Determine the value of kkk such that f(x)=kx2+2f(x) = kx^2 + 2f(x)=kx2+2 for x≤1x \le 1x≤1 and f(x)=x+kf(x) = x + kf(x)=x+k for x>1x > 1x>1 is continuous at x=1x=1x=1.
k=1k = 1k=1
k=2k = 2k=2
k=−1k = -1k=−1
k=0k = 0k=0