A) 10
B) 12
C) 14
D) 18
Correct Answer: A
Solution :
\[\frac{\sqrt{3+1}}{\sqrt{3-1}}=\frac{\sqrt{3+1}}{\sqrt{3-1}}\times \frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{{{(\sqrt{3}+1)}^{2}}}{3-1}\] \[=\frac{3+1+2\sqrt{3}}{2}=2+\sqrt{3}\] \[\therefore \] \[\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}\] \[\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{{{(\sqrt{2}+1)}^{2}}}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{2+1+2\sqrt{2}}{2-1}=3+2\sqrt{2}\]\[\therefore \] \[\frac{\sqrt{2}-1}{\sqrt{2}+1}=3-2\sqrt{2}\] \[\therefore \] Expression \[=2+\sqrt{3}+3+2\sqrt{2}-\sqrt{3}+3-2\sqrt{2}=10\]
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