Find kkk such that f(x)={tan(kx)xx<03x+2k2x≥0f(x) = \begin{cases} \frac{\tan(kx)}{x} & x < 0 \\ 3x + 2k^2 & x \ge 0 \end{cases}f(x)={xtan(kx)3x+2k2x<0x≥0 is continuous at x=0x=0x=0.
k=1k = 1k=1
k=−1k = -1k=−1
k=0k = 0k=0
k=1/2k = 1/2k=1/2