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Q.
If $(y^2 +5y+3)(x^2+ x +1) < 2x$ for all $x\in R,$ , then y lies in the interval
Linear Inequalities
Solution:
Given, $\left(y^2-5 y+3\right)\left(x^2+x+1\right)<2 x, \forall x \in R$
$
\Rightarrow y ^2-5 y +3<\frac{2 x }{ x ^2+ x +1}\left(\because x ^2+ x +1>0 \forall x \in R \right)
$
LHS must be less than the least value RHS.
Let $\frac{2 x }{ x ^2+ x +1}= p \Rightarrow px ^2+( p -2) x + p =0$
Since $x$ is real, we have
$
( p -2)^2-4 p ^2 \geq 0 \Rightarrow-2 \leq p \leq \frac{2}{3}
$
The minimum value of $\frac{2 x}{\left(x^2+x+1\right)}$ is $-2$.
$\text { So, } y ^2-5 y +3<-2 $
$\Rightarrow y ^2-5 y +5<0 $
$\Rightarrow y \in\left(\frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)$