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: $2$
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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