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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]
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]
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