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