The equation $| x + 1 | | x - 1 | = a^{2} - 2 a - 3$ can have real solution for $x$ if $a$ belongs to

(- \infty, - 1] \cup [3, + \infty)

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[1 - \sqrt{5}, 1 + \sqrt{5}]

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[1 - \sqrt{5}, - 1] \cup [3, 1 + \sqrt{5}]

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none of these

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Solution

The correct options are
C $[1 - \sqrt{5}, - 1] \cup [3, 1 + \sqrt{5}]$
D $(- \infty, - 1] \cup [3, + \infty)$
$| x + 1 | | x - 1 | = a^{2} - 2 a - 3$
since on LHS both have mod so RHS should also be +ve
$a^{2} - 2 a - 3 \leq 0$
$a \in (- \infty, - 1] ⋃ [3, \infty)$

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