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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
admin
2014-08-13
14
问题
选项
答案
A
解析
In this question, you are given that x < 1 and x
0, and you are asked to compare x
2
+ 1 with x
3
+ 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.
x
2
+ 1
x
3
+ 1
Then simplify the comparison.
Step 1: Subtract 1 from both sides to get
x
2
x
3
Step 2: Since x
0, you can divide both sides by the positive quantity x
2
to get
1
x
Since you are given that x < 1, or 1 > x, you can conclude that the placeholder
in the simplified comparison 1
x 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 x
2
>x
3
because multiplying by the positive number x
2
retains the inequality greater than(>). Also, x
2
>x
3
implies x
2
+ 1 >x
3
+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
0
GRE QUANTITATIVE
GRE
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