Please submit your concern

Explanation:

Let us evaluate each option.

Option (a): As x < 0 and z > 1, let x = – 1 and z = 3, then (x² –z²) = – 8. Hence, this option is not true.

Option (b): As 0 < y < 1 and z > 1, let y = $\frac{1}{4}$ and z = 2, then yz = $\frac{1}{4}\times 2=\frac{1}{2}.$
Therefore, yz can be less than 1.

Option (c): Since, none of the x and y is equal to zero, therefore xy can never be zero.

Option (d): 0 < y < 1 and z > 1, therefore (y² – z²) is always negative.

Hence, option (a).