If $| | 2 x - x^{2} + 8 | - | x^{2} + 5 | | = | 2 x + 13 |$ , then $x$ lies in

(- \infty, - \frac{13}{2}]

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(- \infty, - 2] \cup [4, \infty)

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[4, \infty)

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(- \infty, - \frac{13}{2}] \cup [4, \infty)

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Solution

The correct option is B $(- \infty, - 2] \cup [4, \infty)$
$| | 2 x - x^{2} + 8 | - | x^{2} + 5 | | = | 2 x + 13 |$
Let, $a = 2 x - x^{2} + 8, b = x^{2} + 5$
and
$a + b = 2 x + 13$
Clearly, $| | a | - | b | | = | a + b |$ when $a b \leq 0$
$\Rightarrow (2 x - x^{2} + 8) (x^{2} + 5) \leq 0$
$\Rightarrow (x^{2} - 2 x - 8) (x^{2} + 5) \geq 0$
$\Rightarrow (x - 4) (x + 2) (x^{2} + 5) \geq 0$
$∵ x^{2} + 5$ is always positive
$\Rightarrow (x - 4) (x + 2) \geq 0$
$\Rightarrow x \in (- \infty, - 2] \cup [4, \infty)$

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