MathRevolution wrote:

Is \(x^2>y^2\)?

1) |x|>y

2) x>|y|

To Find Is \(x^2>y^2\) ?

=> OR \(x^2-y^2>0\)

=> OR \((x-y)(x+y)>0\)

Thus \(x^2>y^2\) is possible if both \((x-y)(x+y)\) are of same sign

PL. NOTE Now IF \(a>b\)----(1) and

=> Case 1

IF \((a+b)>0\) then SQUARING both the side of the inequality (1) the inequality sign (here '>') remains

SAME i.e (2) becomes \(a^2>b^2\)

=> Case 2

IF \((a+b)<0\) then SQUARING both the side of the inequality (1) the inequality sign (here '>')

REVERSE. i.e (2) becomes \(a^2<b^2\)

One can verify the above OUTCOMES by plugging values

Stat 1 \(|x|>y\) ---(2)

So from (2)we have LHS '|x|' is ALWAYS +ve and RHS 'y' is less than |x|

=> Therefore 'y'can be +ve OR -ve and SO \((|x|+y)\) can be >0 OR <0. Thus SQUARING both the side of the inequality (2) will give 2 OUTCOMES i.e

=> Case 1 \((|x|+y)>0\) then \(|x|^2>y^2\) or \(x^2>y^2\)

=> Case 2 \((|x|+y)<0\) then \(|x|^2<y^2\) or \(x^2<y^2\)

Since no unique answer, therefore NOT SUFFICIENT

Stat 2 \(x>|y|\)

Since RHS '|y| is ALWAYS +ve and LHS 'x' is greater than |y|

=> Therefore 'x' is ALSO +ve and SO \((x+|y|)\) is >0. Thus SQUARING both the side of the inequality (2) will give ONLY 1 OUTCOMES

=> Case 1 \((x+|y|)>0\) then \(x^2>|y|^2\) or \(x^2>y^2\)

Since unique answer, therefore SUFFICIENT

Thus Option "B"

Regards

Dinesh