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Question

Let a, b and c be positive real numbers such that a+b+c=6 Then range of ab2c3 is

A
(0,∞)
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B
(0, 1)
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C
(0, 108]
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D
(6, 108]
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Solution

The correct option is C (0, 108]

a+b2.2+c36(a(b2)2(c3)3)1/3

1(ab2c32233)1/3

ab2c3108

(0,108]


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