Let y = sqrt{frac{(x + 1)(x - 3)}{(x - 2)}}. | vedclass

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(A) For $y$ to be a real value,the expression under the square root must be non-negative:$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$We use the sign scheme

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$[-1, 2) \cup [3, \infty)$

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(A) For $y$ to be a real value,the expression under the square root must be non-negative:$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$We use the sign scheme (Wavy Curve Method) for the expression $f(x) = \frac{(x + 1)(x - 3)}{(x - 2)}$.The critical points are $x = -1, 2, 3$.Testing the intervals:$1$. For $x \in (-1, 2)$,let $x = 0$: $\frac{(1)(-3)}{(-2)} = \frac{3}{2} > 0$ (Valid).$2$. For $x \in [3, \infty)$,let $x = 4$: $\frac{(5)(1)}{(2)} = \frac{5}{2} > 0$ (Valid).$3$. At $x = -1$,$y = 0$ (Valid).$4$. At $x = 3$,$y = 0$ (Valid).$5$. At $x = 2$,the expression is undefined.Combining these,the solution is $x \in [-1, 2) \cup [3, \infty)$.

Let $y = \sqrt{\frac{(x + 1)(x - 3)}{(x - 2)}}$. Then,the set of all real values of $x$ for which $y$ takes real values is:

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