Question

Mark the correct alternative in each of the following:
If $\left|x+2\right|$$\le$9, then
(a) x$\in$($-$7, 11)
(b) x$\in$[$-$11, 7]
(c) x$\in$($-$$\infty$, $-$7) $\cup$ (11, $\infty$)
(d) x$\in$($-$$\infty$, $-$7) $\cup$ [11, $\infty$)

Answer 1

$$\begin{gathered} \left|x+2\right|\le 9 \\ \Rightarrow -9\le x+2\le 9 \\ \Rightarrow -9-2\le x+2-2\le 9-2 \\ \Rightarrow -11\le x\le 7 \\ \Rightarrow x\in \left[-11,7\right] \end{gathered}$$

Hence, the correct option is (b).