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If $\frac{x^2+a x+3}{x^2+x+1}$ takes real values for all real values of $x$ then $a$ lies in the interval
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Not possible for y to have real values for all real values of x.
Let $\frac{x^2+a x+3}{x^2+x+1}=y$, where $y \in \mathbf{R}$
Now, $(y-1) x^2+(y-a) x+(y-3)=0$
$\begin{aligned} & \therefore \quad x \in \mathbf{R} \text {, so } D \geq 0 \\ & \Rightarrow \quad(y-a)^2-4(y-1)(y-3) \geq 0 \\ & \Rightarrow\left(y^2-2 a y+a^2\right)-4\left(y^2-4 y+3\right) \geq 0 \\ & \Rightarrow-3 y^2+(16-2 a) y+\left(a^2-12\right) \geq 0 \\ & \Rightarrow \quad 3 y^2+(2 a-16) y+\left(12-a^2\right) \leq 0 \\ & \end{aligned}$
Which is not possible for every real value of $y$.
Now, $(y-1) x^2+(y-a) x+(y-3)=0$
$\begin{aligned} & \therefore \quad x \in \mathbf{R} \text {, so } D \geq 0 \\ & \Rightarrow \quad(y-a)^2-4(y-1)(y-3) \geq 0 \\ & \Rightarrow\left(y^2-2 a y+a^2\right)-4\left(y^2-4 y+3\right) \geq 0 \\ & \Rightarrow-3 y^2+(16-2 a) y+\left(a^2-12\right) \geq 0 \\ & \Rightarrow \quad 3 y^2+(2 a-16) y+\left(12-a^2\right) \leq 0 \\ & \end{aligned}$
Which is not possible for every real value of $y$.
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