Let f(x) = {x3+x2−16x+20(x−2)2, if x≠2k, if x=2. if f(x) be continuous for all x, then k =
For continuous limx→2f(x)=f(2)=k ⇒k=limx→2x3+x2−16x+20(x−2)2 =limx→2(x2−4x+4)(x+5)(x−2)2=7.