Determine the value of kkk such that f(x)={cos(kx)−1x2x<02x+k2−6x≥0f(x) = \begin{cases} \frac{\cos(kx) - 1}{x^2} & x < 0 \\ 2x + k^2 - 6 & x \ge 0 \end{cases}f(x)={x2cos(kx)−12x+k2−6x<0x≥0 is continuous at x=0x = 0x=0.
k=2k = 2k=2
k=3k = 3k=3
k=−2k = -2k=−2
k=−3k = -3k=−3