If $c$ is a real number with $0 < c < 1$, then show that the values taken by the function $y = \frac{x^2 + 2x + c}{x^2 + 4x + 3c}$, as $x$ varies over real numbers, range over all real numbers.
$$y = \frac{x^2 + 2x + c}{x^2 + 4x + 3c}$$ $\Rightarrow x^2 y + 4xy + 3cy = x^2 + 2x + c$ $\Rightarrow (y-1)x^2 + 2x(2y-1) + c(3y-1) = 0 \quad [\because x \text{ is real}]$ $\therefore \{2(2y-1)\}^2 - 4(y-1)\cdot c(3y-1) \geq 0$ $\Leftrightarrow c \leq \frac{(2y-1)^2}{(y-1)(3y-1)}$ Since $0 < c < 1$, we get $\frac{1}{3} < y < 1$.
If $c$ is a real number with $0 < c < 1$, then show that the values taken by the function $y = \frac{x^2 + 2x + c}{x^2 + 4x + 3c}$, as $x$ varies over real numbers, range over all real numbers.