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

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Question

(a) $y = \sqrt {\frac{{(x + 1)(x - 3)}}{{(x - 2)}}} $

Here $x$ cannot be $2$.

==> Either both ${N^r}$ and ${D^r}$ are positive

$x \ge - 1,

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) $y = \sqrt {\frac{{(x + 1)(x - 3)}}{{(x - 2)}}} $ Here $x$ cannot be $2$. ==> Either both ${N^r}$ and ${D^r}$ are positive $x \ge - 1,x \ge 3$ and $x > 2 \Rightarrow x \ge 3$.....$(i)$ Or ${N^r}$ is negative and ${D^r}$ is negative $x \ge - 1$ and $x < 2 \Rightarrow - 1 \le x < 2$…..$(ii)$ From $(i)$ and $(ii),$ $ - 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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