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

Options

A. 6
B. 4
C. 2
D. 0

Answer

D

Explanation

Let `f(x)=x^(2)-8kx+16(k^(2)-k+1)` brgt
`:.Dgt0`
`implies64k^(2)-4.16(k^(2)-k+1)gt0`
`implieskgt1` …….i
`implies(-b)/(2a)gt4implies(8k)/2gt4`
`implieskgt1`………….ii
and `f(4)ge0`
`implies16-32k+16(k^(2)-k+1)ge0`
`impliesk^(2)-3k+2ge0`
`implies(k-1)(k-2)ge0`
`kle1` or `kge2`...........iii
From Eqs i, ii and iii we get
`kge2`
`k_("min")=2`