The smallest value of k, for which both the roots of the equation x2-8kx+16(k2-k+1)=0 are real, distinct and have values at least 4, is [2009]
(2)
The given equation is x2-8kx+16(k2-k+1)=0
∵ Both the roots are real and distinct.
∴ D>0 ⇒ (8k)2-4×16(k2-k+1)>0
⇒k>1 ...(i)
∵ Both the roots are greater than or equal to 4
∴ α+β>8 and f(4)≥0
⇒k>1 ...(ii)
and 16-32k+16(k2-k+1)≥0
⇒k2-3k+2≥0 ⇒(k-1)(k-2)≥0
⇒k∈(-∞,1]∪[2,∞) ...(iii)
Combining (i), (ii) and (iii), we get k≥2
∴ Smallest value of k=2