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 inside the square root must be non-negative,i.e.,$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$.Also,the denomin

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$ - 1 \le x < 2$ or $x \ge 3$

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(A) For $y$ to be a real value,the expression inside the square root must be non-negative,i.e.,$\frac{(x + 1)(x - 3)}{(x - 2)} \ge 0$.Also,the denominator cannot be zero,so $x 
eq 2$.Let $f(x) = \frac{(x + 1)(x - 3)}{(x - 2)}$. We use the wavy curve method (sign scheme) to find the intervals where $f(x) \ge 0$.The critical points are $x = -1, 2, 3$.Testing the intervals:$1$) For $x \in (-1, 2)$,let $x = 0$: $f(0) = \frac{(1)(-3)}{(-2)} = \frac{3}{2} > 0$.$2$) For $x \in (2, 3)$,let $x = 2.5$: $f(2.5) = \frac{(3.5)(-0.5)}{(0.5)} = -3.5 $3$) For $x \ge 3$,let $x = 4$: $f(4) = \frac{(5)(1)}{(2)} = 2.5 > 0$.$4$) For $x \le -1$,let $x = -2$: $f(-2) = \frac{(-1)(-5)}{(-4)} = -1.25 Including the points where the numerator is zero $(x = -1, 3)$,the inequality holds for $-1 \le x < 2$ or $x \ge 3$.

Let $y = \sqrt {\frac{{(x + 1)(x - 3)}}{{(x - 2)}}} $,then all real values of $x$ for which $y$ takes real values,are

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