Question
let a,b,c be three distinct real numbers such that each of expression ax2+bx+c,bx2+cx+a,cx2+ax+b are positive for all x ε R and let
α=bc+ca+ab/a2+b2+c2 then
(A) α<4 (B) α<1 (C) α>1/4 (D) α>1
THIS IS MULTIPLE CHOICE QUESTION
Joshi sir comment
if ax2+bx+c is positive for real x
then b2< 4ac similarly others
on adding all inequalities we get b2 + c2 + a2< 4(ac+ab+bc)
now get answer
Read 1 Solution.
ax² + bx + c = y (let)
or, ax² + bx + (c-y) = 0
since, x is real, b² - 4a(c-y) ≥0
or, y ≥ c-b²/4a
by question, y>0
c-b²/4a >0
b² < 4ac
similarly, a² < 4bc and c² < 4ab
a² + b² + c² < 4(ab + bc+ ca)
α > 1/4
DIPANKAR GUPTA 10 year ago
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