Let y- -x-1 x-3 x-2 - If y takes real values- then x can lie in

Y = √((x+1 ) (x 3 nbsp; )(x 2 )) will be real if (x+1 ) (x 3 ) x 2≥ 0 Hence from above number line x∈[ 1,2)∪[3,∞) ...

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Inequality

Let y= √x+1 ...

Question

$Let y = \sqrt{\frac{(x + 1) (x - 3)}{(x - 2)}}$ . If $y$ takes real values, then $x$ can lie in

(2, \infty)

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[- 1, 3] - {2}

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[- 1, 2)

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[3, \infty)

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