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Question

A function with a period 2π is shown below. The Fourier series fo the function is given by

A
f(x)=12+n=12nπsin(nπ2)cosnx
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B
f(x)=n=12nπsin(nπ2)cosnx
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C
f(x)=12+n=12nπsinnπ2sinnx
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D
f(x)=n=12nπsinnπ2sinnx
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Solution

The correct option is A f(x)=12+∑∞n=12nπsin(nπ2)cosnxf(x)={1,−π2≤×≤π20,otherwise It is given that T = 2π so, a0=1π∫π−πf(x)dx =1π∫π/2−π/2(1)dx=1 an=1π∫π−πf(x)cosnxdx =2xπsinnπ2 bn=1π∫π−πf(x)sinnxdx=0 (∴ integrand is an odd function) So, Fourier series representation is f(x)=a02+∑∞n=1ancosnx+∑∞n=1bnsinnx =12+∑∞n=12nπsin(nπ2)cosnx

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