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: $a<b<c$
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Correct Answer: $a<b<c$
Solution : Given: $a=\sqrt{6}–\sqrt{5}$, $b=\sqrt{5}–2$, $c=2–\sqrt{3}$ Rationalising the given values, $a=\sqrt{6}–\sqrt{5}×\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}$ ⇒ $a=\frac{1}{\sqrt{6}+\sqrt{5}}$ -----(1) $b=\sqrt{5}–2×\frac{\sqrt{5}+2}{\sqrt{5}+2}$ ⇒ $b=\frac{1}{\sqrt{5}+2}$ ⇒ $b=\frac{1}{\sqrt{5}+\sqrt{4}}$------(2) $c=2–\sqrt{3}×\frac{2+\sqrt{3}}{2+\sqrt{3}}$ ⇒ $c=\frac{1}{2+\sqrt{3}}$ ⇒ $c=\frac{1}{\sqrt{4}+\sqrt{3}}$------(3) By comparing $(a,b,c)$ we can find $a<b<c$. Hence, the correct answer is $a<b<c$.
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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 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 $x=(\sqrt{6}-1)^{\frac{1}{3}}$, 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 : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Option 1: $1$
Option 2: $2$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{5}$
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