If x is a positive odd integer and y is a negative even integer

To simplify calculation let’s assume

positive odd integer x = 1 or 3

Negative even integer y = -2

Must be true should hold for all values.

A. \(x^3 + y\) is a positive odd integer.

case 1: x =1, y=-2 : 1-2 = -1 not a positive odd integer

case 2: x =3, y=-2 3-2 = 1 a positive odd integer.

B. \(x^2 + y^2\) is a negative odd integer

case 1: x =1, y=-2 : (1)^2+ (-2)^2 = 1+4 = 5. Positive odd integer.

case 2: x =3, y=-2 : (3)^2+ (-2)^2 = 9+4 = 13 positive odd integer.

C. \(x^0 + y^{11}\) is a negative odd integer

case 1: x =1, y=-2 : 1^0 + (-2)^11

= (-2)^odd power will result in a negative number greater than 1.

Thus 1 - even number (larger) = Negative odd integer.

case 2: x =3, y=-2 : (3)^0 + (-2)^11 . This is will also yield a negative odd integer.


D. \(x + y\) is a positive odd integer

case 1: x =1, y=-2 : 1-2 = -1 negative odd integer.

case 2: x =3, y=-2 : 3-2 = 1 positive odd integer.

E. \(x + y\) is a negative odd integer

case 1: x =1, y=-2 : 1-2 = -1 negative odd integer.

case 2: x =3, y=-2 : 3-2 = 1 positive odd integer.

Option C is a must be true statement .

Here what I thought to solve this that in option C, X^0+Y^11; X^0 always +1 for any value and Y^11 gives negative value for all even integers because power is an odd number.
So here difference of +1 from any even negative integers would always give negative Odd integers.