JEE MAIN - Mathematics (2022 - 27th July Evening Shift - No. 1)
The domain of the function $$f(x)=\sin ^{-1}\left[2 x^{2}-3\right]+\log _{2}\left(\log _{\frac{1}{2}}\left(x^{2}-5 x+5\right)\right)$$, where [t] is the greatest integer function, is :
$$
\left(-\sqrt{\frac{5}{2}}, \frac{5-\sqrt{5}}{2}\right)
$$
$$
\left(\frac{5-\sqrt{5}}{2}, \frac{5+\sqrt{5}}{2}\right)
$$
$$
\left(1, \frac{5-\sqrt{5}}{2}\right)
$$
$$
\left[1, \frac{5+\sqrt{5}}{2}\right)
$$
Explanation
$$ - 1 \le 2{x^2} - 3 < 2$$
or $$2 \le 2{x^2} < 5$$
or $$1 \le {x^2} < {5 \over 2}$$
$$x \in \left( { - \sqrt {{5 \over 2}} , - 1} \right] \cup \left[ {1,\sqrt {{5 \over 2}} } \right)$$
$${\log _{{1 \over 2}}}({x^2} - 5x + 5) > 0$$
$$0 < {x^2} - 5x + 5 < 1$$
$${x^2} - 5x + 5 > 0$$ & $${x^2} - 5x + 4 < 0$$
$$x \in \left( { - \infty ,{{5 - \sqrt 5 } \over 2}} \right) \cup \left( {{{5 + \sqrt 5 } \over 2},\infty } \right)$$
& $$x \in ( - \infty ,1) \cup (4,\infty )$$
Taking intersection
$$x \in \left( {1,{{5 - \sqrt 5 } \over 2}} \right)$$
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