1. Problem A: Verify the biconditional "\(\sqrt{x} \geq x \text{ if and only if } 0 < x < 1\)."\n - The biconditional means two conditionals must be true:\n a) If \(\sqrt{x} \geq x\), then \(0 < x < 1\).\n b) If \(0 < x < 1\), then \(\sqrt{x} \geq x\).\n - Since \(y=\sqrt{x}\) and \(y=x\), for \(0 < x < 1\), \(\sqrt{x} > x\) because the square root function grows slower than the linear function in this interval.\n - For \(x \geq 1\), \(\sqrt{x} \leq x\).\n - Negative \(x\) are excluded because \(\sqrt{x}\) is not real for \(x<0\).\n 2. Problem B: Verify the biconditional "\(x^2 < x^3 \text{ if and only if } x > 0\)."\n - Two conditionals:\n a) If \(x^2 < x^3\), then \(x > 0\).\n b) If \(x > 0\), then \(x^2 < x^3\).\n - For \(x > 0\), since \(x^3 = x \cdot x^2\) and \(x > 1\) or \(0 < x < 1\), the inequality holds.\n - For \(x \leq 0\), the inequality does not hold.\n 3. Problem C: Verify the biconditional "A whole number is divisible by 3 if and only if it is divisible by 6."\n - Two conditionals:\n a) If divisible by 3, then divisible by 6.\n b) If divisible by 6, then divisible by 3.\n - This is false because numbers divisible by 3 are not necessarily divisible by 6 (e.g., 3 is divisible by 3 but not by 6).\n - Therefore, the biconditional is false.\n 4. Problem D: Verify the biconditional "A whole number is even if and only if it is divisible by 2."\n - Two conditionals:\n a) If even, then divisible by 2.\n b) If divisible by 2, then even.\n - By definition, even numbers are exactly those divisible by 2, so both conditionals are true.\n Final answers:\n- A: True biconditional.\n- B: True biconditional.\n- C: False biconditional.\n- D: True biconditional.