Sunday, April 28, 2013

find k such that for every real x we have (1 + kx)/(1 + x^2) < k

Assuming k is any real number... 
%281+%2B+kx%29%2F%281+%2B+x%5E2%29+%3C+k 
leads to
%281+%2B+kx%29%2F%281+%2B+x%5E2%29+-+k+%3C+0 
then 
%281%2Bkx%29%2F%281%2Bx%5E2%29-k%28%281%2Bx%5E2%29%2F%281%2Bx%5E2%29%29+%3C+0 
and combining them we get 
%28%281%2Bkx%29-k%281%2Bx%5E2%29%29%2F%281%2Bx%5E2%29+%3C+0 
after simplifying the numerator... 
%281%2Bkx-k-kx%5E2%29%2F%281%2Bx%5E2%29+%3C+0 
Notice that 1%2Bx%5E2 is always positive, so to determine when the sign of the expression will be less than zero we only need to look at the numerator: 
-kx%5E2%2Bkx%2B%281-k%29 
using the quadratic formula to find x... 
x+=+%28-k+%2B-+sqrt%28+k%5E2-4%2A%28-k%29%2A%281-k%29+%29%29%2F%282%2A%28-k%29%29+ 
then... 
x+=+%28-k+%2B-+sqrt%28+k%5E2-4%2A%28-k%2Bk%5E2%29%29%29%2F%28-2k%29+ 
then... 
x+=+%28-k+%2B-+sqrt%28+k%284-3k%29+%29%29%2F%28-2k%29+ 
which only has a positive real value when 
4-3k%3C=0 and k%3C=0 

or 
4-3k%3E=0 and k%3E=0 
So we notice x is only real on the interval 
0%3C=k%3C=4%2F3%5D 
This interval gives us boundary conditions to test. After testing (using a graphing utility) we see only the values k>4/3 satisfies the inequality for ALL values of x

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