Answer is : A. k > 3
Given f(x) = kx3 - 9x2+ 9x+3
⇒ f’(x) = 3kx2 - 18x+9
⇒ f’(x) = 3(kx2 – 6x + 3) > 0
⇒ kx2 – 6x + 3 > 0
For quadratic equation to be greater than 0. a > 0 and D < 0
⇒ k > 0 and (-6)2- 4(k)(3) < 0
⇒ 36 – 12k < 0
⇒ 12k > 36
⇒ k > 3
∴ k > 3.