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Vaibhav Mathur Grade: 12
        

If logcos x sin x>=2 and 0<=x<=3π then sin x is in the interval


(a)[(51/2-1)/2,1]


(b)(0,51/2-1)/2]


(c)[0,1/2]


(d)none

7 years ago

Answers : (1)

sahil doshi
8 Points
										

logcos xsin x >=2


sin x >= cos2x


sin x >= 1 - sin2x


sin2x + sin x - 1 >= 0


[sin x - (51/2-1) / 2] * [sin x - (-51/2-1) / 2] >= 0


now, sin x > -1


therefore [sin x - (-51/2-1)/2] > 0       { 51/2 ≈ 2        therefore    -(-51/2-1)/2 ≈ 1.5}


therefore [sin x - (51/2-1) / 2] should be > 0


sin x > (51/2-1) / 2]


and sin x lies between [-1,1]


(a)[(51/2-1)/2,1]

7 years ago
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