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  • Solution set of  isOption: 1 <img alt="(-\infty,-3]\cup[-1,\infty)" s

Solution set of  |x+2|\geq 1 is

Option: 1

(-\infty,-3]\cup[-1,\infty)

 

 

 


Option: 2

\left [ -1,\infty \right ]


Option: 3

\left [ -\infty,-3 \right ]


Option: 4

\left [ -\infty,-3 \right ]\cup \left ( -1,\infty \right )


Answers (1)

best_answer

For |f(x)| \geqslant a \quad(a \geqslant 0) \Rightarrow f(x) \leqslant-a \;or f(x) \geqslant a

So, |x+2|\geq 1

\begin{aligned} &\Rightarrow \quad x+2 \leqslant-1 \quad \text { or } \quad x+2 \geqslant1 \\ &\Rightarrow \quad x \leqslant-3\quad \text { or } \quad x \geqslant-1 \end{aligned}

Posted by

manish painkra

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