Any non-zero real numbers x, y such that y ≠ 3 and xy < x+3y-3, will satisfy the condition
Explanation:
Given, xy < x+3y-3
⇒ xy - x+3y-3 < 0
⇒ x(y-3)-y(x+3)y(y-3) < 0
⇒ xy-3x-yx-3yy(y-3) < 0
⇒ -3(x+y)y(y-3) < 0
⇒ (x+y)y(y-3) > 0 ...(1)
Now, checking options.
Option (a): If y > 10, y(y - 3) > 0 From (1): ⇒ (x + y) > 0 ⇒ -x < y Hence, option (b) is incorrect.
Options (b) and (c) are slightly difficult to check. We can come back to them if required.
Option (d): If y < 0, y(y - 3) > 0 From (1): ⇒ (x + y) > 0 ⇒ -x < y Hence, option (d) is correct.
Hence, option (d).
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