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

For continuity, $$LHL = RHL = f(0) = 4$$.

1. Left Hand Limit (LHL):

$$\lim_{x \to 0^-} \frac{2}{x} (\sin((k_1+1)x) + \sin((k_2-1)x))$$

Using $$\lim_{x \to 0} \frac{\sin ax}{x} = a$$:

$$LHL = 2((k_1+1) + (k_2-1)) = 2(k_1 + k_2)$$.

Set $$2(k_1 + k_2) = 4 \implies \mathbf{k_1 + k_2 = 2}$$.

2. Right Hand Limit (RHL):

$$\lim_{x \to 0^+} \frac{2}{x} \log_e \left( \frac{2+k_1x}{2+k_2x} \right) = \lim_{x \to 0^+} \frac{2}{x} \log_e \left( \frac{1+\frac{k_1}{2}x}{1+\frac{k_2}{2}x} \right)$$

Using $$\lim_{x \to 0} \frac{\log(1+ax)}{x} = a$$:

$$RHL = 2(\frac{k_1}{2} - \frac{k_2}{2}) = k_1 - k_2$$.

Set $$k_1 - k_2 = 4$$.

Solving for $$k_1, k_2$$:

Adding the equations: $$2k_1 = 6 \implies k_1 = 3$$.

Subtracting: $$2k_2 = -2 \implies k_2 = -1$$.

Result: $$k_1^2 + k_2^2 = 3^2 + (-1)^2 = 9 + 1 = \mathbf{10}$$. (Option D