For two distinct real numbers x and y, which of them is bigger?
Statement I : $x^2 < y < 1$
Statement II : $y < \sqrt{x} <$ 1

Correct Option: (d) Select this option if the question cannot be answered even using any of the statements

We are given two distinct real numbers x and y, and need to determine which is bigger using the statements. Statement I: x^2 < y < 1. This implies y is positive and less than 1. Since x^2 < y, x lies between -1 and 1. For example, if y = 0.5 and x = 0.4, then x < y. But if y = 0.5 and x = -0.7, then x^2 = 0.49 < 0.5, giving x = -0.7 and y = 0.5, where y > x. However, if y = 0.5 and x = -0.7, y is still bigger. Let's test y = 0.5, x = -0.8 (x^2 = 0.64 > 0.5, invalid). What if y = 0.8, x = -0.85 (x^2 = 0.7225 < 0.8)? Then x = -0.85 and y = 0.8, so y > x. Statement II: y < sqrt(x) < 1. This implies x must be positive (since sqrt(x) is real) and 0 < x < 1, so sqrt(x) < 1 means x < 1. Also y < sqrt(x). However, this does not establish the direct relationship between x and y definitively across all valid values. Combining both statements: x^2 < y < sqrt(x) < 1. Since 0 < x < 1, we know that x^2 < x and sqrt(x) > x. Thus, y is bounded between x^2 and sqrt(x). Since x can be greater than y or less than y depending on precise values, neither statement alone nor together is sufficient to determine which is bigger.