Answer :

f(x)=x3-18x2+96x


f(x)=3x2-36x+96


f’(x)=0


3x2-36x+96=0


x2-12x+32=0


x2-8x-4x+32=0


x(x-8)-4(x-8)=0


Now we have f(0)=0,f(4)=160,f(9)=135


Hence in the given interval least value is f(0)=0


Option(D)

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