Bunuel wrote:
If \(x > y\) and \(x < 0\), then which of the following must be true?
I. \(\frac{1}{x} < \frac{1}{y}\)
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
III. \(\frac{1}{x+1} < \frac{1}{y+1}\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
We're told that
x is NEGATIVE, and we're told that y < x, so
y is also NEGATIVETake each statement and manipulate it (by applying the rules of inequalities) to see which one(s) obey the restriction that y < x
I. \(\frac{1}{x} < \frac{1}{y}\)
Multiply both sides by x to get: \(1 > \frac{x}{y}\)
[since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]Multiply both sides by y to get: \(y < x\)
[since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]PERFECT! This obeys the restriction that y < x
Statement I is true. Check the answer choices . . . ELIMINATE B and C
II. \(\frac{1}{x−1} < \frac{1}{y−1}\)
NOTE: Since x is NEGATIVE, we know that x-1 is NEGATIVE
Likewise, since y is NEGATIVE, we know that y-1 is NEGATIVE
Multiply both sides by \(x-1\) to get: \(1 > \frac{x-1}{y−1}\)
[since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]Multiply both sides by \(y-1\) to get: \(y-1 < x-1\)
[since we multiplied by a NEGATIVE value, we REVERSED the inequality symbol]Add 1 to both sides to get: \(y < x\)
PERFECT! This obeys the restriction that y < x
Statement II is true. Check the remaining answer choices . . . ELIMINATE A and E
By the process of elimination, the correct answer is D
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