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 $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:

Option 1: $\frac{2 \sqrt{6}-6}{5}$

Option 2: $\frac{4 \sqrt{6}-6}{5}$

Option 3: $\frac{4 \sqrt{6}-6}{3}$

Option 4: $\frac{4 \sqrt{3}-6}{5}$

Question : If 5x + 3y = 15 and 2xy = 6, then the value of 5x – 3y is:

Option 1: $3\sqrt{3}$

Option 2: $3\sqrt{5}$

Option 3: $3\sqrt{2}$

Option 4: $3\sqrt{4}$

Question : If $x=5–\sqrt{21}$, the value of $\frac{\sqrt{x}}{\sqrt{32–2x}–\sqrt{21}}$ is:

Option 1: $\frac{1}{\sqrt2}(\sqrt{3}–\sqrt{7})$

Option 2: $\frac{1}{\sqrt2}(\sqrt{7}–\sqrt{3})$

Option 3: $\frac{1}{\sqrt2}(\sqrt{7}+\sqrt{3})$

Option 4: $\frac{1}{\sqrt2}(7–\sqrt{3})$

Question : Directions: By interchanging the given two numbers (not digits) which of the following equations will not be correct?
4 and 6

Option 1: 2 + 6 × 5 – 4 = 16

Option 2: 4 + 5 – 6 = 7

Option 3: 6 ÷ 2 – 4 × 5 = –28

Option 4: 6 × 3 – 4 ÷ 1 = 3