Which of the following statement(s) is/are correct?
​$$\text{I.} \quad \frac{3}{7} < \frac{5}{9} < \frac{7}{11} \\\text{II.} \quad \sqrt{6} > \sqrt[3]{12}$$​​

Correct option is D

Given:
​$$\text{I.} \quad \frac{3}{7} < \frac{5}{9} < \frac{7}{11} \\\ \\\text{II.} \quad \sqrt{6} > \sqrt[3]{12}$$​​
Solution:
Statement I:
$$\text{I.} \quad \frac{3}{7} < \frac{5}{9} < \frac{7}{11} \\$$​​
LCM of (7, 9 , 11) = 693
​$$\frac{3}{7} = \frac{3 \times 99}{693} = \frac{297}{693} \\\ \\\frac{5}{9} = \frac{5 \times 77}{693} = \frac{385}{693} \\\ \\\frac{7}{11} = \frac{7 \times 63}{693} = \frac{441}{693} \\\ \\\text{Thus, } \frac{3}{7} < \frac{5}{9} < \frac{7}{11} \\\ \\$$​​
Statement 1 is correct.
Statement 2:
$$\text{II.} \quad \sqrt{6} > \sqrt[3]{12}$$​​
​$$(\sqrt{6})^6 = 6^{\frac{1}{2} \times 6} = 6^3 = 216 \\\ \\(\sqrt[3]{12})^6 = 12^{\frac{1}{3} \times 6} = 12^2 = 144 \\\ \\$$
Thus,
$$\text{Since } 216 > 144,\ \text{therefore } \sqrt{6} > \sqrt[3]{12}$$​​

Statement 2 is correct.
∴ Both statements are correct.