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Question

If 1|sinx|1+|sinx|23, then sin x lies in

A
(,12][12,)
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B
(12,12)
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C
[12,12]
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D
none of these
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Solution

The correct option is C [12,12]
|sinx|0|sinx||1+|sinx|<1
1|sinx||1+|sinx|>0
So the given equation becomes
1|sinx||1+|sinx|23
13|sinx||1+|sinx||
|1+|sinx||3|sinx|
|sinx|12
12sinx12sinx[12,12]

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