Q52 / 160 · AP EAMCET 2013 · Mathematics
The set of solutions satisfying both $x^2+5 x+6 \geq 0$ and $x^2+3 x-4 < 0$ is
- A$(-4,1)$
- Correct answerB$(-4,-3] \cup[-2,1)$
- C$(-4,-3) \cup(-2,1)$
- D$[-4,-3] \cup[-2,1]$
Solution
Given, $x^2+5 x+6 \geq 0$ and $x^2+3 x-4 < 0$
$$
\begin{aligned}
& \Rightarrow x \in(-\infty,-3] \cup[-2, \infty) \text { and } x \in(-4,1) \\
&
\end{aligned}
$$
Common condition is
$$
x \in(-4,-3] \cup[-2,1)
$$
Hence, the correct answer is option: 'B'.