If g(x)=x∫0(|sint|+|cost|)dt, then the value of g(x+nπ2),( where n∈N) is
A
g(x)+3n
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B
g(x)+n
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C
g(x)+2n
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D
g(x)+n2
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Solution
The correct option is Cg(x)+2n We have, g(x)=x∫0(|sint|+|cost|)dt
Now, g(x+nπ2)=x+nπ2∫0(|sint|+|cost|)dt=x∫0(|sint|+|cost|)dt+x+nπ2∫x(|sint|+|cost|)dt
Using the property: a+nT∫af(x)dx=nT∫0f(x)dx ⇒g(x+nπ2)=g(x)+nπ2∫0(|sint|+|cost|)dt
As (|sint|+|cost|) has a period π2 ⇒g(x+nπ2)=g(x)+g(nπ2)⇒g(x+nπ2)=g(x)+nπ2∫0(|sinx|+|cosx|)dx⇒g(x+nπ2)=g(x)+nπ2∫0(sinx+cosx)dx⇒g(x+nπ2)=g(x)+n[sinx−cosx]π20∴g(x+nπ2)=g(x)+2n