Detailed Solution:

a(i) Finding the zero of f(x):

To find the zero, set \( f(x) = 0 \): \[ x + \frac{12}{x^2} = 0 \] Multiply through by \( x^2 \) (since \( x \neq 0 \)): \[ x^3 + 12 = 0 \] \[ x^3 = -12 \] \[ x = \sqrt[3]{-12} \approx -2.289 \] The function has one real zero at approximately \( x = -2.29 \)

a(ii) Finding the local minimum:

First, find the derivative: \[ f'(x) = \frac{d}{dx}\left(x + 12x^{-2}\right) = 1 – \frac{24}{x^3} \] Set \( f'(x) = 0 \) for critical points: \[ 1 – \frac{24}{x^3} = 0 \] \[ x^3 = 24 \] \[ x = \sqrt[3]{24} \approx 2.884 \] Verify it’s a minimum using second derivative test: \[ f”(x) = \frac{72}{x^4} > 0 \text{ for all } x \neq 0 \] Find y-coordinate: \[ f(2.884) \approx 2.884 + \frac{12}{(2.884)^2} \approx 4.326 \] Local minimum at approximately \( (2.88, 4.33) \)

b) Solving \( f(x) = g(x) \):

Set the functions equal: \[ x + \frac{12}{x^2} = 3 – x \] \[ 2x + \frac{12}{x^2} – 3 = 0 \] Multiply through by \( x^2 \): \[ 2x^3 – 3x^2 + 12 = 0 \] Using numerical methods (graphical or Newton-Raphson): The only real solution is approximately \( x = -1.433 \) Solution: \( x \approx -1.43 \)

Markscheme:

a.

(i) \( x = -2.29 \) (accept \( -\sqrt[3]{12} \) or -2.28942…) A1

(ii) \( (2.88, 4.33) \) (accept \( (\sqrt[3]{24}, \sqrt[3]{24}+\frac{12}{24^{2/3}}) \) or (2.88449…, 4.32647…)) A1A1

Note: Coordinates must be exact or to at least 2 decimal places.

[4 marks]

b.

\( x = -1.43 \) (accept exact solution or -1.4328…) A1

Note: Must show understanding of solving the equation.

[2 marks]

Total: [6 marks]