Question

If $(y^2-5y+3)(x^2+x+1)<2x$ for $x\in R$, then find the interval in which $y$ lies

• A. $(\frac{5-\sqrt{3}}{2},\frac{5+\sqrt{3}}{2})$
• B. $(\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2})$
• C. $(\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2})$
• D. none of these

Answer 1

The correct option is C $(\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2})$

Given, $(y^2-5y+3)(x^2+x+1)<2x,\forall x\in R$
$\Rightarrow y^2-5y+3<\frac{2x}{x^2+x+1}(\because x^2+x+1>0\forall x\in R)$
LHS must be less than the least value RHS.

Let $\frac{2x}{x^2+x+1}=p\Rightarrow p{x}^2+(p-2)x+p=0$
Since $x$ is real, we have
$(p-2{)}^2-4p^2\ge 0\Rightarrow -2\le p\le \frac{2}{3}$
The minimum value of $\frac{2x}{(x^2+x+1)}$ is $-2$.

So,$y^2-5y+3<-2$

$\Rightarrow y^2-5y+5<0$

$\Rightarrow y\in (\frac{5-\sqrt{5}}{2},\frac{5+\sqrt{5}}{2})$