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A function is defined over (0,3) by f(3) = 12 +1. We then extend it to an even periodic function of period 6 and its graph is

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ANSWER: Since the graph of resulting function is symmetrical about Y-axis hence the given function has been extended oeriodically to an even function. Thus

b_n=\frac{1}{L}\int_{-L}^Lf(x)\sin\left ( n\pi x/L \right )dx\\ \implies {\color{Blue} b_n=0} \\ {\color{Red} (\because f(x)\sin\left ( n\pi x/L \right )\ is\ odd\ function)}\\ \text{and}\\ {\color{Magenta} a_0=\frac{1}{L}\int_{-L}^Lf(x)dx}\\ \implies a_0=\frac{1}{3}\int_{-3}^3f(x)dx\\ \implies a_0=\frac{1}{3}\int_{-3}^3\left ( \frac{x}{4}+1 \right )dx=\color{blue}\frac{11}{4}\\ {\color{Magenta} a_n=\frac{1}{L}\int_{-L}^Lf(x)\cos\left ( n\pi x/L \right )dx}\\ \implies a_n=\frac{1}{3}\int_{-3}^3(x/4+1)\cos(n\pi x/3)dx\\ \implies a_n=\frac{3}{2n^2\pi^2}\left [ 1-(-1)^n \right ]\\ \text{Put n=2k+1}\\ a_{2k+1}=\frac{3}{2(2k+1)^2\pi^2}\left [ 1-(-1)^{2k+1} \right ]=\frac{3}{(2k+1)^2\pi^2}\\ \text{Put n=2k}\\ a_{2k}=0

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