x2−8kx+16(k2−k+1) = 0 Since roots are real and distinct, D > 0 ⇒ 64k2−64(k2−k+1) > 0 ... (1) ⇒ k > 1 Also, from the figure, we have −
b
2a
> 4 ⇒
8k
2
> 4 ⇒ k > 1 ... (2) Further f (4) ≥ 0 ⇒ 16 - 32k + 16 (k2 - k + 1) ≥ 0 ⇒ k2 - 3k + 2 ≥ 0 ⇒ k ≤ 1 or k ≥ 2 ... (3) From (1), (2) and (3) K min = 2