Which of the following statement(s) is/are TRUE?
$$\eqalign{ & {\text{I}}.\sqrt {12} > \root 3 \of {16} > \root 4 \of {24} \cr & {\text{II}}.\root 3 \of {25} > \root 4 \of {32} > \root 6 \of {48} \cr & {\text{III}}.\root 4 \of 9 > \root 3 \of {15} > \root 6 \of {24} \cr} $$

Solution (By Examveda Team)

$$\eqalign{ & {\text{I}}.\sqrt {12} > \root 3 \of {16} > \root 4 \of {24} \cr & {12^{\frac{1}{2}}},\,{16^{\frac{1}{3}}},\,{24^{\frac{1}{4}}} \cr & 2{\left( 3 \right)^{\frac{1}{2}}} > 2{\left( 2 \right)^{\frac{1}{3}}} > 2{\left( {\frac{3}{2}} \right)^{\frac{1}{4}}}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{I}} \right){\text{ True}} \cr & {\text{II}}.\root 3 \of {25} > \root 4 \of {32} > \root 6 \of {48} \cr & {25^{\frac{1}{3}}},\,{32^{\frac{1}{4}}},\,{48^{\frac{1}{6}}} \cr & 5{\left( {\frac{1}{5}} \right)^{\frac{1}{3}}} > 2{\left( 2 \right)^{\frac{1}{4}}} > 2{\left( {\frac{3}{4}} \right)^6}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{II}}} \right){\text{ True}} \cr & {\text{III}}.\root 4 \of 9 > \root 3 \of {15} > \root 6 \of {24} \cr & {9^{\frac{1}{4}}},\,{15^{\frac{1}{3}}},\,{24^{\frac{1}{6}}} \cr & {\text{3}}{\left( {\frac{1}{9}} \right)^{\frac{1}{4}}} > {\left( {15} \right)^{\frac{1}{3}}} > {\left( {24} \right)^{\frac{1}{6}}}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{III}}} \right){\text{ False}} \cr} $$