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 $x=0$, then $k_1^2+k_2^2$ is equal to :

<p>$$\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\ & \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\ & \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=4 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2\right) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=2 \\ & \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \\ & \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \\ & \therefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \\ & \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10 \end{aligned}$$</p>