    Question

# If [cosx]+[sinx+1]=0, then value of x satisfying f(x) where x∈[0,2π], and [] denotes greatest integer function,

A
(π2,π][3π2,2π]
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B
(0,π2)
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C
[π2,π][5π2,2π]
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D
[π,2π]
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Solution

## The correct option is A (π2,π]∪[3π2,2π][cosx]+[sinx+1]=0The range of cosx is [−1,1] and thus the range of [cosx] becomes −1,0,1Similarly, since the range of 1+sinx is [0,2]; the range of [1+sinx] becomes 0,1,2Now, since the R.H.S. is 0, one of the terms on L.H.S will be 1 and the other -1 and the other possibility says both terms to be 0!a) If [cosx]=0,cosx∈[0,1); also, [1+sinx]=0,1+sinx∈[0,1)⇒sinx∈[−1,0)This is possible in the fourth quadrant.b) If [cosx]=−1,cosx∈[−1,0) and also, [1+sinx]=1 so 1+sinx∈[1,2) i.e. sinx∈[0,1)This is possible in the second quadrant.Hence option A is correct!  Suggest Corrections  0      Similar questions
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