2. 外语
  3. AIn this question, you are given that x < 1 and x0, and you are asked to compare x2 + 1 with x3+ 1. One way to approach this pro

AIn this question, you are given that x < 1 and x0, and you are asked to compare x2 + 1 with x3+ 1. One way to approach this pro

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答案A
解析 In this question, you are given that x < 1 and x0, and you are asked to compare x2 + 1 with x3+ 1. One way to approach this problem is to set up a comparison between the two quantities using a placeholder symbol to represent the relationship between them as follows.
x2+ 1x3+ 1
Then simplify the comparison.
Step 1: Subtract 1 from both sides to get
x2x3
Step 2: Since x0, you can divide both sides by the positive quantity x2to get
1x
Since you are given that x < 1, or 1 > x, you can conclude that the placeholderin the simplified comparison 1x represents greater than(>). Note that the strategy of simplifying the comparison requires you to consider whether the steps in the simplification are reversible. This is because you must arrive at a conclusion about the initial comparison, not the simplified comparison. If you follow the simplification steps in reverse, you can see that the placeholder in each step remains unchanged: 1 >x implies x2 >x3 because multiplying by the positive number x2 retains the inequality greater than(>). Also, x2 >x3 implies x2 + 1 >x3 +1. Therefore, Quantity A is greater than Quantity B, and the correct answer is Choice A.
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本试题收录于： GRE QUANTITATIVE题库GRE分类
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