The set of all real values ' $a$ ' for which $-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \lt 3$ holds for all real values of $x$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \qquad-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \lt 3 \ & 	ext { Consider }-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \Rightarrow 3 x^2+(a+1) x+3\gt0 \ & \Rightarrow \mathrm{D} \lt 0 \Rightarrow(a+1)^2-36 \lt 0 \Rightarrow a^2+2 a-35 \lt 0\end{aligned}$ $\Rightarrow(a+7)(a-5) \lt 0 \Rightarrow a \in(-7,5)$ ...(i) $\begin{aligned} & 	ext { Also, } \frac{2 x^2+a x+2}{x^2+x+1} \lt 3 \Rightarrow-x^2+(a-3) x-1 \lt 0 \ & \Rightarrow \mathrm{D} \lt 0 \Rightarrow(a-3)^2-4 \lt 0\end{aligned}$ $\Rightarrow a^2-6 a+5 \lt 0 \Rightarrow(a-1)(a-5) \lt 0 \Rightarrow a \in(1,5)$ ...(ii) So, from (i) and (ii), $a \in(1,5)$

The set of all real values ' $a$ ' for which $-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \lt 3$ holds for all real…

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