C-03-Q080mediumsingle_mcqIf a=3+2a = \sqrt{3} + \sqrt{2}a=3+2, a+1a=?a + \dfrac{1}{a} = ?a+a1=?a232\sqrt{3}23b5\sqrt{5}5c222\sqrt{2}22d3\sqrt{3}3ব্যাখ্যাRationalizing, 1a=13+2=3−2\frac1a=\frac1{\sqrt3+\sqrt2}=\sqrt3-\sqrt2a1=3+21=3−2, so a+1a=(3+2)+(3−2)=23a+\frac1a=(\sqrt3+\sqrt2)+(\sqrt3-\sqrt2)=2\sqrt3a+a1=(3+2)+(3−2)=23.