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The smallest value of k, for which both ...

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

A

2

B

3

C

4

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
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.
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