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Which of the following inequalities is equivalent to –2 < x < 4 ?
- a)| x – 2 | < 4
- b)| x – 1 | < 3
- c)| x + 1 | < 3
- d)| x + 2 | < 4
- e)None of the above
Correct answer is option 'B'. Can you explain this answer?
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Which of the following inequalities is equivalent to –2 < x &...
A: |x - 2| < 4
If we simplify this inequality, we get -4 < x - 2 < 4. Adding 2 to all parts of the inequality, we have -2 < x < 6. This is not equivalent to the given range of -2 < x < 4. So, option A is not correct.
If we simplify this inequality, we get -4 < x - 2 < 4. Adding 2 to all parts of the inequality, we have -2 < x < 6. This is not equivalent to the given range of -2 < x < 4. So, option A is not correct.
B: |x - 1| < 3
Simplifying this inequality, we get -3 < x - 1 < 3. Adding 1 to all parts of the inequality, we have -2 < x < 4. This is exactly the same as the given range -2 < x < 4. Therefore, option B is correct.
Simplifying this inequality, we get -3 < x - 1 < 3. Adding 1 to all parts of the inequality, we have -2 < x < 4. This is exactly the same as the given range -2 < x < 4. Therefore, option B is correct.
C: |x + 1| < 3
Simplifying this inequality, we get -3 < x + 1 < 3. Subtracting 1 from all parts of the inequality, we have -4 < x < 2. This range is not equivalent to the given range -2 < x < 4. Hence, option C is not correct.
Simplifying this inequality, we get -3 < x + 1 < 3. Subtracting 1 from all parts of the inequality, we have -4 < x < 2. This range is not equivalent to the given range -2 < x < 4. Hence, option C is not correct.
D: |x + 2| < 4
Simplifying this inequality, we get -4 < x + 2 < 4. Subtracting 2 from all parts of the inequality, we have -6 < x < 2. This range is not equivalent to -2 < x < 4. Therefore, option D is not correct.
Simplifying this inequality, we get -4 < x + 2 < 4. Subtracting 2 from all parts of the inequality, we have -6 < x < 2. This range is not equivalent to -2 < x < 4. Therefore, option D is not correct.
Based on the analysis, the only option that is equivalent to -2 < x < 4 is option B: |x - 1| < 3.
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