(a) Second Derivative:

For \( y = x^4 – 3x^3 + 3x \):

\[ \frac{dy}{dx} = 4x^3 – 9x^2 + 3 \quad (M1) \]

\[ \frac{d^2y}{dx^2} = 12x^2 – 18x = 6x(2x – 3) \quad (A1) \]

[2 marks]

(b) Coordinates of B and C:

Set \( \frac{d^2y}{dx^2} = 6x(2x – 3) = 0 \):

\[ x = 0 \quad \text{or} \quad x = \frac{3}{2} \quad (M1) \]

Verify concavity change:

At \( x = -1 \): \( \frac{d^2y}{dx^2} = 6(-1)(2(-1) – 3) = 30 > 0 \).

At \( x = 1 \): \( \frac{d^2y}{dx^2} = 6(1)(2 – 3) = -6 < 0 \).

At \( x = 2 \): \( \frac{d^2y}{dx^2} = 6(2)(4 – 3) = 12 > 0 \quad (A1) \).

Coordinates:

At \( x = 0 \): \( y = 0 \), so B: \( (0, 0) \).

At \( x = \frac{3}{2} \): \( y = \left( \frac{3}{2} \right)^4 – 3 \left( \frac{3}{2} \right)^3 + 3 \left( \frac{3}{2} \right) = \frac{81}{16} – \frac{162}{16} + \frac{72}{16} = -\frac{9}{16} \), so C: \( \left( \frac{3}{2}, -\frac{9}{16} \right) \quad (A1) \).

[3 marks]

(c) Line Through B and C:

Slope: \( m = \frac{-\frac{9}{16}}{\frac{3}{2}} = -\frac{9}{16} \cdot \frac{2}{3} = -\frac{3}{8} = -0.375 \quad (M1) \)

Line through \( (0, 0) \): \( y = -0.375x \quad (A1) \).

[2 marks]

(d) \( x \)-Coordinate of D:

Substitute \( y = -0.375x \) into \( x^4 – 3x^3 + 3x \):

\[ x^4 – 3x^3 + 3.375x = 0 \quad (M1) \]

\[ x (x^3 – 3x^2 + \frac{27}{8}) = 0 \]

Roots: \( x = 0 \), \( x^3 – 3x^2 + \frac{27}{8} = 0 \). Test \( x = \frac{3}{2} \): \( \frac{27}{8} – \frac{27}{4} + \frac{27}{8} = 0 \quad (A1) \).

Factor: \( (x – \frac{3}{2}) (4x^2 – 6x – 9) = 0 \).

Solve: \( x = \frac{6 \pm \sqrt{36 + 144}}{8} = \frac{3 \pm 3\sqrt{5}}{4} \).

\[ x_D = \frac{3 + 3\sqrt{5}}{4} \approx 2.427 \quad (A1) \]

[3 marks]

(e) Inflection Points in Terms of \( m \):

For \( y = x^4 – mx^3 + nx \):

\[ \frac{d^2y}{dx^2} = 12x^2 – 6mx = 6x(2x – m) = 0 \quad (M1) \]

\[ x = 0, \frac{m}{2} \quad (A1) \]

[2 marks]

(f) Coordinates of B and C:

(i) At \( x = 0 \): \( y = 0 \), so B: \( (0, 0) \quad (A1) \).

[1 mark]

(ii) At \( x = \frac{m}{2} \):

\[ y = \frac{m^4}{16} – m \cdot \frac{m^3}{8} + n \cdot \frac{m}{2} = \frac{-m^4 + 8nm}{16} \quad (M1) \]

\[ C = \left( \frac{m}{2}, \frac{-m^4 + 8nm}{16} \right) \quad (A1) \]

[2 marks]

(g) Line Through B and C:

Slope: \( \frac{\frac{-m^4 + 8nm}{16}}{\frac{m}{2}} = \frac{-m^4 + 8nm}{8m} = -\frac{m^3}{8} + n \quad (M1) \)

Line: \( y = \left( -\frac{m^3}{8} + n \right)x \quad (A1) \)

[2 marks]

(h) Find \( x_A \):

Substitute \( y = \left( -\frac{m^3}{8} + n \right)x \) into \( x^4 – mx^3 + nx \):

\[ x^4 – mx^3 + \frac{m^3}{8} x = 0 \quad (M1) \]

\[ x (x^3 – mx^2 + \frac{m^3}{8}) = 0 \]

Roots: \( x = 0 \), \( x^3 – mx^2 + \frac{m^3}{8} = 0 \). Test \( x = \frac{m}{2} \): satisfies (C) (A1).

Factor: \( (x – \frac{m}{2}) (x^2 – \frac{m}{2} x – \frac{m^2}{4}) = 0 \).

Solve: \( x = \frac{\frac{m}{2} \pm \sqrt{\frac{m^2}{4} + m^2}}{2} = \frac{m}{4} (1 \pm \sqrt{5}) \).

\[ x_A = \frac{m}{4} – \frac{m}{4} \sqrt{5} \quad (A1) \]

[3 marks]

(i) Ratio of Lengths:

\[ x_A = \frac{m}{4} – \frac{m}{4} \sqrt{5}, \quad x_B = 0, \quad x_C = \frac{m}{2} \]

\[ x_B – x_A = \frac{m}{4} (\sqrt{5} – 1) \quad (M1) \]

\[ x_C – x_B = \frac{m}{2} \]

\[ \frac{x_B – x_A}{x_C – x_B} = \frac{\frac{m}{4} (\sqrt{5} – 1)}{\frac{m}{2}} = \frac{\sqrt{5} – 1}{2} \quad (A1) \]

[2 marks]