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Question

The maximum value of the function f(x)=3x318x2+27x40 on the set S={xR:x2+3011x} is :

A
122
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B
122
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C
222
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D
222
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Solution

The correct option is A 122
S={xR:x2+3011x}
x2+3011x
x211x+300
x26x5x+300
(x5)(x6)0
x[5,6]

Now, f(x)=3x318x2+27x40
f(x)=9x236x+27
f(x)=9(x24x+3)
f(5)=72>0 (1)

f′′(x)=9(2x4)
f′′(x)>0 x[5,6]
f(x) is strictly increasing in the interval [5,6] (2)

From (1) and (2), we conclude
f(x)>0 x[5,6]
f(x) is strictly increasing in the interval [5,6]
Maximum value of f(x) when x[5,6] is f(6)=122

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