The smallest value of `k` for which both roots of the equation `x^(2)-8kx+16(k^(2)-k+1)=0` are real distinct and have value at least 4, is

Options

A. 2
B. 3
C. 4
D. none of these

Answer

A

Explanation

Let `f(x) = x^(2) - 8kx + 16 (k^(2)-k+1)`. Both the roots of the equation f(x) = 0 will be real, distinct and have values at least 4, if
(i) `D gt 0`
(ii) `f(4) ge 0`
(iii) Vertex of u = f(x) lies on the right side of (4, 0)
Now,
(i) `D gt 0 rArr 64k^(2) - 64(k^(2) - k + 1) gt 0 rArr k - 1 gt 0 rArr k gt 1`
(ii) `f(4) ge 0`
`rarr" "16 - 32 k + 16 (k^(2)-k+1) ge 0`
`rArr" "k^(2) - 3k + 2 ge 0 rArr k le 1 or k ge 2`
(iii) x-coordinate of vertex `gt` 0
i.e.`" "4k gt 4 rArr k gt 1" "[because "x coordinate" = -(b)/(2a)]`
Taking intersection of these, we have `k in [2, oo)`.
Hence, the least value of k is 2.