If ${log}_{(x + 3)} (x^{2} - x) < 1$ , then the value of x lies in

(- 3, - 2)

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

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(- 3, - 1)

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(- 1, 3)

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Solution

The correct options are
A $(- 3, - 2)$
B $(- 1, 0) \cup (1, 3)$
$Given that, {log}_{(x + 3)} (x^{2} - x) < 1 For {log}_{(x + 3)} (x^{2} - x) to be defined, x^{2} - x > 0 \Rightarrow x (x - 1) > 0 \Rightarrow x < 0 or x > 1 . . . [1] If x + 3 > 1 \Rightarrow x > - 2 then, x^{2} - x < x + 3 \Rightarrow x^{2} - 2 x - 3 < 0 \Rightarrow (x - 3) (x + 1) < 0 \Rightarrow - 1 < x < 3 . . . . [2] From [1] and [2], we get x \in (- 1, 0) \cup (1, 3) If 0 < x + 3 < 1 \Rightarrow - 3 < x < - 2 . . . [3] then, x^{2} - x > x + 3 \Rightarrow x^{2} - 2 x - 3 > 0 \Rightarrow (x - 3) (x + 1) > 0 \Rightarrow x < - 1 or x > 3 . . . . [4] From [3] and [4], we get x \in (- 3, - 2)$

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