For real values of $x$ and $a$, if the expression $\frac{x+a}{2 x^2-3 x+1}$ assumes all real values, then

1. $a \lt -1$ or $a\gt-\frac{1}{2}$
2. $-1 \lt a \lt -\frac{1}{2}$
3. $\frac{1}{2} \lt a \lt 1$
4. $a \lt \frac{1}{2}$ or $a\gt1$

Solution

$\begin{aligned} & \text { Let } \frac{x+a}{2 x^2-3 x+1}=y, \text { where } y \in \mathrm{R} \\ \Rightarrow & x+a=2 y x^2-3 y x+y\end{aligned}$ $\begin{aligned}& \Rightarrow 2 y x^2-(3 y+1) x+y-a=0 \\ & (3 y+1)^2-4.2 y(y-a) \geq 0\end{aligned}$ $(\because x \in \mathrm{R})$ $\begin{aligned} & \Rightarrow 9 y^2+1+6 y-8 y^2+8 a y \geq 0 \\ & \Rightarrow y^2+(8 a+\text { b) } y+1 \geq 0\end{aligned}$ $\begin{aligned} & 1\gt0 \text { and }(8 a+b)^2-4.1 .1 \lt 0 \\ & \Rightarrow 64 a^2+36+96 a-4 \lt 0 \\ & \Rightarrow(2 a+1)(a+1) \lt 0\end{aligned}$
$\Rightarrow-1 \lt \mathrm{a} \lt -\frac{1}{2}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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