Is: (x^2-y^2) > (x-y)
True: (x^2-y^2) > 1
False: (x^2-y^2) <= 1
(1) alone:
x-y < 0 (or) x < y
Take x = 0, y = 1 which satisfies (1)
is (x^2-y^2) > (x-y)?
is( 0 -1) > (0-1) FALSE
Take x = -1, y = 2
is (x^2-y^2) > (x-y)?
is( 1 -4) > (-1-3)?
is -3 > -4 TRUE
One false and one true is enough to say (1) alone cannot identify that x^2 - y^2 greater than x - y.
A and D is gone.
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(2) alone:
(x+y) < 0
Take x = 0, y = -1 which satisfies (2).
is (x^2-y^2) > (x-y)?
is (0-1) > 0+1?
is -1 > 1 FALSE
Take x = -1 y = 0 which satisfies (2)
is (x^2-y^2) > (x-y)?
is(1-0) > (-1-0) ?
is 1 > -1 TRUE
One false and one true is enough to say (2) alone cannot identify that x^2 - y^2 greater than x - y.
B is gone. Only C & E left.
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(1) & (2) together we have:
(x-y < 0 OR x<y) and (x+y) < 0 -----(3)
Take x = -1 and y = 0 which satisifies (3)
Then, is (x^2-y^2) > (x-y)?
is (1-0) > (-1-0)?
is 1 > -1 ? TRUE
Take x = -2 and y = -1 which satisfies (3)
Then, is (x^2-y^2) > (x-y)?
(4 -1) > -2+1 TRUE
We can't arrive a FALSE condition. => We say that (x^2-y^2) > (x-y) is always true.
Hence C is the correct answer choice.