If $- 3 < x < 1$ , then the value of $3 - 2 x - x^{2}$ is:

zero

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positive

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negative

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not determined

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Solution

The correct option is A positive
$3 - 2 x - x^{2}$ $6 - 3 < x < 1$
$\Rightarrow (x + 3) (1 - x)$ $6 - 3 < x < 1$
At $x < - 3 (x < - 3) (1 - x) < 0$
$(∵ (x + 3)$ and $(1 - x)$ has odd power $∴$ sign change at $- 3$ and $1$ )
$∴$ for $- 3 < x < 1 (3 - 2 x - x^{2})$ is $+ v e$

Suggest Corrections

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