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