Explanation

(c)
Let $y = \ \frac{x^{2} - 3x + 4}{x^{2} + 3x + 4}$
$\Rightarrow (y - 1)x^{2} + 3(y + 1)x + 4(y - 1) = 0$
$\because x$ is real.
$\therefore D \geq 0$
$\Rightarrow 9(y + 1)^{2} - 16(y - 1)^{2} \geq 0$
$\Rightarrow \ - 7y^{2} + 50y - 7 \geq 0$
$\Rightarrow \ - 7y^{2} - 50y + 7 \leq 0$
$\Rightarrow (y - 7)(7y - 1) \leq 0$ ...(i)
$\Rightarrow y \leq 7\ \ and\ y \geq \frac{1}{7}\ \Rightarrow \frac{1}{7} \leq y \leq 7$
Hence, maximum value is 7 and minimum value is $\frac{1}{7}$.