JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 13)
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 :
$$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$$ is :
$$( - \infty ,1) \cup (2,\infty )$$
$$(2,\infty )$$
$$\left[ { - {1 \over 2},1} \right) \cup (2,\infty )$$
$$\left[ { - {1 \over 2},1} \right) \cup (2,\infty ) - \left\{ 3,{{{3 + \sqrt 5 } \over 2},{{3 - \sqrt 5 } \over 2}} \right\}$$
Explanation
$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ 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)
$$ x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\} $$
$$ \frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0 $$
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)
$$ x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\} $$
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