We just have to factorise it. It's so simple! Or is it?
∣ ∣ ∣ ∣ x 2 + x + 1 x 2 − 3 x − 1 ∣ ∣ ∣ ∣ ≥ 3 A = max ( x ) B = min ( x )
Find A B ( A − B ) .
The answer is 2.00.
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1 solution
The root of x 2 + 3 x + 2 a r e − 1 a n d − 2 i n s t e a d o f 1 a n d 2 . That is by luck you got right : )
root\quad of\quad { x }^{ 2 }-3x-1=0\\ \\ x=\cfrac { 3\pm \sqrt { 13 } }{ 2 } =3.30277,-0.30277\\ \\ So\quad { x }^{ 2 }-3x-1=\left\{ +ve\quad for\quad x\notin \left( -0.30277,3.30277 \right) \\ \\ -ve\quad for\quad x\in \left( -0.30277,3.30277 \right) \right \\ \\ \\ { x }^{ 2 }+x+1\quad always\quad +ve,so\\ \\ \\ 1.If\quad { x }^{ 2 }-3x-1>0\quad or\quad x\notin \left( -0.30277,3.30277 \right) \\ \\ \left| \cfrac { { x }^{ 2 }-3x-1 }{ { x }^{ 2 }+x+1 } \right| =\cfrac { { x }^{ 2 }-3x-1 }{ { x }^{ 2 }+x+1 } \\ \\ \Rightarrow \cfrac { { x }^{ 2 }-3x-1 }{ { x }^{ 2 }+x+1 } \ge 3\\ \\ \Rightarrow { x }^{ 2 }+3x+2\le 0\\ \\ \Rightarrow x\in \left[ -2,-1 \right] \\ \\ which\quad is\quad not\quad in\quad \left( -0.30277,3.30277 \right) \\ \\ 2.If\quad { x }^{ 2 }-3x-1<0\quad or\quad x\in \left( -0.30277,3.30277 \right) \\ \\ \left| \cfrac { { x }^{ 2 }-3x-1 }{ { x }^{ 2 }+x+1 } \right| =-\cfrac { { x }^{ 2 }-3x-1 }{ { x }^{ 2 }+x+1 } \\ \\ \Rightarrow -\cfrac { { x }^{ 2 }-3x-1 }{ { x }^{ 2 }+x+1 } \ge 3\\ \\ \Rightarrow 4{ x }^{ 2 }+2\le 0\quad not\quad possible\\ \\ so\quad by\quad we\quad see\quad that\quad given\quad condition\quad satisfied\quad only\quad for\quad x\in \left[ -2,-1 \right] \\ \\ A-1,B=-2\Rightarrow Ans=\left( -1 \right) \times \left( -2 \right) \left( -1+2 \right) =2