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```				   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

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```										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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