1. The problem states that the function \( f(X) = aX^3 - 2X^2 + 4 \) has an inflection point at \( x = \frac{1}{3} \). We need to find the value of \( \frac{f(1)}{f'(1)} \). 2. Start by finding the first derivative \( f'(X) \) and the second derivative \( f''(X) \): \[ f'(X) = 3aX^2 - 4X \] \[ f''(X) = 6aX - 4 \] 3. Since \( x = \frac{1}{3} \) is an inflection point, the second derivative equals zero at this point: \[ f''\left(\frac{1}{3}\right) = 6a \times \frac{1}{3} - 4 = 0 \] Simplify: \[ 2a - 4 = 0 \] \[ 2a = 4 \] \[ a = 2 \] 4. Substitute \( a = 2 \) back into the original function and first derivative: \[ f(X) = 2X^3 - 2X^2 + 4 \] \[ f'(X) = 3 \times 2 X^2 - 4X = 6X^2 - 4X \] 5. Calculate \( f(1) \) and \( f'(1) \): \[ f(1) = 2(1)^3 - 2(1)^2 + 4 = 2 - 2 + 4 = 4 \] \[ f'(1) = 6(1)^2 - 4(1) = 6 - 4 = 2 \] 6. Find \( \frac{f(1)}{f'(1)} \): \[ \frac{f(1)}{f'(1)} = \frac{4}{2} = 2 \] Answer: 2, which corresponds to option (d).