The minimum value of the expression 9x2sin2x+4xsinx for xε(0,π) is
A
163
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B
12
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C
6
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D
83
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Solution
The correct option is D12 We know that, AP≥GP⇒3xsinx+22≥√(3xsinx)2 Squaring both sides 9x2sin2x+4+12xsinx4≥6xsinx⇒9x2sin2x+4+12xsinx≥24xsinx⇒9x2sin2x+4≥12xsinx⇒9x2sin2x+4xsinx≥12