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If \(a\#b = a^2 - ab\), for values of a and b, which of the following expressions cannot be negative?
I. |x|#|y|
II. |x-y|#-2
III. (-x- 2)#0
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
I. |x|#|y|
II. |x-y|#-2
III. (-x- 2)#0
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
yashikaaggarwal
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23 Jun 2020, 10:07
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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
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
18 Jul 2021, 09:41
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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
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
