IIT JEE , AIEEE Forum - Mathematics , Algebra

cygnus7

If x is real and 0<b<c, prove that (x²-bc)/(2x-b-c) cannot lie between b and c.

nadeemoidu

Hint : x will not lie between p and q iff
(x-p)(x-q) >= 0

tarinbansal

Nadeemoidu is right
Let the function given above be denoted by F.
Find (F-b)(F-c)
You will it as (x-b)^2(x-c)^2/(2x-b-c)^2.
As in this new function, all terms are squares, it is always positive.
Hence F does not lie between b and c.

Rate if satisfied.

feynmann

This is a very simple qn. on the theory of quadratic eqns. Here is the solution

let y=(x^2-bc)/(2x-b-c);

simplifying , we get the following quadratic eqn of x -

x^2-2xy+(b+c)y-bc=0

now, x, being real , the discriminant of the eqn should be>=0 , which gives

y^2-(b+c)y + bc>=0 .

which gives

(y-b)(y-c)>=0 .

hence y can't lie between b&c . (proved)

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