Bunuel wrote:
If \(xy\neq{0}\), is \(x^y>0\)?
(1) \(\sqrt[x]{32}=\frac{1}{2}\)
(2) \(\sqrt[-y]{16}=\frac{1}{2}\)
Kudos for a correct solution.
Given \(xy\neq{0}\) i.e \(x\neq{0}\) & \(y\neq{0}\)
To Find is \(x^y>0\)
=> \(x^y>0\) possible if
=> Case 1 x > 0 and y > 0 or y < 0
=> Case 2 x < 0 , y= MUST be EVEN (+ve or -ve)
=> Thus if 'y' is EVEN 'x' can be >0 or <0 then \(x^y>0\) always true
So the Ques can be Re-fraced as
is 'y' EVENStat 1 \(\sqrt[x]{32}=\frac{1}{2}\)
=> \(32=(\frac{1}{2})^x\)
=> \(32=2^{-x}\)
=> \(2^5=2^{-x}\)
=> Equating powers of 2 we have \(-x=5\)
=> or \(x=-5\)
=> Thus 'x' is <0 but no information of 'y'. Therefore NOT SUFFICIENT
Stat 2 \(\sqrt[-y]{16}=\frac{1}{2}\)
=> \(16^{\frac{1}{-y}}=2^{-1}\);
=> Since \(16=2^4\), Thereforet \(2^{\frac{4}{-y}}=2^{-1}\);
=> Equating powers of 2 we have \(\frac{4}{-y}=-1\).
=> Thus \(y=4\) i.e y=even
=> Since \(x\neq{0}\) and
y=even. Therefore SUFFICIENT
Therefore
"B"Regards
Dinesh