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(A) Given function is $f(x) = 2x^3 - 9x^2 + 12x + 5$.To find local maxima and minima,we find the derivative $f'(x)$:$f'(x) = 6x^2 - 18x + 12$.Setting

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(A) Given function is $f(x) = 2x^3 - 9x^2 + 12x + 5$.To find local maxima and minima,we find the derivative $f'(x)$:$f'(x) = 6x^2 - 18x + 12$.Setting $f'(x) = 0$ for critical points:$6(x^2 - 3x + 2) = 0 \Rightarrow 6(x - 1)(x - 2) = 0$.So,the critical points are $x = 1$ and $x = 2$.Now,we check the second derivative $f''(x) = 12x - 18$.At $x = 1$,$f''(1) = 12(1) - 18 = -6 $M = f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10$.At $x = 2$,$f''(2) = 12(2) - 18 = 6 > 0$,so $x = 2$ is a local minimum.$m = f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9$.Therefore,$M - m = 10 - 9 = 1$.

Let $M$ and $m$ be respectively the local maximum and the local minimum values of the function $f(x) = 2x^3 - 9x^2 + 12x + 5$ in the interval $[0, 3]$. Then $M - m$ is equal to

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