If the function f(x) = begin{cases} frac{2}{x | vedclass

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(D) For the function to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = 4$.First,conside

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$10$

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(D) For the function to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = 4$.First,consider the left-hand limit: $\lim_{x \rightarrow 0^-} \frac{2}{x} \{\sin(k_1+1)x + \sin(k_2-1)x\} = 4$.Using $\lim_{x \rightarrow 0} \frac{\sin(ax)}{x} = a$,we get $2(k_1+1) + 2(k_2-1) = 4$,which simplifies to $2k_1 + 2k_2 = 4$,or $k_1 + k_2 = 2$.Next,consider the right-hand limit: $\lim_{x \rightarrow 0^+} \frac{2}{x} \ln \left(\frac{2+k_1x}{2+k_2x}\right) = 4$.This can be rewritten as $\lim_{x \rightarrow 0^+} \frac{2}{x} \{\ln(1 + \frac{k_1x}{2}) - \ln(1 + \frac{k_2x}{2})\} = 4$.Using $\lim_{x \rightarrow 0} \frac{\ln(1+ax)}{x} = a$,we get $2(\frac{k_1}{2} - \frac{k_2}{2}) = 4$,which simplifies to $k_1 - k_2 = 4$.Solving the system of equations $k_1 + k_2 = 2$ and $k_1 - k_2 = 4$,we add them to get $2k_1 = 6 \Rightarrow k_1 = 3$.Substituting $k_1 = 3$ into $k_1 + k_2 = 2$,we get $3 + k_2 = 2 \Rightarrow k_2 = -1$.Finally,$k_1^2 + k_2^2 = (3)^2 + (-1)^2 = 9 + 1 = 10$.

If the function $f(x) = \begin{cases} \frac{2}{x} \{\sin(k_1+1)x + \sin(k_2-1)x\} & , x < 0 \\ 4 & , x = 0 \\ \frac{2}{x} \log_e \left(\frac{2+k_1x}{2+k_2x}\right) & , x > 0 \end{cases}$ is continuous at $x = 0$,then $k_1^2 + k_2^2$ is equal to

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