If positive numbers `a ,b ,c` are in H.P., then equation `x^2-k x+2b^(101)-a^(101)-c^(101)=0(k in R)` has both roots positive both roots negative one positive and one negative root both roots imaginary

Options

A. both roots positive
B. both roots
C. one positive and one negative root
D. both roots imaginary

Answer

C

Explanation

a,b,c are in H.P. Hence, H.M of a and c is b.
`:. sqrt(ac) ge b`
Since, A.M `gt` G.M., so
`(a^(101) + c^(101))/(2) gt (sqrt(ac))^(101) ge b^(101):' sqrt(ac) ge b`
or `2b^(101) - a^(101) - c^(101 lt 0`
Let `f(x) = x^(2) - kx + 2b^(101) - a^(101) - c^(101)`
`:. f(0) = 2b^(101) - a^(101) - c^(101) lt 0`
Hence, equation f(x) = 0 has one root in `(- prop, 0)` and other in `(0, prop)`