Explanation

(a)
Let $y = \frac{x^{2} - 2x + 4}{x^{2} + 2x + 4}$
Then, $x^{2}(y - 1) + 2x(y + 1) + 4(y - 1) = 0$
Since, $x$ is real, therefore
Discriminant, $4(y + 1)^{2} - 16(y - 1)^{2} \geq 0$
$\Rightarrow \ \ (y + 1)^{2} - \left\lbrack 2(y - 1) \right\rbrack^{2} \geq 0$
$\Rightarrow \ \ (3 - y)(3y - 1) \geq 0\ \ \ \Rightarrow \ \ \ \frac{1}{3} \leq y \leq 3$