What will be the half-range Fourier series for the given function:

\frac{2 l^2 \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \sin \frac{\pi \mathrm{x}}{l}+\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \sin \frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \sin \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]

\frac{2 l \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \sin \frac{\pi \mathrm{x}}{l}+\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \sin \frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \sin \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]

\frac{2 l^2 \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \sin \frac{\pi \mathrm{x}}{l}-\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \sin \frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \sin \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]

\frac{2 l^2 \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \cos \frac{\pi \mathrm{x}}{l}+\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \cos\frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \cos \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]

\text { If } f(x)= \begin{cases}1+\frac{2 x}{\pi} & \text { if }-\pi \leq x \leq 0 \\ 1-\frac{2 x}{\pi} & \text { if } 0 \leq x \leq \pi\end{cases}, then

f(x)=\frac{8}{\pi^2}\left[\cos x+\frac{1}{3^2} \cos 3 x+\frac{1}{5^2} \cos 5 x+\cdots\right]

f(x)=\frac{\pi^2}{8}\left[\cos x-\frac{1}{3^2} \cos 3 x+\frac{1}{5^2} \cos 5 x+\cdots\right]

f(x)=\frac{8}{\pi^2}\left[\cos x-\frac{1}{3^2} \cos 3 x+\frac{1}{5^2} \cos 5 x+\cdots\right]

f(x)=\frac{\pi^2}{8}\left[\cos x+\frac{1}{3^2} \cos 3 x+\frac{1}{5^2} \cos 5 x+\cdots\right]

The general solution of (x-y)p + (x+y)q = 2xz is

\Phi (x+y - log z, (x^2 + y^2)e^{-2 tan^{-1}\left(\frac{y}{x}\right)})=0.

\Phi (x+y - log z, (x^2 + y^2)log(-2 tan^{-1}\frac{y}{x})=0.

\Phi (x+y + log z, (x^2 + y^2)e^{-2 tan^{-1}\left(\frac{y}{x}\right)})=0.

\Phi (x+y - log z, (x^2 - y^2)e^{-2 tan^{-1}\left(\frac{x}{y}\right)})=0.

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Q 1/4
Score 0
What will be the half-range Fourier series for the given function:
300
$\frac{2 l \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \sin \frac{\pi \mathrm{x}}{l}+\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \sin \frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \sin \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]$
$\frac{2 l^2 \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \sin \frac{\pi \mathrm{x}}{l}-\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \sin \frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \sin \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]$
$\frac{2 l^2 \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \sin \frac{\pi \mathrm{x}}{l}+\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \sin \frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \sin \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]$
$\frac{2 l^2 \mathrm{~d}}{\mathrm{a}(l-\mathrm{a}) \pi^2}\left[\sin \frac{\pi \mathrm{a}}{l} \cos \frac{\pi \mathrm{x}}{l}+\frac{1}{2^2} \sin \frac{2 \pi \mathrm{a}}{l} \cos\frac{2 \pi \mathrm{x}}{l}+\frac{1}{3^2} \frac{\sin 3 \pi \mathrm{a}}{l} \cos \frac{3 \pi \mathrm{x}}{l}+\ldots .\right]$
Q 2/4
Score 0
300

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