For the function $f(x) = 2x^3-3x^2 - 12x+5$, the difference of maximum and minimum value of f(x) is

27

The correct answer is Option (3) → 27

$f(x) = 2x^3 - 3x^2 - 12x + 5$

$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$

$\text{Critical points: } x = -1,\ x = 2$

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$

$f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$

$\text{Maximum value} = 12,\quad \text{Minimum value} = -15$

$\text{Difference} = 12 - (-15) = 27$