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

Question

Aicharya · Accepted Answer

$\begin{aligned} & \lim _{x ightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1ight) \mathrm{x}+\sin \left(\mathrm{k}_2-1ight) \mathrm{x}ight\}=4 \ & \Rightarrow 2\left(\mathrm{k}_1+1ight)+2\left(\mathrm{k}_2-1ight)=4 \ & \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \ & \Rightarrow \lim _{\mathrm{x} ightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}ight)=4 \ & \Rightarrow \lim _{\mathrm{x} ightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2ight) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}ight)=2 \ & \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \ & \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \ & 	herefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \ & \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10\end{aligned}$

If the function $f(x)=\left\{\begin{array}{l}\frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin…

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