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  • Define the mod : .Option: 1 <img alt="|x+2|=\left\{\begin{array}{cc}x+2, a

Define the mod : \mid x+2\mid.

Option: 1

|x+2|=\left\{\begin{array}{cc}x+2, & x \geqslant-2 \\ -x-2, & x<-2\end{array}\right.


Option: 2

|x+2|=x+2


Option: 3

|x+2|=\pm(x-2)


Option: 4

|x+2|= \begin{cases}x+2, & x \geqslant 2 \\ -x-2, & x < 2\end{cases}


Answers (1)

best_answer

Defining means removing mod

As |f(x)|=\left\{\begin{array}{l} f(x), f(x) \geqslant 0 \\ -f(x), f(x)<0 \end{array}\right.

So,

\begin{aligned} |x+2| &=\left\{\begin{array}{cc} x+2, & x+2 \geqslant 0 \\ -(x+2), & x+2<0 \end{array}\right.\\ &=\left\{\begin{array}{cc} x+2, & x \geqslant-2 \\ -x-2, & x<-2 \end{array}\right. \end{aligned}

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manish

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