kiran120680 wrote:
If f(x) = 100 - x^2, is f(a) > f(b)?
(1) a^2 > b^2
(2) a/b > 1
Let us modify the original condition...\(f(x) = 100 - x^2....... f(a)=100-a^2\) and \(f(b)=100-b^2\)
Thus f(a) > f(b) means \(100-a^2>100-b^2.......b^2>a^2\)...
(1) \(a^2 > b^2\)
Exactly opposite of what we were looking for. So, answer is No
(2) a/b > 1
This means \(\frac{a}{b} > 1........\frac{a}{b} - 1>0..........\frac{a-b}{b} >0\)
So two cases
(a) b<0.... a-b will also be<0... a-b<0.....a<b... So a<b<0, thus \(a^2>b^2\), ans is NO
(b) b>0.... a-b will also be>0... a-b>0.....a>b... So a>b>0, thus \(a^2>b^2\), ans is NO
Thus, sufficient
D
first when b<0 , it is not always that a-b will always be <0 . for eg. a could be -2 and b could be -3 . so a-b>0.