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Zorluk: ZorInequalities, Absolute Values, and Number Ranges in Data Sufficiency

If xx is a real number and x≠−1x \neq -1, is ∣x−3∣x+1<1\frac{|x - 3|}{x + 1} < 1?

(1) ∣2x−1∣>3|2x - 1| > 3
(2) x(x−1)>0x(x - 1) > 0

  1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.Cevap
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  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.

Cevap

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Rephrasing the question stem shows that ∣x−3∣x+1<1\frac{|x - 3|}{x + 1} < 1 is satisfied when x<−1x < -1 or x>1x > 1. Statement (1) solves to x<−1x < -1 or x>2x > 2. Since (2,∞)(2, \infty) is completely contained inside (1,∞)(1, \infty), every value satisfying Statement (1) yields a definitive YES to the question. Statement (2) solves to x<0x < 0 or x>1x > 1. Choosing x=2x = 2 gives a YES answer, while choosing x=−0.5x = -0.5 gives a NO answer. Thus, Statement (1) alone is sufficient, but Statement (2) alone is not sufficient.

Adım Adım Çözüm

1
Rephrase the question stem by analyzing cases for the denominator x+1x + 1.
Case 1 (x>−1x > -1): x+1>0x + 1 > 0, so ∣x−3∣<x+1  ⟺  −(x+1)<x−3<x+1|x - 3| < x + 1 \iff -(x + 1) < x - 3 < x + 1. The right inequality −3<1-3 < 1 is always true, and the left inequality −x−1<x−3-x - 1 < x - 3 simplifies to 2x>2  ⟺  x>12x > 2 \iff x > 1.
Case 2 (x<−1x < -1): x+1<0x + 1 < 0. The numerator ∣x−3∣>0|x - 3| > 0 while the denominator is negative, making the ratio negative, which is always <1< 1.
Combining both cases: The target inequality holds if and only if x<−1x < -1 or x>1x > 1.
Rephrasing the stem target simplifies complex absolute value expressions into clear number line intervals.
2
Evaluate Statement (1): ∣2x−1∣>3|2x - 1| > 3.
2x−1>3  ⟹  2x>4  ⟹  x>22x - 1 > 3 \implies 2x > 4 \implies x > 2, or 2x−1<−3  ⟹  2x<−2  ⟹  x<−12x - 1 < -3 \implies 2x < -2 \implies x < -1.
Range: x∈(−∞,−1)∪(2,∞)x \in (-\infty, -1) \cup (2, \infty). Every value in this range satisfies x<−1x < -1 or x>1x > 1, guaranteeing a definitive YES to the question.
Statement 1 specifies a subset of the valid target range, making it sufficient alone.
3
Evaluate Statement (2): x(x−1)>0x(x - 1) > 0.
Range: x<0x < 0 or x>1x > 1. Testing values within this range:
- If x=2x = 2: 2(1)=2>02(1) = 2 > 0, and ∣2−3∣2+1=13<1\frac{|2 - 3|}{2 + 1} = \frac{1}{3} < 1 (YES).
- If x=−0.5x = -0.5: −0.5(−1.5)=0.75>0-0.5(-1.5) = 0.75 > 0, but ∣−0.5−3∣−0.5+1=3.50.5=7≮1\frac{|-0.5 - 3|}{-0.5 + 1} = \frac{3.5}{0.5} = 7 \not< 1 (NO).
Since both YES and NO outcomes are possible, Statement (2) is insufficient.
Finding a counterexample within the statement's solution set proves insufficiency.

Anahtar Kavram

Data Sufficiency evaluation for rational expressions containing absolute values and rephrasing inequality ranges.
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Inequalities, Absolute Values, and Number Ranges in Data Sufficiency

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