The range of the function $f(x)=\frac{x^2+x+1}{x^2+4 x+3}$ is

none of these

$y=\frac{x^2+x+1}{x^2+4 x+3}⇒y(x^2+4x+3)=x^2+x+1$

$⇒x^2(1-y)+x(1-4y)+1-3y=0$

$f(x) =\frac{x^2+x+1}{x^2+4 x+3} \Rightarrow x^2(1-y)+x(1-4 y)+1-3 y=0$

Since x is real ⇒ discriminant ≥ 0

$\Rightarrow 4 y^2+8 y-3 \geq 0$

$\left.\Rightarrow y \in-\infty, \frac{-2-\sqrt{7}}{2} \cup \frac{-2+\sqrt{7}}{2}, \infty\right)$

Hence (4) is the correct answer.