number properties

[sud21]

y is not zero, is (2^x/y)^x<1?
1). x>y
2). y< 0

[neelgandham]

y is not zero, is (2^x/y)^x<1?
Assuming (2^x/y) = (2^(x/y)),

If y!=0, Is 2^(x*x/y)< 1?

1). x>y

If x = 4 and y = 2, then 2^(x*x/y) = 2^(8) > 1
If x = 4 and y = -2, then 2^(x*x/y) = 2^(-8) < 1
If x = 0 and y = -2, then 2^(x*x/y) = 2^(0) = 1
Hence Insufficient!

2). y< 0

If x = 4 and y = -2, then 2^(x*x/y) = 2^(-8) < 1
If x = 0 and y = -2, then 2^(x*x/y) = 2^(0) = 1
Hence Insufficient!

From 1 and 2
If x = 4 and y = -2, then 2^(x*x/y) = 2^(-8) < 1
If x = 0 and y = -2, then 2^(x*x/y) = 2^(0) = 1
Hence Insufficient!

IMO E

[ArunangsuSahu]

(E)

Just plug in the data x=+ive and x=-ive