Let f(x) and g(x) be two functions given by f | vedclass

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(B) Given $f(x) = \frac{2\sin(\pi x)}{x}$.Then $g(x) = f(1 - x) + f(x) = \frac{2\sin(\pi(1 - x))}{1 - x} + \frac{2\sin(\pi x)}{x}$.Since $\sin(\pi - \

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$1/4$

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(B) Given $f(x) = \frac{2\sin(\pi x)}{x}$.Then $g(x) = f(1 - x) + f(x) = \frac{2\sin(\pi(1 - x))}{1 - x} + \frac{2\sin(\pi x)}{x}$.Since $\sin(\pi - \pi x) = \sin(\pi x)$,we have $g(x) = 2\sin(\pi x) \left( \frac{1}{1 - x} + \frac{1}{x} \right) = 2\sin(\pi x) \left( \frac{x + 1 - x}{x(1 - x)} \right) = \frac{2\sin(\pi x)}{x(1 - x)}$.Using the identity $\sin(\pi x) = 2\sin(\frac{\pi x}{2})\cos(\frac{\pi x}{2}) = 2\sin(\frac{\pi x}{2})\sin(\frac{\pi(1 - x)}{2})$,we get:$g(x) = \frac{2 \cdot 2\sin(\frac{\pi x}{2})\sin(\frac{\pi(1 - x)}{2})}{x(1 - x)} = 4 \cdot \frac{\sin(\frac{\pi x}{2})}{x} \cdot \frac{\sin(\frac{\pi(1 - x)}{2})}{1 - x}$.Since $f(\frac{x}{2}) = \frac{2\sin(\pi x / 2)}{x/2} = \frac{4\sin(\pi x / 2)}{x}$,we have $\frac{\sin(\pi x / 2)}{x} = \frac{1}{4} f(\frac{x}{2})$.Similarly,$\frac{\sin(\pi(1 - x) / 2)}{1 - x} = \frac{1}{4} f(\frac{1 - x}{2})$.Substituting these into $g(x)$:$g(x) = 4 \cdot \left( \frac{1}{4} f(\frac{x}{2}) \right) \cdot \left( \frac{1}{4} f(\frac{1 - x}{2}) \right) = \frac{1}{4} f(\frac{x}{2}) f(\frac{1 - x}{2})$.Wait,re-evaluating: $g(x) = 4 \cdot \frac{1}{4} f(\frac{x}{2}) \cdot \frac{1}{4} f(\frac{1 - x}{2}) = \frac{1}{4} f(\frac{x}{2}) f(\frac{1 - x}{2})$.Actually,$g(x) = \frac{4 \sin(\pi x / 2) \sin(\pi(1 - x) / 2)}{x(1 - x)} = \frac{4 (\frac{x}{4} f(x/2)) (\frac{1-x}{4} f((1-x)/2))}{x(1-x)} = \frac{4}{16} f(x/2) f((1-x)/2) = \frac{1}{4} f(x/2) f((1-x)/2)$.Thus,$k = 1/4$.

Let $f(x)$ and $g(x)$ be two functions given by $f(x) = \frac{2\sin(\pi x)}{x}$ and $g(x) = f(1 - x) + f(x)$. If $g(x) = k f(\frac{x}{2}) f(\frac{1 - x}{2})$,then the value of $k$ is

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