akirah wrote:
Despite reading the question and explanation again and again I still can't understand
" ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0." Bunuel could you please help?
Let me try to explain.
When you are given that ab+a/b > 0 ---> \(a* \frac{b^2+1}{b} >0\) , now realise that \(b^2+1\) is ALWAYS > 0 for all values of b and thus for making \(a* \frac{b^2+1}{b} >0\) you only need to worry about a/b > 0
Once you see that the given statement 2 is reduced to a/b > 0 ---> this can only mean that both a and b have the SAME signs. Either both are <0 or both are >0. You can also see it this way:
when a/b >0--> consider 4 cases:
a>0 and b>0 ---> YES for a/b >0
a>0 and b<0 ---> NO for a/b >0
a<0 and b>0 ---> NO for a/b >0
a<0 and b<0 ---> YES for a/b >0
Thus, you get "YES" only when both a and b are of the SAME sign.
You do not have to worry about the actual sign of a or b as shown below:
Once you see that statements need to be combined, you see that the original equation is ab > a/b ? ---> ab-a/b >0 ---> a(b-1/b) > 0 ---> \(\frac{a*(b^2-1)}{b} > 0\)
---> \(\frac{a*(b+1)(b-1)}{b} > 0\), this is ONLY dependent on the sign of (b+1)(b-1) as a/b will ALWAYS be positive for all values of a and b (as both a and b are of the SAME sign).
Statement 1 mentions that |b| >1 --> b>1 or b<-1 , for these ranges, (b+1)(b-1) is > 0 ALWAYS and hence you get a definite "YES", making C as the correct answer.
Hope this helps.
So far, I was so confused with ab, a/b, a and b. This explanation is perfect. It cant get better than this. Thank you so much