The domain of the function [f(x) = ^ - 1( x^2 - 5x + 6 x^2 - 9 ) _e(x^2 - 3x + 2)] is :

Question

The domain of the function

$$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$$ is :

Accepted Answer

$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ and $x^{2}-3 x+2>0, \neq 1$

$$
\frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0
$$

The solution to this inequality is

$x \in\left[\frac{-1}{2}, \infty\right)-\{3\}$

for $x^{2}-3 x+2>0$ and $\neq 1$

$$
x \in(-\infty, 1) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
$$

Combining the two solution sets (taking intersection)

$$
x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
$$

Solutions · 1