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$f(x) = \dfrac{x^{2} + 2x – 15}{x^{2} – 7x – 18}$ is negative if and only if

1. $– 2 < x < 3 \; \text{or} \; x > 9$
2. $x < – 5 \; \text{or} \; 3 < x < 9$
3. $– 5 < x < – 2 \; \text{or} \; 3 < x < 9$
4. $x < – 5 \; \text{or} \; – 2 < x < 3$

Given that, $f(x) = \dfrac{x^{2} + 2x – 15}{x^{2} – 7x – 18}$

Now, if $f(x) < 0$

Then, $\dfrac{x^{2} + 2x – 15}{x^{2} – 7x – 18} < 0$

$\Rightarrow \dfrac{x^{2} + 5x – 3x – 15}{x^{2} – 9x + 2x – 18} < 0$

$\Rightarrow \dfrac{(x+5)(x-3)}{(x-9)(x+2)} < 0$

We can draw the number line.

$x \in (-5, -2) \cup (3, 9)$

$\boxed{-5 < x < -2 \; \text{(or)} \; 3 < x < 9}$

Correct Answer $:\text{C}$

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