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Let $f(x)$ and $g(x)$ be two functions given by $f\left( x \right) = \frac{{2\sin \pi x}}{x}$ and $g\left( x \right) = f\left( {1 - x} \right) + f\left( x \right).$ If $g\left( x \right) = kf(\frac{x}{2})f\left( {\frac{{1 - x}}{2}} \right)$,then the value of $k$ is
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{6}$
$\frac{1}{8}$
Solution
$f(1-\mathrm{x})+f(\mathrm{x})$
$=2\left(\frac{\sin \pi x}{1-x}+\frac{\sin \pi x}{x}\right)$
$=2 \frac{\sin \pi x}{x(1-x)}=4 \cdot \frac{\frac{\sin \pi x}{2} \cdot \cos \frac{\pi x}{2}}{x(1-x)}$
$=\frac{2 \sin \frac{\pi \mathrm{x}}{2}}{4 \cdot \frac{\mathrm{x}}{2}} \cdot \frac{\sin \frac{\pi(1-\mathrm{x})}{2}}{\frac{(1-\mathrm{x})}{2}}$
$=\frac{1}{2} \cdot \frac{1}{2} f\left(\frac{\mathrm{x}}{2}\right) \cdot \frac{1}{2} f\left(\frac{1-\mathrm{x}}{2}\right)$
$=\frac{1}{8} f\left(\frac{\mathrm{x}}{2}\right) \cdot f\left(\frac{1-\mathrm{x}}{2}\right)$
