Step-by-step Solution:

the function given is: \[ f(x) = \begin{cases} \frac{x-x^2}{2x}, & x \neq 0 \\ K, & x = 0 \end{cases} \] For a function to be continuous at \( x=0 \), it must satisfy: \[ \lim_{x \to 0} f(x) = f(0) \] which means: \[ \lim_{x \to 0} f(x) = K \] First, simplify the expression for \( x \neq 0 \): \[ f(x) = \frac{x - x^2}{2x} = \frac{x(1-x)}{2x} = \frac{1-x}{2} \] (as long as \( x \neq 0 \)). \[\] Thus, for \( x \neq 0 \): \[ f(x) = \frac{1-x}{2} \] Now, calculate the limit as \( x \to 0 \): \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1-x}{2} = \frac{1-0}{2} = \frac{1}{2} \] Thus, for continuity at \( x=0 \): \[ K = \frac{1}{2} \] Final Answer: \({\frac{1}{2}} \)