If a b, is 1/(a-b) > ab ? (1) |a| > |b| (2) a < b

While solving DS inequality questions, the best approach is to always breakdown the question stem if possible. To breakdown the question stem, there are three hygiene factors, that if followed will simplify your analysis of the question.

1. Always keep the RHS of the inequality as 0
2. Simplify the LHS to a product or division of values (product and division of terms are easier to analyze)
3. Always try and maintain even powered terms (as the sign of them will always be 0 or positive)

The question here is, Is 1/(a - b) > ab?

1/(a - b) - ab > 0 ----> (1 - ab(a - b))/(a - b) > 0

For this equation to hold both the numerator and denominator need to have the same sign.

Statement 1 : |a| > |b|

Squaring both sides we get a^2 - b^2 > 0 ------> (a - b)(a + b) > 0.

This means that (a - b) and (a + b) can both be positive or both negative. This though answers the question with both a YES and a NO.

Statement 2 : a < b

This tells us that a - b < 0. This makes the denominator negative, but again we do not know the sign of ab in the numerator, ab. Insufficient.

Combining the statements we have (a - b)(a + b) > 0 and a - b < 0. This implies that (a + b) is also negative. But we still are unsure about the sign of ab, ab can either be positive or negative depending on the magnitudes we take for a and b. Insufficient.

E

Hope this helps!

Aditya
CrackVerbal Academics Team