Solution set of $(x^{2} - x - 1) (x^{2} - x - 7) < - 5$ is

(1, 5)

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(- 2, 3)

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(- \infty, - 1) \cup (2, \infty)

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(- 2, - 1) \cup (2, 3)

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Solution

The correct option is D $(- 2, - 1) \cup (2, 3)$
Let $x^{2} - x - 1 = t .$
Then $t (t - 6) + 5 < 0$
$\Rightarrow t^{2} - 6 t + 5 < 0 \Rightarrow (t - 1) (t - 5) < 0 \Rightarrow 1 < t < 5$

Case $1 :$ $t > 1$
$\Rightarrow x^{2} - x - 1 > 1$
$\Rightarrow x^{2} - x - 2 > 0$
$\Rightarrow (x + 1) (x - 2) > 0$
$\Rightarrow x \in (- \infty, - 1) \cup (2, \infty) . . . (1)$

Case $2 :$ $t < 5$
$\Rightarrow x^{2} - x - 1 < 5$
$\Rightarrow x^{2} - x - 6 < 0$
$\Rightarrow (x + 2) (x - 3) < 0$
$\Rightarrow x \in (- 2, 3) . . . (2)$

From $(1)$ and $(2),$
$x \in (- 2, - 1) \cup (2, 3)$

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