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 :
Solution
$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ and $x^{2}-3 x+2>0, \neq 1$
<br/><br/>
$\frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0$
<br/><br/>
The solution to this inequality is
<br/><br/>
$x \in\left[\frac{-1}{2}, \infty\right)-\{3\}$
<br/><br/>
for $x^{2}-3 x+2>0$ and $\neq 1$
<br/><br/>
$$
x \in(-\infty, 1) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
$$
<br/><br/>
Combining the two solution sets (taking intersection)
<br/><br/>
$$
x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
$$
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
This question is part of PrepWiser's free JEE Main question bank. 65 more solved questions on Inverse Trigonometric Functions are available — start with the harder ones if your accuracy is >70%.