Exhaustive values of $x$ satisfying the equation $| x^{4} - x^{2} - 12 | = | x^{4} - 9 | - | x^{2} + 3 |$ is -

x \in [1, \infty)

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x \in (- \infty, - 2] \cup [2, \infty)

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x \in [- 2, 2]

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x \in (- \infty, - 1] \cup [1, \infty)

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Solution

The correct option is B $x \in (- \infty, - 2] \cup [2, \infty)$
$| x^{2} - 4 | | x^{2} + 3 | = | x^{2} - 3 | | x^{2} + 3 | - | x^{2} + 3 |$
$| x^{2} - 4 | = | x^{2} - 3 | - 1$
$\Rightarrow | x^{2} - 4 | + 1 = | x^{2} - 3 |$
$a = x^{2} - 4, b = 1$
$| a | + | b | = | a + b |$
$\Rightarrow$ $a b = | a | | b |$
[By squaring both sides]
$\Rightarrow a b \geq 0$
$\Rightarrow (x^{2} - 4) \geq 0$
$\Rightarrow x^{2} \geq 4$
So $x ϵ (- \infty, - 2] \cup [2, \infty)$

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