Q51 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
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
- A$(-7,5)$
- B$(5, \infty)$
- Correct answerC$(1,5)$
- D$(-\infty, 1)$
Solution
$\begin{aligned} & \qquad-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \lt 3 \\ & \text { Consider }-1 \lt \frac{2 x^2+a x+2}{x^2+x+1} \Rightarrow 3 x^2+(a+1) x+3\gt0 \\ & \Rightarrow \mathrm{D} \lt 0 \Rightarrow(a+1)^2-36 \lt 0 \Rightarrow a^2+2 a-35 \lt 0\end{aligned}$
$\Rightarrow(a+7)(a-5) \lt 0 \Rightarrow a \in(-7,5)$ ...(i)
$\begin{aligned} & \text { Also, } \frac{2 x^2+a x+2}{x^2+x+1} \lt 3 \Rightarrow-x^2+(a-3) x-1 \lt 0 \\ & \Rightarrow \mathrm{D} \lt 0 \Rightarrow(a-3)^2-4 \lt 0\end{aligned}$
$\Rightarrow a^2-6 a+5 \lt 0 \Rightarrow(a-1)(a-5) \lt 0 \Rightarrow a \in(1,5)$ ...(ii)
So, from (i) and (ii), $a \in(1,5)$
Hence, the correct answer is option: 'C'.