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Difficulty: Very hardInequalities, Absolute Values, and Number Ranges in Data Sufficiency

If xx is a real number, is ∣x3−x∣>x2−1|x^3 - x| > x^2 - 1?

(1) ∣x−1∣>0|x - 1| > 0
(2) ∣x+2∣<1|x + 2| < 1

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.Answer
  3. C
    BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
The correct option is the one stating that Statement (2) alone is sufficient, but Statement (1) alone is not sufficient. Simplifying the stem target ∣x3−x∣>x2−1|x^3 - x| > x^2 - 1 into ∣x∣∣x2−1∣>x2−1|x||x^2 - 1| > x^2 - 1 reveals that the inequality holds for every real number except x=1x = 1 and x=−1x = -1 (where both sides equal zero). Statement (1) states ∣x−1∣>0|x - 1| > 0, meaning x≠1x \neq 1, but it permits x=−1x = -1, which yields a 'No' answer, making Statement (1) insufficient. Statement (2) states ∣x+2∣<1|x + 2| < 1, which expands to −3<x<−1-3 < x < -1. Because this interval strictly excludes both x=−1x = -1 and x=1x = 1, every value of xx in this range satisfies the stem inequality, yielding a definitive 'Yes'.

Step-by-Step Solution

1
Rephrase the question stem target algebraically.
The target inequality ∣x3−x∣>x2−1|x^3 - x| > x^2 - 1 simplifies to ∣x(x2−1)∣>x2−1|x(x^2 - 1)| > x^2 - 1, which is equivalent to ∣x∣⋅∣x2−1∣>x2−1|x| \cdot |x^2 - 1| > x^2 - 1.
Factoring allows analysis of the critical values where x2−1=0x^2 - 1 = 0.
2
Analyze the conditions under which ∣x∣⋅∣x2−1∣>x2−1|x| \cdot |x^2 - 1| > x^2 - 1 holds true.
Case 1: If x2−1=0x^2 - 1 = 0 (meaning x=1x = 1 or x=−1x = -1), LHS = 00 and RHS = 00, giving 0>00 > 0, which is FALSE. Case 2: If x2−1<0x^2 - 1 < 0 (meaning −1<x<1-1 < x < 1), LHS is positive and RHS is negative, so non-negative > negative is TRUE. Case 3: If x2−1>0x^2 - 1 > 0 (meaning x>1x > 1 or x<−1x < -1), dividing both sides by x2−1>0x^2 - 1 > 0 yields ∣x∣>1|x| > 1, which is TRUE. Thus, the inequality holds for ALL real numbers EXCEPT x=1x = 1 and x=−1x = -1. The question simplifies to: Is x≠1x \neq 1 and x≠−1x \neq -1?
Simplifying the target reveals that the inequality is true everywhere except at the two boundary roots x=1x = 1 and x=−1x = -1.
3
Evaluate Statement (1): ∣x−1∣>0|x - 1| > 0.
This implies x≠1x \neq 1. However, xx could equal −1-1. If x=−1x = -1, ∣−1−1∣=2>0|-1 - 1| = 2 > 0 is satisfied, but ∣(−1)3−(−1)∣=0|(-1)^3 - (-1)| = 0 and (−1)2−1=0(-1)^2 - 1 = 0, giving 0>00 > 0 (NO). If x=2x = 2, ∣2−1∣=1>0|2 - 1| = 1 > 0 is satisfied, and ∣8−2∣=6>3|8 - 2| = 6 > 3 (YES). Since both 'Yes' and 'No' are possible, Statement (1) is NOT sufficient.
Statement (1) rules out x=1x = 1 but permits x=−1x = -1.
4
Evaluate Statement (2): ∣x+2∣<1|x + 2| < 1.
Solving the absolute value inequality gives −1<x+2<1-1 < x + 2 < 1, which simplifies to −3<x<−1-3 < x < -1. In this range, xx cannot be 11 or −1-1 because the upper boundary is strictly less than −1-1. Thus, for all x∈(−3,−1)x \in (-3, -1), x≠1x \neq 1 and x≠−1x \neq -1 is guaranteed, yielding a definitive 'YES'. Statement (2) is SUFFICIENT.
The open interval (−3,−1)(-3, -1) excludes both x=−1x = -1 and x=1x = 1.

Key Concept

Question Stem Simplification with Absolute Value Inequalities
Estimated Time:2m 0s
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Topics

Inequalities, Absolute Values, and Number Ranges in Data Sufficiency

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