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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

 May 17, 2020
 #1
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d, just plug the numbers in and yull get it

 May 17, 2020
 #2
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The graph of  y  =  x2 + 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  =  - x2 + 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.

 May 17, 2020

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