If (2−x2)(x−3)3(x+1)(x2−3x−4)≥0 then x belong to

(2−x2)(x−3)3(x+1)(x2−3x−4)≥0 (2+x)(2−x)(x−3)3(x+1)2(x−4)≥0 ⇒x∈[−2,−1)∪(−1,2]∪[3,4)

If (2−x2)(x−3)3(x+1)(x2−3x−4)≥0 then x belong to

(2−x2)(x−3)3(x+1)(x2−3x−4)≥0 (2+x)(2−x)(x−3)3(x+1)2(x−4)≥0 ⇒x∈[−2,−1)∪(−1,2]∪[3,4)

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If $\frac{(2 - x^{2}) {(x - 3)}^{3}}{(x + 1) (x^{2} - 3 x - 4)} \geq 0$ then $x$ belong to

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$[- \sqrt{2}, 0) \cup (- 1, \sqrt{2}]$

$[- \sqrt{2}, 0) \cup (- 1, \sqrt{2}] \cup [3, 4)$

$[- \sqrt{2}, - 1) \cup (- 1, \sqrt{2}] \cup [3, 4)$

answer is D.

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$\frac{(2 - x^{2}) {(x - 3)}^{3}}{(x + 1) (x^{2} - 3 x - 4)} \geq 0$
$\frac{(\sqrt{2} + x) (\sqrt{2} - x) {(x - 3)}^{3}}{{(x + 1)}^{2} (x - 4)} \geq 0$
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$\Rightarrow x \in [- \sqrt{2}, - 1) \cup (- 1, \sqrt{2}] \cup [3, 4)$

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If $\frac{(2 - x^{2}) {(x - 3)}^{3}}{(x + 1) (x^{2} - 3 x - 4)} \geq 0$ then $x$ belong to