Question : $2+\frac{6}{\sqrt{3}}+\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}-2}$ equals to:
Option 1: $+\left ( 2\sqrt3 \right )$
Option 2: $-\left ( 2+\sqrt3 \right )$
Option 3: $1$
Option 4: $2$
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Correct Answer: $2$
Solution : On rationalising the denominator of each term, we get $2+\frac{6}{\sqrt{3}}+\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}-2}$ $=2+\frac{6}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}+\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{1}{\sqrt{3}-2} \times \frac{\sqrt{3}+2}{\sqrt{3}+2}$ $=2+2\sqrt{3}+\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}+2}{3-4}$ $=2+2\sqrt{3}+2-\sqrt{3}-\sqrt{3}-2$ $=2$ Hence, the correct answer is $2$.
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