Question : Let $a=\frac{1}{2-\sqrt{3}}+\frac{1}{3-\sqrt{8}}+\frac{1}{4-\sqrt{15}}$ then we have:

Option 1: $a<18 \text{ but } a\neq 9$

Option 2: $a>18$

Option 3: $a=18$

Option 4: $a=9$

Question : If $a+b :\sqrt{ab} = 4:1 $ where $ a > b > 0$, then $ a:b$ is:

Option 1: $\left (2+\sqrt{3} \right):\left (2-\sqrt{3} \right)$

Option 2: $\left (2-\sqrt{3} \right):\left (2+\sqrt{3} \right)$

Option 3: $\left (3+\sqrt{2} \right):\left (3-\sqrt{2} \right)$

Option 4: $\left (3-\sqrt{2} \right):\left (3+\sqrt{2} \right)$

Question : If $a=\sqrt{6}–\sqrt{5}$, $b=\sqrt{5}–2$, and $c=2–\sqrt{3}$, then point out the correct alternative among the four alternatives given below.

Option 1: $b<a<c$

Option 2: $a<c<b$

Option 3: $b<c<a$

Option 4: $a<b<c$

Question : Which of the following is true?

Option 1: $\sqrt 5 + \sqrt 3 > \sqrt 6 + \sqrt 2$

Option 2: $\sqrt 5 + \sqrt 3 < \sqrt 6 + \sqrt 2$

Option 3: $\sqrt 5 + \sqrt 3 = \sqrt 6 + \sqrt 2$

Option 4: $(\sqrt 5 + \sqrt 3 ) (\sqrt 6 + \sqrt 2 )= 1$

Question : If $a=\frac{1}{a-\sqrt{6}}$ and $(a>0)$, then the value of $\left(a+\frac{1}{a}\right)$ is:

Option 1: $\sqrt{6}$

Option 2: $\sqrt{10}$

Option 3: $\sqrt{15}$

Option 4: $\sqrt{7}$