If x!=0, is (x^2 + 1)/x > y?
(1) x = y
(2) y > 0
What will be the correct answer to this question? I am confused.
I tried to simplify this question as follows
x^2 + 1 - xy > 0
x(y-x) < 1
But we can't be sure of the sign of the inequality as I have cross multiplied by x (whose sign I do now know).
So in essence the question can be rephrased as:x(y-x) [greater than, equal to or less than] 1 --> This is what is actually being asked.
I want to know if i have made any fundamental mistake when doing this rephrasing. As I have already mentioned I know that the sign of the inequality should be reversed when multiplied by a negative number.
So going by this approach the answer is A
I will consider both cases of x before cross multiplying.
Case 1: x is negative
x^2+1 < xy
x^2-xy + 1 < 0x(x-y) + 1 <0
Case 2: x is positive
x^2+1 > xy
x^2-xy + 1 > 0x(x-y) + 1 > 0
So based on these 2 equations we find that statement (1) alone is not sufficient because it gives different answers for Case 1 and 2 i.e. 1 < 0 and 1 > 0.
Statement 2 is also insufficient because it does not say anything about x.
Combining 1 and 2 -
y > 0
x = y
Therefore, x > 0x(x-y) + 1 > 0
Therefore C is the answer.
The second approach looks more methodical but I want to understand why the first approach is INVALID (if it is so).