Bunuel wrote:

Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.

Hi

Bunuel.

I understood Statement-2 but confused with statement-1, my logic is as below

From question we can figure out that if xy>0, then lXYl= XY ; so XY>XY so we have to neglect this option

if xy<0, then l XY l = -(xy), So question becomes Is -(xy)>xy, this is what we have to look out from the question set.

So, -(xy)>xy is possible when we have

different signs of x and Y (one + and another -); then we can definitely say

"Yes" to the question Is -(xy)>xy

If X and Y have

same signs then -(xy)<xy ; then a definite

"NO" (to our question which is -(xy)>xy)

now in statement-1

X and Y both can be positive/negative; or X negative and Y positive (or vice versa) as the square will negate the effect of positivity/negativity of X/Y

So we can't be sure from statement 1 whether X/Y have same or different signs. Hence insufficient

I hope you understood my thinking and plz help to correct where me going wrong with my understanding

Thanks

For \(\sqrt{x^2y^2}=xy\) to be true x and y cannot have different signs because if they do, then the right hand side would become negative: \(xy=negative\) and it cannot equal to the square root of any number (\(\sqrt{x^2y^2}\)), which is always positive or 0.