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Find the exhaustive interval of [tex]\(x\)[/tex] for which [tex]\(3 \leq |x-2| \leq 5\)[/tex]:

A. [tex]\([5, 7]\)[/tex]
B. [tex]\([-2, -1] \cup [5, 6]\)[/tex]
C. [tex]\([-3, -1] \cup [5, 7]\)[/tex]
D. [tex]\([-1, 5]\)[/tex]

Sagot :

To determine the exhaustive interval of [tex]\( x \)[/tex] for which [tex]\( 3 \leq |x-2| \leq 5 \)[/tex], we need to break down the inequality [tex]\( 3 \leq |x-2| \leq 5 \)[/tex] into two parts and solve step by step.

First, let's handle the inequality [tex]\( 3 \leq |x-2| \)[/tex]:

1. [tex]\( 3 \leq |x-2| \)[/tex] implies either [tex]\( x-2 \leq -3 \)[/tex] or [tex]\( x-2 \geq 3 \)[/tex]:
- [tex]\( x-2 \leq -3 \)[/tex] leads to [tex]\( x \leq -1 \)[/tex].
- [tex]\( x-2 \geq 3 \)[/tex] leads to [tex]\( x \geq 5 \)[/tex].

So from [tex]\( 3 \leq |x-2| \)[/tex], we get [tex]\( x \leq -1 \)[/tex] or [tex]\( x \geq 5 \)[/tex].

Next, let's handle the inequality [tex]\( |x-2| \leq 5 \)[/tex]:

2. [tex]\( |x-2| \leq 5 \)[/tex] implies [tex]\( -5 \leq x-2 \leq 5 \)[/tex]:
- Adding 2 to all parts of the inequality [tex]\( -5 \leq x-2 \leq 5 \)[/tex], we get [tex]\( -3 \leq x \leq 7 \)[/tex].

So from [tex]\( |x-2| \leq 5 \)[/tex], we get [tex]\( -3 \leq x \leq 7 \)[/tex].

Now, we need to determine the intersection of the two sets of intervals obtained:

- From [tex]\( x \leq -1 \)[/tex] or [tex]\( x \geq 5 \)[/tex], we have two intervals: [tex]\( x \leq -1 \)[/tex] and [tex]\( x \geq 5 \)[/tex].
- From [tex]\( -3 \leq x \leq 7 \)[/tex], we have the interval [tex]\( -3 \leq x \leq 7 \)[/tex].

Combining these results:

- The interval [tex]\( x \leq -1 \)[/tex] intersects with [tex]\( -3 \leq x \leq 7 \)[/tex] gives us the interval [tex]\( -3 \leq x \leq -1 \)[/tex].
- The interval [tex]\( x \geq 5 \)[/tex] intersects with [tex]\( -3 \leq x \leq 7 \)[/tex] gives us the interval [tex]\( 5 \leq x \leq 7 \)[/tex].

Combining these intersected intervals, we get the union of [tex]\( -3 \leq x \leq -1 \)[/tex] and [tex]\( 5 \leq x \leq 7 \)[/tex].

Therefore, the exhaustive interval of [tex]\( x \)[/tex] for which [tex]\( 3 \leq |x-2| \leq 5 \)[/tex] is:
[tex]\[ [-3, -1] \cup [5, 7] \][/tex]

Thus, the correct option is:
(3) [tex]\( [-3, -1] \cup [5, 7] \)[/tex]