Determine the value of kkk such that f(x)={ln(1+kx)xx<0x2+4x≥0f(x) = \begin{cases} \frac{\ln(1+kx)}{x} & x < 0 \\ x^2 + 4 & x \ge 0 \end{cases}f(x)={xln(1+kx)x2+4x<0x≥0 is continuous at x=0x=0x=0.
k=4k = 4k=4
k=2k = 2k=2
k=−4k = -4k=−4
k=−2k = -2k=−2