number properties

[sud21]

Is 2^[(x/y)x]<1?
1). x < 0
2). y < 0


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[neelgandham]

The value of 2^n is less than(<) 1 if the value of n is less than(<) 0.
2^[(x/y)x]<1, implies (x/y)x < 0, (x^2)/y<0 ? So the question can be rephrased to

Is(x^2)/y<0

1). x < 0

Let x = -2, y = 2, then (x^2)/y > 0.
Let x = -2, y = -2, then (x^2)/y < 0.
Let x = -2, y = 0, then (x^2)/y is undefined. Hence insufficient to answer the question

2). y < 0

Let x = -2, y = -2, then (x^2)/y < 0
Let x = +2, y = -2, then (x^2)/y < 0.
Let x = 0, y = -2, then (x^2)/y = 0. Hence insufficient to answer the question

From 1 and 2, x<0 and y<0, Sufficient to answer the question.(e.g.Let x = -2, y = -2, then (x^2)/y < 0))

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