rampuria wrote:

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

1). x>y

2). y< 0

Please explain your approach

agree with B.

(2^x/y)^x = 2^(x^2/y)

1) x > y : not sure about y, which could be -ve or +ve. Either is suff but we need to know y whether it is -ve or +ve.

2) y< 0 : y is -ve. suff. the power of 2 becomes -ve, yeilding a +vwe fraction value.

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