x and y are integers.

from 1:
|x| = |y+1|

x=3, y=2, |x| > |y|
x=-3, y=-4, |x| < |y|
Not Sufficient

from 2:
x^y = x! + |y|

from RHS:
factorial is defined only for 0 and positive integers, therefore x>=0. absolute value is always >0 and least value of a factorial =0! =1

therefore RHS >=1
therefore LHS >=1
therefore on LHS y>=0

also, x cannot be 0, b'cos 0^int is not defined.

So,
if x=1, y=0, |x| > |y|
if x=2, y=2, |x| = |y|
Not Sufficient

combining 1 and 2:
x^y = x! + |y| and |x| = |y+1|
therefore x=1 and y=0
Sufficient

IMO answer C

iamseer .... i disagree with you on one point

2) if x=2, y=2, |x| = |y|

this is not true because putting this back into


x^y = x! + |y| it doesn't hold so this is not true ...

neoreaves wrote:iamseer .... i disagree with you on one point

2) if x=2, y=2, |x| = |y|

this is not true because putting this back into


x^y = x! + |y| it doesn't hold so this is not true ...

x^y = x! + |y|

x=2, y=2, |x| = |y|


2^2= 2! +2

LHS equal to RHS!!

Nero bhai..Iamseer is correct!! Picking C shuld be fine!!

From the question, to prove if x^2-y^2>0

From 1,
x^2-y^2 = 2y+1
So x^2-y^2>0 only if y>=0

But no other info given hence Insufficient

From 2
x^y = x! +|y|

Just looking at this piece of info, I feel that this information is more to help decide if y>=0 than to actually prove x^2-y^2>0
Since RHS is +ve i.e >=1
This is only possible if y >=0

Hence
1 and 2 together are sufficient. Whats the OA