If −4≤x<2 then ||x+2|−3| liesintheinterval

We have −4≤x<2⇒ −2≤x+2<4⇒ 0≤|x+2|<4⇒ −3≤|x+2|−3<1⇒ 0≤∥x+2|−3|≤3

If −4≤x<2 then ||x+2|−3| liesintheinterval

We have −4≤x<2⇒ −2≤x+2<4⇒ 0≤|x+2|<4⇒ −3≤|x+2|−3<1⇒ 0≤∥x+2|−3|≤3

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If $- 4 \leq x < 2$ then $| | x + 2 | - 3 |$ liesintheinterval

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(1, 3]

[1, 3]

[0, 3]

$[0, \infty)$

answer is C.

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Detailed Solution

We have $- 4 \leq x < 2$

$\begin{array}{l} \Rightarrow - 2 \leq x + 2 < 4 \\ \Rightarrow 0 \leq | x + 2 | < 4 \\ \Rightarrow - 3 \leq | x + 2 | - 3 < 1 \\ \Rightarrow 0 \leq ∥ x + 2 | - 3 | \leq 3 \end{array}$

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If $- 4 \leq x < 2$ then $| | x + 2 | - 3 |$ liesintheinterval