If 3+23−2=a+b6, then (a,b) =

3+23−2=a+b6R.F of 3−2 is 3+2(3+2)(3+2)(3−2)(3+2)=a+b6⇒(3+2)2(3)2−(2)2=a+b63+2+263−2=a+b65+26=a+b6⇒a=5,b=2(a,b)=(5,2)

If 3+23−2=a+b6, then (a,b) =

3+23−2=a+b6R.F of 3−2 is 3+2(3+2)(3+2)(3−2)(3+2)=a+b6⇒(3+2)2(3)2−(2)2=a+b63+2+263−2=a+b65+26=a+b6⇒a=5,b=2(a,b)=(5,2)

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If $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = a + b \sqrt{6}$ , then (a,b) =

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(5,2)

(2,5)

(3,2)

(1,5)

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = a + b \sqrt{6} \\ R . F of \sqrt{3} - \sqrt{2} is \sqrt{3} + \sqrt{2} \\ \frac{(\sqrt{3} + \sqrt{2}) (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2}) (\sqrt{3} + \sqrt{2})} = a + b \sqrt{6} \Rightarrow \\ \frac{(\sqrt{3} + \sqrt{2})^{2}}{(\sqrt{3})^{2} - (\sqrt{2})^{2}} = a + b \sqrt{6} \\ \frac{3 + 2 + 2 \sqrt{6}}{3 - 2} = a + b \sqrt{6} \\ 5 + 2 \sqrt{6} = a + b \sqrt{6} \Rightarrow a = 5, b = 2 \\ (a, b) = (5, 2) \end{array}$