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If $\mathrm{A}=\left\{\mathrm{x} \in \mathrm{R}: \mathrm{x}^{2}+6 \mathrm{x}-7 < 0\right\}$ and
$\mathrm{B}=\left\{\mathrm{x} \in \mathrm{R}: \mathrm{x}^{2}+9 \mathrm{x}+14>0\right\}$, then which of the following
is/are correct?
1.$\mathrm{A} \cap \mathrm{B}=\{\mathrm{X} \in \mathrm{R}:-2 < \mathrm{x} < 1\}$
2.$A \backslash B=\{x \in R:-7 < x < -2\}$
Select the correct answer using the code given below:
Options:
$\mathrm{B}=\left\{\mathrm{x} \in \mathrm{R}: \mathrm{x}^{2}+9 \mathrm{x}+14>0\right\}$, then which of the following
is/are correct?
1.$\mathrm{A} \cap \mathrm{B}=\{\mathrm{X} \in \mathrm{R}:-2 < \mathrm{x} < 1\}$
2.$A \backslash B=\{x \in R:-7 < x < -2\}$
Select the correct answer using the code given below:
Solution:
1685 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
We have :
$x^{2}+6 x-7 < 0 \quad$ \& $\quad x^{2}+9 x+14>0$
$\Rightarrow(x-1)(x+7) < 0 \& \quad \Rightarrow(x+2)(x+7)>0$
$\Rightarrow x \in(-7,1) \quad \& \quad \Rightarrow x \in(-\infty,-7) \cup(-2, \infty)$
$A \cap B=\{x \in R:-2 < x < 1\} \rightarrow$ It is true.
$A \backslash B=A-B=\{x \in R:-7 < x < -2\} \rightarrow$ It is also true.
$x^{2}+6 x-7 < 0 \quad$ \& $\quad x^{2}+9 x+14>0$
$\Rightarrow(x-1)(x+7) < 0 \& \quad \Rightarrow(x+2)(x+7)>0$
$\Rightarrow x \in(-7,1) \quad \& \quad \Rightarrow x \in(-\infty,-7) \cup(-2, \infty)$
$A \cap B=\{x \in R:-2 < x < 1\} \rightarrow$ It is true.
$A \backslash B=A-B=\{x \in R:-7 < x < -2\} \rightarrow$ It is also true.
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