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