A#B = A^2 - AB

I. |x|#|y|
A#B = A^2 - AB
|X|#|Y| = X^2 - |X|*|Y|
Value of |X|*|Y| can be greater than X^2 or less than X^2 (insufficient)

II. |x-y|#-2
A#B = A^2 - AB
|X-Y|#-2 = |X-Y|^2 - |X-Y|*(-2)
|X-Y|#-2 = |X-Y|^2 + |X-Y|*(2)
Irrespective of Value of X or Y |X-Y| will always be positive.
Positive + positive = Positive (Sufficient)

III. (-x- 2)#0
A#B = A^2 - AB
(-x-2)#0 = (-x-2)^2 - (-x-2)*0
(-x-2)#0 = (x+2)^2
Irrespective of Value of X , the square of (-X-2) will always be positive. (Sufficient)

Only statement 2&3 are true.

IMO D

Posted from my mobile device
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\(a#b = a^2 - ab\) and we need to find which of the following expressions cannot be negative

I. |x|#|y|
to find |x|#|y| we need to substitute a = |x| and b=|y| in \(a#b = a^2 - ab\)
=> |x|#|y| = \(|x|^2 - |x|*|y|\) = |x| *(|x| - |y|)
Now, this can be negative if |y| > |x|
=> I. CAN be negative

II. |x-y|#-2
to find |x-y|#-2 we need to substitute a = |x-y| and b=-2 in \(a#b = a^2 - ab\)
=> |x-y|#-2 = \(|x-y|^2 - |x-y|*-2\) = \(|x-y|^2 + 2*|x-y|\)
Now, this will always be >=0 as this is sum of absolute value of two numbers. (It will be 0 only when x=y)
=> II. CANNOT be negative

III. (-x- 2)#0
to find (-x- 2)#0 we need to substitute a = (-x- 2) and b=0 in \(a#b = a^2 - ab\)
=> (-x- 2)#0 = \((-x- 2)^2 - (-x- 2)*0\) = \((-(x+ 2))^2 + 0\) = \((x+2)^2\)
Now, this will always be >0 as this is square of a number
=> III. CANNOT be negative

So, answer will be D
Hope it helps!

Watch the following video to learn the Basics of Functions


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Quick method:

First test (I):
Take a to be 2 and b to be 3.
\(2^2\) - \(2^3\) results in -4. This rules out A, C, and E answer choices.

Looking at B and D, you see that both include (II) so you do not have to test this case. Move on to testing case (III).

\((-x-2)^2\) - \((-x-2)^0\)

Factorize the right side and the left side = 1.

results in \(x^2\) + 4x + 3.

Resulting in (x+1)(x+3). Which means x = -1, -3. Plugging both back into the original equation, it is proven that both will be positive. Thus, both (II) and (III) will work and the correct answer is D.