Question

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

• A. $(−∞−2]$
• B. $\left(\frac{5−\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}\right)$
• C. $[\frac{5+\sqrt{5}}{2},∞)$
• D. $\left(\frac{5+\sqrt{5}}{2},−\frac{2}{3}\right)$

Answer 1

Answer: (B)

Given, $\left(y^2−5y+3\right)\left(x^2+x+1\right)<2x,∀x∈R$
$⇒y^2−5y+3<\frac{2x}{x^2+x+1}\left(âˆ\mu x^2+x+1>0∀x∈R\right)$
LHS must be less than the least value RHS.
Let $\frac{2x}{x^2+x+1}=p⇒p{x}^2+(p−2)x+p=0$
Since $x$ is real, we have
$(p−2{)}^2−4p^2â‰\yen 0⇒−2≤p≤\frac{2}{3}$
The minimum value of $\frac{2x}{\left(x^2+x+1\right)}$ is $−2$.
$\text{ So, }{y}^2−5y+3<−2$ $⇒y^2−5y+5<0$ $⇒y∈\left(\frac{5−\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}\right)$