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  • Solution set of  isOption: 1 <img alt="\left [ -3,-2 \right ]" src="/

Solution set of  |x+2|\leq 1 is

Option: 1

\left [ -3,-2 \right ]

 

 


Option: 2

\left [ -3,-1 \right ]


Option: 3

\left [ 2,3 \right ]


Option: 4

\Phi


Answers (1)

best_answer

As |f(x)| \leqslant a \quad(a \geqslant 0) \Rightarrow-a \leqslant f(x) \leqslant a

\begin{aligned} &\Rightarrow \quad|x+2| \leqslant 1 \\ &\Rightarrow-1 \leqslant x+2 \leqslant 1 \\ &\Rightarrow-1-2 \leqslant x \leqslant 1-2 \\ &\Rightarrow-3 \leqslant x \leqslant-1 \end{aligned}

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chirag

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