Is 1/(a - b) < b - a ?

(1) a > b --> we can rewrite this as \(a-b>0\) and \(0>b-a\) so \(LHS=\frac{1}{a-b}>0\) and \(RHS=b-a<0\) --> \((LHS=positive)>(negative=RHS)\), so the answer to the question is NO. Sufficient.

(2) (a + b)(a – b) > 0 --> either both multiples are positive or both are negative, so \(a-b\) can be negative (answer YES) as well as positive (answer NO, see above). Not sufficient.

Answer: A.
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Qtn: Is \(\frac{1}{a-b}<b-a\).
==> Is \(\frac{1}{a-b}-(b-a)<0\).
==> Is \(\frac{1}{a-b}+(a-b)<0\).
==> Is \(\frac{[1+(a-b)^2]}{a-b}<0\).
Notice that the numerator in the above ration is always positive.
for the ratio to be -ve (<0) the denominator has to be < 0
==> \(a-b\)has to be < 0 ==> \(a\) has to be < \(b\)
==> the qtn becomes Is \(a < b\)

Stmnt 1: \(a > b\) Directly gives the answer "NO". Sufficient
Stmnt2: \((a+b)(a-b)>0\)
==> \(a^2-b^2>0\)
==> \(a^2>b^2\)
\(a\) could be \(<\) or \(>\) \(b\) ..Not Suff.

Hence answer is "A".

Regards,
Murali.
Kudos?

Bumping for review and further discussion.
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Here the above equation can be reduced to is A-B <0
statement 1 is hence sufficient
but statement is not as we cannot tell whether A>B OR B>A
Hence A
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Is \(\frac{1}{(a - b)} < b - \)a ?

(1) \(a > b\)

If a > b then \(\frac{1}{(a - b)}\) will be positive.

\(\frac{1}{(a - b) } > 0\)

\(b - a < 0\)

The question can be rephrased as \(is positive value < negative value\)? We can answer with certainty: no.

Sufficient.

(2) \((a + b)(a – b) > 0\)

(a+b) and (a-b) are both either negative or positive. We can't determine which scenario, however.

We don't know if a < b or a > b. Insufficient.

Answer is A.
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