(a) To find the stationary point \(R\), we need to find \(\frac{dy}{dx}\) and set it to zero. The derivative is \(\frac{dy}{dx} = -\frac{8}{(2x-1)^3} + 1\).

Substitute \(x = \frac{3}{2}\) into the derivative: \(\frac{dy}{dx} = -\frac{8}{8} + 1 = 0\). Hence, the \(x\)-coordinate of \(R\) is \(\frac{3}{2}\).

When \(x = \frac{3}{2}\), \(y = \frac{2}{4} + \frac{3}{2} = 2\). Thus, the \(y\)-coordinate of \(R\) is 2.

(b) The \(y\)-coordinate of \(P\) is 3, and the \(y\)-coordinate of \(Q\) is \(\frac{20}{9}\).

The area under the curve from \(x = 1\) to \(x = 2\) is given by:

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

Evaluating this integral gives \(\frac{13}{6}\).

The area of the trapezium formed by the lines \(x = 1\), \(x = 2\), and the line \(y = x\) is \(\frac{1}{2} \left( 3 + \frac{20}{9} \right) = \frac{47}{18}\).

Thus, the exact value of the area of the shaded region is \(\frac{47}{18}\).

(a) To find the points of intersection, set the equations equal: \((3x - 2)^{\frac{1}{2}} = \frac{1}{2}x + 1\).

Square both sides: \(3x - 2 = \left( \frac{1}{2}x + 1 \right)^2\).

Expand the right side: \(3x - 2 = \frac{1}{4}x^2 + x + 1\).

Rearrange to form a quadratic equation: \(\frac{1}{4}x^2 - 2x + 3 = 0\).

Multiply through by 4 to clear the fraction: \(x^2 - 8x + 12 = 0\).

Factorize: \((x - 6)(x - 2) = 0\).

Thus, \(x = 2\) or \(x = 6\).

Substitute back to find \(y\):

For \(x = 2\), \(y = \frac{1}{2}(2) + 1 = 2\).

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

Coordinates are \((2, 2)\) and \((6, 4)\).

(b) The area between the curve and the line is given by the integral:

\(\text{Area} = \int_{2}^{6} \left( (3x - 2)^{\frac{1}{2}} - \left( \frac{1}{2}x + 1 \right) \right) \, dx\).

Integrate each part:

\(\int (3x - 2)^{\frac{1}{2}} \, dx = \frac{2}{9}(3x - 2)^{\frac{3}{2}}\).

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

Evaluate from 2 to 6:

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

Calculate each part:

\(\frac{2}{9}(16)^{\frac{3}{2}} - \frac{2}{9}(4)^{\frac{3}{2}} - \left( \frac{1}{4}(36) + 6 - \left( \frac{1}{4}(4) + 2 \right) \right)\).

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

(a) To find the area between the curves, we calculate the definite integral of the difference between the two functions from \(x = \frac{1}{4}\) to \(x = 4\):

\(\int_{1/4}^{4} \left( \frac{5}{2} - x^{-\frac{1}{2}} - x^{-\frac{1}{2}} \right) \, dx\)

\(= \int_{1/4}^{4} \left( \frac{5}{2} - 2x^{-\frac{1}{2}} \right) \, dx\)

Integrating term by term, we have:

\(\frac{5}{2}x - \frac{2}{3}x^{\frac{3}{2}} \bigg|_{1/4}^{4}\)

Substituting the limits:

\(\left( \frac{5}{2} \times 4 - \frac{2}{3} \times 4^{\frac{3}{2}} \right) - \left( \frac{5}{2} \times \frac{1}{4} - \frac{2}{3} \times \left( \frac{1}{4} \right)^{\frac{3}{2}} \right)\)

\(= \left( 10 - \frac{16}{3} \right) - \left( \frac{5}{8} - \frac{1}{12} \right)\)

\(= \frac{9}{8}\)

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

At \(x = 1\), the gradient \(m = -\frac{1}{2}\).

The equation of the normal at \((1, 1)\) is:

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

Substituting \(x = 0\) to find \(p\):

\(y - 1 = 2(0 - 1)\)

\(y - 1 = -2\)

\(y = -1\)

Thus, \(p = -1\).

(a) Differentiate the curve: \(\frac{dy}{dx} = \frac{1}{2} + \frac{1}{3(x-2)^{\frac{4}{3}}}\).

Evaluate \(\frac{dy}{dx}\) at \(x = 3\): \(\frac{1}{2} + \frac{1}{3(3-2)^{\frac{4}{3}}} = \frac{5}{6}\).

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

Equation of the normal: \(y - \frac{6}{5} = -\frac{6}{5}(x - 3)\).

Simplify: \(y = -\frac{6}{5}x + \frac{24}{5}\).

Set \(y = 0\) to find \(x\): \(0 = -\frac{6}{5}x + \frac{24}{5} \Rightarrow x = 4\).

(b) Integrate the curve from \(x = 2.5\) to \(x = 3\):

\(\int \left( \frac{1}{2}x + \frac{7}{10} - \frac{1}{(x-2)^{\frac{1}{3}}} \right) \, dx\).

Calculate the definite integral: \(\left[ \frac{1}{4}x^2 + \frac{7}{10}x - 3(x-2)^{\frac{2}{3}} \right]_{2.5}^{3}\).

Evaluate: \(\left( \frac{9}{4} + 2.1 - \frac{3}{2} \right) - \left( \frac{6.25}{4} + 1.75 - 3 \times 0.5^{\frac{2}{3}} \right) = 0.48\).

Area of triangle: \(0.6\).

Total area: \(0.48 + 0.6 = 1.08\).

(a) Differentiate \(y = x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}\) to find \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}k^2 x^{-\frac{3}{2}}\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(\frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2}k^2 x^{-\frac{3}{2}}\).

Simplify to get \(x = k^2\). Substitute back into the original equation to find \(y = 2k\).

Thus, the minimum point is \((k^2, 2k)\).

(b) At \(x = 4k^2\), find \(\frac{dy}{dx} = \frac{1}{4k} - \frac{1}{16k} = \frac{3}{16k}\).

The y-coordinate at \(x = 4k^2\) is \(y = 2k + \frac{k}{2} = \frac{5k}{2}\).

The equation of the tangent is \(y - \frac{5k}{2} = \frac{3}{16k}(x - 4k^2)\).

When \(x = 0\), \(y = \frac{7k}{4}\).

(c) Integrate \(\int (x^{\frac{1}{2}} + k^2 x^{-\frac{1}{2}}) \, dx\) to get \(\frac{2}{3}x^{\frac{3}{2}} + 2k^2 x^{\frac{1}{2}}\).

Apply limits from \(x = \frac{9}{4}k^2\) to \(x = 4k^2\).

Calculate the definite integral: \(\left[ \frac{16k^3}{3} + 4k^3 \right] - \left[ \frac{9k^3}{4} + 3k^3 \right] = \frac{49k^3}{12}\).

To find the area under the curve \(y = 2\sqrt{3x+4} - x\) from \(x = 0\) to \(x = 4\), we need to integrate the function with respect to \(x\).

The integral is:

\(\int (2\sqrt{3x+4} - x) \, dx\)

First, integrate \(2\sqrt{3x+4}\):

Let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{1}{3} \, du\).

\(\int 2\sqrt{u} \, \frac{1}{3} \, du = \frac{2}{3} \int u^{0.5} \, du\)

\(= \frac{2}{3} \cdot \frac{2}{3} u^{1.5} = \frac{4}{9} (3x+4)^{1.5}\)

Next, integrate \(-x\):

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

Thus, the integral of the function is:

\(\frac{4}{9} (3x+4)^{1.5} - \frac{1}{2} x^2\)

Evaluate from \(x = 0\) to \(x = 4\):

\(\left[ \frac{4}{9} (3(4)+4)^{1.5} - \frac{1}{2} (4)^2 \right] - \left[ \frac{4}{9} (3(0)+4)^{1.5} - \frac{1}{2} (0)^2 \right]\)

\(= \left[ \frac{4}{9} (16)^{1.5} - 8 \right] - \left[ \frac{4}{9} (4)^{1.5} \right]\)

\(= \left[ \frac{256}{9} - 8 \right] - \left[ \frac{32}{9} \right]\)

\(= \frac{256}{9} - 8 - \frac{32}{9}\)

\(= \frac{256}{9} - \frac{72}{9} - \frac{32}{9}\)

\(= \frac{152}{9}\)

(a) Set \(y = 0\) to find the x-coordinate of A:

\(9(x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}) = 0\)

\(9x^{-\frac{3}{2}}(x - 4) = 0\)

\(x = 4\) only.

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

\(\frac{dy}{dx} = 9\left(-\frac{1}{2}x^{-\frac{3}{2}} + 6x^{-\frac{5}{2}}\right)\)

At \(x = 4\), gradient = \(9\left(-\frac{1}{16} + \frac{6}{32}\right) = \frac{9}{8}\)

Equation of tangent: \(y = \frac{9}{8}(x - 4)\)

(c) Set \(\frac{dy}{dx} = 0\) to find maximum point:

\(9x^{-\frac{5}{2}}\left(-\frac{1}{2}x + 6\right) = 0\)

\(x = 12\)

(d) Integrate to find the area:

\(\int_4^9 9(x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}) \, dx = 9\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}\right]_4^9\)

\(= 9\left[(6 + \frac{8}{3}) - (4 + 4)\right] = 6\)

(a) To find \(\frac{dy}{dx}\), differentiate \(y = \frac{2}{(3 - 2x)^2} - x\). Using the chain rule, \(\frac{d}{dx} \left( \frac{2}{(3 - 2x)^2} \right) = 8x(3-2x)^{-3}\). The derivative of \(-x\) is \(-1\). Thus, \(\frac{dy}{dx} = \frac{8}{(3-2x)^3} - 1\).

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) again: \(\frac{d^2y}{dx^2} = -3 \times 8 \times (3-2x)^{-4} \times (-2) = \frac{48}{(3-2x)^4}\).

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

(b) To find the \(x\)-coordinate of \(M\), set \(\frac{dy}{dx} = 0\): \(0 = \frac{8}{(3-2x)^3} - 1\). Solving gives \(x = \frac{1}{2}\).

(c) The area under the curve is \(\int_{0}^{1/2} \left( \frac{2}{(3 - 2x)^2} - x \right) \, dx\). Using the integral from part (a) and evaluating from 0 to \(\frac{1}{2}\), the area is \(\frac{1}{24}\).

(a) To find the x-coordinates of \(B\) and \(C\), set the equations equal: \(4x^{\frac{1}{2}} - 2x = 3 - x\).

Rearrange to form a quadratic: \(4x^{\frac{1}{2}} - x - 3 = 0\).

Let \(u = x^{\frac{1}{2}}\), then \(4u^2 - u - 3 = 0\).

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

Thus, \(x = 1^2 = 1\) or \(x = 3^2 = 9\).

(b) Differentiate the curve: \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - 2\).

Set \(\frac{dy}{dx} = 0\) for a stationary point: \(2x^{-\frac{1}{2}} - 2 = 0\).

Solve: \(x^{-\frac{1}{2}} = 1\), so \(x = 1\).

Thus, \(B\) is a stationary point.

(c) The area of the shaded region is the area under the line minus the area under the curve from \(x = 4\) to \(x = 9\).

Area under the line: \(\int_4^9 (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_4^9 = 18 - 14 = 4\).

Area under the curve: \(\int_4^9 (4x^{\frac{1}{2}} - 2x) \, dx = \left[ \frac{8}{3}x^{\frac{3}{2}} - x^2 \right]_4^9 = 18 - 14\frac{1}{3} = 3\frac{2}{3}\).

Shaded area = \(4 - 3\frac{2}{3} = 3\frac{2}{3}\).

(a) Differentiate the curve equation: \(\frac{dy}{dx} = 3x^2 - 4bx + b^2\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 4bx + b^2 = 0\).

Factorize: \((3x - b)(x - b) = 0\), giving \(x = \frac{b}{3}\) or \(x = b\).

Since \(x = a\) at maximum, \(a = \frac{b}{3}\), hence \(b = 3a\).

(b) Find the area under the curve from \(0\) to \(a\):

\(\int (x^3 - 6ax^2 + 9a^2x) \, dx = \left[ \frac{x^4}{4} - 2ax^3 + \frac{9a^2x^2}{2} \right]_0^a\).

Evaluate: \(\frac{a^4}{4} - 2a^4 + \frac{9a^4}{2} = \frac{11a^4}{4}\).

Area under line \(OA\) is \(\frac{1}{2} a \times 4a^3 = 2a^4\).

Shaded area: \(\frac{11a^4}{4} - 2a^4 = \frac{3a^4}{4}\).

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

(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 \(3 - \frac{1}{9} = \frac{8}{9}\).

(i) To find \(\frac{dy}{dx}\), differentiate \(y = 3\sqrt{4x + 1} - 2x\).

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

For \(\int y \, dx\), integrate \(y = 3\sqrt{4x + 1} - 2x\).

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

(ii) At \(M\), \(\frac{dy}{dx} = 0\).

Set \(\frac{dy}{dx} = 0\) and solve for \(x\):

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

Solve to find \(x = 2\).

Substitute \(x = 2\) into \(y = 3\sqrt{4x + 1} - 2x\) to find \(y = 5\).

Coordinates of \(M\) are \((2, 5)\).

(iii) To find the area of the shaded region, calculate the area under the curve and subtract the area under the chord.

Area under the curve from \(x = 0\) to \(x = 2\):

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

Area under the chord (trapezium):

\(\frac{1}{2} \times 2 \times (3 + 5) = 8\)

\(Shaded area = 9 - 8 = 1\)

(i) To find \(k\), use the point \(A(2, 2)\) on the curve:

\(2 = k(2^3 - 7 \times 2^2 + 12 \times 2)\)

\(2 = k(8 - 28 + 24)\)

\(2 = k \times 4\)

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

(ii) Verify the curve meets \(y = x\) at \(x = 5\):

Substitute \(x = 5\) into the curve equation:

\(y = \frac{1}{2}(5^3 - 7 \times 5^2 + 12 \times 5)\)

\(y = \frac{1}{2}(125 - 175 + 60)\)

\(y = \frac{1}{2}(10) = 5\)

Since \(y = 5\) when \(x = 5\), the curve meets the line \(y = x\).

(iii) Find the area of the shaded region:

The area is given by the integral:

\(\int_0^2 \left( \frac{1}{2}(x^3 - 7x^2 + 12x) - x \right) \, dx\)

Integrate term by term:

\(\int_0^2 \left( \frac{1}{2}x^3 - \frac{7}{2}x^2 + 5x \right) \, dx\)

\(= \left[ \frac{1}{8}x^4 - \frac{7}{6}x^3 + \frac{5}{2}x^2 \right]_0^2\)

Evaluate the definite integral:

\(= \left( \frac{1}{8}(16) - \frac{7}{6}(8) + \frac{5}{2}(4) \right) - 0\)

\(= 2 - \frac{28}{3} + 10\)

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

To find the area under the curve \(y = x^3 - 2x^2 + 5x\) from \(x = 0\) to \(x = 6\), we need to integrate the function with respect to \(x\).

The integral of \(y = x^3 - 2x^2 + 5x\) is:

\(\int (x^3 - 2x^2 + 5x) \, dx = \frac{x^4}{4} - \frac{2x^3}{3} + \frac{5x^2}{2} + C\)

We evaluate this definite integral from \(x = 0\) to \(x = 6\):

\(\left[ \frac{x^4}{4} - \frac{2x^3}{3} + \frac{5x^2}{2} \right]_0^6\)

Substitute \(x = 6\):

\(\frac{6^4}{4} - \frac{2 \times 6^3}{3} + \frac{5 \times 6^2}{2} = \frac{1296}{4} - \frac{2 \times 216}{3} + \frac{5 \times 36}{2}\)

\(= 324 - 144 + 90\)

\(= 270\)

Thus, the area of the region is 270.

(i) Differentiate the curve equation \(y = 1 - 2x - (1 - 2x)^3\) with respect to \(x\):

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

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

The gradient of the line \(y = 1 - 2x\) is \(-2\), hence \(AB\) is a tangent.

(ii) The shaded region is given by:

\(\int_0^{\frac{1}{2}} (1 - 2x) \, dx - \int_0^{\frac{1}{2}} (1 - 2x - (1 - 2x)^3) \, dx\).

This simplifies to \(\int_0^{\frac{1}{2}} (1 - 2x)^3 \, dx\).

(iii) Evaluate the integral:

\(\int_0^{\frac{1}{2}} (1 - 2x)^3 \, dx = \left[ \frac{(1-2x)^4}{-8} \right]_0^{\frac{1}{2}}\).

Calculate the limits:

\(0 - \left(-\frac{1}{8}\right) = \frac{1}{8}\).

(i) To find the \(x\)-coordinates of \(A\) and \(B\), set the equations equal: \(3 - 2x = 4 - \frac{3}{\sqrt{x}}\).

Rearrange to form a quadratic: \(2x - \frac{3}{\sqrt{x}} + 1 = 0\).

Let \(k = \sqrt{x}\), then \(2k^2 - 3k + 1 = 0\).

Solving this quadratic gives \(k = \frac{1}{2}\) or \(k = 1\).

Thus, \(x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) or \(x = 1\).

(ii) To find the area of the shaded region, calculate the area under each curve and subtract.

Area under \(y = 3 - 2x\) from \(x = \frac{1}{4}\) to \(x = 1\):

\(\int_{1/4}^{1} (3 - 2x) \, dx = [3x - x^2]_{1/4}^{1} = \left(3(1) - 1^2\right) - \left(3\left(\frac{1}{4}\right) - \left(\frac{1}{4}\right)^2\right) = \frac{21}{16}\).

Area under \(y = 4 - \frac{3}{\sqrt{x}}\) from \(x = \frac{1}{4}\) to \(x = 1\):

\(\int_{1/4}^{1} \left(4 - 3x^{1/2}\right) \, dx = [4x - 2x^{3/2}]_{1/4}^{1} = \left(4(1) - 2(1)^{3/2}\right) - \left(4\left(\frac{1}{4}\right) - 2\left(\frac{1}{4}\right)^{3/2}\right) = \frac{5}{4}\).

Required area = \(\frac{21}{16} - \frac{5}{4} = 0.0625\).

(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).

(ii) To find the distance \(AB\), use the distance formula:

\(AB^2 = \left(3/2\right)^2 + \left(3/4\right)^2\)

\(AB = \sqrt{\frac{45}{4}} \approx 1.68\)

(iii) To find the area of the shaded region, calculate the area under the curve \(y = 2x + (x+1)^{-2}\) from \(x = -\frac{1}{2}\) to \(x = 1\).

The area under the curve is:

\(\int_{-1/2}^{1} \left( x^2 - (x+1)^{-1} \right) \, dx = \left[ \frac{1}{4}x^2 + \frac{11}{4}x \right]_{-1/2}^{1} - \left[ x^2 - (x+1)^{-1} \right]_{-1/2}^{1}\)

Apply limits \(-\frac{1}{2} \to 1\) to both integrals:

\(= \frac{27}{16} \approx 1.69\)

(i) To find the equation of the tangent, we first find the derivative of the curve \(y = 9 + 6x - 3x^2\), which is \(\frac{dy}{dx} = 6 - 6x\). At \(x = 2\), the gradient is \(6 - 6(2) = -6\). The equation of the tangent at A (2, 9) is \(y - 9 = -6(x - 2)\), which simplifies to \(y = -6x + 21\). To find the x-coordinate of C, set \(y = 0\) in the tangent equation: \(0 = -6x + 21\), giving \(x = \frac{21}{6} = \frac{7}{2}\).

(ii) The area under the curve from \(x = 2\) to \(x = 3\) is calculated by integrating \(9 + 6x - 3x^2\), which gives \(9x + 3x^2 - x^3\). Evaluating from 2 to 3, we get \((27 + 27 - 27) - (18 + 12 - 8) = 5\). The area under the tangent from \(x = 2\) to \(x = \frac{7}{2}\) is \(\frac{1}{2} \times \frac{3}{2} \times 9 = \frac{27}{4}\). The area of the shaded region ABC is \(\frac{27}{4} - 5 = \frac{7}{4}\).

(i) The equation of the curve is \(y = \frac{8}{\sqrt{3x+4}}\). The derivative is \(\frac{dy}{dx} = \frac{-4}{(3x+4)^{\frac{3}{2}}} \times 3\). At \(x = 0\), \(\frac{dy}{dx} = -\frac{3}{2}\). The slope of the normal is the negative reciprocal, \(\frac{2}{3}\).

The equation of the normal at \(A\) is \(y - 4 = \frac{2}{3}(x - 0)\). Solving for \(x = 4\), we find \(B \left( 4, \frac{20}{3} \right)\).

(ii) The area \(P\) is given by \(\int_{0}^{4} \frac{8}{\sqrt{3x+4}} \, dx = 8 \int_{0}^{4} (3x+4)^{-\frac{1}{2}} \, dx\). This evaluates to \(\frac{32}{3}\).

The area \(Q\) is the area of the trapezium minus \(P\). The area of the trapezium is \(\frac{1}{2} \times (4 + \frac{20}{3}) \times 4 = \frac{64}{3}\). Thus, \(Q = \frac{64}{3} - \frac{32}{3} = \frac{32}{3}\).

Therefore, the areas of \(P\) and \(Q\) are both \(\frac{32}{3}\).

(i) Set the equations equal: \(x + 2 = 3\sqrt{x}\).

Rearrange: \(x - 3\sqrt{x} + 2 = 0\).

Let \(k = \sqrt{x}\), then \(k^2 - 3k + 2 = 0\).

Factorize: \((k-1)(k-2) = 0\).

So, \(k = 1\) or \(k = 2\).

Thus, \(\sqrt{x} = 1\) or \(\sqrt{x} = 2\).

Therefore, \(x = 1\) or \(x = 4\).

For \(x = 1\), \(y = 3\).

For \(x = 4\), \(y = 6\).

Coordinates of \(A\) are \((1, 3)\) and \(B\) are \((4, 6)\).

(ii) Find the area between the curves from \(x = 1\) to \(x = 4\).

Area = \(\int_1^4 (3\sqrt{x} - (x+2)) \, dx\).

Integrate: \(\int_1^4 (3x^{1/2} - x - 2) \, dx\).

\(= \left[ 2x^{3/2} - \frac{1}{2}x^2 - 2x \right]_1^4\).

Calculate: \(\left[ 16 - 8 - 8 \right] - \left[ 2 - \frac{1}{2} - 2 \right]\).

\(= 0 - (-1) = 1\).

Area = \(\frac{1}{2}\).

(i) For \(y = (4x + 1)^{\frac{1}{2}}\), the derivative is \(\frac{dy}{dx} = \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \times 4 = \frac{2}{\sqrt{4x + 1}}\).

When \(x = 2\), \(\frac{dy}{dx} = \frac{2}{3}\).

For \(y = \frac{1}{2}x^2 + 1\), the derivative is \(\frac{dy}{dx} = x\).

When \(x = 2\), \(\frac{dy}{dx} = 2\).

The angle \(\alpha\) is given by \(\alpha = \arctan(m_2) - \arctan(m_1)\), where \(m_1 = \frac{2}{3}\) and \(m_2 = 2\).

\(\alpha = \arctan(2) - \arctan\left(\frac{2}{3}\right) \approx 63.43 - 33.69 = 29.7\) degrees.

(ii) The area of the shaded region is found by integrating the difference between the two curves from \(x = 0\) to \(x = 2\).

\(\int (4x + 1)^{\frac{1}{2}} \, dx = \left[ \frac{(4x+1)^{\frac{3}{2}}}{\frac{3}{2} \times 4} \right] = \left[ \frac{(4x+1)^{\frac{3}{2}}}{6} \right]\).

\(\int \left(\frac{1}{2}x^2 + 1\right) \, dx = \frac{1}{6}x^3 + x\).

Evaluate from 0 to 2:

\(\left[ \frac{(4 \times 2 + 1)^{\frac{3}{2}}}{6} \right] - \left[ \frac{(4 \times 0 + 1)^{\frac{3}{2}}}{6} \right] = \frac{27}{6} - \frac{1}{6} = \frac{26}{6} = \frac{13}{3}\).

\(\left[ \frac{1}{6}(2)^3 + 2 \right] - \left[ \frac{1}{6}(0)^3 + 0 \right] = \frac{8}{6} + 2 = \frac{10}{3}\).

The area is \(\frac{13}{3} - \frac{10}{3} = 1\).

First, find the points of intersection by setting the equations equal: \(2x + 1 = -x^2 + 12x - 20\).

Rearrange to get: \(x^2 - 10x + 21 = 0\).

Factorize to find \((x - 3)(x - 7) = 0\), so \(x = 3\) and \(x = 7\).

Calculate the area under the line \(y = 2x + 1\) from \(x = 3\) to \(x = 7\):

Area of trapezium = \(\frac{1}{2} \times 4 \times (7 + 15) = 44\).

Calculate the area under the curve \(y = -x^2 + 12x - 20\) from \(x = 3\) to \(x = 7\):

Integrate to find \(\int_{3}^{7} (-x^2 + 12x - 20) \, dx = \left[ -\frac{x^3}{3} + 6x^2 - 20x \right]_{3}^{7}\).

Evaluate the integral: \(\left[ -\frac{343}{3} + 294 - 140 \right] - \left[ -\frac{27}{3} + 54 - 60 \right] = 54\frac{2}{3}\).

The shaded area is the difference: \(44 - 54\frac{2}{3} = 10\frac{2}{3}\).

(i) To find \(\frac{dy}{dx}\), differentiate \(y = 8 - \sqrt{4-x}\). Using the chain rule, \(\frac{dy}{dx} = -\frac{1}{2}(4-x)^{-\frac{1}{2}} \times (-1)\).

For \(\int y \, dx\), integrate \(y = 8 - \sqrt{4-x}\). The integral is \(8x - \frac{(4-x)^{\frac{3}{2}}}{\frac{3}{2}} \times (-1)\).

(ii) The equation of the tangent at \(P(3, 7)\) is found using the point-slope form. The slope \(m = \frac{1}{2}\), so the equation is \(y - 7 = \frac{1}{2}(x - 3)\), which simplifies to \(y = \frac{1}{2}x + \frac{5}{2}\).

(iii) The area under the curve from 0 to 3 is \(\int_{0}^{3} (8 - \sqrt{4-x}) \, dx = \frac{58}{3}\). The area under the line is \(\frac{1}{2}(\frac{5}{2} + 7) \times 3 = \frac{75}{4}\). The area of the shaded region is \(\frac{58}{3} - \frac{75}{4} = \frac{7}{12}\).

To find the points of intersection, set the equations equal: \(2x + 11 = 8 - 2x - x^2\).

Rearrange to form a quadratic equation: \(x^2 + 4x + 3 = 0\).

Factorize: \((x + 1)(x + 3) = 0\).

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

To find the area between the curves, integrate the difference between the curve and the line from \(x = -3\) to \(x = -1\):

\(\int_{-3}^{-1} (8 - 2x - x^2) \, dx - \int_{-3}^{-1} (2x + 11) \, dx\).

Integrate each part:

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

\(\int (2x + 11) \, dx = x^2 + 11x\).

Evaluate from \(x = -3\) to \(x = -1\):

\([8(-1) - (-1)^2 - \frac{(-1)^3}{3}] - [8(-3) - (-3)^2 - \frac{(-3)^3}{3}]\).

\([8(-1) - 1 + \frac{1}{3}] - [8(-3) - 9 + \frac{27}{3}]\).

Calculate the definite integrals and subtract: \(\frac{4}{3}\).

(a) To find the maximum point, differentiate the curve equation: \(\frac{dy}{dx} = 9 - \frac{3}{2}(2x+1)^{\frac{1}{2}} \times 2\).

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

\(9 - 3(2x+1)^{\frac{1}{2}} = 0\)

\(2x+1 = 9 \Rightarrow x = 4\)

Substitute \(x = 4\) into the original equation to find \(y\):

\(y = 9(4) - (2(4) + 1)^{\frac{3}{2}} = 36 - 9 = 27\)

Thus, the maximum point is \((4, 9)\).

(b) The gradient of the curve at \(A\) is found by substituting \(x = 1\frac{1}{2}\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 3\)

The gradient of \(AB\) is:

\(\frac{5\frac{1}{2} - 3\frac{1}{2}}{1\frac{1}{2} - 7\frac{1}{2}} = \frac{2}{-6} = -\frac{1}{3}\)

Since the product of the gradients is \(-1\), \(AB\) is the normal to the curve at \(A\).

(c) To find the area of the shaded region, integrate the curve equation from \(x = 1\frac{1}{2}\) to \(x = 7\frac{1}{2}\):

\(\int_{1\frac{1}{2}}^{7\frac{1}{2}} \left(9x - (2x+1)^{\frac{3}{2}}\right) \, dx\)

Calculate the definite integral and subtract the area of the trapezium formed by \(AB\):

\(Shaded area = 44.6 - 27 = 17.6\)

(i) To find the equation of the tangent, we first need the derivative of the curve \(y = (3 - 2x)^3\).

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

At \(x = \frac{1}{2}\), \(\frac{dy}{dx} = -6(3 - 2 \times \frac{1}{2})^2 = -6(2)^2 = -24\).

The slope of the tangent is \(-24\). Using the point-slope form, \(y - 8 = -24(x - \frac{1}{2})\).

Simplifying gives \(y = -24x + 20\).

(ii) To find the area of the shaded region, calculate the area under the curve and subtract the area under the tangent.

Area under the curve from \(x = 0\) to \(x = \frac{1}{2}\):

\(\int_0^{1/2} (3 - 2x)^3 \, dx = \left[ \frac{(3 - 2x)^4}{-8} \right]_0^{1/2} = \left[ \frac{(2)^4}{-8} - \frac{(3)^4}{-8} \right] = \frac{81}{8}\).

Area under the tangent from \(x = 0\) to \(x = \frac{1}{2}\):

\(\int_0^{1/2} (-24x + 20) \, dx = \left[ -12x^2 + 20x \right]_0^{1/2} = -3 + 10 = 7\).

Shaded area = \(\frac{81}{8} - 7 = \frac{9}{8}\) or 1.125.

(i) To find the coordinates of \(C\), we first find the derivative of the curve \(y = \frac{8}{\sqrt{x}} - x\). The derivative is \(\frac{dy}{dx} = -4x^{-\frac{3}{2}} - 1\). Substituting \(x = 4\) into the derivative gives \(\frac{dy}{dx} = -\frac{3}{2}\). The equation of the tangent at \(B(4, 0)\) is \(y - 0 = -\frac{3}{2}(x - 4)\). Solving for \(y\) when \(x = 1\), we find \(C(1, \frac{4}{2})\).

(ii) To find the area of the shaded region, we calculate the area under the curve from \(x = 1\) to \(x = 4\). The integral is \(\int \left( \frac{8}{\sqrt{x}} - x \right) \, dx\), which evaluates to \(8x^{\frac{1}{2}} - \frac{1}{2}x^2\). Using limits 1 to 4, the area under the curve is \(8\frac{1}{2}\). The area under the tangent is \(\frac{1}{2} \times \frac{4}{2} \times 3 = 6\frac{3}{4}\). The shaded area is \(\frac{13}{4}\).

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

At \(B(0, 1)\), the gradient of the tangent is \(2\). Therefore, the gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

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

(ii) To find the area of the shaded region, we first find the x-coordinate of \(A\) by setting \(y = 0\) in \(y = \sqrt{1 + 4x}\), giving \(x = -\frac{1}{4}\).

The integral of \(\sqrt{1 + 4x}\) is \(\int \sqrt{1 + 4x} \, dx = \frac{(1 + 4x)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{4}\).

Evaluating from \(-\frac{1}{4}\) to \(0\), the area under the curve is \(\frac{1}{6}\).

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

Thus, the shaded area is \(1 + \frac{1}{6} = \frac{7}{6}\).

(i) Differentiate \(y = (x - 2)^4\) to find the gradient of the tangent: \(\frac{dy}{dx} = 4(x - 2)^3\). At \(x = 1\), the gradient is \(-4\). The equation of the tangent is \(y - 1 = -4(x - 1)\), which gives \(B \left( \frac{5}{4}, 0 \right)\). The gradient of the normal is \(\frac{1}{4}\), so the equation of the normal is \(y - 1 = \frac{1}{4}(x - 1)\), which gives \(C \left( 0, \frac{3}{4} \right)\).

(ii) Use the distance formula: \(AC^2 = 1^2 + \left( \frac{1}{4} \right)^2\), so \(AC = \frac{\sqrt{17}}{4}\).

(iii) Integrate \((x - 2)^4\) from 1 to 2: \(\int (x - 2)^4 \, dx = \frac{(x - 2)^5}{5}\). Evaluate from 1 to 2 to get \(\frac{1}{5}\). The area of the triangle formed by the tangent and normal is \(\frac{1}{2} \times 1 \times \frac{5}{4} = \frac{5}{8}\). The area of the shaded region is \(\frac{5}{8} - \frac{1}{5} = \frac{3}{40}\) or \(0.075\).

(i) The minimum point occurs when \(y = 0\). Solving \(x(x-2)^2 = 0\) gives \(x = 0\) or \(x = 2\). Thus, \(a = 2\).

(ii) To find \(b\), differentiate \(y = x(x-2)^2\) to get \(\frac{dy}{dx} = 3x^2 - 8x + 4\). Setting \(\frac{dy}{dx} = 0\) gives \((x-2)(3x-2) = 0\), so \(x = 2\) or \(x = \frac{2}{3}\). The maximum point is at \(x = \frac{2}{3}\), so \(b = \frac{2}{3}\).

(iii) The area of the shaded region is \(\int_0^2 x(x-2)^2 \, dx\). Expanding gives \(x^3 - 4x^2 + 4x\). Integrating, we get \(\left[ \frac{x^4}{4} - \frac{4x^3}{3} + 2x^2 \right]_0^2 = \frac{4}{3}\).

(iv) The second derivative is \(\frac{d^2y}{dx^2} = 6x - 8\). Setting \(\frac{d^2y}{dx^2} = 0\) gives \(x = \frac{4}{3}\). At this point, \(\frac{dy}{dx} = -\frac{4}{3}\), so \(m = -\frac{4}{3}\).

(i) To find the intersection points, equate \(y^2 = 2x - 1\) and \(3y = 2x - 1\).

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

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

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

\((2x - 1)^2 = 9(2x - 1)\)

\(4x^2 - 4x + 1 = 18x - 9\)

\(4x^2 - 22x + 10 = 0\)

\(2x^2 - 11x + 5 = 0\)

Solving this quadratic equation using the quadratic formula:

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

\(x = \frac{11 \pm \sqrt{121 - 40}}{4}\)

\(x = \frac{11 \pm \sqrt{81}}{4}\)

\(x = \frac{11 \pm 9}{4}\)

\(x = 5\) or \(x = \frac{1}{2}\)

Thus, \(a = 5\).

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

The area under the curve \(y^2 = 2x - 1\) is:

\(\int_{1/2}^{5} \sqrt{2x - 1} \, dx\)

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

\(\int \sqrt{u} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/2} \, du\)

\(= \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2}\)

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

Evaluate from \(x = \frac{1}{2}\) to \(x = 5\):

\(\left[ \frac{1}{3} (2x - 1)^{3/2} \right]_{1/2}^{5}\)

\(= \frac{1}{3} [(9)^{3/2} - (0)^{3/2}]\)

\(= \frac{1}{3} [27 - 0] = 9\)

The area under the line \(3y = 2x - 1\) is:

\(\int_{1/2}^{5} \frac{2x - 1}{3} \, dx\)

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

\(= \frac{1}{3} \left[ x^2 - x \right]_{1/2}^{5}\)

\(= \frac{1}{3} \left[ 25 - 5 - \left( \frac{1}{4} - \frac{1}{2} \right) \right]\)

\(= \frac{1}{3} \left[ 20 - \left( -\frac{1}{4} \right) \right]\)

\(= \frac{1}{3} \times 20.25 = \frac{27}{4}\)

The area of the shaded region is:

\(9 - \frac{27}{4} = \frac{9}{4}\)

(i) To find the coordinates of \(A\), we need to find the maximum point of the curve \(y = -x^2 + 8x - 10\). The derivative is \(\frac{dy}{dx} = -2x + 8\). Setting \(\frac{dy}{dx} = 0\) gives \(-2x + 8 = 0\), so \(x = 4\). Substituting \(x = 4\) into the equation of the curve gives \(y = -(4)^2 + 8(4) - 10 = 6\). Thus, \(A(4, 6)\).

The gradient of line \(BA\) is 2, so the equation of line \(AB\) is \(y - 6 = 2(x - 4)\). Simplifying gives \(y = 2x - 2\).

To find \(B\), solve the system of equations: \(y = -x^2 + 8x - 10\) and \(y = 2x - 2\). Equating gives \(-x^2 + 8x - 10 = 2x - 2\), which simplifies to \(x^2 - 6x + 8 = 0\). Solving this quadratic equation gives \(x = 2\) or \(x = 4\). For \(x = 2\), \(y = 2(2) - 2 = 2\). Thus, \(B(2, 2)\).

(ii) To find the area of the shaded region, evaluate \(\int_{2}^{4} (-x^2 + 8x - 10) \, dx\). The integral is \(\int (-x^2 + 8x - 10) \, dx = -\frac{x^3}{3} + 4x^2 - 10x\).

Evaluate from 2 to 4: \(\left[-\frac{(4)^3}{3} + 4(4)^2 - 10(4)\right] - \left[-\frac{(2)^3}{3} + 4(2)^2 - 10(2)\right]\).

Calculating gives \(\left[-\frac{64}{3} + 64 - 40\right] - \left[-\frac{8}{3} + 16 - 20\right] = \left[-\frac{64}{3} + 24\right] - \left[-\frac{8}{3} - 4\right]\).

This simplifies to \(\frac{28}{3} + \frac{20}{3} = \frac{48}{3} = 16\).

Thus, the area of the shaded region is \(9\frac{1}{3}\).

(i) To find the points of intersection, set the equations equal: \(9 - x^3 = \frac{8}{x^3}\). Multiply through by \(x^3\) to get \(9x^3 - x^6 = 8\), which simplifies to \(x^6 - 9x^3 + 8 = 0\). Factor this as \((X - 1)(X - 8) = 0\) where \(X = x^3\). Thus, \(x^3 = 1\) or \(x^3 = 8\), giving \(x = 1\) or \(x = 2\). Therefore, \(a = 1\) and \(b = 2\).

(ii) The area between the curves is given by the integral \(\int_{1}^{2} \left( 9 - x^3 - \frac{8}{x^3} \right) \, dx\). Integrate to find \(\left[ 9x - \frac{x^4}{4} + \frac{4}{x^2} \right]_{1}^{2}\). Evaluating this gives \(18 - 4 + 1 - (9 - \frac{1}{4} + 4) = 2\frac{1}{4}\).

(iii) The derivatives are \(\frac{dy}{dx} = -3x^2\) for \(y = 9 - x^3\) and \(\frac{dy}{dx} = -\frac{24}{x^4}\) for \(y = \frac{8}{x^3}\). Set \(-3c^2 = -\frac{24}{c^4}\) to find \(c\). Solving gives \(c^6 = 8\), so \(c = \sqrt[6]{8}\) or approximately \(1.41 \ldots\).

First, find the points of intersection by solving the equations simultaneously:

Set \(y = (x-2)^2\) equal to \(y = 7 - 2x\):

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

Expand and rearrange:

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

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

Factorize:

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

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

Find corresponding \(y\) values:

For \(x = 3\), \(y = (3-2)^2 = 1\), so \(B(3, 1)\).

For \(x = -1\), \(y = (-1-2)^2 = 9\), so \(A(-1, 9)\).

Calculate the area under the line from \(x = -1\) to \(x = 3\):

\(\int_{-1}^{3} (7 - 2x) \, dx = [7x - x^2]_{-1}^{3}\)

\(= (21 - 9) - (-7 - 1) = 12 + 8 = 20\)

Calculate the area under the curve from \(x = -1\) to \(x = 3\):

\(\int_{-1}^{3} (x-2)^2 \, dx = \int_{-1}^{3} (x^2 - 4x + 4) \, dx\)

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

\(= \left( \frac{27}{3} - 18 + 12 \right) - \left( \frac{-1}{3} - 2 - 4 \right)\)

\(= (9 - 18 + 12) - (-\frac{1}{3} + 2 + 4)\)

\(= 3 - \frac{5}{3} = \frac{4}{3}\)

Find the area of the shaded region:

\(20 - \frac{4}{3} = \frac{60}{3} - \frac{4}{3} = \frac{56}{3} = 10\frac{2}{3}\)

The curve \(y = 6x - x^2\) intersects the line \(y = 5\) when:

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

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

Factoring gives:

\((x - 1)(x - 5) = 0\)

Thus, \(x = 1\) and \(x = 5\).

The area under the curve from \(x = 1\) to \(x = 5\) is found by integrating:

\(\int_{1}^{5} (6x - x^2) \, dx = \left[ 3x^2 - \frac{1}{3}x^3 \right]_{1}^{5}\)

Calculating the definite integral:

\(\left( 3(5)^2 - \frac{1}{3}(5)^3 \right) - \left( 3(1)^2 - \frac{1}{3}(1)^3 \right)\)

\(= (75 - \frac{125}{3}) - (3 - \frac{1}{3})\)

\(= \frac{225}{3} - \frac{125}{3} - \left( \frac{9}{3} - \frac{1}{3} \right)\)

\(= \frac{100}{3} - \frac{8}{3}\)

\(= \frac{92}{3}\)

The area of the rectangle formed by the line \(y = 5\) from \(x = 1\) to \(x = 5\) is:

\(5 \times (5 - 1) = 20\)

The shaded area is the area under the curve minus the area of the rectangle:

\(\frac{92}{3} - 20 = \frac{92}{3} - \frac{60}{3} = \frac{32}{3}\)

Thus, the shaded area is \(\frac{10}{3}\).

To find the area of the shaded region, we need to integrate the function \(y = 10x^{\frac{1}{2}} - \frac{5}{2}x^{\frac{3}{2}}\) from \(x = 0\) to \(x = 4\).

The integral is:

\(\int_{0}^{4} \left( 10x^{\frac{1}{2}} - \frac{5}{2}x^{\frac{3}{2}} \right) \, dx\)

Integrating term by term, we have:

\(\int 10x^{\frac{1}{2}} \, dx = \frac{10}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{20}{3} x^{\frac{3}{2}}\)

\(\int -\frac{5}{2}x^{\frac{3}{2}} \, dx = -\frac{5}{2} \cdot \frac{2}{5} x^{\frac{5}{2}} = -x^{\frac{5}{2}}\)

Thus, the integral becomes:

\(\left[ \frac{20}{3} x^{\frac{3}{2}} - x^{\frac{5}{2}} \right]_{0}^{4}\)

Evaluating at the bounds:

\(\left( \frac{20}{3} \times 8 - 32 \right) - \left( 0 \right) = \frac{160}{3} - 32\)

\(= \frac{160}{3} - \frac{96}{3} = \frac{64}{3}\)

Therefore, the area of the shaded region is \(\frac{64}{3}\) or approximately \(21.3\).

(i) To find the stationary point, we first find the derivative of the curve: \(\frac{dy}{dx} = 4x^3 + 4\). Setting this to zero gives \(4x^3 + 4 = 0\), which simplifies to \(x^3 = -1\), so \(x = -1\). Substituting \(x = -1\) back into the original equation gives \(y = (-1)^4 + 4(-1) + 9 = 1 - 4 + 9 = 6\). Thus, the stationary point is \((-1, 6)\).

To determine the nature, we find the second derivative: \(\frac{d^2y}{dx^2} = 12x^2\). Substituting \(x = -1\) gives \(12(-1)^2 = 12\), which is positive, indicating a minimum point.

(ii) To find the area, we integrate the curve from \(x = 0\) to \(x = 1\). The integral of \(y = x^4 + 4x + 9\) is \(\int (x^4 + 4x + 9) \, dx = \frac{x^5}{5} + 2x^2 + 9x\). Evaluating from 0 to 1 gives:

\(\left[ \frac{x^5}{5} + 2x^2 + 9x \right]_0^1 = \left( \frac{1^5}{5} + 2(1)^2 + 9(1) \right) - \left( \frac{0^5}{5} + 2(0)^2 + 9(0) \right) = \frac{1}{5} + 2 + 9 = 11.2\).

(i) To find the coordinates of \(A\) and \(B\), we first find the derivative \(\frac{dy}{dx} = 3x^2 - 12x + 9\). Setting \(\frac{dy}{dx} = 0\) gives the critical points. Solving \(3x^2 - 12x + 9 = 0\) gives \(x = 1\) and \(x = 3\). Substituting back into the original equation, \(y = x^3 - 6x^2 + 9x\), we find \(A(1, 4)\) and \(B(3, 0)\).

(ii) At \(C(2, 2)\), the slope of the tangent is \(m = -3\). The slope of the normal is \(\frac{1}{3}\). The equation of the normal is \(y - 2 = \frac{1}{3}(x - 2)\), which simplifies to \(3y = x + 4\).

(iii) To find the area of the shaded region, integrate \(y = x^3 - 6x^2 + 9x\) from \(x = 0\) to \(x = 3\). The integral is \(\frac{1}{4}x^4 - 2x^3 + \frac{9}{2}x^2\). Evaluating from 0 to 3 gives \(\frac{3}{4}\). The area of the trapezium is \(\frac{1}{2} \times 1 \times (2 + \frac{2}{3}) = \frac{13}{6}\). Subtracting gives the shaded area \(\frac{17}{12}\).

(i) Integrate \(\frac{dy}{dx} = -\frac{k}{x^3}\) to get \(y = -k \frac{x^{-2}}{-2} + c = \frac{k}{2x^2} + c\).

Substitute \((1, 18)\):

\(18 = \frac{k}{2} + c\)

Substitute \((4, 3)\):

\(3 = \frac{k}{32} + c\)

Solving these equations gives \(k = 32\) and \(c = 2\).

Thus, \(y = \frac{16}{x^2} + 2\).

(ii) The area of the shaded region is given by:

\(\int_{1}^{1.6} \left( \frac{16}{x^2} + 2 \right) \, dx\)

Calculate:

\(\left[ -\frac{16}{x} + 2x \right]_{1}^{1.6} = \left[ -10 + 3.2 \right] - \left[ -16 + 2 \right] = 7.2\).

The area under the curve \(y = 2\sqrt{x}\) from \(x = 1\) to \(x = 4\) is given by the definite integral:

\(\int_{1}^{4} 2\sqrt{x} \, dx\)

First, rewrite \(2\sqrt{x}\) as \(2x^{0.5}\).

The integral of \(2x^{0.5}\) is:

\(\frac{2}{1.5} x^{1.5} = \frac{4}{3} x^{1.5}\)

Evaluate this from \(x = 1\) to \(x = 4\):

\(\left[ \frac{4}{3} x^{1.5} \right]_{1}^{4} = \frac{4}{3} (4^{1.5}) - \frac{4}{3} (1^{1.5})\)

Calculate \(4^{1.5} = 8\) and \(1^{1.5} = 1\):

\(\frac{4}{3} (8) - \frac{4}{3} (1) = \frac{32}{3} - \frac{4}{3} = \frac{28}{3}\)

The area under the curve from \(x = 1\) to \(x = 2\) is given by the integral:

\(\int_{1}^{2} \left( 2x + \frac{8}{x^2} \right) \, dx\)

We can split this into two separate integrals:

\(\int_{1}^{2} 2x \, dx + \int_{1}^{2} \frac{8}{x^2} \, dx\)

The first integral is:

\(\int 2x \, dx = x^2\)

Evaluated from 1 to 2:

\([x^2]_{1}^{2} = 2^2 - 1^2 = 4 - 1 = 3\)

The second integral is:

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

Evaluated from 1 to 2:

\(\left[-\frac{8}{x}\right]_{1}^{2} = -\frac{8}{2} + \frac{8}{1} = -4 + 8 = 4\)

Adding both results gives the total area:

\(3 + 4 = 7\)

(i) The curve is given by \(y = x(x-1)(x-2) = x^3 - 3x^2 + 2x\).

Differentiate to find the gradient: \(\frac{dy}{dx} = 3x^2 - 6x + 2\).

At \(A(1, 0)\), the gradient \(m = -1\), so the tangent is \(y = -1(x-1)\).

At \(B(2, 0)\), the gradient \(m = 2\), so the tangent is \(y = 2(x-2)\).

Solving the equations \(y = -x + 1\) and \(y = 2x - 4\) gives \(x = \frac{5}{3}\).

(ii) The area \(R_1\) is given by \(\int_0^1 (x^3 - 3x^2 + 2x) \, dx\).

Integrate: \(\left[ \frac{x^4}{4} - \frac{3x^3}{3} + \frac{2x^2}{2} \right]_0^1 = \frac{1}{4}\).

The area \(R_2\) is given by \(\int_1^2 (x^3 - 3x^2 + 2x) \, dx\).

Integrate: \(\left[ \frac{x^4}{4} - \frac{3x^3}{3} + \frac{2x^2}{2} \right]_1^2 = -\frac{1}{4}\).

Thus, \(R_1 = R_2 = \frac{1}{4}\).

(i) Differentiate the curve: \(\frac{dy}{dx} = 3x^2 - 6x - 9\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 6x - 9 = 0\).

Solving gives \(x = 3\) or \(x = -1\).

Since the minimum point is on the \(x\)-axis, \(y = 0\) at \(x = 3\).

Substitute \(x = 3\) into the curve equation: \(0 = 3^3 - 3(3)^2 - 9(3) + k\).

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

Thus, \(k = 27\).

(iv) Integrate \(y\) to find the area: \(\int (x^3 - 3x^2 - 9x + 27) \, dx\).

The integral is \(\frac{x^4}{4} - x^3 - \frac{9x^2}{2} + 27x\).

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[ \frac{3^4}{4} - 3^3 - \frac{9(3)^2}{2} + 27(3) \right] - \left[ \frac{0^4}{4} - 0^3 - \frac{9(0)^2}{2} + 27(0) \right]\).

Calculate: \(\frac{81}{4} - 27 - 40.5 + 81 = 33.75\).

Thus, the area of the shaded region is 33.75.

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{16}{x^3}\):

\(y = \int \frac{16}{x^3} \, dx = \int 16x^{-3} \, dx = \frac{16x^{-2}}{-2} + c = \frac{-8}{x^2} + c\).

Using the point \((1, 4)\), substitute to find \(c\):

\(4 = \frac{-8}{1^2} + c \Rightarrow c = 12\).

Thus, the equation of the curve is \(y = \frac{-8}{x^2} + 12\).

(ii) The normal has gradient \(-\frac{1}{2}\), so the perpendicular gradient is 2. Set \(\frac{16}{x^3} = 2\) to find \(x\):

\(16 = 2x^3 \Rightarrow x = 2\).

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

\(y = \frac{-8}{2^2} + 12 = 10\).

The equation of the normal is \(y - 10 = -\frac{1}{2}(x - 2)\).

Simplify to \(2y + x = 22\).

(iii) Find the area under the curve from \(x = 1\) to \(x = 2\):

\(A = \int_{1}^{2} \left( \frac{-8}{x^2} + 12 \right) \, dx\).

\(= \left[ \frac{8}{x} + 12x \right]_{1}^{2}\).

\(= \left( \frac{8}{2} + 12 \times 2 \right) - \left( \frac{8}{1} + 12 \times 1 \right)\).

\(= (4 + 24) - (8 + 12) = 28 - 20 = 8\).

To find the area under the curve \(y = \frac{4}{\sqrt{x}}\) from \(x = 1\) to \(x = 4\), we need to evaluate the definite integral:

\(\int_{1}^{4} 4x^{-0.5} \, dx\)

First, find the antiderivative of \(4x^{-0.5}\):

\(\int 4x^{-0.5} \, dx = \frac{4x^{0.5}}{0.5} = 8x^{0.5}\)

Now, evaluate this antiderivative from 1 to 4:

\(\left[ 8x^{0.5} \right]_{1}^{4} = 8 \sqrt{4} - 8 \sqrt{1}\)

\(= 8 \times 2 - 8 \times 1\)

\(= 16 - 8 = 8\)

Thus, the area of the region is 8.

To find the area enclosed by the curve \(y = \sqrt{5x + 4}\), the \(x\)-axis, the \(y\)-axis, and the line \(x = 1\), we need to integrate the function from \(x = 0\) to \(x = 1\).

The integral to evaluate is:

\(\int_0^1 \sqrt{5x + 4} \, dx\)

First, rewrite \(\sqrt{5x + 4}\) as \((5x + 4)^{1/2}\).

Using the power rule for integration, the integral becomes:

\(\int (5x + 4)^{1/2} \, dx = \frac{(5x + 4)^{3/2}}{\frac{3}{2} \cdot 5} = \frac{2}{15} (5x + 4)^{3/2}\)

Evaluate this from \(x = 0\) to \(x = 1\):

\(\left[ \frac{2}{15} (5x + 4)^{3/2} \right]_0^1 = \frac{2}{15} ((5 \cdot 1 + 4)^{3/2} - (5 \cdot 0 + 4)^{3/2})\)

\(= \frac{2}{15} (9^{3/2} - 4^{3/2})\)

\(9^{3/2} = 27\) and \(4^{3/2} = 8\), so:

\(\frac{2}{15} (27 - 8) = \frac{2}{15} \cdot 19 = \frac{38}{15} \approx 2.53\)

To find the area of the shaded region, we need to integrate the difference between the two curves from \(x = 0\) to \(x = 4\).

The area \(A\) is given by:

\(A = \int_0^4 \left( 18 - \frac{3}{8}x^{\frac{5}{2}} - \left( \frac{9}{4}x^2 - 12x + 18 \right) \right) \, dx\)

This simplifies to:

\(A = \int_0^4 \left( 18 - \frac{3}{8}x^{\frac{5}{2}} - \frac{9}{4}x^2 + 12x - 18 \right) \, dx\)

\(A = \int_0^4 \left( -\frac{3}{8}x^{\frac{5}{2}} - \frac{9}{4}x^2 + 12x \right) \, dx\)

Integrating each term separately:

\(\int -\frac{3}{8}x^{\frac{5}{2}} \, dx = -\frac{3}{28}x^{\frac{7}{2}}\)

\(\int -\frac{9}{4}x^2 \, dx = -\frac{9}{12}x^3 = -\frac{3}{4}x^3\)

\(\int 12x \, dx = 6x^2\)

Evaluating from 0 to 4:

\(A = \left[ -\frac{3}{28}x^{\frac{7}{2}} - \frac{3}{4}x^3 + 6x^2 \right]_0^4\)

Substitute \(x = 4\):

\(A = -\frac{3}{28}(4)^{\frac{7}{2}} - \frac{3}{4}(4)^3 + 6(4)^2\)

\(A = -\frac{3}{28} \times 128 - \frac{3}{4} \times 64 + 6 \times 16\)

\(A = -\frac{384}{28} - 48 + 96\)

\(A = -13.7142857 + 48\)

\(A = \frac{240}{7} \approx 34.3\)

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

At \(x = 4\), \(\frac{dy}{dx} = \frac{1}{2}\). The slope of the normal is the negative reciprocal, so \(m = -2\).

The equation of the normal line at point B (4, 4) is \(y - 4 = -2(x - 4)\), which simplifies to \(y = -2x + 12\).

(ii) To find the area of the shaded region, we integrate the curve from \(x = 1\) to \(x = 4\). The integral is \(\int 2\sqrt{x} \, dx = \int 2x^{0.5} \, dx = \frac{2}{1.5}x^{1.5}\).

Evaluating from 1 to 4 gives \(\left[ \frac{2}{1.5}x^{1.5} \right]_1^4 = \frac{2}{1.5}(4^{1.5} - 1^{1.5}) = \frac{2}{1.5}(8 - 1) = \frac{2}{1.5} \times 7 = \frac{28}{3}\).

(i) To find the coordinates of \(P\), set \(y = 3\sqrt{x} = x\). Solving for \(x\), we have:

\(3\sqrt{x} = x\)

\(3 = \sqrt{x}\)

\(x = 9\)

Thus, \(y = x = 9\), so \(P\) is (9, 9).

(ii) To find the area of the shaded region, calculate the area under the curve \(y = 3\sqrt{x}\) from \(x = 0\) to \(x = 9\), and subtract the area under the line \(y = x\) over the same interval.

The area under the curve \(y = 3\sqrt{x}\) is given by:

\(\int_0^9 3\sqrt{x} \, dx = \int_0^9 3x^{1/2} \, dx\)

\(= \left[ 3 \cdot \frac{2}{3} x^{3/2} \right]_0^9\)

\(= \left[ 2x^{3/2} \right]_0^9\)

\(= 2(9^{3/2}) - 2(0^{3/2})\)

\(= 2(27)\)

\(= 54\)

The area under the line \(y = x\) is the area of a right triangle with base and height of 9:

\(\frac{1}{2} \times 9 \times 9 = 40.5\)

Thus, the area of the shaded region is:

\(54 - 40.5 = 13.5\)

(a) To find the area between the curves, integrate the difference of the functions from \(x = 0\) to \(x = 4\):

\(\int (2x^{\frac{1}{2}} + 1) - \left(\frac{1}{2}x^2 - x + 1\right) \, dx\)

Calculate the integrals:

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

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

Evaluate from 0 to 4:

\(\left(\frac{32}{3} + 32 - 0\right) - \left(\frac{44}{3} - 20 + 0\right) = 8\)

(b) Differentiate both curves to find the gradients at \(x = 4\):

Upper curve: \(\frac{dy}{dx} = x^{-\frac{1}{2}}\), at \(x = 4\), gradient = \(\frac{1}{2}\).

Lower curve: \(\frac{dy}{dx} = x - 1\), at \(x = 4\), gradient = 3.

Find \(\alpha\) using the inverse tangent:

\(\alpha = \arctan(3) - \arctan\left(\frac{1}{2}\right)\)

\(\alpha = 45^\circ\)

(a) To find the equation of the tangent, first differentiate the curve equation \(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\).

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

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

The equation of the tangent at \(A(1, 5)\) is \(y - 5 = -\frac{3}{2}(x - 1)\).

Simplifying gives \(y = -\frac{3}{2}x + \frac{13}{2}\).

(b) To find the area of the shaded region, integrate the curve equation from \(x = 1\) to \(x = 16\).

The integral of \(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\) is \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 8x^{\frac{1}{2}}\).

Evaluating from 1 to 16 gives \(\left[ \frac{128}{3} + 32 \right] - \left[ \frac{2}{3} + 8 \right] = 66\).

The area under the line \(y = 5\) from \(x = 1\) to \(x = 16\) is \(15 \times 5 = 75\).

The required area is \(75 - 66 = 9\).

First, find the points of intersection by equating the line and the curve:

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

Rearrange to form a quadratic equation:

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

Let \(u = x^{\frac{1}{2}}\), then \(u^2 = x\). Substitute to get:

\(2u^2 - 5u + 2 = 0\)

Factor the quadratic equation:

\((2u - 1)(u - 2) = 0\)

Thus, \(u = \frac{1}{2}\) or \(u = 2\)

So, \(x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) or \(x = 2^2 = 4\)

Now, calculate the area between the curves from \(x = \frac{1}{4}\) to \(x = 4\):

Area = \(\int_{\frac{1}{4}}^{4} (5x^{\frac{1}{2}} - (2x + 2)) \, dx\)

Integrate each term separately:

\(\int 5x^{\frac{1}{2}} \, dx = \frac{10}{3}x^{\frac{3}{2}}\)

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

Evaluate from \(x = \frac{1}{4}\) to \(x = 4\):

\(\left[ \frac{10}{3}x^{\frac{3}{2}} - (x^2 + 2x) \right]_{\frac{1}{4}}^{4}\)

Calculate:

At \(x = 4\): \(\frac{10}{3}(4^{\frac{3}{2}}) - (4^2 + 2 \times 4) = \frac{80}{3} - 24\)

At \(x = \frac{1}{4}\): \(\frac{10}{3}\left(\left(\frac{1}{4}\right)^{\frac{3}{2}}\right) - \left(\left(\frac{1}{4}\right)^2 + 2\left(\frac{1}{4}\right)\right) = \frac{10}{3} \times \frac{1}{8} - \left(\frac{1}{16} + \frac{1}{2}\right)\)

Combine results:

\(\frac{80}{3} - 24 - \left(\frac{10}{24} - \frac{9}{16}\right)\)

Simplify to get the exact area:

\(\frac{45}{16}\)