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Question

If esinxesinx=a has alteast one real solution, then


A
|a|(e21e,)
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B
|a|[0,e21e]
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C
|a|(e,)
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D
|a|[0,e]
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Solution

The correct option is B |a|[0,e21e]

esinxesinx=a..............(1)
Let esinx=tt[1e,e]
(1)t1ta=t2at1=0
Discriminant =a2+4>0 a R.
Also product of roots =-1
i.e. both the roots can't belong to [1e,e]
Let it have only one root in [1e,e], then
(e2ae1)(1e2ae1)0=[(e21)ae][(e21)+ae]0|a|e21e


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