If the function $f(x)=\begin{cases}\frac2x\{\sin(k_1+1)x+\sin(k_2-1)x\},&x0\end{cases}$ is continuous at $x=0$, then $k_1^2+k_2^2$ is equal to

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Left limit: $2\left[(k_1+1)+(k_2-1)\right]=2(k_1+k_2)=4\Rightarrow k_1+k_2=2$. Right limit: $\frac2x\cdot\frac{(k_1-k_2)x}2=k_1-k_2=4$. So $k_1=3,\ k_2=-1$ and $k_1^2+k_2^2=10$. Answer (4).

Q. If the function $f(x)=\begin{cases}\frac2x\{\sin(k_1+1)x+\sin(k_2-1)x\},&x<0\\4,&x=0\\\frac2x\log_e\left(\frac{2+k_1x}{2+k_2x}\right),&x>0\end{cases}$ is continuous at $x=0$, then $k_1^2+k_2^2$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?