives \(y = 4\sqrt{3x - 2} + c\).

Using the point \((2, 11)\) to find \(c\):

\(11 = 4\sqrt{3(2) - 2} + c\)

\(11 = 4\sqrt{4} + c\)

\(11 = 8 + c\)

\(c = 3\)

Thus, the equation of the curve is \(y = 4\sqrt{3x - 2} + 3\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 6\).

\(y = \int (3x^{\frac{1}{2}} - 6) \, dx\)

\(y = \int 3x^{\frac{1}{2}} \, dx - \int 6 \, dx\)

\(y = 2x^{\frac{3}{2}} - 6x + c\)

Using the point (9, 2) to find \(c\):

\(2 = 2(9)^{\frac{3}{2}} - 6(9) + c\)

\(2 = 54 - 54 + c\)

\(c = 2\)

Thus, the equation of the curve is \(y = 2x^{\frac{3}{2}} - 6x + 2\).

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

\(3x^{\frac{1}{2}} - 6 = 0\)

\(3x^{\frac{1}{2}} = 6\)

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

\(x = 4\)

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

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

At \(x = 4\), \(\frac{d^2y}{dx^2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}\)

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

(i) At \(x = 2\), the gradient \(m = k - 4\). At \(x = 3\), the gradient \(m = k - 6\). Since the tangents are perpendicular, \((k - 4)(k - 6) = -1\).

Solving \((k - 4)(k - 6) = -1\):

\(k^2 - 10k + 24 = -1\)

\(k^2 - 10k + 25 = 0\)

\((k - 5)^2 = 0\)

\(k = 5\)

(ii) Integrate \(\frac{dy}{dx} = 5 - 2x\) to find \(y\):

\(y = 5x - x^2 + c\)

Substitute the point (4, 9):

\(9 = 5(4) - 4^2 + c\)

\(9 = 20 - 16 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = 5x - x^2\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 4 - x\):

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

Using the point \(P(2, 9)\), substitute to find \(c\):

\(9 = 4(2) - \frac{1}{2}(2)^2 + c\)

\(9 = 8 - 2 + c\)

\(c = 3\).

Thus, the equation of the curve is \(y = 4x - \frac{1}{2}x^2 + 3\).

(ii) The gradient of the tangent at \(P\) is \(\frac{dy}{dx} = 4 - 2 = 2\).

The gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

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

(iii) Substitute \(y = 4x - \frac{1}{2}x^2 + 3\) into the normal equation:

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

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

\(x^2 - 9x + 14 = 0\)

Solving the quadratic equation gives \(x = 7\).

Substitute \(x = 7\) back into the curve equation:

\(y = 4(7) - \frac{1}{2}(7)^2 + 3\)

\(y = 28 - 24.5 + 3\)

\(y = 6.5\).

Thus, the coordinates of \(Q\) are \((7, 6.5)\).

(i) At \(x = 1\), the gradient \(m\) of the tangent is \(2\). The gradient of the normal is \(-\frac{1}{2}\) since \(m \cdot m_\text{normal} = -1\).

The equation of the normal is \(y - 8 = -\frac{1}{2}(x - 1)\), which simplifies to \(2y + x = 17\).

The normal meets the x-axis when \(y = 0\), giving \(x = 17\), so \(R(17, 0)\).

The normal meets the y-axis when \(x = 0\), giving \(y = 8.5\), so \(Q(0, 8.5)\).

The mid-point of \(QR\) is \(\left( \frac{0 + 17}{2}, \frac{8.5 + 0}{2} \right) = \left( \frac{17}{2}, \frac{9}{2} \right)\).

(ii) Integrate \(\frac{dy}{dx} = \frac{4}{\sqrt{6 - 2x}}\) to find \(y\).

\(y = \int \frac{4}{\sqrt{6 - 2x}} \, dx = 4(6 - 2x)^{\frac{1}{2}} \cdot \left(-\frac{1}{2}\right) + c\).

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

Substitute \(P(1, 8)\) into the equation: \(8 = \frac{1}{2}(1) - 2 + c\).

Solving gives \(c = 16\).

Thus, the equation of the curve is \(y = \frac{1}{2}x - 2 + 16\).

(i) To find where \(f\) is decreasing, we need \(f'(x) < 0\).

\(3(2x-1)^{\frac{1}{2}} - 6 < 0\)

\(3(2x-1)^{\frac{1}{2}} < 6\)

\((2x-1)^{\frac{1}{2}} < 2\)

\(2x-1 < 4\)

\(2x < 5\)

\(x < \frac{5}{2}\)

Since \(x > \frac{1}{2}\) is given, the set of values is \(\frac{1}{2} < x < \frac{5}{2}\).

(ii) We integrate \(f'(x) = 3(2x-1)^{\frac{1}{2}} - 6\) to find \(f(x)\).

\(f(x) = \int (3(2x-1)^{\frac{1}{2}} - 6) \, dx\)

\(f(x) = \left[ \frac{3}{2} (2x-1)^{\frac{3}{2}} \cdot \frac{1}{2} \right] - 6x + c\)

\(f(x) = (2x-1)^{\frac{3}{2}} - 6x + c\)

Given \(f(1) = -3\), substitute \(x = 1\):

\(-3 = (2 \cdot 1 - 1)^{\frac{3}{2}} - 6 \cdot 1 + c\)

\(-3 = 1 - 6 + c\)

\(c = 2\)

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

(i) To find the equation of the normal, first find the gradient of the tangent at \(P(3, 3)\). The gradient \(m\) is given by \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\). At \(x = 3\), \(\frac{dy}{dx} = \frac{6}{\sqrt{4(3) - 3}} = 2\). The gradient of the normal is the negative reciprocal, \(m = -\frac{1}{2}\). The equation of the normal is \(y - 3 = -\frac{1}{2}(x - 3)\), which simplifies to \(x + 2y = 9\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\). Let \(u = 4x - 3\), then \(du = 4dx\), so \(dx = \frac{1}{4}du\). The integral becomes \(\int \frac{6}{\sqrt{u}} \cdot \frac{1}{4} \, du = \frac{3}{2} \int u^{-1/2} \, du\). This integrates to \(3u^{1/2} + c\). Substituting back \(u = 4x - 3\), we get \(y = 3(4x - 3)^{1/2} + c\). Using the point \((3, 3)\), substitute to find \(c\): \(3 = 3(4(3) - 3)^{1/2} + c\), which gives \(c = -6\). Therefore, the equation of the curve is \(y = 3(4x - 3)^{1/2} - 6\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^2 - 4x + 1\).

\(y = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + c\).

Use the point (1, 5) to find \(c\):

\(5 = 1^3 - 2(1)^2 + 1 + c\)

\(5 = 1 - 2 + 1 + c\)

\(5 = 0 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = x^3 - 2x^2 + x + 5\).

(ii) To find where the gradient is positive, solve \(3x^2 - 4x + 1 > 0\).

Factorize or use the quadratic formula to find the roots:

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

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}\)

\(x = \frac{4 \pm \sqrt{16 - 12}}{6}\)

\(x = \frac{4 \pm 2}{6}\)

\(x = 1\) or \(x = \frac{1}{3}\)

The inequality \(3x^2 - 4x + 1 > 0\) holds for \(x < \frac{1}{3}\) and \(x > 1\).

(i) To find the equation of the curve, integrate the gradient function:

\(\frac{dy}{dx} = \sqrt{1 + 2x} = (1 + 2x)^{1/2}\)

Integrate: \(y = \int (1 + 2x)^{1/2} \, dx\)

Using substitution, let \(u = 1 + 2x\), then \(du = 2 \, dx\), so \(dx = \frac{du}{2}\).

\(y = \int u^{1/2} \cdot \frac{1}{2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{3}(1 + 2x)^{3/2} + C\)

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

Use the point \((4, 11)\) to find \(C\):

\(11 = \frac{2}{3}(1 + 2 \cdot 4)^{3/2} + C\)

\(11 = \frac{2}{3}(9)^{3/2} + C\)

\(11 = \frac{2}{3} \cdot 27 + C\)

\(11 = 18 + C\)

\(C = 2\)

Thus, the equation of the curve is \(y = \frac{2}{3}(1 + 2x)^{3/2} + 2\).

(ii) To find the y-intercept, set \(x = 0\):

\(y = \frac{2}{3}(1 + 2 \cdot 0)^{3/2} + 2\)

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

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

\(y = \frac{2}{3} + \frac{6}{3}\)

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

The curve intersects the y-axis at \((0, \frac{8}{3})\).

(i) At \(P(1,5)\), the gradient \(m\) of the tangent is \(\frac{4}{3}\). The gradient of the normal is \(-\frac{3}{4}\) (since \(m_1 \cdot m_2 = -1\)). The equation of the normal is \(y - 5 = -\frac{3}{4}(x - 1)\). Setting \(y = 0\) to find where it crosses the x-axis, we solve \(0 - 5 = -\frac{3}{4}(x - 1)\), giving \(x = \frac{23}{3}\). Thus, \(Q \left( \frac{23}{3}, 0 \right)\).

(ii) Integrate \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\) to find \(y\). Let \(u = 2x+1\), then \(du = 2dx\), so \(dx = \frac{du}{2}\). The integral becomes \(\int \frac{12}{u^2} \cdot \frac{du}{2} = -\frac{6}{u} + C\). Substituting back, \(y = -\frac{6}{2x+1} + C\). Using \(P(1,5)\), \(5 = -\frac{6}{3} + C\), so \(C = 7\). Therefore, \(y = -\frac{6}{2x+1} + 7\).

(iii) Given \(\frac{dx}{dt} = 0.3\), find \(\frac{dy}{dt}\) using \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). At \(x = 1\), \(\frac{dy}{dx} = \frac{4}{3}\). Thus, \(\frac{dy}{dt} = \frac{4}{3} \times 0.3 = 0.4\) units per second.

Given \(\frac{dy}{dx} = \frac{k}{\sqrt{x}}\), integrate to find \(y\):

\(y = \int \frac{k}{\sqrt{x}} \, dx = k \int x^{-1/2} \, dx = k \cdot \frac{x^{1/2}}{1/2} + c = 2k\sqrt{x} + c\).

Substitute the point \(P(1, -1)\):

\(-1 = 2k\sqrt{1} + c \Rightarrow -1 = 2k + c\).

Substitute the point \(Q(4, 4)\):

\(4 = 2k\sqrt{4} + c \Rightarrow 4 = 4k + c\).

Solving the system of equations:

\(-1 = 2k + c\)

\(4 = 4k + c\)

Subtract the first equation from the second:

\(4 - (-1) = 4k - 2k\)

\(5 = 2k \Rightarrow k = \frac{5}{2}\).

Substitute \(k = \frac{5}{2}\) into \(-1 = 2k + c\):

\(-1 = 2 \times \frac{5}{2} + c \Rightarrow -1 = 5 + c \Rightarrow c = -6\).

The equation of the curve is:

\(y = 5\sqrt{x} - 6\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = (5x - 1)^{\frac{1}{2}} - 2\).

Integrate \((5x - 1)^{\frac{1}{2}}\) using substitution:

Let \(u = 5x - 1\), then \(du = 5 \, dx\), so \(dx = \frac{du}{5}\).

\(\int (5x - 1)^{\frac{1}{2}} \, dx = \int u^{\frac{1}{2}} \cdot \frac{du}{5} = \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{15} (5x - 1)^{\frac{3}{2}} + C\).

Integrate \(-2\) to get \(-2x + D\).

Thus, \(y = \frac{2}{15} (5x - 1)^{\frac{3}{2}} - 2x + C\).

Using the point (2, 3):

\(3 = \frac{2}{15} (5 \times 2 - 1)^{\frac{3}{2}} - 2 \times 2 + C\).

\(3 = \frac{2}{15} \times 27 - 4 + C\).

\(3 = \frac{54}{15} - 4 + C\).

\(3 = \frac{18}{5} - 4 + C\).

\(C = 7 - \frac{18}{5}\).

\(C = \frac{17}{5}\).

Thus, \(y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\).

(ii) Differentiate \(\frac{dy}{dx} = (5x - 1)^{\frac{1}{2}} - 2\) to find \(\frac{d^2y}{dx^2}\).

\(\frac{d^2y}{dx^2} = \frac{d}{dx}((5x - 1)^{\frac{1}{2}})\).

Using the chain rule: \(\frac{1}{2}(5x-1)^{-\frac{1}{2}} \times 5\).

\(\frac{d^2y}{dx^2} = \frac{5}{2}(5x-1)^{-\frac{1}{2}}\).

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

\((5x - 1)^{\frac{1}{2}} - 2 = 0\).

\((5x - 1)^{\frac{1}{2}} = 2\).

\(5x - 1 = 4\).

\(5x = 5\).

\(x = 1\).

Substitute \(x = 1\) into \(y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\):

\(y = \frac{2(5 \times 1 - 1)^{\frac{3}{2}}}{15} - 2 \times 1 + \frac{17}{5}\).

\(y = \frac{2 \times 8}{15} - 2 + \frac{17}{5}\).

\(y = \frac{16}{15} - 2 + \frac{17}{5}\).

\(y = \frac{16}{15} - \frac{30}{15} + \frac{51}{15}\).

\(y = \frac{37}{15}\).

Stationary point is \((1, \frac{37}{15})\).

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

\(\frac{d^2y}{dx^2} = \frac{5}{2}(5 \times 1 - 1)^{-\frac{1}{2}} = \frac{5}{2} \times \frac{1}{2} = \frac{5}{4}\).

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

Given \(\frac{dy}{dx} = 3x^2 + ax + b\), stationary points occur where \(\frac{dy}{dx} = 0\).

For \(x = -1\), \(3(-1)^2 + a(-1) + b = 0\) gives:

\(3 - a + b = 0\)

For \(x = 3\), \(3(3)^2 + a(3) + b = 0\) gives:

\(27 + 3a + b = 0\)

Solving these equations simultaneously:

1. \(3 - a + b = 0\)

2. \(27 + 3a + b = 0\)

Subtract equation 1 from equation 2:

\(27 + 3a + b - (3 - a + b) = 0\)

\(27 + 3a + b - 3 + a - b = 0\)

\(24 + 4a = 0\)

\(a = -6\)

Substitute \(a = -6\) into equation 1:

\(3 - (-6) + b = 0\)

\(3 + 6 + b = 0\)

\(b = -9\)

Integrate \(\frac{dy}{dx} = 3x^2 - 6x - 9\) to find \(y\):

\(y = x^3 - 3x^2 - 9x + c\)

Use point \((-1, 2)\):

\(2 = (-1)^3 - 3(-1)^2 - 9(-1) + c\)

\(2 = -1 - 3 + 9 + c\)

\(c = -3\)

Use point \((3, k)\):

\(k = 3^3 - 3(3)^2 - 9(3) - 3\)

\(k = 27 - 27 - 27 - 3\)

\(k = -30\)

(i) Integrate \(\frac{d^2y}{dx^2} = 2x - 5\) to find \(\frac{dy}{dx} = x^2 - 5x + c\).

Since the curve has a stationary point at \(x = 3\), \(\frac{dy}{dx} = 0\) when \(x = 3\).

Substitute \(x = 3\) into \(x^2 - 5x + c = 0\):

\(9 - 15 + c = 0\)

\(c = 6\)

Integrate again to find \(y = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + d\).

Use the point (3, 6) to find \(d\):

\(6 = \frac{27}{3} - \frac{45}{2} + 18 + d\)

\(6 = 9 - 22.5 + 18 + d\)

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

Thus, the equation of the curve is \(y = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + \frac{1}{2}\).

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

\(x^2 - 5x + 6 = 0\)

Factorize to get \((x - 2)(x - 3) = 0\).

The other stationary point is at \(x = 2\).

(iii) Determine the nature of the stationary points using \(\frac{d^2y}{dx^2} = 2x - 5\):

At \(x = 3\), \(\frac{d^2y}{dx^2} = 2(3) - 5 = 1\) (positive, so minimum).

At \(x = 2\), \(\frac{d^2y}{dx^2} = 2(2) - 5 = -1\) (negative, so maximum).

To find \(f(x)\), integrate \(f'(x) = kx^2 - 2x\):

\(f(x) = \frac{k}{3}x^3 - x^2 + c\)

Use the point \((0, 2)\):

\(2 = \frac{k}{3}(0)^3 - (0)^2 + c\)

\(c = 2\)

Use the point \((3, -1)\):

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

\(-1 = 9k - 9 + 2\)

\(-1 = 9k - 7\)

\(9k = 6\)

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