If the function [f(x)=cases2x((k_1+1)x)+((k_2-1)x), & x

Question

If the function
$$f(x)=\begin{cases}\dfrac2x\{\sin((k_1+1)x)+\sin((k_2-1)x)\}, & x<0\ 4, & x=0\ \dfrac2x\log_e\left(\dfrac{2+k_1x}{2+k_2x}\right), & x>0\end{cases}$$
is continuous at $x=0$, then $k_1^2+k_2^2$ is equal to:

Accepted Answer

LHL: $\displaystyle\lim_{x\to0^-}\frac2x\{\sin((k_1+1)x)+\sin((k_2-1)x)\}=2(k_1+1)+2(k_2-1)=4\implies k_1+k_2=2$

RHL: For small $x$, $\ln\left(\dfrac{2+k_1x}{2+k_2x}\right)=\ln\left(1+\dfrac{(k_1-k_2)x}{2+k_2x}\right)\approx\dfrac{(k_1-k_2)x}{2}$, so
$$\lim_{x\to0^+}\frac2x\cdot\frac{(k_1-k_2)x}{2}=k_1-k_2=4$$

Solving $k_1+k_2=2$ and $k_1-k_2=4$: $k_1=3,\ k_2=-1$.

$k_1^2+k_2^2=9+1=10$

Solutions · 1