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Q.

Thedomainoff(x)=1|sinx|+sinx  is  

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a

R

b

UnZ((2n+1)π,2(n+1)π)

c

ϕ

d

UnZ(2nπ,(2n+1)π)

answer is C.

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Detailed Solution

f(x)=1|sinx|+sinxisdefinediff|sinx|+sinx0|sinx|+sinx={2sinxif2nπ<x<(2n+1)π0Otherwise  Domain  =nZ(2nπ,  (2n+1)π)

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The domain of f(x)=1|sinx|+sinx  is