If f(x) = begin{cases} frac{1-cos Kx}{x sin x | vedclass

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(D) Given that $f(x)$ is continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Therefore,$\lim_{x 	o 0} \frac{1-\cos Kx}{x \sin x} = \frac{1}

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$\pm 1$

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(D) Given that $f(x)$ is continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0)$.Therefore,$\lim_{x 	o 0} \frac{1-\cos Kx}{x \sin x} = \frac{1}{2}$.Using the trigonometric identity $1-\cos 	heta = 2 \sin^2(\frac{	heta}{2})$,we get:$\lim_{x 	o 0} \frac{2 \sin^2(\frac{Kx}{2})}{x \sin x} = \frac{1}{2}$.Dividing numerator and denominator by $x^2$,we get:$\lim_{x 	o 0} \frac{2 \left(\frac{\sin(Kx/2)}{x}ight)^2}{\frac{\sin x}{x}} = \frac{1}{2}$.Multiplying and dividing by $(\frac{K}{2})^2$ in the numerator:$\lim_{x 	o 0} \frac{2 \cdot (\frac{K}{2})^2 \cdot (\frac{\sin(Kx/2)}{Kx/2})^2}{\frac{\sin x}{x}} = \frac{1}{2}$.Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,we have:$2 \cdot \frac{K^2}{4} \cdot \frac{1^2}{1} = \frac{1}{2}$.$\frac{K^2}{2} = \frac{1}{2} \implies K^2 = 1$.Thus,$K = \pm 1$.

If $f(x) = \begin{cases} \frac{1-\cos Kx}{x \sin x}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then the value of $K$ is

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