Let $y=\sqrt{\frac{(x+1)(x-3)}{(x-2)}}$.
Find all the real values of $x$ for which $y$ takes real values.
$(1980,2 M)$
Correct Answer: 17. $x \in[-1,2) \cup[3, \infty)$
Solution:
$ \text { when } \quad \frac{(x+1)(x-3)}{(x-2)} \geq 0 $
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