The domain of f (x) = ^ - 1 ( x ^ 2 - 5x + 6 x ^ 2 - 9 ) ( x ^ 2 -...

We know that, $\cos^{- 1}x$ is defined when $- 1 ⩽ x ⩽ 1$

So, $\cos^{- 1}\left(\dfrac{{x}^{2} - 5x + 6}{{x}^{2} - 9}\right)$ is defined when

$- 1 ⩽ \dfrac{{x}^{2} - 5x + 6}{{x}^{2} - 9} ⩽ 1$ and ${x}^{2} - 9 \ne 0$

$⇒ - 1 \le \dfrac{\left(x - 3\right)\left(x - 2\right)}{\left(x - 3\right)\left(x + 3\right)} \le 1$ and $x \ne \pm 3$

$⇒ - 1 \le \dfrac{x - 2}{x + 3} \le 1$ and $x \ne \pm 3$, $⇒ - 1 \le \dfrac{x + 3 - 5}{x + 3} \le 1$ and $x \ne \pm 3$

$⇒ - 1 \le 1 - \dfrac{5}{x + 3} \le 1$ and $x \ne \pm 3$,

$⇒ - 2 \le - \dfrac{5}{x + 3} ⩽ 0$ and $x \ne \pm 3$

$⇒ 0 \le \dfrac{5}{x + 3} ⩽ 2$ and $x \ne \pm 3$, $⇒ x + 3 \ge \dfrac{5}{2}$ and $x \ne \pm 3$

$⇒ x \ge \dfrac{- 1}{2}$ and $x \ne \pm 3$

$∴ x \in \left[\dfrac{- 1}{2}, \infty\right) - \left\{3\right\} ⋯ \left(1\right)$

Now, we know that $\log \left(x\right)$ is defined for $x > 0$.

So, $\log \left({x}^{2} - 3x + 2\right)$ is defined for ${x}^{2} - 3x + 2 > 0$, $⇒ \left(x - 1\right)\left(x - 2\right) > 0$

$⇒ x > 2$ or $x < 1 ⋯ \left(2\right)$

from equations \left(1\right)& \left(2\right)

$∴ x \in \left[- \dfrac{1}{2}, 1\right) \cup \left(2, \infty\right) - \left\{3\right\}$