If x^2+kx+1<0 for all real values of x, then k belongs to

a.(-2,2)

b.Null set

c.R-[-2,2]

d.None of these

Guest May 17, 2020

#2**+1 **

The graph of y = x^{2} + kx + 1 is a parabola with its minimum point at its vertex and rises, on both sides, from there.

There is no way that this parabola will always be below zero; that is, will always be below the x-axis. So, the answer is the null set.

If the graph were: y = - x^{2} + kx + 1, which is a parabola with a maximum point and falling, it would be possible for the graph to be always below the x-axis.

geno3141 May 17, 2020