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Question

The maximum value of f(x)=2x39x2+12x3 in the interval 0x3 is
  1. 6

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Solution

The correct option is A 6
f(x)=2x39x2+12x3,xϵ[0,3]
f(x)=6x218x+12=0
x=1,2ϵ(0,3)
x 0 1 2 3
f(x) 3 2 1 6

Maximum value of f(x)=6.

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