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Question

The inequality 2cosx1+sin2x1sin2x2 where x[0,2π] holds true in the interval [aπ2,bπ2]. Then the value of a+b is

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Solution

1+sin2x1sin2x=|cosx+sinx||cosxsinx|=2min{|cosx|,|sinx|}

Now,
2cosx2min{|cosx|,|sinx|}2cosxmin{|cosx|,|sinx|}12

Clearly, from the graph,
x[π4,7π4]a+b=12+72=4

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