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: $(\sqrt 5 + \sqrt 3 ) (\sqrt 6 + \sqrt 2 )= 1$
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Correct Answer: $\sqrt 5 + \sqrt 3 > \sqrt 6 + \sqrt 2$
Solution : Squaring both terms $\sqrt 5 + \sqrt 3$, $\sqrt 6 + \sqrt 2$ So, $(\sqrt 5 + \sqrt 3)^2 = 5 + 3 + 2\sqrt 15 = 8 + 2\sqrt 15$ Also, $(\sqrt 6 + \sqrt 2)^2 = 6 + 2 + 2\sqrt 12 = 8 + 2\sqrt 12$ Clearly, $\sqrt 5 + \sqrt 3 > \sqrt 6 + \sqrt 2$ Hence, the correct answer is $\sqrt 5 + \sqrt 3 > \sqrt 6 + \sqrt 2$.
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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 : Let $\sqrt[3]{a}=\sqrt[3]{26}+\sqrt[3]{7}+\sqrt[3]{63}$. Then:
Option 1: $a<729$ but $a>216$
Option 2: $a<216$
Option 3: $a>729$
Option 4: $a = 729$
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 $7+3x>3+2x$; and $x+3(x-7)<5-x$; then $x$ can take which of the following values?
Option 1: –5
Option 2: 5
Option 3: 6
Option 4: –6
Question : What is the third proportion of $2 \sqrt{3}$ and $6 \sqrt{5}$?
Option 1: $50 \sqrt{6}$
Option 2: $40 \sqrt{3}$
Option 3: $20 \sqrt{6}$
Option 4: $30 \sqrt{3}$
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