Find kkk such that f(x)=ekx−1sin(2x)f(x) = \frac{e^{kx}-1}{\sin(2x)}f(x)=sin(2x)ekx−1 is continuous at x=0x=0x=0, given f(0)=2f(0)=2f(0)=2.
k=4k=4k=4
k=2k=2k=2
k=1k=1k=1
k=3k=3k=3