Question

If $(y^2-5y+3)(x^2+x+1)<2x$ for all $xϵR,$ then y lies in the interval

• A. $(-\infty ,-2]$
• B. $[\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}]$
• C. $[\frac{5+\sqrt{5}}{2},\infty )$
• D. $[\frac{5+\sqrt{5}}{2},-\frac{2}{3}]$

Answer 1

The correct option is B $[\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}]$

$Given(y^2-5y+3)(x^2+x+1)<2x$
$\Rightarrow y^2-5y+3<\frac{2x}{x^2+x+1}$ ...(1)
$[\because x^2+x+1>0forallxϵR]$
$Let\frac{2x}{x^2+x+1}=z\Rightarrow z{x}^2+(z-2)x+z=0$
$\because xϵR\Rightarrow (z-2{)}^2-4z.z\ge 0$
$\Rightarrow 3z^2+4z-4\le 0\Rightarrow -2\le z\le \frac{2}{3}\Rightarrow -2\le \frac{2x}{x^2+x+1}\le \frac{2}{3}$ clearly the inequality (1) holds
if $y^2-5y+3<-2\Rightarrow y^2-5y+5<0\Rightarrow \frac{5-\sqrt{5}}{2}<y<\frac{5+\sqrt{5}}{2}$