If the function f(x)= {2/x{ sin (k_1+1) x+ sin (k_2-1) x}, x<0; 4, x=0; 2/x log_e((2+k_1 x)/(2+k_2 x)), x>0} is continuous at x =0, then k_1^2+ k_2^2…

If the function $$f(x)=\left\{\begin{array}{l} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x<0 \\ 4, \quad x=0 \\ \frac{2}{x} \log _e\left(\frac{2+k_1 x}{2+k_2 x}\right), \quad x>0 \end{array}\right. $$ is continuous at $\displaystyle \mathrm{x}=0$, then $\displaystyle \mathrm{k}_1{ }^2+\mathrm{k}_2{ }^2$ is equal to