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Question:
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If $\mathrm{A}=\mathrm{x} \in$
$\mathrm{B}=\left\{\mathrm{x} \in \mathbb{R}: \mathrm{x}^{2}+9 \mathrm{x}+14>0\right\}$, then which of the following
is are correct?
1.$(A \cap B)=(-2,1)$
2.$(\mathrm{A} \mathrm{B})=(-7,-2)$
Select the correct answer using the code given below:
Options:
$\mathrm{B}=\left\{\mathrm{x} \in \mathbb{R}: \mathrm{x}^{2}+9 \mathrm{x}+14>0\right\}$, then which of the following
is are correct?
1.$(A \cap B)=(-2,1)$
2.$(\mathrm{A} \mathrm{B})=(-7,-2)$
Select the correct answer using the code given below:
Solution:
2469 Upvotes
Verified Answer
The correct answer is:
1 only
$x^{2}+6 x-7 < 0$
$\Rightarrow(x+7)(x-1) < 0$
$\Rightarrow \mathrm{x}=(-7,1)$
$\Rightarrow \mathrm{A}=\{-6,-5,-4,-3,-2,-1,0\}$
$\Rightarrow x^{2}+9 x+14>0$
$\Rightarrow(x+7)(x+2)>0$
$\Rightarrow \mathrm{x}=(-\infty,-7) \cup(-2, \infty)$
So $A \cap B=(-2,1)$
$\Rightarrow(x+7)(x-1) < 0$
$\Rightarrow \mathrm{x}=(-7,1)$
$\Rightarrow \mathrm{A}=\{-6,-5,-4,-3,-2,-1,0\}$
$\Rightarrow x^{2}+9 x+14>0$
$\Rightarrow(x+7)(x+2)>0$
$\Rightarrow \mathrm{x}=(-\infty,-7) \cup(-2, \infty)$
So $A \cap B=(-2,1)$
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