Question:

Which is greater, when \(-1<x<0\)?
Quantity A: \(|x|\)
Quantity B: \(x^2\)

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For \(|x|\) and \(x^2\), when \(|x|<1\), the square is always smaller.
Updated On: Oct 3, 2025
  • The two quantities are equal
  • Quantity B is greater
  • Quantity A is greater
  • The relationship cannot be determined from the information given
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The Correct Option is B

Solution and Explanation

Step 1: Analyze the range.

We are given \(-1<x<0\). Thus, \(x\) is a negative fraction between \(-1\) and \(0\).
Step 2: Compare \(|x|\) and \(x^2\).

- For example, let \(x = -\tfrac{1}{2}\). Then \(|x| = \tfrac{1}{2}\) and \(x^2 = \tfrac{1}{4}\).
So here \(|x|>x^2\).
- But if \(x = -\tfrac{1}{4}\), then \(|x| = \tfrac{1}{4}\) and \(x^2 = \tfrac{1}{16}\).
Again, \(|x|>x^2\).
Wait — check carefully: actually for all \(-1<x<0\), we have \(|x| = -x\) and since \(|x|<1\), squaring makes it even smaller (\(x^2<|x|\)).
Step 3: Conclusion.

Therefore, \(|x|>x^2\). So Quantity A is greater.
Final Answer: \[ \boxed{\text{Quantity A is greater.}} \]
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