The number of integral value(s) of $x,$ satisfying the inequality $2 | x + 2 | - | x + 5 | \leq 4$ is

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Solution

$2 | x + 2 | - | x + 5 | \leq 4$

Case $1 :$ If $x < - 5$
then $- 2 x - 4 + x + 5 \leq 4$
$\Rightarrow - x + 1 \leq 4$
$\Rightarrow x \geq - 3$ , which is not possible.

Case $2 :$ If $- 5 \leq x < - 2$
then $- 2 x - 4 - x - 5 \leq 4$
$\Rightarrow - 3 x - 9 \leq 4$
$\Rightarrow x \geq \frac{- 13}{3}$
$∴ x \in [\frac{- 13}{3}, - 2)$

Case $3 :$ If $x \geq - 2$
then $2 x + 4 - x - 5 \leq 4$
$\Rightarrow x - 1 \leq 4$
$\Rightarrow x \leq 5$
$∴ x \in [- 2, 5]$

From all the above cases, we can conclude that
$x \in [\frac{- 13}{3}, - 2) \cup [- 2, 5]$
i.e., $x \in [\frac{- 13}{3}, 5]$
Integral values of $x$ are $- 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$
Number of integral values of $x$ is $10$

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