VJesus12 wrote:If |x|>0, is x^y<1?
(1) x=1/2
(2) y=0
The OA is the option B .
I don't know how to use that |x|>0. Can anyone explain this DS question to me? Please.
We are given that |x| > 0
=> x is a non-zero number; it can be positive or negative but not 0.
We have to find out whether x^y < 1.
Let's see each statement one by one.
(1) x = 1/2
Case 1: Say y = 0
=> x^0 = 1; not less than 1. The answer is no.
It is not important to know the value of x; whether x is negative or positive, x^0 = 1. Note that 0^0 is undefined, but the question states that |x| > 0, implying that x ≠0.
Case 2: Say y = 1
=> (1/2)^1 = 1/2 < 1; The answer is yes.
No unique answer. Insufficient.
(2) y = 0
=> x^0 = 1; not less than 1. The answer is no. A unique answer.
The correct answer:
B
Hope this helps!
-Jay
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