The correct option is A 2 lt; x lt; 3 ⇒ x^2 x 6 lt;0 ⇒ (x 3)(x+2) lt;0 At x lt; 2(x 3)(x+2) gt;0 ∴ 2 lt; x lt;3 ...

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Absolute Value Function

Solve: 6+x-...

Question

Solve: $6 + x - x^{2} > 0$

- 2 < x < 3

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2 < x < 3

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- 2 < x < - 3

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None of these

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Solution

The correct option is A $- 2 < x < 3$
$\Rightarrow x^{2} - x - 6 < 0$
$\Rightarrow (x - 3) (x + 2) < 0$
At $x < - 2 (x - 3) (x + 2) > 0$
$∴ - 2 < x < 3$
$(∴ (x - 3)$ and $(x - 2)$ has odd power $∴$ sign change at $- 2$ and $3$ )

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