Question

The number of integral value(s) of $y$ for which $(y^2-5y+3)(x^2+x+1)-2x<0$ for all $x\in R$ is

Answer 1

$(y^2-5y+3)(x^2+x+1)<2x,\forall x\in R$
$\Rightarrow y^2-5y+3<\frac{2x}{x^2+x+1}(\because x^2+x+1>0\forall x\in R)$

L.H.S. must be less than the least value of R.H.S.
Now, finding the least value of R.H.S.

Let $\frac{2x}{x^2+x+1}=p$
$\Rightarrow p{x}^2+(p-2)x+p=0$
As $x\in R$, so
$D\ge 0\Rightarrow (p-2{)}^2-4p^2\ge 0\Rightarrow -3p^2-4p+4\ge 0\Rightarrow 3p^2+4p-4\le 0\Rightarrow 3p^2+6p-2p-4\le 0\Rightarrow (3p-2\left)\right(p+2)\le 0\Rightarrow p\in [-2,\frac{2}{3}]$

Therefore, the least value of R.H.S. is $-2$, so
$y^2-5y+3<-2\Rightarrow y^2-5y+5<0\therefore y\in (\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2})$

Hence, the possible intergral values are $2,3$