Let y - x -1 x -3- x -2- If y takes real values- then x can lie in A- 2- -B- -1-3-2C- -1-2D- -3- -

The correct options are C [ 1,2) D [3,∞)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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Standard XII

Mathematics

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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Solution

The correct options are
C $[- 1, 2)$
D $[3, \infty)$
$y = \sqrt{\frac{(x + 1) (x - 3)}{(x - 2)}}$ will be real if
$\frac{(x + 1) (x - 3)}{x - 2} \geq 0$

Hence from above number line
$x \in [- 1, 2) \cup [3, \infty)$

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Let y=√(x+1)(x−3)(x−2) . If y takes real values, then x can lie in

The correct options are C [−1,2) D [3,∞)y=√(x+1)(x−3)(x−2) will be real if (x+1)(x−3)x−2≥0 Hence from above number line x∈[−1,2)∪[3,∞)

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

The correct options are
C $[- 1, 2)$
D $[3, \infty)$
$y = \sqrt{\frac{(x + 1) (x - 3)}{(x - 2)}}$ will be real if
$\frac{(x + 1) (x - 3)}{x - 2} \geq 0$

Hence from above number line
$x \in [- 1, 2) \cup [3, \infty)$